º #º Nº ||||||||||||}| | : E- NH ERISPEninsulawardosº Pº Q A + S \ . . . " \s. \scº i /... . A , -- 4 / 7 J cº-...-- Ş.," *~ \,..., EUCLID's ELEMENTS OF GEOMETRY, THE FIRST SIX BOOKS, AND THE PORTIONS OF THE ELEVENTH AND TWELFTH BOOKS READ AT CAMBRIDGE: CHIEFLY FROM THE TEXT OF DR. SIMSON, WITH EXPLANATORY NOTES AND QUESTIONS: TOGETHER WITH A SELECTION OF GEOMETRICAL EXERCISES FROM THE SENATE-HOUSE AND COLLEGE EXAMINATION PAPERS: DESIGNED FOR THE USE OF THE HIGHER FORMS IN PUBLIC SCHOOLS AND STUDENTS IN THE UNIVERSITIES. BY ROBERT POTTS, M.A. TRINITY COLLEGE, CAMBRIDGE. Kabapploi puxic Aoyukic sial v ai Labmplarukai £rtarſipat. Hierocles. CORRECTED AND IMPROVED, * L ON DO N : LONGMAN, GREEN, LONG MAN, ROBERTS, & GREEN. M.DCCC.LXV. , -, | (. 2 INTERNATIONAL EXHIBITION, 1862, A Medal has been awarded to R. Potts, “For the excellence of his Works on Geometry.” Jury Awards Class xxix, p. 313. Gº *_ PRE FA C E. THE favourable notices of the First Edition of this work which appeared on its publicătion, and the reception it has met with in the Principal Schools and Universities of this Country and the Colonies, claim the Editor's grateful acknowledgements, and afford the satisfaction, that this his first though imperfect essay in Editorship has not been made in vain. - - One of the Reviewers was pleased to observe in reference to the Selection of the Geometrical Exercises, “As a series of judi- cious exercises, we do not think there exists one at all compa- rable to it in our language, viewed either in reference to the student or teacher”. This opinion may receive some confirma- tion from the fact, that a Translation in German by Hans H. von Aller, of the Geometrical Exercises, was published at Hanover in 1860, with a preface by Dr. Wittstein. In this Edition, the text of Dr. Simson is retained as the authorised English Version of Euclid's Elements, with the ex- ception of a few verbal emendations. Those portions of the Eleventh and Twelfth Books which are now read at Cambridge are retained, and the rest omitted. - - The explanatory notes have been considerably augmented and improved; and Algebraical proofs of the propositions of the Fifth Book have been added to the notes on that Book. A Selection of questions on the Elements, chiefly taken from the College Examination Papers, has been added to each of the first Six Books after the notes on each Book. Considerable additions have been made to the Geometrical Exercises, and those which involve the consideration of Loci, Maxima and Minima, and Tangencies have been arranged in three separate classes. As the design of the Editor has been to exhibit the teaching at Cambridge of Elementary Geometry, as it might be learned from the Exercises proposed in the College and University Examinations, he considers it advisable in this New Edition to replace many of the Geometrical Exercises by others which have appeared in recent examinations since the publication of the First Edition. - a- The “hints, &c., for the solution of the Problems, &c.” have been included in the Volume itself. - * iv. ~. - PREFACE. It was intended to add some account of the Extensions of the Euclidean Geometry, including the Porisms, Transversals, Poles and Polars, &c. &c., but as the volume has grown to larger dimen- sions than was anticipated, the Editor is compelled to reserve that part of his work for a future publication. The sketch of the History of Geometry prefixed to the First Edition is also reserved; but the Editor trusts that the following brief notices and remarks, may not be without interest to the student. Euclid's Elements of Geometry are invested with an interest which belongs to no other Elementary work on Pure Science. Having survived the age in which it was composed for the long period of upwards of twenty centuries, it has maintained its superiority in Schools and Universities, as the best introduc- tion to the Science of Pure Geometry. In ancient times this work sustained its reputation in the Schools of Athens and Alex- andria, the chief places of resort of philosophers and their hearers. In the age succeeding that of Euclid, the Science was greatly advanced by Archimedes, Apollonius, Theodosius, and others; and during the long stationary period which followed, down to the times of Proclus in the fifth century of the Christian era, no superior work appeared to take the place of Euclid's Elements. The appropriate opinion of Professor De Morgan may be here quoted:—“There never has been, and till we see it, we never shall believe there can be, a system of Geometry worthy of the name, which has any material departures (we do not speak of corrections, or extensions or developements) from the plan laid down by Euclid. If there be one worthy of consideration, it is the commencing with a strict theory of proportion. But it may very well be doubted whether any complete treatment of the Fifth Book of Euclid could be made intelligible to students of our day, before they have had some familiarity with demonstra- tion applied to particular species of magnitude. We say of our day, because it is impossible to foresee what the advance of edu- cation may do. It is perfectly conceivable that the rapid advance of demonstrative Arithmetic as a study preliminary to that of Geometry, may ultimately render the change desirable”. * The rise of the Mahommedan power in the Seventh Century, and the rapid and desolating effects which followed, hastened the extinction of Grecian Science. In the conquest of Egypt, the great library of Alexandria was committed to the flames by the ignorance and fanaticism of the Arabian conquerors; and the PREFACE. - V learned men there congregated for the cultivation of Science and Philosophy, either fell by the sword, or escaped by flight, carrying with them some remains of the Sciences. It is remarkable that in little more than a century after this infatuated deed of destruction, the Arabians became most zealous cultivators of the Science and Philosophy of the Greeks. The few Manuscripts of the Mathematical and Philosophical writings of the Greeks which had escaped the general ruin, were diligently sought and translated into Arabic, and commen- taries were written to elucidate and explain them. - About the middle of the Twelfth Century, in the reign of Henry I., the Elements of Euclid were introduced into England through a Latin translation from the Arabic. Manuscript copies of this most ancient translation of Euclid's Elements are pre- served at Oxford in the Library of Trinity College, and in the Bodleian Library. It was long after this period, the fact became known in Western Europe, that Euclid's Elements were originally written in Greek. .- The revival of ancient literature in Europe about the middle of the Fifteenth Century, contributed to extend the knowledge of the Mathematical writings of the Greeks; and the discovery of the art of printing, was the commencement of a new era in literature and science. The writings of the Greeks were no longer con- fined to the few who had the means of purchasing or procuring manuscript copies; and both the Greek text of Euclid's Elements and Latin translations of it were printed in different countries. The Latin translation of the Elements made from the Arabic by Campanus of Novara, was printed at Venice in 1482, and was the earliest Edition printed in that language. The ori- ginal Greek was first printed at Basle in 1533. Translations also of the Elements soon after appeared in the modern languages of Western Europe. The first English translation of Euclid's Elements was made by Henry Billingsley, a citizen of London, and was published in London in the year 1570. A learned preface was prefixed to the translation by a Fellow of Trinity. College, one of those who were first elected at the foundation of the College by Henry VIII. It does not appear that this English Version of Euclid's Elements was adopted at either of the English Universities. Latin being at that time the language of the learned, translations in that language were preferred, as text-books in Elementary Geometry, as in all other subjects of Academical study. . - w - vi - PREFACE. In the year 1655, the first Lucasian Professor of Mathe- matics, Isaac Barrow, M.A., Fellow, and afterwards Master of Trinity College, published a Latin translation of Euclid's Elements; and five years afterwards an English translation, of which more than one Edition was printed; the latest bears the date of 1751. A new Edition of Dr. Barrow's Mathematical writings, edited by Dr. Whewell, has recently been printed at the University Press. In the year 1756, Dr. Robert Simson, when he had been Pro- fessor of Mathematics in the University of Glasgow for upwards of forty years, published a Latin and also an English transla- tion of the first Six and the Eleventh and Twelfth books of Euclid's Elements. Dr. Simson's new translations of the Elements were favourably received at Oxford, and the Rev. Robert Smith, D.D., then Master of Trinity College, Cambridge, entertained so high an opinion of their superiority, that he earnestly pro- moted the introduction of them at Trinity College. There is ex- tant a letter in the hand-writing of Dr. Simson, in which he refers to thirty-six copies subscribed for by the Rev. Dr. Smith at Cambridge; and two letters of Dr. Smith, from which it appears, he paid to Dr. Simson's publisher in London forty pounds within a few shillings, for copies of the work which he had received. Dr. Smith was also the founder of two annual prizes which bear his name, for the two most distinguished proficients in Mathe- matics and Natural Philosophy among the commencing Bachelors of Arts in the University. Dr. Simson's translation having been thus introduced by Dr. Smith at Trinity College, it appears to have been preferred in the University, until at length it became established, as the authorised text of the English Version of Euclid's Elements of Geo- metry. Time, usage and experience have concurred to confirm this estimation. The translation, though not faultless, having now maintained its character for superiority at the English |Universities for more than a century, is entitled, at least to the careful and cautious consideration of any editor in making amendments and alterations. Some years ago Simson's Euclid was published in “the Symbolical form”, and became extensively used both at Cambridge and elsewhere. This form is open to very serious objections. A more correct taste has of late years returned, and the Symbolical form of Euclid's Elements is not deemed at Cambridge, the most appropriate for introducing the | PREFACE. - vii student to a correct knowledge of the Science of Pure Geometry. It may be added, that this opinion has been publicly recognised in the University. In the Examination papers in Pure Geo- metry both in the Senate-House and in College examinations, a caution has, of late years, been printed at the head of the papers against the use of Algebraical Symbols in writing the demonstra- tions of Euclid's propositions. There is still among mathemati- cians a difference of opinion with respect to the desirableness of exhibiting the demonstrations of Pure Geometry in Algebraical language. It is by some urged that the symbolical demonstra- tions are shorter and save time; by others it is replied, that “attempts at abbreviation have caused endless confusion”. It may perhaps be granted, that when a learner understands clearly the distinction between Pure Geometry and Algebraic Geometry, a certain latitude may be admitted in the use of symbolical lan- guage in Pure Geometry, which if allowed at an early stage of his progress, would most probably have led him into error, by the interchange of the terms of one Science with the written notation of the other. . i A distinguished Geometer of the last century has thus fairly stated one side of the question:—“The frequent use of symbols, common to the algebraic notation, may perhaps be looked upon as repugnant to the rigour and strictness of Geometry. But it is not the use of Symbols (which some more scrupulous than discerning, have condemned) but the ideas annexed to them, that render the consideration Geometrical or un-Geometrical. In Pure Geometry, regard is always had to absolute quantity of some one of the three kinds of extension, abstractedly considered; and whatever symbols are used here, are to be considered as expressive of the quan- tities themselves, and not as any measures, or numerical values of them. Thus, by A x B taken in a Geometrical sense, we have an idea, not of the product of two numbers (as in the Alge- braic Notation,) but of a real, rectangular space, comprehended under two right lines, represented by A and B, and two others equal to them. So likewise, ** C in the light of an algebraic fraction, but as a right line which is a fourth proportional to three other right lines, represented by A, B and C. These distinctions are absolutely necessary to those who would have an accurate idea of the subject.” - Professor De Morgan, on the other side maintains, that “Those who introduce Algebraical symbols into Elementary Geometry, is not to be understood here º viii * PREFACE. . destroy the peculiar character of the latter to every student who has any mechanical associations connected with those symbols; that is, to every student who has previously used them in ordi- nary Algebra. Geometrical reasoning, and Arithmetical process, have each its own office: to mix the two in elementary instruc- tion, is injurious to the proper acquisition of both.” & The following opinion of Sir Isaac Newton will be regarded as not without weight on this point: “Equations are expressions of Arithmetical computation, and properly have no place in Geometry, except as far as quantities truly Geometrical (that is lines, sur- faces, solids, and proportions) may be said to be some equal to others. Multiplications, Divisions, and such sort of computations, are newly received into Geometry, and that unwarily, and con- trary to the first design of this Science. For whosoever considers the construction of Problems by a right line and a circle, found out by the first Geometricians, will easily perceive that Geometry was invented that we might expeditiously avoid, by drawing lines, the tediousness of computation. Therefore these two Sciences ought not to be confounded. The ancients did so industriously distinguish them from one another, that they never introduced Arithmetical terms into Geometry. And the moderns, by con- founding both, have lost the simplicity in which all the elegance of Geometry consists. Wherefore that is arithmetically more. simple which is determined by the more simple Equations; but that is geometrically more simple which is determined by the more simple drawing of lines; and in Geometry, that ought to be reckoned best which is geometrically most simple.” The judgment of Sir Isaac Newton therefore is decided and clear, that the two Sciences of Pure Geometry and Algebraic Geometry ought not to be confounded. The remark of La Place is also worth notice:–“Cependant, les considérations géométriques ne doivent point étre abandonnées; elles sont de la plus grande utilité dans les arts. D'ailleurs, il est curieux de se figurer dans l'espace les divers résultats de l'analyse; et réciprogue- ment, de lire toutes les affections des lignes et des surfaces, et toutes les variations du mouvement des corps, dans les equa- tions qui les expriment. Ce rapprochement de la géométrie et de l'analyse répand un nouveau jour sur ces deux sciences; les opérations intellectuelles de celle-ci, rendues sensibles par les images de la première, sont plus faciles à Saisir, plus intéressantes à Suivre.” - PREFACE. & ix At the present day, the school studies of Elementary Geometry and Arithmetic occupy the time of students at the University which ought to be applied to studies, which with greater pro- priety may be termed Academical. The University of Cam- bridge by its recent legislation has attempted to enlarge and extend the sphere of Academical study by the institution of the Natural Science, and the Moral Science Triposes, and that of Con- stitutional History and Law. The limited success which has followed these laudable measures, may be mainly, if not entirely attributed to the imperfect and otherwise defective elementary knowledge with which a large number of students commence resi- dence in the University. It is essential that those students who desire to secure the advantages of a course of study at Cambridge, should have previously mastered the Elements of Geometry and Arithme- tic, as well as those of the Classical languages. Indeed it would be no detriment to the character of the University of Cambridge, if it were to adopt the restriction (umbels áyeop Tpmtos eiotro) which is reported to have been maintained by Plato with respect to his disciples before they were admitted to pursue the more advanced intellectual studies. The Ancient Statutes of the Uni- versity of Cambridge, set forth by Edward VI. ordained;—“that a student's first year should be devoted to Arithmetic, Geometry, and as much as he could manage of Astronomy and Cosmography.” At this period, students were considerably younger at the time of their admission than at present, and at the end of their first year, the studies of Geometry and Arithmetic were mastered and completed, before they had actually attained the age at which students now commence their residence at the University. At a later period in the history of the University, however, when Geometry and other elementary subjects had fallen into neglect, it was found necessary to remind the Candidates for Academical honours, that a competent knowledge of Euclid's Elements was essential, as will be seen from the following caution published by the Vice-Chancellor on May 20, 1774:—“That unless a person be found to have a competent knowledge of Euclid's Elements, and of the plainer parts of the four branches of Natural Philosophy, no attention will be paid to his other Mathematical knowledge. And that in every branch of science, the clearest and most accurate knowledge, rather than the most extensive, will be regarded as the best claim to Academical Honors.” Before closing these remarks, the Editor desires to direct the x PREFACE. student's attention to the opinions of two distinguished men, on the utility and importance of the Geometrical Element in Mathe- matical Studies. The late Rev. Dr. Chalmers has stated:—“I am not aware that as an expounder to the people of the lessons of the Gospel, I am much the better for knowing that the three angles of a triangle are equal to two right angles; or that the square of the hypotenuse is equal to the squares of the two containing sides in a right-angled triangle. But I have a strong persuasion, that both the power to apprehend and the power to convince, may be mightily strengthened—that the habit of clear and con- secutive reasoning, may be firmly established by the successive journeys which the mind is called on to perform along the pathway of Geometrical Demonstration. The truth is, that, as a prepara- tive, whether for the bar or for the pulpit, I have more value in Mathematics for the exercise which the mind takes as it travels along the road, than for all the spoil which it gathers at the landing-place”. And Dr. Whewell's opinion is to the same effect:- “Yet perhaps it may sometimes appear, both to teachers and to students, that it is a waste of time and a perverseness of judg- ment to adhere to the ancient kinds of Mathematics [Arithmetic and Geometry], when we have, in the modern Analysis, an in- strument of greater power and range for the solution of Problems; giving us the old results by more compendious methods; an instrument, too, in itself admirable for its beauty and generality. But to this we reply, that we require our Permanent Mathematical Studies, not as an instrument, but as an exercise of the intellectual powers; that it is not for their results, but for the intellectual habits which they generate, that such studies are pursued. To this we may add, as we have already stated, that in most minds, the significance of Analytical Methods is never fully understood, except when a foundation has been laid in Geometrical Studies. There is no more a waste of time in studying Geometry before we proceed to solve questions by the Differential Calculus, than there is a waste of time in making ourselves acquainted with the grammar of a language before we try to read its philosophical or poetical Literature.” R. P. TRINITY Col.I.EGE, October 1, 1861. PREFACE TO THE FIRST EDITION. THIS new edition of Euclid's Elements of Geometry will be found to differ considerably from those at present in general use in Academical Education. The text is taken from Dr. Simson's approved edition, with occasional alterations; but so arranged as to exhibit to the eye of the student the successive steps of the de- monstrations, and to facilitate his apprehension of the reasoning. No abbreviations or symbols of any kind are employed in the text. The ancient Geometry had no symbols, nor any notation beyond ordinary language and the specific terms of the science. We may question the propriety of allowing a learner, at the com- mencement of his Geometrical studies, to exhibit Geometrical demonstrations in Algebraical symbols. Surely it is not too much to apprehend that such a practice may occasion serious confusion of thought. It may be remarked that the practice of exhibiting the demonstrations of Elementary Geometry in an Algebraical form, is now generally discouraged in this University. To each book are appended explanatory notes, in which especial care has been taken to guard the student against the common mistake of confounding ideas of number with those of magnitude. The work contains a selection of problems and theorems from the Senate-house and College Examination Papers, for the last forty-five years. These are arranged as Geometrical exercises to the several books of the Elements, and to a few only in each book the solutions are given. An Introduction is prefixed, giving a brief outline of the history and progress of Geometry. - | The analysis of language, together with the sciences of number and magnitude, have been long employed as the chief elements of intellectual education. At a very early period, the study of Geometry was regarded as a very important mental discipline, as may be shewn from the seventh book of the Republic of Plato. To his testimony may be added that of the celebrated Pascal, (CEuvres, Tom. I. p. 66,) which Mr. Hallam has quoted in his History of the Literature of the Middle Ages. “Geometry, Pascal observes, is almost the only subject as to which we find truths wherein all men agree; and one cause of this is, that geometers álone regard the true laws of demonstration. These xii Pitºr’ACE. are enumerated by him as eight in number. 1. to define nothing which cannot be expressed in clearer terms than those in which it is already expressed. 2. To leave no obscure or equivocal terms undefined. 3. To employ in the definition no terms not already known. 4. To omit nothing in the principles from which we argue, unless we are sure it is granted. 5. To lay down no axiom which is not perfectly self-evident. 6. To demonstrate nothing which is as clear already as it can be made. 7. To prove every thing in the least doubtful, by means of self-evident axioms, or of propositions already demonstrated. 8. To substitute mentally the definition instead of the thing defined. Of these rules he says the first, fourth, and sixth are not absolutely requisite to avoid erroneous conclusions; but the other five are indispensable. He also re- marks, that although they may be found in our ordinary books of logic, yet none but geometers have recognised their importance, or been guided by them.” - - If we consider the nature of Geometrical and Algebraical reasoning, it will be evident that there is a marked distinction between them. To comprehend the one, the whole process must be kept in view from the commencement to the conclusion; while in Algebraical reasonings, on the contrary, the mind loses the distinct perception of the particular Geometrical magnitudes com- pared; the attention is altogether withdrawn from the things signified, and confined to the symbols, with the performance of certain mechanical operations, according to rules of which the rationale may or may not be comprehended by the student. It must be obvious that greater fixedness of attention is required in the former of these cases, and that habits of close and patient observation, of careful and accurate discrimination will be formed by it, and the purposes of mental discipline more fully answered. In these remarks it is by no means intended to undervalue the methods of reasoning by means of symbolical language, which are no less important than Geometry. It appears, however, highly desirable that the provinces of Geometrical and Algebraical rea- soning were more definitely settled than they are at present, at least in those branches of science which are employed as a means of mental discipline. The boundaries of Science have been ex- tended by means of the higher analysis; but it must not be forgotten that this has been effected by men well skilled in Geometry and fully able to give a geometrical interpretation of the results of their operations; and though it may be admitted PREFACE. xiii that the higher analysis is the more powerful instrument for that purpose, it may still be questioned whether it be well suited to form the chief discipline of ordinary intellects without a previous knowledge of the principles of Geometry, and some skill in their application. Though the method of Geometrical analysis is very greatly inferior in power to the Algebraical, yet as supplementary to the Elements of Euclid, it is of great importance. It may be added, that a sound knowledge of the ancient geometry is the best introduction to the pursuits of the higher analysis and its extensive applications. On this subject the judgment of Sir Isaac Newton has been recorded by Dr. Pemberton, in the preface to his view of Sir Isaac Newton's Discoveries. He says: “Newton censured the handling of geometrical subjects by algebraical calculations. He used to commend the laudable attempt of Hugo d'Omerique (in his “Analysis Geometrica,') to restore the ancient analysis, and very much esteemed the tract of “Apollonius De Sectione Rationis,' for giving us a clearer notion of that analysis than we had before. The taste and mode of geometrical demon- stration of the ancients he professed to admire, and even censured himself for not having more closely followed them than he did and spoke with regret of his mistake, at the beginning of his mathe- matical studies, in applying himself to the works of Descartes and other algebraic writers, before he had considered the Elements of Euclid with that attention which so excellent a writer deserves.” Regarding the study of Geometry as a means of mental discipline, it is obviously desirable that the student should be accustomed to the use of accurate and distinct expressions, and even to formal syllogisms. In most sciences our definitions of things are in reality only the results of the analysis of our own imperfect conceptions of the things; and in no science, except that of number, do the conceptions of the things coincide So exactly (if we may use the expression) with the things them- selves, as in Geometry. Hence, in geometrical reasonings, the comparison made between the ideas of the things, becomes almost a comparison of the things themselves. The language of pure Geometry is always precise and definite. The demonstrations are effected by the comparison of magnitudes which remain unaltered, and the constant use of terms whose meaning does not on any occasion vary from the sense in which they were defined. It is this peculiarity which renders the study so valuable as a mental discipline: for we are not to suppose that the habits of thought xty ºncer ACE. thus acquired, will be necessarily confined to the consideration of lines, angles, surfaces and solids. The process of deduction pur- sued in Geometry from certain admitted principles and possible constructions to their consequences, and the rigidly exact com- parison of those consequences with known and established truths, can scarcely fail of producing such habits of mind as will influence most beneficially our reasonings on all subjects that may come before us. - In support of the views here maintained, that Geómetrical studies form one of the most suitable and proper introductory elements of a scientific education, we may add the judgment of a distinguished living writer, the author of “The History and Philosophy of the Inductive Sciences,” who has shºwn, in his “Thoughts on the Study of Mathematics,” that mathematical studies judiciously pursued, form one of the most effective means of developing and cultivating the reason: and that “the object of a liberal education is to develope the whole mental system of man;–to make his speculative inferences coincide with his practical convictions;–to enable him to render a reason for the belief that is in him, and not to leave him in the condition of Solomon's sluggard, who is wiser in his own conceit than seven men that can render a reason.” To this we may sub- join that of Mr. John Stuart Mill, which he has recorded in his invaluable System of Logic, (Vol. II. p. 180) in the following terms. “The value of Mathematical instruction as a preparation for those more difficult investigations (physiology, society, govern- ment, &c.) consists in the applicability not of its doctrines, but of its method. Mathematics will ever remain the most perfect type of the Deductive Method in general; and the applications of Mathe- matics to the simpler branches of physics, furnish the only school in which philosophers can effectually learn the most difficult and important portion of their art, the employment of the laws of simpler phenomena for explaining and predicting those of the more complex. These grounds are quite sufficient for deeming mathe- matical training an indispensable basis of real scientific education, and regarding, with Plato, one who is dyeoplétpmtos, as wanting in one of the most essential qualifications for the successful culti- vation of the higher branches of philosophy.” • . - - R. P. , TRINITY College, October 1, 1845. C O N T ENTS, PAGE PREFACE . . . . . * e e s a e e s = e º e s C tº e º e s e s e s = e s a e © e º 'º r s g c e e is e g º a s g c tº - a tº e s is . iii Preface to the First Edition . . . . . . tº G e º 'º º ſº tº e e º e º e o 'º a * * * * * * * * * *. . . . . . . . , xi Contents . . . . . . . . . . . . . . . . . . . . . . . . . • * * * * * * * * * .* * * * * * * * * * & e º e s e a e s is a a s XV First Book of the Elements .......... • * * * * * * * * * * * * * * * * * * * * * * * * * … 1 Notes to the First Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº ſº tº º O p ∈ G → • 4 41 Questions on the First Book. . . . . . . . . . . . . . . tº e o ſº t e º e º e e º a e s a e e s a s e s = . . . 61 Second Book of the Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º 'º º e & 1 s e a e a 67 Notes to the Second Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . tº e º gº º ºs • * * * e º º ºs e e 81 Note on the Algebraical Symbols and Abbreviations used in Geometry... 91 Questions on the Second Book. . . . . . . . . . . . . . . . . • . . . . . . . . . . . . . . . . . . . . . . 92 Third Book of the Elements . . . . . . . . . . . . . . . . . . . . . . . g is tº e C & G is e s s e º e . . . . . 96 Notes to the Third Book . . . . . . . . . . . . . . Q & G G E G s a e º e º e s s a e u e s s e s a u s a e s e 127 Questions to the Third Book. . . . . . . . . . . . . . e º e º º ºn e e º e º a º e º a s e º a e º e s º Gº & & 132 Fourth Book of the Elements . . . . . . . . . . . . a • * * * * * . . . . . . . . . . . . . . . _ s a ſe e s & 135 Notes to the Fourth Book. . . . . . . . . . . . . . . . . . . . . . . . . . . . . * G e º is © . . . . . . . . . 150 Questions on the Fourth Book. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . … 152 Fifth Book of the Elements . . . . . . • * * * * * * - - - - - - - - - - - - - - - . . . . . . . . . . . . . . 155 Notes to the Fifth Book . . . . . . . . . . . . . . . . . . . . • * * * * * * * - - - - - - - - - - - - - • * * * * 184 Questions on the Fifth Book.... . . . . . . . . . . . . . . G e, tº º C Q & s & s 6 s tº e s a s e n e º 'º e a 206 Sixth Book of the Elements ...... • . . . . . . . . . . . . . . . . . . . tº º G & e º e º e º gº © tº 0 ... 209 Notes to the Sixth Book .............. & © tº is ſº tº s e º ºs e º 'º • . . . . . . . . . . . . . . . 244 Questions on the sixth Book ............... ............. * - - - - - - - - - - 248 Eleventh Book of the Elements . 2 * * a s º e º e º e º e e & O º e º C G & © e º e a c e º G & 8 . . . 251 Notes to the Eleventh Book . . . . . . . . . . . tº nº e o dº e º e º e & © e º a e s ∈ G & & © e e s e e a a e 271 Twelfth Book of the Elements .............. C & C G & © tº ºi e º g º ºs e c tº e º e º e º ſº º is tº 274 Notes to the Twelfth Book . . . . . . . . . . . . . * : C → C & C º e C C C C C C e º ºs e º e º e u e º ſº tº º 280 On the Propositions in the Elements. . . . . . . . . . . . . . . . . . . . .............. 281 On the Ancient Geometrical Analysis . . . . . . . . . . . e & e º e º G e º a G e º ºr º tº e º is e e 288 The Analysis of Theorems......... e º e e º e º e e g º e s c e º ſº e º 'º dº ſº tº tº º e s tº a . . . . . ºb. The Analysis of Problems ... . . . . . . . . . . . . . . . tº ſº ºn tº tº º is tº tº @ ſº tº & 0 s s e º s º ºs e e º 'º a 291 Geometrical Exercises on Book I. . . . . . . . . . . . . . . . . . . . . . . . . . . … . , 294 $6 & 6 ( & II. . . . . . . . . . . . . . . . . . . . . . . . . . © e g º e º e G & u ę. 310 66 & 6 & 6 III. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 “ & 6 66 IV. . . . . . . . . . . & G G G º ºs e º e s is e e s s a • a s e o e s e e 339 66 & © . “ VI. .................................... 356 xvi CONTENTs. f PAGE loci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C & © tº a C & © tº 380 Maxima and Minima . . . . . . . . Q S C C e s e º e e s e e o s º e s e e s e º e s s © tº S W C C e a e e s s a . 387 Tangencies . . . . . . . . . . . . . . . . . . . . . . • * * * . . . . . . . . . . . . . . . . . . . . ............ 394 Geometrical Exercises on Book XI........... . . . . . . . . . . . . . . . . . . . . . . . . . 401 66 66 & 6 XII. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 Solutions, Hints, &c. to the Geometrical Exercises on Book I. … 417 66 - & 6 & 6 6 (; 46 66 II. * G - © e º e a 427 &G 66 66 66 66 (; 6 III. . . . . . . . . 429 66 - 6& 66 tº 66 $6 IV. . . . . . . . . 444 &G (; ; & 6 © (; 66 66 VI. tº º e º 'o º tº º 456 Solutions, Hints, &c. on the Loci . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474 “ “ “ Maxima and Minima. . . . . . . . . . . . • . . . . . . . . . 478 e 6 & C “ Tangencies . . . . . . . . . . . • * @ tº º e º O © e a s e e a . . . . 481 & © & 6 “ on Book XI . . . . . . . . • a e s e e e s • s s . . . . . . . . . . . 486 66 6 & “ on Book XII . . . . . . . . . . . * * * 0 & © e s e º e º e e s e 490 Index to the Geometrical Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 EUCLIDS ELEMENTS OF GEOMETRY. BOOK I. DEFINITIONS. I. A PoſNT is that which has no parts, or which has no magnitude. II. A line is length without breadth. III. The extremities of a line are points. IV. A straight line is that which lies evenly between its extreme points. W. A superficies is that which has only length and breadth. VI. The extremities of a superficies are lines. VII. A plane superficies is that in which any two points being taken, the straight line between them lies wholly in that superficies. WHEI. e -- . A plane angle is the inclination of two lines to each other in a plane, which meet together, but are not in the same direction. IX A plane rectilineal angle is the inclination of two straightlines to one another, which meet together, but are not in the same straight line. D A 2 - EUCLID's ELEMENTS. N.B. If there be only one angle at a point, it may be expressed by a letter placed at that point, as the angle at E.: but when several angles are at one point B, either of them is expressed by three letters, of which the letter that is at the vertex of the angle, that is, at the point in which the straight lines that contain the angle meet one another, is put between the other two letters, and one of these two is somewhere upon one of these straight lines, and the other upon the other line. Thus the angle which is contained by the straight lines AB, CB, is named the angle ABC, or CBA ; that which is contained by AB, DB, is named the angle ABD, or DBA.; and that which is contained by DB, CB, is called the angle DBC, or CBD. . X. When a straight line standing on another straight line, makes the adjacent angles equal to one another, each of these angles is called a right angle; and the straight line which stands on the other is called a perpendicular to it. sº XI. An obtuse angle is that which is greater than a right angle. XII. An acute angle is that which is less than a right angle. XIII. A term or boundary is the extremity of any thing. - XIV. A figure is that which is enclosed by one or more boundaries. XV. . A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point Within the figure to the circumference, are equal to one another. DEFINITIONS. 3 And this point is called the center of the circle. . A diameter of a circle is a straight line drawn through the center, and terminated both ways by the circumference. XVIII A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. XTX. The center of a semicircle is the same with that of the circle. Rectilineal figures are those which are contained by straight lines. Trilateral figures, or triangles, by three straight lines. Quadrilateral, by four straight lines. Multilateral figures, or polygons, by more than four straight lines. XXIV. - Of three-sided figures, an equilateral triangle is that which has three equal sides. - An isosceles triangle is that which has two sides equal. B 2 4 - * EUCLID'S ELEMENTS. A scalene triangle is that which has three unequal sides. ^ XXVII. A right-angled triangle is that which has a right angle. XXVIII. An obtuse-angled triangle is that which has an obtuse angle. XXIX. º An acute-angled triangle is that which has three acute angles. 2 XXX. Of quadrilateral or four-sided figures, a square has all its sides equal and all its angles right angles. - - * XXXI. ... An oblong is that which has all its angles right angles, but has not all its sides equal. - - XXXII. A rhombus has all its sides equal, but its angles are not right angles. DEFINITIONS. § A rhomboid has its opposite sides equal to each other, but all its sides are not equal, nor its angles right angles. . \ \ XXXIV. All other four-sided figures besides these, are called Trapeziums. XXXV. Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways, do not meet. A. A parallelogram is a four-sided figure, of which the opposite sides are parallel: and a diameter, or a diagonal is a straight line joining two of its opposite angles. POSTULATES. I. LET it be granted, that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. And that a circle may be described from any center, at any distance from that center. - AXIOMS. I. THINGs which are equal to the same thing, are equal to one another. - - II. If equals be added to equals, the wholes are equal. **. 6 EUCLID'S ELEMENTS: If equals be taken from equals, the remainders are equal. IV. If equals be added to unequals, the wholes are unequali. - W. - - If equals be taken from unequals, the remainders are unequal. - WI. Things which are double of the same, are equal to one another. Things which are halves of the same, are equal to one another. VIII. Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. - . DX. The whole is greater than its part. X. Two straight lines cannot enclose a space. XI. All right angles are equal to one-another. XII. If a straight line meets two straight lines, so as to make the two. interior angles on the same side of it taken together less than two right angles; these straight lines being continually produced, shall at length meet upon that side on which are the angles which are less than two right angles. - - - BOOK I. PROP. I, II. 7 PROPOSITION I. PROBLEM. To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line. It is required to describe an equilateral triangle upon A.B. C /\, From the center A, at the distance AB, describe the circle BCD; (post. 3.) from the center B, at the distance B.A, describe the circle A CE; and from C, one of the points in which the circles cut one another, draw the straight lines CA, CB to the points A, B. (post. 1.) Then ABC shall be an equilateral triangle. Because the point A is the center of the circle BCD, therefore AC is equal to AB; (def. 15.) and because the point B is the center of the circle ACE, therefore BC is equal to AB; but it has been proved that AC is equal to AB; therefore AC, BC are each of them equal to Af ; but things which are equal to the same thing are equal to one another; therefore AC is equal to BC; (ax. 1.) wherefore AB, BC, CA are equal to one another: and the triangle ABC is therefore equilateral, and it is described upon the given straight line A.B., Which was required to be done. PROPOSITION II. PROBLEM. From a given point, to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw from the point 4 a straight line equal to BC, From the point A to B draw the straight line AB; (post. 1.) upon AB describe the equilateral triangle ABD, (1.1.) and produce the straight lines DA, DB to E and F; (post. 2.) from the center B, at the distance BC describe the circle CGH, (post. 3.) cutting DF in the point G : and from the center D, at the distance DG, describe the circle GKL, cutting DE in the point L. Then the straight line AL shall be equal to BC. Because the point B is the center of the circle CGH, 8 - EUCLID's ELEMENTS. therefore BC is equal to BG; (def. 15.) and because D is the center of the circle GRTL, therefore D.L. is equal to DG, and DA, DB parts of them are equal; (I. 1.) therefore the remainder AL is equal to the remainder BG; (ax. 3.) but it has been shewn that BC is equal to BG, wherefore AL and BC are each of them equal to BG; and things that are equal to the same thing are equal to one another; therefore the straight line AL is equal to B.C. (ax. 1.) Wherefore from the given point A, a straight line AL has been drawn equal to the given straight line B.C. Which was to be done. PROPOSITION III. PROBLEM. From the greater of two given straight lines to out off a part equal to the less. Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off A.B. the greater, a part equal to C, the less. From the point A draw the straight line AD equal to C; (r. 2.) and from, the center A, at the distance A.D, describe the circle. DEF (post. 3.) cutting AB in the point E. Then A.E shall be equal to C. Because A is the center of the circle DEF. therefore AE is equal to AD; (def, 15.) but the straight line C is equal to AD; (constr.) whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to C. (ax. 1.) And therefore from AB the greater of two straight lines, a part A.E has been cut off equal to C, the less. Which was to be done. PROPOSITION IV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each ; and have likewise the angles contained by those sides equal to each other; they shall likewise have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite. Let ABC, DEF be two triangles, which have the two sides AB, A C equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to D.F, and the included angle BAC equal to the included angle EDF, Then shall the base BC be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles to which the equal sides are opposite shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE. BOOK I. PROP. Iv, V. 9. D B C E F Tor, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line A B on DE; then the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, the straight line A C shall fall on DF, because the angle BAC is equal to the angle EDF'; therefore also the point C shall coincide with the point F, because A C is equal to DF; but the point B was shewn to coincide with the point E; wherefore the base BC shall coincide with the base EF; because the point B coinciding with E, and C with F. if the base BC do not coincide with the base EF, the two straight lines BC and EF would enclose a space, which is impossible. (ax. 10.) Therefore the base BC does coincide with EF, and is equal to it; and the whole triangle ABC coincides with the whole triangle JDEF, and is equal to it; also the remaining angles of one triangle coincide with the remaining angles of the other, and are equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides, &c. Which was to be demonstrated. PROPOSITION V. THEOREM. The angles at the base of an isosceles triangle are equal to each other; and ºf the equal sides be produced, the angles on the other side of the base shall be equal. Let ABC be an isosceles triangle of which the side AB is equal to AC, and let the equal sides AB, A C be produced to D and E. Then the angle ABC shall be equal to the angle ACB, and the angle DBC to the angle ECB. A. F G. D IE In BD take any point F; from AF the greater, cut off AG equal to AF the less, (I. 3.) and join FC, G.B. Because AF is equal to AG, (constr.) and AB to AC; (hyp.) the two sides FA, AC, are equal to the two GA, AB, each to each; 1() • EUCLID'S ELEMENTS. and they contain the angle FAG common to the two triangles AFC, .46 B; therefore the base FC is equal to the base GB, (I. 4.) and the triangle AFC is equal to the triangle AGB, also the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal sides are opposite; viz. the angle ACF to the angle A.BG, * and the angle AFC to the angle AGB. And because the whole AF is equal to the whole AG, of which the parts AB, A C, are equal; therefore the remainder BF is equal to the remainder CG ; (ax. 3.) and FC has been proved to be equal to GB; hence because the two sides B.F, FC are equal to the two CG, GB, each to each ; and the angle BFC has been proved to be equal to the angle CGB, also the base BC is common to the two triangles BFC, CGB; wherefore these triangles are equal, (r. 4. and their remaining angles, each to each, to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. And, since it has been demonstrated, that the whole angle ABG is equal to the whole ACF. the parts of which, the angles CBG, BCF are also equal; therefore the remaining angle ABC is equal to theremaining angle ACB, which are the angles at the base of the triangle ABC; and it has also been proved, that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore the angles at the base, &c. Q.E.D. CoR. Hence an equilateral triangle is also equiangular. PROPOSITION VI. THEOREM. If two angles of a triangle be equal to each other, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. Let ABC be a triangle having the angle ABC equal to the angle A CB. Then the side A B shall be equal to the side A.C. A D B C For, if A B be not equal to AC, one of them is greater than the other. If possible let A B be greater than AC; and from BA cut off BD equal to CA the less, (I. 3.) and join DC. Then, in the triangles DBC, ABC, - because DB is equal to AC, and BC is common to both triangles, the two sides DB, BC are equal to the two sides AC, CB, each to each; BOOK I. PROP. VI, VII. 11 and the angle DBC is equal to the angle A CB; (hyp.) therefore the base DC is equal to the base AB, (I. 4.) and the triangle DBC is equal to the triangle ABC, the less equal to the greater, which is absurd. (ax. 9.) Therefore AB is not unequal to AC, that is, AB is equal to A. C. Wherefore, if two angles, &c. Q.E.D. CoR. Hence an equiangular triangle is also equilateral. PROPOSITION VII. THEOREM, Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other eactremity. If it be possible, on the same base AB, and upon the same side of it, let there be two triangles A CB, ADB, which have their sides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their sides, CB, DB, that are terminated in B. C D A. B Join CD. First. When the vertex of each of the triangles is without the other triangle. Because A C is equal to AD in the triangle A CD, therefore the angle ADC is equal to the angle A CD; (1. 5.) but the angle A CD is greater than the angle BCD; (ax. 9.) therefore also the angle ADC is greater than B CD; much more therefore is the angle BDC greater than BCD. Again, because the side BC is equal to BD in the triangle BCD, (hyp.) therefore the angle BDC is equal to the angle BCD; (I. 5.) but the angle BDC was proved greater than the angle BCD, hence the angle BDC is both equal to, and greater than the angle BCD; which is impossible. Secondly. Let the vertex D of the triangle ADB fall within the triangle A.C.B. w Produce AC to E and AD to F, and join CD. Then because AC is equal to AD in the triangle A CD, therefore the angles ECD, FDC upon the other side of the base, CD, are equal to one another; (I. 5.) but the angle ECD is greater than the angle BCD; (ax. 9.) therefore also the angle FDC is greater than the angle BCD; 12 EUCLID'S ELEMENTS. much more then is the angle BDC greater than the angle BCD. Again, because BC is equal to BD in the triangle BCD, therefore the angle BDC is equal to the angle BCD ; (I. 5.) but the angle BDC has been proved greater than BCD, wherefore the angle BDC is both equal to, and greater than the angle BCD; § which is impossible. Thirdly. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base and on the same side of it, &c. Q.E.D. IPROPOSITION VIII. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which as contained by the two sides of the one, shall be equal to the angle contained by the two sides equal to them, of the other. Let ABC, DEF be two triangles, having the two sides AB, A C, equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and also the base B C equal to the base EF. HD G | B C JE F Then the angle BAC shall be equal to the angle EDF. Eor, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC on EF; then because BC is equal to EF, (hyp.) therefore the point C shall coincide with the point F; wherefore BC coinciding with EF, BA and AC shall coincide with ED, DF; for, if the base BC coincide with the base EF, but the sides B.A., A C, do not coincide with the sides ED, DF, but have a different situation as EG, GF: - Then, upon the same base, and upon the same side of it, there can be two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those side which are terminated in the other extremity; | but this is impossible. (I. 7.) Therefore, if the base BC coincide with the base EF. the sides B.A, A C cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. (ax. 8.) Therefore, if two triangles have two sides, &c. Q.E.D. BOOK 1. PROP. IX, X. 13 PROPOSITION IX. PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal angles. Let BAC be the given rectilineal angle. It is required to bisect it. - A T) E / \ B C In AB take any point D; from AC cut off A.E. equal to A.D., (I. 3.) and join DE; on the side of DE remote from A, describe the equilateral triangle DEF (I. 1.) and join A.F. Then the straight line AF'shall bisect the angle B.A. C. Because AD is equal to A.E, (constr.) and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two sides EA, A.F. each to each; and the base DF is equal to the base EF; (constr.) therefore the angle DAF is equal to the angle E.A.F. (I. 8.) Wherefore the angle BAC is bisected by the straight line A.F. Q. E. F. PROPOSITION X. PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line. It is required to divide AB into two equal parts. Upon AB describe the equilateral triangle ABC; (1.1.) and bisect the angle ACB by the straight line CD meeting AB in the point D. (1.9.) - Then AB shall be cut into two equal parts in the point D. Because AC is equal to CB, (constr.) and CD is common to the two triangles A CD, BCD; the two sides AC, CD are equal to the two BC, CD, each to each; and the angle A CD is equal to BCD; (constr.) therefore the base AD is equal to the base B.D. (r. 4.) Wherefore the straight line AB is divided into two equal parts in the point D. Q. E. F. 14 EUCLID's ELEMENTS. IPROPOSITION XI. PROBLEM. To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be the given straight line, and C a given point in it. It is required to draw a straight line from the point C at right angles to A.B. A. L) C E In A C take any point J), and make CE equal to CD; (I. 3.) upon DE describe the equilateral triangle DEF, (I. 1.), and join CF. Then CF drawn from the point C shall be at right angles to A.B. Because DC is equal to EC, and FC is common to the two tri- angles DCF, ECF; the two sides DC, CF are equal to the two sides EC, CF, each to each; and the base DF is equal to the base EF; (constr.) therefore the angle DCF is equal to the angle ECF: (I. 8.) and these two angles are adjacent angles. But when the two adjacent angles which one straight line makes with another straight line, are equal to one another, each of them is called a right angle: (def. 10.) Therefore each of the angles DCF, ECF is a right angle. Wherefore from the given point C, in the given straight line AB, FC has been drawn at right angles to A.B. Q.E. F. CoR. By help of this problem, it may be demonstrated that two straight lines cannot have a common segment. If it be possible, let the segment AB be common to the two straight lines ABC, ABD. - E ID _-T A B C From the point B, draw B.E at right angles to AB; (I. 11.) then because ABC is a straight line, therefore the angle ABE is equal to the angle E.B.C. (def. 10.) Similarly, because ABD is a straight line, therefore the angle ABE is equal to the angle EBD; but the angle ABE is equal to the angle EBC, wherefore the angle EBD is equal to the angle EBC, (ax. 1.) the less equal to the greater angle, which is impossible. Therefore two straight lines cannot have a common segment. PROPOSITION XII. PROBLEM. . To draw a straight line perpendicular to a given straight line of un- limited length, from a given point without it. Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it. BOOK I. PROP. XII, XIII. 15 It is required to draw a straight line perpendicular to AB from the point C. Upon the other side of AB take any point D, and from the center C, at the distance CD, describe the circle EGF meeting AB, produced if necessary, in F and G.; (post. 3.) bisect FG in H (I. 10.), and join C.H. Then the straight line CH drawn from the given point C, shall be perpendicular to the given straight line A.B. - Join FC, and C.G. Because FH is equal to HG, (constr.), - and HC is common to the triangles FHC, GHC; the two sides FH, HC, are equal to the two GH, HC, each to each; and the base CF is equal to the base CG ; (def. 15.) therefore the angle FHC is equal to the angle GHC; (I. 8.) and these are adjacent angles. But when a straight line standing on another straight line, makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpen- dicular to it. (def. 10.) Therefore from the given point C, a perpendicular CH has been drawn to the given straight line A.B. Q. E. F. PROPOSITION XIII. THEOREM. The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD. Then these shall be either two right angles, or, shall be together, equal to two right angles. |E A A ID B C D H-—a For if the angle CBA be equal to the angle ABD, each of them is a right angle. (def. 10.) - But if the angle CBA be not equal to the angle ABD, from the point B draw BE at right angles to CD. (1.1.1.) Then the angles CBE, EBD are two right angles. (def. 10.) And because the angle CBE is equal to the angles CBA, 4.B.E, add the angle EBD to each of these equals; therefore the angles CBE, EBI) are equal to the three angles CBA, ABE, EBD. (ax. 2.) 16 EUCLID'S ELEMENTS. Again, because the angle DBA is equal to the two angles DBE, EBA, add to each of these equals the angle ABC; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. But the angles CBE, EBD have been proved equal to the same three angles; and things which are equal to the same thing, are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC; but the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are together equal to two right angles. (ax. 1.) Wherefore when a straight line, &c. Q.E.D. PROPOSITION XIV. THEOREM. If at a point ºn a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles together equal to two right angles ; then these two straight lines shall be in one and the same straight line. At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles. , . Then BD shall be in the same straight line with BC. (— C B I) For, if BD be not in the same straight line with BC, if possible, let BE be in the same straight line with it. Then because A B meets the straight line CBE; therefore the adjacent angles CBA, ABE are equal to two right angles; (I. 13.) but the angles CBA, ABD are equal to two right angles; (hyp.) º º angles CBA, ABE are equal to the angles CBA, ABD: ax. 1. take away from these equals the common angle CBA, therefore the remaining angle ABE is equal to the remaining angle ABD; (ax. 3.) the less angle equal to the greater, which is impossible: therefore B.E is not in the same straight line with BC. And in the same manner it may be demonstrated, that no other can be in the same straight line with it but BD, which therefore is in the Same straight line with BC. Wherefore, if at a point, &c. Q.E.D. PROPOSITION XV. THEOREM. If two straight lines cut one another, the vertical, or opposite angles shall be equal. f Let the two straight lines AB, CD cut one another in the point E. Book I. PROP. xv, xvi. 17 Then the angle AEC shall be equal to the angle DEB, and the angle CEB to the angle A.E.D. Decause the straight line A.E makes with CD at the point E, the adjacent angles CEA, AED; these angles are together equal to two right angles. (I. 13.) Again, because the straight line D.E makes with AB at the point E, the adjacent angles A.E.D, DEB; these angles also are equal to two right angles; but the angles CEA, AED have been shewn to be equal to two right angles; wherefore the angles CEA, AED are equal to the angles AED, DEB; take away from each the common angle AED, and theremaining angle CEA is equaltotheremaining angle D.E.B. (ax. 3.) In the same manner it may be demonstrated, that the angle CEB is equal to the angle A.E.D.- Therefore, if two straight lines cut one another, &c. Q.E.D. CoR. 1. From this it is manifest, that if two straight lines cut each other, the angles which they make at the point where they cut, are together equal to four right angles. CoR. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles. PROPOSITION XVI. THEOREM. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let the side BC be produced to D. Then the exterior angle A CD shall be greater than either of the interior opposite angles CBA or BAC. - Bisect AC in E, (I. 10.) and join BE; produce B.E to F, making EF equal to B.E, (I. 3.) and join FC. Because AE is equal to EC, and BE to EF; (constr.) the two sides A.E, EB are equal to the two CE, E.F, each to each, in the triangles A.B.E, CFE; and the angle A.EB is equal to the angle CEF, because they are opposite vertical angles; (I. 15.) therefore the base AIB is equal to the base CF, (r. 4.) and the triangle AEB to the triangle CEF, C 18 EUCLID'S ELEMENTS. and the remaining angles of one triangle to the remaining angles of the other, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD or A CD is greater than the angle ECF; therefore the angle A CD is greater than the angle BAE or B.A. C. In the same manner, if the side BC be bisected, and AC be pro- duced to G; it may be demonstrated that the angle BCG, that is, the angle A CD, (I. 15.) is greater than the angle ABC. . Therefore, if one side of a triangle, &c. Q.E.D. PROPOSITION XVII. THEOREM. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle. Then any two of its angles together shall be less than two right angles. A - Produce any side BC to D. Then because A CD is the exterior angle of the triangle ABC; therefore the angle A CD is greater than the interior and opposite angle ABC; (I. 16.) to each of these unequals add the angle A CB; therefore the angles A CD, ACB are greater than the angles ABC, ACB ; but the angles A CD, ACB are equal to two right angles; (I, 13.) therefore the angles ABC, ACB are less than two right angles. Tn like manner it may be demonstrated, \ that the angles BAC, ACB are less than two right angles; as also the angles CAB, ABC. Therefore any two angles of a triangle, &c. Q.E.D. PROPOSITION XVIII. THEOREM. The greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of which the side AC is greater than the side AB. Then the angle ABC shall be greater than the angle ACB. A /> B C Since the side A C is greater than the side AB, (hyp.) make A.D equal to AB, (I. 3.) and join BID. Then, because AD is equal to AB, in the triangle ABD, therefore the angle ABD is equal to the angle ADB, (I. 5.) BOOK I. PROP. XVIII—XX. J.9 but because the side CD of the triangle BDC is produced to 4, therefore the exterior angle ADB is greater than the interior and opposite angle DCB; (I. 16.) *~ but the angle ADB has been proved equal to the angle ABD, therefore the angle ABD is greater than the angle DOB; wherefore much more is the angle ABC greater than the angle 4 CB. Therefore the greater side, &c. Q.E.D. - PROPOSITION XIX. THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. - Let ABC be a triangle of which the angle ABC is greater than the angle BCA. Then the side A C shall be greater than the side AB. A. - { º C For, if A C be not greater than AB, AC must either be equal to, or less than AB; if A C were equal to AB, then the angle ABC would be equal to the angle ACB; (I. 5.) e but it is not equal; (hyp.) - therefore the side AC is not equal to A.B. Again, if A C were less than AB, then the angle ABC would be less than the angle A CB; (I. 18.) but it is not less, (hyp.) - therefore the side A C is not less than AB; and AC has been shewn to be not equal to AB; therefore AC is greater than A.B. Wherefore the greater angle, &c. Q.E.D. PROPOSITION XX. THEOREM. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle. Then any two sides of it together shall be greater than the third side, viz. the sides BA, AC greater than the side BC; AB, BC greater than AC; and BC, CA greater than AB. D - * B C Produce the side BA to the point D, make AD equal to AC, (I. 3.) and join DC. C 2 20 EUCLID's ELEMENTS. º Then because AD is equal to AC, (constr.) therefore the angle A CD is equal to the angle ADC; (I. 5.) but the angle BCD is greater than the angle A CD; (ax. 9.) therefore also the angle BCD is greater than the angle ADC. And because in the triangle DBC, the angle BCD is greater than the angle BDC, and that the greater angle is subtended by the greater side; (I. 19.) therefore the side DB is greater than the side BC; but DB is equal to BA and A C, therefore the sides B.A. and AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA; also that BC, CA are greater than A.B. Therefore any two sides, &c. Q. E. D. PROPOSITION XXI. THEOREM, If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle ; these shall be less than the other two sides of the triangle, but shall contain a greater angle. Det ABC be a triangle, and from the points B, C, the ends of the side BC, let the two straight lines BD, CD be drawn to a point D within the triangle. - - Then BD and DC shall be less than BA and AC the other two sides of the triangle, - but shall contain an angle BDC greater than the angle BAC. Produce B D to meet the side A C in E. Because two sides of a triangle are greater than the third side, (I. 20.) therefore the two sides BA, AE of the triangle ABE are greater than B E ; to each of these unequals add EC; therefore the sides B.A, AC are greater than B.E., E.C. (ax. 4.) Again, because the two sides CE, ED of the triangle CED are greater than DC; (I. 20.) add DB to each of these unequals; therefore the sides CE, EB are greater than CD, D.B. (ax. 4.) But it has been shewn that BA, AC are greater than B.E, EC; - much more then are BA, A C greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle; (I. 16.) therefore the exterior angle BDC of the triangle CDE is greater than the interior and opposite angle CED; for the same reason, the exterior angle CED of the triangle ABE is greater than the interior and opposite angle BAC; and it has been demonstrated, that the angle BDC is greater than the angle CEB; much more therefore is the angle BDC greater than the angle BAC. Therefore, if from the ends of the side, &c. q. E. D. BOOK F. PROP. xxii, XXIII. . 21 PROPOSITION XXII. PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third. - Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, (I. 20.) namely, A and B greater than C; A and C greater than B; and B and C greater than A. - It is required to make a triangle of which the sides shall be equal to A, B, C, each to each. EC A. D / 2\\ C L – Take a straight line D.E terminated at the point D, but unlimited towards E, make DF equal to A, FG equal to B, and GH equal to C; (I. 3.) from the center F, at the distance FD, describe the circle DJſ L.; (post. 3.) -- from the center G, at the distance GH, describe the circle HLRſ; from K where the circles cut each other draw KF, KG to the points F, G; t then the triangle KFG shall have its sides equal to the three straight lines A, B, C. Because the point Fis the center of the circle DKL, therefore FD is equal to FK; (def. 15.) but FD is equal to the straight line A ; therefore FK is equal to 4. - Again, because G is the center of the circle HKL; therefore GH is equal to G.K., (def. 15.) - but GH is equal to C; therefore also GK is equal to C; (ax. 1.) and FG is equal to B; therefore the three straight lines KF, F6, GK, are respectively equal to the three, A, B, C: and therefore the triangle KFG has its three sides KF, FG, GH, equal to the three given straight lines A, B, C. Q. E. F. PROPOSITION XXIII. PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. *- Let AB be the given straight line, and A the given point in it, - and DCE the given rectilineal angle. It is required, at the given point A in the given straight line AB, to make an angle that shall be equal to the given rectilineal angle DCE. 22. EUCLID's ELEMENTS. C - E A. \ / B In CD, CE, take any points D, E, and join DE; on AB, make the triangle AFG, the sides of which shall be equal' to the three straight lines CD, DE, EC, so that AF be equal to, CD, A G to CE, and FG to D.E. (I. 22.) Then the angle FAG shall be equal to the angle DCE. Because FA, AG are equal to DC, CE, each to each, and the base FG is equal to the base DE; therefore the angle FA G is equal to the angle DCE. (I. 8.) Wherefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q.E.F. I) /* PROPOSITION XXIV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other; the base of that which has the greater angle, shall be greater than the base of the other. - Let ABC, DEF be two triangles, which have the two sides AB, A C, equal to the two DE, DF, each to each, namely, AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF Then the base BC shall be greater than the base EF. D A B C F Of the two sides DE, DF, let DE be not greater than DF, at the point D, in the line DE, and on the same side of it as DF, make the angle EDG equal to the angle BAC; (I. 23.) make DG equal to D.F or AC, (I. 3.) and join EG, GF - - Then, because DE is equal to AB, and DG to AC, the two sides DE, DG are equal to the two AB, AC, each to each, and the angle EDG is equal to the angle BAC; therefore the base EG is equal to the base B.C. (I. 4.) And because DG is equal to DF in the triangle DFG, therefore the angle DFG is equal to the angle DGF; (I. 5.) but the angle DGF is greater than the angle EGF; (ax.9.) therefore the angle DFG is also greater than the angle EGF; much more therefore is the angle EFG greater than the angle EGF- And because in the triangle EFG, the angle EFG is greater than the angle EGF, and that the greater angle is subtended by the greater side; (I, 19.) therefore the side EG is greater than the side EF; - but EG was proved equal to BC; - therefore B0 is greater than EF. Wherefore, if two triangles, &c. Q. E. P. Book I. PROP. xxv, xxvi. 23 PROPOSITION XXV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other; the angle contained by the sides of the one which has the greater base, shall be greater than the angle contained by the sides, equal to them, of the other. Iet ABC, DEFbe two triangles which have the two sides AB, AC, equal to the two sides DE, DF, each to each, namely, AB equal to DE, and AC to DF'; but the base BC greater than the base EF. Then the angle BAC shall be greater than the angle EDF A. D \\ E} C E. F For, if the angle BAC be not greater than the angle EDF, it must either be equal to it, or less than it. If the angle BAC were equal to the angle EDF, then the base BG would be equal to the base EF; (I. 4.) but it is not equal, (hyp.) *~, therefore the angle BAC is not equal to the angle EDF. Again, if the angle BAC were less than the angle EDF, then the base BC would be less than the base EF; (I. 24.) - but it is not less, (hyp.) therefore the angle BAC is not less than the angle EDF'; and it has been shewn, that the angle BAC is not equal to the angle EDF; therefore the angle BAC is greater than the angle EDF Wherefore, if two triangles, &c. Q.E.D. PROPOSITION XXVI. THEORFM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal angles in each, or the sides opposite to them; then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other. - Let ABC, DEF be two triangles which have the angles ABC, BCA, equal to the angles DEF, EFD, each to each, namely, ABC to DEF, and BCA to EFD; also one side equal to one side. First, let those sides be equal which are adjacent to the angles tha. are equal in the two triangles, namely, BC to EF. Then the other sides shall be equal, each to each, namely, AB to DE, and AC to D.F, and the third angle BAC to the third angle EDF. A D G - N **-º-º-º- S. B C * 24 EUCLID's ELEMENTS: For, if A B be not equal to DE, one of them must be greater than the other. If possible, let AB be greater than DE, make BG equal to ED, (I. 3.) and join G.C. Then in the two triangles GBC, DEF, because GB is equal to DE, and BC to EF, (hyp.) the two sides. GB, BC are equal to the two DE, EF, each to each;, and the angle GBC is equal to the angle DEF'; therefore the base GC is equal to the base D.F. (I. 4.) and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; - therefore the angle GCB is equal to the angle DFE; but the angle ACB is, by the hypothesis, equal to the angle DFE; wherefore also the angle GCB is equal to the angle ACB ; (ax. 1.) the less angle equal to the greater, which is impossible; therefore AB is not unequal to DE, that is, AB is equal to D.E. Hence, in the triangles ABC, DEF; because AB is equal to DE, and BC to EF, (hyp.) and the angle ABC is equal to the angle DEF; (hyp.) therefore the base AC is equal to the base DF, (I. 4.) and the third angle BAC to the third angle EDF Secondly, let the sides which are opposite to one of the equal angle in each triangle be equal to one another, namely, AB equal to D.I. Then in this case likewise the other sides shall be equal, AC to DF, and BC to EF, and also the third angle BAC to the third angle EDF A D For if BC be not equal to EF, one of them must be greater than the other. If possible, let BC be greater than EF; make BH equal to EF, (I. 3.) and join A.H. Then in the two triangles ABH, DEF, because AB is equal to DE, and BH to EF, and the angle ABH to the angle DEF; (hyp.) therefore the base AH is equal to the base DF, (1. 4.) and the triangle ABH to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite ; therefore the angle BHA is equal to the angle EFD; but the angle EFD is equal to the angle BCA; (hyp.) therefore the angle BHA is equal to the angle BCA, (ax. 1.) that is, the exterior angle BHA of the triangle AHC, is "equal to its interior and opposite angle BCA; which is impossible; (I. 16.) wherefore BC is not unequal to EF, that is, B0 is equal to EF. Hence, in the triangles ABC, DEF; because AB is equal to DE, and BC to EF, (hyp.) BOOK I. PROP. KXVI—XXVIII. 25. and the included angle ABC is equal to the included angle DEF; (hyp.) therefore the base AC is equal to the base DF, (I. 4.) and the third angle BAC to the third angle EDF, Wherefore, if two triangles, &c. Q.E.D. PROPOSITION xxvii. THEOREM. If a straight line falling on two other straight lines, make the alter- nate angles equal to each other; these two straight lines shall be parallel. Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another, Then, A B shall be parallel to CD. For, if A B be not parallel to CD, then AB and CD being produced will meet, either towards A and C, or towards B and D. 4 Let AB, CD, be produced and meet if possible, towards B and D, in the point G; then GEF is a triangle. And because a side GE of the triangle GEF is produced to A, therefore its exterior angle AEF is greater than the interior and opposite angle EFG ; (I. 16.) but the angle AEF is equal to the angle EFG ; (hyp.) therefore the angle AEF is greater than, and equal to, the angle EFG ; which is impossible. Therefore AB, CD being produced, do not meet towards B, D. In like manner, it may be demonstrated, that they do not meet when produced towards A, C. - But those straight lines in the same plane, which meet neither way, though produced ever so far, are parallel to one another; (def. 35.) - therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION XXVIII. THEOREM. If a straight line falling upon two other straight lines, make the octerior angle equal to the interior and opposite upon the same side of the line; or make the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another. - Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD, upon the same side of the line EF; or make the two interior angles. BGH, G HD on the same side together equal to two right angles. Then AB shall be parallel to CD. 26 EUCLID'S ELEMENTS. Because the angle EGB is equal to the angle GHD, (hyp.) and the angle EGB is equal to the angle A GH, (I, 15.) , . therefore the angle AGH is equal to the angle GHD; (ax. 1.) and they are alternate angles, therefore AB is parallel to CD. (I. 27.) Again, because the angles BGH, G HD are together equal to two right angles, (hyp.) and that the angles A GH, BGH are also together equal to two right angles; (I. 13.) therefore the angles A GH, BGH are equal to the angles BGH, GHD; (ax. 1.) take away from these equals, the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; (ax. 3.) and they are alternate angles; therefore AB is parallel to CD. (I. 27.) Wherefore, if a straight line, &c. Q.E.D. PROPOSITION XXIX. THEOREM. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; and the eacterior angle equal to the winterior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles. * Let the straight line EF'fall upon the parallel straight lines AB, CD. Then the alternate angles A GH, G HD shall be equal to one another; the exterior angle EGB shall be equal to the interior and opposite angle GHD upon the same side of the line EF; - and the two interior angles BGH, G HD upon the same side of EF, shall be together equal to two right angles. First. For, if the angle AGH be not equal to the alternate angle GHD, one of them must be greater than the other; if possible, let A GH be greater than GHD, then because the angle A GH is greater than the angle GHD, add to each of these unequals the angle BGH; therefore the angles A GH, BGH are greater than the angles BGH, GHD; (ax. 4.) but the angles A GH, BGH are equal to two right angles; (I, 13.) therefore the angles BGH, G HD are less than two right angles; but those straight lines, which with another straight line falling upon them, make the two interior angles on the same side less than two right angles, will meet together if continually produced; (ax. 12.) therefore the straight lines AB, CD, if produced far enough, will Imeet towards B, D ; but they never meet, since they are parallel by the hypothesis; therefore the angle AGH is not unequal to the angle GHD, that is, the angle A GH is equal to the alternate angle GHD. Secondly, because the angle A GH is equal to the angle EGB; (I, 15.) and the angle A GH is equal to the angle GHD, **- BOOK I. PROP. XXIX—XXXI. 27. therefore the exterior angle EGB is equal to the interior and opposite, * . angle GHD on the same side of the line. Thirdly. Because the angle EGB is equal to the angle GHD, add to each of them the angle BGH; therefore the angles EGB, BGH are equal to BGH, G HD; (ax. 2.) but EGB, BGH are equal to two right angles ; (I. 13.) therefore also the two interior angles BGH, G HD on the same side. of the line are equal to two right angles. (ax. 1.) Wherefore, if a straight line, &c. Q.E.D. PROPOSITION XXX. THEOREM. Straight lines which are parallel to the same straight line are parallel to each other. Let the straight lines AB, CD, be each of them parallel to EF. Then shall A.B be also parallel to CD. Let the straight line GHK cut AB, EF, CD. Then because GHIſ cuts the parallel straight lines AB, EF, in G, II; therefore the angle AGH is equal to the alternate angle GHF (I. 29.) Again, because GHR cuts the parallel straight lines EF, CD, in H, K ; therefore the exterior angle GHF is equal to the interior angle HKD;. and it was shewn that the angle A GH is equal to the angle GHF'; therefore the angle AGH is equal to the angle GKD; and these are alternate angles; therefore AB is parallel to CD. (I. 27.) Wherefore, straight lines which are parallel, &c. Q.E.D. PROPOSITION XXXI. PROBLEM. To draw a straight line through a given point parallel to a given. straight line. Let A be the given point, and BC the given straight line. It is required to draw, through the point A, a straight line parallel, to the straight line B.C. E A F B D C In the line BC take any point D, and join AD, at the point A in the straight line AD, make the angle DAE equal to the angle ADC, (I. 23.) on the opposite side of AD; and produce the straight line E.A to F. * Then EF shall be parallel to BC. Because the straight line AD meets the two straight lines E.F, BC, and makes the alternate angles EAD, ADC, equal to one another, therefore EF is parallel to B.C. (I. 27.) Wherefore, through the given point A, has been drawn a straight line EAF parallel to the given straight line B.C. Q.E. F. 28. - EUCLID'S ELEMENTS. PROPOSITION XXXII. THEOREM. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are together equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D. Then the exterior angle A CD shall be equal to the two interior and opposite angles CAB, ABC: and the three interior angles ABC, BCA, CAB shall be equal to. two right angles. - A. , E. B C D Through the point C draw CE parallel to the side B.A. (I. 31.) Then because CE is parallel to BA, and A C meets them, r therefore the angle ACE is equal to the alternate angle B.A. C. (I. 29.) Again, because CE is parallel to AB, and BD falls upon them, therefore the exterior angle ECD is equal to the interior and op- posite angle ABC; (I. 29.) but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle A CD is equal to the two interior and opposite angles CAB, ABC. (ax. 2.) - Again, because the angle A CD is equal to the two angles ABC, BA-C, to each of these equals add the angle ACB, \ therefore the angles A CD and ACB are equal to the three angles ABC, BAC, and A.C.B.; (ax. 2.) but the angles A CD, ACB are equal to two right angles, (I. 13.) therefore also the angles ABC, BAC, ACB are equal to two right angles. (ax. 1.) Wherefore, if a side of any triangle be produced, &c. Q.E.D. CoR. 1. All the interior angles of any rectilineal figure together with four right angles, are equal to twice as many right angles as ther figure has sides. /* For any rectilineal figure ABCDE can be divided into as many tri- angles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. * r Then, because the three interior angles of a triangle are equal to two right angles, and there are as many triangles as the figure has sides, therefore all the angles of these triangles are equal to twice as many right angles as the figure has sides; but the same angles of these triangles are equal to the interior angles of the figure together with the angles at the point F: BOOK I. PROP. xxx11, xxxHII. 29 and the angles at the point F, which is the common vertex of all the triangles, are equal to four right angles; (I. 15. Cor. 2.) therefore the same angles of these triangles are equal to the angles of the figure together with four right angles; but it has been proved that the angles of the triangles are equal to twice as many right angles as the figure has sides; therefore all the angles of the figure together with four right angles, are equal to twice as many right angles as the figure has sides. . CoR. 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. X N C, D B Since every interior angle ABC, with its adjacent exterior angle ABD, is equal to two right angles, (I. 13.) * • therefore all the interior angles, together with all the exterior angles, are equal to twice as many right angles as the figure has sides; but it has been proved by the foregoing corollary, that all the inte- rior angles together with four right angles are equal to twice as many right angles as the figure has sides; therefore all the interior angles together with all the exterior angles, are equal to all the interior angles and four right angles, (ax. 1.) take from these equals all the interior angles, therefore all the exterior angles of the figure are equal to four right angles. (ax. 3.) PROPOSITION XXXIII. THEOREM. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel. . . Let 4B, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, B'D. - Then AC, BD shall be equal and parallel. A B V\ - Join BC. g Then because AB is parallel to CD, and BC meets them, therefore the angle ABC is equal to the alternate angle BCD; (r. 29.) and because AB is equal to CD, and BC common to the two tri- angles 4 BC, DOB; the two sides AB, BC, are equal to the two DC, 05, each to each, and the angle ABC was proved to be equal to the angle BCD: therefore the base AC is equal to the base BD, (1. 4.) 30 - EUCLID's ELEMENTS. and the triangle ABC to the triangle BCD, and the other angles to the other angles, each to each, to which the equal sides are opposite; - therefore the angle A CB is equal to the angle CBD. And because the straight line BC meets the two straight lines AC, JB D, and makes the alternate angles ACB, CBD equal to one another; therefore AC is parallel to BD; (I. 27.) e and AC was shewn to be equal to B.D. Therefore, straight lines which, &c. Q.E.D. PROPOSITION XXXIV. THEOREM. The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it into two equal parts. - Let A CDB be a parallelogram, of which BC is a diameter. Then the opposite sides and angles of the figure shall be equal to One another; and the diameter BC shall bisect it. Because AB is parallel to CD, and BC meets them, . therefore the angle ABC is equal to the alternate angle BCD. (I. 29.) And because A C is parallel to BD, and BC meets them, therefore the angle ACB is equal to the alternate angle CBD. (I. 29.) Hence in the two triangles ABC, CBD, because the two angles ABC, BCA in the One, are equal to the two angles BCD, CBD in the other, each to each; and one side BC, which is adjacent to their equal angles, common to - the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, (I. 26.) namely, the side AB to the side CD, and 40 to BD, and the angle BAC to the angle BDC. - And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle A CD; ax. 2. *-, ( and i. angle BAC has been shewn to be equal to BDC; therefore the opposite sides and angles of a parallelogram are equal to one another. - Also the diameter BC bisects it. - For since AB is equal to CD, and BC common, - the two sides A,B, BC, are equal to the two DC, CB, each to each, and the angle ABC has been proved to be equal to the angle BCD; , therefore the triangle ABC is equal to the triangle BCD; (I. 4.) and the diameter BC divides the parallelogram A CDB into two equal parts. Q. E. D. - * BOOK I. PROP. xxxv, xxxvi. 31 PROPOSITION XXXV. THEOREM. Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, B.C. Then the parallelogram ABCD shall be equal to the parallelogram EBOF. * piz vºz' wº ETT6 B C B C If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D; then it is plain that each of the parallelograms is double of the triangle BDC; (I. 34.) - and therefore the parallelogram ABCD is equal to the parallelogram DBCF. (ax, 6.) But if the sides AD, EF, opposite to the base BC, be not termi- nated in the same point; Then, because ABCD is a parallelogram, therefore AD is equal to BC; (I. 34.) and for a similar reason, EF is equal to BC; wherefore AD is equal to EF; (ax. 1.) - and DE is common : therefore the whole, or the remainder A.E, is equal to the whole, or the remainder DF; (ax. 2 or 3.) and AB is equal to DC; (I. 34.) hence in the triangles EAB, FDC, because FD is equal to EA, and DC to AB, and the exterior angle FDC is equal to the interior and opposite angle JEAB; (I. 29.) - therefore the base FC is equal to the base EB, (I. 4.) and the triangle FDC is equal to the triangle E.A.B. From the trapezium ABCF take the triangle FDC, and from the same trapezium take the triangle EAB, - and the remainders are equal, (ax. 3.) therefore the parallelogram ABCD is equal to the parallelogram. EBCF. Therefore, parallelograms upon the same, &c. Q.E.D. PROPOSITION XXXVI. THEOREM. Parallelograms upon equal bases and between the same parallels, are equal to one another. g - Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels A.H., B.G. zº the parallelogram ABCD shall be equal to the parallelogram A D E H 32 ‘EUCLID's ELEMENTS. dº Join BE, C.H. Then because BC is equal to FG, (hyp.) and FG to EH, (1.34.) therefore BC is equal to EH ; (ax. 1.) and these lines are parallels, and joined towards the same parts by the straight lines B.E, CH; - but straight lines which join the extremities of equal and parallel straightlines towards the same parts, are themselves equal and parallel; (I. 33. ) therefore B.E, CH are both equal and parallel ; wherefore EBCH is a parallelogram. (def. A.) - And because the parallelograms ABCD, EBCH, are upon the same base BC, and between the same parallels BC, API; therefore the parallelogram A.BCD is equal to the parallelogram EBCH. (I. 35.) For the same reason, the parallelogram. EFGH is equal to the parallelogram. EBCH; therefore the parallelogram ABCD is equal to the parallelogram Therefore, parallelograms upon equal, &e. Q.E.D. PROPOSITION XXXVII. THEOREM. Triangles upon the same base and between the same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC, .." and between the same parallels AD, B.C. . Then the triangle ABC shall be equal to the triangle DBC. E A D F N/\!/ B C Troduce AD both ways to the points E, F; through B draw B.E parallel to CA, (I. 31.) and through C draw CF parallel to B.D. Then each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DB CF. (I. 35.) because they are upon the same base BC, and between the same parallels BC, EF. And because the diameter AB bisects the parallelogram. EBCA, therefore the triangle ABC is half of the parallelogram. EBCA; (I. 34.) also because the diameter DC bisects the parallelogram DB CF, therefore the triangle DBC is half of the parallelogram DBCF, but the halves of equal things are equal; (ax. 7.) therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c. Q.E.D. PROPOSITION XXXVIII. THEOREM. Triangles upon equal bases and between the same parallels, are equal to one another. - BOOK I. PROP. xxxvii.1, xxxix. 33 Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, A.D. Then the triangle ABC shall be equal to the triangle DEF: G A D H V\|N| ‘B -C E JF Produce AD both ways to the points G, H.; through B draw. BG parallel to CA, (I. 31.) and through F draw FH parallel to ED. Then each of the figures GBCA, DEFEI is a parallelogram; and they are equal to one another, (r. 36.) because they are upon equal bases BC, EF, and between the same parallels BF, GH. And because the diameter AB bisects the parallelogram GBCA, therefore the triangle ABC is the half of the parallelogram GBCA ; 1. 34. ( it. because the diameter DF'bisects the parallelogram DEFH, therefore the triangle DEF is the half of the parallelogram DEFH: but the halves of equal things are equal; (ax. 7.) therefore the triangle ABC is equal to the triangle DEF Wherefore, triangles upon equal bases, &c. Q.E.D. PROPOSITION XXXIX. THEOREM. JEqual triangles upon the same base and upon the same side of it, are between the same parallels. Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it. Then the triangles ABC, DBC shall be between the same parallels. .A I) B C Join AD; then AD shall be parallel to BC. ... - For if AD be not parallel to BC, if possible, through the point A, draw AE parallel to BC, (I. 31.) meeting BD, or BD produced, in E, and join EC. Then the triangle ABC is equal to the triangle EBC, (I. 37.) because they are upon the same base BC, and between the same parallels BC, AE: but the triangle ABC is equal to the triangle DBC; (hyp.) therefore the triangle DBC is equal to the triangle EBC, the greater triangle equal to the less, which is impossible: therefore AE is not parallel to BC. In the same manner it can be demonstrated, that no other line drawn from A but AD is parallel to BC; AD is therefore parallel to BC. Wherefore, equal triangles upon, &c. Q.E.D. - D \ 34 EUCLID'S ELEMENTS. PROPOSITION XI. THEOREM. Equal triangles upon equal bases in the same straight line, and towards the same parts, are between the same parallels. Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight line BF, and towards the same parts. Then they shall be between the same parallels. A D VT-J. /\ºs B C E F. Join AD; then AD shall be parallel to B.F. For if AD be not parallel to B.F. * if possible, through. A draw AG parallel to BF (I. 31.) * meeting ED, or ED produced in G, and join GF. Then the triangle ABC is equal to the triangle GEF, (I. 38.) because they are upon equal bases BC, EF, and between the same parallels BF, A G ; but the triangle ABC is equal to the triangle DEF; (hyp.) therefore the triangle DEF is equal to the triangle GEF, (ax. 1.) the greater triangle equal to the less, which is impossible: - therefore AG is not parallel to B.F. N And in the same manner it can be demonstrated, that there is no other line drawn from A parallel to it but AD; 2- AD is therefore parallel to B.F. Wherefore, equal triangles upon, &c. Q.E.D. PROPOSITION XLI. THEOREM. If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle. Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AF. Then the parallelogram ABCD shall be double of the triangle EBC. A D E 2^ F8 C Join A. C. Then the triangle ABC is equal to the triangle EBC, (I. 37.) because they are upon the same base BC, and between the same parallels BC, A.E. - But the parallelogram ABCD is double of the triangle ABC, : because the diameter AC bisects it; (I. 34.) wherefore ABCD is also double of the triangle EBC. . Therefore, if a parallelogram and a triangle, &c. Q.E.D. BOOK I. PROP. XLII, XLIII. 35 PROPOSITION XLII. PROBLEM. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D. W . Bisect BC in E, (I. 10.) and join AE; at the point E in the straight line EC, 'make the angle CEF equal to the angle D; (I. 23.) ‘through C draw CG parallel to EF, and through A draw AFG parallel to BC, (I. 31.) meeting EF in F, and CG in G. Then the figure CEFG is a parallelogram. (def. A.) And because the triangles A.B.E, AEC are on the equal bases B.E, JEC, and between the same parallels BC, AG; they are therefore equal to one another; (I. 38.) & and the triangle ABC is double of the triangle AEC; but the parallelogram FECG is double of the triangle AEC, (I. 41.) because they are upon the same base EC, and between the same parallels EC, AG; therefore the parallelogram FECG is equal to the triangle ABC, (ax. 6.) and it has one of its angles CEF equal to the given angle D. Wherefore, a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle B. Q. E. F. • PROPOSITION XLIII. THEOREM, The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another. Let ABCD be a parallelogram, of which the diameter is A C: and JEH, GF the parallelograms about AC, that is, through which AC passes: also BK, KD the other parallelograms which make up the whole _, figure ABCD, which are therefore called the complements. {Then the complement BK shall be equal to the complement KD. A H D s K TNT BTG C Because ABCD is a parallelogram, and AC its diameter, therefore the triangle ABC is equal to the triangle ADC. (I. 34.) Again, because EKHA is a parallelogram, and A Kits diameter, D 2 36 EUCLID'S ELEMENTS. therefore the triangle AEK is equal to the triangle AITK; (I. 34.) and for the same reason, the triangle KGC is equal to the triangle KFC. Wherefore the two triangles AEK, KGC are equal to the two triangles AHK, KFC, (ax. 2.) but the whole triangle ABC is equal to the whole triangle ADC; therefore the remaining complement BK is equal to the remaining complement KD. (ax. 3.) Wherefore the complements, &c. Q.E.D. - PROPOSITION XLIV. PROBLEM. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. - It is required to apply to the straight line AB, a parallelogram equal to the triangle C, and having an angle equal to the angle D. F E JK N ºf H. A L Make the parallelogram BEFG equal to the triangle C, and having the angle EB G equal to the angle D, (I. 42.) so that BE be in the same straight line with AB; produce FG to H, through A draw AH parallel to BG or EF, (I. 31.) and join H.B. Then because the straight line IIF falls upon the parallels AH, EF, therefore the angles A.H.F, HFE are together equal to two right angles; (I. 29.) wherefore the angles BHF, HFE are less than two right angles: but straight lines which with another straight line, make the two interior angles upon the same side less than two right angles, do meet if produced far enough: (ax. 12.) therefore HB, FE shall meet if produced; let them be produced and meet in K, through K draw KL parallel to EA or FH, and produce HA, GB to meet Iſ L in the points L, Mſ. Then HL/CR' is a parallelogram, of which the diameter is IIIſ; and AG, ME are the parallelograms about HK; s also LB, BF are the complements; therefore the complement LB is equal to the complement BF'; (1.43.) but the complement B F is equal to the triangle C; (constr.) wherefore LB is equal to the triangle C. And because the angle GBE is equal to the angle ABM, (I. 15.) and likewise to the angle D; (constr.) therefore the angle ABM is equal to the angle D. (ax. 1.) Therefore to the given straight line AB, the parallelogram LB has been applied, equal to the triangle C, and having the angle ABIſ equal to the given angle D. Q. E. F. BOOK I. PROP. XIV. 37 PROPOSITION XIV. PROPLEM. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given recti- lineal angle. It is required to describe a parallelogram that shall be equal to the figure ABCD, and having an angle equal to the given angle B. D G L A →T F | ſº B C T KTHTM Join D.B. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E.; (I. 42.) - * ' to the straight line GH, apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. I. 44. . Tº the iºn FKML shall be the parallelogram required. Because each of the angles FKH, GHM, is equal to the angle E, therefore the angle FICH is equal to the angle GHM; add to each of these equals the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal to two right angles; (I. 29.) therefore also KHG, GHM are equal to two right angles; and because at the point H, in the straight line GH, the two straight lines KH, HIM, upon the opposite sides of it, make the ad- jacent angles JCHG, GHM equal to two right angles, - therefore III is in the same straight line with H.M. (I. 14.) And because the line HG meets the parallels KM, FG, therefore the angle MHG is equal to the alternate angle HGF; (I. 29.) add to each of these equals the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGI, ; but the angles MHG, HGL are equal to two right angles; (I. 29.) therefore also the angles HGF, HGL are equal to two right angles, and therefore FG is in the same straight line with G.L. (I, 14.) And because KP is parallel to HG, and HG to ML, therefore KF is parallel to ML; (I. 30.) and FL has been proved parallel to KM, wherefore the figure FKML is a parallelogram; and since the parallelogram HF is equal to the triangle ABD, and the parallelogram GM to the triangle BDC; therefore the whole parallelogram KFLM is equal to the whole rectilineal figure ABCD. w Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FICM equal to the given angle E. Q.E. F. CoR. From this it is manifest how, to a given straight line, to apply a parallelogram, which shall have an angle equal to a given rectilineal 38 EUCLID'S ELEMENTS. angle, and, shall be equal to a given rectilineal figure; viz. by applying to the given straight line, a parallelogram, equal to, the first triangle: ABD, (I, 44.) and having an angle equal to the given angle. PROPOSITION XLVI. PROBLEM. To describe a square upon a given straight line. Let AB be the given straight line. It is required to describe a square upon A.B. From the point A draw A-C at right angles to AB; (I, 11 ) make AiD equal to AB; (I. 3.) through the point D draw. DE parallel to AB; (I. 31.) and through B, draw B.E parallel to AD, meeting DE in E; therefore A BED is a parallelogram; whence AB is equal to DE, and AD to BE; (1.34.) but AD is equal to A:B, therefore the four lines AB, B.E., ED, D.A. are equal to...one another, and the parallelogram A BED is equilateral. It has likewise all its angles right angles; since AD meets the parallels AB, DE, therefore the angles B.A.D, ADE are equal to two right angles; (I. 29.) but B.A.D is a right angle; (constr.) therefore also ADE is a right angle. But the opposite angles of parallelograms are equal; (I. 34.) therefore each of the opposites angles A.B.E, BED is a right angle; wherefore the figure ABED is rectangular, and it has been proved to be equilateral; therefore the figure ABED is a square, (def. 30.) and it is described upon the given straight line A.B. Q. E. F. CoR. Hence, every parallelogram that has one of its angles a right angle, has all its angles right angles. PROPOSITION, XLVII. THEOREM. Jh any, right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain, the right angle. & Let ABC be a right-angled triangle, having the right angle BAC. Then the square described upon the side BC, shall be equal to the squares described upon BA, A C. BOOK I. PROP. XLVII, XLVIII. 39 G º B R D L E Tpon BC describe the square BDEC, (1.46.) and upon BA, AC the squares GB, HC ; through. A draw AL parallel to BD or CE; (I. 31.) and join AD, FC. Then because the angle BAC is a right angle, (hyp.) and that the angle BAG is a right angle, (def. 30.) the two straight lines AC, AG upon the opposite sides of AB, make with it at the point A, the adjacent angles equal to two right angles; therefore CA is in the same straight line with A.G. (I. 14.) For the same reason, BA and AH are in the same straight line. And because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each of these equals the angle ABC, therefore the whole angle ABD is equal to the whole angle FBC. (ax. 2.) And because the two sides AB, BD are equal to the two sides º'B, BC, each to each, and the included angle ABD is equal to the included angle FBC, - therefore the base AD is equal to the base FC, (I. 4.) and the triangle ABD to the triangle FBC. Now the parallelogram B L is double of the triangle ABD, (I. 41.) because they are upon the same base BI), and between the same parallels BD, AL; - e also the square GB is double of the triangle FBC, because these also are upon the same base FB, and between the same parallels FB, G.C. But the doubles of equals are equal to one another; (ax. 6.) therefore the parallelogram B L is equal to the square G.B. Similarly, by joining AE, BR, it can be proved, that the parallelogram CL is equal to the square H.C. Therefore the whole square BDEC is equal to the two squares GB, IIC; (ax. 2.) and the square BDEC is described upon the straight line BC, and the squares GB, HC, upon AB, A C : therefore the square upon the side BC, is equal to the squares upon the sides AB, A. C. Thereforg, in any right-angled triangle, &c. Q.E.D. K z PROPOSITION XLVIII. THEOREM. If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle. Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other two sides, AB, 40. 40 EUCLID's ELEMENTS. Then the angle BAC shall be a right angle. From the point A draw AD at right angles to AC, (I. 11.) make AD equal to AB, and join DC. Then, because AD is equal to AB, the square on AD is equal to the square on AB; to each of these equals add the square on A. C.; therefore, the squares on AD, A C are equal to the squares on AB, A C: but the squares on AD, A C are equal to the square on DC, (I. 47.) because the angle DA-C is a right angle; and the square on BC; by hypothesis, is equal to the Squares on BA, A C; therefore the square on DC is equal to the square on BC; and therefore the side D0 is equal to the side B.C. And because the side AD is equal to the side AB, and A C is common to the two triangles DAC, BAC; the two sides DA, AC, are equal to the two B.A, AC, each to each; and the base DC has been proved to be equal to the base BC; therefore the angle DAC is equal to the angle BAC; (I. 8.) but DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square described upon, &c. Q. E. D. NOTES TO BOOK I. ON THE DEFINITIONS. GEOMETRY is one of the most perfect of the deductive Sciences, and seems to rest on the simplest inductions from experience and observation. The first principles of Geometry are therefore in this view consistent hypotheses founded on facts cognizable by the senses, and it is a subject of primary importance to draw a distinetion between the conception of things and the things themselves. These hypotheses do not involve any property contrary to the real nature of the things, and consequently cannot be regarded as arbitrary, but in certain respects, agree with the conceptions which the things themselves suggest to the mind through the medium of the senses. The essential definitions of Geometry there- fore being inductions from observation and experience, rest ultimately on the evidence of the senses. \ It is by experience we become acquainted wità the existence of individual forms of magnitudes ; but by the mental process of abstraction, which begins with a particular instance, and proceeds to the general idea of all objects of the same kind, we attain to the general conception of those forms which come under the same general idea. - The essential definitions of Geometry express generalized conceptions of real existences in their most perfect ideal forms: the laws and appearances of nature, and the operations of the human intellect being supposed uniform and consistent. * - It has been maintained: by some writers both of ancient and modern times, that Geometry is a perfectly abstract Science, a body of truths completely independent of all human observation and experience. The truths of Geometry may, possibly, be a portion of absolute and universal truths, such as change not in all time, and which maintain the same constant universality under every conceivable state of existence. Still it must be admitted that however abstract and indepen- dent of experience such truths may be in themselves, it was not as such that they were originally discovered, or in that form that they are apprehended by the human mind. - The natural process of the human mind in the acquirement of knowledge and in the discovery of truth, is, to proceed from the particular to the general, from the sensible to the abstract. It is perhaps not too much to affirm, that the human mind would never have speculated on the abstract properties of circles and triangles unless some visible forms of such figures had first been exhibited to the senses. It does seem more probable and analogous to the rise and progress of other branches of human knowledge, that the fundamental truths of Geometry should have been first discovered from suggestions made to the senses; and this opinion, too, is not repugnant to the earliest historical notices existing on the subject. The human mind is so constituted that exact knowledge in any Science can only be acquired progressively. Every successive step in advance must be taken as a sequel to, and dependent upon, the previous acquirements; and some intelligible facts and first principles must form the basis of all human Science. Now the only possible way of explaining terms denoting simple perceptions is to excite those simple perceptions. The impossibility of defining a word expressive of a simple perception is well known to every one who has paid any attention to his own intellectual progress. The only way of rendering a simple term intelligible is to exhibit the object of which it is the sign, or some sensible representation of it. A straight line therefore must be drawn, and by R. 42 EUCLID's ELEMENTS. drawing a curved line and a crooked line, the distinction will be perfectly understood. Again, the definition of a complex term consists merely in the enumeration of the simple ideas for which it stands, and it will be found that all definitions must have some term or terms equally requiring definition or explanation with the one defined. TEut in cases where the subject falls under the class of simple ideas, the terms of the definitions so called, are no more than merely equivalent expressions. The simple idea described by a proper term or terms, does not in fact admit of definition properly so called. The definitions in Euclid's Elements may be divided into two classes, those which merely explain the meaning of the terms employed, and those, which, besides explaining the meaning of the terms, suppose the existence of the things described in the definitions. - Definitions in Geometry are intended to designate the things named, and not to explain the nature and properties of the figures defined: it is sufficient that they give marks whereby the thing defined may be distinguished from every other of the same kind. It will at once be obvious, that the definitions of Geometry, one of the pure sciences, being abstractions of space, are not like the definitions in any one of the physical sciences. - The discovery of any new physical facts may render necessary some alteration or modification in the definitions of the latter. - • - The definitions of Euclid appeal directly to the senses; and the fundamental theorem (Euc. I. 4) which forms the basis of all the succeeding propositions, is demonstrated by one of the simplest appeals to experience. At every step there is a reference made to something exhibited to the senses, the coincidence of the lines, the angles, and lastly, the surfaces of the two triangles; and by shewing a perfect coincidence, their equality is inferred. The instance exhibited, and the proof applied to it, is equally valid for any triangles whatever which have the same specified conditions given in the hypothesis. The same reasoning may be applied to any similar case which can be conceived, and thus from a single instance demonstrated by appeal to the senses, we are led to admit the statement contained in the general enunciation. These considerations appear to support “ the opinion, that the truths of Geometry, as a portion of human Science, rest ultimately on the evidence of the senses. It may also be suggested, whether it be not a point of considerable importance to be able to discriminate, where human Science begins, and how certainty is acquired. Def. I. Simson has adopted Theon's definition of a point. Euclid's definition is a musiov ša riv og uépos ovéév, “A point is that, of which there is no part,” or which cannot be parted or divided, as the line, the angle, the surface, and the solid.” The word point in Geometry is not employed in the sense in which the Earth is called a point in respect of the Universe. The Greek term a musiou, literally means, a visible sign or mark on a surface, in other words, a physical point. The English term point, means the sharp end of any thing, or a mark made by it. The word point comes from the Latin punctum, through the French word point. Neither of these terms, in its literal sense, appears to give a very exact notion of what is to be understood by a point in Geometry. Euclid's definition of a point merely expresses a negative property, which excludes the properand literal meaning of the Greek term, as appliéd to denote a physical point, or a mark which is visible to the senses. - ~ . Pythagoras defined a point to be uova's 6éorivºxova'a, “a monad having position.” By uniting the positive idea of position, with the negative idea of defect of magni- tude, the conception of a point in Geometry may be rendered intelligible, so that a point has position only, but no magnitude. 3. NOTES TO BOOK I. - 43 Def. II. Every visible line has both length and breadth, and it is impossible to draw any line whatever which shall have no breadth. The definition requires the conception of the length only of the line to be considered, abstracted from, and independently of, all idea of its breadth. * I)ef. III. This definition renders more intelligible the exact meaning of the definition of a point: and we may add, that, in the Elements, Euclid supposes that the intersection of two lines is a point, and that two straight lines can intersect each other in one point only. . Def. Iv. The straight line or right line is a term so clear and intelligible as to be incapable of becoming more so by formal definition. Euclid's definition is Eü08ta. Ypappui àorruv, fires é8 to ov 'rois Éq,’ &avriis ormuetous keirau, wherein he states it to lie evenly, or equally, or upon an equality (ść toov) between its ex- tremities, and which Proclus explains as being stretched between its extremities, 1j čar' drpov terapiávn. * (, If the line be conceived to be drawn on a plane surface, the words āś lorov may mean, that no part of the line which is called a straight line deviates either from one side or the other of the direction which is fixed by the extremities of the line; and thus it may be distinguished from a curved line, which does not lie, in this sense, evenly between its extreme points. If the line be conceived to be drawn in space, the words ič toov must be understood to apply to every direction on every side of the line between its extremities. - Every straight line situated in a plane, is considered to have two sides; and when the direction of a line is known, the line is said to be given in position; also when the length is known or can be found, it is said to be given in magnitude. From the definition of a straight line, it follows, that two points fix a straight line in position, which is the foundation of the first and second postulates. Hence straight lines which are proved to coincide in two or more points, are called, “ one and the same straight line,” Prop. 14, Book I, or which is the same thing, viz. “two straight lines cannot have a common segment.” . - The following definition of straight lines has also been proposed. “Straight lines are those which, if they coincide in any two points, coincide as far as they are produced.” But this is rather a criterion of straight lines, and analo- gous to the eleventh axiom, which states that, “all right angles are equal to one another,” and suggests that all straight lines may be made to coincide wholly; if the lines be equal; or partially, if the lines be of unequal lengths. A definitions should properly be restricted to the description of the thing defined, as it exists, independently of any comparison of its properties or of tacitly assuming the ex- istence of axioms. I)ef, v11. Euclid's definition of a plane surface is 'Earírečos étruqdustd: ágriv. #tts #8 to ov rais Éq' éavriis ei'0stats keitat, “A plane surface is that which lies. evenly or equally with the straight lines in it;” instead of which Simson has given. the definition which was originally proposed by Hero the Elder. A plane super- ficies may be supposed to be situated in any position, and to be continued in every, direction to any extent. - * Def. VIII. Simson remarks that this definition seems to include the angles formed by two curved lines, or a curve and a straight line, as well as that formed by two straight lines. The latter only are treated of in Elementary Geometry. Def. IX. An angle is a species of magnitude; for one angle may be greater. than, equal to, or less than another angle. It is of the highest importance to . attain a clear conception of an angle, and of the sum and difference of two. angles. The literal meaning of the term angulus suggests the Geometrical conception of an angle, which may be regarded as formed by the divergence of 44 - EUCLID's ELEMENTs. two straight lines from a point. In the definition of an angle, the magnitude of the angle is independent of the lengths of the two lines by which it is included; their mutual divergence from the point at which they meet, is the criterion of the magnitude of an angle, as it is pointed out in the succeeding definitions. The point at which the two lines meet is called the angular point, or the vertex of the angle, and must not be confounded with the magnitude of the angle itself. The right angle is fixed in magnitude, and, on this account, it is made the standard with which all other angles are compared. Two straight lines which actually intersect one another, or which when produced would intersect, are said to be inclined to one another, and the inclination of the two lines is determined by the angle which they make with one another. .* Two straight lines are said to be conterminous which have a common termination, or when one extremity of each line coincides in the same point, and both the lines may be, or may not be, in the same direction. Def. x. It may here be observed that in the Elements, Euclid always assumes that when one line is perpendicular to another line, the latter is also perpendicular to the former; and always calls a right angle, épôni youta ; but a straight line, sw8sia ypapºuri. - Def. xvi. Euclid in defining a circle does not give any method by which it may be described. The definition simply states what a circle is, and one property which distinguishes it from other figures. In laying down his principles Euclid avoids entering into any method of showing how a straight line or a circle may be conceived to be generated. A circle might be defined in the following manrer:—If a finite straight line be supposed to revolve in a plane about one of its extremities which remains fixed, until it return to its original position, the surface which the revolving line has passed over is called a circle, and the linear space which the moving extremity of the revolving line has. traced out is called the circumference of the circle. The straight line which revolves is called the radius, and the fixed point about which it revolves is called the center. The adoption of such a form of definition would have introduced the consideration of a locus. Instead of adopting such a mechanical method of defining a circle, he assumes that a circle may be described, and the third postulate seems to suggest that the method above stated is the assumption made by Euclid, which he does not place among the principles of his First Book. In the Eleventh Book, the sphere, the cone, and the cylinder are defined by plane figures which revolve about one side which is supposed to remain fixed. Def. xix. This has been restored from Proclus, as it seems to have a meaning in the construction of Prop. 14, Book II. ; the first case of Prop. 38, Book III., and Prop. 13, Book VI. The definition of the segment of a circle is not once alluded to in Book I., and is not required before the discussion of the properties of the circle in Book III. Proclus remarks on this definition : “Eſence you may collect that the center has three places: for it is either within the figure, as in the circle; or in its perimeter, as in the semicircle; or without the figure, as in certain conic lines.” - Def. xxv-xxix. Triangles are divided into three classes, by reference to: the relations of their sides; and into three other classes, by reference to their angles. A further classification may be made by considering both the relation of the sides and angles in each triangle. In Simson's definition of the isosceles triangle, the word only must be omitted, as in the Cor. Prop. 5, Book I., an isosceles triangle may be equilateral, and an equilateral triangle is considered isosceles in Prop. 15, Book iv. Objection. NOTES To Book I. 45 has been made to the definition of an acute-angled triangle. It is said that. It cannot be admitted as a definition, that all the three angles of a triangle are acute, which is supposed in Def. 29. It may be replied, that the definitions of the three kinds of angles point out and seem to supply a foundation for a similar distinction of triangles. - . Def. xxx.-xxxiv. The definitions of quadrilateral figures are liable to ob- jection. All of them, except the trapezium, come under the general idea of a parallelogram; but as Euclid defined parallel straight lines after he had defined four-sided figures, no other arrangement could be adopted than the one he has followed; and for which there appeared to him, without doubt, some probable reasons. Sir Henry Savile, in his Seventh Lecture, remarks on some of the definitions of Euclid, “Nec dissimulandum aliquot harum in manibus exiguum esse usum in Geometriá.” A few verbal emendations have been proposed in some of them. A square is a four-sided plane figure having all its sides equal, and one angle a right angle: because it is proved in Prop. 46, Book I, that if a parallelogram have one angle a right angle, all its angles are right angles. An oblong is a plane figure of four sides, having only its opposite sides equal, and one of its angles a right angle. A rhomboid is a four-sided plane figure having only its opposite sides equal to one another and its angles not right angles. Sometimes an irregular four-sided figure which has two sides parallel, is called a trapezoid. -> Def. xxxv. It is possible for two straight lines never to meet when produced, and not be parallel. - Def. A. The term parallelogram literally implies a figure formed by parallel straight lines, and may consist of four, six, eight, or any even number of sides, where every two of the opposite sides are parallel to one another. In the Elements, however, the term is restricted to four-sided figures, and includes the square, the oblong, the rhombus, and the rhomboid. - * The synthetic method is followed by Euclid not only in the demonstrations of the propositions, but also in laying down the definitions. He commences with the simplest abstractions, defining a point, a line, an angle, a superficies, and their different varieties. This mode of proceeding involves the difficulty, almost in- surmountable, of defining satisfactorily the elementary abstractions of Geometry. It has been observed, that it is necessary to consider a solid, that is, a magnitude which has length, breadth, and thickness, in order to understand aright the de- finitions of a point, a line, and a superficies. A solid or volume considered apart from its physical properties, suggests the idea of the surfaces by which it is bounded: a surface, the idea of the line or lines which form its boundaries: and a finite line, the points which form its extremities. A solid is therefore bounded by surfaces; a surface is bounded by lines; and a line is terminated by two points. A point marks position only: a line has one dimension, length only, and defines distance: a superficies has two dimensions, length and breadth, and defines ex- tension: and a solid has three dimensions, length, breadth, and thickness, and defines some portion of space. It may also be remarked that two points are sufficient to determine the position of a straight line, and three points not in the same straight line, are neeessary to fix the position of a plane. - ** ON THE POSTULATES. EcoLio prescribes no instruments as sufficiently accurate or sufficiently ex- tensive for drawing the straight lines and describing the circles required in his demonstrations. He postulates, so to speak, both the drawing of straight lines 46 - - EUCLID'S ELEMENTS. and the description of circles. The ruler and compasses may be preferred to other instruments for performing the operations allowed by the postulates, only taking care that the compasses are confined to their proper use, namely, to de- scribe circles, not to measure distances. The following opinion of Sir Isaac Newton accords with this view. “Nam et linearum rectarum et circulorum descriptiones, in quibus Geometria fundatur, ad Mechanicam pertinent. Has lineas describere Geometria non docet sed postulat. Postulat enim ut tyro easdem accurate describere prius didicerit, quam limen attingat Geometriae : dein, quomodo per quas operationes problemata solvantur, docet : ..rectas et circulos describere problemata sunt, sed non Geometrica. Ex Mechanica postulaturhorum solutio, in Geometria docetur solutorum usus. Ac gloriatur Geometria quod, tam paucis principiis aliunde petitis, tam multa præstet.” * The definitions assume the possible existence of straight lines and circles, and the postulates predicate the possibility of drawing and of producing straight lines, and of describing circles. The postulates form the principles of construction. assumed in the Elements; and are, in fact, problems, the possibility of which is admitted, not only because the description of them may be readily conceived, but also, because it is impossible to draw a perfectly straight line, or to describe an exact circle by any methods consistent with the definitions of them. The second postulate admits that a straight line may be produced in either direction or in both directions. - It must, however, be carefully remarked, that the third postulate only admits, that when any line is given in position and magnitude, a circle may be described from either extremity of the line as a center, and with a radius equal to the length of the line, as in Euc. I. 1. It does not admit the description of a circle with any other point as a center than one of the extremities of the given line. . The third postulate does not admit that the true length of a straight line imay be taken by a pair of compasses, and that length so taken transferred to another place. Fuc. I. 2, shews how, from any given point, to draw a ‘straight line equal to another straight line which is given in magnitude and Aposition. - ON THE AXIOMS. fe Axioms are usually defined to be self-evident truths, which cannot be rendered ‘more evident by demonstration; in other words, the axioms of Geometry are theorems, the truth of which is admitted without proof. It is by experience we 'first become acquainted with the different forms of geometrical magnitudes, and the axioms, or the fundamental ideas of their equality or inequality rest on the same basis. The conception of the truth of the axioms does not appear to be more. removed from experience than the conception of the definitions. These axioms, or first principles of demonstration, are such theorems as can- se - Let the straight line A B be divided into two equal parts in the point C, and into two unequal parts in the point D. -. Then the rectangle AD, DB, together with the square on CD, shall be equal to the square on CB. - A C D B. E G F Dpon CB describe the square CEFB, (1, 46.) join BE, through D draw DHG parallel to CE or B.F. (I. 31.) meeting BE in H, and EF in G, and through H draw KLM parallel to CB or EF, meeting CE in L, and BF'in M; also through A draw AK parallel to CD or BM, meeting MLK in K. Then because the complement CH is equal to the complement HF (I. 43.) to each of these equals add DM; - therefore the whole CM is equal to the whole DF'; but because the line AC is equal to CB, ^e therefore AL is equal to CM, (I. 36.) therefore also AL is equal to DF; to each of these equals add CH, BOOK II. PROP. vi. 71 and therefore the whole A His equal to DF and CH: - but AH is the rectangle contained by AD, DB, for DH is equal to DB; and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, D.B.: to each of these equals add LG, which is equal to the square on CD; (II. 4. Cor.) . therefore the gnomon CMG together with LG, is equal to the rectangle A.D, DB, together with the square on CD : - but the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB; … therefore the rectangle AD, DB, together with the square on CD is equal to the square on CB. - Wherefore, if a straight line, &c. Q.E.D. *CoR. From this proposition it is manifest, that the difference of the squares on two unequal lines AC, CD, is equal to the rectangle contained by their sum AD and their difference D.B. - PROPOSITION VI. THEOREM, If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D. Then the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD. -- ºr - - : w A C B D al- –54& M E. G F Upon CD describe the square CEFD, (1, 46.) and join DE, through B draw BHG parallel to CE or D.F. (I. 31.) meeting DE in H, and EF'in G: - through H draw KLM parallel to AD or EF, meeting DF in - M, and CE in L: and through. A draw AK parallel to CL or DM, meeting MLK in K. Then because the line AC is equal to CB, therefore the rectangle AL is equal to the rectangle CH, (I. 36.) - but CH is equal to HF; (1.43.) therefore AL is equal to HF; to each of these equals add CM; - therefore the whole AM is equal to the gnomon CMG : but AM is the rectangle contained by AD, DB, for DM is equal to DB : (II. 4. Cor.) gº therefore the gnomon CMG is equal to the rectangle AD, D.B.: to each of these equals add LG which is equal to the square on CB: therefore the rectangle AD, DB, together with the square on CB, is equal to the gnomon CMG, and the figure LG; but the gnomon CMG and LG make up the whole figure CEFD, which is the square on CD; 72. EUCLIB's ELEMENTs. therefore the rectangle AD, DB, together with the square on CB, is equal to the square on CD. Wherefore, if a straight line, &c., Q.E.D. PROPOSITION VII. THEOREM. . If a straight line be divided into any two parts, the squares on the whole lºwe, and on one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square on the other part. Let the straight line AB be divided into any two parts in the point C. Then the squares on AB, BC shall be equal to twice the rectangle 4:B, BC, together with the square on A.C. A. G. B. / H!—%—K I) F. E. Upon AB describe the square ADEB, (r. 46.) and join BD; through C draw CF parallel to AD or BE (I. 31.) meeting BD in G, and DE in F; • - ſe through G draw HGH parallel to AB or DE, meeting AD in H, and BE in J.C. - - Then because AG is equal to GE, (r. 43.) - add to each of them CPC; therefore the whole. A K is equal to the whole CE; and therefore AK, CE, are double of A.K. - lout AIK, CE, are the gnomon AKF and the square C.Iſ; therefore the gnomon ART" and the square CK are double of A.K. but twice the rectangle AB, BC, is double of A.K., ... -- for BK is equal to BC; (II.4. Cor.) therefore the gnomon AKF and the square C.K., are equal to twice the rectangle AB, BC; -> * * to each of these equals add HF, which is equal to the square on AC, - therefore the gnomon AKF, and the squares C.K., HF, are equal to twice the rectangle AB, BC, and the square on A.C.; but the gnomon AKF, together with the squares CK, HF make up the whole figure ADEB and CK, which are the squares on AB and 80; * - therefore the squares on AB and BC are equal to twice the rectangle AB, BC, together with the square on A. C. - - Wherefore, if a straight line, &c., Q.E.D. PROPOSITION VIII. THEOREM, If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square on the other part, is equal to the square on the straight line, which is made up of the whole and that part. - Let the straight line AB be divided into any two parts in the point C. BOOK IT. PROP. VIII. 73. Then four times the rectangle AB, BC, together with the square on AC, shall be equal to the square on the straight line made up of 4B and BC together. - - - Produce AB to D, so that BD be equal to CB, (r. 3.) upon AD describe the square AEFD, (1, 46.) and join DE, through B, C, draw BL, CH parallel to A.E or DF, and cutting DE in the points K, Prespectively, and meeting EF in L, H: through K, P, draw MGKN, XPRO parallel to AD or EF. Then because CB is equal to BD, CB to GK, and BD to KV; therefore GK is equal to KN; for the same reason, PR is equal to R0; and because CB is equal to BD, and GK to KN, therefore the rectangle CK is equal to BN, and GR to RN; (I. 36.) but CK is equal to RN, (1.43.) - because they are the complements of the parallelogram 00; therefore also BN is equal to GR; and the four rectangles BN, CK, GR, RN, are equal to one another, and so are quadruple of one of them CK. Again, because CB is equal to BD, and BD to BK, that is, to CG ; and because CB is equal to GK, that is, to GP; therefore CG is equal to GP. And because CG is equal to GP, and PR to R0, therefore the rectangle AG is equal to MP, and PL to RF; but the rectangle MP is equal to PL, (I. 43.) because they are the complements of the parallelogram ML: - wherefore also AG is equal to RF': . - therefore the four rectangles AG, MP, PL, RF, are equal to on another, and so are quadruple of one of them A.G. - And it was demonstrated, that the four C.R., BN, GR, and RN, are quadruple of CK., - therefore the eight rectangles which contain the gnomon AOH, are quadruple of A K. And because AK is the rectangle contained by AB, BC, - * for BK is equal to BC; - g therefore four times the rectangle AB, BC is quadruple of AK : but the gnomon. A 0B1 was demonstrated to be quadruple of AK ; therefore four times the rectangle AB, BC is equal to the gnomon AOH ; to each of these equals add XH, which is equal to the square on AC; therefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOH and the square XHT; but the gnomon AOH and XII make up the figure AEFD, which is the square on AD; - - , therefore four times the rectangle AB, BC together with the square On AC, is equal to the square on AD, that is, on AB and BC added together in one straight line. \ - Wherefore, if a straight line, &c. Q.E.D. 74 EUCLID's ELEMENTs. PROPOSITION IX. THEOREM. If a straight line be divided into two equal, and also into two un- equal parts; the squares on the two unequal parts are together double of the square on half the line, and of the square on the line between the points of 8ection. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D. Then the squares on AD, DB together, shall be double of the squares on 40, CD. Zºš A C D B From the point C draw CE at right angles to AB, (1. 11.) make CE equal to AC or CB, (I. 3.) and join EA, EB; through D draw DF'parallel to CE, meeting EB in F, (I. 31.) .** through F draw FG parallel to BA, and join AF. - Then, because AC is equal to CE, . -. therefore the angle AEC is equal to the angle EAC; (I. 5.) and because ACE is a right angle, * therefore the two other angles AEC, EAC of the triangle, are together equal to a right angle; (I 32.) and since they are equal to one another; therefore each of them is half a right angle. For the same reason, each of the angles CEB, EBC is half a right angle, - 3. and therefore the whole AEB is a right angle. And because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the interior and opposite angle ECB, (I. 29.) therefore the remaining angle EFG is half a right angle; wherefore the angle GEF is equal to the angle EFG, and the side GF equal to the side EG. (I. 6.) Again, because the angle at B is half a right angle, and FDB a right angle, ** for it is equal to the interior and opposite angle ECB, (I. 29.) therefore the remaining angle BFD is half a right angle; wherefore the angle at B is equal to the angle BFD, and the side DF'equal to the side D.B. (I. 6.) And because AC is equal to CE, the square on A C is equal to the square on CE; therefore the squares on AC, CE are double of the square on A C; but the square on AE is equal to the squares on 40, CE, (I. 47.) because ACE is a right angle; therefore the square on AE is double of the square on A. C. Again, because EG is equal to GF, the square on EG is equal to the square on GF; ... therefore the squares on EG, GF are double of the square on GF; but the square on EF is equal to the squares on EG, GF; (I. 47.) therefore the square on EF is double of the square on GF; - BOOK II. PROP. IX, x. 75. * and GF is equal to CD; (r. 34.) - therefore the square on EF is double of the square on CD; but the square on AE is double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD; but the square on AF'is equal to the squares on AE, EF, because AEF is a right angle: (1.47.) therefore the square on A Fis double of the squares on AC, CD: but the squares on AD, DF are equal to the square on AF; because the angle ADF is a right angle; (1.47.) . therefore the squares on AD, DF’ are double of the squares on AC, CD; and DF is equal to D.B.; therefore the squares on AD, DB are double of the squares on AC, CD. If therefore a straight line be divided, &c. Q.E.D. PROPOSITION X. THEOREM. If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected, and of the square on the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to the point D. Then the squares on 4D, DB, shall be double of the squares on AC, CD. * F. F C a g G From the point C draw CE at right angles to AB, (1.11.) make CE equal to AC or CB, (I. 3.) and join AE, EB; through E. draw EF parallel to AB, (I. 31.) and through D draw DF parallel to CE, meeting EF in F. Then because the straight line EF meets the parallels CE, FD, - therefore the angles CEF, EFD are equal to two right angles; (I. 29.) and therefore the angles BEF, EFD are less than two right angles. But straight lines, which with another straight line make the in- terior angles upon the same side of a line, less than two right angles, will meet if produced far enough; (I. ax. 12.) - -- therefore EB, FD will meet, if produced towards B, D.; let them be produced and meet in G, and join AG. Then, because AC is equal to CE, therefore the angle CEA is equal to the angle EAC; (ii. 5.) and the angle ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle. (I. 32.) B'or the same reason, each of the angles CEB, EBC is half a right angle; therefore the whole AEB is a right angle. And because EBC is half a right angle, therefore DBG is also half a right angle, (I. 15.) for they are vertically opposite; , but BDG is a right angle, 76. EUCLID's ELEMENTS. because it is equal to the alternate angle DCE; (I. 29.) ... therefore the remaining angle DGB is half a right angle; and is therefore equal to the angle DBG; wherefore also the side BD is equal to the side DG. (I, 6.) - Again, because EGF is half a right angle, and the angle at F is a right angle, being equal to the opposite angle ECD, (I. 34.) therefore the remaining angle FEG is half a right angle, and therefore equal to the angle EGF; wherefore also the side GF'is equal to the side F.E. (I. 6.) And because EC is equal to CA; the square on EC is equal to the Square on CA ; therefore the squares on EC, CA are double of the square on CA: but the square on EA is equal to the squares on EC, CA; (I. 47.) therefore the square on EA is double of the square on A. C. - Again, because GF is equal to FE, the square on GF'is equal to the square on FE; therefore the squares on GF, FE are double of the square on FE; but the square on EG is equal to the squares on GF, FE; (I. 47.) therefore the square on EG is double of the square on FE; - and FE is equal to CD; (I. 34.) - wherefore the square on EG is double of the square on CD; but it was demonstrated - - - - that the square on EA is double of the square on AC; therefore the squares on EA, EG are double of the squares on A C, CD; but the square on AG is equal to the squares on EA, EG ; (1.47.) therefore the square on AG is double of the squares on AC, CD: but the squares on AD, DG are equal to the square on A G ; therefore the squares on AD, D6, are double of the squares on AC, CD; . but DG is equal to DB ; therefore the squares on AD, DB are double of the squares on AC, CD. - Wherefore, if a straight line, &c. Q.E.D. - 4 PROPOSITION XI. PROBLEM. To divide a given straight line into two parts, so that the rectangle con- tained by the whole and one of the parts, shall be equal to the square on the other part. Let AB be the given straight line. It is required to divide AB into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square on the other part. F G I. LT C b. D Tpon AB describe the square A CDB ; (r. 46.) bisect 4 C in E, (r. 10.) and join B.E, - produce CA to F, and make EF equal to EB, (1. 3.) BOOK II. PROP. XI, XII. 77 upon AF describe the square FGHA. (I. 46.) Then AB shall be divided in H, so that the rectangle AB, BH is equal to the square on AH. * º * º Produce GH to meet CD in K. Then because the straight line AC is bisected in E, and produced to F, therefore the rectangle CF, FA together with the square on A.E, is equal to the square on EF; (II. 6.) • . but EF is equal to EB; therefore the rectangle CF, FA together with the square on AE, is equal to the square on EB; - but the squares on BA, AE are equal to the square on EB, (1.47.) because the angle EAB is a right angle; - therefore the rectangle CF, FA, together with the square on A.E, is equal to the squares on BA, AE; . - take away the square on AE, which is common to both; therefore the rectangle contained by CF, F.A is equal to the square on B.A. - But the figure FK is the rectangle contained by CF, FA, for FA is equal to FG ; and AD is the square on AB; therefore the figure FK is equal to AD; - take away the common part A.K., therefore the remainder FH is equal to the remainder HD; but HD is the rectangle contained by AB, BH, Aſ for AB is equal to BD; - and FH is the square on A.H.; therefore the rectangle AB, BH, is equal to the square on A.H. Wherefore the straight line AB is divided in H, so that the rectangle AB, BFI is equal to the square on A.H. Q.E.F. PROPOSITION XII. THEOREM. In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square on the side sub- tending the obtuse angle, is greater than the squares on the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle. Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and from the point A, let AD be drawn perpendicular to BC produced. . Then the square on AB shall be greater than the squares on AC, CB, by twice the rectangle BC, CD. - A -* l; C |D Because the straight line BD is divided into two parts in the point C. therefore the square on BD is equal to the squares on BQ, CD, and twice the rectangle BC, CD; (II. 4.) to each of these equals add the square on DA; 78 EUCLID's ELEMENTs. therefore the squares on BD, DA are equal to the squares on BC, CD, DA, and twice the rectangle BC, CD; - but the square on BA is equal to the squares on BD, DA, (1.47.) because the angle at D is a right angle; and the square on CA is equal to the squares on CD, D.A.; therefore the square on BA is equal to the squares on BC, CA, and twice the rectangle BC, CD; that is, the square on BA is greater than the squares on BC, CA, by twice the rectangle BC, CD. Therefore in obtuse-angled triangles, &c. Q.E.D. PROPOSITION XIII. THEOREM, In every triangle, the square on the side subtending either of the acute angles, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line inter- cepted between the acute angle and the perpendicular let fall upon it from the opposite angle. - - Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular AD from the opposite angle. (I. 12.) Then the square on AC opposite to the angle B, shall be less than the squares on CB, BA, by twice the rectangle CB, B.D. A --- B D C. First, let AD fall within the triangle ABC. Then, because the straight line CB is divided into two parts in D, the squares on CB, BD are equal to twice the rectangle contained by UB, BD, and the square on DC; (II.7.) - to each of these equals add the square on AD; therefore the squares on CB, BD, DA, are equal to twice the rectangle CB, BD, and the squares on A.D., DC; but the square on AB is equal to the squares on BD, D4, (r. 47.) because the angle BDA is a right angle; and the square on AC is equal to the squares on AID, DC; , , therefore the squares on CB, BA are equal to the square on 40, and twice the rectangle CB, BD: - that is, the square on AC alone is less than the squares on CB, B4, by twice the rectangle CB, B.D. - Secondly, let AD fall without the triangle ABC. A B C I) Then, because the angle at D is a right angle, - the angle ACB is greater than a right angle; (I, 16.) BOOK II. PROP. XIII, xiv. 79 and therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle BC, CD; (II. 12.) . . to each of these equals add the square on BC; therefore the squares on AB, BC are equal to the square on AC, twice the square on BC, and twice the rectangle BC, CD; but because BD is divided into two parts in C, - therefore the rectangle DB, BC is equal to the rectangle BC, CD, and the square on BC; (II. 3.) and the doubles of these are equal: that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC: therefore the squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC: s, - wherefore the square on AC alone is less than the squares on AB, BC; by twice the rectangle DB, BC. Lastly, let the side AC be perpendicular to BC. A IB C Then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest, that the squares on AB, BC, are equal to the square on AC, and twice the square on B.C. (I. 47.) Therefore in any triangle, &c. Q.E.D. PROPOSITION XIV. PROBLEM. Z0 describe a square that shall be equal to a given rectilineal figure. Let A be the given rectilineal figure. It is required to describe a square that shall be equal to A. (Z). Describe the rectangular parallelogram BCDE equal to the recti- lineal figure A. (I. º p gr Q. Then, if the sides of it, BE, ED, are equal to one another, it is a square, and what was required is now done. But if B.E., ED, are not equal, produce one of them BE to F, and make EF equal to ED, bisect BF in G.; (I. 10.) -- - from } center G, at the distance GB, or GF, describe the semicircle HF, and produce DE to meet the circumference in H. The square described upon EH shall be equal to the given recti- lineal figure A. - 80 - EUCLID's ELEMENTs. Join GH. * - Then because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E ; therefore the rectangle B.E., EF, together with the square on EG, is equal to the square on GF; (II.5.) . - but GF is equal to GH; (def. 15.) - therefore the rectangle B.E, JEF, together with the square on EG, is - equal to the square on GH; - but the squares on HE, EG are equal to the square on GH; (I. 47.) therefore the rectangle B.E., E.F, together with the square on EG, is equal to the squares on HE, EG ; take away the square on EG, which is common to both ; therefore the rectangle B.E., EF'is equal to the square on H.E. But the rectangle contained by B.E., EF is the parallelogram B.D, because EF is equal to ED; - therefore BD is equal to the square on EH; but BD is equal to the rectilineal figure A; (constr.) therefore the square on EH is equal to the rectilineal figure A. - Wherefore a square has been made equal to the given rectilineal figure A, namely, the square described upon EH, Q.E.F. NOTES TO BOOK II. IN Book I, Geometrical magnitudes of the same kind, lines, angles and sur- faces, more particularly triangles and parallelograms, are compared, either as being absolutely equal, or unequal to one another. *- In Book II, the properties of right-angled parallelograms, but without reference to their magnitudes, are demonstrated, and an important extension is made of Euc. 1. 47, to acute-angled and obtuse-angled triangles. Euchid has given no definition of a rectangular parallelogram or rectangle : probably, because the Greek expression trapax)\nxóypapºpov ćpfloyaſutov, or épôoyaſutov simply, is a definition of the figure. In English, the term rectangle, formed from rectus angulus, ought to be defined before its properties are demonstrated. A rectangle may be defined to be a parallelogram having one angle a right angle, or a right-angled parallelogram ; and a square is a rectangle having all its sides equal. - - - As the squares in Euclid's demonstrations are squares described or supposed to be described on straight lines, the expression “the sguare on AB,” is a more appropriate abbreviation for “the square described on the line AB,” than “the square of A.B.” The latter expression mere fitly expresses the arithmetical or algebraical equivalent for the square on the line AB, * Ö In Euc; 1.35, it may be seen that there may be an indefinite number of paral- lelograms on the same base and between the same parallels whose areas are always equal to one another ; but that one of them has all its angles right angles, and the length of its boundary less than the boundary of any other parallelogram upon the same base and between the same parallels. The area of this rectangular parallelogram is therefore determined by the two lines which contain one of its right angles. Hence it is stated in Def. 1, that every right-angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles. No distinction is made in Book II, between equality and identity, as the rectangle may be said to be contained by two lines which are equal respec- tively to the two which contain one right angle of the figure. It may be remarked that the rectangle itself is bounded by four straight lines. It is of primary importance to discriminate the Geometrical conception of a rectangle from the Arithmetical or Algebraical representation of it. The subject of Geometry is magnitude not number, and therefore it would be a departure from strict reasoning on space, to substitute in Geometrical demonstrations, the Arithmetical or Algebraical representation of a rectangle for the rectangle itself. It is however, absolutely necessary that the connexion of number and magnitude be clearly understood, as far as regards the representation of lines and areas. All lines are measured by lines, and all surfaces by surfaces. Some one line of definite length is arbitrarily assumed as the linear unit, and the length of every other line is represented by the number of linear units, contained in it. The square is the figure assumed for the measure of surfaces. The square unit or the unit of area is assumed to be that square, the side of which is one unit in length, and the magnitude of every surface is represented by the number of square units contained in it. But here it may be remarked, that the properties of rect- angles and squares in the Second Book of Euclid are proved independently of the consideration, whether the sides of the rectangles can be represented by any multiples of the same linear unit. If, however, the sides of rectangles are supposed to be divisible into an exact number of linear units, a numerical representation for the area of a rectangle may be deduced. - *~ G 82 EUCLID's, ELEMENTs. p On two lines at right angles to each other, take AB equal to 4, and AD equal to 3 linear units. Complete the rectangle ABCD, and through the points of division of AB, AD, draw EL, FM, GN parallel to AD; and HP, KQ parallel to AB respectively. A E. F. G. B H|— ––. R R. —Hº DTL-M N C Then the whole rectangle AC is divided into squares, all equal to each other. And AC is equal to the sum of the rectangles AL, EM, FN, GC; (II. 1.) also these rectangles are equal to one another, (I. 36.) therefore the whole AC is equal to four times one of them AL. Again, the rectangle A L is equal to the rectangles EH, HR, RD, and these rectangles, by construction, are squares described upon the equal lines AH, H.K., RD, and are equal to one another. - Therefore the rectangle AL is equal to 3 times the square on AH, - but the whole rectangle AC is equal to 4 times the rectangle AL, therefore the rectangle AC is 4 x 3 times the square on AH, or 12 square units: that is, the product of the two numbers which express the number of linear units in the two sides, will give the number of square units in the rectangle, and therefore will be an arithmetical representation of its area. - And generally, if AB, AD, instead of 4 and 3, consisted of a and b linear units respectively, it may be shewn in a similar manner, that the area of the rectangle AC would contain ab square units; and therefore the product ab is a proper representation for the area of the rectangle A.C. Hence, it follows, that the term rectangle in Geometry corresponds to the term product in Arithmetic and Algebra, and that a similar comparison may be made between the products of the two numbers which represent the sides of rectangles, as between the areas of the rectangles themselves. This forms the basis of what are called Arithmetical or Algebraical proofs of Geometrical properties. If the two sides of the rectangle be equal, or if b be equal to a, the figure is a square, and the area is represented by aa or a”. Also, since a triangle is equal to the half of a parallelogram of the same base and altitude; - - Therefore the area of a triangle will be represented by half the rectangle which has the same base and altitude as the triangle : in other words, if the length of the base be a units, and the altitude be b units ; Then the area of the triangle is algebraically represented by #ab. The demonstrations of the first eight propositions, exemplify the obvious axiom, . that, “the whole area of every figure in each case, is equal to all the parts of it taken together.” - Def. 2. If through any point F in the diameter BD of a parallelogram, ABCD, there be drawn two straight lines EFG, HFK, parallel respectively to the sides of the figure, these lines will divide the figure into four parallelograms. The two parallelograms EK, HG, through which the diameter BD passes, are said to be “about the diameter of the parallelogram”, and the other two parallelograms A F, FC, which complete the figure, are called “the complements of the paralleló- grams about a diameter”, also either of the parallelograms about a diameter together with the two complements is called a gnomon, so that the parallelogram EK together with the complements AF, FC, is also a gnomon, as well as the NOTEs To Book II. 83 parallelogram HG with the same complements. Also if the other diameter AC be drawn, two gnomons are formed by the lines drawn through any point in AC parallel to the sides. - Prop. 1. For the sake of brevity of expression, “the rectangle contained by the straight lines AB, BC,” is called “the rectangle AB, BC;” and sometimes “the rectangle ABC.” To this proposition may be added the corollary: If two straight fines be divided into any number of parts, the rectangle contained by the two straig.t lines, is equal to the rectangles contained by the several parts of one line and the several parts of the other respectively. - * The method of reasoning on the properties of rectangles, by means of the products which indicate the number of square units contained in their areas, is foreign to Luclid's ideas of rectangles, as discussed in his Second Book, which have no re- ference to any particular unit of length or measure of surface. Prop. I. The figures BH, BK, DL, EH are rectangles, as may readily be shewn. Por, by the parallels, the angle CEL is equal to EDK; and the angle EDK is equal to BDG (Euc. I. 29.). But BDG is a right angle. Hence one of the angles in each of the figures BH, BK, DL, EH is a right angle, and therefore (Euc. 1, 46, Cor.) these figures are rectangular. Prop I. Algebraically. (fig. Prop. 1.) Let the line BC contain a linear units, and the line A, b linear units of the same length. Also suppose the parts BD, DE, EC to contain m, n, p linear units respectively. Then a = m + n + p, multiply these equals by b, therefore ab = bn + brº + bp. That is, the product of two numbers, one of which is divided into any number of parts, is equal to the sum of the products of the undivided number, and the several parts of the other; or, if the Geometrical interpretation of the products be restored, The number of square units expressed by the product ab, is equal to the number of square units expressed by the sum of the products bm, bn, bp, Prop. I. may be exhibited arithmetically as well as algebraically: If a be equal to 12 units, and b to 5 units, also if m, n, p, be equal to 6, 4, 2 units respectively, then 12 = 6 + 4 + 2. Multiply these equals by 5, ..", 5 x 12 = 5 x (6 + 4 + 2), or, 5 x 12 = 5 x 6 + 5 x 4 + 5 x 2, In a similar way, the following propositions may be exhibited, by taking any numbers whatever for the algebraic expressions assumed for the lines. - Prop. II. Algebraically. (fig. Prop. II.) Let AB contain a linear units, and AC, CB, m and n linear units respectively. Then m + n = a, multiply these equals by a, - \ therefore am + an = a”. That is, if a number be divided into any two parts, the sum of the products of the whole and each of the parts is equal to the square of the whole number. Prop. III. In the construction, BC is the part of the line taken. The other part AC of the line may be taken, and it is equally true, that the rectangle contained by AB, 42 is equal to the rectangle contained by AC, CB together with the square on A.C. - - G 2 84 - EUCLID's ELEMENTS. g Prop. III. Algebraically. (fig. Prop. III.) g Let AB contain a linear units, and let BC contain m, and AC, n linear units. - - Then a = m + n, multiply these equals by m, . therefore ma = m” + mn. - That is, if a number be divided into any two parts, the product of the whole number and one of the parts, is equal to the square of that part, and the product of the two parts. Prop. Iv. might, have been deduced from the two preceding propositions; but Euclid has preferred the method of exhibiting, in the demonstrations of the second book, the equality of the spaces compared. In the corollary to Prop. xlvi. Book I, it is stated that a parallelogram which has one right angle, has all its angles right angles. By applying this corollary, the demonstration of Prop. Iv, may be considerably shortened. - If the two parts of the line be equal, then the square on the whole line is equal to four times the square on half the line. e Also, if a line be divided into any three parts, the square on the whole line is equal to the squares on the three parts, and twice the rectangles contained by every two parts. And generally, If a line be divided into any number of parts, the square on the whole line, is equal to the squares on the several parts, together with twice the rectangles contained by every two parts. Prop. Iv. Algebraically. (fig. Prop. Iv.) . * Let the line AB contain a linear units, and the parts of it AC and BC, m and m linear units respectively. *- Then a = m + n, squaring these equals, ... a' = (m + n)", * or a* = m” + 2mm + n°. That is, if a number be divided into any two parts, the square of the number is equal to the squares of the two parts together with twice the product of the two parts. - From Euc. II. 4, may be deduced a proof of Euc. I, 47. In the fig. take DL on DE, and EM on EB, each equal to BC, and join CH, HL, LM, MC. Then the figure HLMC is a square, and the four triangles CAH, HDL, LEM, MBC are equal to one another, and together are equal to the two rectangles AG, GE. Now AG, GE, FH, CK are together equal to the whole figure ADEB; and HLMC, with the four triangles C.A.H, HDL, LEB, MBC also make up the whole figure A DEB; * * Hence AG, GE, FH, CK are equal to HLMC together with the four triangles: but AG, GE are equal to the four triangles, wherefore FH, CK are equal to HLMC, that is, the squares on AC, AHare together equal to the square on CH. Prop. v. When a straight line is divided into two parts, the rectangle contained by the two parts is the greatest possible, and the sum of the squares of the two parts is the least possible, when the two parts are equal, or when the line is bisected. - - - The proof of the Corollary to this proposition may be easily proved from the diagram. 4: It must be kept in mind, that the sum of two straight lines in Geometry, means the straight line formed by joining the two lines together, so that both may be in the same straight-line. NOTES TO Book II. . . . 85 The following simple properties respecting the equal and unequal division of a line are worthy of being remembered. I. Since AB = 2BC = 2(BD + DC) = 2BD + 2DC. (fig. Prop. v.) - - and AB = AD + DB ; * ..", 20I) + 2 DB = AD + DB, and by subtracting 2DR from these equals, ... 2CD = AD – DB, and CD = 3 (AD – D.B). asº - That is, if a line A B be divided into two equal parts in C, and into two unequal parts in D, the part CD of the line between the points of section is equal to half the difference of the unequal parts AD and D.B. - - - II. Here AD = AC + CD, the sum of the unequal parts, (fig. Prop. v.) - and DB = AC — CD their difference. Hence by adding these equals together, ..'. AD + DB = 2AC, or the sum and difference of two lines AC, CD, are together equal to twice the greater line. And the halves of these equals are equal, .*.*. AD + #. DB = AC, - - or, half the sum of two unequal lines AC, CD added to half their difference, is equal to the greater line A.C. III. Again, since A D = AC + CD, and DB = AC – CD, by subtracting these equals, ..'. AD – DB = 20 D, or, the difference between the sum and difference of two unequal lines, is equal to twice the less line. And the halves of these equals are equal, ... 3. AD – #.DB = CD, ~ * or, half the difference of two lines subtracted from half their sum is equal to the less of the two lines. sº IV. Since A C – CD = DB the difference, y ..". AC = CD + DB, and adding CD the less to each of these equals, ..". AC + CD = 2CD + DB, or, the sum of two unequal lines, is equal to twice the less line together with the difference between the lines. - Prop. v. The rectangle contained by AD and DB, and the square on BC are each bounded by the same extent of lime, but the spaces enclosed differ by the square on CD. e Bearing in mind that AC, CD are respectively half the sum and half the difference of the two lines AD, DB ; the corollary to this proposition may be expressed in the following form: “The rectangle contained by two straight lines is equal to the difference of the squares on half their sum and half their difference.” Prop. v. Algebraically. - , Let AB contain 2a linear units, its half BC will contain a linear units. a And let CD the line between the points of section contain m linear units. Then AD the greater of the two unequal parts, contains a + m linear units; and DB the less contains a — m units. / 86 EUCLID's ELEMENTS. Also m is half the difference of a + m and a — m ; - ... (a + m). (a — m). = a” — mº, to each of these equals add m”;. ... (a + m) (a — m). H m? = a”. W. That is, If a number be divided into two equal parts, and also into two unequal parts, the product of the unequal parts together with the square of half their difference, is equal to the square of half the number, - Prop. vi. A given straight line is said to be produced when it has its length in- creased in either direction, and the increase it receives, is called the part produced, If a point be taken in a line or in a line produced, the line is said to be divided internally or externally, and the distances of the point from the ends of the line are called the internal or external segments of the line, according as the point of section is in the line or the line produced. Props. v. and VI., also.Ix. and x., * are, in fact, the same, if the division of the lines be regarded with respect to their external and, internal division. Prop. VI. Algebraically, & Let AB contain 2a linear units, then its half BC contains a units; and let BD contain munits. - Then A B contains 2a + 'm units, and ... (2a + m) m = 2am + m”;. to each of these equals add a”, ... (2a + m) m + a' = a + 2am + m”. But a* + 2am + n’ = (a + m)*, ... (2a + m) m + a” = (a + m)”. - That is, If a number be divided into two equal numbers, and another number be added to the whole and to one of the parts ; the product of the whole number thus increased and the other number, together with the square of half the given number, is equal to the square of the number which is made up of half the given number increased. , -- The algebraical results of Prop. v. and Prop. vi. are identical, as it is obvious that the difference of a + m and a — m in Prop. v. is equal to the difference of 22 + m and m, in Prop. VI., and one algebraical result expresses the truth of both propositions. wº This arises from the two ways in which the difference between two unequal lines may be represented geometrically, when they are in the same direction. In the diagram (fig...to Prop. v.), the difference DB of the two unequal lines AC and CD is exhibited by producing the less line CD, and making CB equal to AC the greater, Then the part produced DB is the difference between AG and CD,. for AC is equal to CB, and taking CD from each, the difference of AC and CD is equal to the difference of CB and CD. In the diagram (fig. to Prop. vi.), the difference DB. of the two unequal lines CD and CA is exhibited by cutting off from CD, the greater, a part CB equal to CA the less. - - * Prop. VII. Either of the two parts AC, CB of the line A.B. may be taken : and it is equally true, that the squares on AB and AC are equal to twice the rectangle AB, AC, together with the square on BC. This Proposition may be enunciated in the following form. The square On the difference of any two lines, is equal to the difference between the sum of the squares on the two lines, and twice their rectangle. a' It also appears that the difference between the squares on the sum, and on the NOTES TO BOOK II. . 87 difference of two lines, is equal to four times the - rectangle contained by the liues themselves. Prop. VII. Algebraically. Let AB contain a linear units, and let the parts AC and CB contain m and n linear units respectively. Then a = m + n : squaring these equals, ... a' = m” + 2mn + n”, add nº to each of these equals, ... a” + n* = m” + 2mm + 2n”. But 2mn + 2n” = 2 (m + n) n = 2an, ... a” + n’ = m^+2an. That is, If a number be divided into any two parts, the squares of the whole number and of one of the parts, are equal to twice the product of the whole number and that part, together with the square of the other part. Prop. VIII. As in Prop. vii., either part of the line may be taken, and it is . also true in this Proposition, that four times the rectangle contained by AB, AC together with the square on BC, is equal to the square on the straight line made up of AB and AC together. The truth of this Proposition may be deduced from Euc. II. 4 and 7. For the square on AD (fig. Prop. 8.) is equal to the squares on AB, BD, and twice the rectangle AB, BD ; (Euc. II. 4.) or the squares on AB, BC, and twice the rectangle AB, BC, because BC is equal to BD: and the squares on AB, BC are equal to twice the rectangle AB, BC with the square on AC: (Euc. II. 7.) therefore the square on AD is equal to four times the rectangle AB, BC together with the square on A.C. - Prop. VIII. Algebraically. Let the whole line AB contain a linear units of which the parts AC, CB contain m, n units respectively. -> Then m + n = a, and subtracting or taking n from each, ..”. ºn - d - ??, squaring these equals, ... m” = a” – 2an + n*, and adding 4an to each of these equals, ..'. 4an + m” = a” + 2an + n”. But a* + 2an + n’ = (a + n)*, ... 4an + m” = (a + n)”. That is, If a number be divided into any two parts, four times the product of the whole number and one of the parts, together with the square of the other part, is equal to the square of the number made of the whole and the part first taken. - - - Prop. v.III. may be put under the following form : The square on the sum of two lines exceeds the square on their difference, by four times the rectangle contained by the lines. - Prop. Ix. The demonstration of this proposition may be deduced from Euc. II. 4 and 7. : For (Euc. II. 4.) the square on AD is equal to the squares on AC, CD and twice the rectangle AC, CD; (fig. Prop. 9) and adding the square on DB to each, therefore the squares on AD, DB are equal to the squares on AC, CD and 88 EUCLID's ELEMENTS. twice the rectangle AC, CD together with the square on BB;- or to the squares on BC, CD and twice the rectangle BC, CD with the square on DB, because Be is equal to A.C. - * - _ –- But the squares on BC, CD are equal to twice the rectangle B6, CD, with the square on D.B. (Euc. II. 7.) - Wherefore the squares on AD, DB are equal to twice the squares on BC and CD. From this Proposition, it appears that the square on the sum of two lines AC, CD and the square on their difference DB, are together equal to double of the sum of the squares on the two lines. AC, CD, Prop. Ix. Algebraically. Let AB contain 2a linear units, its half Ac or BC will contain a units; and ..let CD the line between the points of section contain. m. units. Then AD the greater of the two unequal parts contains a + m units, and DB the less contains a — m units;, - ... (a + m)* = a + 2am + m”, * and (a — m)* = a” – 2am + m”, Hence by adding these equals, ... (a + m)* + (a — m)* = 2a" + 2m”. That is, If a number be divided into two equal parts, and also into, two unequal parts, the sum of the squares of the two unequal parts, is equal to twice the square of half the number itself, and twice the square of half the difference of the unequal parts. The proof of Prop. x. may be deduced from Euc, II, 4, 7, as Prop. Ix. Prop. x. Algebraically. w Let the line AB contain 2a linear units, of which its half AC or CB will contain a units; - and let BD contain: m units. - - Then the whole line and the part produced will contain 2a-i- m units, and half the line and the part produced will contain a + m units, ... (2a + m)* = 4a2 + 4am + m”; *~ add mº to each of these equals, . (2a + m)* + m” = 4a” + 4am + 2m”. Again, (a -- m)* = a” + 2am + m”, add a” to each of these equals, ... (a + m)” + a* = 2a" + 2am + m", and doubling these equals, ... 2(a + m)* + 2a” = 4a" + 4am + 2m”. But (2a + m)* + m2 = 4a" + 4am + 2m”. Hence ... (2a + m)* + m” = 2a" + 2 (a + m)”. That is, If a number-be divided into two equal parts, and the whole number and one of the parts be increased by the addition of another number, the squares of the whole number thus increased, and of the number by which it is increased, are equal to double the squares of half the number, and of half the number increased. - The algebraical results of Prop. Ix., and Prop. x., are identical, (the enuncia- tions of the two Props. arising, as in Prop. v. and Prop. vi., from the two ways of exhibiting the difference between two lines); and both may be included under the following proposition: The square on the sum of two lines and the square on their difference, are together equal to double the sum of the squares on the two lines. - NOTES To BOOK II. - 89. Prop. x1. The following problem is contained in the construction: Produce a straight line, so that the rectangle contained by the whole line thus produced. and the part produced, shall be equal to the square on the original line. For the rectangle contained by A.F, FA is manifestly equal to the square on CA or AB. Two series of lines, one series decreasing, and the other series increasing in, magnitude, and each line divided in the same manner may be found by means of this proposition. - (1) To find the decreasing series. º In the fig. Euc. II. 11, AB = AH -- BH, and since AB. BH = AH*, ... (AH + BH). BH = AH*, ... BH” = AH* – AH. BH = AH. (AH – BH). If now in HA, HL be taken equal to BH, then HL* = AH (AH – HL), or AH.A.L = HL*: that is, AH is divided in L, so that the rectangle contained by the whole line. A Hº and one part, is equal to the square on the other part H.L. By a similar process, HL may be so divided; and so on, by always taking from the greater part of the divided line, a part equal to the less. (2) To find the increasing series. I'rom the fig. it is obvious that CF. FA = CA*. Hence CF is divided in A, in the same manner as AB is divided in H, by adding AF a line equal to the greater segment, to the given line CA or AB. And by successively adding to the last line thus divided, its greater segment, a series of lines increasing in magnitude may be found similarly divided to AB. It may also be shewn that the squares on the whole line and on the less segment are equal to three times the square on the greater segment. (Euc. xIII.4.) To solve Prop. xI. algebraically, or to find the point H in AB such that, the ‘rectangle contained by the whole line AB and the part HB, shall be equal to the square on the other part A.H. . Let AB contain a linear units, and AH one of the unknown parts, contain. a units, { then the other part HB contains a - a units. ‘And ... a (a – w = z*, by the problem, or a " + az = a”, a quadratic equation. Whence a = +++* & The former of these values of a determines the point H. W5 – 1 2 3 – V5 2 It may be observed, that the parts AH and HB cannot be numerically expressed by any rational number. Approximation to their true values in terms of AB, may be made to any required degree of accuracy, by extending the extraction of the square root of 5 to any number of decimals. - v 5 + 1 –––.a. So that a = ...AB = AH, one part, and a — a = a – AH = . AB = HB, the other part. To ascertain the meaning of the other result a' = In the equation a (a — ac) = z*, for a write — w, then a (a + æ) =a", which, when translated into words, gives the following problem: To find the length to which a given line must be produced, so that the 90 EUCLID'S ELEMENTS. rectangle contained by the given line and the line made up of the given line and the part produced, may be equal to the square on the part produced. Or, the problem may also be expressed as follows: •, To find two lines having a given difference, such that the rectangle contained by the difference and one of them, may be equal to the square on the other. It may here be remarked, that Prop. XI. Book II. affords a simple Geometrical construction for a quadratic equation. Prop. xII. The perpendicular may be drawn from either of the acute angles. In Euclid's construction, the perpendicular is drawn from the acute angle A to meet the side BC produced in D: the other construction is omitted, as the perpendicular also may be drawn from the acute. angle B to meet the side AD produced in some point E. Prop. xII. Algebraically. Assuming the truth of Euc. I.47. Let BC, CA, AB contain a, b, c linear units respectively, and let CD, DA, contain m, n units, then BD contains a + m units. And therefore, c* = (a + m)* + n°, from the right-angled triangle ABD, also b% = m^+ n” from ACD ; ... cº — b% = (a + m)* — m” - = a” + 2am + m” – mº a” + 2am, ... c” = b% + a” + 2am, that is, c* is greater than bº + a” by 22m. Prop. xIII. Case II. may be proved more simply as follows: Since BD is divided into two parts in the point D, therefore the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on CD; (II.7.) - add the square on AD to each of these equals; - therefore the squares on CB, BD, DA are equal to twice the rectangle CB, BD, --- and the squares on CD and DA, but the squares on BD, DA are equal to the square on AB, (1.47.) and the squares on CD, DA are equal to the square on AC, therefore the squares on CB, BA are equal to the square on AC, and twice the rectangle CB, BD. That is, &c. The first and second cases of this Proposition may be included in the same proof. - g Prop. XIII. Algebraically. - - - Let BC, GA, AB contain respeetively a, b, c linear units, and let BD and AD also contain m and n units. s Case I. Then DC contains a - m units. Therefore c’ = n” -- m^ from the right-angled triangle ABD, and bº = n°4 (a — m)” from ADC; - ... cº — bº = m” — (a — m)* - m” – a” + 2am — m” = — a” + 2am, ... a” + c” = bº + 2am, or bº + 2am = a” + cº, that is, b’ is less than a” + c” by 2am. NO TES TO BOOK II. 91 Case II. DC = m — a units, ... cº = m” + n° from the right-angled triangle ABD, and b* = (m — a)* + n° from ACD, ... cº — b% = m” – (m — a)*, = m? — mº-E 2am — a” = 2am — a”, ... a” + c = bº. H 2am, g or b% + 2am = a” + cº, that is, b” is less than a” + cº by 2am. Case III. Here m is equal to a. And bº + a”= c”, from the right-angled triangle ABC. Add to each of these equals a”, ... b” + 2a” = c^* + a”, that is, 5* is less than cº -i- a” by 2a", or 2aa. These two propositions, Euc. II. 12, 13, with Euc. I.47, exhibit the relations which subsist between the sides of an obtuse-angled, an acute-angled, and a right- angled triangle respectively. * NOTE ON THE ABBREVIATIONS AND ALGEBRAICAL SYMBOLS EMPLOYED IN GEOMETRY. THE ancient Geometry of the Greeks admitted no symbols bêsides the diagrams and ordinary language. In later times, after symbols of operation had been devised by writers on Algebra, they were very soon adopted and employed on account of their brevity and convenience, in writings purely geometrical. Dr. Barrow was one of the first who introduced algebraical symbols into the language of Elementary Geometry, and distinctly states in the preface to his Euclid, that his object is “to content the desires of those who are delighted more with symbolical than verbal demonstrations.” As algebraical symbols are employed in almost all works on the mathematics, whether geometrical or not, it seems proper in this place to give some brief account of the marks which may be regarded as the alphabet of symbolical language. - The mark=was first used by Robert Recorde, in his treatise on Algebra entitled, “The Whetstone of Witte,” 1557. He remarks; “And to avoide the tediouse. repetition of these woordes: is equalle to : I will sette as I doe often in woorke use, a paire of paralleles, or Gemowe lines of one lengthe, thus: = , bicause noe 2 thynges can be more equalle.” It was employed by him as simply affirming the equality of two numerical or algebraical expressions. Geometrical equality is not. exactly the same as numerical equality, and when this symbolis used in geometrical; reasonings, it must be understood as having reference to pure geometrical equality. The signs of relative magnitude, > meaning, is greater than, and <, is less than, were first introduced into algebra by Thomas Harriot, in his “Artis Analyticae Praxis,” which was published after his death in 1631. The signs + and — were first employed by Michael Stifel, in his “Arithmetica Integra,” which was published in 1544. The sign + was employed by him for the word plus, and the sign —, for the word minus. These signs were used by Stifel strictly as the arithmetical or algebraical signs of addition and subtraction, 92. - - EUCLID's ELEMENTS. * The sign of multiplication x was first introduced by Oughtred in his “Clavis Mathematica,” which was published in 1631. In algebraical multiplication he either connects the letters which form the factors of a product by the sign x , or writes them as words without any sign or mark between them, as had been done before by Harriot, who first introduced the small letters to designate known and unknown quantities. However concise and convenient the notation AB x BC or A.B. BC may be in practice for “the rectangle contained by the lines AB and BC’’; the student is cautioned against the use of it, in the early part of his geometrical studies, as its use is likely to occasion a misapprehension of Euclid's meaning, by confounding the idea of Geometrical equality with that of Arithmetical equality. Later writers on geometry who employed the Latin language, explain the notation A B × BC by “A B ductum in B6”; that is, if the line AB be carried along the line BC in a normal position to it, until it come to the end C, it will, then form with BC, the rectangle contained by A B and BC. Dr. Barrow sometimes expresses “the rec- tangle contained by AB and BC’’ by “the rectangle ABC.” Michael Stifel was the first who introduced integral exponents to denote the powers of algebraical symbols of quantity, for which, he employed capital letters. Vieta afterwards used the vowels to denote known, and the consonants, unknown quantities, but used words to designate the powers. Simon Stevin, in his treatise on Algebra, which was published in 1605, improved the notation of Stifel, by placing the figures that indicated the powers within small circles. Peter Ramus adopted the initial letters l, g, c,..bg of latus, quadratus, cubus, biquadratus, as the no- tation of the first four powers. Harriot exhibited the different powers of algebraical symbols by repeating the symbol, two, three, four, &c. times, according to the order of the power. Descartes restored the numerical exponents of powers, placing them at the right of the numbers, or symbols of quantity, as at the present time. Jr. Barrow employed the notation A.Bg, for “the square on the line AB,” in his edition of Euclid. The notations AB", AB3, for “the square and cube on the line whose extremities are A and B,” as well as AB x BC, for “the rectangle contained by AB and BC,” are used as abbreviations in almost all works on the Mathe- matics, though not wholly consistent with the algebraical notations a” and a”. - The symbol V, being originally the initial letter of the word radiz, was first used by Stifel to denote the square root of the number, or of the symbol, before which it is placed. - ** - The Hindus, in their treatises on Algebra, indicated the ratio of two numbers, or of two algebraical symbols, by placing one above the other, without any line of separation. The line was first introduced by the Arabians, from whom it passed to the Italians, and from them to the rest of Europe. This notation has been employed for the expression of geometrical ratios by almost all writers on the Mathematics, on account of its great convenience. Oughtred first used points to indicate proportion ; thus, a b : ; c : d, means that a bears the same proportion to b, as c does to d. f QUESTIONS ON BOOK II. 1. DEFINE the word magnitude in all the senses in which it can be used in Geometry, and clearly explain the species of magnitudes considered in the Second Eook of Euclid. - 2. How may a rectangular parallelogram be conceived to be generated? Is ...the conception recognised in any of the demonstrations of the Second Book of Euclid QUESTIONS ON BOOK II. 93 - 3. Is reotangle the same as rectus angulus? Explain the distinction, and give the corresponding Greek terms. - - 4. What is meant by the sum of two, or of more than two straight lines in Creometry? : 5. Is there any difference between the straight lines by which a rectangle is said to be contained, and those by which it is bounded? - 6. Define a gnomon. How may gnomons appear from the same construction in the same rectangle? Find the difference between them. Nº. 7. What axiom is assumed in proving the first eight propositions of the Second Book of Euclid * - & 8. Of equal squares and of equal rectangles, which must necessarily coincide? 9. Distinguish between the square on a line and the square of a line. What objection exists to the use of the notation AB", or A.B. BC in a system of pure geometry? - - 10. In a given square, shew how a gnomon may be drawn equal in area to any part, (as a half, a third, or a fourth) of the given square. ** 11. When the adjacent sides of a rectangle are commensurable, the area of the rectangle is properly represented by the product of the number of units in two adjacent sides of the rectangle. Illustrate this by considering the case when the two adjacent sides contain 3 and 4 units respectively, and distinguish between the units of the factors and the units of the product. Shew generally that a rectangle whose adjacent sides are represented by the integers a and b, is represented by ab. Also shew, that in the same sense, the rectangle is represented by ; if the sides - a b be represented by m ’ n ° * • 12. Why may not Algebraical or Arithmetical proofs be substituted (as being shorter) for the demonstrations of the Propositions in the Second Book of Euclid 13. In what sense is the area of a triangle said to be equal to half the product of its base and its altitude? What two propositions of Euclid may be adduced to plove it? . 14. How do you shew that the area of a rhombus is equal to half the rectangle contained by the diagonals? d 15. How may a rule be deduced for finding a numerical expression for the area of any parallelogram, when two adjacent sides are given 16. The area of a trapezium which has two of its sides parallel, is equal t that of a rectangle contained by its altitude and half the sum of its parallel sides. What propositions of the First and Second Books of Euclid are employed to prove this? Of what service is the above in the mensuration of fields with irregular borders : - 17. From what propositions of Euclid may be deduced the following rule for finding the area of any quadrilateral figure :—“Multiply the sum of the perpen- diculars drawn from opposite angles of the figure upon the diagonal joining the other two angles, and take half the product '' 18. From Euc. II. 3, shew that the difference between the rectangles con- tained by the whole line AB and each of the parts, AC and BC, is equal to the difference of the squares on the parts BC, AC. - 19. If a straight line be divided into any number of parts, the difference be- tween the square on the whole line, and the sum of the squares on the several parts, is equal to twice the sum of all the rectangles that can be formed by the several parts. 20. How may the demonstration of Euclid I.i.4, be legitimately shortened? *J 94 EUCLID's ELEMENTS. wº Give the Algebraical proof, and state on what suppositions it can be regarded as a proof. * - * 21. Shew that the proof of Eue. II.4, can be deduced from the two previous propositions without any geometrical construction. *. 22. Shew that if the two complements be together equal to the two squares, the given line is bisected. - 23. Prove geometrically that if a straight line be trisected, the square on the whole line equals nine times the square on a third part of it. - 24. If the line AB, as in Euc. II. 4, be divided into any three parts, enunciate and prove the analogous proposition. 25. Thraw two-gnomons to a given square, so that the inner square may be one half of the given square. - - 26. Deduce from Euc. II. 4, a proof of Euc. I.47. 27. If a straight line be divided into two parts, when is the rectangle con-, tained by the parts, the greatest possible? and when is the sum of the squares of the parts, the least possible? - 28. Shew that if a line be divided into two equal parts and into two unequal parts; the part of the line between the points of section is equal to half the diffe- rence of the unequal parts. p 29. If half the sum of two unequal lines be increased by half their difference, the sum will be equal to the greater line: and if the sum of two lines be dimi- nished by half their difference, the remainder will be equal to the less line. 30. Explain what is meant by the internal and external segments of a line. Why is this extension of the term segment made? Shew that the sum of the ex- ternal segments of a line, or the difference of the internal segments, is double the distance between the points of section and bisection of the line. 31. Shew how Euc. II. 6, may be deduced immediately from the preceding Proposition, Euc. 11. 5. 32. Prove Geometrically that the squares on the sum and difference of two lines are equal to twice the squares on the lines themselves. 33. A given rectangle is divided by two straight lines into four rectangles. Given the areas of the two which have not common sides: find the areas of the other two. - . 34. In how many ways may the difference of two lines be exhibited? Enun- ciate the propositions in Book II. which depend on that circumstance. 35. How may a series of lines be found similarly divided to the line AB in Euc. II. 11 ? -* 36. Divide Algebraically a given line a into two parts, such that the rect- angle contained by the whole and one part may be equal to the square on the other part. Deduce Euclid's construction from one solution, and explain the other. 37. Given the less segment of a line, divided as in Euc. II. 11, find the greater. - ~, 38. Enunciate the Arithmetical theorems expressed by the following Alge- braical formulae: tº - (a + b)* = a” + 2a5 + 5° : a* – bº = (a + b)(a – 5): (a — b)* = a” – 2ab +8°, and state the corresponding Geometrical propositions. 39. Shew that the first of the Algebraical propositions, (a + 2) (a — a + 4* = a”: (a + æ)* + (a – a.)* = 2a" + 2*, is equivalent to the two propositions v. and vi., and the second of them, to the two propositions Ix. and x. of the Second Book of Euclid. 40. Prove Euc. II, 12, when the perpendicular BE is drawn from B on AG QUESTIONS ON BOOK II. 95 produced to E, and shew that the rectangle BC, CD is equal to the rectangle AC, CE. . * - #1. Include the first two cases of Euc. II. 13, in one proof. 42. In the second case of Euc. II. 13, draw a perpendicular CE from the obtuse angle C upon the side AB, and prove that the square on AB is equal to the rectangle AB, AE together with the rectangle BC, BD. 43. Enunciate Euc. II. 13, and give an Algebraical or Arithmetical proof of it. 44. The sides of a triangle are as 3, 4, 5. Determine whether the angles be- tween 3, 4; 4, 5; and 3, 5; respectively are greater than, equal to, or less than, a right angle. •. z- 45. Two sides of a triangle are 4 and 5 inches in length, if the third side be 6% inches, the triangle is acute-angled, but if it be 63% inches, the triangle is obtuse-angled. ...sº - - 46. A triangle has its sides 7, 8, 9 units respectively; a strip of breadth 2 units being taken off all round from the triangle ; find the area of the remainder. 47. If the original figure, Euc. II. 14, were a right-angled triangle, whose sides were represented by 8 and 9, what number would represent the side of a square of the same area? Shew that the perimeter of the square is less than the perimeter of the triangle. * 48. If the sides of a rectangle are 8 feet and 2 feet, what is the side of the equivalent square - - 49. “All plane rectilineal figures admit of quadrature.” Point out the suc- cession of steps by which Euclid establishes the truth of this proposition. 50. Explain the construction (without proof) for making a square equal to a plane polygon. - - &= 51. Shew from Euc. II. 14, that any algebraical surd as Va can be represented by a line, if the unit be a line. 52. How may a rectangle be dissected so as to form an equivalent rectangle of any proposed length - 53. If the sides of the rectangle in Prop. 14, Book II, be given numerically, what is the condition that the side of the square may be expressed by a rational number * 54. Shew that the area of every rectangular parallelogram, is equal to the rectangle contained by the diagonal and the perpendicular drawn from one of the angles upon the diagonal. - - 55. Classify all the properties of triangles demonstrated in the First and Second Books of Euclid. - - 56. Could any of the propositions of the Second Book be made corollaries to other propositions, with advantãge? Point out any such propositions, and give your reasons for the alterations you would make. BOOK III. DEFINITIONS. I. Equal, circles are those of which the diameters are equal, or from the centers of which the straight lines to the circumferences are equal. w This is not a definition, but a theorem, the truth of which is evident ; for, if the circles be applied to one another, so that their centers coincide, the . must likewise coincide, since the straight lines from the centers are equal. - II. A straight line is said to touch a circle when it meets the circle, and being produced does not cut it. * - III. Circles are said to touch one another, which meet, but do not cut one another. - IV. Straight lines are said to be equally distant from the center of a circle, when the perpendiculars drawn to them from the center are equal. - And the straight line on which the greater perpendicular falls, is said to be further from the center. * VI. A segment of a circle is the figure contained by a straight line, and the arc or the part of the circumference which it cuts off. * ^ BOOK IFI. . * - 97 WII. The angle of a segment is that which is contained by a straight line and a part of the circumference. An angle in a segment is any angle contained by two straight lines drawn from any point in the arc of the segment, to the extremities of the straight line which is the base of the segment. IX. An angle is said to insist or stand upon the part of the circum- ference intercepted between the straight lines that contain the angle. X. A sector of a circle is the figure contained by two straight lines drawn from the center and the arc between them. - Similar segments of circles are those in which the angles are equal, or which contain equal angles. 98 EUCLID's ELEMENTS. sº PROPOSITION I. PROBLEM. To find the center of a given circle. Let ABC be the given circle: it is required to find its center. Draw within it any straight line AB to meet the circumference in A, B ; and bisect. AB in D; (I. 10.) from the point D draw DC at right angles to AB, (I. 11.) meeting the circumference in C, produce CD to E to meet the circumference again in E, and bisect CE in F. . Then the point F shall be the center of the cirèle ABC. For, if it be not, if possible, let G be the center, and join GA, GD, GB. * Then, because DA is equal to DB, (constr.) and DG common to the two triangles ADG, BDG, the two sides AD, DG, are equal to the two BD, DG, each to each; and the base GA is equal to the base G.B., (I. def. 15.) because they are drawn from the center G : therefore thé angle ADG is equal to the angle GDB : (I. 8.) but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle; (I. def. 10.) . . - therefore the angle GDB is a right angle: but FDB is likewise a right angle; (constr.) wherefore the angle FDB is equal to the angle GDB, (ax. 11.) the greater angle equal to the less, which is impossible; therefore G is not the center of the circle ABC. . - In the same manner it can be shewn that no other point out of the line CE is the center; - and since CE is bisected in F, - g any other point in CE divides CE into unequal parts, and cannot be the center. Therefore no point but Fis the center of the circle ABC. Which was to be found. CoR. From this it is manifest, that if in a circle a straight line bisects another at right angles, the center of the circle is in the line which bisects the other. PROPOSITION II. THEOREM. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. - - Let ABC be a circle, and A, B any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle. C Yº E B BOOK III. PROP. II, III. $99 For if AB do not fall within the circle, - let it fall, if possible, without the circle as AEB; find D the center of the circle ABC, (III. 1.) and jein DA, D.B.; in the circumference A,B take any point F, join DF, and produce it to meet AB in E. Then, because DA is equal to DB, (I. def. 15.) therefore the angle DBA is equal to the angle DAB ; (I. 5.) and because A E, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle BAE; (I. 16.) but DAE was proved to be equal to the angle DBE; therefore the angle DEB is greater than the angle DBE; but to the greater angle the greater side is opposite, (I. 19.) therefore DB is greater than DE: but DB is equal to DF; (I. def. 15.) wherefore DF is greater than DE, the less than the greater, which is impossible; - therefore the straight line drawn from A to B does not fall without the circle. - In the same manner, it may be demonstrated that it does not f upon the circumference; therefore it, falls within it. Wherefore, if any two points, &c. Q.E.D. IPROPOSITION III. THEOREM. If a straight line drawn through the center of a circle bisect a straight line in it which does not pass through the center, it shall cut it at right angles; and conversely, if it cut it at right angles, it shall bisect it. Let ABC be a circle; and let CD, a straight line drawn through the center, bisect any straight line AB, which does not pass through the center, in the point F. - Then CD shall cut A.B at right angles. - - ID Take E the center of the circle, (III. 1.) and join EA, EB. Then, because AF is equal to FB, (hyp.) and FE common to the two triangles AFE, BFE, there are wº sides in the one equal to two sides in the other, each to each; and the base EA is equal to the base EB; (I. def. 15.) therefore the angle AFE is equal to the angle BFE; (1.8.) but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of them is a right angle; (I. def. 10.) therefore each of the angles AFE, BFE, is a right angle: wherefore the straight line CD, drawn through the center, bisecting another AB that does not pass through the center, cuts the same at right angles. 4 - s, * * * fy nº Q r º £ - * f; [. 3, . c 9 r ſº j º $' Téſ g J ..", 0 } & * J * 100 EUCLID's ELEMENTS. Wºr Conversely, let CD cut AB at right angles. r Then CD shall also bisect AB, that is, AF shall be equal to FB. The same construction being made, * because, EB, EA, from the center are equal to one another; (I. def. 15.) - therefore the angle EAF is equal to the angle EBF; (1. 5.) and the right angle AFE is equal to the right angle BFE; (I. def. 10.) therefore, in the two triangles, EAF, EBF, there are two angles in the one equal to two angles in the other, each to each; and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal; (I. 26.) therefore AF is equal to F.B. Wherefore, if a straight line, &c. Q.E.D. FROPOSITION IV. THEOREM. If in a circle two straight lines out one another, which do not both pass through the center, they do not bisect each other. Let ABCD be a circle, and AC, BD two straight lines in it which cut one another in the point E, and do not both pass through the center. Then AC, BD, shall not bisect one another. For, if it be possible, let A.E be equal to EC, and BE to ED. If one of the lines pass through the center, it is plain that it cannot be bisected by the other which does not pass through the center: but if neither of them pass through the center, find F the center of the circle, (III. 1.) and join EF. Then because FE, a straight line drawn through the center, bisects another AC which does not pass through the center, (hyp.) therefore FE cuts AC at right angles: (III. 3.) wherefore FEA is a right angle. - Again, because the straight line F.E. bisects the straight line BD, which does not pass through the center, (hyp.) - therefore FE cuts BD at right angles: (III. 3.) wherefore FEB is a right angle : 'but FEA was shewn to be a right angle; therefore the angle FEA is equal to the angle FEB, (ax. 1.) the less equal to the greater, which is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q.E.D. PROPOSITION v. THEOREM. - If two circles out one another, they shall not have the same center. r Let the two circles ABC, CDG, cut one another in the points B, C, ... . . They shall not have the same center. - & ** , M J ** 1 & , , , ) : ) b ! J. f J : ) BOOK III. PROP. V, VI, VII. 101 If possible, let E be the center of the two circles; join EC, and draw any straightline EFG meeting the circumferences in F and G. . And because E is the center of the circle ABC, therefore EF is equal to EC: (I. def. 15.) again, because E is the center of the circle CDG, therefore EG is equal to EC: (I. def. 15.) but EF was shewn to be equal to EC; s" therefore EF is equal to EG, (ax. 1.) the less line equal to the greater, which is impossible. Therefore E is not the center of the circles ABC, CDG. Wherefore, if two circles, &c. Q.E.D. PROPOSITION VI. THEOREM. If one circle touch another internally, they shall not have the same center. Let the circle CDE touch the circle ABC internally in the point C. They shall not have the same center. C "Spy ID If possible, let F be the center of the two circles: join FC, and draw any straight line FEB, meeting the circumferences in E and B. And because Fis the center of the circle ABC, FB is equal to FC; (I. def. 15.) - also, because Fis the center of the circle CDE, FE is equal to FC: (I. def. 15.) but FB was shewn to be equal to FC; therefore FE is equal to FB, (ax. 1.) the less line equal to the greater, which is impossible: therefore F is not the center of the circles ABC, CDE. Therefore, if two circles, &c. Q.E.D. PROPOSITION VII. THEOREM. If any point be taken in the diameter of a circle which is not the center, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the center is, and the other part of that diameter is the least; and, of the rest, that which is nearer to the line which passes through the center, is always greater than one more remote : and from the same point, there can be drawn only two equal straight lines to the circumference, one upon each side of the diameter. - Let ABCD be a circle, and AD its diameter, in which let any point Fbe taken which is not the center: let the center be E. 102 EUCLID's ELEMENTs. Then, of all the straight lines FB, FC, FG, &c. that can be drawn . from F to the circumference, - FA, that in which the centeris, shall be the greatest, and FD, the other part of the diameter AD, shall be the least: and of the rest, FB, the nearer to FA, shall be greater than FC the more remote, and FC greater than F6. G iſ - Join BE, CE, GE. Because two sides of a triangle are greater than the third side, (I. 20.) therefore B.E, JEF, are greater than BF: but AE is equal to BE; (I. def. 15.) therefore AE, E, F, that is, AF is greater than B.F. Again, because B.E is equal to CE, and FE common to the triangles BEF, CEF, the two sides B.E., E.F. are equal to the two CE, EF, each to each; but the angle BEF is greater than the angle CEF; (ax. 9.) therefore the base BF is greater than the base CF. (I. 24.) For the same reason CF is greater than G.F. Again, because GF, FE are greater than EG, (I. 20.) and EG is equal to ED; therefore GF, FE are greater than ED: - take away the common part F.E, and the remainder GF is greater than the remainder FD. (ax. 5.) Therefore, FA is the greatest, and FD the least, of all the straight lines from F to the circumference; and BF is greater than 0P, and CF than G.F. Also, there can be drawn only two equal straight lines from the point Fto the circumference, one upon each side of the diameter. At the point E, in the straight line EF, make the angle FEH equal to the angle FEG, (I. 23.) and join FH: * Then, because GE is equal to EH, (1. def. 15.) and EF common to the two triangles GEF, HEF; - the two sides. GE, JEF are equal to the two HE, EF, each to each, and the angle GEF is equal to the angle HEF; (constr.) - therefore the base FG is equal to the base-FH: (I. 4.) but, besides FH, no other straight line can be drawn from F to the circumference equal to Fé: ~ i. for, if possible, let it be FK: and because FK is equal to FG, and FG to FH, therefore FK is equal to FH: (ax. 1.) that is, a line nearer to that which passes through the center, is equal to one which is more remote; which has been proved to be impossible. Therefore, if any point be taken, &c., Q.E.D. BOOK III. PROP. VIII. . . 103 PROPOSITION VIII. THEOREM. If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the center; of those which fall upon the concave part of the circumference, the greatest is that which passes through the center; qnd of the rest, that which is nearer to the one zassing through the center is always greater than one more remote : but of those which fall upon the convex part of the circumference, the least is that between the point without the circle and the diameter; and of the rest, that which is nearer to the least is always less than one more remote; and only two equal straight lines can be drawn from the same point to the circumference, one upon each side of the line which passes through the center. Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the circumference, whereof DA passes through the center. - D E. A. * Of those which fall upon the concave part of the circumference AEFC, the greatest shall be DA, which passes through the center; and any line nearer to it shall be greater than one more remote, viz. DE shall be greater than DF, and DF greater than DC; but of those which fall upon the convex part of the circumference HLKG, the least shall be DG between the point D and the diameter AG; and any line nearer to it shall be less than one more remote, viz. DJſ less than DL, and DL less than D.H. Take M the center of the circle ABC, (III. 1.) and join ME, ME, MC, MK, ML, MH. And because AM is equal to ME, add MD to each of these equals, therefore AD is equal to EM, MD: (ax. 2.) but EM, MD are greater than ED; (I. 20.) therefore also AD is greater than ED. . Again, because ME is equal to MF, and MD common to the triangles JEMD, FMD; EM, MD, are equal to FM, MD, each to each; but the angle EMD is greater than the angle FMD; (ax. 9.) therefore the base ED is greater than the base F.D. (I. 24.) In like manner it may be shewn that FD is greater than CD. Therefore DA is the greatest; and DE greater than DF, and DF greater than DC. And, because MK, KD are greater than M.D., (I. 20.) and MK is equal to MG, (1. def. 15.) the remainder KD is greater than the remainder GD, (ax. 5.) that is, GD is less than KD : and because MLD is a triangle, and from the points M, D, the extremities of its side MD, the straight lines MK, DK are drawn to the point K within the triangle, - 104 ...” EUCLID's ELEMENTs. therefore MK, KD are less than MZ, LD: (1.21.) but MK is equal to, MI, ; (I. def. 15.) . - therefore; the remainder DK is less than the remainder D.L. (ax. 5.). In like manner it may be shewn, that DL is less than D.H. Therefore, DG is the least, and DJſ less than DL, and DL less than D.H. Also, there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the line which, passes through the center. At the point M, in the straight line MD, make the angle DMB equal to the angle DMK, (I. 23.) and join D.B. And because MK is equal to MB, and MD common to the triangles JKMD, BMD, the two sides KM, MD are equal to the two BM, MD; each to each; and the angle KMD is equal to the angle BMD; (constr.) - therefore the base DK is equal to the base DB : (r. 4.) but, besides DB, no straight line equal to DK can be drawn from D to the circumference, for, if possible, let it be DN; and because DK is equal to DN, and also to DB, ,- therefore DB is equal to DN; : that is, a line nearer to the least, is equal to one more remote, which has been proved to be impossible. - If therefore, any point, &c., Q.E.D., PROPOSITION IX. THEOREM. If a point be taken within a circle, from which there fall more, than two equal straight lines to the circumference, that point is the center of the circle. - - Let the point D'be taken within the circle ABC, from which to the circum- ference there fall more than two equal straight lines, viz. DA, DB, IJ.C. Then the point D. shall be the center of the circle. Eor; if not, let E, if possible, be the center : join DE, and produce it to meet the circumference in F, G : then FG is a diameter of the circle ABC: (I. def. 17.) * and because in FG, the diameter of the Gircle ABC, there is taken the point D, which is not the center; therefore DG is the greatest line drawn from it to the circumference, and DC is greater than DB, and DB greater than DA::. (III, 7.) but these lines are likewise equal, (hyp.) which is impossible: therefore E is not the center of the circle AB6. In like manner it may be demonstrated, that no other point but D is the center; D therefore is the center. Wherefore, if a point be taken, &c. Q.E.D. Book III. PROP. X, XI. f05 PROPOSITION X. THEOREM. One circumference of a circle cannot out another in more than two points. If it be possible, let the circumference ABC cut the circumference, DEF in more than two points, viz, in B, 6, F. - *2+s º G. ~ E C * * Take the center K of the circle ABC, (III. 1.) and join KB, KG, KF. - Then because K is the center of the circle ABC, therefore KB, KG, KF are all equal to each other: (I. def. 15.) and because within the circle DEF'there is taken the point K, from which to the circumference DEF fall more than two equal straight lines KB, KG, KF; therefore the point K is the center of the circle DEF: (III. 9.) but K is also the center of the circle ABC; (constr.) . therefore the same point is the center of two circles that cut one another, which is impossible. (III. 5.) - .. * Therefore, one circumference of a circle cannot cut another in more than two points. Q.E.D. * PROPOSITION XI. THEOREM, If one circle touch another internally in any point, the straight line. which joins their centers being produced, shall pass through that point of contact. - Let the circle ADE touch the circle ABC internally in the point A: ; and let F be the center of the circle ABC, and G the center of the circle ADE; - then the straight line which joins the centers F, G, being produced, shall pass through the point of contact A. • A \ C - B For, if FG produced do not pass through the point A, let it fall otherwise, if possible, as FGDH, and join AF; AG. Then, because two sides of a triangle are together greater than the third side, (I. 20.) - * - therefore FG, GA are greater than FA : but FA is equal to FH; (I. def. 15.) therefore FG, GA are greater than FH: take away from these unequals the common part FG ; therefore the remainder AG is greater than the remainder GH; (ax. 5.) but AG is equal to GD; (I. def. 15.) - therefore GD is greater than GH, the less than the greater, which is impossible. 106 EUCLID'S ELEMENTS. Therefore the straight line which joins the points F, G, being produced, cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if one circle, &c. Q.E.D. PROPOSITION XII. THEOREM. , If two circles touch each other easternally in any point, the straight line which joins their centers, shall pass through that point of contact. Let the two circles ABC, ADE, touch each other externally in the point A ; t and let F be the center of the circle ABC, and G the center of A.D.E. Then the straight line which joins the points F, G, shall pass through the point of contact A. º E *S) D If not, let it pass otherwise, if possible, as FCDG, and join FA, AG. And because F is the center of the circle ABC, , - FA is equal to FC: also, because G is the center of the circle A.D.E. GA is equal to GD: therefore FA, AG are equal to FC, DG ; (ax. 2.) wherefore the whole FG is greater than FA, AG: |but FG is less than FA, AG; (I. 20.) which is impossible: therefore the straight line which joins the points F, G, cannot pass otherwise than through. A the point of contact, - that is, FG must pass through the point A. Therefore, if two circles, &c. Q.E.D. PROPOSITION XIII. THEOREM. One circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF touch the circle ABC in more points than one, * and first on the inside, in the points B, D. H Join BD, and draw GH bisecting BD at right angles. . (I. 11.) Because the points B, D are in the circumferences of each of the circles, therefore the straight line BD falls within each of them; (III. 2.) therefore their centers are in the straight line GH which bisects BD at right angles; (III. 1. Cor.) Cº. BOOK III. PROP. XIII, XIV. - 107. therefore GH passes through the point of contact: (III. 11.) ; but it does not pass through it, because the points B, D are without the straight line G.II; which is absurd.: therefore one circle cannot touch another on the inside in more points than one. Nor can two circles touch one another on the outside in more than, one point. For, if it be possible, let the circle ACIſ touch the circle ABC in the points A, C ; join A. C. . C Because the two points A, C are in the circumference of the circle. ACK, - - therefore the straight line AC which joins them, falls within the circle. ACK: (III. 2.) — ` but the circle A CK is without the circle ABC; (hyp.) therefore the straight line AC is without this last circle : but, because, the points A, Care in the circumference of the circle ABC, the straight line AC must be within the same circle, (III. 2.) which is absurd; therefore one circle cannot touch another on the outside in more than. one point: and it has been shewn, that they cannot touch on the inside in more, points than one. . . . Therefore, one circle, &c. Q.E.D. PROPOSITION XIV. THEOREM. - Equal straight lines in a circle are equally distant from the center; and conversely, those which are equally distant from the center, are equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another. Then AB and CD shall be equally distant from the center. * C KSº …' Nº. B Take E the center of the circle ABDC, (III. 1.) - from E. draw EF, EG perpendiculars to AB, CD, (I. 12.) and join. EA, EC. } Then, because the straight line EF' passing through the center, cuts AB, which does not pass through the center, at right angles; JEF bisects AB in the point F: (III. 3.) therefore AF is equal to FB, and AB double of AF. 108 EUCLID's ELEMENTS. Eor the same reason CD is double of CG: but AB is equal to CD: (hyp.) therefore AF is equal to CG. (ax. 7.) And because AE is equal to EC, (I. def. 15.) the square on AE is equal to the square on EC: but the squares on AF, FE are equal to the square on A.E, (I. 47.) because the angle AFE is a right angle; and for the same reason, the squares on EG, GC are equal to the square on EC; - therefore the squares on A F, FE are equal to the squares on CG, GE: (ax. 1.) e f - but the square on AF'is equal to the square on CG, because AF is equal to CG ; therefore the remaining square on EF"is equal to the remaining square on EG, (ax. 3.) and the straight line EF is therefore equal to EG: but straight lines in a circle are said to be equally distant from the center, when the perpendiculars drawn to them from the center are equal: (III. def. 4.) therefore AB, CD are equally distant from the center. Conversely, let the straight lines AB, CD be equally distant from the center, (III. def. 4.) that is, let FE be equal to EG ; then AB shall be equal to CD. For the same construction being made, it may, as before, be demonstrated, * that AB is double of AF, and CD double of CG, and that the squares on FE, AF are equal to the squares on EG, GC: but the square on FE is equal to the square on EG, because FE is equal to EG ; (hyp.) therefore the remaining square on AF'is equal to the remaining square on CG: (ax. 3.) and the straight line AF is therefore equal to CG : but A B was shewn to be double of AF, and CD double of CG; wherefore AB is equal to CD. (ax. 6.) Therefore equal straight lines, &c. Q.E.D. PROPOSITION XV. THEOREM. The diameter is the greatest straight line in a circle; and of the rest, that which is nearer to the center is always greater than one more remote : and conversely, the greater is nearer to the center than the less. Let ABCD be a circle of which the diameter is AD, and the center E; and let BC be nearer to the center than FG. Then AD shall be greater than any straight line BC, which is not a diameter, and BC shall be greater than FG. ſº A B. BOOK III. PROP. xv, xvi. 109 From E draw EH, perpendicular to BC, and EK to FG, (1.12.) - - and join EB, EC, EF. And because AE is equal to EB, and ED to EC, (I. def. 15.) therefore AD is equal to EB, EC: (ax. 2.) but EB, EC are greater than BC; (I. 20.) wherefore also AD is greater than BC. And, because BC is nearer to the center than FG, (hyp.) therefore EH is less than EK: (III. def. 5.) but, as was demonstrated in the preceding proposition, JBC is double of BH, and FG double of FK, and the squares of EH, HB are equal to the squares on EK, KF: but the square on EHis less than the square on EK, because EH is less than EK; therefore the square on BH is greater than the square on FK, and the straight line BH greater than FK, and therefore BC is greater than FG. Next, let BC be greater than FG; then BC shall be nearer to the center than FG, that is, the same con- struction being made, EH shall be less than EK. (III. def. 5.). Because BC is greater than FG, * BH likewise is greater than KF': * and the squares on BB, HE are equal to the squares on FK, KE, of which the square on BH is greater than the square on FK, because BH is greater than FK: therefore the square on EH is less than the square on EK, and the straight line EH less than EK: - and therefore BC is nearer to the center than FG. (III. def. 5.) Wherefore the diameter, &c. Q.E.D. PROPOSITION XVI. THEOREM. The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn from the extremity between that straight line and the circumference, so as not to out the circle : or, which is the same thing, no straight line can make so great an acute angle with the diameter at its extremity, or so small an angle with the straight line which is at right angles to it, as not to cut the circle. Let ABC be a circle, the center of which is D, and the diameter A.B. Then the straight line drawn at right angles to AB from its ex- tremity 4, shall fall without the circle. For, if it does not, let it fall, if possible, within the circle, as AC; and draw DC to the point C, where it meets the circumference. And because DA is equal to DC, (1. def. 15.) the angle DAC is equal to the angle A CD: (1.5.) 110 EUCLID'S ELEMENTS. but DAC is a right angle; (hyp.) therefore A CD is a right angle; and therefore the angles DAC, A CD are equal to-two right angles, which is impossible: (I. 17.) . therefore the straight line drawn from A at right angles to BA, does not fall within the circle. In the same manner it may be demonstrated, that it does not fall upon the circumference; - therefore it must fall without the circle, as A.E. Also, between the straightline AE and the circumference, no straight line can be drawn from the point A which does not cut the circle. JFor, if possible, let AF'fall between them, and from the point D, let DG be drawn perpendicular to AF, (I, 12.) - and let it meet the circumference in H. And because A G D is a right angle, and DAG less than a right angle, (I, 17.) therefore DA is greater than DG. (I. 19.) but DA is equal to D.H.; (I. def. 15.) therefore DH is greater than DG, the less than the greater, which is impossible: therefore no straight line can be drawn from the point A, between A B, and the circumference, which does not cut the circle: or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point A, or however small an angle it makes with A.E, the circumference must pass be- tween that straight line and the perpendicular AB. Q.E.D. • CoR. From this it is manifest, that the straight line which is drawn at right angles to the diameter, of a circle from the extremity of it touches the circle; (III, def. 2.) and that it touches it only in one point, because, if it did meet the circle in two, it would fall within it. (III. 2.) “Also, it is evident, that there can be but one straight line, which touches the circle in the same point.” - PROPOSITION XVII. PROBLEM. To draw a straight line from a given point, either without or in the cir- cumference, which shall touch a given circle. Eirst, let A be a given point without the given circle BCD; * it is required to draw a straightline from 4 which shall touch the circle. A Find the center E of the circle, (III. 1.) and join AE; BOOK III. PROP. XVII, XVIII. 111 and from the center E, at the distance EA, describe the circle AFG ; from the point D draw DF’ at right angles to EA, (I. 11.) meeting the circumference of the circle AFG in F; and join EBF, AB. Then AB shall touch the circle BCD in the point B. Because E is the center of the circles BCD, AFG, (I. def. 15.) therefore EA is equal to EF, and ED to EB; therefore the two sides A.E, EB, are equal to the two FE, ED, each to each : *- and they contain the angle at E common to the two triangles AEB, FED; - therefore the base DF is equal to the base AB, (1. 4.) and the triangle FED to the triangle AEB, and the other angles to the other angles: therefore the angle EBA is equal to the angle EDF: but EDF is a right angle, (constr.) wherefore EBA is a right angle : (ax. 1.) & and EB is drawn from the center: but a straight line drawn from the extremity of a diameter, at right angles to it, touches the circle: (III. 16. Cor.) therefore AB touches the circle ; - and it is drawn from the given point A. Secondly, if the given point be in the circumference of the circle, t - - as the point D, & draw DE to the center E, and DF at right angles to DE: then DF touches the circle. (III. 16. Cor.) Q.E.F. PROPOSITION XVIII. THEOREM, If a straight line touch a circle, the straight line drawn from the center to the point of contact, shall be perpendicular to the line touching the circle. Let the straight line DE touch the circle ABC in the point C; take the center F, and draw the straight line FC. (III. 1.) Then FC shall be perpendicular to D.E. D C G. E. If FC be not perpendicular to DE; from the point F, if possible, let FBG be drawn perpendicular to D.E. And because FGC is a right angle, therefore GCF is an acute angle; (r. 17.) r and to the greater angle the greater side is opposite: (r. 19.) therefore FC is greater than FG: * but FC is equal to FB; (1. def. 15.) therefore FB is greater than FG, the less than the greater, which is impossible: therefore FG is not perpendicular to D.E. 112 EUCLID's ELEMENTS. In the same manner it may be shewn, - that no other line is perpendicular to DE besides FC, that is, FC is perpendicular to D.E. Therefore, if a straight line, &c. Q.E.D. PROPOSITION XIx. THEOREM. If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the touching line, the center of the circle shall be ºn that line. - Let the straight line D.E. touch the circle ABC in C, and from C let CA be drawn at right angles to D.E. Then the center of the circle shall be in CA. A ID 'C TE For, if not, let F be the center, if possible, and join CF. Because DE touches the circle ABC, and FC is drawn from the center to the point of contact, therefore FC is perpendicular to DE; (III. 18.) therefore FCE is a right angle: but ACE is also a right angle; (hyp.) therefore the angle FCE is equal to the angle ACE, (ax. 1.) the less to the greater, which is impossible: therefore F is not the center of the circle ABC. In the same manner it may be shewn, that no other point which is not in CA, is the center; that is, the center of the circle is in C.A. Therefore, if a straight line, &c. Q.E.D. PROPOSITION XX. THEOREM. The angle at the center of a circle is double of the angle at the circumfer- ence upon the same base, that is, upon the same part of the circumference. Let ABC be a circle, and BEC an angle at the center, and BAC an angle at the circumference, which have BC the same part of the circumference for their base. * * - - Then the angle BEC shall be double of the angle BAC. A \ B K ſº " F & C} sº Join AE, and produce it to F. First, let the center of the circle be within the angle BAC. Pecause EA is equal to EB, therefore the angle EBA is equal to the angle EAB; (I. 5.) Book III. PROP. xx, xxi. . 113 therefore the angles EAB, EBA are double of the angle EAB : but the angle BEF is equal to the angles EAB, EBA ; (I, 32.) therefore also the angle BEF is double of the angle EAB : for the same reason, the angle FEC is double of the angle EAC: therefore the whole angle B.EC is double of the whole angle B.A. C. Secondly, let the center of the circle be without the angle B4C. It may be demonstrated, as in the first case, that the angle FEC is double of the angle FA C, and that FEB, a part of the first, is double of FAB, a part of the other; therefore the remaining angle BEC is double of the remaining angle BAC. Therefore the angle at the center, &c. Q.E.D. ^- |PROPOSITION XXI. THEOREM. The angles in the same segment of a circle are equal to one another. - . Let ABCD be a circle; - and BAD, BED angles in the same segment BAED. Then the angles B.A.D., BED shall be equal to one another. First, let the segment BAED be greater than a semicircle. A E (; - Take F, the center of the circle ABCD, (III. 1.) and join BF, F.D. Because the angle BFD is at the center, and the angle BAD at the circumference, and that they have the same part of the circumfer- ence, viz. the arc BCD for their base; - therefore the angle BFD is double of the angle BAD: (III. 20.) for the same reason the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle B.E.D. (ax. 7.) . Next, let the segment BAED be not greater than a semicircle. A E - Draw AF to the center, and produce it to C, and join CE. Because A C is a diameter of the circle, - therefore the segment BADC is greater than a semicircle; and the angles in it BAC, BEC are equal, by the first case: I ^ 114 EUCLID's ELEMENTS. ,’ for the same reason, because CBED is greater than a semicircle, - the angles CAD, CED, are equal: …- therefore the whole angle BAD is equal to the whole angle B.E.D. (ax. 2.) Wherefore the angles in the same segment, &c. Q.E.D. PROPOSITION XXII. THEOREM. The opposite angles of any quadrilateral figure insoribed in a circle, are together equal to two right angles. Let ABCD be a quadrilateral figure in the circle AB CD. Then any two of its opposite angles shall together be equal to two right angles. t Join A C, BD. And because the three angles of every triangle are equal to two right angles, (I. 32.) - the three angles of the triangle CAB, viz. the angles CAB, ABC, • BCA, are equal to two right angles: but the angle CAB is equal to the angle CDB, (III. 21.) because they are in the same segment CDAB; and the angle ACB is equal to the angle ADB, because they are in the same segment ADCB: , therefore the two angles CAB, ACB are together equal to the whole angle ADC : (ax. 2.) - - to each of these equals add the angle ABC; therefore the three angles ABC, CAB, BCA are equal to the two angles ABC, ADC: (ax. 2.) * but ABC, CAB, BCA, are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles. In the same manner, the angles B.A.D, DCB, may be shewn to be equal to two right angles. - Therefore, the opposite angles, &c. Q.E.D. PROPOSITION XXIII. THEOREM. Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with one another. If it be possible, upon the same straight line AB, and upon the same side of it, let there be two similar segments of circles, ACB, ADB, not coinciding with one another. ** - D - B A Then, because the circumference ACB cuts the circumference ADB - - - - - - - Book III. PROP. xxIII, XXIV, xxv. . . 115 . . Lºri in the two points A, B, they cannot cut one another in any other point: (III, 10.) . . . . . - ... -- *--> therefore one of the segments must fall within the other: let ACB fall within ADH: - * draw the straight line BCD, and join CA, D.A. . . Because the segment ACB is similar to the segment ADB, (hyp.) and that similar segments of circles contain equal angles; (III. def. 11.) - therefore the angle ACB is equal to the angle ADB, the exterior angle to the interior, which is impossible. (I. 16.) Therefore, there cannot be two similar segments of circles upon the same side of the same line, which do not coincide. Q.E.D. PROPOSITION XXIV. THEOREM, Similar segments of circles upon equal straight lines, are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD. - - - Then the segment AEB shall be equal to the segment CFD. * E F (~, - £-> For if the segmemt AEB be applied to the segment CFD, so that the point 4 may be on G, and the straight line AB upon CD, then the point B shall coincide with the point D, .” because 4B is equal to CB: therefore, the straight line AB coinciding with CD, the segment 4 EB must coincide with the segment CFD, (III. 23.) and therefore is equal to it. (I. ax. 8.) Wherefore similar segments, &c. Q.E.D. PROPOSITION XXV. PROBLEM. 4 segment of a circle being given, to describe the circle of which it is the 8egment. - - Let ABC be the given segment of a circle. - It is required to describe the circle of which it is the Segment. - Bisect 40 in D, (r. 10.) and from the point D draw ºf at right angles to AC, (I, 11.) and join A.B. w First, let the angles ABD, BAD be equal to one another: then the straight line DA is equal to DB, (1.6.) and therefore, to Do; and because the three straight lines DA, D#, DC are all equal 7 ...therefore D is the center of the circle. (III. 9.) & From the center D, at the distance of any of the three DA, DB DC, describe a circle; - f 2 I 2 , 116 EUCLID's ELEMENTs. * - this shall pass through the other points; . . . . . and the circle of which ABC is a segment has been described: and because the center D is in AC, the segment ABC is a semicircle. But if the angles ABD, B.AD are not equal to one another: T} AéPSc SL’ / | E E A D C at the point A, in the straight line AB, make the angle BAE equal to the angle ABD, (I. 23.) and produce BD, if necessary, to meet AE in E, and join EC. Because the angle ABE is equal to the angle BAE, therefore the straight line EA is equal to EB : (I, 6.) and because AD is equal to DC, and DE common to the triangles ADE, CDE, - - the two sides AD, DE, are equal to the two CD, DE, each to each; - and the angle ADE is equal to the angle CDE, for each of them is a right angle; (constr.) therefore the base EA is equal to the base EC: (I. 4.) but EA was shewn to be equal to EB : - * wherefore also EB is equal to EC : (ax. 1.) and therefore the three straight lines EA, EB, EC are equal to one another: f wherefore E is the center of the circle. (III. 9.) •, From the center E, at the distance of any of the three EA, EB, J'C, describe a circle; - this shall pass through the other points; and the circle of which ABC is a segment, is describe. And it is evident, that if the angle ABD be greater than the angle BAD, the center E falls without the segment ABC, which therefore is less than a semicircle : but if the angle ABD be less than BAD, the center E falls within the segment ABC, which is therefore greater than a semicircle. Wherefore a segment of a circle being given, the circle is described of which it is a segment. Q.E. F. - PROPOSITION XXVI. THEOREM, In equal circles, equal angles stand upon equal arcs, whether the angles be at the centers or circumferences. Det ABC, DEF be equal circles, and let the angles BGC, EHF at their centers, and BAC, EDF at their circumferences be equal to each other. Then the arc BKC shall be equal to the arc ELF. A D BOOK III. PROP. xxv.1, xxvii. 117 . Join BC, EF. And because the circles ABC, DEF are equal, the straight lines drawn from their centers are equal: (III. def. 1.) therefore the two sides BG, GC, are equal to the two EH, HF, each to each : . – and the angle at G is equal to the angle at H; (hyp.) therefore the base BC is equal to the base EF. (I. 4.) And because the angle at A is equal to the angle at D, (hyp.) the segment BAC is similar to the segment EDF: (III, def. 11.) and they are upon equal straight lines BC, EF: • but similar segments of circles upon equal straight lines, are equal to one another, (III. 24.) - therefore the segment BAC is equal to the segment EDF: but the whole circle ABC is equal to the whole DEF; (hyp.) therefore the remaining segment BKC is equal to the remaining segment ELF. (L. ax. 3.) - * ~ * and the arc BKC to the arc ELF. Wherefore, in equal circles, &c. Q.E.D. PROPOSITION xxvii. THEOREM. In equal circles, the angles which stand upon equal arcs, are equal to one another, whether they be at the centers or circumferences. º Let ABC, DEFbe equal circles, and let the angles BGC, EHF at their centers, and the angles BAC, EDF at their circumferences, stand upon the equal arcs BC, EF. Then the angle BGC shall be equal to the ºngle EHF, and the angle BAC to the angle b1) F. A. T) K If the angle BGC be equal to the angle EHF, it is manifest that the angle BAC is also equal to EDF (III. 20. and I. ax. 7.) But, if not, one of them must be greater than the other: if possible, let the angle BGC be greater than EHF, and at the point G, in the straight line BG, * make the angle BGK equal to the angle EHF (I. 23.) Then because the angle BGK is equal to the angle EHF and that equal angles stand upon equal arcs, when they are at the centers; (III. 26. * therefore the arc BK is equal to the are EF: but the arc EF is equal to the arc BC; (hyp.) therefore also the arc BK is equal to the arc BC, the less equal to the greater, which is impossible: (I. ax. 1.) therefore the angle BGC is not unequal to the angle EHF; - - that is, it is equal to it: '- - * 118 - EUCLID's ELEMENTs. \, but the angle at 4 is half of the angle BGC, (III. 20.) and the angle at D, half of the angle EHF'; therefore the angle at A is equal to the angle at D. (I. ax. 7.) Wherefore, in equal circles, &c., Q.E.D. - PROPOSITION XXVIII. THEOREM. In equal circles, equal straight lines out off equal arcs, the greater equal to the greater, and the less to the less. *. Tiet ABC, DEFbe equal circles, and BC, EFequal straightlines in them, which cut off the two greater arcs BAC, EDF, and the two less BGC, EHF. - Then the greater arc BAC shall be equal to the greater EDF, . and the less arc BGC to the less EHF. g A & D. B Q E © F. G - Take K, L, the centers of the circles, (IIT. 1.) and join BK, RTC, EL, L.F. - t Decause the circles ABC, DEF" are equal, the straight lines from their centers are equal: (III. def. 1.) therefore BK, KC are equal to EL, LF, each to each : . and the base BC is equal to the base EF, in the triangles BCK, EFL, therefore the angle BKC is equal to the angle ELF: (r. 8.) but equal angles stand upon equal arcs, when they are at the centers: (III. 26.) - - therefore the arc BGC is equal to the arc EHF: - but the whole circumference ABC is equal to the whole EDF; (hyp.) therefore the remaining part of the circumference, viz. the arc BAC, is equal to the remaining part EDF (I. ax. 3. A r Therefore, in equal circles, &c. Q.E.D. . - PROPOSITION XXIX. THEOREM. In equal circles, equal aros are subtended by equal straight lines. Let ABC, DEF be equal circles, and let the arcs BGC, EHF also be equal, and joined by the straight lines BC, EF. Then the straight line BC shall be equal to the straight line E.F. Take K, E, (III. 1.) the centers of the circles, and join BK, KC, EL, L.F. Because the arc BGC is equal to the arc EHF, - therefore the angle BKC is equal to the angle ELF: (III. 27.) BOOK III, PROP. xxix., xxx, xxxi. - 119 and because the circles ABC, DEF, are equal, the straight lines from their centers are equal ; (III, def. 1.) therefore BK, KC, are equal to EL, LF, each to each : and they contain equal angles in the triangles BCK, EFI, ; therefore the base BC is equal to the base EF. (I. 4.) - Therefore, in equal circles, &c. Q. E.D.' PROPOSITION XXX. PROBLEM. To bisect a given are, that is, to divide it into two equal parts. Let ADB be the given arc; it is required to bisect it. D £TS JB Join AB, and bisect it in C; (r. 10.) - from the point C draw CD at right angles to A.B., (I. 11.) Then the arc ADB shall be bisected in the point D. * Join AD, D.B. And because AC is equal to CB, and CD common to the triangles 4 CD, BCD, the two sides AC, CD, are equal to the two BC, CD, each to each; and the angle A CD is equal to the angle BCD, because each of them is a right angle: therefore the base AD is equal to the base B.D. (I. 4.) But equal straight lines cut off equal arcs, (III. 28.) the greater arc equal to the greater, and the less arc to the less; and the arcs AD, DB, are each of them less than a semicircle;- because DC, if produced, passes through the center: (III. 1. Cor.) therefore the arc AD is equal to the arc D.B. Therefore the given arc ADB is bisected in D. Q.E.F. A PROPOSITION XXXI. THEOREM. In a circle, the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle &n a segment less thqn a semicircle is greater than a right angle. Let ABCD be a circle, of which the diameter is BC, and center E. and let CA be drawn, dividing the circle into the segments ABC, ADC. - Join B.A., A.D, D.C. - Then the angle in the semicircle BAC shall be a right angle; and the angle in the segment ABC, which is greater than a semicircle, shall be less than a right angle ; - and the angle in the segment ADC, which is less than a semicircle, shall be greater than a right angle. 120 a EUCLID's ELEMENTS. Join AE, and produce BA to F. First, because EB is equal to EA, (I. def. 15.) the angle EAB is equal to EBA ; (I. 5.) also, because EA is equal to EC, the angle ECA is equal to EA 0; wherefore the whole angle BAC is equal to the two angles EBA, JECA ; (I. ax. 2.) - - but FA C, the exterior angle of the triangle ABC, is equal to the two angles EBA, ECA; (I. 32.) therefore the angle BAC is equal to the angle FA C; (ax. 1.) and therefore each of them is a right angle: (E. def. 10.) wherefore the angle BAC in a semicircle is a right angle. Secondly, because the two angles ABC, BAC of the triangle ABC are together less than two right angles, (I. 17.) - and that BAC has been proved to be a right angle; - therefore ABC must be less than a right angle: and therefore the angle in a segment ABC greater than a semicircle, is less than a right angle. - | And lastly, because ABCD is a quadrilateral figure in a cirele, any two of its opposite angles are equal to two right angles: (III. 22.) therefore the angles ABC, ADC, are equal to two right angles: and 4 BC has been proved to be less than a right angle; wherefore the other ADC is greater than a right-angle. Therefore, in a circle the angle in a semicircle is a right angle, &c. Q.E.D. - CoR. From this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle: because the angle adjacent to it is equal to the same two; (1. 32.) and when the adjacent angles are equal, they are right angles. (I. def. 10.) - PROPOSITION xxxii. THEOREM, If a straight line touch a circle, and from the point of contact a straight line be drawn meeting the circle; the angles which this line makes with the line touching the curole shall be equal to the angles which are in the alternate segments of the circle. Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn, meeting the circumference in D, and dividing it into the segments DCB, DAB, of which DCB is less than, and DAB greater than a semicircle. Then the angles which BD makes with the touching line EF, shall be equal to the angles in the alternate segments of the circle; that is, the angle DBF shall be equal to the angle which is in the segment DAB, - - and the angle DBE shall be equal to the angle in the alternate segment DCB, - A. SD *-x H-E-r From the point B draw BA at right angles to EF, (r. 11.) meeting the circumference in A; BOOK III. PROP. XXXII, XXXIII. 121 take any point C in the arc DB, and join AD, DC, CB. Because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles to the touching line from the point of contact B, - the center of the circle is in BA : (III. 19.) therefore the angle ADB in a semicircle is a right angle: (III. 31.) and consequently the other two angles B.AD, ABD, are equal to a right angle; (I. 32.) but ABF is likewise a right angle; (constr.) - therefore the angle ABF is equal to the angles B.A.D., ABD: (I, ax. 1.) take from these equals the common angle ABD: therefore the remaining angle DBF is equal to the angle BAD, (I, ax. 3.) - which is in BDA, the alternate segment of the circle. º And because ABCD is a quadrilateral figure in a circle, the opposite angles B.A.D., BCD are equal to two right angles: (III. 22.) but the angles DBF, DBE are likewise equal to two right angles; (I. 13.) therefore the angles DBF, DBE are equal to the angles B.AD, BCD, (I. ax. 1.) * and DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD, in BDC the alternate segment of the circle. (I. ax. 3.) . - - Wherefore, if a straight line, &c. Q.E.D. PROPOSITION XXXIII. PROBLEM. Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle. - * Let AB be the given straight line, - and the angle C the given rectilineal angle. It is required to describe upon the given straight line AB, a segment of a circle, which shall contain an angle equal to the angle C. S - First, let the angle C be a right angle. - - 2– - - E (2 N A F B Bisect AB in F, (r. 10.) and from the center F, at the distance FB, describe the semicircle AHB, and draw A.H, BH to any point H in the circumference.’ - Therefore the angle AHB in a semicircle is equal to the right angle C. (III. 31.) - But if the angle C be not a right angle: at the point A. in the straight line AB, make the angle BAD equal to the angle C, (1.23.) 122 , * EUCLID's ELEMENTs. and from the point A draw A.E at right angles to AD; (I, 11.) - e bisect AB in F, (I. 10.) and from F draw FG at right angles to AB, (1.1.1.) and join G.B. Because AF is equal to FB, and FG common to the triangles, A FG, BFG, - - the two sides AF, FG are equal to the two BF, FG, each to each, and the angle AFG is equal to the angle BFG ; (I. def. 10.) therefore the base AG is equal to the base GB; (I. 4.) - and the circle described from the center G, at the distance GA, shall pass through the point B : - let this be the circle A.H.B. -_ The segment AHB shall contain an angle equal to the given rec- tilineal angle C. - Because from the point A the extremity of the diameter AE, AD. is drawn at right angles to AE, sº - t therefore AD touches the circle: (III. 16. Cor.) and because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to the angle in the alternate segment. AHB : (III. 32.) - but the angle DAB is equal to the angle C; (constr.) therefore the angle C is equal to the angle in the segment AITB. Wherefore, upon the given straight line AB, the segment AHB. of a circle is described, which contains an angle équal to the given, angle C. Q.E.F. PROPOSITION xxxiv. PROBLEM. From a given circle to out off a segment, which shall contain an angle equal to a given rectilineal angle. Let ABC be the given circle, and D the given rectilineal angle. It is required to cut off from the circle ABC a segment that shall Contain an angle equal to the given angle D. - E B F - ( º, the straight line EF touching the circle ABC in any point B, and at the point B, in the straight line B.F. make the angle FBC equal to the angle D. (r. 23.) º the segment BAC shall contain an angle equal to the given angle D. º - - - - Because the straight line EF touches the circle ABC, and BC is drawn from the point of contact B, therefore the angle FBC is equal to the angle in the alternate segment BAC of the circle: (III. 32.) * , but the angle FBC is equal to the angle D; (constr.) : therefore the angle in the segment BAC is equal to the angle D. Wherefore from the given circle ABC, the segment BAC is cut off containing an angle equal to the given angle D. Q.E.F. ** BOOK III. PROP. XXXV. - 123 PROPOSITION xxxv. THEOREM. If two straight lines cut one another within a circle, the rectangle contained by the segments of one of them, is equal to the rectangle contained by the segments of the other. >, - - Let the two straight lines AC, BD, cut one another in the point E, within the circle ABCD. - Then the rectangle contained by AE, EC shall be equal to the rectangle contained by B.E., ED. . - A. s E. º D. B. YC - \ ^ - ar First, if AC, BD pass each of them through the center, so that E. is the center; , it is evident that since AE, EC, BB, ED, are all equal, (I. def. 15.) therefore the rectangle A.E, EC is equal to the rectangle B.E., E.D. Secondly, let one of them BD pass through the center, and cut the other AC, which does not pass through the center, at right angles, in the point E. B . Then, if BD be bisected in F, Fis the center of the circle ABCD Join AF. Because BD which passes through the center, cuts the straight Hine AC, which does not pass through the center, at right angles in E, therefore AE is equal to EC: (III. 3.) - and because the straight line BD is cut into two equal parts in the point F, and into two unequal parts in the point E, therefore the rectangle B.E., ED, together with the square on EF, is equal to the square on FB; (II. 5.) that is, to the square on FA : but the squares on A. E. E.F. are equal to the square on FA : (1.47.) therefore the rectangle B.E., ED, together with the square on EF, is equal to the squares on AE, EF: (I. ax. 1.) take away the common square on EF, *x and the remaining rectangle B.E., ED is equal to the remaining square on AE; (I. ax. 3.) that is, to the rectangle AE, EC. Thirdly, let B.D, which passes through the center, cut the other AC, which does not pass through the center, in E, but not at right angles. Then, as before, if BD be bisected in F, Fis the center of the circle. Join AF, and from F draw FG perpendicular to AC; (1.12.) therefore AG is equal to GC; (III. 3.) - 124 EUCLID's ELEMENTs. wherefore the rectangle AE, EC, together with the square on EG, is equal to the square on 4G : (II. 5.) • *- to each of these equals add the square on GF; - therefore the rectangle A E, EC, together with the squares on EG, GF, is equal to the squares on AG, GF; (I. ax. 2.) but the squares on EG, GF, are equal to the square on EF; (I. 47.) and the squares on AG, GF, are equal to the square on AF: therefore the rectangle A E, EC, together with the square on EF, is equal to the square on AF; that is, to the square on FB : but the square on FB is equal to the rectangle B.E., ED, together with the square on EF; (II.5.) therefore the rectangle AE, EC, together with the square on EF, is equal to the rectangle B.E., ED, together with the square on JEF; (I. ax. 1.) - i - - take away the common square on EF, and the remaining rectangle AE, EC, is therefore equal to the re- - maining rectangle B.E., E.D. (ax. 3.) * Lastly, let neither of the straight lines AC, BD pass through the center, Take the center F, (III. 1.) and through E the - intersection of the straight lines AC, DB, draw the diameter GEFH. And because the rectangle AE, EC is equal, as has been shewn, to the rectangle GE, EH ; and for the same reason, the rectangle B.E., ED is equal to the same rectangle. GE, EH ; * . therefore the rectangle A E, EC is equal to the rectangle B.E., ED. I. ax. 1. - * t ) Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION XXXVI. THEOREM. If from any point without a circle two straight lines be drawn, one of which outs the circle, and the other touches it; the rectangle contained by the whole line which outs the circle, and the part of it without the circle, shall be equal to the square on the line which touches it. Let D be any point without the circle ABC, and let DCA, DB be two straight lines drawn from it, of which DCA cuts the circle, and DB touches the same. ,43 Then the rectangle AD, DC shall be equal to the square on D.B. Either DCA passes through the center, or it does not: g first, let it pass through the center E. Join EB, .* D - BOOK III. PROP. XXXVI. 125 & therefore the angle EBD is a right angle. (III. 18.) . And because the straight line AC is bisected in E, and produced to the point D, - - therefore the rectangle AD, DC, together with the square on EC, is equal to the square on ED: (II. 6.) « - . but CE is equal to EB; therefore the rectangle AD, DC, together with the square on EB, - is equal to the square on ED: but the square on ED is equal to the squares on EB, BD, (I. 47.) because EBD is a right angle: *- therefore the rectangle AD, DC, together with the square on EB, is equal to the squares on EB, BD: (ax. 1.) take away the common square on EB; therefore the remaining rectangle A.D, DC is equal to the square on the tangent D.B. (ax. 3.) { . - - Next, if DCA does not pass through the center of the circle ABC. D Take E the center of the circle, (III. 1.) - draw EF perpendicular to AC, (I. 12.) and join EB, EC, ED. - Because the straight line EF, which passes through the center, cuts the straight line AC, which does not pass through the center, at right angles; it also bisects AC, (III. 3.) therefore A Fis equal to FC; and because the straight line AC is bisected in F, and produced to D, the rectangle A.D, DC, together with the square on FC, - is equal to the square on FD: (II. 6.) to each of these equals add the square on FE; - therefore the rectangle AD, DC, together with the squares on CF, F.E, is equal to the squares on D.F, FE: (I. ax. 2.) but the square on ED is equal to the squares on D.F, F.E, (I. 47.) because EFD is a right angle; and for the same reason, the square on EC is equal to the squares on CF, FE; therefore the rectangle AD, DC, together with the square on EC, is equal to the square on ED: (I. ax. 1.) - - but CE is equal to EB; - "a therefore the rectangle AD, DC, together with the square on EB, is equal to the square on ED: . - but the squares on EB, BD, are equal to the square on ED, (1.47.) because EBD is a right angle: therefore the rectangle AD, DC, together with the square on EB, - is equal to the squares on EB, BD ; take away the common square on EB; and * ºws rectangle AD, DC is equal to the square on D.B. I. ax. 3. - Wherefore, if from any point, &c. Q.E.D. 126 EUCLID's ELEMENTs. * A C; B - CoR. If from any point without a circle, there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle, are equal to one another, viz. the rectangle BA, AE, to the rectangle, CA, AF': for each of them is equal to the square on the straight line AD, which touches the circle. - PROPOSITION XXXVII. THEOREM, If from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which outs the circle, and the part of it without the circle, be equal to the square on the line which meets it, the line which meets, shall touch the circle. Let any point D be taken without the circle ABC, and from it let 2 two straight lines DCA and DB be drawn, of which DCA cuts the circle in the points C, A, and DB meets it in the point B. If the rectangle AD, DC be equal to the square on D.B.; then DB shall touch the circle. ID - * Draw the straight line DE, touching the circle ABC, in the point E; (III. 17.) find F, the center of the circle, (III. 1.) and join FE, FB, FD. Then FED is a right angle: (III. 18.) * and because D.E. touches the circle ABC, and DCA cuts it, therefore the rectangle AD, DC is equal to the square on DE: (III. 36.) but the rectangle AD, DC, is, by hypothesis, equal to the square on DB : therefore the square on DE is equal to the square on D.B.; (I. ax. 1.) and the straight line D.E. equal to the straight line DB : - and FE is equal to FB; (I. def. 15.) wherefore DE, EF are equal to DB, BF, each to each; and the base FD is common to the two triangles DEF, DBF; therefore the angle DEF is equal to the angle DBF: (1.8.) but DEF was shewn to be a right angle; therefore also DBF is a right angle: (I. ax. 1.) and BF, if produced, is a diameter; r − = and the straight line which is drawn at right angles to a diameter, from the extremity of it, touches the circle; (III. 16. Cor.) therefore DB touches the circle ABC. Wherefore, if from a point, &c. Q.E.D. NOTES To BOOK III. In the Third Book of the Elements are demonstrated the most elementary pro- perties of the circle, assuming all the properties of figures demonstrated in the First and Second Books. - It may be worthy of remark, that the word circle will be found sometime taken to mean the surface included within the circumference, and sometimes the circumference itself. Euclid has employed the word "reptºpépeta, periphery, both for the whole, and for a part of the circumference of a circle. If the word circum- ference were restricted to mean the whole circumference, and the word arc to mean a part of it, ambiguity might be avoided when speaking of the circumference of a circle, where only a part of it is the subject under consideration. A circle is said to be given in position, when the position of its center is known, and in magnitude, when its radius is known. *. - Def. 1. And to which may be added, “ or of which the circumferences are equal.” And conversely: if two circles be equal, their diameters and radii are equal; as also their circumferences. - r Def. T. states the criterion of equal circles. Simson calls it a theorem ; and Euclid seems to have considered it as one of those self-evident theorems, or axioms, which might be admitted as a basis for reasoning on the equality of circles. Def. II. There seems to be tacitly assumed in this definition, that a straight line, when it meets a circle and does not touch it, must necessarily, when pro- duced, cut the circle. . A straight line which touches a circle, is called a tangent to the circle; and a straight line which cuts a circle is called a secant. Def. III. One circle may touch another externally or internally, that is, the convex circumference of one circle may touch the convex or concave circum- ference of another circle. Two circles may be said to touch each other externally, but it cannot with propriety be said that two circles touch each other internally. Def. Iv. The distance of a straight line from the center of a circle is the dis- tance of a point from a straight line. See Note on Euc. I. 11, p. 54. Def. vi. When any Geometrical magnitude is divided into parts, the parts are called segments, as if a straight line be divided into any parts, any one of the parts, is a segment of the line: but the term segment of a circle is not any part of a circle, but is limited to any part of a circle which can be cut off by one straight line. Def. vi. x. An are of a circle is any portion of the circumference; and a chord is the straight line joining the extremities of an arc. Every chord except a dia- meter divides a circle into two unequal segments, one greater than, and the other less than, a semicircle. And in the same manner, two radii drawn from the center to the circumference, divide the circle into two unequal sectors, which become equal when the two radii are in the same straight line. As Euclid, however, does Irot notice re-entering angles, a sector of the circle seems necessarily restricted to he figure which is less than a semicircle. A quadrant is a sector whose radii are erpendicular to one another, and which contains a fourth part of the circle. Def. vii. No use is made of this definition in the Elements. Def. xi. The definition of similar segments of circles as employed in the Third Book is restricted to such segments as are also equal. Props. xxiii. and xxiv. are the only two instances, in which reference is made to similar segments of circles. Prop. 1. The expression “lines drawn in a circle,” always means in Euclid, such lines only as are terminated at their extremities by the circumference. 128 - EGCLED's ELEMENTs. If the point G be in the diameter CE, but not coinciding with the point F, the demonstration given in the text does not hold good. At the same time, it is obvi- ous that G cannot be the center of the circle, because GC is not equal to GE. The best practical method of finding the center of a circle is to bisect any two chords in the circle, and at the points of bisection, to draw perpendiculars to the chords; the intersection of these perpendiculars is the center, as is seen in the construction of Euc. Iv. 5. r - Indirect demonstrations are more frequently employed in the Third Book than in the First Book of the Elements. Of the demonstrations of the forty-eight pro- positions of the First Book, nine are indirect: but of the thirty-seven of the Third IBook, no less than fifteen are indirect demonstrations. The indirect is, in general, less readily appreciated by the learner, than the direct form of demonstration. The indirect form, however, is equally satisfactory, as it excludes every assumed hypothesis as false, except that which is made in the enunciation of the proposi- tion. It may be here remarked that Euclid employs three methods of demon- strating converse propositions. First, by indirect demonstrations as in Euc. I. 6 : III. 1, &c. Secondly, by shewing that neither side of a possible alternative can be true, and thence inferring the truth of the proposition, as in Euc. I. 19, 25. Thirdly, by means of a construction, thereby avoiding the indirect mode of demonstration, as in Euc. 1. 48 : III. 37. £ºr - Prop. II. In this proposition, the circumference of a circle is proved to be essentially different from a straight line, by shewing that every straight line joining any two points in the arc falls entirely within the circle, and can neither coincide with any part of the circumference, nor meetit except in the two assumed points. It excludes the idea of the circumference of a circle being flexible, or capable under any circumstances, of admitting the possibility of the line falling outside the circle. If the line could fall partly within and partly without the circle, the circum- ference of the circle would intersect the line at some point between its extremities, and any part without the circle has been shewn to be impossible, and the part within the circle is in accordance with the enunciation of the Proposition. If the line could fall upon the circumference and coincide with it, it would follow that a straight line coincides with a curved line. - From this proposition follows the corollary, that “a straight line cannot cut the circumference of a circle in more points than two.” * Commandine's direct demonstration of Prop. II, depends on the following axiom: “If a point be taken nearer to the center of a circle than the circumference, that point falls within the circle.” Take any point E in AB, and join DA, DE, DB. (fig. Euc. III. 2.) Then be- cause DA is equal to DB in the triangle DAB ; therefore the angle DAB is equal to the angle DBA ; (1. 5.) but since the side AE of the triangle DAE is produced to B, therefore the exterior angle DEB is greater than the interior and opposite angle DAE ; (I. 16.) but the angle DAE is equal to the angle DBE, therefore the angle DEB is greater than the angle DBE. And in every triangle, the greater side is subtended by the greater angles therefore the side DB is greater than the side DE ; but D.B from the center meets the circumference of the circle, therefore DE does not meet it. Wherefore the point E falls within the circle : and E is any point in the straight line AB : therefore the straight line AB falls within the circle. Prop. IV. The only case in which two chords in a circle can bisect each other, is when they pass through the center, and in that case, the chords are dia- Imeters. - - Prop. VII, and Prop. VIII, exhibit the same property; in the former, the point * ** NOTEs To BOOK III. 129 * Ap - - - - ! staken in the diameter, and in the latter in the diameter produced; and exhibit an instance of the division of the diameter into internal and external segments. From this proposition the following corollary may be deduced:—If two chords of a circle intersect each other and make equal angles with a diameter at the point of intersection, the two chords are equal to one another. This is obvious if GF, HF be produced to meet the circumference in M, N, Then MF is equal to NF, whence GM is equal to HN. Also, If chords be drawn through the same point in a circle, they are cut less and less unequally in that point, as the angle formed with the diameter passing through that point approaches a right angle. Prop. viri. An arc of a circle is said to be conver or concave with respect to a •point, according as the straight lines drawn from the point meet the outside or inside of the circular are : and the two points found in the circumference of a circle by two straight lines drawn from a given point to touch the circle, divide the circumference into two portions, one of which is convez and the other concave, with respect to the given point. This is the first proposition in which mention is made ºf the convexity and con- cavity of circular arcs. These terms have not been noticed in the definitions, as the expressions ºrpds riv kotAmv reptºpépstav and ºrpós riv kvptiv reptºpépetav in the original, are sufficiently explanatory of the meaning. If two chords of a circle intersect each other when produced, and make equal angles with the diameter produced and passing through the point of intersection, the two chords may be shewn to be equal. Let DB be produced to meet the cir- cumference in P, the chord BIP may be shewn to be equal to the chord K.E. Prop. Ix. This appears to follow as a corollary from Euc. III. 7. Prop. XI. and Prop. xII. In the enunciation it is not asserted that the contact of two circles is confined to a single point. The meaning appears to be, that supposing two circles to touch each other in any point, the straight line which joins their centers being produced, shall pass through that point in which the circles touch each other. In Prop. xIII. it is proved that a circle cannot touch another in more points than one, by assuming two points of contact, and proving that this is impossible. © / - Both Prop. xI. and Prop. xII. might have been proved directly if they had been placed after Prop. xviii. For let a straight line touch both circles at the point A. If F, G be the centers of the circles, join FA, GA; then each of these lines is at right angles to the line which touches the circles at A, whence FAG is a straight line, (Euc. 1, 14.) if one circle lie outside the other; or AF coincides partly with AG, if one circle be within the other. - Prop. xIII. The following is Euclid's demonstration of the case, in which one circle touches another on the inside. * - If possible, let the circle EBF touch the circle ABC on the inside, in more points than in one point, namely in the points B, D. (fig. Euc. III. 13.) Let P be the center of the circle ABC, and Q the center of EBF. Join P, Q; then Po produced shall pass through the points of contact B, D. For since P is the . center of the circle ABC, PB is equal to PD, but PB is greater than QD, much more then is QB greater than QD. Again, since the point Q is the center of the * circle EBF, QB is equal to QD; but QB has been shewn to be greater than QD, which is impossible. One circle therefore cannot touch another, &c. ..” “. Prop. xvi. may be demonstrated, directly by assuming the following axiom: “If a point be taken further from the center of a circle than the circumference, that point falls without the circle.” - If one circle touch another, either internally or easternally, the two circles can have, at the point of contact, only one common tangent. º K 130 - EUCLID's ELEMENTS. Prop. xvii. When the given point is without the circumference of the given circle, it is obvious that two equal tangents may be drawn from the given point to touch the circle. - Produce FD to meet the circumference of the circle AKF in K, draw ER cutting the circumference of the circle DBC in H: join AH, then AH is also a tangent drawn from A, and equal to AB. The tangent AB does not necessarily terminate in the point B, in the figure, it may be produced to any length. The best practical method of drawing a tangent to a circle from a given point without the circumference, is the following: join the given point and the center of the circle, upon this line describe a semicircle cutting the given circle, then the line drawn from the given point to the intersection will be a tangent. Circles are called concentric circles when they have the same center. - A tangent may be drawn from any point in the circumference of a circle without finding the center. From D the given point, take equal arcs DB, BC on the circumference, join DC, with center D and radius DB, describe a circle FBH cutting DC in F. Take BH equal to BF, and join HD. Then HD touches the circle. The proofrequires Euc. III. 32. Prop. xv.1II. appears to be the converse of Prop. xvi., because a tangent to any point in the circumference of a circle, is a straight line at right angles at the ex- tremity of the diameter which meets the circumference in that point. Prop. xx. If in the circumference of a circle, two points A, B be taken, and from them be drawn two lines to the center C, and two other lines to any point in the circumference; two angles are formed, one ACB called the angle at the center, and the other ADB the angle at the circumference. This proposition is proved by Euclid only in the case in which the angle at the circumference is less than a right angle, and the demonstration is free from objection. If, however, the angle at the circumference be a right angle, the angle at the center disappears, by the two straight lines from the center to the extremities of the arc, becoming one straight line. And, if the angle at the circumference be an obtuse angle, the angle formed by the two lines from the center, does not stand on the same arc, but upon the arc which the assumed arc wants of the whole circumference. If Euclid's definition of an angle be strictly observed, Prop. xx. is geome- trically true, only when the angle at the center is less than two right angles. If, however, the defect of an angle from four right angles may be regarded as an angle, the proposition is universally true, as may be proved by drawing a line from the angle in the circumference through the center, and thus forming two angles at the center, in Euclid's strict sense of the term. In case 1, it is assumed that, if there be four magnitudes, such that the first is double of the second, and the third double of the fourth, then the first and third together shall be double of the second and fourth together: also in case 2, that, if one magnitude be double of another, and a part taken from the first be double of a part taken from the second, the remainder of the first shall be double the remainder of the second, which is, in fact, a particular case of Euc. v. 5. Prop. XXI. Hence, the locus of the vertices of all triangles upon the same base, and which have the same vertical angle, is a circular arc. Prop. xxii. The converse of this Proposition, namely: If the opposite angles of a quadrilateral figure be equal to two right angles, a circle can be described about it, is not proved by Euclid. It is obvious from the demonstration, that if any side of the inscribed figure be produced, the exterior angle is equal to the opposite angle of the figure. . NOTES TO BOOK III. - 131 Prop. xxi.1). It is obvious from this proposition that of two circular segments upon the same base, the larger is that which contains the smaller angle. Prop. xxiv. This Proposition is supplemental to the assumption, that if the radii of two circles are equal, the circles themselves are equal, and also their circumferences are equal. Prop. xxv. The three cases of this proposition may be reduced to one, by drawing any two contiguous chords to the given arc, bisecting them, and from the points of bisection drawing perpendiculars. The point in which they meet will be the center of the circle. This problem is equivalent to that of finding a point equally distant from three given points, which are not in the same straight line. Prop. xxvi. The remaining segment BKC of the circle ABC is proved equal to the remaining segment ELF of the circle DEF; and since the equal segments BKC, ELF are upon the equal straight lines BC, EF, the arc BKC is equal to the arc ELF, by Euc. III, 24. - - ~. Props. xxv.1—xxIx. The properties predicated in these four propositions with respect to equal circles, are also true when predicated of the same circle. In IEuc. Iv. 12, it is admitted that angles at the center of the same circle on equal arcs, are equal. - Prop. xxx. It is manifest from the demonstration of this proposition, that the straight line drawn through the center of a circle to bisect the chord, bisects the arc: and if it bisects the arc, it also bisects the chord. Prop. xxxI. suggests a method of drawing a line at right angles to another, at a given point, when the given point is at the extremity of the given line. And that if the diameter of a circle be one of the equal sides of an isosceles triangle, the base is bisected by the circumference. - It is also obvious from the figure that the middle point of the hypotenuse of a right-angled triangle, is equally distant from the three angular points. Prop. xxxII. The converse of this proposition is not proved by Euclid. Prop. xxxv. The most general case of this Proposition might have been first demonstrated, and the other more simple cases deduced from it. But this is not Euclid's method. He always commences with the more simple case, and proceeds to the more difficult afterwards. The following process is the reverse of Euclid's method. - Assuming the construction in the last fig. to Euc. III. 35. Join FA, FD, and draw FK perpendicular to AC, and FL perpendicular to BD. Then (Euc. II. 5.) the rectangle A E, EC with the square on EK is equal to the square on AK : add to these equals the square on FK: therefore the rectangle AE, EC, with the squares on EK, FK, is equal to the squares on AK, FK. But the squares on EK, FK are equal to the square on EF, and the squares on A.K, FK are equal to the square on A.F. Hence the rectangle AE, EC, with the square on EF is equal to the square on A.F. . In a similar way may be shewn, that the rectangle B.E., ED with the square on EF is equal to the square on FD. And the square on FD is equal to the square on A.F. Wherefore the rectangle AE, EC with the square on EF is equal to the rectangle B.E., ED with the square on EF. Take from these equals the square on EF, and the rectangle AE, EC is equal to the rectangle B.E., E.D. The other more simple cases may easily be deduced from this general case. The converse is not proved by Euclid; namely,–If two straight lines intersect one another, so that the rectangle contained by the parts of one, is equal to the rectangle contained by the parts of the other; then a circle may be described passing through the extremities of the two lines. Or, in other words:—If the K 2 132 EUCLID's ELEMENTS. diagonals of a quadrilateral figure intersect one another, so that the rectangle contained by the segments of one of them, is equal to the rectangle contained by the segments of the other; then a circle may be described about the quadrilateral. - Prop. xxxvi. The converse of the corollary to this proposition may be thus stated:—If there be two straight lines, such that, when produced to meet, the rectangle contained by one of the lines produced, and the part-produced, be equal to the rectangle contained by the other line produced and the part produced; then a circle can be described passing through the extremities of the two straight lines. Or, If two opposite sides of a quadrilateral figure be pro- duced to meet, and the rectangle contained by one of the sides produced and the part produced, be equal to the rectangle contained by the other side produced and the part produced; then a circle may be described about the quadrilateral figure. z Prop. xxxvri. The demonstration of this theorem may be made shorter by a reference to the mote on Euclid III. Def. 2: for if DB meet the circle in B and do not touch it at that point, the line must, when produced, cut the circle in two points. *. - It is a circumstance worthy of notice, that in this proposition, as well as in Prop. xlvii.I. Book I. Euclid departs from the ordinary ea absurdo mode of proof of converse propositions. —r QUESTIONS ON BOOK III. 1. DEFINE accurately the terms radius, arc, circumference, chord, secant. 2. How does a sector differ in form from a segment of a circle? Are they in any case coincident? - 3. What is Euclid's criterion of the equality of two circles? What is meant by a given circle? How many points are necessary to determine the magnitude and position of a circle? ve 4. When are segments of circles said to be similar? Enunciate the proposi- tions of the Third Book of Euclid, in which this definition is employed. Is it employed in a restricted or general form * 5. In how many points can a circle be cut by a straight line, and by another circle? *_ - - 6. When are straight lines equally distant from the center of a circle? 7. Shew the necessity of an indirect demonstration in Euc. III. 1. 8. Find the center of a given circle without bisecting any straight line. 9. Shew that if the circumference of one of two equal circles pass through the center of the other, the portions of the two circles, each of which lies without the circumference of the other circle, are equal. 10. If a straight line passing through the center of a circle bisect a straight line in it, it shall cut it at right angles. Point out the exception; and shew that if a straight line bisect the are and base of a segment of a circle, it will, when produced, pass through the center. 11. If any point be taken within a circle, and a straight line be drawn from it to the circumference; how many lines can generally be drawn equal to it? Draw them. - - * * , 12. Find the shortest distance between a circle and a given straight line without it. QUESTIONS ON Book III. 133 ar - - N t {\ - 13. Shew that a circle can only have one center, stating the axioms upon which your proof depends. - 14. Why would not the demonstration of Euc. III. 9, hold good, if there were only two sueh equal straight lines? - 15. Two parallel chords in a circle are respectively six and eight inches in length, and one inch apart; how many inches is the diameter in length 16. Which is the greater chord in a circle whose diameter is 10 inches; that whose length is 5 inches, or that whose distance from the center is 4 inches? 17. What is the locus of the middle points of all equal straight lines in a circle? - t 18. The radfus of a circle BCDGF, (fig. Euc. III. 15.) whose center is E, is equal to five inches. The distance of the line FG from the center is four inches, and the distance of the kine BC from the center is three inches, required the lengths of the lines FG, BC. - 19. If the chord of an arc be twelve inches long, and be divided into two segments of eight and four inches by another chord: what is the length of the latter chord, if one of its segments be two inches? 20. What is the radius of that circle of which the chords of an arc and of double the arc are five and eight inches respectively - 21. If the chord of an arc of a circle whose diameter is 8% inches, be five inches, what is the length of the chord of double the arc of the same circle? 22. When is a straight line said to touch a circle? Shew from your defini- tion that a straight line cannot be drawn to touch a circle from a point within it. 23. Can more circles than one touch a straight line in the same point? 24. Shew from the construction, Euc. III. 17, that two equal straight lines, and only two, can be drawn touching a given circle from a given point without it: and one, and only one, from a point in the circumference. . - 25. What is the locus of the centers of all the circles which touch a straight line in a given point s 26. How may a tangent be drawn at a given point in the circumference of a circle, without knowing the center? - . 27. In a circle place two chords of a given length at right angles to each other. 28. From Euc. III. 19, shew how many circles equal to a given circle may be drawn to touch a straight line in the same point. 29. Enunciate Euc. III. 20. Is this true, when the base is greater than a semicircle? If so, why has Euclid omitted this case? .-- † 30. The angle at the center of a circle is double of that at the circumference. How will it appear hence that the angle in a semicircle is a right angle 31. How is it shewn that any angle exterior to a circle is less than, and any angle interior to a circle is greater than, half the angle at the center on the same base or arc, and towards the same parts? sº .* 32. What conditions are essential to the possibility of the inscription and cir- cumscription of a circle in and about a quadrilateral figure ? - 33. What conditions are requisite in order that a parallelogram may be in- scribed in a circle Are there any analogous conditions requisite that a parallelo- gram may be described about a circle? - 34. Define the angle in a segment of a circle, and the angle on a segment; and shew that in the same circle, they are together equal to two right angles. 35. State and prove the converse of Euc. III. 22. - - 36. All circles which pass through two given points have their centers in a certain straight line. - - 37. Describe the circle of which a given segment is a part. Give Euclid's 134 - EUCLID's ELEMENTs. more simple method of solving the same problem independently of the magnitude of the given segment. - - - 38. In the same circle equal straight lines cut off equal circumferences. If these straight lines have any point common to one another, it must not be in the tircumference. Is the enunciation given complete 39. Enunciate Euc. III, 31, and deduce the proof of it from Euc. III. 20. 40. What is the locus of the vertices of all right-angled triangles which can be described upon the same hypotenuse * - 41. How may a perpendicular be drawn to a given straight line from one of its extremities without producing the line? *. - 42. If the angle in a semicircle be a right angle; what is the angle in a quadrant? - 43. The sum of the squares of any two lines drawn from any point in the arc of a semicircle to the extremity of the diameter is constant. Express that canstant in terms of the radius. * - 44. In the demonstration of Euc. III, 30, it is stated that “equal straight lines cut off equal circumferenees, the greater equal to the greater, and the less to the less:” explain by reference to the diagram the meaning of this statement. 45. How many circles may be described so as to pass through one, two, and three given points? In what case is it impossible for a circle to pass through three given points - 46. Compare the circumference of the segment (Euc. III. 33.) with the whole circumference when the angle contained in it is a right angle and a half. 47. Include the four cases of Euc. III. 35, in one general proof. 48. Enunciate the propositions which are converse to Props. 32, 35 of Book III. 49. If the position of the center of a circle be known with respect to a given point outside a circle, and the distance of the circumference to the point be ten inches: what is the length of the diameter of the circle, if a tangent drawn from the given point be fifteen inches 2 50. If two straight lines be drawn from a point without a circle, and be both terminated by the concave part of the circumference, and if one of the lines pass through the center, and a portion of the other line intercepted by the circle, be equal to the radius : find the diameter of the circle, if the two lines meet the con- vex part of the circumference, a, b, units respectively from the given point. 51. Upon what propositions depends the demonstration of Euc, III. 35 Is any extension made of this proposition in the Third Book? 52. What conditions must be fulfilled that a circle may pass through four given points? - 53. Why is it considered necessary to demonstrate all the separate cases of TEuc. III. 35, 36, geometrically, which are comprehended in one formula, when ex- pressed by Algebraic symbols? . - 54. Enunciate the converse propositions of the Third Book of Euclid which are not demonstrated ea: absurdo: and state the three methods which Euclid employs in the demonstration of converse propositions in the First and Third Books of the Elements. 55. What supposition is made in Euc, III. 2, with respect to the form of the circumference of the circle, so that the straight line may apparently fall outside the circle Is such a supposition consistent with Euclid's definition of a circle? 56. Could any of the Propositions of the Third Book have been dispensed with, or modified with advantage, if Euclid had recognized in his definition of a circle any method for the generation of it, as he has done for the generation of a sphere in the fourteenth definition of the Eleventh Book of the Elements BOOK IV. DEFINITIONS, E. A Recritinear figure is said to be inscribed in another rectilineal figure, when all the angular points of the inscribed figure are upon the sides of the figure in which it is inscribed, each upon each. O II. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each. A rectilineal figure is said to be inscribed in a circle, when all the angular points of the inscribed figure are upon the circumference of the circle. - IV. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle. G. N ? W. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle touches each side of the figure. WI. a" A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described. WII. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle. 136 - EUCLID'S ELEMENTS. PROPOSITION I. PROBLEM. In a given circle to place a straight line, equal to a given straight line which is not greater than the diameter of the circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. it is required to place in the circle ABC a straight line equal to D. A. © Draw BC the diameter of the circle ABC. Then, if B C is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D. - But, if it is not, BC is greater than D; (hyp.) º - make CE equal to D, (I. 3.) and from the center C, at the distance CE, describe the circle AEF, and join C4. - Then CA shall be equal to D. Because C is the center of the circle AEF, therefore CA is equal to CE: (I. def. 15.) but CE is equal to D; (constr.) - therefore D is equal to CA. (ax. 1.) - Wherefore in the circle ABC, a straight line OA is placed equal to the given straight line D. which is not, greater than the diameter of the circle. Q.E.F. - PROPOSITION II. PROBLEM. Ph a given oirole to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF: Draw the straightline & A.H. touching the circle in the point A, (III. 17.) and at the point A, in the straight line AH, º make the angle HAC equal to the angle DEF; (I. 23.) and at the point A, in the straight fine A6, make the angle GAB equal to the angle BFE; and join BC: then ABC shall be the triangle required. Because HAG touches the circle ABC, and 40 is drawn from the point of contact, * therefore the angle HAC is equal to the angle ABC in the alternate, segment of the circle: (III. 32.) - - but HAC is equal to the angle DEF; (constr.) therefore also the angle AB0 is equal to DEF: (ax. 1.) S Book IV. PROP. II, III. 137 for the same reason, the angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal to the remaining angle EDF: (1. 32. and ax. 3.) - - wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Q.E.F. PROPOSITION III. PROBLEM. About a given circle to describe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle. It is required to describe a triangle about the circle ABC equian- gular to the triangle DEF: M B N Produce EFboth ways to the points G, H, find the center K of the circle ABC, (III. 1.) and from it draw any straight line KB; at the point Kin the straight line KB, make the angle BKA equal to the angle DEG, (1.23.) and the angle BKC equal to the angle DFH: and through the points A, B, C, draw the straight lines LAM, MBM, NCL, touching the circle ABC. (III. 17.) Then LMW shall be the triangle required. - . Because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the center are drawn K.A., KB, KC, therefore the angles at the points A, B, C are right angles: (III. 18.) and because the four angles of the quadrilateral figure AMBR are equal to four right angles, - for it can be divided into two triangles; and that two of them. KAM, KBM are right angles, therefore the other two AKB, AMB are equal to two right angles: (ax. 3.) - but the angles DEG, DEF are likewise equal to two right angles; (I, 13.) - therefore the angles AKB, AMB are equal to the angles DEG, DEF; (ax. 1. of which AKB is equal to DEG.; (constr.) wherefore the remaining angle AMB is equal to the remaining angle DEF. (ax. 3.) - - In like manner, the angle LNM may be demonstrated to be equal to DFE; - P and therefore the remaining angle MLN is equal to the remaining angle EDF: (1.32 and ax. 3.) therefore the triangle LMWis equiangular to the triangle DEF: and it is described about the circle ABC, Q.E.F. ... [38. EUCLID'S ELEMENTS. PROPOSITION IV. PROBLEM. To inscribe a circle in a given triangle. Let the given triangle be ABC. . It is required to inscribe a circle in ABC. A Bisect the angles ABC, BCA, by the straight lines BD, CD meeting one another in the point D, (I. 9.) - . from which draw DE, DF, DG perpendiculars to AB, BC, CA. (I, 12.) And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that the right angle BED is equal to the right angle BFD; (ax. 11.) therefore the two triangles EBD, FBD have two angles of the one, equal to two angles of the other, each to each; : and the side BD, which is opposite to one of the equal angles in each, is common to both ; therefore their other sides are equal; (I. 26.) wherefore DE is equal to DF: for the same reason, DG is equal to DF: - - therefore DE is equal to DG : (ax. 1.) therefore the three straight lines DE, DF, DG are equal to one another; - and the circle described from the center D, at the distance of any of them, will pass through the extremities of the other two, and touch the straight lines AB, BC, CA, t º because the angles at the points E, F, G are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches the circle: (III. 16.) therefore the straight lines AB, BC, CA do each of them touch the circle, - and therefore the circle EFG is inscribed in the triangle ABC. Q.E.F. PROPOSITION V. PROBLEM. To describe a circle about a given triangle. Let the given triangle be ABC. It is required to describe a circle about ABC. Bisect AB, A Cin the points D, E, (I.10.) and from these points draw DF, EF at right angles to AB, AC; (I-11.) • - BOOK Iv. PROP. v, vi. 139 -* DF, EF produced meet one another: - for, if they do not meet, they are parallel, - wherefore AB, AC, which are at right angles to them, are parallel; which is absurd.: . - . . let them meet in F, and join FA ; also, if the point F be not in BC, join BF, CF. Then, because AD is equal to DB, and DF common, and at right angles to AB, •. therefore the base AF is equal to the base F.B. (I. 4.) In like manner, it may be shewn that CF is equal to FA ; and therefore BF is equal to FC; (ax. 1.) and FA, FB, FC are equal to one another: wherefore the circle described from the center F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Q.E. F. CoR.—And it is manifest, that when the center of the circle falls within the triangle, each of its angles is less than a right angle, (III. 31.) each of them being in a segment greater than a semicircle; but when the center is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, (III. 31.) is a right angle; and if the center falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, (III. 31.) is greater than a right angle: therefore, conversely, if the given triangle-be acute- angled, the center of the circle falls within it; if it be a right-angled triangle, the center is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the center falls without the triangle, be- yond the side opposite to the obtuse angle. .* PROPOSITION VI. PROBLEM. Zo inscribe a square in a given oircle. . Let ABCI) be the given circle. It is required to inscribe a square in ABCD. A ./ E) C * * diameters AC, BD, at right angles to one another, (III.1. BI101 I. 1 1. - and join AB, BC, CD, D.A. The figure ABCD shall be the square required. Because BF is equal to ED, for E is the center, and that EA is common, and at right angles to BD; , the base BA is equal to the base AD: (1. 4.) and, º, ; same reason, BC, CD are each of them equal to BA, OT ; - * therefore the quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD being the diameter of the circle AB CD, - - - 140 EUCLID's ELEMENTS. BA D is a semicircle; - wherefore the angle BAD is a right angle: (III. 31.) for the same reason, each of the angles ABC, BCD, CD4 is a right angle: - flºor. the quadrilateral figure ABCD is rectangular: and it has been shewn to be equilateral, - therefore it is a square: (I, def. 30.) and it is inscribed in the circle ABCD. Q.E.E. PROPOSITION VII. PROBLEM. To describe a square about a given circle. - Let ABCD be the given circle. It is required to describe a square about it. G. A. E B —ip H C K Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D, draw FG, GH, HK, KF touching the circle. (III. 17.) - The figure FGHK shall be the square required. - Because FG touches the circle ABCD, and EA is drawn from the center E to the point of contact A, t therefore the angles at A are right angles: (III. 18.) for the same reason, the angles at the points B, C, D, are right angles; and because the angle AEB is a right angle, as likewise is EBG, therefore GH is parallel to AC: (I. 28.) for the same reason A0 is parallel to FK: and in like manner GF, HK may each of them be demonstrated to be parallel to BED: * : therefore the figures G.K., GC, A.K, FB, BK are parallelograms; ... and therefore GF'is equal to HK, and GH to FK: (1.34.) and because AC is equal to BD, and that AC is equal to each of the two GH, FK; - - and BD to each of the two GF, HK: GH, FK are each of them equal to GF, or HK; . therefore the quadrilateral figure FGHK is equilateral. - It is also rectangular; . for GBEA being a parallelogram, and AEB a right angle, therefore AGB is likewise a right angle: (I. 34.) and in the same manner it may be shewn that the angles at H, K, F, are right angles: - . * - therefore the quadrilateral figure FGHK is rectangular: and it was demonstrated to be equilateral; . - therefore it is a square; (I. def. 30.) and it is described about the circle ABCD. Q.E.F. Book Iv. PROP. VIII, IX. 141 PROPOSITION VIII. PROBLEM. To inscribe a circle in a given square. Let ABCD be the given square. It is required to inscribe a circle in ABCD. A E TO F K B H. C. Bisect each of the sides AB, A.D in the points F, E, (I, 10.) and through E. draw EH parallel to 4B or DC, (r. 31.) and through F draw FK parallel to AD or BG: . . . . therefore each of the figures A.K, KB, A.H., HD, AG, GC, BG, GD, is a right-angled parallelogram; and their opposite sides are equal: (I. 34.) and because AD is equal to AB, (I., def. 30.) and that AE is the half of AD, and AF'the half of AB, - therefore AE is equal to AF; (ax. 7.) wherefore the sides opposite to these are equal, viz. FG to GE: in the same manner it may be demonstrated that GH, GK are each of them equal to FG or GE: - . therefore the four straight lines GE, GF, GH, GK are equal to one another; - . and the circle described from the center G at the distance of one of them, will pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA; because the angles at the points E, F, H, K are ‘right angles, (r. 29.) and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle: (ITI. 16. Cor.) therefore each of the straight lines AB, BC, CD, DA, touches the circle, which therefore is inscribed in the square ABCD. Q.E.F. PROPOSITION IX. PROBLEM. To describe a circle about a given square. Let ABCD be the given square. gº It is required to describe a circle about ABCB. A D 2. | Nº. B C Join AC, BD, cutting one another in E: º and because DA is equal to AB, and AC common to the triangles DAC, BAC, (I. def. 30.) - the two sides DA, AC are equal to the two BA, AC, each to each; * , and the base DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC; (1.8.) 142 EUCLID's ELEMENTS. and the angle DAB is bisected by the straight line A C : in the same manner it may be demonstrated that the angles ABC, BCD, CDA, are severally bisected by the straight lines BD, A C : therefore, because the angle DAB is equal to the angle ABC, I. def. 30.) - *s and * the angle EAB is the half of DAB, and EBA the half of ABC; therefore the angle EAB is equal to the angle EBA ; (ax. 7.) wherefore the side EA is equal to the side EB : (I. 6.) in the same manner it may be demonstrated, that the straight lines BC, ED, are each of them equal to EA or EB : therefore the four straight lines EA, EB, EC, ED, are equal to one another; - and the circle described from the center E, at the distance of one of them, will pass through the extremities of the other three, and be described about the square ABCD. Q.E. F. PROPOSITION X. PROBLEM. To describe an isosceles triangle, having each of the angles at the base double of the third angle. Take any straight line AB, and divide it in the point C, (II. 11.) so that the rectangle AB, BC may be equal to the square on CA; and from the center A, at the distance AB, describe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; (IV. 1.) and join DA. Then the triangle ABD shall be such as is required, that is, each of the angles ABD, ADB shall be double of the angle B.A.D. - -- Join DC, and about the triangle ADC describe the circle A CD. (IV. 5.) And because the rectangle AB, BC is equal to the square on AC, and that AC is equal to BD, (constr.) - the rectangle AB, BC, is equal to the square on BD: (ax. 1.) and because from the point B, without the circle A CD, two straight lines BCA, BD, are drawn to the circumference, one of which cuts, and E A C B D the other meets the circle, and that the rectangle AB, BC, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square on BD which meets it; a therefore the straight line BD touches the circle A CD: (III. 37.) and because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equal to the angle DAC in the alternate segment of the circle: (III. 32.) - to each of these add the angle CDA ; f BOOK Iv. PROP. x, xI. 143' * = therefore the whole angle BDA is equal to the two angles CDA, DAC : (ax. 2.) . Sº, - - but the exterior angle BCD is equal to the angles CDA, DAC; (I. 32.) therefore also BDA is equal to BCD: (ax. 1.) but BDA is equal to the angle CBD, (I. 5.) because the side AD is equal to the side AIB ; therefore CBD, or DBA, is equal to BCD; (ax. 1.) - and consequently the three angles BDA, D.B.A., BCD are equal to . one another: and because the angle DBC is equal to the angle BCD, the side BD is equal to the side DC: (I. 6.) but B.D was made equal to CA; therefore also CA is equal to CD, (ax. 1.) and the angle CDA equal to the angle DAC; (I. 5.) therefore the angles CDA, DAC together, are double of the angle JDA C : . but BCD is equal to the angles CDA, DAC; (I. 32.) - therefore also BCD is double of DAC: - - and BCD was proved to be equal to each of the angles B.D.A, DBA.; therefore each of the angles B10A, DBA is double of the angle D.A.B. Wherefore an isosceles triangle ABD has been described, having each of the angles at the base double of the third angle. Q.E.F. PROPOSITION XI. PROBLEM. * To inscribe an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle. It is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Describe an isosceles triangle FGH, having each of the angles at G, H double of the angle at F; (IV. 10.) - and in the circle ABCDE inscribe the triangle AOD equiangular to the triangle FGH, (IV. 2.) - - so that the angle CAD may be equal to the angle at F, - and each of the angles A CD, CDA equal to the angle at G or H; wherefore each of the angles A CD, CDA is double of the angle CAD. Bisect the angles A CD, CDA by the straight lines CE, DB; (1.9.) - and join AB, BC, DE, E.A. G H C’s Tºp * Then ABCDE shall be the pentagon required. Because each of the angles A CD, CDA is double of CAD, and that they are bisected by the straight lines CE, D.B.; therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another: - - but equal angles stand upon equal circumferences; (III. 26.) 144 EUCLID's ELEMENTS. therefore the five circumferences AB, BC, CD, DE, EA are equal to one another: - and equal circumferences are subtended by equal straight lines; (III. 29.) therefore the five straight lines A,B, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. iſt is also equiangular: for, because the circumference AB is equal to the circumference D.E, if to each be added BCD, f the whole ABCD is equal to the whole EDCB: (ax. 2.). but the angle AED stands on the circumference ABCD; and the angle BAE on the circumference EDCB; therefore the angle BAE is equal to the angle AED: (III. 27.) - for the same reason, each of the angles A.B.C., BCD, CDE is equal to the angle B.A.E, or AED: - therefore the pentagon ABCDE is equiangular; and it has been shewn that it is equilateral: wherefore, in the given circle, an equilateral and equiangular pentagon has been described. Q.E.F. . PROPOSITION XII. PROBLEM. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE be the given circle. *s It is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angular points of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumferences AB, BC, CD, DE, EA are equal; (IV. 11.) and through the points A, B, C, D, E draw GH, HK, JCL, LM, JMG touching the circle; (III. 17.) the figure GHR.I.M. shall be the pentagon required. Take the center F, and join FB, FK, FC, FL, FD. And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the center F, - FC is perpendicular to KL, (III. 18.) therefore each of the angles at C is a right angle: for the same reason, the angles at the points B, D, are right angles: K C L and because FCK is a right angle, & the square on FK is equal to the squares on FC, CK: (1.47.). for the same reason, the square on FK is equal to the squares on FB, BK: - * *. -º- Book IV. PROP. XII. 145 therefore the squares, on FC, CK are equal to the squares on FB, BIſ; ax. 1. g * ( of ºła the square on FC is equal to the square on FB; therefore the remaining square on CK is equal to the remaining Square on J3 K, (ax. 3.) and the straight line .CH equal to B.K. and because FB is equal to FC, and FK common to the triangles BFK, CFK, * the two B.F, FK are equal to the two CF, FK, each to each: and the base BK was proved equad to the base KC: therefore the angle BFK is equal to the angle KFC, (I. 8.) and the angle BKF to FKC: (I. 4.) wherefore the angle BFC is double of the angle KFC, and BKC double of FKO : for the same reason, the angle CFD is double of the angle CFL, . and CLD double of CLF: and because the circumference BC is equal to the circumference CD, the angle BFC is equal to the angle CFD; (III. 27.) and BFG is double of the angle KFC, and CFD double of CFL; - therefore the angle KFC is equal to the angle CFL: (ax. 7.) and the right angle FCK is equal to the right angle J'CL; therefore, in the two triangles FKC, FLC, there;are two angles of the one equal to two angles of the other, each to each; and the side FC which is adjacent to the equal angles in each, is common to both ; - - therefore the other sides are equal to the other sides, and the third angle to the third angle: (I. 26.) - - - therefore the straight line KC is equal to CL, and the angle FICC to the angle FLC: -- - and because KG is equal to C.L, JCL is double of KC. In the same manner it may be shewn that HK is double of BIſ: and because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, therefore HK is equal to KL: (ax. 6.) - in like manner it may be shewn that GH, GM, MI, are each of them . equal to HK, or KL : - - therefore the pentagon GHKLM is equilateral. It is also equiangular : . for, since the angle FKC is equal to the angle JFLC, and that the angle HKL is double-of-the angle FKC, and KLM double of FLC, as was before demonstrated; therefore the angle HKL is equal to KLM. (ax. 6.) and in like manner it may be shewn, - that each of the angles KHG, HGM, GML is equal to the angle HKI, or KLM: :- - therefore the five angles GHK, HKI, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular: and it is equilateral, as was demonstrated; and it is described about the circle ABCDE, Q.E.F. * 146 EUCLID's ELEMENTS. *- - PROPOSITION XIII. PROBLEM. To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular pentagón. It is required to inscribe a circle in the pentagon ABCDE. C K D Bisect the angles BCD, CDE by the straight lines CF, DF, (I. 9.) and from the point F, in which they meet, draw the straight lines FB, FA, FE: - therefore since BC is equal to CD, (hyp.) and CF common to the triangkºs BCF, DCF, the two sides BC, CF are equal to the two DC, CF, each to each; and the angle BCF is equal to the angle DCF'; (constr.) therefore the base B.Fis equal to the base FD, (I. 4.) and the other angles to the other angles, to which the equal sides are opposite: . therefore the angle CBF is equal to the angle CDF: and because the angle CDE is double of CDF, and that CDE is equal to CBA, and CDF to CBF'; CBA is also double of the angle CBF'; therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the straight line BF': . in the same manner it may be demonstrated, that the angles BAE, AED are bisected by the straight lines AF, F.E. From the point F, draw FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA : (I. 12.) - and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; therefore in the triangles FHC, FKC, there are two angles of the one equal to two angles of the other, each to each ; e and the side FC, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal, each to each ; (I. 26.) wherefore the perpendicular FH is equal to the perpendicular FK: in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK: therefore the five straight lines FG, FH, FK, FL, FM are equal to one another: - wherefore the circle described from the center F, at the distance of one of these five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA, - because the angles at the points G, H, K, L, M are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches the circle; (III. 16.) therefore each of the straight lines AB, BC, CD, DE, EA touches the circle: t wherefore it is inscribed in the pentagon ABCDE, Q.E.F. Book IV. PROP. xiv, xv. - 147 PROPOSITION XIV. PROBLEM. To describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilateral and equiangular Dentagon. It is required to describe a circle about ABCDE. . C ID Bisect the angles BCD, CDE by the straight lines CF, FD, (1.9.) and from the point F, in which they meet, draw the straight lines FB, FA, FE, to the points B, A, E. • * It may be demonstrated, in the same manner as the preceding proposition, * that the angles CBA, BAE, AED are bisected by the straight lines FB, FA, FE. And because the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of 2CD E; - therefore the angle FCD is equal to FDC; (ax. 7.) . . wherefore the side CF is equal to the side FD: (I. 6.) in like manner it may be demonstrated, that FB, FA, FE, are each of them equal to FC or FD: therefore the five straight lines FA, FB, FC, FD, FE are equal to one another; e - - *A and the circle described from the center F, at the distance of one of them, will pass through the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Q. E. F. PROPOSITION xv. PROBLEM. To inscribe an equilateral and equiangular hea agon in a given circle. & Let ABCDEF be the given circle. It is required to inscribe an equilateral and equiangular hexagon in it. ſº º Find the center 6 of the circle ABCDEF, and draw the diameter AGD; (III. 1.) - and from D, as a center, at the distance DG, describe the circle EG CH, L 2 148 EUCLID'S ELEMENTS. join EG, CG, and produce them to the points B, F; - and join AB, BC, CD, DE, EF, FA : the hexagon ABCDEF shall be equilateral and equiangular. - Recause G is the center of the circle ABCDEF, GE is equal to GD: and because D is the center of the circle EGCH, DE is equal to GD : wherefore GE is equal to DE, (ax. 1.) and the triangle EGD is equilateral; - and therefore its three angles EGD, GDE, DEG are equal to one another: (I. 5. Cor.) but the three angles of a triangle are equal to two right angles; (I, 32.) therefore the angle EGD is the third part of two right angles: in the same manner it may be demonstrated, that the angle DGC is also the third part of two right angles: and because the straight line GC makes with EB the adjacent angles JEGC, CGB equal to two right angles; (I, 13.) - the remaining angle CGB is the third part of two right angles: therefore the angles EGD, IOGC, CGB are equal to one another: and º these are equal the vertical opposite angles BGA, AGF, FGE: I. 15.) - , - therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE, are equal to one another: g but equal angles stand upon equal circumferences; (III. 26.) therefore the six circumferences AB, BC, CD, DE, EF, FA, are equal to one another: - * º circumferences are subtended by equal straight lines: III. 29. - therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular: for, since the circumference AF is equal to ED, to each of these equals add the circumference ABCD; therefore the whole circumference FABCD is equal to the whole JEDOBA : and the angle FED stands upon the circumference FABCD, and the angle AFE upon EDCBA; therefore the angle AFE is equal to FED: (III. 27.) in the same manner it may be demonstrated, that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED: therefore the hexagon is equi- angular; and it is equilateral, as was shewn; and it is inscribed in the given circle ABCDEF. Q.E.F. CoR.—From this it is manifest, that the side of the hexagon is equal to the straight line from the center, that is, to the semi-diameter of the circle. - - And if through the points A, B, C, D, E, F, there be drawn straight lines touching the circle, an equilateral and equiangular hexagon will be described about it, which may be demonstrated from what has been said of the pentagon: and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. d BOOK IV. PROP. XVI. 149 PROPOSITION XVI. PROBLEM. To inscribe an equilateral and equiangular quindecagon in a given circle. - Let ABCD be the given circle. It is required to inscribe an equilateral and equiangular quin- decagon in the circle ABC.D. A Let AC be the side of an equilateral triangle inscribed in the circle, (iv. 2.) and AB the side of an equilateral and equiangular pentagon inscribed in the same; (IV. 11.) - therefore, of such equal parts as the whole circumference ABCDF contains fifteen, - the circumference ABC, being the third part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, con- tains three; therefore BC, their difference, contains two of the same parts: - •. bisect BC in E; (III. 30.) - therefore BE, EC are, each of them, the fifteenth part of the whole circumference ABCD: - therefore if the straight lines BE, EC be drawn, and straight lines, equal to them be placed round in the whole circle, (IV. 1.) an equilateral and equiangular quindecagon will be inscribed in it. Q. E. F. And in the same manner as was done in the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon will be described about it: and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quin- decagon, and circumscribed about it. MOTES TO BOOK IV. THE Fourth Book of the Elements contains some particular cases of four general problems on the inscription and the circumscription of triangles and regular figures in and about circles. Euclid has not given any instance of the inscription or circumscription of rectilineal figures in and about other rectilineal figures. * : Any rectilineal figure, of five sides and angles, is called a pentagon; of seven sides and angles, a heptagon; of eight sides and angles, an octagon; of nine sides and angles, a nonagon; of ten sides and angles, a decagon; of eleven sides and angles, an undecagon; of twelve sides and angles, a duodecagon; of fifteen sides and angles, a quindecagon, &c. - These figures are included under the general name of polygons; and are called equilateral, when their sides are equal; and equiangular, when their angles are equal; also when both their sides and angles are equal, they are called regular polygons. - Prop. III. An objection has been raised to the construction of this problem. It is said that in this and other instances of a similar kind, the lines which touch the circle at A, B, and C, should be proved to meet one another. This may be done by joining AB, and then since the angles KAM, KBM are equal to two right angles (III. 18.), therefore the angles BAM, ABM are less than two right angles, and consequently (ax. 12.), AM and BM must meet one another, when produced far enough. Similarly, it may be shewn that AL and CL, as also CN and BN meet one another. - Prop. Iv. This Problem is one case of the more general Problem:-To de- scribe a circle which shall touch three straight lines. If the triangle be equilateral, the center of the inscribed circle is equidistant from the three angular points of the triangle. The centers of the circles inscribed in, and circumscribed about an equilateral triangle coincide, and the radius of one is double the radius of the other. sº Prop. v. is the same as “To describe a circle passing through three given points, provided that they are not in the same straight line.” - The corollary to this proposition appears to have been already demonstrated in Prop. 31. Book III. - It is obvious that the square described about a circle is equal to double the square inscribed in the same circle. Also that the circumscribed Square is equal to the square on the diameter, or four times the square on the radius of the circle. Prop. VII. It is manifest that a square is the only right-angled parallelogram which can be circumscribed about a circle, but that both a rectangle and a square may be inscribed in a circle. gº Prop. x. By means of this proposition, a right angle may be divided into five equal parts. - Prop. xvi. The arc subtending a side of the quindecagon, may be found by placing in the circle from the same point, two lines respectively equal to the sides of the inscribed regular hexagon and pentagon. The centers of the inscribed and circumscribed circles of any regular polygon are coincident. - - Besides the circumscription and inscription of triangles and regular polygons about and in circles, some very important problems are solved in the constructions respecting the division of the circumferences of circles into equal parts. By inscribing an equilateral triangle, a square, a pentagon, a hexagon, &c. NOTEs To Book Iv. 151 in a circle, the circumference is divided into three, four, five, six, &c. equal parts. In Prop. 26, Book III., it has been shewn that equal angles at the centers of equal circles, and therefore at the center of the same circle, subtend equal arcs; by bisecting the angles at the center, the arcs which are subtended by them are also bisected, and hence, a sixth, eighth, tenth, twelfth, &c. part of the circumference of a circle may be found. - e If the right angle be considered as divided into 90 degrees, each degree into 60 minutes, and each minute into 60 seconds, and so on, according to the sexagesimal division of a degree ; by the aid of the first corollary to Prop. 32, Book I., may be found the numerical magnitude of an interior angle of any regular polygon whatever. - Let 0 denote the magnitude of one of the interior angles of a regular polygon of n sides, - then nô is the sum of all the interior angles. But all the interior angles of any rectilineal figure together with four right. angles, are equal to twice as many right angles as the figure has sides, that is, if it be assumed to designate two right angles, ... n.0 + 2 tree nar, and n0 = nºr – 27 = (n − 2).ºr, (n − 2) 22, * 6 = tº gº , T, the magnitude of an interior angle of a regular polygon of n sides. By taking n = 3, 4, 5, 6, &c. may be found the magnitude in terms of two right angles, of an interior angle of any regular polygon whatever. Pythagoras was the first, as Proclus informs us in his commentary, who discovered that a multiple of the angles of three regular figures only, namely, the trigon, the square, and the hexagon, can fill up space round a point in a plane. - - It has been shewn that the interior angle of any regular polygon of n sides in terms of two right angles, is expressed by the equation 9 = * * • Te Let 0, denote the magnitude of the interior angle of a regular figure of three sides, in which case, n = 3. -- 3 – 2 Then 63 = —a-ºr = º: = one-third of two right angles, ..". 303 - Tr, and 60s == 2ar, that is, six angles, each equal to the interior angle of an equilateral triangle, are equal to four right angles, and therefore six equilateral triangles may be placed so as completely to fill up the space round the point at which they meet in a plane. - In a similar way, it may be shewn that four squares and three hexagons may be placed so as completely to fill up the space round a point in a plane. Also it will appear from the results deduced, that no other regular figures besides these three, can be made to fill up the space round a point; for any multiple of the interior angles of any other regular polygon, will be found to be in excess above, or in defect from four right angles. The equilateral triangle or trigon, the square or tetragon, the pentagon, and the hexagon, were the only regular polygons known to the Greeks, capable of being inscribed in circles, besides those which may be derived from them. x= 152 ,” EUCLID's EI.EMENTS. . M. Gauss in his Disquisitiones Arithmeticae, has extended the number by shewing that in general, a regular polygon of 2" + 1 sides is capable of being inscribed in a circle by means of straight lines and circles, in those cases in which 2" + 1 is a prime number. - - The case in which n = 4, in 2" + 1, was proposed by Mr. Lowry of the Royal Military College, to be answered in the seventeenth number of Leybourn's Mathematical Repository, in the following form:— - Required a geometrical demonstration of the following method of constructing a regular polygon of seventeen sides in a circle : -- Draw the radius CO at right angles to the diameter AB; on OC and OB, take OQ equal to the half, and OD equal to the eighth part of the radius; make DE and DF each equal to DQ, and, EG, and FH respectively equal to EQ and FQ ; take OK a mean proportional between OH and OQ, and, through K, draw KM parallel to AB, meeting the semicircle described on OG in M, draw MN parallel to OC cutting the given circle in N, the arc AN is, the seventeenth part of the whole circumference. A demonstration of the truth of this construction has been given by Mr. Lowry himself, and will be found in the fourth volume of Leybourn's , Repository. The demonstration including the two lemmas occupies more than eight pages, and is by no means of an elementary character. QUESTIONS ON BOOK IV. 1. WHAT is the general object of the Fourth Book of Euclid? 2. What consideration renders necessary the first proposition of the Fourth IBook of Euclid P 3. When is a circle said to be inscribed within, and circumscribed about a rectilineal figure? . 4. Shew that a circle cannot be described about a rhombus. Can a circle be inscribed in a rhombus - -> 5. When is one rectilineal figure said to be inscribed in, and circumscribed about another rectilineal figure ?' 6. Modify the construction of Euc. Iv. 4, so that the circle may touch one side of the triangle and the other two sides produced. 7. The sides of a triangle are 5, 6, 7 units-respectively; find the radii of the inscribed and circumscribed circles. f - - 8. Give the constructions by which the centers of circles described about, and . inscribed in triangles are found. In what triangles will they coincide?. 9. How is it shewn that the radius of the circle inscribed in, an equilateral' triangle, is half the radius described about the same triangle 2. 10. The equilateral triangle inscribed in a circle is one-fourth of the equilateral; triangle circumscribed about the same circle. 11. If a circle be inscribed in an equilateral triangle, the triangle formed by joining the points of contact will also be equilateral. 12. What relation subsists between the square inscribed in, and the square circumscribed about the same circle - 13. Enunciate Euc. III. 22:...and extend this property to any inscribed polygon having an even number of sides. - 14. If any triangle be inscribed in a circle, the sum of the three angles in the segments cut off by the sides, is equal tº four right angles. - QUESTIONS ON BOOK IV. 1:53 15. Trisect a quadrantal arc of a circle, and show that every arc which is an. th part of a quadrantal arc, may be trisected geometrically: m and n being whole numbers. 16. If one side of a quadrilateral figure inscribed in a circle be produced, the exterior angle is equal to the interior and opposite angle of the figure. Is this property true of any inscribed polygon having an even number of sides - 17. In what parallelograms can circles be inscribed P- 18. What Geometrical Problem must be solved as a condition precedent to. the construction of a regular pentagon? 19. Shew that in the figure Euc. Iv. 10, there are two triangles possessing the required property. - - 20. How is it made to appear that the line BD is the side of a regular decagon inscribed in the larger circle, and the side of a regular pentagon inscribed in the smaller circle?, fig. Euc. IV. 10. 21. In the construction of Euc. Iv. 3, Euclid has omitted to shew that the tangents drawn through the points A and B will meet in some point M. How may this be shewn P. --- - 22. Shew that if the points of intersection of the circles in Euclid's figure, Book Iv. Prop. 10, be joined with the vertex of the triangle and with each other, another triangle will be formed 'equiangular and equal to the former. 23. Divide a right angle into five equal parts. How may an isosceles triangle be described upon a given base, having each angle at the base one- third of the angle at the vertex? - - 24. What regular figures may be inscribed in a circle by the help of Euc. Iv. 10 ° . . - 25. The difference of the squares described on the straight lines joining the extremities of the base of the constructed triangle in the figure of Euc. Iv. 10, with the other point of intersection of the circles, is equal to the square on the side of the triangle. * - 26. What is Euclid's definition of a regular pentagon Would the stellated figure, which is formed by joining the alternate angles of a regular pentagon, as described in the Fourth Book, satisfy this definition ? 27. Shew that each of the interior angles of a regular pentagon inscribed in a circle, is equal to three-fifths of two right angles. - 28. If two sides not adjacent, of a regular pentagon, be produced to meet; what is the magnitude of the angle contained at the point where they meet? 29. Is there any method more direct than Euclid's for inscribing a regular pentagon in a circle s 30. State briefly the mode of describing an equilateral and equiangular hexagon about a given circle. * 31. In what sense is a regular hexagon also a parallelogram * Would the same observation apply to all regular figures with an even number of sides 3 32. Why has Euclid not shewn how to inscribe an equilateral triangle in a circle, before he requires the use of it in Prop. 16, Book Iv. 33. An equilateral triangle is inscribed in a circle by joining the first, third, and fifth angles of the inscribed hexagon. - 34. If the sides of a hexagon be produced to meet, the angles formed by these lines will be equal to four right angles. 85. Shew that the area of an equilateral triangle inscribed in a circle, is one- half of a regular hexagon inscribed in the same circle. * 154 EUCLID'S ELEMENTS. 36. If a side of an equilateral triangle be six inches: what is the radius of the inscribed circle • t 37. Find the area of a regular hexagon inscribed in a circle whose diameter is twelve inches. What is the difference between the inscribed and the circum- scribed hexagon? i 38. Which is the greater, the difference between the side of the square and the side of the regular hexagon inscribed in a circle whose radius is unity; or the difference between the side of the equilateral triangle and the side of the regular pentagon inscribed in the same circle? - 39. The regular hexagon inscribed in a circle, is three-fourths of the regular circumscribed hexagon. 40. Assuming the construction of Euc. Iv. 6; how may a regular octagon be inscribed in a circle : 41. The sum of the squares on the three diagonals of a regular hexagon that terminate in one angular point is equal to ten times the square on one side. 42. All the interior angles of an octagon are equal to twelve right angles. 43. What figure is formed by the production of the alternate sides of a regular octagon 44. How many square inches are in the area of a regular octagon whose side is eight inches 2 - 45. If an irregular octagon be capable of having a circle described about it, shew that the sums of the angles taken alternately are equal. 46. Find an algebraical formula for the number of degrees contained by a interior angle of a regular polygon of n sides. 47. What are the three regular figures which can be used in paving a plane area Shew that no other regular figures but these will fill up the space round a point in a plane. - - 48. ..Into what number of equal parts may a right angle be divided geome- trically What connection has the solution of this problem with the possibility of inscribing regular figures in circles? 49. Assuming the demonstrations in Euc. Iv, shew that any equilateral figure of 3.2", 4.2”, 5.2”, or 15.2” sides may be inscribed in a circle, when n is any of the numbers, 0, 1, 2, 3, &c. - . . - 50. With a pair of compasses only, shew how to divide the circumference of a given circle into twenty-four equal parts. - 51. Shew that if any polygon inscribed in a circle be equilateral, it must also be equiangular. Is the converse true? 52. The area of a regular polygon of n sides is equal to n times the area of the triangle whose base is a side of the polygon and altitude equal to the radius of the inscribed circle. - 53. Shew that if the circumference of a circle pass through three angular points of a regular polygon, it will pass through all of them. 54. Similar polygons are always equiangular: is the converse of this proposi- tion true * t 55. What are the limits to the Geometrical inscription of regular figures in circles? What does Geometrical mean when used in this way ? 56. What is the difficulty of inscribing geometrically an equilateral and equiangular undecagon in a circle? Why is the solution of this problem said to be beyond the limits of plane geometry? Why is it so difficult to prove that the geo- metrical solution of such problems is impossible? º * BOOK W. DEFINITIONS. I. A LEss magnitude is said to be a part of a greater magnitude, when the less measures the greater; that is, “when the less is contained a certain number of times exactly in the greater.’ II. A greater magnitude is said to be a multiple of a less, when the greater is measured by the less, that is, “when the greater contains the less a certain number of times exactly.” III. “Ratio is a mutual relation of two magnitudes of the same kind to one another, th respect of quantity.” - IV. Magnitudes are said to have a ratio to one another, when the less can be multiplied so as to exceed the other. W. The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth : if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth : or, if the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth : or, if the multiple of the first be greater than that of the second, the multiple of the third is also greater than that of the fourth. Magnitudes which have the same ratio are called proportionals. N.B. “When four magnitudes are proportionals, it is usually expressed by saying, the first is to the second, as the third to the fourth.’ WTI. When of the equimultiples of four magnitudes (taken as in the fifth definition), the multiple of the first is greater than that of the Second, but the multiple of the third is not greater than the multiple of the fourth; then the first is said to have to the second a greater ratio than the third magnitude has to the fourth : and, on the contrary, the third is said to have to the fourth a less ratio than the first has to the second. - - º VIII. “Analogy, or proportion, is the similitude of ratios.”. 156 - EUCLID's ELEMENTS. IX. Proportion consists in three terms at least. X. When three magnitudes are proportionals, the first is said to have to the third, the duplicate ratio of that which it has to the second. . XI. When four magnitudes are continual proportionals, the first is said to have to the fourth, the triplicate ratio of that which it has to the second, and so on, quadruplicate, &c. increasing the denomination still by unity, in any number of proportionals. Definition A, to wit, of compound ratio. When there are any number of magnitudes of the same kind, the first is said to have to the last of them the ratio compounded of the ratio which the first has to the second, and of the ratio which the . second has to the third, and of the ratio which the third has to the fourth, and so on unto the last magnitude. For example, if A, B, C, D be four magnitudes of the same kind, the first A is said to have to the last D, the ratio compounded of the ratio of A to B, and of the ratio of B to C, and of the ratio of C to D ; or, the ratio of A to D is said to be compounded of the ratios of A to B, B to C, and C to D. - And if A has to B the same ratio which E has to F; and B to C the same ratio that G has to H; and C to D the same that K has to L; then, by this definition, A is said to have to D the ratio compounded of ratios which are the same with the ratios of E to F, G to H, and K to L. And the same thing is to be understood when it is more briefly expressed by saying, A has to D the ratio compounded of the ratios of E to F, G to H, and K to L. In like manner, the same things being supposed, if M has to N the same ratio which A has to D ; then for shortness' sake, M is said to have to N the ratio compounded of the ratios of E to F, G to H, and K to L. XII. In proportionals, the antecedent terms are called homologous to one another, as also the consequents to one another. “Geometers make use of the following technical words, to signify certain ways of changing either the order or magnitude of proportionals, so that they continue still to be proportionals.’ XIII. Permutando, or alternando, by permutation, or alternately. This word is used when there are four proportionals, and it is inferred that the first has the same ratio to the third which the Second has to the fourth; or that the first is to the third as the second to the fourth : as is shewn in Prop. XVI. of this Fifth Book. * XIV. - Invertendo, by inversion; when there are four proportionals, and it is inferred, that the second is to the first, as the fourth to the third. I’rop. B. Book v. DEFINITIONS. - 157. * XV. Componendo, by composition; when there are four proportionals, and it is inferred that the first together with the second, is to the second, as the third together with the fourth, is to the fourth. Prop. 18, Book V. " . - - XVI. Dividendo, by division; when there are four proportionals, and it is inferred, that the excess of the first above the second, is to the second, as the excess of the third above the fourth, is to the fourth. Prop. 17, Book V. - - - Convertendo, by conversion; when there are four proportionals, and it is inferred, that the first is to its excess above the second, as the third to its excess above the fourth. Prop. E. Book V. XVIII. Ex aequali (sc. distantiá), or ex æquo, from equality of distance: when there is any number of magnitudes more than two, and as many others such that they are proportionals when taken two and two of each rank, and it is inferred, that the first is to the last of the first rank of magnitudes, as the first is to the last of the others: “Of this there are the two following kinds, which arise from the different order in which the magnitudes are taken, two and two.’ XIX. --> Ex aequali, from equality. This term is used simply by itself, when the fifst magnitude is to the second of the first rank, as the first to the second of the other rank; and as the second is to the third of the first - rank, so is the second to the third of the other; and so on in order: and the inference is as mentioned in the preceding definition; whence this is called ordinate proportion. It is demonstrated in Prop. 22, Book v. - XX. Ex aequali in proportione perturbatā seu inordinatá, from equality in perturbate or disorderly proportion.* This term is used when the first magnitude is to the second of the first rank, as the last but one is to the last of the second rank; and as the second is to the third of the first rank, so is the last but two to the last but one of the second rank: and as the third is to the fourth of the first rank, so is the third from the last to the last but two of the second rank; and so on in a cross order: and the inference is as in the 18th definition. It is demon- strated in Prop. 23, Book v. AXIOMS. I. EQUIMULTIPLEs of the same, or of equal magnitudes, are equal to One another. - II. Those magnitudes, of which the same or equal magnitudes a.T0 equimultiples, are equal to one another. * Prop. 4, Lib. II. Archimedis de sphaera et cylindro. * 158 EUCLID's ELEMENTs. III. A multiple of a greater magnitude is greater than the same mul- tiple of a less. * * -- IV. That magnitude, of which a multiple is greater than the same multiple of another, is greater than that other magnitude. PROPOSITION I. THEOREM. If any number of magnitudes be equimultiples of as many, each of each: what multiple soever any one of them is of its part, the same multiple shall all the first magnitudes be of all the other. Let any number of magnitudes AB, CD be equimultiples of as many others E, F, each of each. - Then whatsoever multiple AB is of E, the same multiple shall AB and CD together be of E and F together. A G B C FI D E— F Because AB is the same multiple of E that CD is of F, as many magnitudes as there are in A.B equal to E, so many are there in CD equal to F. g Divide AB into magnitudes equal to E, viz. AG, GB; and CD into CH, HD, equal each of them to F; - ...therefore the number of the magnitudes CH, HD shall be equal to the number of the others AG, GB ; - - and because AG is equal to E, and CH to F, . therefore AG and CH together are equal to E and Ftogether: (I. ax. 2.) for the same reason, because GB is equal to E, and HD to F; GB and HD together are equal to E and F together: wherefore as many magnitudes as there are in AB equal to E. , so many are there in AB, CD together, equal to E and F together: therefore, whatsoever multiple AB is of E, the same multiple is AB and CD together, of E and F together. Therefore, if any magnitudes, how many soever, be equimultiples of as many, each of each; whatsoever multiple any one of them is . of its part, the same multiple shall all the first magnitudes be of all the others; ‘For the same demonstration holds in any number of magnitudes, which was here applied to two.’ Q.E.D. PROPOSITION II. THEOREM, If the first magnitude be the same multiple of the second that the third ts of the fourth, and the fifth the same multiple of the second that the sixth ts of the fourth; then shall the first together with the fifth be the same multiple of the second, that the third together with the sixth is of the fourth. Let AB the first be the same multiple of C the second, that DE the third is of F the fourth : º - Book v. PROP. II, III. 159 and BG the fifth the same multiple of C the second, that EH the sixth is of F the fourth. , : Then shall AG, the first together with the fifth, be the same mul- tiple of C the second, that DH, the third together with the sixth, is of F the fourth. - A B G D E H C— F— Because AB is the same multiple of C that DE is of F; there are as many magnitudes in AB equal to C, as there are in DE equal to F. In like manner, as many as there are in BG equal to C, so many are there in EH equal to F: . - therefore as many as there are in the whole AG equal to C, so many are there in the whole DH equal to F: p therefore AG is the same multiple of C that DH is of F; that is, AG, the first and fifth together, is the same multiple of the second C, - - that DH, the third and sixth together, is of the fourth F. If therefore, the first be the same multiple, &c. Q. E. D. CoR. From this it is plain, that if any number of magnitudes AB, BG, GH be multiples of another C; , . and as many DE, EK, KL be the same multiples of F, each of each : then the whole of the first, viz. AH, is the same multiple of C, that the whole of the last, viz. DL, is of F. A B G H D E K - L ſº | - t | C— F — |PIROPOSITION III. THEOREM. . If the first be the same multiple of the second, which the third is of the fourth; and if of the first and third there be taken equimultiples ; these shall be equimultiples, the one of the second, and the other of the fourth. Let A the first be the same multiple of B the second, that C the third is of D the fourth : ... • and of A, C let equimultiples EF, GH be taken. Then EF shall be the same multiple of B, that GH is of D. E. }{ F G L H J ſº A C B— D— Because EF is the same multiple of A, that GH is of C, there are as many magnitudes in EF equal to 4, as there are in GH equal to C: - let EF be divided into the magnitudes EK, KF, each equal to A; - and GH into GL, LH, each equal to C : . - therefore the number of the magnitudes E.K, KF, shall be equal to the number of the others GL, LH ; * and because A is the same multiple of B, that C is of D, and that EK is equal to A, and GL equal to C: therefore EK is the same multiple of B, that GL is of D: - C - *- (i. - tº 9 * * ... t 9 a. * } t a º º 166) EUCLID's ELEMENTS. -- for the same reason KF is the same multiple of B, that LII is of D; and so, if there be more parts in EF, GH, equal to A, C:- therefore, because the first EK is the same multiple of the second B, which the third GL is of the fourth D, • and that the fifth KF is the same multiple of the second B, which the sixth LH is of the fourth D; EF º º, * with the fifth, is the same multiple of the second , (W. 2.) which G.H the third, together with the sixth, is of the fourth D. If, therefore, the first, &c. Q.E.D. - PROPOSITION IV. THEOREM. . If the first of four magnitudes has the same ratio to the second which the third has to the fourth; then any equimultiples whatever of the first and third shall have the same ratio to any equimultiples of the second and fourth, viz, ‘the £4%multiple ºf the first shall have the same ratio to that of the second which the equimultiple of the third has to that of the fourth.” 2 Let 4 the first have to B the second, the same ratio which the third - C has to the fourth D; and of 4 and Clet there be taken any equimultiples whatever JE, F, and of B and D any equimultiples whatever G, H. - Then E shall have the same ratio to G, which Fhas to H. K— M E G A — B — C— D — F ºr H– L— N - Take of E and F any equimultiples whatever Iſ., L, and of G, H any equimultiples whatever M, N: then because E is the same multiple of A, that F is of C; and of E and F have been taken equimultiples Jſ, L; therefore K is the same multiple of A, that L is of C: (v. 3.) for the same reason, Mis the same multiple of B, that Wis of D. And because, as A is to B, so is C to D, (hyp.) . and of A and C have been taken certain equimultiples K, L, and of B and D have been taken certain equimultiples M. W.; - therefore if K be greater than M, L is greater than W; - and if equal, equal; if less, less: (V. def. 5.) - but K, L are any equimultiples whatever of E, F, (constr.) ºw and M, W any whatever of G, H.; therefore as E is to G, so is F to H. (v. def. 5.) Therefore, if the first, &c. Q.E.D. - - CoR. Likewise, if the first has the same ratio to the second, which the third has to the fourth, then also any equimultiples whatever of the first and third shall have the same ratio to the second and fourth; and in like manner, the first and the third shall have the same ratio to any equimultiples whatever of the second and fourth. Let A the first have to B the second the same ratio which the third C has to the fourth D, - - and of A and C let E and F be any equimultiples whatever, Book v. PROP. IV, V, VI. 161 Then Eshall be to B as F to D. - Take of E, F any equimultiples whatever, K. L.; and of B, JO any equimultiples whatever G, H: then it may be demonstrated, as before, that K is the same multiple of A, that L is of C: < - * ~ and because A is to B, as C is to D, (hyp.) and of A and C certain equimultiples have been taken viz. K and L; and of B and D certain equimultiples G, H, - therefore, if K be greater than G, L is greater than H: and if equal, equal; if less, less: (V, def. 5.) * but K, L are any equimultiples whatever of E, F, (constr.) and G, H any whatever of B, D; therefore as E is to B, so is F to D. (V. def. 5.) And in the same way the other case is demonstrated. PROPOSITION v. THEOREM. If one magnitude be the same multiple of another, which a magnitude taken from the first is of a magnitude taken from the other; the remainder shall be the same multiple of the remainder, that the whole is of the whole. Let the magnitude A B be the same multiple of CD, that AE taken from the first, is of CF taken from the other. The remainder EB shall be the same multiple of the remainder JFD, that the whole AB is of the whole CD. G. A E B ‘C F D Take AG the same multiple of FD, that AE is of CF: - therefore AE is the same multiple of CF, that EG is of CD: (v. 1.) but AE, by the hypothesis, is the same multiple of CF, that AB is of CD; - r therefore EG is the same multiple of CD that AB is of CD; wherefore EG is equal to AB : (v. ax. 1.) take from each of them the common magnitude AE; and the remainder AG is equal to the remainder JEB. - wº since AE is the same multiple of CF, that AG is of FB, Constr. - and that AG has been proved equal to EB; * therefore AE is the same multiple of CF, that EB is of FD: but AE is the same multiple of CF that A3 is of CD: (hyp.) therefore EB is the sama multiple of FD, that AB is of CD. Therefore, if one magnitude, &c. Q.E.D. PROPOSITION VI. THEOREM. If two magnitudes be equimultiples of two others, and if equimultiples of these be taken from the first two; the remainders are either equal to these others, or equimultiples of them. - Let the two magnitudes AB, CD be equimultiples of the two E, F, and let AG, CH taken from the first two, be equimultiples of the same E, F. - M 162 EUCLID's ELEMENTs. Then the remainders GP, HD shall be either equal to E, F, or equimultiples of them. º A G B K C H D F – E— First, let GB be equal to E: JHD shall be equal to F. app - Make CK equal to F: & * and because AG is the same multiple of E, that CH is of F: (hyp.) and that GB is equal to E, and C/ſ to F; - therefore AB is the same multiple of E, that KH is of F: but AB, by the hypothesis, is the same multiple of E, that CD is of F: therefore KH is the same multiple of F, that CD is of F: wherefore KHis equal to CD: (v. ax. 1.) take away the common magnitude CH, then the remainder KC is equal to the remainder HD: but KC is equal to F: (constr.) therefore HD is equal to F. - Next let GB be a multiple of E. Then HD shall be the same multiple of F. A G B E— K. C. H. P F – Make CK the same multiple of F, that GB is of E: and because AG is the same multiple of E, that CH is of F: (hyp.) and GB the same multiple of E, that CK is of F; therefore AB is the same multiple of E, that KH is of F: (v. 2.) but AB is the same multiple of E, that CD is of F; (hyp.) therefore KHis the same multiple of F, that CD is of F; wherefore KHis equal to CD: (v. ax. 1.) take away CH from both; therefore the remainder KC is equal to the remainder HD : and because GB is the same multiple of E, that KC is of F, (constr.) . and that KC is equal to HD; therefore HD is the same multiple of F, that GB is of E. If, therefore, two magnitudes, &c. Q.E.D. - PROPOSITION A. THEOREM. - If the first of four magnitudes has the same ratio to the second, which the third has to the fourth; then, if the first be greater than the second, the third is also greater than the fourth; and if equal, equal; if less, less. Take any equimultiples of each of them, as the doubles of each; then, by def, 5th of this book, if the double of the first be greater . than the double of the second, the double of the third is greater than the double of the fourth : & * but if the first be greater than the second, the double of the first is greater than the double of the second; . wherefore also the double of the third is greater than the double of the fourth ; - BOOK v. PROP. A, B, C. 163 therefore the third is greater than the fourth : in like manner, if the first be equal to the second, or less than it, the third can be proved to be equal to the fourth, or less than it, Therefore, if the first, &c. Q.E.D. PROPOSITION B. THEOREM, If four magnitudes are proportionals, they are proportionals also when taken inversely. - - Het A be to B, as C is to D. Then also inversely, B shall be to A, as D to C. A— B C —-- D G - E— H F Take of B and D any equimultiples whatever E and F: and of A and C any equimultiples whatever G and H. First, let E be greater than G, then G is less than E: and because A is to B, as C is to D, (hyp.) and of A and C, the first and third, G and H are equimultiples; and of B and D, the second and fourth, E and F are equimultiples; and that G is less than E, therefore His less than F"; (v. def. 5.) - that is, F is greater than H; if, therefore, E be greater than G, Fis greater than H; - in like manner, if E be equal to G, JF may be shewn to be equal to Hſ; - and if less, less; ,” but E, F, are any equimultiples whatever of B and D, (constr.) and G, H any whatever of A and C; therefore, as B is to A, so is D to C. (v. def. 5.) Therefore, if four magnitudes, &c. Q.E.D. PROPOSITION C. THEOREM. If the first be the same multiple of the second, or the same part of #, that the third is of the fourth; the first is to the second, as the third is to the fourth. º Let the first A be the same multiple of the second B, that the third C is of the fourth B. Then A shall be to B as C is to B). A— B— C— D— E G ‘F— H Take of A and C any equimultiples whatever E and F; and of B and D any equimultiples whatever G and H. Then, because A is the same multiple of B that C is of D; (hyp.) and that Eis the same multiple of A, that Fis of C; (constr.) therefore E is the same multiple of B, that F is of D; (v. 3.) that is, E and F are equimultiples of B and D: but G and H are equimultiples of B and D; (constr.) therefore, if E be a greater multiple of B than G is of B, Fis a greater multiple of D than His of D; that is, if E be greater than G, Fis greater than H: M 2. I64 - EUCLID's ELEMENTs. in like manner, if E be equal to G, or less than it, F may be shewn to be equal to H, or less than it, but E, F are equimultiples, any whatever, of A, C; (constr.) and G, H any equimultiples whatever of B, D; j . therefore A is to B, as C is to D. (v. def. 5.) * Next, let the first A be the same part of the second B, that the third C is of the fourth D. - - Then A shall be to B, as C is to D. A — B C— D For since A is the same part of B that C is of D, therefore B is the same multiple of A, that D is of C: - wherefore, by the preceding case, B is to A, as D is to C; and therefore inversely, A is to B, as C is to D. (v. B.) Therefore, if the first be the same multiple, &c. Q.E.D. PROPOSITION D. THEOREM. If the first be to the second as the third to the fourth, and if the first be a multiple, or a part of the second; the third is the same multiple, or the same part of the fourth. Tet A be to B as C is to D: and first, let A be a multiple of B. Then C shall be the same multiple of D. A— B C— D– Take E equal to A, and whatever multiple A or E is of B, make F the same multiple of D : -* then, because A is to B, as C is to D; (hyp.) and of B the second, and D the fourth, equimultiples have been taken, E and F; - therefore A is to E, as C to F. (v. 4. Cor.) but A is equal to E, (constr.) therefore C is equal to F: (v. A.) and Fis the same multiple of D, that A is of B; (constr.) therefore C is the same multiple of D, that A is of B. Next, let A the first be a part of B the second. Then C the third shall be the same part of D the fourth. Because A is to B, as C is to D; (hyp.) then, inversely, B is to A, as D to C. (V. B.) A— B C— D— lout A is a part of B, therefore B is a multiple of A : (hyp.) therefore, by the preceding case, D is the same multiple of C; that is, C is the same part of D, that A is of B. Therefore, if the first, &c. Q.E.D. Book v. PROP. VII, VIII. 165 PROPOSITION VII. THEOREM. Equal magnitudes have the same ratio to the same magnitude : and the same has the same ratio to equal magnitudes. | - . Let A and B be equal magnitudes, and C any other. - Then A and B shall each of them have the same ratio to 0: and C shall have the same ratio to each of the magnitudes A and B. A — - B— C— D E— F Take of A and B any equimultiples whatever D and E, and of 0 any multiple whatever F. ſº Then, because D is the same multiple of A, that E is of B, (constr.) . and that A is equal to B: (hyp.) ~. therefore D is equal to E; (v. ax. 1.) therefore, if D be greater than F, E is greater than F; and if equal, equal; if less, less: but D, E are any equimultiples of A, B, (constr.) and F is any multiple of 0; therefore, as A is to C, so is B to C. (v. def. 5.) Eikewise C shall have the same ratio to A, that it has to B. For having made the same construction, D may in like manner be shewn to be equal to E; therefore, if F be greater than D, it is likewise greater than E; and if equal, equal; if less, less; but Fis any multiple whatever of C, and D, E are any equimultiples whatever of A, B ; therefore, C is to A as Cis to B. (v. def. 5.) Therefore, equal magnitudes, &c. Q.E.D. PROPOSITION VIII. THEOREM. Of two unequal magnitudes, the greater has a greater ratio to any other magnitude than the less has : and the same magnitude has a greater ratio to the less of two other magnitudes, than it has to the greater. Let 4B, BC be two unequalmagnitudes, of which AB is the greater, and let D be any other magnitude. Then AB shall have a greater ratio to D than BC has to D: and D shall have a greater ratio to BC than it has to A.B. Fig. 1. - Fig. 2. Fig. 3. : IE Tº F'- A F}- || * -- G |G|. F|- A - I. K. H. D. B cº H | | I, K H. D. G B ! . | L. K D 166 - EUCLID's ELEMENTs. If the magnitude which is not the greater of the two AC, CB, be not less than D, - -- - take EF, FG, the doubles of AC, CB, (as in fig. 1) but if that which is not the greater of the two AC, CB, be less than D, (as in fig. 2 and 3) this magnitude can be multiplied, so as to become greater than D, whether it be AC, or C.B. Let it be multiplied until it become greater than D, and let the other be multiplied as often; and let EF be the multiple thus taken of AC, and FG the same multiple of CB : therefore EF and FG are each of them greater than D: - and in every one of the cases, - take H the double of D, Kits triple, and so on, - till the multiple of D be that which first becomes greater than FG: let L be that multiple of D which is first greater than FG, and K the multiple of D which is next less than L. Then because L is the multiple of D, which is the first that becomes greater than FG, g the next preceding multiple Kis not greater than FG ; - that is, FG is not less than K : and since EF"is the same multiple of AC, that FG is of CB; (constr.) therefore FG is the same multiple of CB, that EG is of AB; (v. 1.) that is, EG and FG are equimultiples of AB and CB ; and since it was shewn, that FG is not less than K, and, by the construction, EF is greater than D; - ºtherefore the whole EG is greater than K and D together: but K together with D is equal to L; (constr.) - therefore EG is greater than L.; - but FG is not greater than L: (constr.) and EG, FG were proved to be equimultiples of AB, BC; . and L is a multiple of D; (constr.) : therefore AB has to D a greater ratio than BC has to D. (v. def. 7.) Also D shall have to BC a greater ratio than it has to AB. For having made the same construction, it may be shewn in like manner; that L is greater than FG, but that it is not greater than EG: g - and L is a multiple of D; (constr.) -> and FG, EG were proved to be equimultiples of CB, AB; therefore D has to CB a greater ratio than it has to A.B. (v. def. 7.) Wherefore, of two unequal magnitudes, &c., Q.E.D. PROPOSITION IX. THEOREM, Magnitudes which have the same ratio to the same magnitude are equal to one another; and those to which the same magnitude has the same ratio are equal to one another. - Let A, B have each of them the same ratio to 0. Then A shall be equal to B. A— D. C— F— Tor, if they are not equal, one of them must be greater than the other: if possible, let A be the greater: BOOK W. PROP. IX, ºx. - 167 * then, by what was shewn in the preceding proposition, there are some equimultiples of A and B, and some multiple of C, such, that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than that of C, let these multiples be taken; and let D, E be the equimultiples of A, B, and F the multiple of C, such that D may be greater than F, but E not greater than F. Then, because A is to C as B is to C, (hyp.) S. and of A, B, are taken equimultiples, D, E, and of C is taken a multiple F; and that D is greater than F; *- therefore E is also greater than F. (v. def. 5.) but E is not greater than F; (constr.) which is impossible: therefore A and B are not unequal; that is, they are equal. Next, let 0 have the same ratio to each of the magnitudes A and B. - Then A shall be equal to B. - - For, if they are not equal, one of them must be greater than the other: - if possible, let A be the greater: therefore, as was shewn in Prop. VIII., there is some multiple F of C, and some equimultiples E and D of B and A such, that F is greater than E, but not greater than D: and because C is to B, as C is to A, (hyp.) - and that F the multiple of the first, is greater than E the multiple of the second; - * therefore F the multiple of the third, is greater than D the multiple of the fourth : (v. def. 5.) - but Fis not greater than D (hyp.); which is impossible: therefore A is equal to B. - & Wherefore, magnitudes which, &c. Q.E.D. eº PROPOSITION X. THEOREM. That magnitude which has a greater ratio than another has unto the 8ame magnitude, is the greater of the two, and that magnitude to which the same has a greater ratio than it has unto another magnitude, is the less of the two. - Let A have to C a greater ratio than B has to C; then A shall be greater than B. - D A * f C— F B— - E . For, because A has a greater ratio to C, than B has to C, - there are some equimultiples of A and B, and some multiple of C such, (v. def. 7.) that the multiple of A is greater than the multiple of C, but the multiple of B is not greater than it: . let them be taken; - * * and let D, E be the equimultiples of A, B, and F the multiple of C; such, that D is greater than F, but E is not greater than F: therefore D is greater than E: • 168 º EUCLID'S ELEMENTS. and, because D and E are equimultiples of A and B, and that D is greater than E; therefore A is greater than B. (V. ax. 4.) Next, let C have a greater ratio to B than it has to A. s Then B shall be less than A. For there is some multiple F of C, (v. def. 7.) and some equimultiples E and D of B and A such, that F is greater than E, but not greater than D: therefore E is less than D: and because E and D are equimultiples of B and A, k and that E is less than D, therefore B is less than A. (V. ax. 4.) Therefore, that magnitude, &c. Q.E.D. PROPOSITION XI. THEOREM. JRatios that are the same to the same ratio, are the same to one another. Let A be to B as Cºis to D.; and as 0 to D, so let E be to F. Thén. A shall be to B, as E to F. G H K— A. C E B— D— F—- L. M. N. * Take of A, C, E, any equimultiples whatever G, H, PC; and of B, D, F any equimultiples whatever L, M, W. Therefore, since A is to B as 0 to D, . and G, H are taken equimultiples of A, C, and E, M, of B, D; T 2. if G be greater than L, H is greater than M: and if equal, equal; and if less, less. (v. def. 5.) Again, because C is to D, as E is to F. and H, K are taken equimultiples of C, E, and M, W, of D, F; if H be greater than M, K is greater than W; and if equal, equal; and if less, less: but if G be greater than L, it has been shewn that H is greater than M: and if equal, equal; and if less, less: therefore, if G be greater than L, Jſ is greater than M ; and if equal, equal; and if less, less: and G, K are any equimultiples whatever of A, E; * and L, N any whatever of B, F; therefore, as A is to B, so is E to F. (v. def. 5.) © Wherefore, ratios that, &c. Q.E.D., * Ae’ PROPOSITION XII. THEOREM, If any number of magnitudes be proportionals, as one of the antecedents. is to its consequent, so shall all the antecedents taken together be to all the consequents. g Let any number of magnitudes A, B, C, D, E, F, be proportionals: that is, as A is to B, so C to D, and E to F. -*. BOOK v. PROP. XII, XIII. - 169 Then as A is to B, so shall A, C, E together, be to B, D, F together. G H. I- A C– B—— B— D— F — L M N Take of A, C, E any equimultiples whatever G, H, Iſ; and of B, D, F any equimultiples whatever L, M, N. Then, because A is to B, as C is to D, and as E to F; and that G, H, K are equimultiples of A, C, E, and L, M, W equimultiples of B, D, F; therefore, if G be greater than L, His greater than M, and K greater than W; and if equal, equal; and if less, less: (V. def. 5.) wherefore if G be greater than L, then G, H, K together, are greater than L, M, N together; and if equal, equal; and if less, less: but 6, and G, H, K together; are any equimultiples of A, and A, C, E together; because if there be any number of magnitudes equimultiples of as many, each of each, whatever multiple one, of them is of its part, the same multiple is the whole of the whole: (v. 1.) - for the same reason L, and L, M, W are any equimultiples of B, and B, D, F: therefore as A is to B, so are A, C, E together to B, D, F together. W. def. 5.) ( ) Wherefore, if any number, &c. Q.E.D. PROPOSITION XIII. THEOREM. If the first has to the second the same ratio which the third has to the fourth, but the third to the fourth, a greater ratio than the fifth has to the sixth; the first shall also have to the second a greater ratio than the fifth has to the sixth. - Let A the first have the same ratio to B the second, which C the third has to D the fourth, but C the third a greater ratio to D the fourth, than E the fifth has to F the sixth. Then also the first A shall have to the second B, a greater ratio • than the fifth E has to the sixth F. M G- H A— C — E— B D— F— N. R L Because C has a greater ratio to D, than E to F. there are some equimultiples of C and E, and some of D and F, such that the multiple of C is greater than the multiple of D, but the mul- tiple of E is not greater than the multiple of F: (v. def. 7.) - - -- let these be taken, and let G, H be equimultiples of C, E, and K, L equimultiples of D, F. such that G may be greater than K, but H. not greater than L: and whatever multiple G is of C, take M the same multiple of A; 170 - EUCLID's ELEMENTs. and whatever multiple K is of D, take N the same multiple of B: then, because A is to B, as C to D, (hyp.) and of A and C, M and G are equimultiples; and of H and D, N and K are equimultiples; therefore, if M be greater than N, G is greater than Iſ; and if equal, equal; and if less, less: (v. def. 5.) but G is greater than K, (constr.) therefore M is greater than N. but H is not greater than L: (constr.) and M, H are equimultiples of A, E; and N, L equimultiples of B, F; - therefore A has a greater ratio to B, than E has to F. (v. def. 7.) Wherefore, if the first, &c. Q.E.D. CoR. And if the first have a greater ratio to the second, than the third has to the fourth, but the third the same ratio to the fourth, which the fifth has to the sixth; it may be demonstrated, in like manner, that the first has a greater ratio to the second, than the fifth has to the sixth. - PROPOSITION XIV. THEOREM. If the first has the same ratio to the second which the third has to the fourth; then, if the first be greater than the third, the second shall be greater than the fourth; and ºf equal, equal; and ºf less, less. Let the first A have the same ratio to the second B which the third C has to the fourth D. r If A be greater than C, B shall be greater than D. (fig. 1.) 1. 2. 3. *- A— A A B B IB C * C C D — D D -— Decause A is greater than C, and B is any other magnitude, A has to B a greater ratio than C has to B : (v. 8.) but, as A is to B, so is C to D; (hyp.) therefore also C has to D a greater ratio than C has to B: (v. 13.) but of two magnitudes, that to which the same has the greaterratio, is the less: (v. 10.) º, - therefore D is less than B; - that is, B is greater than D. Secondly, if A be equal to C, (fig. 2.) then B shall be equal to D. For A is to B, as C, that is, A to D: * therefore B is equal to D. (v. 9.) Thirdly, if A be less than C, (fig. 3.) then B shall be less than D. For C is greater than A; - and because C is to D, as A is to B, - therefore D is greater than B, by the first case; that is, B is less than D. Therefore, if the first, &c. Q.E.D. Book v. PROF. xv, xvi. 171 PROPOSITION XV. THEOREM. Magnitudes have the same ratio to one another which their equimultiples have. * - - Let AB be the same multiple of C, that DE is of F. Then C shall be to F, as AB is to D.E. A G H B D K L E c— F— Because AB is the same multiple of C, that DE is of F; there are as many magnitudes in AB equal to C, as there are in DE equal to F: - let AB be divided into magnitudes, each equal to C, viz. A G, GH, HB; and DE into magnitudes, each equal to F, viz. D.K., KL, LE: then the number of the first AG, GH, HB, is equal to the number of the last DK, KL, LE: - - and because AG, GH, HB are all equal, and that DK, KL, LE, are also equal to one another; therefore AG is to DK, as GH to KL, and as HB to LE: (v. 7.) but as one of the antecedents is to its consequent, so are all the antecedents together to all the consequents together, (v. 12.) wherefore, as AG is to DK, so is AB to DE: but AG is equal to C, and DK to F: therefore, as C is to F, so is AB to D.E. Therefore, magnitudes, &c. Q.E.D. PROPOSITION XVI. THEOREM. If four magnitudes of the same kind be proportionals, they shall also is proportionals when taken alternately. - Let A, B, C, D be four magnitudes of the same kind, which are. . proportionals, viz. as 4 to B, so C to D. They shall also be proportionals when taken alternately: that is, A shall be to C, as B to D. • E— G A— C— B — D— F. H Take of A and B any equimultiples whatever E and F; and of C and D take any equimultiples whatever G and H. : and because E is the same multiple of A, that Fis of B, and that magnitudes have the same ratio to one another which, their equimultiples have; (v. 15.) • therefore A is to B, as E is to F: but as A is to B, so is C to D; (hyp.) wherefore as C is to D, so is E to F: (v. 11.) again, because G, H are equimultiples of C, D, therefore as C is to D, so is G to H. (v. 15.) but it was proved that as C is to D, so is E to F; therefore, as E is to F, so is G to H. (v. 11.) But when four magnitudes are proportionals, if the first be greater. than the third, the second is greater than the fourth : A. and if equal, equal; if less, less; (v. 14.) T72 * - EUCLID's ELEMENTS. therefore, if E be greater than G, Flikewise is greater than Hſ; and if equal, equal; if less, less: and E, F are any equimultiples whatever of A, B ; (constr.) • and G, H any whatever of C, D : therefore A is to C, as B to D. (V. def. 5.) If then four magnitudes, &c. Q.E.D. PROPOSITION XVII. THEOREM, If magnitudes, taken jointly, be proportionals, they shall also be propor- tionals when taken separately: that is, ºf two magnitudes together have to one of them, the same ratio which two others have to one of these, the re- maining one of the first two shall have to the other the same ratio which the remaining one of the last two has to the other of these. Let AB, BE, CD, DF be the magnitudes, taken jointly which are &P proportionals; - - - , that is, as AB to BE, so let CD be to D.F. Then they shall also be proportionals taken separately, - viz. as AE to EB, so shall CF be to FD. G EI K X I, M. N. P º b - U U. A E B - C F D —H. Take of AE, EB, CF, FD any equimultiples whatever GH, HK, IM, MN: - - and again, of EB, FD take any equimultiples whatever KX, N.P. Then because GH is the same multiple of AE, that HK is of EB, therefore GH is the same multiple of A.E, that GK is of AB; (v. 1.) but GH is the same multiple of AE, that LM is of CF; therefore GK is the same multiple of AB, that LM is of CF. Again, because LM is the same multiple of CF, that MN is of FD; therefore LM is the same multiple of UF that LW is of CD: (v. 1.) but LM was shewn to be the same multiple of CF, that GK is of AB; therefore GK is the same multiple of AB, that LW is of CD; that is, GK, LN are equimultiples of AB, CD. Next, because HK is the same multiple of EB, that MW is of FD; and that KX is also the same multiple of EB, that NP is of FD; therefore HX is the same multiple of EB, that MP is of FD. (v. 2.) And because AB is to BE, as CD is to D.F. (hyp.) and that of AB and CD, GK and LN are equimultiples, and of EB and FD, HX and MP are equimultiples; therefore if GK be greater than HX, then LN is greater than MP; and if equal, equal; and if less, less: (v. def. 5.) but if GH be greater than KX, then, by adding the common part HIſ to both, GK is greater than HX; (I. ax. 4.) wherefore also LN is greater than MP; and by taking away MN from both, LM is greater than WP: (I. ax. 5.) therefore, if GH be greater than KX, LM is greater than NP. Book v. PROP. XVII, XVIII. 173 In like manner it may be demonstrated, that if GH be equal to KX, LM is equal to NP; and if less, less: - but GH, LM are any equimultiples whatever of AE, CF, (constr.) and KX, WP are any whatever of EB, FD: therefore, as AE is to EB, so is CF to FD. (V. def. 5.) If then magnitudes, &c. Q.E.D. PROPOSITION XVIII. THEOREM. If magnitudes, taken separately, be proportionals, they shall also be pro- portionals when taken jointly: that is, if the first be to the second, as the third to the fourth, the first and second together shall be to the second, as the third and fourth together to the fourth. * Let A.E, JEB, CF, FD be proportionals; that is, as AE to EB, so let CF be to FD. j .* Then they shall also be proportionals when taken jointly; that is, as AB to B.E, so shall CD be to D.F. G K O H. L. N. P. M. A E B | O C F D D - Take of AB, BE, CD, DF’ any equimul LM, MN; - tiples whatever GH, HK, and again, of BE, DF take any equimultiples whatever KO, NP: - and because KO, NP are equimultiples of BE, DF; and that KH, NM are likewise equimultiples of BE, DF; therefore if KO, the multiple of B.E., be greater than KH, which is a multiple of the same B.E, then NP, the multiple of D.F, is also greater than WM, the mul- tiple of the same DF'; - * - - and if KO be equal to KH, NP is equal to NM; and if less, less. First, let KO be not greater than KH; • therefore WP is not greater than WM: and because GH, HI(, are equimultiples of AB, B.E, and that AB is greater than B.E, therefore GH is greater than HK; (v. ax. 3.) but KO is not greater than KH; therefore GH is greater than KO. In like manner it may be shewn, that LM is greater than WP. Therefore, if KO be not greater than KH, • then GH, the multiple of AB, is always greater than KO, the multiple of BE; and likewise LM, the multiple of CD, is greater than NP, the multiple of D.F. - - - Next, let KO be greater than KH; \. therefore, as has been shewn, N.P is greater than NM. - G. K. H. O L N M P | I A E B C F D ! D And because the whole GH is the same multiple of the whole .4B, that HK is of B.E, - - t 174 EUCLID's ELEMENTs. therefore the remainder G.K. is the same multiple of the remainder AE that GH is of AB, (v. 5.) - - -> - which is the same that LM is of CD. * In like manner, because LM is the same multiple of CD, that MN is of D.E, therefore the remainder LN is the same multiple of the remainder CF, that the whole LM is of the whole CD: (v. 5.) - , but it was shewn that LM is the same multiple of CD, that GK is of AE; . therefore GK is the same multiple of AE, that LN is of CF; that is, GK, LN are equimultiples of A.E, C.F. And because KO, NP are equimultiples of BE, DF, therefore if from KO, NP there be taken KH, NM, which are likewise equimultiples of BE, DF, the remainders HO, MP are either equal to BE, DF, or equi- multiples of them. (v. 6.) - First, let HO, MP be equal to BE, DF: - ‘then because AE is to EB, as CF to FD, (hyp.) *- and that GK, LN are equimultiples of A.E, CF; therefore GK is to EB, as LW to FD : (v. 4. Cor.) but HO is equal to EB, and MP to FD; I - wherefore GK is to HO, as LN to MP; therefore if GK be greater than HO, LV is greater than MP; (v. A.) and if equal, equal; and if less, less. But let HO, MP be equimultiples of EB, FD. Then because A E is to EB, as CF to FD, (hyp.) ‘G. K. H. O. L N M P —H- D | A E C F D and that of AE, CF are taken equimultiples GK, LN; and of EB, FD, the equimultiples HO, MP; if GK be greater than HO, LV is greater than MP; and if equal, equal; and if less, less; (v. def. 5.) which was likewise shewn in the preceding case. But if GH be greater than KO, • . taking KH from both, GK is greater than HO; (I. ax. 5.) wherefore also LN is greater than MP; - and consequently adding NM to both, LM is greater than NP: (I. ax. 4.) therefore, if GH be greater than K0, . LM is greater than WP. } In like manner it may be shewn, that if GH be equal to KO, LM is equal to NP; and if less, less. And in the case in which KO is not greater than KH, it has been shewn that GH is always greater than KO, and likewise LM greater than WP: but GH, LM are any equimultiples whatever of AB, CD, (constr.) - and KO, NP are any whatever of BE, DF; - therefore, as AB is to B.E., so is CD to D.F. (v. def. 5.) If then magnitudes, &c. Q.E.D. * BOOK v. PROP. XIX, xx. 175 K PROPOSITION XIX. THEOREM. If a whole magnitude be to a whole, as a magnitude taken from thefirst ts to a magnitude taken from the other; the remainder shall be to the remainder as the whole to the whole. - - Let the whole AB be to the whole CD, as AE a magnitude taken from AB is to CF'a magnitude taken from CD. - . Then the remainder EB shall be to the remainder FD, as the whole .A.B to the whole CD. - f - A E B *m. C - F D IBecause AB is to CD, as AE to CF: therefore alternately, BA is to A.E, as DC to CF. (v. 16.) and because if magnitudes taken jointly be proportionals, they are also proportionals, when taken separately; (v. 17.) therefore, as BE is to EA, so is DF to FC; and alternately, as BE is to D.F, so is EA to FC: , but, as AE to CF, so, by the hypothesis, is AB to CD; therefore also B.E. the remainder is to the remainder DF, as the whole AB to the whole CD. (v. 11.) - Wherefore, if the whole, &c. Q.E.D. - x CoR.—If the whole be to the whole, as a magnitude taken from the first is to a magnitude taken from the other; the remainder shall likewise be to the remainder, as the magnitude taken from the first to that taken from the other. The demonstration is contained in the preceding. - PROPOSITION E. THEOREM, If four magnitudes be proportionals, they are also proportionals by con- tersion ; that is, the first is to its eaccess above the second, as the third to its 6&cess above the fourth. Let AB be to B.E, as CD to D.F. Then BA shall be to AE, as DC to CF. A. E. B C F D | Because AB is to BE, as CD to DF, therefore by division, AE is to EB, as CF to FD; (v. 17.) -------" and by inversion; BE is to EA, as DF is to CF; (v. B.) - wherefore, by composition, BA is to AE, as DC is to CF. (v. 18.) If therefore four, &c. Q.E.D. PROPOSITION XX. THEOREM. If there be three magnitudes, and other three, which, taken two and two, have the same ratio; then if the first be greater than the third, the fourth shall be greater than the sixth; and if equal, equal; and if less, less. Let A, B, C be three magnitudes, and D, E, Fother three, which taken two and two have the same, ratio, - - 176 EUCLID's ELEMENTS. viz. as A is to B, so is D to E; and as B to C, so is E to F. - If A be greater than C, D shall be greater than F; and if equal, equal; and if less, less. A— B C D— JE - F—— Because A is greater than C, and B is any other magnitude, and that the greater has to the same magnitude a greater ratio than the less has to it; (v. 8.) - therefore A has to B a greater ratio than C has to B : but as D is to E, so is A to B; (hyp.) º therefore D has to E a greater ratio than C to B : (v. 13.) - and because B is to C, as E to F, by inversion, C is to B, as Fis to E: (v. B.) and D was shewn to have to E a greater ratio than C to B : therefore D has to E a greater ratio than F to E: (v. 13. Cor.) but the magnitude, which has a greater ratio than another to the same magnitude, is the greater of the two; (v. 10.) - - therefore D is greater than F. . Secondly, let A be equal to C. Then D shall be equal to F. A. - 33– C— IB JE— F JBecause A and C are equal to one another, A is to B, as C is to B: (v. 7.) but A is to B, as D to E; (hyp.) and C is to B, as F to E; (hyp.) wherefore D is to E, as F to E; (v. 11. and v. B.) and therefore D is equal to F. (v. 9.) Next, let A be less than C. Then D shall be less than F. A B— C D — E – I' Eor C is greater than A; and as was shewn in the first case, C is to B, as F to E, and in like manner, B is to A, as E to D ; therefore F is greater than D, by the first case; that is, D is less than F. Therefore, if there be three, &c. Q.E.D. PROPOSITION XXI. THEOREM. If there be three magnitudes, and other three, which have the same ratio taken two and two, but in a cross order; then ºf the first magnitude be greater than the third, the fourth shall be greater than the sixth; and if equal, equal; and if less, less. - - Let A, B, C be three magnitudes, and D, E, F other three, which have the same ratio, taken two and two, but in a cross order, viz. as A is to B, so is E to F, BOOK v. PROP. xx1, xxii. 177 : and as B is to -C, so is D to JE. If A be greater than C, D shall be greater than F; and if equal, equal; and if less, less. A B C D— E— F Because A is greater than C, and B is any other magnitude, A has to B a greater ratio than C has to B: (v. 8.) but as E to F, so is A to B; (hyp.) therefore E has to F a greater ratio than C to B: (v. 13.) - and because B is to C, as D to E; (hyp.) A. by inversion, C is to B, as E to D: and E was shewn to have to F a greater ratio than C has to B; therefore E has to F a greater ratio than E has to D: (v. 13. Cor.) but the magnitude to which the same has a greater ratio than it has to another, is the less of the two: (v. 10.) - therefore F is less than D; that is, D is greater than F. Secondly, let A be equal to C; D shall be equal to F. A , B- C i.2 JD—- JE— F– Because A and C are equal, A is to B, as C is to B : (v. 7.) 'but A is to B, as E to F; (hyp.) and C is to B, as E to D;. wherefore E is to F, as E to D; (v. 11.) and therefore D is equal to F. (v. 9.) Next, let A be less than G: - JD shall be less than F. A— B C --—- £)— Tº F For C is greater than A; and as was shewn, C is to B, as E to J2, and in like manner, B is to A, as F to E; therefore Fis greater than D, by case first : , that is, D is less than F. Therefore, if there be three, &c. Q.E.D. pROPOSITION XXII. THEOREM. If there be any number of magnitudes, and as many others, which taken two and two in order, have the same ratio; the first shall have to the last of the first magnitudes, the same ratio which the first has to the last of the others. N.B. This is usually cited by the words “ex aequali,” or “ex aequo.” First, let there be three magnitudes A, B, C, and as many others D, E, F, which taken two and two in order, have the same ratio, that is, such that A is to B, as D to E; and as B is to C, so is E to F. - Then A shall be to C, as D to F. 178 EUCLID'S ELEMENTS. M G R A — B C— D — E— F— H- I- N Take of A and D any equimultiples whatever G and H, and of B and D any equimultiples whatever K and L; and of C and F any whatever M and N: then because A is to B, as D to E, and that G, H are equimultiples of A, D, - and K, L equimultiples of B, E, therefore as G is to K, so is H to L: (v. 4.) for the same reason, K is to M as L to N: and because there are three magnitudes G, K, M, and other three II, L, N, which two and two, have the same ratio; therefore if G be greater than M. His greater than N; and if equal, equal; and if less, less; (v. 20.) but G, H are any equimultiples whatever of A, D, and M, N are any equimultiples whatever of C, F, (constr.) therefore, as A is to C, so is D to F. (v. def. 5.) i Next, let there be four magnitudes A, B, C, D, and other four E, F, G, H, which two and two have the sameratio, • . - viz. as A is to B, so is E to ſº; and as B to C, so F to G.; and as 0 to D, so G to H. Then A shall be to D, as E to H. • -ºr A. B. C. D E. F. G. H. Because A, B, C are three magnitudes, and E, F, G other three, which taken two and two, have the same ratio; therefore by the foregoing case, A is to C, as E to G: but 0 is to D, as G is to H; - wherefore again, by the first case A is to D, as E to H. and so on, whatever be the number of magnitudes. Therefore, if there be any number, &c. Q. E. D. PROPOSITION XXIII. THEOREM, If there be any number of magnitudes, and as many others, which taken two and two in a cross order, have the same ratio : the first shall have to the last of the first magnitudes the same ratio which the first has to the last of the others. W. B. This is usually cited by the words “ex aequali in proportione perturbatā;” or “ex aequo perturbato.” IFirst, let there be three magnitudes A, B, C, and other three D, JE, F, which taken two and two in a cross order have the same ratio, that is, such that A is to B, as E to F, - and as B is to C, so is D to E. ' Then A shall be to C, as D to F. G— H. I- A– IB C— D— - E- F— R M— N BOOK v. PROP. XXIII, xxiv. 179 Take of A, B, D any equimultiples whatever G, H, K ; and of C, E, F, any equiñnultiples whatever L, M, N: and because G, H are equimultiples of A, B, and that magnitudes have the same ratio which their equimultiples have; (v. 15.) - - therefore as A is to B, so is G to H : and for the same reason, as E is to F, so is M to Nº: but as A is to B, so is E to F; (hyp.) therefore as G is to H, so is M to N: (v. 11.) - and because as B is to C, so is D to E, (hyp.) ‘and that H, K are equirmultiples of B, D, and L, M of C, E'; therefore as H is to L, so is K to M. (v. 4.) and it has been shewn that G is to JH, as M to N: therefore, because there are three magnitudes G, H, L, and other three K, M, AV, which have the same ratio taken two and two in a cross order; - - * if G be greater than Z, K is greater than W: and if equal, equal; and if less, less: (v. 21.) but 6, K are any equimultiples whatever of A, D; (constr.) - and L, N any whatever of C, F; - therefore as A is to C, so is D to F. (v. def. 5.) - Next, let there be four magnitudes A, B, C, D, and other four E, F, G, H, which taken two and two in a cross order have the same ratio, g viz. A to B, as G to Hº; JB to C, as F to G; - and C to B, as E to F. Then A shall be to D, as E to H. A . . E) E. F. G. H. § B. ..] E. F. Because A, B, C are three magnitudes, and F, G, H other three which taken two and two in a cross order, have the same ratio; - by the first case, A is to C, as F to H; - ‘but C is to D, as E is to F; wherefore again, by the first case, A is to D, as E to H; and so on, whatever be the number of magnitudes. Therefore, if there be any number, &c. Q.E.D. PROPOSITION XXIV. THEOREM. If the first has to the second the same ratio which the third has to the Jourth; and the fifth to the second the same ratio which the sixth has to the fourth; the first and fifth together shall have to the second, the same ratio which the third and sixth together have to the fourth. Let AB the first have to C the second the same ratio which DE the third has to F the fourth; and let BG the fifth have to C the second the same ratio which EH the sixth has to F the fourth. - * Then AG, the first and fifth together, shall have to C the second, .* ratio which DH, the third and sixth together, has to F the O * > IN 2 180 EUCLID's ELEMENTs. A B G D E HI c— F— Because BG is to C, as EH to F; . loy inversion, C is to BG, as F to EH. (v. B.) and because, as AB is to C, so is DE to F; (hyp.) and as C to BG, so is F to EH; ex aequali, AB is to BG, as DE to EH: (v. 22.) - and because these magnitudes are proportionals when taken separately, they are likewise proportionals when taken jointly; (v. 18.) - therefore as AG is to GB, so is DH to HE: i but as GB to C, so is HE to F: (hyp.) therefore, ex æquali, as AG is to C, so is DH to F. (v. 22.) Wherefore, if the first, &c. Q.E.D. CoR. 1.—If the same hypothesis be made as in the proposition, the excess of the first and fifth shall be to the second, as the excess of the third and sixth to the fourth. The demonstration of this is the same with that of the proposition, if division be used instead of composition. CoR. 2.—The proposition holds true of two ranks of magnitudes, whatever be their number, of which each of the first rank has to the second magnitude the same ratio that the corresponding one of thes second rank has to a fourth magnitude: as is manifest. PROPOSITION xxv. THEOREM. gº If four magnitudes of the same kind are proportionals, the greatest and least of them together are greater than the other two together. t Let the four magnitudes AB, CD, E, F be proportionals, - viz. AB to CD, as E to F; and let AB be the greatest of them, and consequently F the least. - (v. 14, and A.) - Then AB together with Fshall be greater than CD together with E. A G. B C H D t F. F— Take AG equal to E, and CH equal to F. Then because as AB is to CD, so is E to F, and that AG is equal to E, and CH equal to F, therefore AB is to CD, as AG to CH: (v. 11, and 7.) and because AB the whole, is to the whole CD, as A G is to CH, likewise the remainder GB is to the remainder HD, as the whole AB is to the whole CD: (v. 19.) - º - but AB is greater than CD; (hyp.) therefore GB is greater than HD; (v. A.) - and because AG is equal to E, and CH to F; AG and F together are equal to CH and E together: (T. ax. 2.) therefore if to the unequal magnitudes GB, HD, of which GB is the greater, there be added equal magnitudes, viz. to GB the two AG and F, and CH and E to HD; . . AB and F together are greater than CD and E. (I. ax. 4.) Therefore, if four magnitudes, &c. Q.E.D. BOOK v. PROP. F., G. 181 * - - PROPOSITION F. THEOREM. JRatios which are compounded of the same ratios, are the same to one another. Let A be to B, as D to E; and B to C, as E to F. Then the ratio which is compounded of the ratios of A to B, and B to O. - - - which, by the definition of compound ratio, is the ratio of A to C, shall be the same with the ratio of D to F, which, by the same definition, is compounded of the ratios of D to E, and E to F. A . B. C D. E . F Because there are three magnitudes A, B, C, and three others D, E, F, which, taken two and two, in order, have the same ratio ; ex aequali, A is to C, as D to F. (v. 22.) Next, let A be to B, as E to F, and B to C, as D to E: A . B. C D. E. F therefore, ea: aquali in proportione perturbata, (v. 23.) A is to C, as D to F; - s that is, the ratio of A to C, which is compounded of the ratios of . A to B, and B to C, is the same with the ratio of D to F, which is compounded of the ratios of D to E, and E to F. - - And in like manner the proposition may be demonstrated, what- ever be the number of ratios in either case. Q.E.D. - w PROPOSITION G, THEOREM. If several ratios be the same to several ratios, each to each; the ratio which is compounded of ratios which are the same to the first ratios, each to each, shall be the same to the ratio compounded of ratios which dre the same to the other ratios, each to each. Let A be to B, as E to F; and C to D, as G to HT: and let A be to B, as K to L; and C to D, as L to M. ww. Then the ratio of K to M, by the definition of compound ratio, is compounded of the ratios of K to L, and L to M, which are the same with the ratios of A to B, and C to D. - Again, as E to F, so let W be to 0; and as G to H, so let 0 be to P. Then the ratio of N to P is compounded of the ratios of W to 0, and O to P, which are the same with the ratios of E to F, and G to H . . . and it is to be shewn that the ratio of K to M, is the same with the ratio of N to P; . or that K is to M, as W to P. A . B. C. D. K. L. M. E. F. G. H. N. O. P Because K is to L, as (A to B, that is, as E to F, that is, as) N to 0: and as L to M, so is (C to D, and so is G to H, and so is) O to P: - - ex aequali, K is to M, as N to P. (v. 22.) - Therefore, if several ratios, &c., Q.E.D. 182 EUCLID's ELEMENTs. PROPOSITION H. THEOREM. If a ratio which is compounded of several ratios be the same to a ratio which ts compounded of several other ratios ; and if one of the first ratios, or the ratio which is compounded of several of them, be the same to one of the last zatios, or to the ratio which is compounded of several of them ; then the re- maining ratio of the first, or, if there he more than one, the ratio compounded of the remaining ratios, shall be the same to the remaining ratio of the last, or, if there be more than one, to the ratio compounded of these remaining ratios. Let the first ratios be those of A to B, B to C, C to D, D to E, and JE to F; - - - and let the other ratios be those of G to H, H to K, K to I, and E to Mſ: i - also, let the ratio of A to F, which is compounded of the first ratios, the the same with the ratio of G to M, which is compounded of the other ratios; and besides, let the ratio of A to D, which is compounded of the ratios of A to B, B to C, C to D, be the same with the ratio of G to Jº, which is compounded of the ratios of G to H, and H to K. Then the ratio compounded of the remaining first ratios, to wit, of the ratios of D to E, and E to F, which compounded ratio is the ration, of D to F, shall be the same with the ratio of K to M, which is com- pounded of the remainingratios of K to L, and L to M of the other ratios. A. B. C. D.E. F G. H. K. L. M. Because, by the hypothesis, A is to D, as G to K, by inversion, D is to A, as K to G.; (v. B.) and as A is to F, so is G to M, (hyp.) therefore, ex aequali, D is to F, as K to M. (v. 22.) - If, therefore, a ratio which is, &c. Q.E.D. PROPOSITION K. THEOREM, If there be any number of ratios, and any number of other ratios, such, that the ratio which is compounded of ratios which are the same to the first zatios, each to each, is the same to the ratio which is compounded of ratios which are the same, each to each, to the last ratios; and if one of the first ratios, or the ratio which is compounded of ratios which are the same to several of the first ratios, each to each, be the same to one of the last ratios, or to the ratio. which is compounded of ratios which are the same, each to each, to several of the last ratios; then the remaining ratio of the first, or, if there be more than. one, the ratio which is compounded of ratios which are the same each to each, to the remaining ratios of the first, shall be the same to the remaining ratio of the last, or, if there be more than one, to the ratio which is compounded of: * ratios which are the same each to each to these remaining ratios, I let the ratios of A to B, C to D, E to F, be the first ratios: ` and the ratios of G to H, K to L, M to N, O to P, Q to R, be the other ratios: - and let A be to B, as S to T; and C to D, as 7" to V; and E to F, as V to X: ! - - - Book v. PROP. K. 183 therefore, by the definition of compound ratio, the ratio of S to X is compounded of the ratios of S to T, T to V, and V to X, which arg the same to the ratios of A to B, C to D, E to. F: each to each. Also, as G to H, so let Y be to Z; and K to L, as Z to a ; M to N, as a to b; 0 to P, as b. to c ; and Q to R, as a to d: therefore, by the same definition, the ratio of Y to d is compounded of the ratios of Y to Z, Z to a, a to b, b to c, and c to d, which are the same, each to each, to the ratios of G to H, K to L, M to N, O to P, and Q to R: - - gº therefore, by the hypothesis, S is to X, as Y to d. Also, let the ratio of A to B, that is, the ratio of S to T, which is one of the first ratios, be the same to the ratio of e to g, which is com- pounded of the ratios of e to f, and f to g, which, by the hypothesis, are the same to the ratios of G. to H, and K to L, two of the other ratios; and let the ratio of h to l be that which is compounded of the ratios of h to k, and k to l, which are the same to the remaining first ratios, viz. of 0 to D, and E to F; also, let the ratio of m to p, be that which is compounded of the ratios of m to n, n to o, and 0 to p, which are the same, each to each, to the remaining other ratios, viz. of M to N, O to P, and Q to R. Then the ratio of h to l shall be the same to the ratio of m to p; or s h shall be to l, as m to p. - h, k, l. A, B ; C, D ; E, F. S, T, V, X. - G, H ; K, L ; M, N ; O, P; Q, R. Y, Z, a, b, c, d. e, f, g. In, In, 0, p. Because e is to f, as (G to H, that is, as) Y to Z; and fis to g, as (K to L, that is, as) Z to a ; therefore, ex æquali, e is to g, as Y to a (v. 22.) and by the hypothesis, A is to B, that is, S to T, as e to g; - wherefore S is to T, as Y to a ; (v. 11.) and by inversion, T is to S, as a to Y: (v. B.) but S is to X, as Y to D; (hyp.) - therefore, ex æquali, T'is to X, as a to d: also, because h is to k, as (C to D, that is, as) T to V; (hyp.) and k is to l as (E to F, that is, as) V to X; therefore, ex æquali, h is to l, as T to X: in like manner, it may be demonstrated, that m is to p, as a to d: and it has been shewn, that T is to X, as a to d: therefore h is to l, as 'm to p. (v. 11.) Q.E.D. The propositions G and K are usually, for the sake of brevity, ex- pressed in the same terms with propositions F and H. and therefore it was proper to shew the true meaning of them when they are so expressed; especially since they are very frequently made use of by geometers. - - * NOTES TO BOOK W. In the first four Books of the Elements are considered, only the absolute equality and inequality of Geometrical magnitudes. The Fifth Book, contains an exposition of the principles whereby a more definite comparison may be instituted of the relation of magnitudes, besides their simple equality or inequality. The doctrine of Proportion is one of the most important in the whole course of mathematical truths, and it appears probable that if the subject were read simul- taneously in the Algebraical and Geometrical form, the investigations of the properties, under both aspects, would mutually assist each other, and both become equally comprehensible; also their distinct characters would be more easily per- ceived. - g Def. I, II, In the first Four Books the word-part is used in the same sense as we find it in the ninth axiom, “The whole is greater than its part:” where the word part means any portion whatever of any whole magnitude:... but in the Fifth Book, the word part is restricted to mean that portion of magnitude which is contained an exact number of times in the whole. For instance, if any straight line be taken two, three, four, or any number of times another straight line, by Euc. I. 3; the less line is called a part, or rather a submultiple of the greater line; and the greater, a multiple of the less line. The multiple is composed of a repetition of the same magnitude, and these definitions suppose that the multiple may be divided into its parts, any one of which is a measure of the multiple. And it is also obvious that when there are two magnitudes, one of which is a multiple of the other, the two magnitudes must be of the same kind, that is, they must be two lines, two angles, two surfaces, or two solids: thus, a triangle is doubled, trebled, &c., by doubling, trebling, &c. the base, and completing the figure. The same may be said of a parallelogram. Angles, arcs, and sectors of equal circles may be doubled, trebled, or any multiples found by Prop. xxvi-xxix, Book III. Two magnitudes are said to be commensurable when a third magnitude of the same kind can be found which will measure both of them ; and this third magnitude is called their common measure; and when it is the greatest magnitude which will measure both of them, it is called the greatest common measure of the two magnitudes: also when two magnitudes of the same kind have no common measure, they are said to be incommensurable. The same terms are also applied to numbers. Unity properly so called, may be assumed to represent that portion of every kind of magnitude which is taken as the measure of all magnitudes of the same kind. The composition of units, cannot produce Geometrical magnitude; three units are more in number than one unit, but still as different from magnitude as unity itself. Numbers may be considered as quantities, for we consider avery thing that can be exactly measured, as a quantity. . Unity is a common measure of all rational numbers, and all numerical reasonings proceed upon the hypothesis that the unit is the same throughout the whole of any, particular process. Euclid has not fixed the magnitude of any unit of length, nor made reference to any unit of measure of lines, angles, surfaces, or volumes. Hence arises an essentialdifference between number and magnitude; unity, being invariable, measures all rational numbers; but though any quantity be assumed as the unit of magnitude, it is impossible to assert that this assumed unit will measure all other magnitudes of the same kind; All whole numbers therefore are commensurable; for unity is their common measure; also all rational fractions proper or improper, are commensurable; for NOTES To Book v. 185 any such fractions may be reduced to other equivalent fractions having one common denominator, and that fraction whose denominator is the common denominator, and whose numerator is unity, will measure any one of the fractions. Two magni- tudes having a common measure can be represented by two numbers which express the number of times the common measure is contained in both the magnitudes: but two incommensurable magnitudes cannot be exactly represented by any two whole numbers or fractions whatever. ; - Incommensurability arises out of the attempt to express numerically the relation of one magnitude to another of the same kind, either by taking one of them as a standard unit to measure the other, or by taking some third magnitude as a standard unit by which to measure both. The possibility of finding a stan- dard unit, however small, which will exactly measure any given magnitudes, is not a likely occurrence: for instance, the side and the diagonal of a square are two lines, and the square on the diagonal is double the square on the side uni- versally and constantly: but what is the relation between the side and diagonal them- selves 2 The answer generally given is, if the side of the square be unity, the diagonal is the square root of 2 or V2: but what is V2 Simply no number at all in the original sense of the word as the ancients understood the term. It cannot be expressed in the decimal scale, nor in any other conceivable scale. An approximation to its value may be made to any assigned degree of accuracy. Numbers may be assigned at any stage of the approximation, between which its value lies, but its true and exact value can never be expressed. For, it may be shewn, numerically, that if the side of a square contain one unit of length, the diagonal contains more than one, but less than two units of length. If the side be divided into 10 units, the diagonal contains more than 14, but less than 15 such units. Also if the side contain 100 units, the diagonal contains more than 141, but less than 142 such units. And again, if the side contain 1000 units, the diagonal contains more than 1414, but less than 1415 such units. It is also obvious, that as the side is successively divided into a greater number of equal parts, the error in the magnitude of the diagonal will be diminished continually, but never can be entirely exhausted; and therefore into whatever number of equal parts the side of a square be divided, the diagonal will never contain an exact number of such parts. Thus the diagonal and side of a square having no common measure, cannot be exactly represented by any two numbers. 3. The term equimultiple in Geometry is to be understood of magnitudes of the same kind, or of different kinds, taken an equal number of times, and implies only a division of the magnitudes into the same number of equal parts. Thus, if two given lines are trebled, the trebles of the lines are equimultiples of the two lines: and if a given line and a given triangle be trebled, the trebles of the line and triangle are equimultiples of the line and triangle : as (vſ. 1. fig.) the straight line HC and the triangle AHC are equimultiples of the line BC and the triangle ABC : and in the same manner, (v1, 33. fig.) the arc EN and the angle EHN are equimultiples of the arc EF and the angle EHF. ~ Def. III. Adºyos écrirl 6 ſo Meyeóðu duo'yevaju i kató arm?Atkörnra arods &AAnNa aroué gyāorts. By this definition of ratio is to be understood the conception of the mutual relation of two magnitudes of the same kind, as two straight lines, two angles, two surfaces, or two solids. To prevent any misconception, Def. Iv. lays down the criterion, whereby it may be known what kinds of magnitudes can have. a ratio to one another; namely, Aéryov #xsiv Trpos d\\mXa playé6m Aéyétat, & 3évaras aroMAarxaatašáugua d\\?\tov Útrepéxeiv. “Magnitudes are said to have a ratio to one another, which, when they are multiplied, can exceed one another;” in other words, the magnitudes which are capable of mutual comparison must be of the 186 - EUCLID'S ELEMENTS: same kind. The former of the two terms is called the antecedent; and the latter, the consequent of the ratio. If the antecedent and consequent are equal, the ratio is called a ratio of equality; but if the antecedent be greater or less than the consequent, the ratio is called a ratio of greater or of less inequality. Care must be taken not to confound the expressions “ratio of equality”, and “equality of ratio:” the former is applied to the terms of a ratio when they, the antecedent and consequent, are equal to one another, but the latter, to two or more ratios, when they are equal. - i The ratios which form the subject of the propositions of the Fifth Book are the ratios of finite magnitudes, which are not supposed to admit of any variation. No property is founded on a single ratio, nor on the magnitudes which constitute a ratio ; but all the reasonings on ratios in this Book are restricted to two or more ratios, as being equal to, greater than, or less than one another. The question— what do ratios become when one of the terms, or both of the terms of the ratios are increased or diminished indefinitely, does not form any part of the subject of the Fifth Book of the Elements. - The simple idea of ratio itself, absolutely considered, could not, in anyway, lead to any conclusion respecting the properties of figures, any more than the mere idea of magnitude. It is by the comparison of two or more magnitudes subjected to some specifted conditions in the first Four Books of the Elements, that all the propo- sitions have been demonstrated: and it is by the comparison of the ratios of two or nore pairs of correlative magnitudes, subject to specified conditions, that the properties of figures depending on ratio are to be established. As each ratio involves the idea of two magnitudes, the least number of magnitudes between which a comparison of ratios is possible, is four, two for each of the ratios; and when these ratios are equal, the sixth definition gives the name of proportionals to the four magnitudes which constitute the two ratios. - Arithmetical ratio has been defined to be the relation which one number bears to another with respect to quotity; the comparison being made by considering what multiple, part or parts, one number is of the other. - An arithmetical ratio, therefore, is represented by the quotient which arises from dividing the antecedent by the consequent of the ratio; or by the fraction which has the antecedent for its numerator and the consequent for its denominator, Hence it will at once be obvious that the properties of arithmetical ratios will be made to depend on the properties of fractions. It must ever be borne in mind that the subject of Geometry is not number, but the magnitude of lines, angles, surfaces, and solids; and its object is to demonstrate their properties by a comparison of their absolute and relative magnitudes. Also, in Geometry, multiplication is only a repeated addition of the same mag- nitude ; and division is only a repeated subtraction, or the taking of a less magnitude successively from a greater, until there be either no remainder, or a remainder less than the magnitude which is successively subtracted. The Geometrical ratio of any two given magnitudes of the same kind will obviously be represented by the magnitudes themselves; thus, the ratio of two lines is represented by the lengths of the lines themselves; and, in the same manner, the ratio of two angles, two surfaces, or two solids, will be properly represented by the magnitudes themselves. - In the definition of ratio as given by Euclid, all reference to a third magnitude of the same geometrical species, by means of which, to compare the two, whose ratio is the subject of conception, has been carefully avoided. The ratio of the two magnitudes is their relation one to the other, without the intervention of any standard unit whatever, and all the propositions demonstrated in the Fifth Book respecting NOTES To BOOK v. 187 - the equality or inequality of two or more ratios, are demonstrated independently of any knowledge of the exact numerical measures of theratios; and their generality includes all ratios, whatever distinctions may be made, as to the terms of them being commensurable or incommensurable. In measuring any magnitude, it is obvious that a magnitude of the same kind must be used; but the ratio of two magnitudes may be measured by every thing which has the property of quantity. Two straight lines will measure the ratio of two triangles, or parallelograms (VI.1. fig.): and two triangles, or two parallelo- grams will measure the ratio of two straight lines. It would manifestly be absurd to speak of the line as measuring the triangle, or the triangle measuring the line. (See notes on Book II.) The ratio of any two quantities depends on their relative and not their absolute magnitudes; and it is possible for the absolute magnitude of two quantities to be changed, and their relative magnitude to continue the same as before ; and thus, the same ratio may subsist between two given magnitudes, and any other two of the same kind. - In this method of measuring Geometrical ratios, the measures of the ratios are the same in number as the magnitudes themselves. It has however two advantages; first, it enables us to pass from one kind of magnitude to another, and thus, independently of any numerical measure, to institute a comparison between such magnitudes as cannot be directly compared with one another ; and secondly, the ratio of two magnitudes of the same kind may be measured by two straight lines, which form a simpler measure of ratios than any other kind of magnitude. - But the simplest method of all would be, to express the measure of the ratio of two magnitudes by one; but this cannot be done, unless the two magnitudes are commensurable. If there be two lines, one of which AB contains 12 units of any length, and the other CD contains 4 units of the same length; then the ratio of the line AB to the line CD, is the same as the ratio of the number 12 to 4. Thus, two numbers may represent the ratio of two lines when the lines are commensurable. In the same manner, two numbers may represent the ratio of two angles, two sur- faces, or two solids. Thus, the ratio of any two magnitudes of the same kind may be expressed by two numbers, when the magnitudes are commensurable. By this means, the consideration of the ratio of two magnitudes is changed to the consideration of the ratio of two numbers, and when one number is divided by the other, the quotient will be a single number, or a fraction, which will be a measure of the ratio of the two numbers, and therefore of the two quantities. If 12 be divided by 4, the quotient is 3, which measures the ratio of the two numbers 12 and 4. Again, if besides the ratio of the lines A B and CD which contain 12 and 4 units respectively, we consider . two other lines EF and GH which contain 9 and 3 units respectiyely; it is obvious that the ratio of the line EF to GH is the same as the ratio of the number 9 to the number 3. And the measure of the ratio of 9 to 3 is 3. That is, the numbers 9 and 3 have the same ratio as the numbers 12 and 4. But this is a numerical measure of ratio, and can only be applied strictly when, the antecedent and consequent are to one another as one number to another. And generally, if the two lines AB, CD contain a and b units respectively, and g. be the quotient which indicates the number of times the number b is contained in a, then q is the measure of the ratio of the two numbers a and b : and if EF and GH contain c and dunits, and the number d be contained a times in c : the number a has to b the same ratio as the number c has to d. This is the numerical definition of proportion, which is thus expressed in Euclid's Elements, Book VII, definition 20. “Four numbers are proportionals when the first. I 188 . EUCLID's ELEMENTS. is the same multiple of the second, or the same part or parts of it, as the third is of the fourth.” This definition of the proportion of four numbers, leads at once to an equation : * * for, since a contains b, g times; # = g : ; •- º tº gº © and since c contains d, g times; # = q : * # = # wheh is the fundamental equation upon which all the reasonings on the proportion of numbers depend. - If four numbers be proportionals, the product of the extremes is equal to the product of the means. - - • For if a, b, c, d be proportionals, or a b :: c : d, therefore (! C. Then # = , ; - bd cbd; Multiply these equals by bà, .”. º: =: f 2 or, ad = be, that is, the product of the extremes is equal to the product of the means. And conversely, If the product of the two extremes be equal to the product of the two means, the four numbers are proportionals. IFor if a, b, c, d be four quantities, such that ad = be, then dividing these equals by bà, therefore; == #. and a . b :: c : d, or the first number has the same ratio to the second, as the third has to the fourth. If c = b, then ad = b%; and conversely if ad = bº: then . = º These results are analogous to Props. 16 and 17 of the Sixth Book. Def. v. This definition lays down a criterion by which two ratios may be known to be equal, or four magnitudes proportionals, without involving any inquiry re- specting the four quantities, whether the antecedents of the ratios contain, or are contained in, their consequents exactly; or whether there are any magnitudes which measure the terms of the two ratios. The criterion only requires, that the relation of the equimultiples expressed should hold good, not merely for any particular multiples, as the doubles or trebles, but for any multiples whatever, whether large or small. - - This criterion of proportion may be applied to all Geometrical magnitudes which can be multiplied, that is, to all which can be doubled, trebled, quadrupled, &c. But it must be borne in mind, that this criterion does not exhibit a definite measure for either of the two ratios which constitute the proportion, but only, an undetermined measure for the sameness or equality of the two ratios. The nature of the proportion of Geometrical magnitudes neither requires nor admits of a numerical measure of either of the two ratios, for this would be to suppose that all magnitudes are commensurable. Though we know not the definite measure of either of the ratios, further than that they are both equal, and one may be taken as the measure of the other, yet particular conclusions may be arrived at by this method : for by the test of proportionality here laid down, it can be proved that one magnitude is greater than, equal to, or less than another: that a third proportional can be found to two, and a fourth proportional to three straight lines, also that a mean proportional can be found between two straight lines: and NOTES TO Book v. 189 . further, that which is here stated of straight lines may be extended to other Geometrical magnitudes. The fifth definition is that of equal ratios. The definition of ratio itself (defs. 3, 4) contains no criterion by which one ratio may be known to be equal to another ratio: analogous to that by which one magnitude is known to be equal to another magnitude (Euc. I. Ax. 8). The preceding definitions (3, 4) only restrict the conception of ratio within certain limits, but lay down no test for comparison, or the deduction of properties. All Euclid's reasonings were to turn upon this com- parison of ratios, and hence it was competent to lay down a criterion of equality and inequality of two ratios between two pairs of magnitudes. In short, his effective definition is a definition of proportionals. The precision with which this definition is expressed, considering the number of conditions involved in it, is remarkable. Like all complete definitions the terms (the subject and predicate) are convertible: that is, (a) If four magnitudes be proportionals, and any equimultiples be taken as prescribed, they shall have the specified relations with respect to “greater, greater,” &c. - - (b) If of four magnitudes, two and two of the same Geometrical Species, it can be shewn that the prescribed equimultiples being taken, the conditions under which those magnitudes exist, must be such as to fulfil the criterion “greater, greater, &c.”; then these four magnitudes shall be proportionals. It may be remarked, that the cases in which the second part of the criterion (“equal, equal”) can be fulfilled, are comparatively few : namely those in which the given magnitudes, whose ratio is under consideration, are both exact multiples of some third magnitude—or those which are called commensurable. When this, however, is fulfilled, the other two will be fulfilled as a consequence of this. When this is not the case, or the magnitudes are incommensurable, the other two criteria determine the proportionality. However, when no hypothesis respecting com- mensurability is involved, the contemporaneous existence of the three cases (“greater, greater; equal, equal; less, less”) must be deduced from the hypothe- tical conditions under which the magnitudes exist, to render the criterion valid. With respect to this test or criterion of the proportionality of four magnitudes, it has been objected, that it is utterly impossible to make trial of all the possible equimultiples of the first and third magnitudes, and also of the second and fourth. It may be replied, that the point in question is not determined by making such trials, but by shewing from the nature of the magnitudes, that whatever be the multipliers, if the multiple of the first exceeds the multiple of the second magni- tude, the multiple of the third will exceed the multiple of the fourth magnitude, and if equal, will be equal; and if less, will be less, in any case which may be taken. - The Arithmetical definition of proportion in Book vii, Def. 20, even if it were equally general with the Geometrical definition in Book v, Def. 5, is by no means universally applicable to the subject of Geometrical magnitudes. The Geome- -trical criterion is founded on multiplication, which is always possible. When the magnitudes are commensurable, the multiples of the first and second may be equal or unequal: but when the magnitudes are incommensurable, any multiples whatever of the first and second must be unequal: but the Arithmetical criterion of propor- tion is founded on division, which is not always possible. Euclid has not shewn in Book v, how to take any part of a line or other magnitude, or that the two terms of a ratio have a common measure, and therefore the numerical definition could not be strictly applied, even in the limited way in which it may be applied. Number and Mognitude do not correspond in all their relations; and hence the 190 EUCLID's ELEMENTs. - - distinction between Geometrical ratio and Arithmetical ratio; the former is a comparison karð armºukórnºra, according to quantity, but the latter, according to quotity. The former gives an undetermined, though definite measure, in magni- tudes; but the latter attempts to give the exact value in numbers. The fifth book exhibits no method whereby two magnitudes may be determined to be commensurable, and the Geometrical conclusions deduced from the multiples of magnitudes are too general to furnish a numerical measure of ratios, being all independent of the commensurability or incommensurability of the magnitudes themselves. s It is the numerical ratio of two magnitudes which will more certainly discover whether they are commensurable or incommensurable, and hence, recourse must be had to the forms and properties of numbers. All numbers and fractions are either rational or irrational. It has been seen that rational numbers and fractions can ea:press the ratios of Geometrical magnitudes, when they are commensurable. Similar relations of incommensurable magnitudes may be expressed by irrational numbers, if the Algebraical expressions for such numbers may be assumed and employed in the same manner as rational numbers. The irrational expressions being considered the exact and definite, though undetermined, values of the ratios, to which a series of rational numbers may successively approximate. Though two incommensurable magnitudes have not an assignable numerical ratio to one another, yet they have a certain definite ratio to one another, and two other magnitudes may have the same ratio as the first two : and it will be found, that, when reference is made to the numerical value of the ratios of four incom- mensurable magnitudes, the same irrational number appears in the two ratios. - The sides and diagonals of squares can be shewn to be proportionals, and though the ratio of the side to the diagonal is represented Geometrically by the , two lines which form the side and the diagonal, there is no rational number or fraction which will measure exactly their ratio. - If the side of a square contain a units, the ratio of the diagonal to the side is numerically as W2 to 1; and if the side of another square contain b units, the ratio of the diagonal to the side will be found to be in the ratio of V2 to 1. Again, the two parts of any number of lines which may be divided in extreme and mean ratio will be found to be respectively in the ratio of the irrational number V6 – 1 to 3 – A/5. Also, the ratios of the diagonals of cubes to the diagonals of one of the faces will be found to be in the irrational or incommensurate ratio of V3 to V2. ... • Thus it will be found that the ratios of all incommensurable magnitudes which are proportionals do involve the same irrational numbers, and these may be used as the numerical measures of ratios in the same manner as rational numbers and fractions. * It is not however to such enquiries, nor to the ratios of magnitudes when ex- pressed as rational or irrational numbers, that Euclid's doctrine of proportion is legitimately directed. There is no enquiry into what a ratio is in numbers, but whether in diagrams formed according to assigned conditions, the ratios between certain parts of the one are the same as the ratios between corresponding parts of the other. Thus, with respect to any two squares, the question that properly be- longs to pure Geometry is:—whether the diagonals of two squares have the same ratio as the sides of the squares? Or whether the side of one square has to its diagonal, the same ratio as the side of the other square has to its diagonal Or again, whether in Euc. vi. 2, when BC and DE are parallel, the line BD has to the line DA, the same ratio that the line CE has to the line AE? There is no purpose on the part of Euclid, to assign either of these ratios in numbers; but only to prove that their universal sameness is inevitably a consequence of the t NOTES TO BOOK W. 191 \ original conditions according to which the diagrams were constituted. There is, consequently, no introduction of the idea of incommensurables: and indeed, with such an object as Euclid had in view, the simple mention of them would have been at least irrelevant and superfluous. If however it be attempted to apply nume- rical considerations to pure geometrical investigations, incommensurables will soon be apparent, and difficulties will arise which were not foreseen. Euclid, however, effects his demonstrations without creating this artificial difficulty, or even recognising its existence. Had he assumed a standard-unit of length, he - would have involved the subject in numerical considerations; and entailed upon the subject of Geometry the almost insuperable difficulties which attach to all ‘such methods. - It cannot, however, be too strongly or too frequently impressed upon the 'learner's mind, that all Euclid's reasonings are independent of the numerical ex- positions of the magnitudes concerned. That the enquiry as to what numerical function any magnitude is of another, belongs not to Pure Geometry, but to another Science—a Science, which because Geometrical magnitudes may be, and for many purposes conveniently are subjected to its mechanism, is yet very liable to have its principles confounded with those of Geometry in the mind of the learner. This liability is increased by the interchange of the terms of the one Science with the written notation of the other. The true distinction is, that into Pure Geo- metry, the intermediate standard-unit does not enter; into Algebraic Geometry, it essentially enters, and indeed constitutes the fundamental idea. The former is wholly free from numerical considerations; the latter is entirely dependent upon them. - Def. VI. The fourth magnitude is called a fourth proportional to the first, second and third magnitudes. The first and second magnitudes must be of the same kind, as also the third and fourth, when the four magnitudes form a proportion. All the four magnitudes may be of the same kind, but there is no absolute necessity requiring that condi- tion, in order that the four magnitudes may be proportionals. Def. VII is analogous to Def. 5, and lays down the criterion whereby the ratio of two magnitudes of the same kind may be known to be greater or less than the ratio of two other magnitudes of the same kind. - Def. VIII. “Analogy or proportion, is the similitude of ratios,” is the render- ing given by Simson of 'Ava\oyia 65 orw i tāv Āóyov Tavrátns. This rendering has been objected to, as similitude or similarity admits of degrees, and cannot with propriety be applied to two ratios which constitute a proportion ; besides j artów Aöywv ravitörns, the sameness or the identity of the ratios, forms the subject of the definition, of which subject &va\oyia is asserted as the name. The word sameness or identity has also been objected to as not clearly expressing Euclid's meaning. It is urged that two things really distinct cannot be said to be the same : it must however be observed, that it is not a sameness of two magnitudes, but the same- ness of the mutual relation of two magnitudes and of two others, which is asserted to constitute an analogy or a proportion. “The equality of two ratios” has been suggested as the most proper rendering; the word equality in Geometry is applied to such magnitudes as can be so placed as to be coincident, or which by a division of their parts may be made to coincide. Besides the equality of ratios, is an equality of mutual relations, and implies the idea of the arithmetical values of the ratios. The word equivalent has also been proposed as less liable to objestion than the word equal. The sameness or identity of the ratios, appears on the whole to be pre- ferable to the similitude of ratios. Def. Ix. states that a proportion consists in three terms at least ; the meaning 192 EUCLID's ELEMENTS. of which is, that the second magnitude is repeated, being made the consequent of the first, and the antecedent of the second ratio. It is also obvious that when a proportion censists of three magnitudes, all three are of the same kind ; and the first is said to have the same ratio to the second as the second has to the third. The second is called a mean proportional between the first and second: and the third is called a third proportional to the first and second magnitudes. The three following convenient definitions have been given, when three quan- tities are in Arithmetical, Geometrical and Harmonical preportion. 1. Three quantities are in Arithmetical proportion when the difference be- tween the first and second is to the difference between the second and third, as the first to the first, *** 2. Three quantities are in Geometrical proportion, when the difference be- tween the first and second, is to the difference between the second and third, as the first is to the second. 3. Three quantities are in Harmonical proportion, when the difference be- tween the first and second is to the difference between the second and third, as the first is to the third. - - * Thus, if a, b, c be three quantities. Then a – b : b – c :: a a, when they are in Arithmetical proportion, " a – b : b – c :: a b, when they are in Geometrical proportion, a – 5 ; b – c :: a e, when they are in Harmonical proportion. T)ef. xI. includes Def. x. as three magnitudes may be continued proportionals, as well as four or more than four. In continued proportionals, all the terms ex- cept the first and last, are made successively the consequent of one ratio, and the -antecedent of the next; whereas in other proportionals this is not the case. A series of numbers or Algebraical quantities in continued proportion, is called a Geometrical progression, from the analogy they bear to a series of Geometrical magnitudes in continued proportion. Def. A. The term compound ratio was devised for the purpose of avoiding cir- cumlocution, and no difficulty can arise in the use of it, if its exact meaning be strictly attended to. ~~ With respect to the Geometrical measures of compound ratios, three straight lines may measure the ratio of four, as in Prop. 23, Book v1. For K to L measures the ratio of BC to CG, and L to M measures the ratio of DC to CE; and the ratio of K to M is that which is said to be compounded of the ratios of K to L, and L to M, which is the same as the ratio which is compounded of the ratios of the sides of the parallelograms. • * Both duplicate and triplicateratio are species of compound ratio: for duplicate ratio is a ratio compounded of two equal ratios; and in the case of three magni- tudes which are continued proportionals, means the ratio of the first magnitude, to a third proportional to the first and second. * Triplicate ratio, in the same manner, is a ratio compounded of three equal' , ratios; and in the case of four magnitudes which are continued proportionals, the triplicate ratio of the first to the second means the ratio of the first to a fourth proportional to the first, second, and third magnitudes. Instances of the compo- sition of three ratios, and of triplicate ratio, will be found in the eleventh and twelfth books. - -- The product of the fractions which represent or measure the ratios of numbers, corresponds to the composition of Geometrical ratios of magnitudes. - It has been shewn that the ratio of two numbers is represented by a fraction whereof the numerator is the antecedent, and the denominator the consequent of the ratio; and if the antecedents of two ratios be multiplied together, as also the NOTES TO BOOK W. . . . 193 consequents, the new ratio thus formed is said to be compounded of these two ratios; and in the same manner, if there be more than two. It is also obvious, that the ratio compounded of two equal ratios is equal to the ratio of the squares of one of the antecedents to its consequent ; also when there are three equal ratios, the ratio compounded of the three ratios is equal to the ratio of the cubes of any one of the antecedents to its consequent. And further, it may be observed, that when several numbers are continued proportionals, the ratio of the first to the last is equal to the ratio of the product of all the antecedents to the product of all the consequents. - It may be here remarked, that, though the constructions of the propositions in Book v. are exhibited by straight lines, the enunciations are expressed of magni- tude in general, and are equally true of angles, triangles, parallelograms, arcs, Sectors, &c. - z The two following arioms may be added to the four Euclid has given. Ax. 5. A part of a greater magnitude is greater than the same part of a less magnitude. - - Ax. 6. That magnitude of which any part is greater than the same part of another, is greater than that other magnitude. The learner must not forget that the capital letters, used generally by Euclid in the demonstrations of the Fifth Book, represent the magnitudes, not any numerical or Algebraical measures of them : sometimes however the magnitude of a line is represented in the usual way by two letters which are placed at the extremities of the line. - Prop. I. Afgebraically. Let each of the magnitudes A, B, C, &c. be equimultiples of as many a, b, c, &c. . that is, let A = m times a = ma, - B = m times b = mb, C = m times c = me, &c. First, if there be two magnitudes equimultiples of two others, then A + B = ma + mb = m (a + b) = m times (a + i). Hence A + B is the same multiple of (a + b), as A is of a, or B of b. Secondly, if there be three magnitudes equimultiples of three others,. then 4 + B + C = ma + mb + me = m (a + b + c) = m times (a + b + c). Hence A + B + C is the same multiple of (a + b + c); as A is of a, B of b, and C of c. Similarly, if there were four, or any number of magnitudes. Therefore, if any number of magnitudes be equimultiples of as many, each of each ; what multiple soever, any one is of its part, the same multiple shall the first magnitudes be of all the other. Prop. II, Algebraically. Let A, the first magnitude, be the same multiple of a, the second, as As the third, is of a, the fourth ; and As the fifth the same multiple of as the second, as A, the sixth, is of a, the fourth. - That is, let A1 = m times as = max, As = m times a, = ma), A. = n times as - naº, - As = n times a 4 = n.a4. Then by addition, A, + A. = ma + na = (m + n) aa = (m + n) times a, and As + As = ma; + na; = (m -- n) as = (m + n) times as: O 194 EUCLID'S ELEMENTS. ~. Therefore A1 + A, is the same multiple of ag, as 43 + 4, is of a. That is, if the first magnitude be the same multiple of the second, as the third is of the fourth, &c. CoR. If there be any number of magnitudes A1, A2, A3, &c. multiples of another a, such that A1 = ma, A2 = n.a., As = pa. &c. - And as many others B1, B2, Bs, &c. the same multiples of another b, such that S B1 = mb, B2 = n.5, Bs = pb, &c. - w Then by addition, A, 4- A, + As + &c. = ma + na + pa + &c. = (m. -- n + p + &c.) a = (m -- n + p + &c.) times a : and B, -- B, 4- Bs + &c. = mb + nb + pb + &c. = (in + n + p + &c.) b * = (m. -- n + p + &c.) times b : that is 41 + 42 + As + &c. is the same multiple of a that - B1 + B2 + Bs + &c., is of b. Prop. III. Algebraically. Let A, the first magnitude, be the same multiple of as the second, -- as A3 the third, is of at the fourth, that is, let A1 = m times az = finae, and A3 = m times aa = maa. If these equals be each taken n times, * then n A1 = mnaz = mn times as, and nA3 = mna; - mn times at, or n A1, nAs each contain as, as respectively mn times. - Wherefore nA, n.4, the equimultiples of the first and third, are respectively equimultiples of as and aa, the second and fourth. - Prop. Iv. Algebraically. . Let A1, as, As, aa, be proportionals according to the Algebraical definition : that is, let 41 : a, , ; 43 : aa, A. A then tº = **, 02 (14 multiply these equals by #, n and n being any integers, ... m.A. m. As na, T maa” or m A1 : nag :: mA3: nas. - That is, if the first of four magnitudes has the same ratio to the second which the third has to the fourth; then any equimultiples whatever of the first and third shall have the same ratio to any equimultiples of the second and fourth. The Corollary is contained in the proposition itself: for if n be unity, then m A. : as :: mA3 : as : - and if m be unity, also A1 : naz :: 43 : naa. Prop. v. Algebraically. \ Let Al be the same multiple of a1, that As a part of A1, is of az, a part of ai. Thén A1 – A2 is the same multiple of al – as, as A1 is of ai. - For let A1 = m times ai = mai, and A2 = m times az = maa, - then A1 – A2 = mai — maz = m (a1 – a,) = m times (a — ae), that is, 41 – A, is the same multiple of (a, - a.), as A, is of ai. Prop. VI. Algebraically. - Let 41, 42 be equimultiples respectively of a1, a2, two others, NOTES To Book v. 195 *~ * that is, let A1 = m times ai = mai, and 42 = m times as = max. Also if B1 a part of A1 = n times ai = nai, and B, a part of A2 = n times as = nas. Then by taking equals from equals, - ‘. 41 – B1 = ma) – naï = (m – n) ai = (m — n) times ai, As – B2 = maz – nax = (m – n) aa = (m — n) times as : that is, the remainders A1 – B1, A2 – B2 are equimultiples of ar, a, respectively. And if m = n = 1, then A1 – B1 = a1, and A2 – B2 = a, : - or the remainders are equal to ai, as respectively, z Prop. A. Algebraically. * Let 41, a2, 4s, as be proportionals, or 41 : a, , ; 43 : a, then 4, = 4s ſº 02 04 And since the fraction # is equal to : , the following relations only can gº 4 subsist between A1 and a2 ; and between As and a1. First, if A, be greater than as ; then As is also greater than a.. Secondly, if Al be equal to as ; then As is also equal to a,. Thirdly, if Al be less than a, ; then Aa is also less than a. 4s Otherwise, the fraction #. could not be equal to the fraction a." 4. Prop. B. Algebraically Let 41, a2, 43, as be proportionals, or 4 ; a As as: w Then shall as : Ai :: a, : As. For since 4; : «» :: As : ae . 41 43 ... ... = a, , and if 1 be divided by each of these equals, , - 41 – 1 - 43 = 1 + → , 02 04 or º == % , and therefore as : Ai :: a, ;4, The student must however bear in mind, that the Algebraical definition is not equally applicable to the Geometrical demonstrations contained in the sixth, eleventh, and twelfth Books of Euclid, where the Geometrical definition is em- ployed. It has been before remarked, that Geometry is the science of magnitude and not of number; and though a sum and a difference of two magnitudes can be represented Geometrically, as well as a multiple of any given magnitude, there is no method in Geometry whereby the quotient of two magnitudes of the same kind can be expressed. The idea of a quotient is entirely foreign to the principles of the Fifth Book, as are also any distinctions of magnitudes as being commensurable or incommensurable. As Euclid in Books VII—x has treated of the properties of proportion according to the Arithmetical definition, and df their application to Geometrical magnitudes ; there can be no doubt that his intention was to exclude all reference to numerical measures and quotients in his treatment of the doctrine of proportion in the Fifth Book; and in his applications of that doctrine in the sixth, eleventh and twelfth Books of the Elements. It may however be shewn 0-2 *196 EUCLID'S ELEMENTS. that the Geometrical definition of proportion is a consequence of the Arithmetical, definition, and conversely. If four magnitudes are proportionals according to the Arithmetical definition, they are also proportionals according to the Geometrical definition of Proportion. For if a, b, c, d represent numerically four magnitudes which are proportional according to the Arithmetical definition; the ratio of a to b is equal to the ratio of c to d, - *. or t = } - - 5 T 3 * If m, n be any integral numbers whatever, Asº 372 CE, 7%6 then nb := rid 9 Now since these fractions are always equal, the following relations only can subsist, between ma, nb, mc, nd. First, Ifna be greater than nē, then me is greater than na. Secondly, If ma be equal to nb, then me is equal to nd. Thirdly, If ma be less than mb, then me is less then nd. That is, of the four terms of a proportion a, b, c, d, if any equimultiples whatever, ma, me of the first and third be taken, and any equimultiples whatever mb, nd of the second and fourth; if ma the multiple of the first be greater than mb the multiple of the second, then me the multiple of the third is greater than nd the multiple of the fourth; if equal, equal; and if less, less. Conversely. If four magnitudes are proportionals according to the Geometrical definition, they are also proportionals according to the Arithmetical definition of proportion. *. - For if a, b, c, d be four magnitudes, and when any equimultiples whatever ma, mc, of a and c the first and third are taken, and any whatever mb, nd, of b and d the second and fourth; it is found that if ma the multiple of the first be greater than nb the multiple of the second, then me the multiple of the third is greater than na the multiple of the fourth; and if equal, equal; if less, less : then sº a, b, c, d are so related that . - à. or that a, b, c, d are proportionals according to the Arithmetical definition of proportion. For if not, let a, b, c, d", be proportionals, . Q. C. so that § - d. ' ma _ me . 7.5 T ind” %0, 7?? 6. *; md If ma be greater than, equal to, or less than nē, then me will be greater than, equal to, or less than nd', © whence nd is equal to nd, or dº is equal to d, - and consequently a, b, c, d are proportionals according to the Arithmetical defini- * tion of proportion. | - The following is the method by which Simson shews that the Geometrical definition of proportion is a consequence of the Arithmetical definition, and conversely. . i s “First, if A, B, C, D, be four magnitudes, such that A is the same multiple, or the same part of B, which C is of D : Multiplying these equals by: ; :". and since the fraction : is always equal to NOTES To Book v. 197 Then A, B, C, D, are proportionals: this is demonstrated in proposition (c). * Secondly, if AB contain the same parts of CD that EF does of GH; in this case likewise AB is to CD, as EF to G.H. A B E F * C. K. D. G - L. H. ~—º-— ſº Let CK be a part of CD, and GL the same part of GH; and let AB be the same multiple of CK, that EF is of GL; therefore, by Prop. c, of Book v, AB is to CK, as EF to GL : and CD, GH, are equimultiples of CK, GL, the second and fourth; wherefore, by Cor. Prop. 4, Book v, AB is to CD, as EF to G.H. And if four magnitudes be proportionals according to the 5th def. of Book v, they are also proportionals according to the 20th def. of Book v11. Y - First, if A be to B, as C to D; then if A be any multiple or part of B, C is the same multiple or part of D by Prop. D, Book v. - - - Next, if AB be to CD, as EF to GH: then if AB contain any part of CD, EF contains the same part of GH: A B E F M. C K D G L H for let CK be a part of CD, and GL the same part of GH, *- and let AB be a multiple of CK; - - EF is the same multiple of GL : take M the same multiple of GL that AB is of CK; therefore, by Prop. c, Book v, AB is to CK, as M to GL : - and CD, GH, are equimultiples of CK, GL; wherefore, by Cor. Prop. 4, Book v, AB is to CD, as M to GH. And, by the hypothesis, AB is to CD, as EF to GH; ſº therefore M is equal to EF by Prop. 9, Book v, and consequently, EF is the same multiple of GL that AB is of CK.” Prop. C. Algebraically. Let 41, a2, As, a, be four magnitudes. * First. Tet A1 = max, and As = max. Then 41 az : : As : aa. o A. 02 and As = mas, ..". 7% = **s O - 04 - A. A Hence : = +8, 02 04 and A1 : a, , ; 4 s : aa. I 1 Secondly. Let A1 = ; %, and A3 = ; *. - A I Then, as before, # = + , and 4s = l Q. - a2 m. a'i m. Hence 41 == #, - * 4. and Ai : ao ; ; 43 : a,. - * T Q 198 EUCLID's ELEMENTs. Prop. D. Algebraically. Let 41, a2, 43, a, be proportionals, or 41 : as :: As : a. First. Let 41 be a multiple of a., or A1 = m times ai = mai. Then shall 43 = ma). - For since A1 A. A. : az : : 43 : aa, ... t1 = f°: 04 º %20. A but since A1 = max, * = **, or m = +, Qa Q4 04 and As = ma; ; therefore the third As is the same multiple of as the fourth. Secondly. If A1 = cºme I in “s, then shall As ; %. tº A A For since tº = f°: 04 A 1 and A1 = - az, .". # T m tº 4s 1. wherefore, the third A3 is the same part of the fourth a 4. Prop. VII. is so obvious that it may be considered axiomatic, when it is considered in reference to the Algebraical definition. A similar remark may be made with respect to the eighth and ninth propositions. - Prop. x. Algebraically. Let A1 have a greater ratio to a, than As has to a. Then 41 > 4s. For the ratio of A1 to a is represented by #, and the ratio of As to a is represented by :* 2 and since 4. > 43; Q. Q, it follows that A, - A, - Secondly. Ilet a have to Aa a greater ratio than a has to Aº, Then A3 < A1. For the ratio of a . As is represented by # 2 3. and the ratio of a ; A1 is represented by #. 9. 1. and since 4 a. dividing these unequals by a, • * ~ * , ſº tº As > 41 ! and multiplying these unequals by 41 x 43, ... 4 × 4, or As & A1. NOTES TO BOOK W. 199 Let the ratio of A. : a, be the same as the ratio of A, ; @4, and the ratio of As: a, be the same as the ratio of A, ; a. Then the ratio of Al: a, shall be the same as the ratio of A Prop. x1. Algebraically. 5 : 0s- gº A. A For since A1: a, ; : As: a, ... + = **, 02 04 º * A A. and since As: a, ::A; : d's, ". f** = ***. 04 4s A. A. Hence # = f°, 02 06 and 41; as :: As : as . . Prop. xII. Algebraically. - Let 41, a2, Ag, aw, A., a, be proportionals, so that A1 : a, ; : As : a, : ;4, 6, Then shall 41: a, ; ; 4 + 4, + A. : a, + as + as. Tor since 41: as :: Aa: as :: A, ; as, . 41 - 4a – As as a, 06 . A. A - And •." :*l == *3 § ..". 41a, =: a24s, 0:2 04 A 4s and * = **, ... Ala, - as A, (22 06 also A1a2 = az41. Hence A1 (a, -i- a, + ad) = az (Ai + As +45), by addition, and dividing these equals by as (az + as + ag), . 41 – 4 + 43 + 4s. ‘a, T as -- a-- a, ’ and A. : as :: A + 43 + 4, ; a + a, + as: Prop. XIII. Algebraically. Let A1, a, As, as, 4s, as, be six magnitudes, such that A1: a, : As: aa, but that the ratio of Aa : a, is greater than the ratio of A, : as: Then the ratio of A, ; as shall be greater than the ratio of A, ; as: 41 – 4s For since 4.1 : as :: As : aa, .". a, a.” c wo 43 4s but since As: a* > As : as .”. † - a. * A. A Hence tº > **. 06 That is, the ratio of A1 : a, is greater than the ratio of 4, as: Prop. xiv. Algebraically. te - Let A1, a2, A, as be proportionals. If A, - A, then a, - a, and if equal, equal; and if less, less, For since A1: aa : 43; as - . 41 – 43 o A &Z Multiply these equals by #, ..". . = #. : - 3 and because these fractions are always equal, 3. * 200 g EUCLID's ELEMENTS. * if A, be > 4. then az must be greater than a4, for if a., were not greater than a4, the fraction * could not be equal to 41 5. a 43 which would be contrary to the hypothesis. In the same manner, if A, be = As, then as must be equal to as, and if Al be < 43, as must be less than aa. IHence, therefore, if &c. Prop. xv. Algebraically. - Let A1, as be any magnitudes of the same kind; then 41 : as :: mA1 : mas; nai and maz being any equimultiples of Ai and as. and since the numerator and denominator of a fraction may be multiplied by the same number without altering the value of the fraction, . . 41 m41. Q r * as T mas and A1 : a, , ; mA, ; mas. Drop. xvi. Algebraically. Let A1, a2, A3, a, be four magnitudes of the same kind, which are proportionals, 41 : az : 4s 3 a. Then these shall be proportionals when taken alternately, that is, A, As is a a For since A. : as :: As : aa, then # 4. (l, ;i. Multiply these equals by a2 . .". * = 43 48 . and 41 : 43 :: aa a. Prop. xv.11. Algebraically. Let A1 + ag, a2, 43 + aa, as be proportionals, then Ai, az, As, a, shall be proportionals. For since A1 + as : a, :: 4. + 24 : dº 41 + q2 = 4s H 24 * 2 0's 04 Or 41 + 1 = 43 + 1, 08. 04 g A1 A3 and taking 1 from each of these equals, .". ... = a, , - 2 * *4 and 41 : a, As a. Prop. xviii, is the converse of Prop. xv.11. The following is Euclid's indirect demonstration Let AE, EB, CF, FD be proportionals, that is, as AE to EB, so let CF be to FD. Then these shall be proportionals also when taken jointly; that is, as AB to BE, so shall CD be to D.F. - NOTES TO BOOK W. - 201 A E B C Q F D For if the ratio of AB to BE be not the same as the ratio of CD to DF; the ratio of AB to BE is either greater than, or less than the ratio of CD to DF. First, let AB have to BE a less ratio than CD has to DF; and let DQ be taken so that AB has to BE the same ratio as CD to DQ': and since magnitudes when taken jointly are proportionals, they are also proportionals when taken separately; (v. 17.) therefore AE has to EB the same ratio as CQ to QD ; but, by the hypothesis, AE has to EB the same ratio as CF to FD; therefore the ratio of CQ to QD is the same as the ratio of CF to FD. (v. 11.) And when four magnitudes are proportionals, if the first be greater than the second, the third is greater than the fourth ; and if equal, equal; and if less, less;, (v. 14.) gº but GQ is less than CF, . therefore QD is less than FD ; which is absurd. Wherefore the ratio of AB to BE is not less than the ratio of CD to DF; that is, AB has the same ratio to BE as CD has to D.F. Secondly. By a similar mode of reasoning, it may likewise be shewn, that AB has the same ratio to BE as CD has to DF, if AB be assumed to have to B.E. a greater ratio than CD has to DF. sº tº Prop. xvi II. Algebraically. Let 41 : aa : As : a,. Then 41 + as as :: 43 + aq.; a 4. Al Aa For since Ai : a, ; ; As : aa, ‘. ... = a, te A A and adding 1 to each of these equals, .". i. + 1 = . + 1, - g 4. A1 + as 43 + as r == 3. 0's 0.4 - and A1 + ao ; as tº 43 + as : aa, O Prop. xIx. Algebraically. Let the whole A1 have the same ratio to the whole A2, as ai taken from the first, is to as taken from the second, that is, let A1 : As :: a. $ 92- Then 41 – at : 42 – as :: Ai i A3. e - A. (Z For since 41 : 42 :: ai : az, .”. #. t= . tº sº tº - - A A. A. di A Multiplying these equals by tº, ... : x +3 = * x #3; W. plying q y al 9 4's l 02 X al y 41 As Or — = — &1 az . . . Al 42 . and subtracting 1 from each of these equals, ...". zº – 1 = z - l, l 2 4. – t- Of 1 di == 42 Qa - 0.1 Q2 -- - (º, 4. – a & ſº ſº 1 1 •". l I s: *1 and multiplying these equals by A, -a,” “A, - a, Ta,' A - but # = *, - - - 42 Q 2 202 EUCLID's ELEMENTs. A1 – at A1 dº e A. Fa, == A. 9 • and A1 – at : A2 – as :: Ai : 43. Cor. If A1 : A, ; ; ai : aa. - Then AI — a 1: As – as :: a, ; az, is found proved in the preceding process. Prop. E. Algebraically. - - Let A1 : as :: As : aa. Then shall A1 : A1 – a, ; : As As - a. For since A1 : as :: As : a, .". 41 := 4s 5, - 02 G4 tº - . . 41 A3 subtracting I from each of these equals, ... as T I = a. T 1, A1 – as _ 43 - as -------, but 41 – 43 e 22 d:1 º Dividing the latter by the former of these equals, *-*- - . 41 . 41 – as 43 . 43 - 2, . g — ; OI’ d2 02 04 04 Cº. A Q, or #4 x -: *— = f° x ++–, 02 41 – az 04 4s - as As and 4, . A – d. 4, . As — as: Prop.xx. Algebraically. Let A1, A2, As be three magnitudes, and a1, a2, as, other three, such that Al A2 : : a, , a, and 42 : As :: a, as: . if A 1 » A3, then shall a > as, and if equal, equal; and if less, less. A1 •- Oll Since 41 : A2 : : ai : aa, * A a.” 2 g Ao ao also since A2 : As :: a, ; as, ... + = . 3 3 tº º tº . A and multiplying these equals, ..'. A, X As emº a. X a.” 41 61 or − = − , 43 as e . A . and since the fraction #. is equal to : , and that A1 - 4a: 3 3 it follows that a1 is > a.s. In the same way it may be shewn, that if A = As, then ai = as ; and if Al be < As then a, “ as: Prop. XXI. Algebraically. Let A1, A2, As, be three magnitudes, and a1, a2, as three others, - such that zł. : A2 :: a, ; as, and 42 : 43 : : a, , az. If A, - As, then shall ai > as, and if equal, equal; and if less, less. © c 41 as For since 41 : 42 ; ; az : as, '' A. " as’ NOTES To Book v. 203 & a1 - A and since A. : As :: a, : ... + = + , 2 § 0.1 02, 43 02 A2 - - - A. 0.2 . , 0. M lt l th l ..". **! - - - *l ultiplying these equals, 42 X 43 as X a.” 41 a. Or 4s - as 9 * . 41. Q, and since the fraction #. is equal to . , and that Al P. As; 3 3. it follows that also ai > a.s. Similarly, it may be shewn, that if A1 = As, then ai = as; and if 41 < As, also ai > a.3. Prop. xxII. Algebraically. Let A1, A2, As be three magnitudes, and a1, a2, as other three, such that A1 : A2 :: ai : ag, and 42 : As :: a, as: Then shall A1 : A's, :: ai : as: For since Ai : 42 :: ai : a 2, ...". -- = + , 42 As Ta, as’ and 4.1 : 43 :: ai : as: - Next, if there be four magnitudes, and other four, such that 41 + 42 tº ai : aa, 42 : 43 :: a 2 : as, 43 : 44 :: as : aa. Then shall 41 : A, ; ; ai : a*. For since Al 42 :: ai : a, ". * = * , 42 a.2 º A 0. and 42 : As :: a, ; a., ... + = ** - 2 3 d2. 0's, As º as * A. Q. also 43 : 4, ; ; as : a, ... + = **. 3 4 § 9 * 4, 04 © ºf ſº 41 .. 42 .. 43 at ... a. ... as Multiplying these equals, ... : x : x #8 = ** x * x 3 plying q}lais, 42 43 44 a. * as “a” A1 at - or i=z. ". 41 : A, a a. And similarly, if there were more than four magnitudes. Prop. xxLII. Algebraically. Let 41, 42, 43 be three magnitudes, and a1, a2, as other three;. such that A : 4, ; ; as : as, and A. : 4, .. 0.1 # 02. - Then shall A1 : As :: ai : as: For since 41 : A, ; ; as : as, ... #=#, 204 - EUCLID'S ELEMENTS. and since 42 : As :: a, ; az, .". # =: e 3 3. - A. A Q., & Multiplying these equals, ... + x tº - “x : plying eq º 42 4s as a2 A1 ai OT As - 3. 9. and 41 : 43 :: ai : aa. If there were four magnitudes, and other four, - - such that A1 : A2 : : as : aa, 42 : 43 : : a, as, 43 : 44 :: ai : a 2. Then shall also 4, ; 4. . . a, as: • e A1 as For since 4; ; 42 :: as : a, '' A. Ta.” A., a 42 : 43 : : az as, ... + =#. As a 43 : 4, ; ; ai : az, .". 2. =#. tº º Gº - 41 .. 42 ... As as ... dº ... di l th als, ... + #. * = ** x * x * Multiplying these equals, ; x 3 X 4 04 X - × a. * Aſ & 44 04 ..". 41 : 44 :: ai : a, and similarly, if there be more than four magnitudes. Prop. xxiv. Algebraically. - - Let 41 : a 43 : a, and As as :: As a. Then shall 41 + As : a 2 :: A3 + 4, ; a. 4s For since Ai : as :: As : aa, ... + = **, 02 04 º A A and since A. : as :: A, ; as, ... + = +. 0. 04 I)ivide the former by the latter of these equals, ... : + +* = Hº -- Hº a d2 04 &4 Or a, X Z. - 04 A. 9 Al Aa or al- i. - sº sº e - A1 As adding 1 to each of these equals, ... + + 1 = + + 1, - - 4s ~ 4s A1 + As A3 + 46 Or - = 2. As 4s and 4s == 4s º - as a 4 A. A., 4- Aa A Multiply these equals together, ... ++x *-###x # * NOTES To BOOK v. 205 T 41 + 4, 4s + 4s a2 a. and ..'. Al + As : a, , ; 4s 4- 4a a. CoR. 1. Similarly may be shewn, that A1 – A, ; as :: 43 – 4s 3 a. Prop. xxv. Algebraically. * Let A1 : a, , ; 4s : a*, and let A, be the greatest, and consequently a, the least. - Then shall A1 + a 4 × as + 43. O 3. o - A Since Al as :: 4s : aa, ". : Multiply these equals by , ... + =#, 3 subtract 1 from each of these equals, .". 41 - 43 da - as 43 *-º-º: 04 º 43 . 41 – 43 4s – as " " " as - a, a.' A A * > but tº - †", Q2 24 as - a, T as ' Or Multiplying these equals by 02 *e e but A, - a, "." A, is the greatest of the four magnitudes, g ... also A1 – A3 - az - a, - add As + as to each of these equals, ..". 41 + a, P ax + 43. “The whole of the process in the Fifth Book is purely logical, that is, the whole of the results are virtually contained in the definitions, in the manner and sense in which metaphysicians (certain of them) imagine all the results of mathe- matics to be contained in their definitions and hypotheses. No assumption is made to determine the truth of any consequence of this definition, which takes for granted more about number or magnitude than is necessary to understand the definition itself. The latter being once understood, its results are deduced by in- spection—of itself only, without the necessity of looking at any thing else. Hence, a great distinction between the fifth and the preceding books presents itself. The first four are a series of propositions, resting on different funda- mental assumptions; that is, about different kinds of magnitudes. The fifth is a definition and its development; and if the analogy by which names have been given in the preceding Books had been attended to, the propositions of that Book would have been called corollaries of the definition.”—Connezion of Number and Magnitude, by Professor De Morgan, p. 56. e - The Fifth Book of the Elements as a portion of Euclid's System of Geometry ought to be retained, as the doctrine contains some of the most important charac- teristics of an effective instrument of intellectual Education. This opinion is favoured by Dr. Barrow in the following expressive terms: “There is nothing in the whole body of the Elements of a more subtile invention, nothing more solidly established, or more accurately handled than the doctrine of proportionals.” .*- 206 EUCLID's ELEMENTs. QUESTIONS ON BOOK v. 1. ExPLAIN and exemplify the meaning of the terms, multiple, submultiple, equimultiple. **. t - 2. What operations in Geometry and Arithmetic are analogous? 3. What are the different meanings of the term measure in Geometry? When are Geometrical magnitudes said to have a common measure? What is meant by the greatest common measure and by the least common measure of two or more magnitudes - 4. When are magnitudes said to have, and not to have, a ratio to one another? What restriction does this impose upon the magnitudes in regard to their species 5. When are magnitudes said to be commensurable or incommensurable to reach other Do the definitions and theorems of Book v, include incommensurable quantities 2 6. What is meant by the term Geometrical ratio 2 How is it represented 7. Why does Euclid give no independent definition of the ratio of two Geo- Imetrical magnitudes 2 - 8. What sort of quantities are excluded from Euclid's idea of ratio, and how does his idea of ratio differ from the Algebraic definition ? 9. How is a ratio represented Algebraically 2 Is there any distinction between the terms, a ratio of equality, and equality of ratio 2 . 10. In what manner are ratios, in Geometry, distinguished from each other as equal to, greater, or less than, one another What objection is there to the use of an independent definition (properly so called) of ratio in a system of Geometry 11. Point out the distinction between the geometrical and algebraical methods of treating the subject of proportion. 12. What is the geometrical definition of proportion? Whence arises the necessity of such a definition as this - 13. Shew the necessity of the qualification “any whatever” in Euclid's definition of proportion. . 14. Must magnitudes that are proportional be all of the same kind 15. To what objection has Euc. v. def. 5, been considered liable 16. Point out the connexion between the more obvious definition of proportion and that given by Euclid, and illustrate clearly the nature of the advantage obtained by which he was induced to adopt it. 17. Why may not Euclid's definition of proportion be superseded in a system of Geometry by the following: “Four quantities are proportionals, when the first is the same multiple of the second, or the same part of it, that the third is of the fourth º'’ * * 18. Apply Euclid's definition of proportion, to shew that if four quantities be. proportional, and if the first and the third be divided into the same arbitrary number of equal parts, then the second and fourth will either be equimultiples of those parts, or will lie between the same two successive multiples of them. 19. The Geometrical definition of proportion is a consequence of the Algebraical definition; and conversely. - 20. What Geometrical test has Euclid given to ascertain that four quantities are not proportionals What is the Algebraical test ? - 21. Shew in the manner of Euclid, that the ratio of 15 to 17 is greater than that of 11 to 13. 2 - QUESTIONS ON Book v. 207 gº 22. How far may the fifth definition of the fifth Book be regarded as an axiom Is it convertible If so, state the converse; if not, state why. 23. Def. -9, Book v. “Proportion consists of three terms at least.” How is this to be understood cº- 24. Define duplicate ratio. How does it appear from Euclid that the duplicate ratio of two magnitudes is the same as that of their squares? - 25. Shew that the ratio compounded of any ratio, and the reciprocal of that ratio, is a ratio of equality. 26. What is meant by “the ratio compounded of the ratio of A to B, and of C to D’’, when A, B, C, D are Geometrical magnitudes of the same species How is the phrase translated into common Arithmetic when A, B, C, D are numbers ? g 27. By what process is a ratio found equal to the composition of two or more given ratios ? Give an example, where straight lines are the magnitudes which express the given ratios. g & 28. What limitation is there to the alternation of a Geometrical proportion ? 29. Explain the construction and sense of the phrases, ez aquali, and ea: aquali in proportione perturbata, used in proportions. 30. Explain the meaning of the word homologous as it is used in the Fifth Book of the Elements. 31. Why, in Euclid v. 11, is it necessary to prove that ratios which are the same with the same ratio, are the same with one another 3 32. Apply the Geometrical criterion to ascertain, whether the four lines of 3, 5, 6, 10 units are proportionals, 33. Prove by taking equimultiples according to Euclid's definition, that the magnitudes 4, 5, 7, 9, are not proportionals. 34. Give the Algebraical proofs of Props. 17 and 18, of the Fifth Book. - 35. What is necessary to constitute an exact definition ? In the demonstration of Euc. v. 18, is it legitimate to assume the converse of the fifth definition of that Book? Does a mathematical definition admit of proof on the principles of the science to which it relates ? - 36. Explain why the properties proved in Book v, by means of straight lines, are true of any concrete magnitudes. *. 37. Enunciate Euc. v. 8, and illustrate it by numerical examples. 38. Prove Algebraically Euc. v. 25. - 39. Shew that when four magnitudes are proportionals, they cannot, when equally increased or equally diminished by any other magnitude, continue to be proportionals. 40. What grounds are there for the opinion that Euclid intended to exclude the idea of numerical measures of ratios in his Fifth Book - - 41. State the particular conclusions that may be arrived at by means of Euclid's definition of Proportion. - '42. Prove from a property of the circle, that if four quantities are propor- tionals, the sum of the greatest and least is greater than the sum of the other two, 43. “Number and magnitude do not correspond in all their relations.” In what relations do they correspond Does any intelligible relation exist between - a point in Geometry and an unit in Arithmetic * 44. If four finite straight lines which are not proportionals be taken in order of magnitude: can any line be found Geometrically or , Algebraically, which when added to, or taken from, each of the four lines, shall make the four lines so increased, or so diminished, to be proportionals? - 45. If arcs of circles be admitted as measures of angles subtended by them * 208 s - EUCLID'S ELEMENTS. at the centers of circles; can an arc, and the angle it subtends at the centre of a circle, have a ratio to each other ? 46. Is the following definition of proportion as comprehensive as Euclid's If not, point out in what respects. “Four magnitudes are proportionals, if when the first and second are multiplied by two such numbers as make the products equal, the third and fourth being respectively multiplied by the same numbers, likewise make equal products.” That is, if a, b, c, d be four magnitudes, and m and n any two numbers, such that ma = nb and me = nd, then a, b, c, d are propor- tionals, or a is to b as c is to d. - 47. What effect is produced on a ratio, if equal magnitudes be taken from both the terms of the ratio: first, when the antecedent is greater than the conse- quent ; secondly, when it is less - 48. Shew Geometrically and Algebraically, that if four magnitudes be pro- portionals, and the first be the greatest; the first and fourth together are greater than the second and third. - - 49. If four magnitudes of the same kind be proportionals, and if the first magnitude be the greatest, the fourth shall be the least; and if the first be the least, the fourth shall be the greatest. 50. If three magnitudes are proportionals, the sum of the extremes is greater than double the mean. - - 51. If a series of magnitudes be in continual proportion, and there be taken any series of equidistant terms; that series will also form a continual proportion. 52. If any number of magnitudes be continual proportionals, the successive sums of the first and second, second and third, &c. terms shall be proportionals: as also the successive differences. - t 53. In every series of magnitudes in continual proportion, the ratio of the first magnitude to the third is the same as the duplicate ratio of the first to the second : the ratio of the first to the fourth, the same as the triplicate ratio of the first to the second ; the ratio of the first to the fifth, the same as the quadruplicate ratio of the first to the second; and so on, whatever may be the number of magnitudes in the series. - 54. If four magnitudes be proportionals, and the first be the greatest, the difference between the first and third shall be greater than the difference between the second and fourth, 55. If the first magnitude of a ratio be greater than the second, the ratio is increased by adding equal magnitudes to both terms of the ratio; but if the first magnitude of a ratio be less than the second, the ratio is diminished by adding ‘equal magnitudes to both terms of the ratio. 56. If the first of four magnitudes of the same kind has a greater ratio to the second than the third has to the fourth; the first shall have to the third a greater ratio than the second has to the fourth. 57. It has been objected to Euclid's definition of Proportion, that “it is cumbrous and difficult of comprehension to a learner.” Examine these objec- tions, and state how far Dr. Barrow's opinion is tenable, namely—“That there is nothing in the whole body of the Elements of a more subtile invention, nothing more solidly established, or more accurately handled, than the doctrine of pro- portionals.” 58, What is the object of the Fifth Book of Euclid's Elements BOOK WI. DEFINITIONS. I. SIMILAR rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles pro- portionals. II “JReciprocal figures, viz. triangles and parallelograms, are such as have their sides about two of their angles proportionals in such a manner, that a side of the first figure is to a side of the other, as the remaining side of the other is to the remaining side of the first.” III. A straight line is said to be cut in extreme and mean ratio, when the whole is to the greater segment, as the greater segment is to the less. - - IV. The altitude of any figure is the straight line drawn from its vertex perpendicular to the base. -A— *- PROPOSITION I. THEOREM, Triangles and parallelograms of the same altitude are one to the other as their bases. . -- Let the triangles ABC, ACD, and the parallelograms EC, CF, have the same altitude, viz. the perpendicular drawn from the point A to BD or BD pro- duced. . . As the base BC is to the base CD, so shall the triangle ABC be to the triangle A CD, . - e - and the parallelogram EC to the parallelogram C.F. P 210 - EUCLID's ELEMENTs. E A F HG B C D K L - Produce BD both ways to the points H, L, and take any number of straight lines BG, GH, each equal to the base BC; (I. 3.) and DJſ, KL, any number of them, each equal to the base CD; - - and join AG, AH, AK, AL. - Then, because CB, BG, GH, are all equal, the triangles AHG, AGB, ABC, are all equal: (1. 38.) therefore, whatever multiple the base HC is of the base B6, the same multiple is the triangle A HC of the triangle ABC: for the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC': and if the base HC be equal to the base CL, the triangle AHC is also equal to the triangle ALC: (1. 38.) and if the base HC be greater than the base CL, likewise the triangle AJIC is greater than the triangle ALC; and if less, less; therefore since there are four magnitudes, viz. the two bases BC, CD, and the two triangles ABC, A CD; and of the base BC, and the triangle ABC, the first and third, any equimultiples whatever have been taken, viz. the base HC and the triangle AHC ; and of the base CD and the triangle A CD, the second and fourth, have been taken any equimultiples whatever, - viz. the base CL and the triangle ALC; w ; and since it has been shewn, that, if the base HC be greater than * the base CL, - - the triangle A.HC is greater than the triangle ALC; and if equal, equal; and if less, less; therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle A CD. (v. def. 5.) . And because the parallelogram CE is double of the triangle ABC, I. 41. ( tº the parallelogram CF double of the triangle A CD, and that magnitudes have the same ratio which their equimultiples have; (v. 15.) - as the triangle ABC is to the triangle A CD, so is the parallelogram FC to the parallelogram CF; and because it has been shewn, that, as the base BC is to the base CD, so is the triangle ABC to the triangle A CD; and as the triangle ABC is to the triangle A CD, so is the paralle- logram EC to the parallelogam CF; therefore, as the base BC is to the base CD, so is the parallelogram JEC to the parallelogram C.F. (v. 11.) - Wherefore, triangles, &c. Q.E.D. CoR. From this it is plain, that triangles and parallelograms that have equal altitudes, are to one another as their bases. Let the figures be placed so as to have their bases in the same straight line; and having drawn perpendiculars from the vertices of -BOOK WI. PROP. I, II. 211 the triangles to the bases, the straight line which joins the vertices is parallel to that in which their bases are, (I. 33.) because the perpen- diculars are both equal and parallel to one another. (I. 28.) Then, if the same construction be made as in the proposition, the demonstration will be the same. PROPOSITION II. THEOREM. If a straight line be drawn parallel to one of the sides of a triangle, it shalf out the other sides, or these produced, proportionally: and conversely, if the sides, or the sides produced, he out proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle. Let DE be drawn parallel to BC, one of the sides of the triangle ABC. and cut AB, AC, or these produced in D, E, Then BD shall be to DA, as CE to E.A. A A F. D IB C D E 5 C Join BE, CD. Then the triangle BDE is equal to the triangle CDE, (I. 37.) because they are on the same base DE, and between the same parallels DE, BC; -> but ADE is another triangle; and equal magnitudes have the same ratio to the same magnitude; (v. 7. & as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE: - but as the triangle BDE to the triangle ADE, so is BD to DA, (VI. 1.) because, having the same altitude, viz. the perpendicular drawn from the point E to AB, they are to one another as their bases; and for the same reason, as the triangle CDE to the triangle ADE, so is CE to EA : - therefore, as BD to DA, so is CE to E.A. (v. 11.) Next let the sides AB, A C of the triangle ABC, or these sides produced, be cut proportionally in the points D, E, that is, so that BD may be to DA as CE to EA, and join D.E. Then DE shall be parallel to BC. The same construction being made, - |because as BD to DA, so is CE to EA ; s and as BD to DA, so is the triangle BDE to the triangle ADE; (VI. 1.) and as CE to EA, so is the triangle CDE to the triangle ADE; therefore the triangle BDE is to the triangle ADE, as the triangle CDE to the triangle ADE; (v. 11.) " . that is, the triangles BDE, CDE have the same ratio to the triangle AD E: * therefore the triangle BDE is equal to the triangle CDE: (v. 9.) and they are on the same base DE: P 2 212 EUCLIo's ELEMENTs. but equal triangles on the same base and on the same side of it, are between the same parallels: (I. 39.) therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q.E.D. PROPOSITION III. THEOREM. If the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same. zatio which the other sides of the triangle have to one another: and con- versely, ºf the segments of the base have the same ratio which the other sides of the triangle have to one another; the straight line drawn from the vertea to the point of section, divides the vertical angle into two equal angles. Let ABC be a triangle, and let the angle BAC be divided into two equal angles by the straight line AD. * . Then BD shall be to DC, as B.A. to A. C. Through the point C draw CE parallel to DA, (I. 31.) and let BA produced meet CE in E. Because the straight line AC meets the parallels AD, EC, the angle ACE is equal to the alternate angle C.A.D.: (I, 29.) but CAD, by the hypothesis, is equal to the angle BAD; wherefore BAD is equal to the angle ACE. (ax. 1.) Again, because the straight line BAE meets the parallels AID, EC, the outward angle BAD is equal to the inward and opposite angle AEC: (I. 29.) * |but the angle ACE has been proved equal to the angle BAD; therefore also ACE is equal to the angle AEC, (ax. 1.) and consequently, the side A E is equal to the side AC: (I. 6.) and because AD is drawn parallel to EC, one of the sides of the triangle BCE, - - therefore BD is to DC, as BA to A.E. (VI. 2.) - but AE is equal to A C; therefore, as BD to DC, so is BA to A.C. (v. 7.) Next, let BD be to DC, as B.A. to AC, and join A.D. - - Then the angle BAC shall be divided into two equal angles by the straight line AID. - - The same construction being made; because, as BD to DC, so is B.A. to A.C.; - and as BD to DC, so is B.A to AE, because AD is parallel to EC; (VI. 2. • º BA is to AC, as BA to AE: (v. 11.) > consequently AC is equal to A.E, (v. 9.) and therefore the angle AEC is equal to the angle ACE: (1. 5.) but the angle AEC is equal to the outward and opposite angle BAD. and the angle ACE is equal o the alternate angle CAD: (1, 29.) BOOK VI. PROP. III, A. 213 wherefore also the angle BAD is equal to the angle CAD; (ax. 1.) that is, the angle BAC is cut into two equal angles by the straight line A.D. - Therefore, if the angle, &c. Q.E.D. PROPOSITION A. THEOREM. If the outward angle of a triangle made by producing one of its sides, be divided into two equal angles, by a straight line, which also cuts the base produced; the segments between the dividing line and the eatremities of the base, have the same ratio which the other sides of the triangle have to one another: and conversely, if the segments of the base produced have the same ratio which the other sides of the triangle have ; the straight line drawn from the vertea to the point of section divides the outward angle of the triangle &nto two equal angles. - Let ABC be a triangle, and let one of its sides BA be produced to E; and let the outward angle CAE be divided into two equal angles by the straight line AD which meets the base produced in D. Then BD shall be to DC, as B.A. to A.C. E A - F - B C D Through C draw CF parallel to AD : (1. 31.) and because the straight line AC meets the parallels AD, FC, the angle ACF is equal to the alternate angle CAD: (I. 29.) but CAD is equal to the angle D.A.E; (hyp.) therefore also DAE is equal to the angle ACF. (ax. 1.) - Again, because the straight line FAE meets the parallels AD, FC, the outward angle D.A.E is equal to the inward and opposite angle CFA : (I. 29.) - - but the angle ACF has been proved equal to the angle DAE; therefore also the angle ACF is equal to the angle CFA ; (ax. 1.) and consequently the side AF'is equal to the side A C: (I. 6.) and because AD is parallel to FC, a side of the triangle BCF, therefore BD is to DC, as BA to AF: (VI. 2.) r but AF is equal to AC; therefore, as BD is to DC, so is B.A. to A.C. (v. 7.) Next, let BD be to DC, as BA to AC, and join AD. The angle CAD, shall be equal to the angle D.A.E. The same construction being made, because BD is to DC, as B.A. to A. C.; and that BD is also to DC, as B.A. to AF; (wr. 2.) therefore BA is to AC, as B.A. to AF; (v. 11.) wherefore AC is equal to A.F. (v. 9.) and the angle AFG equal to the angle A CF: (I. 5.) - but the angle AFC is equal to the outward angle EAD, (I. 29.) and the angle A CF to the alternate angle CAD; therefore also EAD is equal to the angle CAD. (ax. 1.) Wherefore, if the outward, &c., Q.E.D. - 214 EUCLID's ELEMENTS. PROPOSITION IV. THEOREM, *. The sides about th equal angles of equiangular triangles are proportionals * and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios. - Tet ABC, DCE be equiangular triangles, having the angle ABC equal to the angle DCE, and the angle ACB to the angle DEC; and consequently the angle BAC equal to the angle CDE, (I. 32.) The sides about the equal angles of the triangles ABC, DCE shall be proportionals; - e - *. - and those shall be the homologous sides which are opposite to the equal angles. F A D E C E Let the triangle DCE be placed, so that its side CE may be con- tiguous to BC, and in the same straight line with it. (I. 22.) Then, because the angle BCA is equal to the angle CED, (hyp.) add to each the angle ABC; therefore the two angles' ABC, BCA are equal to the two angles ABC, CED: (ax. 2.) - but the angles ABC, BCA are together less than two right angles; I. 17.) arº the angles ABC, CED are also less than two right angles: wherefore BA, ED if produced will meet: (I. ax. 12.) let them be produced and meet in the point F: then because the angle ABC is equal to the angle DCE, (hyp.) -- BF is parallel to CD; (I. 28.) and because the angle ACB is equal to the angle DEC, AC is parallel to FE: (I. 28.) - therefore FA CD is a parallelogram; and consequently AF is equal to CD, and AC to FD: (I. 34.) and because AC is parallel to FE, one of the sides of the triangle F.B.E, i #3A is to AF, as BC to CE: (VI. 2.) - but AIF is equal to CD; therefore, as BA to CD, so is BC to CE: (v. 7.) and alternately, as AB to BC, so is DC to CE; (v. 16.) again, because CD is parallel to BF, as BC to CE, so is FD to DE: (VI. 2.) - but FD is equal to A C : therefore, as BC to CE, so is A0 to DE; (v. 7.) and alternately, as BC to CA, so CE to ED: (v. 16.) - therefore, because it has been proved - that AB is to BC, as DC to CE, * and as BC to CA, so CE to ED, ex aequali, BA is to AC, as CD to D.E. (v. 22.) * Therefore the sides, &c. Q.E.D. BOOK WI. PROP. v, v1. . 215 PROPOSITION v. THEOREM. If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular; and the equal angles shall be those which are opposite to the homologous sides. Let the triangles ABC, DEF have their sides proportionals, - so that AB is to BC, as DE to EF; and BC to CA, as EF to FD; and, consequently, ex æquali, B.A. to AC, as ED to D.F. w Then the triangle ABC shall be equiangular to the triangle DEF, . and the angles which are opposite to the homologous sides shall be èqual, viz. the angle ABC equal to the angle DEF, and BCA to JEFD, and also BAC to EDF. - A D º . B C G. - . - At the points E, F, in the straight line EF, make the angle FEG equal to the angle ABC, and the angle EFG equal to BCA; (I. 23.) wherefore the remaining angle EGF, is equal to the remaining angle BAC, (I. 32.) * and the triangle GEF is therefore equiangular to the triangle ABC: consequently they have their sides opposite to the equal angles pro- portional: (VI.4. - - wherefore, as AB to BC, so is GE to EF; but as AB to BC, so is DE to EF: (hyp.) therefore as DE to EF, so GE to EF; (v. 11.) that is, DE and GE have the same ratio to EF, and consequently are equal. (V. 9.) For the same reason, DF is equal to FG ; and because in the triangles DEF, GEF, DE is equal to EG, and JFF is common, . , the two sides DE, EF are equal to the two GE, EF, each to each; and the base DF is equal to the base GF; therefore the angle DEF is equal to the angle GEF, (r. 8.) and the other angles to the other angles which are subtended by the equal sides; (I. 4.) • . therefore the angle DFE is equal to the angle GFE, and EDF to EGF': and because the angle DEF is equal to the angle GEF, t and GEF equal to the angle ABC; (constr.) therefore the angle ABC is equal to the angle DEF: (ax. 1.). for the same reason, the angle ACB is equal to the angle DFE, and the angle at A equal to the angle at D : therefore the triangle ABC is equiangular to the triangle DEF. - Wherefore, if the sides, &c. Q.E.D. * PROPOSITION VI. THEOREM. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous stales. . - 216. EUCLID's ELEMENTs. Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles pro- portionals; that is, B.A. to AC, as ED to D.F. - Then the triangles ABC, DEF shall be equiangular, and shall have, the angle ABC equal to the angle DEF, and ACB to DFE. - A. ID G N IV - - iſ 3 At the points D, F, in the straight line DF, make the angle. FDG equal to either of the angles BAC, EDF; (I. 23.) l and the angle DFG equal to, the angle ACB : wherefore the remaining angle at B is equal to the remaining angle at G: (I. 32.) - - and consequently the triangle DGF'is equiangular to the triangle ABC; therefore as B.A. to AC, so is G.D to DF: (VI. 4.) |but, by the hypothesis, as BA to AC, so is ED to DF; therefore as EI) to D.F, so is GD to DF; (v. 11.) wherefore ED is equal to DG ; (v. 9.) and DF is common to the two triangles EDF, GDF: - therefore the two sides ED, DF’ are equal to the two sides G.D, DF, each to each ; * and the angle. EDF is equal to the angle. GDF; (constr.) wherefore the base. EF is equal to the base FG, (I. 4.) and the triangle EDF to the triangle. GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides: - therefore the angle DFG is equal to the angle, DFE, and the angle at G to the angle at E; - |but the angle DFG is equal to the angle A CB; (constr.) therefore the angle ACB is equal to the angle DFE; (ax. 1.) and the angle BAC is equal to the angle EDF: (hyp.) * wherefore also the remaining angle at B is equal to the remaining angle at E; (I. 32.) - therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, &c. Q.E.D. PROPOSITION VII. THEOREM. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals; then, if each of the Yemaining angles be either less, or not less, than a right angle, or if one of them be a right angle; the triangles shall be equiangular; and shall have. ' those angles equal about which the sides are proportionals. Tet the two triangles ABC, DEF have one angle in the one equal. to one angle in the other, - viz. the angle BAC to the angle EDF, and the sides about two other angles A.B.C, DEF proportionals, so that AB is to BC, as DE to EF; g a" . ,” / s BOOK WI. PROP. VII. 217. and in the first case, Iet each of the remaining angles at C, F be less than a right angle. - * The triangle ABC shall be equiangular to the triangle DEF, viz. the angle ABC shall be equal to the angle DEF, - and the remaining angle at C equal to the remaining angle at F. A. IO, 23, 2\ B C E E. For if the angles ABC, DEF be not equal, one of them must be greater than the other: let ABC be the greater, - and at the point B, in the straight line AB, make the angle ABG equal to the angle DEF; (I. 23.) and because the angle at A is equal to the angle at D, (hyp.) - and the angle ABG to the angle DEF; - the remaining angle AGB is equal to the remaining, angle DFE: (I. 32. - - therefore #. triangle ABG is equiangular to the triangle DEF: wherefore as AB is to BG, so is DE to EF: (VI. 4.) but as DE to EF, so by hypothesis, is AB to BC; therefore as AB to BC, so is AB to BG: (v. 11.) and because AB has the same ratio to each of the lines BC, BG, JBC is equal to BG; (v. 9.) and therefore the angle BGC is equal to the angle BCG : (T. 5.) but the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle; and therefore the adjacent angle A G B must be greater than a right angle ; (I. 13.) * but it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle; but, by the hypothesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal, - that is, they are equal: , and the angle at A is equal to the angle at D: (hyp.) wherefore the remaining angle at C is equal to the remaining angle at F: (1. 32.) therefore the triangle ABC is equiangular to the triangle D.E.F. Next, let each of the angles at C, F be not less than a right angle. Then the triangle ABC shall also in this case be equiangular to the triangle DEF. - - A. D 2G 21 - B C. E F - The same construction being made, - it may be proved in like manner that BC is equal to BG, , and therefore the angle at C equal to the angle BGC: but the angle at C is not less than a right angle; (hyp.) f therefore the angle BGC is not less than a right angle: wherefore two angles of the triangle BGC are together not less than two right angles; º g - which is impossible: (I. 17.) 218. EUCLID's ELEMENTs. and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. Lastly, let one of the angles at C, F, viz. the angle at C, be a right angle: in this case likewise the triangle ABC shall be equiangular to the triangle DEF. A A T) C E F B C B à Eor, if they be not equiangular, at the point B in the straight line AB, make the angle ABG equal to the angle DEF; - then it may be proved, as in the first case, that BG is equal to B0: and therefore the angle BCG is equal to the angle BGC: (r. 5.) but the angle BCG is a right angle, (hyp.) e therefore the angle BGC is also a right angle; (ax. 1.) whence two of the angles of the triangle BGC are together not less than two right angles; *~ - - which is impossible: (I. 17.) therefore the triangle ABC is equiangular to the triangle DEF: Wherefore, if two triangles, &c. Q.E.D. PROPOSITION VIII. THEOREM. In a right-angled triangle, if a perpendicular be drawn from the right angle to the base; the triangles on each side of it are similar to the whole triangle, and to one another. * Let ABC be a right-angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base B.C. Then the triangles ABD, ADC shall be similar to the whole tri- angle ABC, and to one another. A Because the angle BAC is equal to the angle ADB, each of them being a right angle, (ax. 11.) * and that the angle at B is common to the two triangles ABC, ABD: the remaining angle ACB is equal to the remaining angle BAD; (I. 32.) - therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal angles are proportionals; (VI. 4.) wherefore the triangles are similar: (VI. def. 1.) in the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC. And the triangles ABD, A CD, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right-angled, &c. Q.E.D. CoR. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean propor- BOOK WI. PROP. VIII, Ix, x. - 219 tional between the segments of the base; and also that each of the sides is a mean proportional between the base, and the segment of it adjacent to that side : because in the triangles BDA, ADC; BD is to I)A, as DA to DC; (VI. 4.) and in the triangles ABC, DBA.; BC is to BA, as BA to BD: (VI. 4.) and in the triangles ABC, A CD; BC is to CA, as CA to CD. (VI. 4.) .* PROPOSITION IX. PROBLEM. From a given straight line to cut off any part required. Let AB be the given straight line. It is required to cut off any part from it. A B C Erom the point A draw a straight line AC, making any angle with AB; r- and in AC take any point D, - and take AC the same multiple of AD, that AB is of the part which is to be cut off from it; # join BC, and draw DE parallel to C.B. Then AE shall be the part required to be cut off. Because ED is parallel to BC, one of the sides of the triangle ABC, as CD is to DA, so is B E to EA ; (VI. 2.) and by composition, CA is to AD, as B.A. to A.E. (v. 18.) but CA is a multiple of AD; (constr.) . therefore BA is the same multiple of A.E. (V. D.) - whatever part therefore AD is of AC, AE is the same part of AB : wherefore, from the straight line AB the part required is cut off. Q. E. F. - - PROPOSITION X. PROBLEM. To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which tha. parts of the divided given straight line have. - Let AB be the straight line given to be divided, and AC the divided. - line. - º It is required to divide AB similarly to A.C. A. F AD. G E B R. C Let A C be divided in the points J), E'; and let AB, A C be placed so as to contain any angle, and join BC, and through the points D, E draw DF, EG parallels to BC. (I. 31.) 220 EUCLID'S ELEMENTS. Then A B shall be divided in the points F, G, similarly to A.C. -- Through D draw DHK parallel to A.B.: - therefore each of the figures, FH, HB is a parallelogram; wherefore DH is equal to FG, and HK to GB: (I. 34.) and because HE is parallel to KC, one of the sides of the triangle I).KC, - - - as CE to ED, so is KH to HD : (VI. 2.) but KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF: (v. 7.) again, because FD is parallel to GE, one of the sides of the triangle A GE, as ED is to DA, so is GF to FA : (VI. 2.) therefore, as has been proved, as CE is to ED, so is BG to GF, • and as ED is to DA, so is GF to FA : ºtherefore the given straight line AB is divided similarly to A.C. Q.E.F. PROPOSITION XI. PROBLEM. To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines. It is required to find a third proportional to AB, 40. f\ l) E Let AB, A C be placed so as to contain any angle: produce AB, A C to the points D, E: and make BD equal to AC; join BC, and through D, draw DE parallel to B.C. (I. 31.) Then CE shall be a third proportional to AB and A.C. Because BC is parallel to DE, a side of the triangle ADE, AB is to BD, as AC to CE: (VI. 2.) but BD is equal to AC; therefore as AB is to AC, so is AC to CE. (v. 7.) Wherefore, to the two given straight lines AB, AC, a third pro- portional CE is found. Q.E.F. - PROPOSITION XII. PROBLEM. To find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines. It is required to find a fourth proportional to A, B, C. Take two straight lines DE, DF, containing any angle EDF: and º º, * DG equal to A, GE equal to B, and DH equal O U : [I. 3. - BOOK WI. PROP. XII, XIII; XIV. -- - 221. join GH, and through E. draw EF parallel to it. (I. 31.) Then HF shall be the fourth proportional to A, B, C. Because GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH to HF; (VI. 2.) but DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. (v. 7.) Wherefore, to the three given straight lines A, B, C, a fourth proportional HF is found. Q. E. F. * - PROPOSITION XIII. PROBLEM. - To find a mean proportional between two given straight lines. Tiet AB, BC be the two given straight lines. •. It is required to find a mean proportional between them. - D Place AB, BC in a straight line, and upon AC describe the semi- circle ADC, - and from the point B draw BD at right angles to A.C. (I. 11.) Then BD shall be a mean proportional between AB and BC. tº. - Join AD, D.C. And because the angle ADC in a semicircle is a right angle, (III. 31.) and because in the right-angled triangle ADC, DB is drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the base: (VI. 8. Cor.) --~~~~ therefore, between the two given straight lines AB, BC, a mean proportional DB is found. Q. E. F. PROPOSITION XIV. THEOREM. JEqual parallelograms which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally pro- portional : and conversely, parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reci- procally proportional, are equal to one another. Let A.B, BC be equal parallelograms, which have the angles at B equal. - The sides of the parallelograms AB, BC about the equal angles, shall be reciprocally proportional; that is, DB shall be to BE, as GB to B.F. A F Fº G C 222 EUCLID'S ELEMENTS. Let the sides DB, BE be placed in the same straight line; wherefore also FB, BG are in one straight line: (I, 14.) A. complete the parallelogram F.E. . And because the parallelogram AB is equal to BC, and that FE is another parallelogram, AB is to FE, as BC to FE: (v. 7.) but as AB to FE, so is the base DB to B.E, (VI. 1.) and as BC to FE, so is the base GB to BF'; therefore, as DB to BE, so is GB to B.F. (v. 11.) Wherefore, the sides of the parallelograms AB, BC about their -- equal angles are reciprocally proportional. - - Next, let the sides about the equal angles be reciprocally proportional, viz. as DB to B.E., so GB to BF': . the parallelogram AB shall be equal to the parallelogram BC. Because, as DB to B.E., so is GB to BF'; and as DB to B.E, so is the parallelogram AB to the parallelogram FE; (VI. 1.) and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FE; (v. 11.) therefore the parallelogram. AB is equal to the parallelogram BC. W. 9. - ( ) Therefore, equal parallelograms, &c. Q.E.D. PROPOSITION xv. THEOREM. Equal triangles which have one angle of the one equal to one angle of 'the other, have their sides about the equal angles reciprocally proportional : and conversely, triangles which have one angle in the one equal to one angle *n the other, and their sides about the equal angles reciprocally proportional, are equal to one another. s - Let ABC, ADE be equal triangles, which have the angle BAC ‘equal to the angle D.A.E. Then the sides about the equal angles of the triangles shall be reciprocally proportional; - - that is, CA shall be to AD, as EA to A.B. B D C E Tet the triangles be placed, so that their sides CA, AD be in one straight line; - - wherefore also EA and AB are in one straight line; (I. 14.) ~ and join BD. 4 - * Because the triangle ABC is equal to the triangle ADE, ~- and that ABD is another triangle; therefore as the triangle CAB, is to the triangle BAD, so is the triangle AED to the triangle DAB; (v. 7.) \ - but as the triangle CAB to the triangle BAD, so is the base CA to the base AD, (VI. 1.) - BOOK WI. PROP. XV, xvi. 223 and as the triangle EAD to the triangle DAB, so is the base EA to the base AB; (VI. 1.) therefore as CA to AD, so is EA to AB : (v. 11.) wherefore the sides of the triangles ABC, ADE, about the equal angles are reciprocally proportional. * Next, let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional, - viz. OA to AID, as EA to AB. Then the triangle ABC shall be equal to the triangle A.D.E. - Join BD as before. Then because, as CA to A.D, so is EA to AB; (hyp.) and as CA to AD, so is the triangle ABC to the triangle BAD: VI. 1. gº º as EA to AB, so is the triangle EAD to the triangle BAD; (VI. 1. & as the triangle BAC to the triangle BAD, so is the tri- angle EAD to the triangle BAD; (v. 11.) that is, the triangles BAC, EAD have the same ratio to the tri- angle BAD: wherefore the triangle ABC is equal to the triangle A.D.E. (v. 9.) - Therefore, equal triangles, &c. Q.E.D. PROPOSITION XVI. THEOREM, If four straight lines be proportionals, the rectangle contained by the eactremes is equal to the rectangle contained by the means: and conversely, &f the rectangle contained by the extremes be equal to the rectangle con- tained by the means, the four straight lines are proportionals. Let the four straight lines AB, CD, E, F be proportionals, .* ^ viz. as AB to CD, so E to F. The rectangle contained by AB, F, shall be equal to the rectangle contained by CD, E. H - G *-*-Eºmº E F - A B - ... • - STE Erom the points A, C draw AG, CH at right angles to AB, CD: I. 11. - º * A G equal to F, and OH equal to E; (I. 3.) and complete the parallelograms BG, D.H. (I. 31.) - Because, as AB to CD, so is E to F; and that E is equal to CH, and F to AG; AB is to CD as CH to AG: (v. 7.) ** therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional; but parallelograms which have their sides about equal angles reci- procally proportional, are equal to one another; (VI. 14.) therefore the parallelogram B G is equal to the parallelogram DH: but the parallelogram B G is contained by the straight lines AB, F; because AG is equal to F; zº 22} EUCLID's ELEMENTS. and the parallelogram D.H is contained by CD and E: because CH is equal to E; - - therefore the rectangle contained by the straight lines AB, F, is - equal to that which is contained by CD and E. * - And if the rectangle contained by the straight lines AB, F, be equal to that which is contained by CD, E: these four lines shall be proportional, viz. A B shall be to CD, as E to F. . The same construction being made, - because the rectangle contained by the straight lines AB, F, is equal to that which is contained by CD, E, and that the rectangle BG is contained by AB, F; - because AG is equal to F; and the rectangle DH by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram D.H.; (ax. 1. . ) and they are equiangular: , - but the sides about the equal angles of equal parallelograms are reciprocally proportional : (v. 14.) . v- wherefore, as AB to CD, so is CH to AG. JBut CH is equal to E, and A G to F; therefore as AB is to CD, so is E to F. (v. 7.) Wherefore, if four, &c. Q.E.D. PROPOSITION XVII. THEOREM, If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square on the mean; and conversely, if the rectangle contained by the extremes be equal to the square on the mean, the three 28traight lines are proportionals. Ilet the three straight lines A, B, C be proportionals, viz. as A to B, so B to C. * The rectangle contained by A, C shall be equal to the square on B. A B-— C | D D— Take D equal to B. And because as A to B, so is B to C, and that B is equal to D; A is to B, as D to C. (v. 7.) - but if four straight lines be proportionals, the rectangle contained by the extremes is equal to that which is contained by the means; (VI. 16.) - theº; the rectangle contained by A, C is equal to that contained y B, D : but the rectangle contained by B, D, is the square on B, because B is equal to D: therefore the rectangle contained by A, C, is equal to the square on B. And if the rectangle contained by A, C, be equal to the square on B, then A shall be to B, as B to 0. The same construction being made, BOOK WI. PROP. XVII, XVIII. 225 * because the rectangle contained by A, C is equal to the square on B, and the square on B is equal to the rectangle contained by B, D, because B is equal to D; -* y therefore the rectangle contained by A, C, is equal to that contained by B, D; . is . b: the rectangle contained by the extremes be equal to that con- tained by the means, the four straight lines are proportionals: (VI. 16.) , therefore A is to B, as D to C. but B is equal to D; wherefore, as A to B, so B to C. Therefore, if three straight lines, &c. Q.E.D. PROPOSITION XVIII. PROBLEM. Upon a given straight line to describe a rectilineal figure similar, and similarly situated, to a given rectilineal figure. Let AB be the given straight line, and CDEF the given rectilineal figure of four sides. - . It is required upon the given straight line AB to describe a rectili- neal figure similar, and similarly situated, to CDEF. H tº E L ſ\}. 8-| t B Join DF, and at the points A, B in the straight line AB, make the angle BAG equal to the angle at C, (I. 23.) and the angle ABG equal to the angle CDF; & therefore the remaining angle AGB is equal to the remaining angle CFD. (I 32 and ax. 3.) - therefore the triangle FCD is equiangular to the triangle GAB. Again, at the points G, B, in the straight line GB, make the angle BGH equal to the angle DFE, (I. 23.) - t and the angle GBH equal to FDE; therefore the remaining angle GHB is equal to the remaining angle FED, and the triangle FDE equiangular to the triangle G.B.H. then, because the angle AGB is equal to the angle CFD, and BGBſ to DFE, the whole angle AGH is equal to the whole angle CFE; (ax. 2.) for the same reason, the angle ABH is equal to the angle CDE: also the angle at A is equal to the angle at C, (constr.) - and the angle GHB to FED: . therefore the rectilineal figure ABHG is equiangular to CDEF: likewise these figures have their sides about the equal angles pro- portionals; because the triangles GAB, FCD being equiangular, BA is to AG, as CD to CF; (VI. 4.) and because AG is to GB, as CF to FD; and as GB is to GH, so is FD to FE, by reason of the equiangular triangles BGH, DFE, therefore, ex aequali, AG is to GH, as CF to F.E. (v. 22.) Q -*. 226 EUCLID's ELEMENTs. In the same manner it may be proved that AB is to BH, as CD to DE: *. and GH is to HB, as FE to ED. (VI. 4.) Wherefore, because the rectilineal figures ABHG, CDEF are equiangular, º - and have their sides about the equal angles proportionals, they are similar to one another. (VI. def. 1.) - Next, let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated, to the rectilineal figure CDKEF of five sides. - Join DE, and upon the given straight line AB describe the rectili- neal figure ABHG similar, and similarly situated, to the quadrilateral figure CDEF, by the former case : - and at the points B, H, in the straight line BH, make the angle HBI, equal to the angle EDK, and the angle BHL equal to the angle DEK; therefore the remaining angle at L is equal to the remaining angle at K. (I. 32 and ax. 3.) - And because the figures ABHG, CDEF are similar, the angle GHB is equal to the angle FED: (VI. def. 1.) and BHL is equal to DEK; wherefore the whole angle GHL is equal to the whole angle FEK: for the same reason the angle ABL is equal to the angle CDPſ: therefore the five-sided figures AG.HLB, CFEKD are equiangular: and because the figures A GHB, CFED are similar, GHis to HB, as FE to ED; (VT, def. 1.) but as HB to HL, so is ED to EK; (VI. 4.) therefore, ex æquali, G H is to HL, as FE to EPſ: (v. 22.) for the same reason, AB is to BL, as CD to DJſ: and BL is to LH, as DK to KE, (VI. 4.) because the triangles BLH, DICE are equiangular: therefore because the five-sided figures AGHLB, GFEICD are equi- angular, - and have their sides about the equal angles proportionals, they are similar to one another. In the same manner a rectilineal figure of six sides may be described upon a given straight line similar to one given, and so on. Q. E. F. PROPOSITION XIX. THEOREM. Similar triangles are to one another in the duplicate ratio of their homo- logous sides. g Let ABC, DEF be similar triangles, having the angle B equal to the angle E, dº - t and let AB be to BC, as DE to EF. so that the side BC may be homologous to EF. (v. def. 12.) Then the triangle ABC shall have to the triangle DEF the dupli- cate ratio of that which BC has to EF. A D 2^ E # B G 3 \ BOOK VI. PROP. XIX, XX. - 227 Take BG a third proportional to BC, EF, (VI. 11.) so that BC may be to EF, as EF to BG, and join G.A. Then, because as AB to BC, so DE to EF'; alternately, AB is to DE, as BC to EF: (v. 16.) but as BC to EF, so is EF to BG; (constr.) - therefore, as AB to DE, so is EF to BG: (v. 11.) therefore the sides of the triangles A.BG, DEF, which are about the equal angles, are reciprocally proportional: but triangles, which have the sides about two equal angles recipro- cally proportional, are equal to one another: (VI. 15.) ©. therefore the triangle ABG is equal to the triangle DEF: and because as BC is to EF, so EF to BG; - and that if three straight lines be proportionals, the first is said to have to the third, the duplicate ratio of that which it has to the second: (v. def. 10. - therefore %u. to BG the duplicateratio of that which BC has to EF: but as BC is to BG, so is the triangle ABC to the triangle ABG ; (VI. 1.) therefore the triangle ABC has to the triangle A.BG, the duplicate ratio of that which BC has to EF: but the triangle ABG is equal to the triangle DEF; - therefore also the triangle ABC has to the triangle DEF, the dupli- cate ratio of that which BC has to EF. ... Therefore, similar triangles, &c. Q.E.D. - CoR. From this it is manifest, that if three straight lines be pro- portionals, as the first is to the third, so is any triangle upon the first, to a similar and similarly described triangle upon the second. PROPOSITION XX. THEOREM. Similar polygons may be divided into the same number of similar tri- angles, having the same ratio to one another that the polygons have ; and the polygons have to one another the duplicate ratio of that which their homo- logous sides have. Let ABCDE, FGHRL be similar polygons, and let AB be the side homologous to FG : * the polygons ABCDE, FGHKL may be divided into the same number of similar triangles, whereof each shall have to each the same ratio which the polygons have ; and the polygon ABCDE shall have to the polygon. FGHRI, the duplicate ratio of that which the side AB has to the side FG. - Join BE, EC, GL, LII. And because the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL, (VI. def. 1.) and BA is to AE, as GF to FL: (VI. def. 1.) therefore, because the triangles ABE, FGL have an angle in one, equal to an angle in the other, and their sides about these equal angles proportionals, Q 2 228 EUCLID'S ELEMENTS. the triangle ABE is equiangular to the triangle FGL: (VI. 6.) and therefore similar to it; (VI. 4.) wherefore the angle ABE is equal to the angle FGL : and because the polygons are similar, the whole angle ABC is equal to the whole angle FGH; (VI. def 1.) therefore the remaining angle EBC is equal to the remaining angle LG H. (I. ax. 3.) - - and because the triangles A.B.E, FGL are similar, JEB is to BA, as LG to GF; (VI. 4.) and also, because the polygons are similar, AB is to BC, as FG to GH; (VI. def. 1.) therefore, ex æquali, EB is to BC, as LG to GH; (v. 22.) that is, the sides about the equal angles EBC, LG H are proportionals; therefore, the triangle EBC is equiangular to the triangle LGH, (VI. 6.) and similar to it; (VI. 4.) ~ - for the same reason, the triangle ECD likewise is similar to the tri- angle LHK: * therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles. Also these triangles shall have, each to each, the same ratio which the polygons have to one another, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHR : - and the polygon ABCDE shall have to the polygon FGHICL the dupli- cate ratio of that which the side AB has to the homologous side FG. Because the triangle ABE is similar to the triangle FGL, ABE has to FGL, the duplicate ratio of that which the side B.E has to the side GL : (VI. 19.) for the same reason, the triangle BEC has to GLH, the duplicate ratio of that which B.E has to GL : therefore, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH. (v. 11.) * * Again, because the triangle EBC is similar to the triangle LGII, JEBC has to LGH, the duplicate ratio of that which the side EC has to the side LH: & - - for the same reason, the triangle ECD has to the triangle LIIIſ, the . duplicate ratio of that which EC has to LH: therefore, as the triangle. EBC is to the triangle LGH, so is the tri- -angle ECD to the triangle LHK: (v. 11.) but it has been proved, that the triangle EBC is likewise to the triangle LGH, as the tri- angle ABE to the triangle FGL : therefore, as the triangle ABE to the triangle FGL, so is the triangle JEBO to the triangle LGH, and the triangle EOD to the triangle LH/ſ: and therefore, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents: (v. 12.) that is, as the triangle ABE to the triangle FGL, so is the polygon ABCDE to the polygon FGHIL: but the triangle ABE has to the triangle FGL, the duplicate ratio of that which the side AB has to the homologous side FG ; (VI. 19.) therefore also the polygon ABCDE has to the polygon FGHKI, the duplicate ratio of that which AB has to the homologous side FG. Wherefore, similar polygons, &c. Q.E.D. CoR. 1. In like manner it may be proved, that similar four-sided * BOOK WI. PROP. xx, XXI, XXII. 229 figures, or of any number of sides, are one to another in the duplicate ratio of their homologous sides: and it has already been proved in tri- angles: (VI. 19.) therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides. CoR. 2. And if to AB, FG, two of the homologous sides, a third proportional M be taken, (VI. 11.) - AB has to M the duplicate ratio of that which AB has to FG : (v. def. 10.) - & but the four-sided figure or polygon upon AB, has to the four- sided figure or polygon upon FG likewise the duplicate ratio of that which AB has to FG : (VI. 20. Cor. 1.) - therefore, as AB is to M, so is the figure upon AB to the figure upon FG: (v. 11.) which was also proved in triangles: (VI. 19. Cor.) therefore, universally, it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any rectilineal figure upon the first, to a similar and similarly described rectilineal figure upon the second. PROPOSITION XXI. THEOREM, Rectilineal figures which are similar to the same rectilineal figure, are also similar to one another. - Let each of the rectilineal figures A, B be similar to the rectilineal figure C. The figure A shall be similar to the figure B. ^ A A Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportional: (VI. def. 1.) - again, because B is similar to C, they are equiangular, and have their sides about the equal angles proportionals: (VI. def. 1.) therefore the figures A, B are each of them equiangular to C, and have the sides about the equal angles of each of them and of C pro- portionals. & Wherefore the rectilineal figures A and B are equiangular, & ax. 1.) and have their sides about the equal angles proportionals: W. 11. ) therefore A is similar to B. (VI. def. 1.) Therefore, rectilineal figures, &c. Q.E.D. PROPOSITION XXII. THEOREM. If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals: and conversely, if the similar rectilineal figures similarly described upon four straight lines be proportiona's, those straight lines shall be proportionals. Z 230 EUCLID's ELEMENTS. Let the four straight lines AB, CD, EF, GH be proportionals, viz. AB to CD, as EF to GH; and upon AB, CD let the similar rectilineal figures ICAB, LCD be similarly described; and upon EF, GH the similar rectilineal figures M.F. W.H., in like IOla, Illſle]" . - the rectilineal figure KAB shall be to LCD, as MF to NH. FC L - M N S 2^N /N = \|\ \D → T] A B G D E F G H P B. To AB, CD take a third proportional X; (VI. 11.) and to EF, GH a third proportional 0: and because AB is to CD, as EF to GH, therefore CD is to X, as GH to 0; (v. 11.) wherefore, ex æquali, as AB to X, so EF to 0: (v. 22.) but as AB to X, so is the rectilineal figure KAB to the rectilineal figure LCD, - - and as EF to 0, so is the rectilineal figure MF to the rectilineal figure WHſ: (VI. 20. Cor. 2.) g therefore, as KAB to LCD, so is MF to WH. (v. 11.) And if the rectilineal figure KAB be to LCD, as MF to WH; the straight line AB shall be to CD, as EF to G.H. Make as AB to CD, so EF to PR, (VI. 12.) . and upon PR describe the rectilineal figure SR similar and simi- larly situated to either of the figures M.F. W.H. (VI. 18.) then, because as AB to CD, so is EF to PR, w ! and that upon AB, CD are described the similar and similarly situated rectilineals KAB, I, CD, and upon EF, PR, in like manner, the similar rectilineals MF, SR; - therefore KAB' is to LCD, as MF to SR : but by the hypothesis K.A.B is to LCD, as MF to WH; and therefore the rectilineal MF having the same ratio to each of the two NH, SR, - - e - these are equal to one another; (v. 9.) * they are also similar, and similarly situated; therefore GH is equal to PR : and because as AB to CD, so is EF to PR, and that PR is equal to GH; AB is to CD, as EF to G.H. (v. 7.) If therefore, four straight lines, &c. Q.E.D. PROPOSITION XXIII. THEOREM. Jºquiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. - Let AC, CF be equiangular parallelograms, having the angle BCD equal to the angle ECG. Then the ratio of the parallelogram AC to the parallelogram CF, º . the same with the ratio which is compounded of the ratios of their sides. - BOOK WI. PROP. xxIII, XXIV. 231 \ KLM E F - Let BC, CG be placed in a straight line; therefore DC and CE are also in a straight line; (I. 14.) and complete the parallelogram DG ; and taking any straight line K, make as BC to CG, so Jº to L; (VI. 12.) and as DC to CE, so make L to M, (VI. 12.) therefore, the ratios of K to L, and L to M, are the same with the ratios of the sides, - xy-4 viz. of BC to CG, and DC to CE: - but the ratio of K to M is that which is said to be compounded of the ratios of K to L, and L to M. (v. def. A.) . - therefore K has to M the ratio compounded of the ratios of the sides: and because as BC to CG, so is the parallelogram AC to the paral- lelogram CH; (VI. 1.) - but as BC to CG, so is K to L.; - therefore K is to L, as the parallelogram AC to the parallelogram CH: (v. 11.) - again, because as DC to CE, so is the parallelogram CH to the parallelogram CF; - |but as DC to CE, so is Z to M ; wherefore L is to M, as the parallelogram CH to the parallelogram CF; (v. 11.) & * - therefore since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; and as L., to M, so is the parallelogram CH to the parallelogram. CF; ex æquali, K is to M, as the parallelogram. A C to the parallelogram CF: (v. 22.) - - .* but É has to M, the ratio which is compounded of the ratios of the Sldes; aº therefore also the parallelogram AC has to the parallelogram CF, the ratio which is compoundéd of the ratios of the sides. & Wherefore, equiangular parallelograms, &c. Q.E.D. PROPOSITION xxiv. 'THEOREM. Aarallelograms about the diameter of any parallelogram, are similar to the whole, and to one another. Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK parallelograms about the diameter. . The parallelograms EG, HK shall-be similar both to the whole parallelogram ABCD, and to one another. - A E B 232 EUCLID's ELEMENTS. Because DC, GF are parallels, . . the angle ADC is equal to the angle AGF: (1.29.) for the same reason, because BC, EF are parallels, the angle ABC is equal to the angle A.EF: and each of the angles BCD, EFG is equal to the opposite angle DAB, (I. 34.) - - - and therefore they are equal to one another : wherefore the parallelograms ABCD, AEFG, are equiangular: and because the angle ABC is equal to the angle AEF. and the angle BAC common to the two triangles B.A. C., E.A.F. they are equiangular to one another; - therefore as AB to BC, so is AE to EF; (VI. 4.) and because the opposite sides of parallelograms are equal to one another, (I. 34.) - AB is to AD, as AE to AG; (v. 7.) and DC to CB, as GF to FE; and also CD to DA, as FG to GA: therefore the sides of the parallelograms ABCD, AEFG about the equal angles are proportionals; and they are therefore similar to one another; (VI. def. 1.) for the same reason, the parallelogram A B CD is similar to the parallelogram FHCK: wherefore each of the parallelograms GE, KH is similar to DB : but rectilineal figures which are similar to the same rectilineal figure, are also similar to one another: (VI. 21.) therefore the parallelogram GE is similar to KH. Wherefore, parallelograms, &c. Q.E.D, PROPOSITION XXV. PROBLEM. To describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure, - , Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. - - It is required to describe a rectilineal figure similar to ABC, and equal to D. l . A - - K - B \9 F [DT 2^ | - G H L E M TJpon the straight line BC describe the parallelogram B.E. equal to the figure ABC; (I. 45. Cor.) * . also upon CE describe the parallelogram CM equal to D, (1, 45. Cor.) and having the angle FCE equal to the angle CBL: therefore BC and CF are in a straight line, as also LE and EM: (I. 29. and I. 14.) * between BC and CF find a mean proportional GH, (VI. 13.) and upon GH describe the rectilineal figure KGH similar and simi- larly situated to the figure ABC. (VI. 18.) . BOOK VI. PROP. xxv, xxvi. 233 - Because BC is to GH as GH to CF, and that if three straight lines be proportionals, as the first is to the third, so is the figure upon the first to the similar and similarly described figure upon the second ; (VI. 20. Cor. 2.) therefore, as BC to CF, so is the rectilineal figure ABC to KGH: but as BC to CF, so is the parallelogram BE to the parallelogram EF; (VI. 1.) * * therefore as the rectilineal figure ABC is to KGH, so is the paral- lelogram B.E to the parallelogram EF: (v. 11.) and the rectilineal figure ABC is equal to the parallelogram BE; constr.) &: the rectilineal figure KGH is equal to the parallelogram JEF: (v. 14.) : (s but EF is equal to the figure D; (constr.) - wherefore also KGH is equal to D: and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D. Q.E. F. --- PROPOSITION XXVI. THEOREM. If two similar parallelograms have a common angle, and be similarly situated; they are about the same diameter. Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common. ABCD and AEFG shall be about the same diameter. B C For if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the paral- lelogram. EG, - and let GF meet A HC in H ; and through H draw HK parallel to AD or BC; therefore the parallelograms ABCD, A KHG being about the same diameter, they are similar to one another; (VI. 24.) - - wherefore as DA to AB, so is GA to A.K. (VI. def. 1.) but because ABCD and AEFG are similar parallelograms, (hyp.) as DA is to AB, so is GA to AE; x therefore as GA to A.E, so GA to AK; (v. 11.) - that is, GA has the same ratio to each of the straight lines A.B., AK and consequently AK is equal to A.F., (V. 9.) the less equal to the greater, which is impossible : - therefore ABCD and AKHG are not about the same diameter: wherefore ABCD and AEFG must be about the same diameter. Therefore, if two similar, &c. Q.E.D., 234 EUCLID'S ELEMENTS. PROPOSITION XXVII. THEOREM. Of all parallelograms applied to the same straight line, and deficient by parallelograms, similar and similarly situated to that which is described wpon the half of the line; that which is applied to the half, and is similar to its defect, is the greatest. Let AB be a straight line divided into two equal parts in C; and let the parallelogram AD be applied to the half AC, which is therefore deficient from the parallelogram upon the whole line A B by the parallelogram CE upon the other half CB : of all the parallelograms applied to any other parts of AB, and deficient by parallelograms that are similar and similarly situated to CE, AD shall be the greatest. - Let AF be any parallelogram applied to A.K, any other part of AB than the half, so as to be deficient from the parallelogram upon the whole line AB by the parallelogram KH similar and similarly situ- ated to CE: - A TC K B AD shall be greater than A.F. First, let AK the base of AF, be greater than AC the half of AB: and because CE is similar to the parallelogram HK, (hyp.) they are about the same diameter: (VI. 26.) draw their diameter DB, and complete the scheme: then, because the parallelogram CF is equal to F.E, (I. 43.) - add KH to both : therefore the whole CH is equal to the whole-ICE: - but CH is equal to CG, (I. 36.) because the base AC is equal to the base CB; therefore CG is equal to KE: (ax. 1.) to each of these equals add CF': then the whole AF is equal to the gnomon CHL: (ax. 2.) therefore CE, or the parallelogram AD is greater than the paral- lelogram A.F. Next, let A K the base of AF be less than AC : G F M H Ll ND E A K C B then, the same construction being made, because BC is equal to CA, therefore HM is equal to MG; (I. 34.) * º parallelogram DH is equal to the parallelogram DG ; I. 36. - wherefore DH is greater than LG: but DH is equal to DK; (I. 43.) therefore DK is greater than LG: BOOK WI. PROP. xxvi.1, xxvii.I. 235 to each of these add AL; then the whole AID is greater than the whole A.F. Therefore, of all parallelograms applied, &c. Q.E.D. PROPOSITION XXVIII. PROBLEM. . To a given straight line to apply a parallelogram equal to a given rectilineal figure, and deficient by a parallelogram similar to a given paral- lelogram: but the given rectilineal figure to which the parallelogram to be applied is to be equal, must not be greater than the parallelogram applied to half of the given line, having its defect similar to the defect of that which is to be applied; that is, to the given parallelogram. . Let AB be the given straightline, and C the given rectilineal figure, to which the parallelogram to be applied is required to be equal, which figure must not be greater (VI. 27.) than the parallelogram applied to the half of the line, having its defect from that upon the whole line similar to the defect of that which is to be applied; and let D be the parallelogram to which this defect is required to be similar. - It is required to apply a parallelogram to the straight line AB, which shall be equal to the figure C, and be deficient from the paral- lelogram upon the whole line by a parallelogram similar to D. Divide AB into two equal parts in the point E, (I. 10.) and upon EB describe the parallelogram. EBFG similar and simi- larly situated to D, (VI. 18.) and complete the parallelogram AG, which must either be equal to C, or greater than it, by the determination. If A G be equal to C, then what was required is already done :- E[ G O F \ L. M. tº- R. N. ' A E S B for, upon the straightline AB, the parallelogram AG is applied equal to the figure C, and deficient by the parallelogram JEF similar to D. But if A G be not equal to C, it is greater than it: - and EF is equal to AG; (I. 36.) therefore EF also is greater than C. Make the parallelogram KLMN equal to the excess of EF above C, and similar and similarly situated to D: (VI. 25.) then, since D is similar to EF, (constr.) Y therefore also KM is similar to EF, (VI. 21.) - let KL be the homologous side to EG, and LM to GF: and because EF is equal to C and KM together, BF is greater than Klſ; . - therefore the straight line EG is greater than K.L, and GF than Llſ: . make GX equal to LK, and G0 equal to LM, (I. 3.) and complete the parallelogram XGOP: (I. 31.) therefore XO is equal and similar to Klſ: but Klſ is similar to EF; x 2^ wherefore also X0 is similar to EF; 236 EUCLID'S ELEMENTS. and therefore X0 and EF are about the same diameter: (VI. 26.) let GPB be their diameter and complete the scheme. Then because EF is equal to C and KM together, - and XO a part of the one is equal to KM a part of the other, the remainder, viz. the gnomon ERO, is equal to the remainder C. ax. 3. - * * OR is equal to XS, by adding SR to each, (I. 43.) the whole OB is equal to the whole XB: but XB is equal to TE, because the base AE is equal to the base JEB; (I. 36.) *. wherefore also TE is equal to OB: (ax. 1.) add XS to each, then the whole TS is equal to the whole, viz. to the gnomon ER 0: but it has been proved that the gnomon ER0 is equal to C; and therefore also TS is equal to C. Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB, deficient by the parallelogram S.R, similar to the given one D, because S.R is similar to EF. (VI. 24.) Q.E.F. PROPOSITION XXIX. PROBLEM. To a given straight line to apply a parallelogram equal to a given rect- lineal figure, eacceeding by a parallelogram similar to another given. Let AB be the given straight line, and C the given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar. It is required to apply a parallelogram to the given straight line AB which shall be equal to the figure C, exceeding by a parallelogram similar to D. - WDV F I, M K H A TE B \O P X G Divide AB into two equal parts in the point E, (I. 10.) and upon . |FB describe the parallelogram EL similar and similarly situated to D: (VI. 18.) *- - and make the parallelogram. GH equal to EL and C together, and similar and similarly situated to D: (VI. 25.) wherefore GH is similar to EL: (VI. 21.) let KH be the side homologous to FL, and KG to FE: and because the parallelogram. GH is greater than EL, therefore the side KH is greater than FL, • and KG than FE': - produce FL and FE, and make FLM equal to KH, and FEN to KG, and complete the parallelogram MN: MN is therefore equal and similar to GHſ: but GH is similar to EL ; wherefore MN is similar to EL; and consequently EL and MN are about the same diameter: (VI. 26.) - draw their diameter FX, and complete the scheme, º BOOK WI. PROP. xxix., xxx. 237 *s Therefore, since GH is equal to EL and C together, and that GH is equal to MN; MN is equal to EL and C: take away the common part EL; then the remainder, viz. the gnomon NOL, is equal to C. | And because A.E. is equal to EB, the parallelogram AW is equal to the parallelogram WB, (I. 36.) that is, to BM: (I, 43.) add WO to each ; * - therefore the whole, viz. the parallelogram Aly, is equal to the gnomon NOL: p but the gnomon WOL is equal to C; therefore also AJY is equal to C. Wherefore to the straight line AB there is applied the parallelogram Aly equal to the given rectilineal figure C, exceeding by the paral- lelogram PO, which is similar to D, because P0 is similar to EL. (VI. 24.) Q.E. F. - PROPOSITION XXX, PROBLEM. To cut a given straight line in extreme and mean ratio. Let AB be the given straight line. It is required to cut it in extreme and mean ratio. U II: B C F Tpon AB describe the square BC, (1.46.) - and to AC apply the parallelogram CD, equal to BC, exceeding by the figure AD similar to BC: (VI. 29.) - then, since BC is a square, therefore also AD is a square: and because BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD : and these figures are equiangular, - therefore their sides about the equal angles are reciprocally propor- tional: (VI. 14.) therefore, as FE to ED, so AE to EB : but FE is equal to AC, (1.34) that is, to AB; (def. 30.) and ED is equal to A.E; therefore as B.A. to AE, so is AE to EB : but AB is greater than AE; wherefore AE is greater than EB : (v. 14.) - therefore the straight line AB is cut in extreme and mean ratio in J. (VI. def. 3.) Q.E. F. - Otherwise, let AB be the given straight line. It is required to cut it in extreme and mean ratio. 238 EUCLID's ELEMENTs. A C B - Divide AB in the point C, so that the rectangle contained by AB, BC, may be equal to the Square on A.C. (II. 11.) Then, because the rectangle AB, BC is equal to the square on AC; as BA to AC, so is AC to CB : (VI. 17.) therefore AB is cut in extreme and mean ratio in C. (VI. def. 3.) Q. E. F. PROPOSITION xxxi. THEOREM. In right-angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle. Let ABC be a right-angled triangle, having the right angle BAC. The rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon B.A., A.C. Draw the perpendicular AD: (1.12.) therefore, because in the right-angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar to the whole triangle ABC, and to one another: (VI. 8.) - and because the triangle ABC is similar to ADB, as CB to BA, so is BA to BD: (VI. 4.) and because these three straight lines are proportionals, as the first is to the third, so is the figure upon the first to the similar and similarly described figure upon the second: (VI. 20. Cor. 2.) therefore as CB to BD, so is the figure upon CB to the similar and similarly described figure upon B.A. : - and inversely, as DB to BC, so is the figure upon B.A. to that upon BC: (v. B.) - - for the same reason, as DC to CB, so is the figure upon CA to that upon CB : - therefore as BD and DC together to BC, so are the figures upon BA, AC to that upon B.C. (v. 24.) - but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described figures on BA, A.C. (v. A.) Wherefore, in right-angled triangles, &c. Q.E.D. PROPOSITION XXXII. THEOREM. If two triangles which have two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to one another, the remaining sides shall be in a straight line. * Book VI. PROP. xxxi.1, xxxHI. 239 Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, - - . viz. B.A to: A C, as CD to DE; and let AB be parallel to DC, and AC to D.E. A \\ i; CTE - Then BC and GE shall be in a straight line. Because AB is parallel to D6, and the straight line AC meets them, the alternate angles BAC, A CD are equal; (I. 29.) for the same reason, the angle CDE is equal to the angle A CD; - wherefore also BAC is equal to CDE: (ax. 1.) and because the triangles ABC, DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. B.A. to AC, as CD to DE, the triangle ABC is equiangular to DCE: (VI. 6.) therefore the angle ABC is equal to the angle DCE: and the angle BAC was proved to be equal to A CD; therefore the whole angle ACE is equal to the two angles ABC, BAC: (ax. 2.) add to each of these equals the common angle A CB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB: but ABC, BAC, ACB are equal to two right angles: (I. 32.) therefore also the angles ACE, ACB are equal to two right angles: and since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles A CE, A CB equal to two right angles; therefore BC and CE are in a straight line. (I. 14.) Wherefore, if two triangles, &c. Q.E.D. - PROPOSITION XXXIII. THEOREM. In equal circles, angles, whether at the centers or circumferences, have the same ratio which the circumferences on which they stand have to one another: so also have the Sectors. - Det ABC, DEF be equal circles; and at their centers the angles BGC, EHF, and the angles BAC, EDF, at their circumferences. As the circumference BC to the circumference EF, so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF'; 3. and also the sector BGC to the sector EHF. A 2-S D. k ys F Take any number of circumferences C.K., KL, each equal to BC, and any number whatever Filſ, MN, each equal to EF: 240 EUCLID'S ELEMENTS. - and join GK, GL, HM, H.N. - Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal: (III. 27.) therefore what multiple soever the circumference BL is of the cir- cumference BC, the same multiple is the angle BGL of the angle- JBG C: - •, for the same reason, whatever multiple the circumference EN is of the circumference EF, the same multiple is the angle EHM of the angle EHF: - and if the circumference BL be equal to the circumference EN, the angle BGL is also equal to the angle EHN; (III. 27.) and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less; therefore, since there are four magnitudes, the two circumferences IRC, EF, and the two angles BGC, EHF'; and that of the circum- ference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples whatever, viz. the circumference EN, and the angle EHN: and since it has been proved, that if the circumference BL be greater than EN ; the angle BGL is greater than EITV; and if equal, equal; and if less, less; • therefore as the circumference BC to the circumference EF, so is th angle BGC to the angle EHF: (v. def. 5.) - but as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF: (v. 15.) for each is double of each; (III. 20.) therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF, - Also, as the circumference BC to EF, so shall the sector BGC be to the sector JEDIF. * D -- 7. D. *- ſ H º \ | M. N Join BC, CK, and in the circumferences, BC, CK, take any poinst Y, 0, and join BX, XC, CO, O.K. - Then, because in the triangles GBC, GCIC, the two sides BG, GC are equal to the two CG, GIſ each to each, and that they contain equal angles; the base BC is equal to the base CK, (I. 4.) and the triangle GBC to the triangle GCK : and because the circumference BC is equal to the circumference C.K., the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle: (ax. 3. - * therefore the angle BYC is equal to the angle COK; (III. 27.) and the segment BJKC is therefore similar to the segment COIſ (III, def. 11.) *- and they are upon equal straight lines, BC, CK : *- Book VI. PROP. xxxi.11, B. 241 but similar segments of circles upon equal straight lines, are equal to one another: (III. 24.) - \ therefore the segment BYC is equal to the segment 00Fſ: and the triangle BGC was proved to be equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGAſ: for the same reason, the sector KGL is equal to each of the sectors BGC, CGK: - Tº in the same manner, the sectors EHF, FHM, MHV may be proved equal to one another: - - therefore, what multiple soever the circumference BZ is of the circum- ference BC, the same multiple is the sector BGL of the sector BGC ; and for the same reason, whatever multiple the circumference EN is of EF, the same multiple is the sector EHM of the sector JºHF: - and if the circumference BL be equal to EW, the sector BGL is ... equal to the sector EHN; and if the circumference BI, be greater than EN, the sector BGL, is greater than the sector EHN; * and if less, less; since then, there are four magnitudes, the two circumferences BC, JEF, and the two sectors BGC, EHF, and that of the circumference BC, and sector BGC, the circumference BL and sector BGL are any equimultiples whatever; and of the circumference EF, and sector JEHF, the circumference EN, and sector EHN are any equimultiples whatever; and since it has been proved, that if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less: therefore, as the circumference BC is to the circumference EF, so is the sector BGC to the sector EHF (v. def. 5.) Wherefore, in equal circles, &c. Q.E.D. PROPOSITION B. THEOREM, If an angle of a triangle be bisected by a straight line which likewise outs the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle. Let 4 BC be a triangle, and let the angle BAC be bisected by the straight line AD. . . º The rectangle BA, AC shall be equal to the rectangle BD, DC, together with the square on A.D. A Describe the circle ACB about the triangle, (rv. 5.) and produce AD to the circumference in E, and join EC. Then because the angle BAD is equal to the angle CAE, (hyp.) and the angle ABD to the angle AEC, (III. 21.) * R B G 242 EUCLID's ELEMENTs. for they are in the same segment; - - the triangles ABD, AEC are equiangular to one another: (I. 32.) therefore as B.A. to AD, so is EA to AC; (VI.4.) - and consequently the rectangle BA, AC is equal to the rectangle EA, A.D., (VI. 16.) that is, to the rectangle ED, DA, together with the square on AD; (II. 3. - but the * ED, DA is equal to the rectangle BD, DC; (III. 35.) therefore the rectangle BA, AC is equal to the rectangle BD, DC, - together with the square on A.D. Wherefore, if an angle, &c. Q.E.D. PROPOSITION C. THEQREM. If from any angle of a triangle, a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle de- scribed about the triangle. - - - Let ABC be a triangle, and AD the perpendicular from the angle A to the base B.C. - - The rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle. A .*. Ú E Describe the circle ACB about the triangle, (IV. 5.) and draw its diameter A.E, and join EC. - Because the right angle BDA is equal to the angle ECA in a semicircle, (III. 31. - Af * angle A3D equal to the angle AEC in the same segment; III. 21.) - : the triangles ABD, AEC are equiangular: therefore as B.A. to AD, so is EA to AC; (VI. 4.) and consequently the rectangle BA, AC is equal to the rectangle EA, AD. (VI. 16.) If therefore, from any angle, &c. Q.E.D. PROPOSITION D. THEOREM. The rectangle contained by the diagonals of a quadrilateral figure inscribed in a circle, is equal to both the rectangles contained by its opposite sides. Let ABCD be any quadrilateral figure inscribed in a circle, and join A C, B.D. • * - -- The rectangle contained by AC, BD shall be equal to the two rectangles contained by AB, CD, and by AD, B.C. Make the angle ABE equal to the angle DBC: (I. 23.) BOOK v1. PROP. D. * 243 add to each of these equals the common angle EBD, then the angle ABD is equal to the angle EBC: and the angle BDA is equal to the angle BCE, because they are in the same segment; (III. 21.) therefore the triangle ABD is equiangular to the triangle BCE: wherefore, as BC is to CE, so is BD to D.A.; (VI. 4.) and consequently the rectangle BC, AD is equal to the rectangle * BD, CE: (VI. 16.) • again, because the angle ABE is equal to the angle BBC, and the angle BAE to the angle BDC, (III, 21.) * . the triangle ABE is equiangular to the triangle BCD: therefore as B.A. to AE, so is BD to DC; - wherefore the rectangle BA, DC is equal to the rectangle BD, AE: but the rectangle BC, AD has been shewn to be equal to the rectangle BD, CE; therefore the whole rectangle AC, BD is equal to the rectangle AB, DC, together with the rectangle AD, B.C. (II. 1.) Therefore the rectangle, &c. Q.E.D. This is a Lemma of c. Ptolemaeus, in page 9 of his MeyáAn 2.ſvrats. R 2 NOTES TO BOOK WI. IN this Book, the theory of proportion exhibited in the Fifth Book, is applied to the comparison of the sides and areas of plane rectilineal figures, both of those which are similar, and of those which are not similar. - T]ef. I. In defining similar triangles, one condition is sufficient, namely, that similar triangles are those which have their three angles respectively equal; as in Prop. 4, Book VI, it is proved that the sides about the equal angles of equiangular triangles are proportionals. But in defining similar figures of more than three sides, both of the conditions stated in Def. I, are requisite, as it is obvious, for instance, in the case of a square and a rectangle, which have their angles respec- tively equal, but have not their sides about their equal angles proportionals. The following definition has been proposed: “Similar rectilineal figures of more than three sides, are those which may be divided into the same number of similar triangles.” This definition would, if adopted, require the omission of a part of Prop. 20, Book v1. Def. III. To this definition may be added the following: A straight line is said to be divided harmonically, when it is divided into three parts, such that the whole line is to one of the extreme segments, as the other extreme segment is to the middle part. Three lines are in harmonical proportion, when the first is to the third, as the difference between the first and second, is to the difference between the second and third ; and the second is called a harmonic mean between the first and third. - The expression ‘harmonical proportion' is derived from the following fact in the Science of Acoustics, that three musical strings of the same material, thick- mess and tension, when divided in the manner stated in the definition, or numeri- cally as 6, 4, and 3, produce a certain musical note, its fifth, and its octave. Def. Iv. The term altitude, as applied to the same triangles and parallelograms, will be different according to the sides which may be assumed as the base, unless they are equilateral. te Prop. 1. In the same manner may be proved, that triangles and parallelograms upon equal bases, are to one another as their altitudes. v. Prop. II. Observe, that the three diagrams include all the possible cases of this important proposition. In the first diagram, the line intersects the sides between the base and the vertex of the triangle: in the second, the line intersects the sides produced below the base: in the third, the line intersects the sides pro- duced beyond the vertex. - Prop. A. When the triangle ABC is isosceles, the line which bisects the exterior angle at the vertex is parallel to the base. In all other cases, if the line which bisects the angle BAC cut the base BC in the point G, then the straight line BD is harmonically divided in the points G, C. For BG is to GC as BA is to AC; (v1. 3.) . - and BD is to DC as BA is to AC, (v1. A.) therefore BD is to DC as BG is to GC, but BG = BD – DG, and GC = GD – DC. Wherefore BD is to DC as BD – DG is to GD – DC. Hence BD, DG, DC, are in harmonical proportion. Prop. Iv is the first case of similar triangles, and corresponds to the third case of equal triangles, Prop. 26, Book I. - .NOTES TO BOOK WI. 245 Sometimes the sides opposite to the equal angles in two equiangular triangles, are called the corresponding sides, and these are said to be proportional, which is simply taking the proportions in Euclid alternately. - The term homologous (GuðMoyos), has reference to the places the sides of the triangles have in the ratios, and in one sense, homologous sides may be considered as corresponding sides. The homologous sides of any two similar rectilineal figures will be found to be those which are adjacent to two equal angles in each figure. Prop. v., the converse of Prop. IV, is the second case of similar triangles, and corresponds to Prop. 8, Book I, the second case of equal triangles. Prop. v1 is the third case of similar triangles, and corresponds to Prop. 4, Book I, the first case of equal triangles. The property of similar triangles, and that contained in Prop. 47, Book I, are the most important theorems in Geometry. Prop. v1.1 is the fourth case of similar triangles, and corresponds to the fourth case of equal triangles demonstrated in the note to Prop. 26, Book I. Prop. VIII. It may be remarked that propositions are introduced in the Second and Third Books under the form of properties of rectangles, and are identical with propositions in the Sixth Book given in the form of proportion : as for in- stance, the first proportion in the Corollary to Euc. vi. 8, is identical to Euc. 11. 14. In the latter (see fig.) the square on EH is equal to the rectangle BE, EF: in the former (see fig.) AD is a mean proportional between BD and DC. Prop. Ix. The learner must not forget the different meanings of the word part, as employed in the Elements. The word here has the same meaning as in Euc. v. def. 1. It may be remarked, that this proposition is a more simple case of the next, namely, Prop. x. Prop. x1. This proposition is that particular case of Prop. xII, in which the second and third terms of the proportion are equal. These two problems exhibit the same results by a geometrical construction, as are obtained by numerical multiplication and division. - - Prop. XIII. The difference in the two propositions Euc. II. 14, and Euc. VI. 13, is this: in the Second Book, the problem is, to make a rectangular figure or square equal in area to an irregular rectilinear figure, in which the idea of ratio is not introduced. In the Prop. in the Sixth Book, the problem relates to ratios only, and it requires to divide a line into two parts, so that the ratio of the whole line to the greater segment may be the same as the ratio of the greater segment to the less. The result in this proposition obtained by a Geometrical construction, is ana- logous to that which is obtained by the multiplication of two numbers, and the extraction of the square root of the product. - It may be observed, that half the sum of AB and BC is called the Arithmetic mean between these lines; also that BD is called the Geometric mean between the same lines. - To find two mean proportionals between two given lines is impossible by the straight line and circle. Pappus has given several solutions of this problem in Book III, of his Mathematical Collections; and Eutocius has given, in his Com- mentary on the Sphere and Cylinder of Arêhimedes, ten different methods of solving this problem. Prop. xrv depends on the same principle as Prop. xv, and both may easily be demonstrated from one diagram. Join DF, FE, EG in the fig. to Prop. xiv, and the figure to Prop. xv is formed. We may add, that there does not appear any reason why the properties of the triangle and parallelogram should be here separated, and not in the first proposition of the Sixth Book. / * 246 EUCLID's ELEMENTS. Prop. xv holds good when one angle of one triangle is equal to the defect from what the corresponding angle in the other wants of two right angles. - This theorem will perhaps be more distinctly comprehended by the learner, if he will bear in mind, that four magnitudes are reciprocally proportional, when the ratio compounded of these ratios is a ratio of equality. - - Prop. xv.11 is only a particular case of Prop. xv.1, and more properly, might appear as a corollary: and both are cases of Prop. xiv. Algebraically. Let AB, CD, E, F, contain a, b, c, d units respectively. Then, since a, b, c, d are proportionals, ... ; == i. Multiply these equals by bd, ... ad = be, or, the product of the extremes is equal to the product of the means. And conversely, If the product of the extremes be equal to the product of the means, - - Yº \ - or ad = be, - then, dividing these equals by 5d, .". ; -: i. or the ratio of the first to the second number, is equal to the ratio of the third to the fourth. ºv Similarly may be shewn, that if ; =} ; then ad = bº. - And conversely, if ad = 5%; then; = %. Trop. xviii. Similar figures are said to be similarly situated, when their homologous sides are parallel, as when the figures are situated on the same straight line, or on parallel lines: but when similar figures are situated on the sides of a triangle, the similar figures are said to be similarly situated when the homologous sides of each figure have the same relative position with respect to one another; that is, if the bases on which the similar figures stand, were placed parallel to one another, the remaining sides of the figures, if similarly situated, would also be parallel to one another. - z Prop. xx. It may easily be shewn, that the perimeters of similar polygons, are proportional to their homologous sides. - Prop. xxi. This proposition is not restricted to triangles, but must be so un- derstood as to include all rectilineal figures whatsoever, which require for the conditions of similarity another condition than is required for the similarity of triangles. See note on Euc. VI. Def. I. Prop. xxIII. The doctrine of compound ratio, including duplicate and triplicate ratio, in the form in which it was propounded and practised by the ancient Geo- meters, has been almost wholly superseded. However satisfactory for the purposes of exactreasoning the method of expressing the ratio of two surfaces, or of two solids by two straight lines, may be in itself, it has not been found to be the form best suited for the direct application of the results of Geometry. Almost all modern writers on Geometry and its applications to every branch of the Mathematical Sciences, have adopted the algebraical notation of a quotient AB; BC; or of a fraction #: for expressing the ratio of two lines AB, BC: as well as that of a product AB x BC, or AB . BC, for the expression of a rectangle. The want of a concise and expressive method of notation to indicate the proportion of Geome- trical Magnitudes in a form suited for the direct application of the results, has doubtless favoured the introduction of Algebraical symbols into the language of Germetry. It must be admitted, however, that such notations in the language of NOTES TO BOOK WI. - 247 * ~ pure Geometry are liable to very serious objections, chiefly on the ground that pure Geometry does not admit the Arithmetical or Algebraical idea of a product or a quotient into its reasonings. On the other hand, it may be urged, that it is not the employment of symbols which renders a process of reasoning peculiarly Geometrical or Algebraical, but the ideas which are expressed by them. If symbols be employed in Geometrical reasonings, and be understood to express the magnitudes themselves and the conception of their Geometrical ratio, and not any measures, or numerical values of them, there would not appear to be any very great objections to their use, provided that the notations employed were such as are not likely to lead to misconception. It is however desirable, for the sake of avoiding confusion of ideas in reasoning on the properties of number and of magnitude, that the language and notations employed both in Geometry and Algebra should be rigidly defined and strictly adhered to, in all cases. At the commencement of his Geometrical studies, the student is recommended not to employ the symbols of Algebra in Geometrical demonstrations. How far it may be necessary or advisable to employ them when he fully understands the nature of the subject, is a question on which some difference-of opinion exists. Prop. xxv. There does not appear any sufficient reason why this proposition should be placed between Prop. XXIV. and Prop. xxvi. Prop. xxvii. To understand this and the three following propositions more easily, it is to be observed: - 1. “That a parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. Ex. gr. the parallelogram AC is said to be applied to the straight line A B. - - 2. But a parallelogram AE is said to be applied to a straight line AB, de- ficient by a parallelogram, when AD the base of AE is less than AB, and therefore AE is less than the parallelogram AC described tupon AB in the same angle, and between the same parallels, by the parallelogram DC; and DC is therefore called the defect of A.E. - - 3. And a parallelogram AG is said to be applied to a straight line AB, ex- ceeding by a parallelogram, when AF the base of AG is greater than AB, and therefore AG exeeeds AC the parallelegram described upon AB in the same angle, and between the same parallels, by the parallelogram B.G.”—Simson. Props. xxv.11—XXIX are not usually read: and Prop. xxx, is read as deduced from Euc. II, 10. Prop. xxxi. This proposition is the general case of Prop. 47, Book 1, for any similar rectilineal figure described on the sides of a right-angled triangle. The demonstration, however, here given is wholly independent of Euc. 1. 47. Prop. xxxi.II. In the demonstration of this important proposition, angles greater than two right angles are employed, in accordance with the criterion of proportionality laid down in Euc. v. def. 5. - - This proposition forms the basis of the assumption of arcs of eircles for the measures of angles at their centers. One magnitude may be assumed as the measure of another magnitude of a different kind, when the two are so connected, that any variation in them takes place simultaneously, and in the same direct proportion. This being the case with angles at the center of a circle, and the arcs subtended by them ; the arcs of circles can be assumed as the measures of the angles they subtend at the center of the circle. Prop. B. The converse of this proposition does not hold good when the triangle is isosceles. •. f 248 EUCLID's, ELEMENTS. ** QUESTIONS ON BOOK WI. 1. DISTINGUISH between similar figures and equal figures. 2. What is the distinction between homologous sides, and equal sides in Geo- metrical figures? - - - 3. What is the number of conditions requisite to determine similarity of figures Is the number of conditions in Euclid's definition of similar figures greater than what is necessary Propose a definition of similar figures which includes no superfluous condition. - 4. Explain how Euclid makes use of the definition of proportion in Euc. v.I. 1. 5. Prove that triangles on the same base are to one another as their altitudes. 6. If two triangles of the same altitude have their bases unequal, and if one of them be divided into m equal parts, and if the other contain n of those parts; prove that the triangles have the same numerical relation as their bases, Why is this Proposition less general than Eue. vi. 1 & 7. Are triangles which have one angle of one equal to one angle of another, and the sides about two other angles proportionals, necessarily similar 8. What are the conditions, considered by Euclid, under which two triangles are similar to each other 3 - 9. Apply Euc, VI. 2, to trisect the diagonal of a parallelogram. 10. When are three lines said to be in harmonical proportion If both the interior and exterior angles at the vertex of a triangle (Euc. vſ. 3, A.) be bisected by lines which meet the base, and the base produced in D, G ; the segments BG GD, GC of the base shall be in harmonical proportion. * - 11. If the angles at the base of the triangle in the figure Euc. vi. 4, be equal to each other, how is the proposition modified ? - - 12. Under what circumstances will the bisecting line in the fig, Euc, VI. A., meet the base on the side of the angle bisected? Shew that there is an indeter- minate case. | 13. State some of the uses to which Euc. v1.4, may be applied. *. 14. Apply Euc. vi. 4, to prove that the rectangle contained by the segments of any chord passing through a given point within a circle is constant. 15. Point out clearly the difference in the proofs of the two latter cases in Euc. VI. 7. & 16. From the corollary of Euc. vi. 8, deduce a proof of Euc. I.47. 17. Shew how the last two properties stated in Euc, VI. 8. Cor. may be deduced from Euc. I. 47; II. 2; VI. 17. - 18. Given the nth part of a straight line, find by a Geometrical construction, the (n + 1)th part. 19. Define what is meant by a mean proportional between two given lines: and find a mean proportional between the lines whose lengths are 4 and 9 units respectively. Is the method you employ suggested by any propositions in any of the first Four Books s > 20. Determine a third proportional to two lines of 5 and 7 units: and a fourth proportional to three lines of 5, 7, 9, units. - z- 21. Find a straight line which shall have to a given straight line, the ratio of 1 to Vå. *s. 22. Define reciprocal figures. Enunciate the propositions proved respecting such figures in the Sixth Book. 23. Give the corollary, Euc. vi. 8, and prove thence that the Arithmetic mean is greater than the Geometric mean between the same extremes. QUESTIONS ON B00K VI. 249 24. If two equal triangles have two angles together equal to two right angles, the sides about those angles are reciprocally proportional. 25. Give Algebraical proofs of Prop. 16 and 17 of Book v1. 26. Enunciate and prove the converse of Euc. vi. 15. 27. Explain what is meant by saying, that “similar triangles are in the dupli- cate ratio of their homologous sides.” 28. What are the data which determine triangles bothin species and magnitude? How are those data expressed in Geometry? 29. If the ratio of the homologous sides of two triangles be as 1 to 4, what is the ratio of the triangles And if the ratio of the triangles be as 1 to 4, what is the ratio of the homologous sides 2 A - 30. Shew that one of the triangles in the figure, Euc. IV. 10, is a mean propur- tional between the other two. 31. What is the Algebraical interpretation of Euc, VI. 19 32. From your definition of Proportion, prove that the diagonals of a square are in the same proportion as their sides. - ſº 33. What propositions does Euclid prove respecting similar polygons 34. The parallelograms about the diameter of a parallelogram are similar to the whole and to one another. Shew when they are equal. 35. Prove Algebraically, that the areas of similar triangles and of similar parallelograms, are proportional to the squares on their homologous sides. 36. How is it shewn that equiangular parallelograms have to one another the ratio which is compounded of the ratios of their bases and altitudes 37. To find two lines which shall have to each other, the ratio compounded of the ratios of the lines A to B, and C to D. - 38. State the force of the condition “similarly described ;” and shew that, on a given straight line, there may be described as many polygons of different magni- tudes, similar to a given polygon, as there are sides of different lengths in the polygon. 9 39. Describe a triangle similar to a given triangle, and having its area double that of the given triangle. 40. The three sides of a triangle are 7, 8, 9 units respectively; determine the length of the lines which meeting the base, and the base produced, bisect the interior angle opposite to the greatest side of the triangle, and the adjacent exterior angle. 41. The three sides of a triangle are 3, 4, 5 inches respectively; find the lengths of the internal and external segments of the sides determined by the lines which bisect the interior and exterior angles of the triangle. 42. What are the segments into which the hypotenuse of a right-angled triangle is divided by a perpendicular drawn from the right angle, if the sides containing it are a and 3a units respectively? 43. If the homologous sides of two triangles be as 3 to 4, and the area of one triangle be known to contain 100 square units; how many square units are contained in the area of the other triangle - 44. Prove that if BD be taken in AB produced (fig. Euc. vi. 30.) equal to the greater segment AC, then AD is divided in extreme and mean ratio in the point B. Shew also, that in the series 1, 1, 2, 3, 5, 8, &c. in which each term is the sum of the two preceding terms, the last two terms perpetually approach to the pro- portion of the segments of a line divided in extreme and mean ratio. Find a general expression (free from surds) for the nth term of this series. 45. The parts of a line divided in extreme and mean ratio are incommensurable with each other. 250 EUCLID's ELEMENTS. - 46. Shew that in Euclid's figure (Euc. II. 11.) four other lines, besides the given line, are divided in the required manner. - .47. Enunciate Euc. vi. 31. What theorem of a previous book is included in this proposition ? - 48. What is thesuperior limit, as to magnitude, of the angle at the circumference in Euc. vi. 33° Shew that the proof may be extended by withdrawing the usually supposed restriction as to angular magnitude; and then deduce, as a corollary, the proposition respecting the magnitudes of angles in segments greater than, equal to, or less than a semicircle. - * - 49. The sides of a triangle inscribed in a circle are a, b, c units respectively : find by Euc. v.I. c, the radius of the circumscribing circle. 50. Enunciate the converse of Euc. VI. D. 51. Shew independently that Euc. v.I. D, is true when the quadrilateral figure is rectangular. - - - - 52. Shew that the rectangles contained by the opposite sides of a quadrilateral figure which does not admit of having a circle described about it, are together greater than the rectangle contained by the diagonals. 53. What different conditions may be stated as essential to the possibility of the inscription and circumscription of a circle in and about a quadrilateral figure ? 54. Find two lines which shall be reciprocally proportional to two given lines. 55. Apply Euc. vi. 11, to find a series of six lines which shall be in con- tinual proportion when the first two lines of the series are given. 56. What problem in the Sixth Book is immediately solved by Euc. II. 14. 57. Find a fourth proportional to three given similar triangles. 58. How many similar triangles are in the figure Euc. vſ. B. : Name them according to their homologous sides. ^, 59. State the proposition from which it is shewn that the side of an isosceles triangle inscribed in a circle is a mean proportional between the altitude of the triangle and the diameter of the circle. - 60. How do you explain the case of Euc. v.I.A., when the triangle is isosceles? 61. Shew by the aid of Euc. vi. 31, that a trapezium may be constructed equal and similar to any number of similar trapeziums. 62. If the sides about each of the angles of two triangles be proportional, the triangles are similar. Is this necessarily true in the case of quadrilateral figures 63. Under certain circumstances the reciprocals of theorems are necessarily true. Illustrate this by examples from the Sixth Book of the Elements. 64. The perimeters of similar polygons are proportional to the homologous sides of the polygons. - - * 65. What are the four forms in which Euclid's definition of Proportion is applied in the Fifth and Sixth Books of the Elements Quote instances. 66. If Euclid had proved first that rectangles and parallelograms of the same altitude are to one another as their bases; and then deduced that triangles of the same altitude are to one another as their bases:—does it appear that any advan- tages yould have arisen in this mode of treating the subject 67. Point out those propositions in the Sixth Book in which Euclid's definition of proportion is directly applied. , 68. In what cases are triangles proved to be equal in Euclid, and in what cases are they proved to be similar * - - > , - BOOK XI. IDEFINITIONS. •. - I. A solid is that which hath length, breadth, and thickness. II. That which bounds a solid is a superficies. A straight line is perpendicular, or at right angles to a plane, when it makes right angles with every straight line meeting it in that plane. - - e IV. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. W. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which a perpen- dicular to the plane drawn from any point of the first line above the plane, meets the same plane. . - - The inclination of a plane to a plane is the acute angle contained by two straight lines drawn from any the same point of their common Section at right angles to it, one upon one plane, and the other upon the other plane. WII. - Two planes are said to have the same, or a like inclination to one another, which two other planes have, when the said angles of inclination are equal to one another. - - - WIII. Parallel planes are such as do not meetone another though produced. IX. A solid angle is that which is made by the meeting, in one point, of more than two plane angles, which are not in the same plane. X Equal and similar solid figures are such as are contained by similar planes equal in number and magnitude. * 252 EUCLID'S ELEMENTS. XI. Similar solid figures are such as have all their solid angles equal, each to each, and are contained by the same number of similar planes. XII. A pyramid is a solid figure contained by planes that are constituted betwixt one plane and one point above it in which they meet. XIEI. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and parallel to one another; and the others parallelograms. - - XIV. A sphere is a solid figure described by the revolution of a semicircle about its diameter, which remains unmoved. - XV. The axis of a sphere is the fixed straight line about which the semicircle revolves. XVI. - The center of a sphere is the same with that of the semicircle. XVII. The diameter of a sphere is any straight line which passes through the center, and is terminated both ways by the superficies of the sphere. XVIII. A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed. - If the fixed side be equal to the other side containing the right angle, the cone is called a right-angled cone; if it be less than the other side, an obtuse-angled ; and if greater, an acute-angled cone. XIX The axis of a cone is the fixed straight line about which the triangle revolves. • XX. The base of a cone is the circle described by that side containing the right angle, which revolves. XXI. A cylinder is a solid figure described by the revolution of a right- angled parallelogram about one of its sides which remains fixed. - - **. XXII. ' The axis of a cylinder is the fixed straight line about which the parallelogram revolves. BOOK XI. PROP. I. 253 XXIII. The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram. XXIV. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals. XXV. - A cube is a solid figure contained by six equal squares. XXVI. A tetrahedron is a solid figure contained by four equal and equi- lateral triangles. An octahedron is a solid figure contained by eight equal and equilateral triangles. - XXVIII. A dodecahedron is a solid figure contained by twelve equal pentagons which are equilateral and equiangular. - XXIX. An icosahedron in a solid figure contained by twenty equal and equilateral triangles. Def. A. A parallelopiped is a solid figure contained by six quadrilateral figures, whereof every opposite two are parallel. PROPOSITION I. THEOREM, One part of a straight line oannot be in a plane, and another part above it. If it be possible, let AB, part of the straight line ABC, be in the plane, and the part BC above it: - * = C and since the straight line AB is in the plane, it can be produced in that plane: - let it be produced to D; • * º and let any plane pass through the straight line AD, and be turned about it until it pass through the point C: and because the points B, C are in this plane, - the straight line BC is in it: (I. def. 7.) therefore there are two straight lines ABC, ABD in the same plane that have a common segment AB; (I, 11, Cor.) which is impossible. Therefore, one part, &c. Q.E.D. 254 EUCLID's ELEMENTs. * PROPOSITION II. THEOREM. Two straight lines which cut one another are in one plane, and three straight lines which meet one another are in one plane. : Let two straight lines AB, CD cut one another in E; - then AB, CD shall be in one plane: and three straight lines FC, CB, B.E, which meet one another, shall be in one plane. ... " e - - Tet any plane pass through the straight line EB, and let the plane be turned about EB, produced, if necessary, until it pass through the point C. Then, because the points E, C are in this plane, the straight line EC is in it: (I. def. 7.) for the same reason, the straight line BC is in the same: - and by the hypothesis, EB is in it: therefore the three straight lines EC, CB, BE are in one plane; but in the plane in which EC, EB are, in the same are CD, AB: (XI. 1.) therefore, AB, CD are in one plane. Wherefore, two straight lines, &c. Q.E.D. BROPOSITION III. THEOREM. If two planes out one another, their common section is a straight line. Let two planes AB, BC cut one another, and let the line DB be their common section. Then DB shall be a straight line. JB F f A |c D If it be not, from the point D to B, draw, in the plane AB, the straight line DEB, (post. 1.) and in the plane BC, the straight line DFB : then two straight lines DEB, DFB have the same extremities, and therefore include a space betwixt them; which is impossible: (I, ax. 10.) - therefore BD, the common section of the planes AB, BC, cannot but be a straight line. - Wherefore, if two planes, &c. Q.E.D. BOOK XI. PROP. IV. ‘s - 255. PROPOSITION IV. THEOREM. If a straight line stand at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane which passes through them, that is, to the plane in which they are. Let the straight line EFstand at right angles to each of the straight lines AB, CD, in E the point of their intersection. Then EF shall also be at right angles to the plane passing through Alb, CD. Take the straightlines A.E, EB, CE, ED all equal to one another; and through E draw, in the plane in which are AB, CD, any straight line G.E.H ; : and join AD, CB; then from any point F, in EF, draw FA, FG, FD, FC, F.H., F.B. And because the two straight lines A.E, JED are equal to the two B.E, EC, each to each, - and that they contain equal angles AED, BEC, (I. 15.) the base AD is equal to the base BC, (I. 4.) and the angle DAE to the angle EBC: and the angle AEG is equal to the angle B.E.H. (I. 15.) therefore the triangles AEG, BEH have two angles of the one equal to two angles of the other, each to each, and the sides A.E, JEB, adjacent to the equal angles, equal to one another: - * - - wherefore they have their other sides equal: (1.26.) therefore GE is equal to EH, and AG to BH: .. and because AE is equal to EB, and FE common and at right angles to them, - the base AF is equal to the base FB; (r. 4.) for the same reason, CF is equal to FD: and because AD is equal to BC, and AF to FB, the two sides FA, AD are equal to the two FB, BC, each to each; and the base DF was proved equal to the base FC; therefore the angle FAD is equal to the angle FBC : (1.8.) : again, it was proved that GA is equal to BH, and also AF to FB; therefore FA and A G are equal to FB and B.H., each to each ; and the angle FAG has been proved equal to the angle FBFI; therefore the base GF'is equal to the base FH: (I. 4.) again, because it was proved that GE is equal to EH, and EFis common; therefore G.E, JEF are equal to HE, EF, each to each; and the base GF'is equal to the base FH; therefore the angle GEF is equal to; he angle HEF; (1.8.) and consequently each of these angles is a right angle. (I. def. 10.) Therefore FE makes right angles with GH, that is, with any straight line drawn through E in the plane passing through AB, CD. - In like manner, it may be proved, that FE makes right angles with every straight line which meets it in that plane. - 256 s EUCLID's ELEMENTS. But a straight line is at right angles to a plane when it makes right angles with every straightline which meets it in that plane: (XI. def. 3.) therefore EF is at right angles to the plane in which are AB, CD. Wherefore, if a straight line, &c. Q. E. D. BROPOSITION W. THEOREM. If three straight lines meet all in one point, and a fºraight line stands at right angles to each of them in that point; these three Straight lines are &n one and the same plane. Let the straightline AB stand at right angles to each of the straight lines BC, BD, B.E., in B the point where they meet. Then BC, BD, BE shall be in one and the same plane. b If not, let, if it be possible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which, with the plane in which BD and BE are, is a straight line; (XI. 3.) let this be BF': therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AB, BC. And because AB stands at right angles to each of the straight lines BD, B.E, it is also at right angles to the plane passing through them: (XI. 4.) and therefore makes right angles with every straight line meeting it in that plane; (XI. def. 3.) but BF, which is in that plane, meets it; therefore the angle ABF is a right angle: but the angle ABC, by the hypothesis, is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane, which is impossible; (I. ax. 9.) therefore the straight line BC is not above the plane in which are BD and B.E.: wherefore the three straight lines BC, BD, BE are in one and the same plane. -- Therefore, if three straight lines, &c. Q.E.D. PROPOSITION VI. THEOREM. If two straight lines be at right angles to the same plane, they shall be parallel to one another. - Let the straight lines AB, CD be at right angles to the same plane. Then AB shall be parallel to CD, BOOK x1. PROP. VI, VII. 257 2. Let themi meet the plane in the points B, D, and draw the straight line BD, to which draw DE at right angles, in the same plane; (I. 11.) and make DE equal to AB, (r. 3.) and join BE, AE, AD. Then, because AB is perpendicular to the plane, it makes right angles with every straight line which meets it, and is in that plane: (XI. def. 3.) - but BD, B.E, which are in that plane, do each of them meet AB; therefore each of the angles ABD,-ABE is a right angle; ,” for the same reason, each of the angles CDB, CDE is a right angle: and because AB is equal to DE, and BD common, - the two sides AB, BD are equal to the two ED, DB, each to each; and they contain right angles: ... - therefore the base AD is equal to the base BE: (I. 4.) again, because AB is equal to DE, and BE to AD; AB, BE are equal to ED, DA, each to each; and, in the triangles A.B.E., EDA, the base AE is common; therefore the angle ABE is equal to the angle EDA : (I. 8-) * but ABE is a right angle; - therefore EDA is also a right angle, and ED perpendicular to DA: but it is also perpendicular to each of the two BD, DC; wherefore ED is at right angles to each of the three straight lines BD, DA, DC in the point in which they meet: - therefore these three straight lines are all in the same plané: (XI. 5.) but AB is in the plane in which are BD, DA, (XI. 2.) because any three straightlines which meet one another are in one plane: therefore AB, BD, DC are in one plane: and each of the angles ABD, BDC is a right angle; therefore AB is parallel to CD. (I. 28.) . Wherefore, if two straight lines, &c., Q.E.D. PROPOSITION VII. THEOREM. If two straight lines be parallel, the straight line drawn from any point in the one to any point in the other, is in the same plane with the parallels. Let AB, CD be parallel straightlines, and take any point E in the one, and the point Fin the other. Then the straight line which joins E and F shall be in the same plane with the parallels. - - a E B Nº FIV\ C F. D. 258 - EUCLID's ELEMENTS. If not, let it be, if possible, above the plane, as EGF; and in the plane ABCD in which the parallels are, draw the straight line EHF from E to F. And since EGF also is a straight line, the two straight lines EHF, JEGF'include a space between them, which is impossible. (I. ax. 10.) Therefore the straight line joining the points E, F is not above the plane in which the parallels AB, CD are, and-is therefore in that plane. Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION VIII. THEOREM, If two straight lines be parallel, and one of them be at right angles to a plane; the other also shall be at right angles to the same plane. Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a plane. - * Then the other CD shall be at right angles to the same plane. A C E Let AB, CD meet the plane in the points B, D, and join BD: therefore AB, CD, BD are in one plane. (xi. 7.) In the plane to which AB is at right angles, draw DE at right angles to B.D., (I. 11.) and make DE equal to AB, (I. 3.) and join B.E, AE, AD. And because AB is perpendicular to the plane, it is perpendicular to every straight line which meets it, and is in that plane ; (x1, def. 3.) therefore each of the angles ABD, ABE is a right angle: and because the straight line BD meets the parallel straight lines AIB, CD, the angles ABD, CDB. are together equal to two right angles: (1, 29.) - and ABD is a right angle; ~ therefore also CDB is a right angle, and CD perpendicular to BD: and because AB is equal to DE, and BD common, the two AB, BD are equal to the two ED, DB, each to each; and the angle ABD is equal to the angle EDB, because each of them is a right angle; therefore the base AD is equal to the base B.E: (1. 4.) again, because AB is equal to DE, and B.E to A.D, the two AB, BE are equal to the two ED, DA, each to each; and the base AE is common to the triangles A.B.E, EDA ; wherefore the angle ABE is equal to the angle EDA : (I. 8.) but ABE is a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: - but it is also perpendicular to BD; (constr.) Book x1. PROP. VIII, Ix, x. 259 therefore ED is perpendicular to the plane which passes through BD, D.A.; (XI. 4.) * , , ; - - and therefore makes right angles with every straight line meeting it in that plane: (XI. def. 3.) but DC is in the plane passing through BD, DA, because all three are in the plane in which are the parallels AB, CD wherefore ED is at right angles to DC; - and therefore CD is at right angles to DE: but CD is also at right angles to DB ; therefore CD is at right angles to the two straight lines DE, DB, in the point of their intersection D; - - - and therefore is at right angles to the plane passing through DJ, DB, (XI. 4.) which is the same plane to which AB is at right angles. Therefore, if two straight lines, &c. Q. E. D. PROPOSITION Ix. THEOREM, Two straight lines which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB, CD be each of them parallel to EF, and not in the same plane with it. Then AB shall be parallel to CD. A H B e—X-—r a-É—i, In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF; (I. 14.) and in the plane passing through EF, CD draw GK at right angles to the same EF. - And because EF is perpendicular both to GH and GK, JEF is perpendicular to the plane HGH passing through them: (XI. 4.) - and EF is parallel to AB; therefore AB is at right angles to the plane HG.K. (XI. 8.) For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGIſ. But if two straight lines are at right angles to the same plane, they are parallel to one another : (XI. 6.) g therefore AB is parallel to CD. Wherefore, two straight lines, &c. Q.E.D. PROPOSITION X. THEOREM. If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two; the first two and the other two shall contain equal angles. Let the two straight lines AB, BC, which meet one another, be g - , S 2 260 EUCLID'S ELEMENTS. parallel to the two straight lines DE, EF, that meet one another, and are not in the same plane with AB, BC. - - • The angle ABC shall be equal to the angle DEF: *ZINo º D F Take BA, BC, ED, EF, all equal to one another; and join AD, CF, B.E, AC, DF, Then, because BA is equal and parallel to ED, therefore AD is both equal and parallel to B.E. (I. 33.) ge For the same reason, CF is equal and parallel to B.E. Therefore AD and CF are each of them equal and parallel to B.E. But straight lines that are parallel to the same straight line, and not in the same plane with it, are parallel to one another: (XI, 9.) therefore AD is parallel to CF; and it is equal to it; (1...ax. 1.) and AC, DFjoin them towards the same parts; and therefore A C is equal and parallel to D.F. (I. 33.) And because AB, BC are equal to DE, EF, each to each, - and the base A C to the base DF'; - the angle ABC is equal to the angle DEF (r. 8.) Therefore, if two straight lines, &c. Q.E.D. PROPOSITION XI. PROBLEM. To draw a straight line perpendicular to a plane, from a given pownt above it. - Let A be the given point above the plane B.H. It is required to draw from the point A a straight line perpen- dicular to the plane B.H. . In the plane draw any straight line BC, - and from the point A draw AD perpendicular to BC. (r. 12.) If then AD be also perpendicular to the plane BII, the thing required is already done: * but if it be not, from the point D draw, in the plane BH, the straight line D.E at right angles to BC: (1. 11.) tº and from the point A draw AF perpendicular to D.E. Then AF'shall be perpendicular to the plane BH. E A G \ | Through F draw GH parallel to BC. (I. 31.) And because BC is at right angles to ED and DA, BC is at right angles to the plane passing through ED, DA: (XI. 4.) - and GH is parallel to B0: * . Book xi. PROP. x1, xII, XIII. 261 but, if two straight lines be parallel, one of which is at right angles to a plane, - * - - the other is at right angles to the same plane; (XI. 8.) wherefore GH is at right angles to the plane through ED, DA ; and is perpendicular to every straight line meeting it in that plane (xI. def. 3. - but AF, *ia. is in the plane through ED, DA, meets it; therefore GH is perpendicular to AF; and consequently AF is perpendicular to GH; * and AF is perpendicular to DE; therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stand at right angles to each of two straight lines in the point of their intersection, it is also at right angles to the plane passing through them: (XI. 4.) but the plane passing through ED, GH is the plane BH; therefore AF is perpendicular to the plane BH: therefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Q.E.F. PROPOSITION XII. PROBLEM. . . To erect a straight line at right angles to a given plane, from a point given in the plane. g Let A be the point given in the plane. - It is required to erect a straight line from the point A at right angles to the plane. * I B. From any point B above the plane draw BC perpendicular to it; (xi. 11.) and from A draw AD parallel to BC. (I. 31.) Because AD, CB, are two parallel straight lines, and one of them BC is at right angles to the given plane, therefore the other AD is also at right angles to it: (XI. 8.) wherefore a straight line has been erected at right angles to a given plane, from a point given in it. Q.E.F. **~ PROPOSITION XIII. THEOREM, From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it: and there can be but one perpendicular to a plane from a point above the plane. - For, if it be possible, let the two straight lines AB, A C be at right angles to a given plane from the same point A in the plane, and upon the same side of it. - Let a plane pass through BA, AC; the common section of this with the given plane is a straight line passing through 4: (xi. 3.) - 262 • EUCLID's ELEMENTS. - B C g D A E. - let DAF be their common section: therefore the straight lines AB, A C, DAE are in one plane: and because CA is at right angles to the given plane, it makes right angles with every straight line meeting it in that plane: (XI. def. 3.) but D.A.E, which is in that plane, meets CA; therefore CAE is a right angle. - For the same reason, BAE is a right angle. Wherefore the angle CAE is equal to the angle BAE; (ax. 11.) and they are in one plane, which is impossible. • . Also, from a point above a plane, there can be but one perpendi- cular to that plane: - - for, if there could be two, they would be parallel to one another, which is absurd. (XI. 6.) Therefore, from the same point, &c. Q.E.D. PROPOSITION XIV. THEOREM. Planes to which the same straight line is perpendicular, are parallel to one another. Let the straight line AB be perpendicular to each of the planes CD, EF. These planes shall be parallel to one another. If not, they shall meet one another when produced: let them meet ; their common section is a straight line GH, in which take any point K, and join AK, BK. - Then, because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK which is in that plane: (XI. def. 3.) w therefore ABK is a right angle. - For the same reason, BAK is a right angle: wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impossible: (r. 17.) *: the planes CD, EF, though produced, do not meet one another; - - .." * - that is, they are parallel. (x1, def. 8.) Therefore, planes, &c. Q.E.D. Book x1. PROP. xv, xvi. 263 PRóposition xv. THEOREM. If two straight lines meeting one another be parallel to two other straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallel to DE, EF, two other straight lines that meet one another, but are not in the same plane with AB, BC. - - The planes through AB, BC, and DE, EF shall not meet, though produced. - E B F C; K A T) H. From the point B draw. BG perpendicular to the plane which passes through DE, EF, (XI.1.1.) . and let it meet that plane in G ; and through G draw GH parallel to ED, and GK parallel to EF (r. 31.) And because BG is perpendicular to the plane through DE, EF, it makes right angles with every straight line meeting it in that plane: (XI. def. 3.) - - but the straight lines GH, GK in that plane meet it; therefore each of the angles BGH, BGK is a right angle: and because BA is parallel to GH (for each of them is parallel to DE, and they are not both in the same plane with it), (XI. 9.) the angles GBA, BGH are together equal to two right angles: (I. 29.) - - and BGH is a right angle; w therefore also GBA is a right angle, and GB perpendicular to B.A. For the same reason, GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines B.A., BC that cut one another in B; GB is perpendicular to the plane through BA, BC: (XI. 4.) and it is perpendicular to the plane through DE, EF; (constr.) therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: - - but planes to which the same straight line is perpendicular, are parallel to one another; (XI. 14.) *; plane through AB, BC is parallel to the plane through , EF. - -º- Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION XVI. THEOREM. If two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes AB, CD be cut by the plane EFH,6, and let their common sections with it be EF, GH. * - . Then EF shall be parallel to GH. 264. EUCLID's ELEMENTs. , K For, if it is not, EF, GH shall meet, if produced, either on the side of FH, or EG. Tirst, let them be produced on the side of FH, and meet in the point K. - Therefore, since EFK is in the plane AB, - every point in EFX is in that plane: (XI. 1.) and K is a point in EFK; therefore K is in the plane A.B.: for the same reason, K is also in the plane CD; wherefore the planes AB, CD produced, meet one another: but they do not meet, since they are parallel by the hypothesis; therefore the straight lines EF, GH, do not meet when produced on the side of FH. . . . In the same manner it may be proved, that EF, GH do not meet when produced on the side of E.G. . But straight lines which are in the same plane, and do not meet, though produced either way, are parallel; therefore EFis parallel to G.H. Wherefore, if two parallel planes, &c. Q.E.D. PROPOSITION XVII. THEOREM. If two straight lines be out by parallel planes, they shall be out in the same ratio. - - Let the straight lines AB, CD be eut by the parallel planes GH, I(L, MN, in the points A, E, B; C, F, D. As AE is to EB, so shall CF be to FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. - - Because the two parallel planes KL, MW are cut by the plane EBDX, the common sections EX, BD are parallel: (xi. 16.) - for the same reason, because the two parallel planes GH, KL are cut by the plane A2(FC, - - - the common sections AC, XF are parallel: - and because EX is parallel to BD, a side of the triangle ABD; as AE to EB, so is Ax to XD: (VI. 2.) . again, because XF is parallel to AC, a side of the triangle ADC; as AlK to XD, so is CF to FD: Book XI. PROP. xvii, XVIII, XIX. 265 - and it was proved that AM is to XD, as 4E to EB; therefore, as A E to EB, so is CF to FD. (v. 11.) Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION XVIII. THEOREM. - If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane. Let the straight line AB be at right angles to the plane C.K. Every plane which passes through AB shall be at right angles to the plane CK. . . - C F B E Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK ; -* take any point Fin CE, from which draw FG in the plane D.E. at right angles to C.E. (I. 11.) And because AB is perpendicular to the plane CK, therefore it is also perpendicular to every straight line in that plane meeting it; (XI. def. 3.) and consequently it is perpendicular to CE: wherefore ABF is a right angle: but GFB is likewise, a right angle; (constr.) therefore AB is parallel to FG: (I. 28.) and AB is at right angles to the plane CJſ; therefore FG is also at right angles to the same plane. (XI. 8.) But one plane is at right angles to another plane, when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other plane; (XI. def. 4.) and any straight line FG in the plane D.E, which is at right angles to CE, the common section of the planes, has been proved to be per- pendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all planes which pass through AB, are at right angles to the plane C.K. Therefore, if a straight line, &c. Q.E.D. PROPOSITION XIx. THEOREM. If two planes which out one another be each of them perpendicular to a third plane; their common section shall be perpendicular to the same plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section of the first two. Then BD shall be perpendicular to the third plane. 266 - EUCLID'S ELEMENTS. B 235 S * . A C . If it be not, from the point D draw, in the plane AB, the straight line D.E at right angles to AD, the common section of the plane AB ‘with the third plane; (I. 11.) and in the plane BC draw DF’ at right angles to CD the common section of the plane BC with the third plane. * And because the plane AB is perpendicular to the third plane, and JDE is drawn in the plane A.B at right angles to AD, their common section, - - DE is perpendicular to the third plane. (XI. def. 4.) In the same manner, it may be proved, that DF is perpendicular to the third plane. - Wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side of it, which is impossible: (XI. 13.) therefore, from the point D there cannot be any straight line at right angles to the third plane, except B.D the common section of the planes AB, BC: - therefore BD is perpendicular to the third plane. Wherefore, if two planes, &c. Q.E.D. PROPOSITION XX. THEOREM. If a solid angle be contained by three plane angles, any two of them are greater than the third. - Let the solid angle at A be contained by the three plane angles JBAC, CA.D, D.A.B. Any two of them shall be greater than the third. ID |B E C If the angles BAC, CAD, DAB be all equal, it is evident, that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, .* make, in the plane which passes through BA, AC, the angle BAE equal to the angle DAB; (I. 23.) and make AE equal to AD, and through E. draw BEC cutting r AB, AC, in the points B, C, and join DB, DC. And because DA is equal to AE, and AB is common, the two DA, AB, are equal to the two EA, AB, each to each; and the angle DAB is equal to the angle EAB; therefore the base DB is equal to the base B.E. (I. 4.) BOOK XI. PROP. xx, XXI. " . 267 ... • - and because BD, DC are greater than CB, (1, 20.) - and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC: (I. ax. 5.) and because DA is equal to A.E, and AC common, but the base DC greater than the base EC; therefore the angle DAC is greater than the angle EAC; (I. 25.) and, by the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC; (I. ax. 4.) but BAC is not less than either of the angles DAB, DAC : therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, &c. Q.E.D. PROPOSITION XXI. THEOREM. Foery solid angle is contained by plane angles, which together are less than four right angles. - First, let the solid angle at A be contained by three plane angles BAC, CAD, DAB. These three together shall be less than four right angles. D As -- B C Take in each of the straight lines AB, AC, AD, any points B, C, D, and join BC, CD, DB. Then, because the solid angle at B is contained by the three plane. angles CBA, ABD, DBC, - any two of them are greater than the third; (XI. 20.) therefore the angles CBA, ABD are greater than the angle DBC: for the same reason, the angles BCA, A CD are greater than the angle DCB; - and the angles CDA, ADB greater than BDC: wherefore the six angles CBA, ABD, BCA, A CD, CDA, ADB, are greater than the three angles DBC, BCD, CDB : - but the three angles DBC, BCD, CDB are equal to two right angles; (I. 32.) - - therefore the six angles CBA, ABD, BCA, A CD, CDA, ADB are greater than two right angles: and because the three angles of each of the triangles ABC, A CD, ADB are equal to two right angles, therefore the nine angles of these three triangles, viz. the angles: CBA, BAC, ACB, A CD, CDA, DAC, ADB, DBA, BAD are equal to, six right angles; . of these the six angles CBA, ACB, A CD, CDA, ADB, DBA are greater than two right angles: - therefore the remaining three angles BAC, CAD, DAB, which contain the solid angle at A, are less than four right angles. Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, D.A.E., E.A.F, F.A.B. - w 268 EUCLID’S, ELEMENTS. --~~ These shall together be less than four right angles. s A C E ID - - Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, F.B. And because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater than the third, (XI. 20.) the angles CBA, ABF, are greater than the angle FBC: for the same reason, the two plane angles at each of the points C, D, JE, F, viz. those angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF: therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon : - and because all the angles of the triangles are together equal to - twice as many right angles as there are triangles; (I. 32.) - that is, as there are sides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygon; (I. 32. Cor. H.) therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles: (I. ax. 1.) but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved; . wherefore the remaining angles of the triangles, viz. those of the vertex, which contain the solid angle at A, are less than four right angles. Therefore, every solid angle, &c. Q.E.D. PROPOSITION XXII. THEOREM. If every two of three plane angles be greater than the third, and if the straight lines which contain them be all equal; a triangle may be made of the straight lines that join the extremities of those equal straight lines. f FROPOSITION XXIII. PROBLEM. _- To make a solid angle which shall be contained by three given plane angles, any two of them being greater than the third, and all three together less than four right angles. PROPOSITION A. THEOREM. If each of two solid angles be contained by three plane angles, which are equal to one another, each to each; the planes in which the equal angles are, have the same inclination to one another. . . . PROPOSITION B. THEOREM, If two solid angles be contained, each by three plane angles which are equal to one another, each to each, and alike situated; these solid angles are equal to one another. * FOOK xi. - - 269 PROPOSITION C. THEOREM. _ = Solid figures which are contained by the same number of equal and similar planes alike situated, and having none of their solid angles con- tained by more than three plane angles, are equal arid similar to one another. - • PROPOSITION XXIV. THEOREM. - If a solid be contained by six planes, two and two of which are parallel; the opposite planes are similar and equal parallelograms. PROPOSITION XXV. THEOREM. If a solid parallelopiped be cut by a plane parallel to two of its op- posite planes; it divides the whole into two solids, the base of one of which shall be to the base of the other, as the one solid is to the other. - PROPOSITION XXVI. PROBLEM. - At a given point in a given straight line, to make a solid angle equal to a given solid angle contained by three plane angles. PROPOSITION XXVII. PROBLEM. To describe from a given straight line a solid parallelopiped similar,. and similarly situated, to one given: - . PROPOSITION XXVIII. THEOREM. z If a solid parallelopiped be cut by a plane passing through the dia- gonals of two of the opposite planes; it shall be cut in two equal parts. PROPOSITION xxix. THEOREM. Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. PROPOSITION XXX. THEOREM. Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. . . . PROPOSITION XXXI. THEOREM. • Solid parallelopipeds, which are upon equal bases, and of the same altitude, are equal to one another.” PROPOSITION XXXII. THEOREM. Solid parallelopipeds which have the same altitude are to one another as their bases. - . . - CoR. From this it is manifest, that prisms upon triangular bases, of the same altitude, are to one another as their bases. PROPOSITION XXXIII. THEOREM. Similar solid parallelopipeds are one to another in the triplicate ratio of their homologous sides. - CoR. From this it is manifest, that, if four straight lines be con- tinual proportionals, as the first is to the fourth, so is the solid parallelopiped described from the first to the similar solid similarly described from the second; because the first straight line has to the fourth the triplicate ratio of that which it has to the second. * * N. 270 ^ EUCLID'S ELEMENTS. PROPOSITION ID. THEOREM. Solid parallelopipeds contained by parallelograms equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio which is the same with the ratio compounded of the ratios of their sides. PROPOSITION XXXIV. THEOREM. The bases and altitudes of equal solid parallelopipeds, are recipro- cally proportional: and conversely, if the bases and altitudes be reci- procally proportional, the solid parallelopipeds are equal. i FROPOSITION XXXV. THEOREM, - If, from the vertices of two equal plane angles, there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to the planes in which the first named angles are; and from the points in which they meet the planes, Straight lines be drawn to the vertices of the angles first named; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles. CoR. From this it is manifest, that if from the vertices of two equal plane angles, there be elevated two equal straight lines containing equal angles with the sides of the angles, each to each ; the perpen- diculars drawn from the extremities of the equal straight lines to the planes of the first angles, are equal to one another. - PROPOSITION XXXVI. THEOREM. If three straight lines be proportionals, the solid parallelopiped de- scribed from all three, as its sides, is equal to the equilateral parallelo- piped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other figure. PROPOSITION XXXVII. THEOREM. If four straight lines be proportionals, the similar solid parallelo- pipeds similarly described from them shall also be proportionals: and, conversely, if the similar parallelopipeds similarly described from four straight lines be proportionals, the straight lines shall be proportionals. PROPOSITION XXXVIII. THEOREM, “If a plane be perpendicular to another plane, and a straight line be drawn from a point in one of the planes perpendicular to the other plane, this straight line shall fall on the common section of the planes.” PROPOSITION XXXIX. THEOREM. In a solid parallelopiped, if the sides of two of the opposite planes be divided, each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the solid parallelopiped, cut each other into two equal parts. PROPOSITION, XL. THEOREM. ‘If there be two triangular prisms of the same altitude, the base of one of which is a parallelogram, and the base of the other a triangle; if the parallelogram be double of the triangle, the prisms shall be equal to one another. . * NOTES TO BOOK XI. THE Eleventh Book of the Elements commences with the definitions of the Geometry of Planes and Solids, and then proceeds to demonstrate the most ele- mentary properties of straight lines and planes, solid angles and parallelopipeds. The solids considered in the eleventh and twelfth books are Geometrical solids, portions of space bounded by surfaces which are supposed capable of penetrating and intersecting one another. - In the first six books, all the diagrams employed in the demonstrations are supposed to be in the same plane, which may lie in any position whatever, and be extended in every direction, and there is no difficulty in representing them roughly on any plane surface; this, however, is not the case with the diagrams employed in the demonstrations in the eleventh and twelfth books, which cannot be so intelligibly represented on a plane surface on account of the perspective. A more exact conception may be attained, by adjusting pieces of paper to represent the different planes, and drawing lines upon them as the constructions may require, and by fixing pins to represent the lines which are perpendicular to, or inclined to any planes. Any plane may be conceived to move round any fixed point in that plane, either in its own plane, or in any direction whatever; and if there be two fixed points in the plane, the plane cannot move in its own plane, but may move round the straight line which passes through the two fixed points in the plane, and may assume every possible position of the planes which pass through that line, and every different position of the plane will represent a different plane; thus, an in- definite number of planes may be conceived to pass through a straight line which will be the common intersection of all the planes. Hence, it is manifest, that though two points fix the position of a straight line in a plane, neither do two points nor a straight line fix the position of a plane in space. If however, three points, not in the same straight line, be conceived to be fixed in the plane, it will be manifest, that the plane cannot be moved round, either in its own plane or in any other direction, and therefore is fixed. Also any conditions which involve the consideration of three fixed points not in the same straight line, will fix the position of a plane in space; as two straightlines which meet or intersect one another, or two parallel straight lines in the plane. Def. v. When a straight line meets a plane, it is inclined at different angles to the different lines in that plane which may meet it; and it is manifest that the inclination of the line to the plane is not determined by its meeting any line in that plane. The inclination of the line to the plane can only be determined by its in- clination to some fixed line in the plane. If a point be taken in the line different from that point where the line meets the plane, and a perpendicular be drawn to meet the plane in another point; then these two points in the plane will fix the position of the line which passes through them in that plane, and the angle con- tained by this line and the given line, will measure the inclination of the line to the plane ; and it will be found to be the least angle which can be formed with the given line and any other straight line in the plane. If two perpendiculars be drawn upon a plane from the extremities of a straight line which is inclined to that plane, the straight line in the plane intercepted be- tween the perpendiculars is called the projection of the line on that plane; and it is obvious that the inclination of a straight line to a plane is equal to the inclina- tion of the straight line to its projection on the plane. If however, the line be parallel to the plane, the projection of the line is of the same length as the line 272 EUCLID's ELEMENTS. itself; in all other cases the projection of the line is less than the line, being the base of a right-angled triangle, the hypothenuse of which is the line itself. The inclination of two lines to each other, which do not meet, is measured by the angle contained by two lines drawn through the same point and parallel to the two given lines. & Def. VI. Planes are distinguished from one another by their inclinations, and the inclinations of two planes to one another will be found to be measured by the acute angle formed by two straight lines drawn in the planes, and perpendicular to the straight line which is the common intersection of the two planes. It is also obvious that the inclination of one plane to another will be measured by the angle contained between two straight lines drawn from the same point, and perpendicular, one on each of the two planes. The intersection of two planes suggests-a new conception of the straight line. Def. IX. XTeps&yovía éotiv i örö arXetévov i öüo yovidiv čarvaré8wv Treplexo- uéun, ui oiladiv iv Tiš airá čtvré84 rods iv. ongstº, ovvurragávov. The render- ing by Simson of this definition may be slightly amended. The word replexouéun is rather comprehended or contained than made : and avvua Tapiévov means joined and fitted together, not meeting. “A Solid angle is that which is contained by more than two plane angles joined together at one point, (but) which are not in the same plane.” When a solid angle is contained by three plane angles, each plane which con- tains one plane angle, is fixed by the position of the other two, and consequently, only one solid angle can be formed by three plane angles. But when a solid angle . is formed by more than three plane angles, if one of the planes be considered fixed in position, there are no conditions which fix the position of the rest of the planes which contain the solid angle, and hence, an indefinite number of solid angles, unequal to one another, may be formed by the same plane angles, when the num- ber of plane angles is more than three. - - Def. x is restored, as it is found in the editions of the Greek text of Euclid. It appears to be universally true, supposing the planes to be similarly situated, in which are contained the corresponding equal plane angles of each figure. Def. xiv. The sphere, as well as the come and the cylinder are defined by a mode in which the figures may be conceived to be generated. Here motion is for the first time introduced in defining of Geometrical figures. In these motions, the successive change of position only is considered which a figure undergoes, and the figure traced out in consequence of the motion. The velocity with which the new figure is traced out, as well as the time and force requisite for effecting it, are considerations which do not enter into the subject of Geometry. & Def. A. Parallelopipeds are solid figures in some respects analogous to paral- lelograms, and remarks might be made on parallelopipeds similar to those which were made on rectangles in the notes to Book II, p. 81; and every right- angled parallelopiped may be said to be contained by any three of the straight lines which contain the three right angles by which any one of the solid angles of the figure is formed; or more briefly, by the three adjacent edges of the parallelopiped. As all lines are measured by lines, and all surfaces by surfaces, so all solids are measured by solids. The cube is the figure assumed as the measure of solids or volumes, and the unit of volume is that cube, the edge of which is one unit in length. If the edges of a rectangular parallelopiped can be divided into units of the same length, a numerical expression for the number of cubic units in the parallelo- piped may be found, by a process similar to that by which a numerical expression for the area of a rectangle was found. - - Let AB, AC, AD be the adjacent edges of a rectangular parallelopiped AG, and let AB contain 5 units, AC, 4 units, and AD, 3 units in length. - NOTEs to Book XI. ºr 273 •es - Then if through the points of division of AB, AC, AD, planes be drawn parallel to the faces BG, BD, AE respectively, the parallelopiped will be divided into cubic units, all equal to one another. Iº C E º And since the rectangle ABEC contains 5 × 4 square units, (note, p. 81.) and that for every linear unit in AD there is a layer of 5 x 4 cubic units corresponding to it; consequently, there are 5 × 4 × 3 cubic units in the whole parallelopiped 4G. That is, the product of the three numbers which express the number of linear units in the three edges, will give the number of cubic units in the parallelopiped, and therefore will be the arithmetical representation of its volume. And generally, if AB, AC, AD; instead of 5, 4 and 3, consisted of a, b, and c linear units, it may be shewn, in a similar manner, that the volume of the paral- lelopiped would contain a be cubie units, and the product abo would be a proper representation of the volume of the parallelopiped. - - If the three sides of the figure were equal to one another, or b and c each equal to a, the figure would become a cube, and its volume would be represented by a a a, or a*. It may easily be shewn Algebraically that the volumes of similar rectangular parallelopipeds are proportional to the cubes of their homologous edges. Let the adjacent edges of the parallelopiped AB contain a, b, c units, and those of another similar parallelopiped A'B' contain a', b', c' units respectively, Also, let V, V' denote their volumes. Then W = abc, and V* = a'b'c'. But since the parallelopipeds are similar, ... + = # = 7 : V abo a b c a a a aº bº cº Hence = <= +H = −. H. F = +...+...+ = H = H = -, . W’ a'b'c' aſ 'b' "c' aſ " aſ a 3 3 T c 3 In a similar manner, it may be shewn that the volumes of all similar solid figures bounded by planes, are proportional to the cubes of their homologous edges. Prop. v1. From the diagram, the following important construction may be made. If from B a perpendicular BF be drawn to the opposite side DE of the triangle DBE, and AF be joined; then A Fshall be perpendicular to DE, and the angle AFB measures the inclination of the planes AED and BED. Prop. xix. It is also obvious, that if three planes intersect one another; and if the first be perpendicular to the second, and the second be perpendicular to the third; the first shall be perpendicular to the third; also the intersections of every two shall be perpendicular to one another. - The Demonstrations of the last Nineteen Propositions of the Eleventh Book have been omitted, as they are not included in the course of reading prescribed for Mathematical Honours at Cambridge. - - - - T .** BOOK XII. LEMMA I. If from the greater of two unequal magnitudes, there be taken more than its half, and from the remainder more than its half; and so on : there shall at length remain a magnitude less than the least of the proposed magnitudes. (Book x. Prop. 1.) : Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken more than its half, and from the remainder more than its half, and so on; there shall at length remain a magnitude less than 0. A D B C E For C may be multiplied so as at length to become greater than AB Let it be so multiplied, and let DE its multiple be greater than AB, and let DE be divided into D.F, FG, GE, each equal to C. *. From AB take BH greater than its half, and from the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in AB be A.K, KH, HB; and the divisions in DE be DF, FG, GE. - And because DE is greater than AB, and that EG taken from DE is not greater than its half, but BPI taken from AB is greater that its half; therefore the remainder GD is greater than the remainder H.A. Again, because GD is greater than HA, and that GF is not greater than the half of G.D, but HK is greater than the half of HA; therefore the remainder FD is greater than the remainder AK: - and FD is equal to C, therefore C is greater than AK; that is, AK is less than C. Q.E.D. . And if only the halves be taken away, the same thing may in the same way be demonstrated. - Book XII. PROP. 1, IL. 275 PROPOSITION I. THEOREM. Similar polygons inscribed in circles, are to one another as the squares on their diameters. . . Let ABCDE, FGHKL be two circles, and in them the similar polygons ABCDE, FGHAZ; - - e and let BM, GW be the diameters of the circles: as the polygon ABCDE is to the polygon FGHICL, so shall the square on BM be to the square on G.N. - A F B E {G} L. M N H K *C I} Join BE, AM, GL, FN. - - - And because the polygon ABCDE is similar to the polygon FGHIJKL, the angle BAE is equal to the angle GFL, and as BA to AE, so is GF to FL: . therefore the two triangles BAE, GFL having one angle in one equal to one angle in the other, and the sides about the equal angles proportionals, are equiangular; and therefore the angle AEB is equal to the angle FLG: but AEB is equal to AMB, because they stand upon the same - circumference: (III. 21.) and the angle FLG is, for the same reason, equal to the angle FWG : therefore also the angle AMB is equal to FNG: and the right angle BAM is equal to the right angle GFN; (III. 31.) wherefore the remaining angles in the triangles ABM, FGM are equal, - and they are equiangular to one another: therefore as BM to GN, so is BA to GF; (VI. 4.) ‘x. and therefore the duplicate ratio of BM to GM, is the same with: the duplicate ratio of B.A to GF: (v. def. 10. and v. 22.) but the ratio of the square on BM to the square on GN, is the duplicate ratio of that which BM has to GN: (VI. 20.) and the ratio of the polygon ABCDE to the polygon FGHKL is the duplicate of that which BA has to GF: (VI. 20.) therefore as the polygon ABCDE is to the polygon FGHKL, so is the square on BM to the square on G.N. - Wherefore, similar polygons, &c. Q.E.D. sº PROPOSITION II. THEOREM. Circles are to one another as the squares on their diameters. Let ABCD, EFGH be two circles, and BD, FH £heir diameters. As the square on BD to the square on FH, so shall the circle ABCD be to the circle EFG H. - - For, if it be not so, the square on BD must be to the square on FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it. - T2 276 EUCLID's ELEMENTS. First, if possible, let it be to a space S less than a circle EFGII; and in the circle EFG Hinscribe the square EFG H. (IV. 6.) This square is greater than half of the circle EFGH; because, if through the points E, F, G, H, there be drawn tangents t to the circle, - - - Q the square EFG His half of the square described about the circle: (1.47.) - and the circle is less than the square described about it; therefore the square EFGH is greater than half of the circle. Divide the circumferences E.F, FG, GH, HE, each into two equal parts in the points K, L, M, W, and join EK, KF, FL, LG, GM, H.M., HW, WE; • : therefore each of the triangles EKF, FL6, GMH, HNE, is greater than half of the segment of the circle in which it stands; because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines E.F. FG, GH, HE be completed, - each of the triangles EKF, FLG, GMH, HNE is the half of the parallelogram in which it is: (I. 41.) but every segment is less than the parallelogram in which it is; wherefore each of the triangles EKF, FLG, GMH, HNE is greater than half the segment of the circle which contains it. Again, if the remaining circumferences be divided each into two equal parts, and their extremities be joined by straight lines, by con- tinuing to do this, there will at length remain segments of the circle, which together are less than the excess of the circle EFG H above the space S; e ſº - because, by the preceding Lemma, if from the greater of two unequal magnitudes there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes. - Iet then the segments EK, KF, FL, LG, GM, MH, HN, NE be those that remain, and are together less than the excess of the circle EFG H above_S : ºtherefore the rest of the circle, viz. the polygon EKFLGMHVis greater than the space S. Describe likewise in the circle ABCD the polygon A.KBOCPDR similar to the polygon EKFLGMHN: * as therefore the square on BD is to the square on FH, so is the polygon AIXBOCPDR to the polygon EKFLGMHN: (XII. 1.) but the square on BD is also to the square on FH, as the circle ABCD is to the space S; (hyp.) . . . therefore as the circle ABCD is to the space S, so is the polygon A.KBOCPDR to the polygon EKFLGMHN: (v. 11.) but the circle ABCD is greater than the polygon contained in it; wherefore the space S is greater than the polygon EKFLGMHV: - (v. 14.) G but it is likewise less, as has been demonstrated; which is impossible. Therefore the square on BD is not to the square on FH, as the circle ABCD is to any space less than the circle EFG H. - • BOOK XII. PROP. II. - 277 In the same manner, it may be demonstrated, that neither is the square on FH to the square on BD, as the circle EFGH is to any space less than the circle At BCD. - - Nor is the square on BD to the square on FH, as the circle ABCD is to any space greater than the circle EFG H. For, if possible, let it be so to T, a space greater than the circle EFGH: A ' * - g 2.É. N therefore, inversely, as the square on FH to the square on BD, so is the space T to the circle ABCD; - but as the space T is to the circle ABCD, so is the circle EFGH . to some space, which must be less than the circle 4 BCD, (v. 14.) because the space T is greater, by hypothesis, than the circle EFGH; therefore as the square on FH is to the square on BI), so is the circle JEFGH to a space less than the circle ABCD, which has been demon- strated to be impossible ; - therefore the square on BD is not to the square on FH, as the circle ABCD is to any space greater than the circle EFGH: and it has been demonstrated, that neither is the square on BD to -the square on FH, as the circle ABCD to any space less than the circle JEFG HT: w - wherefore, as the square on BB to the square on FH, so is the circle A B CD to the circle EFG H. * Circles, therefore, are, &c. Q.E.D. PROPOSITION III. THEOREM. - . Every pyramid having a triangular base, may be divided into two equal and similar pyramids having triangular bases, and which are : similar to the whole pyramid, and into two equal prisms which together are greater than half of the whole pyramid. - - PROPOSITION IV. THEOREM. If there be two pyramids of the same altitude, upon triangular bases, and each of them be divided into two equal pyramids similar to the whole pyramid, and also into two equal prisms; and if each of these pyramids be divided in the same manner as the first two, and so on; as the base of one of the first two pyramids is to the base of the other, so shall all the prisms in one of them be to all the prisms in the other, that are produced by the same number of divisions. - PROPOSITION V. THEOREM. . Pyramids of the same altitude which have triangular bases, are to one another as their bases. - . - PROPOSITION VI. THEOREM. - Pyramids of the same altitude which have polygons for their bases, are to one another as their bases. PROPOSITION VII. THEOREM. - Every prism having a triangular base may be divided into three pyramids that have triangular bases, and are equal to one another. .. 278 - EUCLID's ELEMENTS. CoR. 1. From this it is manifest, that every pyramid is the third part of a prism which has the same base, and is of an equal altitude with it: for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases. CoR. 2. Prisms of equal altitudes are to one another as their bases; because the pyramids upon the same bases, and of the same altitude, are to one another as their bases. - JPROPOSITION VIII. THEOREM. Similar pyramids, having triangular bases, are one to another in the triplicate ratio of that of their homologous side. CoR. From this it is evident, that similar pyramids which have multangular bases, are likewise to one another in the triplicate ratio of their homologous sides: for they may be divided into similar pyra- mids having triangular bases, because the similar polygons which are their bases, may be divided into the same number of similar triangles homologous to the whole polygons: therefore, as one of the triangular pyramids in the first multangular pyramid is to one of the triangular pyramids in the other, so are all the triangular pyramids in the first to all the triangular pyramids in the other; that is, so is the first mul- tangular pyramid to the other : but one triangular pyramid is to its similar triangular pyramid, in the triplicate ratio of their homologous sides; and therefore the first multangular pyramid has to the other, the triplicate ratio of that which one of the sides of the first has to the homologous side of the other. - PROPOSITION IX. THEOREM. The bases and altitudes of equal pyramids having triangular bases are reciprocally proportional: and, conversely, triangular pyramids of which the bases and altitudes are reciprocally proportionals, are equal to one another: - PROPOSITION X. THEOREM. -- Every cone is the third part of a cylinder which has the same base, and is of an equal altitude with it. PROPOSITION XI. THEOREM, Cones and cylinders of the same altitude are to one another as thoir Vases. - PROPOSITION XII. THEOREM. Similar cones and cylinders have to one another the triplicate ratio of that which the diameters of their bases have. PROPOSITION XIII. THEOREM. If a cylinder be eut by a plane parallel to its opposite planes, or bases; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other. - PROPOSITION XIV. THEOREM. Cones and cylinders upon equal bases are to another as their altitudes. - PROPOSITION XV. THEOREM. The bases and altitudes of equal cones and cylinders, are reciprocally proportional; and, conversely, if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another. º BOOK XII. PROP. XVI. 279 P PROPOSITION XVI. PROBLEM. In the greater of two circles that have the same center, to inscribe a polygon of an even number of equal sides, that shall not meet the less circle. -- 4 LEMMA II. If two trapeziums ABCD, EFGH be inscribed in the circles, the centers of which are the points K, L; and if the sides AB, DC be arallel, as also EF, HG; and the other four sides AD, BC, EH, FG, #. all equal to one another; but the side AB greater than EF, and DC greater than HG; the straight line K4 from the center of the circle in which the greater sides are, is greater than the straight line L.E. drawn from the center to the circumference of the other circle. PROPOSITION XVII. PROBLEM. In the greater of two spheres which have the same center, to describe a solid polyhedron, the superficies of which shall not meet the less sphere. , P CoR. And if in the less sphere there be inscribed a solid polyhe- dron, by drawing straight lines betwixt the points in which the straight lines from the center of the sphere drawn to all the angles of the solid polyhedron in the greater sphere meet the superficies of the less; in the same order in which are joined the points in which the same lines from the center meet the superficies of the greater sphere; the solid poly- hedron in the sphere BCDE has to this other solid polyhedron the triplicate ratio of that which the diameter of the sphere BCDE has to the diameter of the other sphere. For if these two solids be divided into the same number of pyramids, and in the same order, the pyramids shall be similar to one another, each to each : because they have the solid angles at their common vertex, the center of the sphere, the same in each pyramid, and their other solid angles at the bases equal to one another, each to each, because they are contained by three plane angles, each equal to each; and the pyramids are contained by the same number of similar planes; and are therefore similar to one another, each to each : but similar pyramids have to one another the triplicate ratio of their homologous sides. Therefore the pyramid of which the base is the quadrilateral KBOS, and vertex A, has to the pyramid in the other sphere of the same order, the triplicate ratio of their homologous sides, that is, of that ratio which AB from the center of the greater sphere has to the straight line from the same center to the Superficies of the less sphere, and in like manner, each pyramid in the greater sphere has to each of the same order in the less, the triplicate ratio of that which AB has to the semi-diameter of the less sphere. And as one antecedent is to its consequent, so are all the antecedents to all the consequents. Wherefore the whole solid polyhedron in the greater sphere has to the whole solid polyhedron in the other, the tri- plicate ratio of that which AB the semi-diameter of the first has to the semi-diameter of the other; that is, which the diameter BD of the greater has to the diameter of the other sphere. PROPOSITION XVIII. THEOREM. Spheres have to one another the triplicate ratio of that which their diameters have. - - NOTES TO BOOK XII. THIs book treats of the properties of prisms and cylinders, pyramids and cones. A new principle is introduced called “the method of Exhaustions,” which may be applied for the purpose of finding the areas and ratios of circles, and the rela- tions of the surfaces and of the volumes of cones, spheres and cylinders. The first comparison of rectilinear areas is made in the First Book of the -Elements by the principle of superposition, where two triangles are coincident in all respects; next, comparison is made between triangles and other rectilinear figures when they are not coincident. e In the Sixth Beok, similar triangles are compared by shewing that they are in the duplicateratio of their homologous sides, and then by dividing similar polygons into the same number of similar triangles, and shewing that the polygons are also in the duplicate ratio of any of their homologous sides. In the Eeventh Book, similar rectilinear solids are compared by shewing that their volumes are to one another in the triplicate ratio of their homologous sides. “The method of Exhaustions” is founded on the principle of eachausting a mag- nitude by continually taking away a part of it, as it is explained in the first Lemma of the Tenth Book of the Elements. - i “The method of Exhaustions” was employed by the Ancient Geometers and was strictly rigorous in its principles; but it was too tedious and operose in its ap- plication to be of extensive utility as an instrument of investigation. It is exem- plified in Euc. xII. 2, where it is proved that the areas of circles are proportional to the squares on their diameters. In demonstrating this truth, it is first shewn by inscribing successively in one of the circles regular polygons of four, eight, sixteen, &c. sides, and thus tending to exhaust the area of the circle, that a polygon may be found which shall differ from the circle by a quantity less than any magnitude which can be assigned : and then since similar polygons inscribed in the circles are as the squares on their diameters (Euc. xII. 1.) the truth of the proposition is established by means of an indirect proof. * “The method of Exhaustions” may be applied to find the circumference and area of a circle. A rectilinear figure may be inscribed in the circle and a similar one cir-" cumscribed about it, and then by continually doubling the number of sides of the inscribed and circumscribed polygons, by this principle, it may be demonstrated, that the area of the circle is less than the area of the circumscribed polygon, but greater than the area of the inscribed polygon; and that as the number of sides of the polygon is increased, and consequently the magnitude of each diminished, the differences between the circle and the inscribed and circumscribed polygons are continually exhausted. r . In a similar way, the principle is applied to the surfaces and volumes of cones, cylinders and spheres. As only the first and second propositions of the Twelfth Book of the Elements are required to be read for Honours at Cambridge, the demonstrations of the re- maining fifteen propositions of this book have been omitted. The second pro- position is perhaps retained merely as an example of the method employed by the ancient Geometers. This method has been replaced by that of prime and ultimate ratios, which is now employed in the proofs of such propositions as were formerly effected by “the method of Exhaustions.” , ' º ON THE PROPOSITIONS IN THE ELEMENTS. THERE are only two forms of Propositions in the Elements, the theorem and the problem. In the theorem, it is asserted, and is to be proved, that if a geometrical figure be constructed with certain speci- fied conditions, then some other specified relations must necessarily exist between the constituent parts of that figure. Thus:—if squares be described on the sides and hypotehuse of a right-angled triangle, the square on the hypotenuse must necessarily be equal to the other two squares upon the sides (Euc. I. 47). In the problem, certain things are given in magnitude, position, or both, and it is required to find certain other things in magnitude, position, or both, that shall neces- sarily have a specified relation to the things given, or to each other, or to both of them. Thus:—a circle being given, it may be required to . construct a pentagon, which shall have its angular points in the cir- cumference, and which shall also have both all its sides equal, and all its angles equal. (Euc. IV. 11.) - - A problem is said to be determinate, when with the prescribed conditions it admits of one definite solution: and it is said to be in- determinate, when it admits of more than one definite solution. This latter circumstance arises from the data not absolutely fixing, but merely Festricting the quaesita, leaving certain points or lines not fixed in one position only. The number of given conditions may be insufficient for a single determinate solution; or relations may subsist among some of , the given conditions, from which one or more than one of the remain- ing given conditions may be deduced. z - It may be remarked in Euclid's propositions, that there is in general, an aim at definiteness, considered in reference to the quaesitum of the problem, and the predicate of the theorem. The quasitum of the problem is either a single thing, as the perpendicular in Euc. I. 11; or at most, two, as the tangents to the circle in the first case of Euc. III. 17; and in the most general problems, even those which transcend the ordinary geometry, the solutions are, in general, re- stricted to a definite number, which can always be assigned a priori for every problem. In certain cases, however, the conditions given in Euclid are not sufficient to fix entirely the quaesitum in all re- spects. For instance, in Euc. I. 2, it has been seen that the position of the line required is not fixed by the conditions of the problem, so that more lines than one can be drawn from the given point fulfilling the required conditions: nor is the direction prescribed in which the line is to be drawn, so that it has been seen, that from the given point, two lines in opposite directions can be drawn from the given point for each of the possible constructions. And in Euc. IV. 10, the magnitude of the triangle is any whatever, and therefore not entirely fixed in all respects: or, again, in Euclid IV. 11, the pentagon may be any whatever, so that its position in the circle is not fixed. To fix the magnitude of the triangle, or the position of the pentagon, some other condition independent of the data, must be added to the conditions of the problem. The length and posi- tion of some line connected with the triangle, (as one of the equal sides, the base, the perpendicular, &c.) would have fixed the triangle in magnitude and position; and the position of one angular point of the pentagon, or the condition that one side of the pentagon should pass 282 ON THE PROPOSITſCNS through a given point (though this point must be subject to a certain restriction as to position, if within the circle), or any other possible conditions, would have confined the pentagon to a single position, or to the alternative of two positions. Such is the only kind of indeter- minateness in the problems of “the Elements.” In the enunciation of the theorems too, the same aim at singleness in the property asserted to. be consequent on the hypothesis, is apparent throughout. There is, however, a remarkable difference in the characters of the hypotheses themselves, in Euclid's theorems: viz. - 1. That in some of them, one thing alone, or a certain definite number, possesses the property which is affirmed in the enunciation. 2. That in others, all the things constituted subject to the hypo- thetical conditions, possess the affirmed property. As instances of the first class, the greater number of theorems in the Elements may be referred to, as Euc. I. 4, 5, 6, 8, which are of the simplest class. In these, only one thing is asserted to be equal to another specified thing. In all the theorems of the Second Book, one thing is asserted to be equal to several other things taken together; and the same occurs in Euc. I. 47, as well as frequently in the other Books. They sometimes also take the form of asserting that no certain magnitude is greater or less than another, as in Euc. I. 16, or that two things together are less than, er greater than, some one thing or several things, as Euc. I. 17. In all cases, however, this class is distinguished by the circumstance, that the things asserted to have the property are of a given finite number. I As instances of the second class, reference may be made to Euc. I. 35, 36, 37, 38, where all the parallelograms in the two former, and all the triangles in the two latter, are asserted to have the property of be- ing equal to one given parallelogram or one given triangle. Or to Euc. III. 14, 20, 21; the lines in the circle in Prop. 14, or the angles at the circumference in Props. 20, 21, are any whatever, and therefore all the lines or angles constituted as in the enunciations, fulfil the conditions. Or again, in Book v. the two pairs of indefinite multiples, which form the basis of Euclid's definition of proportionals; or his propositions “ea: aquo,” and “ea: aquo perturbato,” and the Proposi- tions F, G, H, K; or, lastly, Euc. VI. 2, in which the property is (really, though not formally,) affirmed to be true when any line is łrawn parallel to any one of the sides of the triangle. The very circumstance, indeed, just noticed parenthetically, prevails so much in Euclid's enunciations, as to render it clear that it was his object as much as possible to render the conditions of the hypothesis formally definite in number; and if these remarks had no prospective reference, the circumstance would scarcely deserve notice. Still, with such prospective reference, it is necessary to insist upon the fact, that however the form of enunciation may be calculated to remove observa- tion from it, the hypothesis itself is indefinite, or includes an indefinite number of things, which an additional condition would, as in the case of the problem, have restricted either to one thing or to a certain number of things. Sometimes too, the theorem is enunciated in the form of a negation of possibility, as Euc. I. 7; III. 4, 5, 6, &c. These offer no occasion for remark, except the ingenious modes of demonstration employed by Euclid. All such demonstrations must necessarily be indirect, assum- ing as an admitted truth the possibility of the fact denied in the enun- ciation. • * - * IN THE ELEMENTS. 283 It may be remarked that the Ancient Geometers most probably arrived at Theorems in their attempts to solve Problems. The first Geometrical enquiries must naturally have arisen in form of questions or problems, in which some things were given, and some things re- quired to be done : and in the attempts to discover the relations between the things given and the things required, many truths would be suggested which afterwards became the subjects of separate demonstration. - Both among the Theorems and Problems, cases occur in which the hypotheses of the one, and the data or quaesita of the other, are re- stricted within certain limits as to magnitude and position. Sometimes it will be found, while some Problems are possible within definite limits, that certain magnitudes involved increase up to a certain value, and then begin to decrease; or decrease down to a certain value, and then begin to increase. This circumstance gives rise to the question of the greatest or least value which certain magnitudes may admit of, in indeterminate Problems and Theorems. The determination of these limits constitutes the doctrine of Mazima and Minima. For instance, the limit of possible diminution of the sum of the two sides of a triangle described upon a given base, is the magnitude of the base itself, Euc. I. 20, 22: And of all the equal triangles upon the same base and between the same parallels (Euc. I. 37.), that triangle which has the greatest vertical angle is an isosceles triangle; and the vertical angles of the other triangles on each side of the vertex of the isosceles triangle, become greater and greater as the vertices of these triangles approach the vertex of the isosceles triangle. When a straight line is divided into two parts, the rectangle contained by the parts is a maxi- mum when the given line is divided into two equal parts. The line AB (Euc. II. 5. fig.) is divided into any two parts in the point D. If BD the smaller part of the line be supposed to increase by the point D moving towards C, it is obvious that as the smaller part B.D in- creases, the larger part AD diminishes, until the point D coincides with C, and both parts are then each equal to half the line AB. And it is clear that so long as BD increases, the rectangle AD, DB increases, and the square on CD decreases: and when D coincides with C, the square on CD vanishes, and the rectangle AD, DB, then becomes the ... Square on DC, or on DB, the square on half the line A.B. If the point D be supposed to move beyond C towards A, the rect- angle AD, DB begins to diminish, and the square on DC to increase, in the same manner as they increased and decreased when the point was considered to move from B to C. Hence it is manifest that when a line is divided into two parts, the rectangle contained by the parts is a maximum or the greatest possible, when the two parts of the line are equal. It also appears that when the rectangle contained by the two parts of a line is a maa'imum, the sum of the squares on the parts is a minimum. For if a line be divided into any two parts (Euc. II. 4.), the square on the whole line is equal to the squares on the two parts and twice the rectangle contained by the parts. Hence it follows that the greater the rectangle contained by the two parts of the line, the less will be the sum of the squares on these parts. Therefore when the rectangle contained by the parts is a maximum, the sum of the squares on the two parts is a minimum. That is, the sum of the squares on the two parts of a line is a minimum when the line is bisected; and the minimum value is double the square on half the line. 284 ON THE PROPOSITIONS > The two propositions, Euc. III. 7, 8, afford instances of the greatest and least lines which can be drawn from a given point to the circum- ference of a circle. ... ---- - The straight line (Euc. III. 8, fig.) drawn from a fixed point with- out a circle to the circumference is a maximum when it passes through the center and meets the concave circumference; but a minimum when it meets the convex circumference and, if produced, would pass through the center. It is obvious, that the two values of the line on each side of the minimum value, are both greater than that value: and the two values of the line on each side of the mazimum value, are both less than that value. In other words, the magnitude of the line as it approaches the minimum, continually decreases till it reaches that value, and then increases: and the value of the line as it approaches the maximum, continually increases till it reaches that value and then decreases. When the given point is within the circle (Euc. III. 7), the greatest line that can be drawn from the point to the circumference is the line which passes through the center, and the least line that can be drawn from the same point, is the part produced of the greatest line between the given point and the circumference. & The theorem Euc. VI. 27 is a case of the mazimum value which a figure fulfilling the other conditions can have ; and the succeeding proposition is a problem involving this fact among the conditions as a part of the data, in truth, perfectly analogous to Euc. I. 20, 22. The doctrine itself was carefully cultivated by the Greek Geometers, and no solution of a Problem or demonstration of a theorem was con- sidered to be complete, in which it was not determined, whether there existed such limitations to the possible magnitudes concerned in it, and how those limitations were to be actually determined. Such Propositions as directly relate to Maasima and Minima, may be proposed either as Theorems or Problems. For the most part, how- ever, it is the more general practice to propose them as Problems; but this has most probably arisen from the greater brevity of the enuncia- tions in the form of a Problem. When proposed as a Problem, there is greater difficulty involved in the solution, as it is required to find the limits with respect to increase and decrease; and then to prove the truth of the construction: whereas in the form of a Theorem, the con- struction itself is given in the hypothesis. - It may be remarked that though the Differential Calculus is always effective for the determination of Maxima and Minima, (in cases where such exist) yet in many cases, where it is applied to the Problems which were cultivated by the Ancient Geometers, it is far less direct and elegant in its determinations than the Geometrical methods. Now if reference be made to what has been stated respecting Theorems, where the hypothesis is indeterminate, or wanting in that completeness which reduces the property spoken of to a single example of the figure in question, a consequence of that peculiarity in such classes of Propositions may be remarked. This peculiarity introduces another class of Propositions, which, though in “the Elements” some- what disguised, formed an importantportion of the Ancient Geometry:— the doctrine of Loci. * e. If the converse of Euc. I. 34, 35, 36, 37, and Euc. III. 20, 21, be taken in the form of Problems, they will become:– '. 1. Given the base and area, to construct the parallelogram. 2. Given the base and area, to construct the triangle. * IN THE ELEMENTs. 285 3. Given the base and vertical angle, to construct the triangle. Now three conditions are necessary to fix the magnitude of a triangle or a parallelogram, and in general, three only are sufficient for the purpose; but here it will be observed that only two are given in each case. The precise triangle or parallelogram, viewed as peculiarly solving the Problem, cannot be separated from all the others, except by adding some third condition to the two already given. - *. The side of the parallelogram in (1), and the vertex of the triangle in (2), opposite to the base, may be in any position in a certain line parallel to the base; and the vertex of the triangle in (3), may be at any point in the circumference of a segment of a certain circle. The parallel line in which the vertices of all the equal triangles are situated, in one case, and the arc of the circle in which the vertices of all the triangles having equal vertical angles are situated, are.each called the locus of the vertex of the triangle, since it occupies, in each case, all the places in which that vertex may be situated so as to fulfil the required conditions. In the same way, the parallel to the base is also the locus of all the positions in which the other two angular points of the paral- lelogram may be situated. These Problems are the simplest instances of that class which is called Local Problems; and their peculiar cha- racter is, that the data are one less than the number of conditions required by the nature of the Problem to restrict the quaesitum to a single or specified number of cases; as in these Problems the data consist of two conditions, while the exactly defining conditions must be three. s Again, viewed as Theorems, they mºy be thus enunciated:— - 1. If the base and area of a parallelogram be given, the locus of the other angular points will be a straight line parallel to the base. 2. If the base and area of a triangle be given, the locus of its vertex is a straight line parallel to the base. - 3. If the base and vertical angle of a triangle be given, the locus of the vertex will be an arc of a circle. - - In the original form of the propositions, the entire meaning, and that justified by Euclid's own reasoning, is that which would result from saying, “all triangles,” “all parallelograms,” &c. It will obvi- ously be the case here, as in the Maxima and Minima, that the propo- sition may be enunciated either as a local theorem or as a local problem ; and the circumstances will be similar as to the comparative brevity of enunciation and difficulty of the solution, when the proposition is given in the form of a Problem. - The great use made of loed by the Ancient Geometers was in the construction of determinate Problems. If a problem relate to the de- termination of a single point, and the data be sufficient to determine the position of that point, the problem is determinate: but if one or more of the conditions be omitted, the data which remain may be sufficient for the determination of more points than one, each of which satisfies the condition of the problem; in that case the problem is &ndeterminate, and in general such points are found to be situated in some line, which satisfies the conditions of the problem. . A certain number of data is required according to the nature of the problem for rendering the quaesitum determinate. The subject will be better illustrated by one or two examples. * For instance, the locus of the centers of all the circles whose cir- cumferences pass through two given points, A, B, is a straight line 286 ON THE PROPOSITIONs drawn perpendicular to AB at the point D, the point of bisection of the line AB; so that an indefinite number of circles may be described having their centers in the line perpendicular to AB drawn through D, and their circumferences passing through the two given points A, B. If a third point C be taken, but not in the same straight line with the first and second points A and B. Let A, C be joined and bisected in E, then the perpendicular to AC drawn through E the point of bisection of AC, will be the locus of the centers of the circles whose circumferences pass through the two points A and C. Hence the point F, the intersection of these two perpendiculars, will be the center of that circle whose circumference passes through the three given points A, B, C (Euc. IV. 5.), and is called the intersection of the two loci. - - Another example may be founded on the second and third theorems already noticed, which will take the following form :— Given the base, the area, and the vertical angle of a triangle, to construct it. - ". When the base and the area of a triangle are given, the locus of its vertex is a straight line which can be determined from these data; and when the base and vertical angle are given, the locus of the vertex is a portion of the circumference of a circle which can be determined from these data. Now the point or points of intersection of these loci, will fulfil both conditions, that the triangle shall have the given area, and the given vertical angle. To express the principle generally:— let there be n conditions requisite for the determination of a point which either constitutes the solution, or upon which the solution of the problem depends. Find the locus of this point subject to (n − 1) of these conditions; and again, the locus of the point subject to any other (n − 1) of these conditions. The intersection of these two loci gives the point required. It may be observed that (n − 2) of the data must be the same in determining the two loci, and no one of the n data . must be a consequent of, or depend upon, the remaining (n − 1) data, in other words, the n data must separately express n independent conditions. There are however cases in which one datum is involved in another, and these are of two different kinds—essential and accidental. To illustrate this distinction, let the following Problems be taken: Given one angle of a triangle a right angle, the base, and the difference of the squares on the hypotenuse and the perpendicular, to construct the triangle. Given the base, the area, and the perpendicular drawn from the vertex to the base of the triangle, to construct it. Given the base, the vertical angle and the sum of the other two angles at the base of the triangle, to construct it. - Now in each of these problems, the third datum is absolutely determined and invariable, in consequence of its essential dependence on the two previous ones. This dependence is universal and essential. Again, suppose the problem were:— Given the base of a triangle and a circle in magnitude and position, and likewise the vertical angle, to construct the triangle which shall have its vertex in the circumference of the given circle. : In this case, the given circle will generally be a different one from that which forms the locus of the vertical angle, and in that case, the intersections, or the point of contact, of the two circles will give either * IN THE ELEMENTs. - - 287 two solutions or one solution of the Problem. But on the other hand, the given circle may coincide with the locus, and thus again render the Problem indeterminate in this particular case. Generally the construc- tion is possible, and only accidentally it becomes indeterminate. The distinction between these two cases is very important. As Pro- blems are generally constructed by the intersections of loci, it is easy to imagine cases and conditions that shall give loci which can never meet. For instance, in the problem just stated, the two circles may never meet; and in the preceding one, the straight line and circle may never meet. In all such cases a problem is impossible with given conditions, when these conditions are incompatible with each other in their nature, or in their magnitude and position, or with the coexistence of that which constitutes the quaesitum. - The importance of the distinction alluded to, when one datum is contained in another, arises from its constituting the foundation of another Class of Propositions. These are called the Porisms. * > Whenever the quaesitum is a point, the problem on being rendere indeterminate, becomes a locus, whether the deficient datum be of the essential or of the accidental kind. When the quaesitum is a straight line or a circle, (which were the only two loci admitted into the ancient Plementary Geometry) the problem may admit of an accidentally *ndeterminate case; but will not invariably or even very frequently do so. This will happen when the line or circle shall be so far arbitrary in its position, as depends upon the deficiency of a single condition to fix it perfectly:—that is, (for instance) one point in the line, or two points in the circle, may be determined from the given conditions, but the remaining one is indeterminate from the accidental relations among the data of the problem. Determinate Problems become indeterminate by the merging of . Some one datum in the results of the remaining ones. This may arise in three different ways; first, from the coincidence of two points; secondly, from that of two straight lines; and thirdly, from that of two circles. These, moreover, are the only three ways in which the accidental coincidence of data can produce this indeterminateness in the problem. - There is a large class of indeterminate Problems which involve loci, and satisfy certain defined conditions. Every indeterminate pro- blem containing a locus may be made to assume the form of a porism, but the converse of this does not hold. Porisms are of a more general nature than indeterminate problems which involve a locus. In conclusion, it may be observed that the Ancient Geometers ap- pear to have undertaken the solution of Problems with a scrupulous and minute attention, which would scarcely allow any of the collateral truths to escape their observation. They never considered a Problem as solved till they had distinguished all its varieties, and evolved separately every different case that could occur, carefully distinguish- ing whatever change might arise in the construction from any change that was supposed to take place among the magnitudes which were given. This cautious method of proceeding would lead them to see that from circumstances in the data, the solution of some Problems would be impossible: that some would be determinate while others would be indeterminate: that some would admit ºf a maximum or a minimum : that some would involve a locus: and lastly, that some would assume the form of a Porism. * - THE ANCIENT GEOMETRICAL ANALYSIS. THE terms Analysis and Synthesis are usually understood, the former to signify the separation of any whole thing into its constituent parts, for the purpose of examining them separately: the latter to signify the composition, or the putting together of the several parts of any thing for the purpose of constituting the whole of it. . These terms used in their strict etymological sense, are not exactly in accordance with the use made of them in Geometry, where they are employed to indicate the direct and reverse order of a construction or demonstration. Synthesis, or the method of composition, is a mode of reasoning in geometry, which commences with the data of a Problem, or the hypo- thesis of a Theorem, and proceeding regularly by a process of construc- tion and reasoning, ends with the quaesitum of the Problem or with the proof of the predicate of the Theorem. The Synthetic method is pur- sued in the demonstrations of the Propositions in Euclid's Elements: certain principles are assumed or admitted, and on these principles are founded the construction of Problems and the demonstration of Theo- rems. This may be termed a direct process, as it leads from principles to their consequences. Analysis, or the method of resolution, is a process, the reverse of Synthesis, which commences with assuming the quaesitum of the Problem as found, or the predicate of the Theorem as proved, and by a process of construction and reasoning, terminates in the data of the Problem or the hypothesis of the Theorem. This may be considered an indirect process, a method of reasoning from consequences to prin- ciples. Hence, “Analysis presents the medium of invention; while Synthesis naturally directs the course of instruction.” The Synthesis begins where the Analysis ends; the last step of the Analysis becomes the first step of the Synthesis, and the other steps of the Analysis are retraced in order to the first, which constitutes the last step of the Synthesis. - - - z -. THE ANALYSIS OF THEOREMS. IT may have been remarked, that in the Elements, Euclid frequently uses the indirect method of demonstration:—that is, of proving the truth of a theorem by demonstrating that a contrary conclusion is in- compatible with the hypothesis of that theorem. To effect this, he supposes the enunciated property to be false; and its contrary to be true. He reasons from the assumed truth of this false property, till he arrives at a conclusion dependent upon that assumption, which is contrary to the original hypothesis; and thence it is inferred that the assumption being incompatible in its consequences with the original conditions of the theorem, those conditions and that assumption cannot coexist. If, then, all the alternatives of the alleged property be thus examined, and shown to be incompatible with the original. hypothesis, it will necessarily follow, that this property itself is true. Thus, in Euc. I. 25, where one included angle BAC is alleged in the enunciation to be greater than the other EDF, under the hypothesis of B4, 40, being respectively equal to ED, DF, but BC greater than J.F.; instead of proving the assertion itself, he admits, that in the first place the angle BAC is equal to the angle EDF, and in the second, that it is less. The consequences of these admissions are both shewn to be incompatible with the hypothesis, and hence it is inferred that the angle B40 can neither be equal to EDF, nor less than it. THE ANALYSIs of THEOREMs. 289 Wherefore as these are the only alternatives to the truth of the enun- ciation, and both these are false, it follows that the alleged relation of the angles BAC, EDF is true. , -- - This method of proof occurs frequently in the first and third books of Euclid; and it may be remarked generally, that it occurs more often in the outset of the development of a system of truths than in the more advanced parts, or in the more recondite theorems. It must naturally have occurred to Geometers, who were familiar with the use of this mode of assumption, to inquire: “What would be the effect of supposing the alleged theorem to be true, instead of false P” He, who first asked this question made the first step in the Geometrical Analysis. He would see at once that the conclusion ought to be con- sistent with the hypothesis, and with all, previously known properties of the hypothetical figure. He may, indeed, find it of little conve- nience, often of none, in suggesting a direct proof of a very elementary theorem, but as he would be of course led to try its efficacy in more complex cases, he would be gradually impressed with the facts:—that in many cases his steps were merely the reversal of the steps which he had employed in the hypothetic demonstration of the theorem; and that in all cases, a reversal in the order of the steps of his analysis would constitute a synthetic demonstration, though perhaps different from any one previously known to him. He would then have discovered the true principle of the Geometrical Analysis of Theorems ; and he would require but little additional skill to reduce the whole process to a com- plete system. It is probable that his discovery might lead to some such rules as the following: 1. Assume that the theorem is true. 2. Proceed to examine any consequences that result from this admission, by the aid of other truths respecting the figure, which have been already proved. g *. - 3. Examine whether any of these consequences be themselves such as are already known to be true, or to be false. 4. If any one of them be false, we have arrived at a reductio ad absurdum, which proves that the theorem itself is false, as in Euc. I. 25. 5. If none of the consequences so deduced be known to be either true or false, proceed to deduce other consequences from all or any of these, as in (2). * 6. Examine these results, and proceed as in (3) and (4); and if still without any conclusive indications of the truth or falsehood of the alleged theorem, proceed still further, until such are obtained. In the case of the theorem being false, we shall ultimately arrive at some result contradictory either to the original hypothesis, or to some truth depending upon it. Euclid's indirect demonstrations always end with a contradiction to the immediate hypothesis; but as the propositions to which he applies the method—are so extremely elementary, this could scarcely happen otherwise, as, so far, deductions would be made from the hypothesis by direct steps. Where, however, we find a contradiction in our results to any one of the consequences of the hypothesis, our conclusion, that the theorem is false, is as legitimate as though the contradiction had immediately been of the hypothesis itself. Nevertheless, if it should be imposed as a rule, that the contra- diction shall be that of the hypothesis itself, it is only requisite to reverse that consequence of the hypothesis which is so contradicted, and to employ the contradiction instead of the conclusion of that consequence; U 290 THE ANALYSIs of THEOREMs. and the result will be thus carried back into direct contradiction to the original hypothesis. It may sometimes happen that our attempts thus to analyse a theorem may be carried on through a considerable number of succes. sive steps, and yet no conclusive evidence of the truth or falsehood of the alleged theorem present itself. Nor can we ever judge, a priori, whether we should succeed by continuing the process further in any one particular direction. Under one aspect this may be considered an inconvenience; but even were it a real inconvenience, it is inevit- able, and must so far be taken as a drawback upon the value of the method. The inconvenience is, however, more apparent than real; or, at least, the inconvenience is amply compensated by the advantages it otherwise confers, not indeed in reference to the demonstration of the proposed theorem, but in its extension of geometrical discovery. A mistake might occur in the synthetic deduction of a proposed theorem, or the theorem might be a mistaken inference from analogy, or from the contemplation of carefully drawn diagrams; but it does not often. happen that a theorem is proposed for solution, of the truth of which the proposer has not satisfied himself. The probabilities then are greatly in favour of such proposition being correct. Now in this case, all the investigations which have been made with a view to the analysis of that theorem, will become so many synthetic demonstrations of the re- sults which have been obtained during those unsuccessful attempts to analyse. It will in general be found, too, that they are of such a châracter as would scarcely have occurred to any Geometer to adopt with pure reference to synthetic purposes. There can, in fact, be little doubt that the greater part of the most profound and original theorems that are found in the writings of the greatest Geometers of ancient and of modern times, have originated in attempts to analyse some proposed theorem; and which have failed merely from the direction which was pursued, lying in that of the more recondite instead of the more simple order of truths connected with the proposed one. Such failures should therefore be always carefully preserved, till the proposition itself, from which they were deduced, be proved either to be false or true. If the course of analysing pursued in the first instance be not found to succeed, and if the conclusions become more and more elementary in their character, some other properties of the figure connected with the assumed truth should be tried in the same manner; and if this also fail to accomplish the immediate object, the investigations should be pursued as before. It has occasionally happened, though extremely seldom, that several such attempts have failed in succession. Yet some mode of deduction must necessarily become a true analysis of the theorem; which will in general result from adequate perseverance. All the results ob-, tained in the preceding efforts to analyse the theorem, will then constitute a number of truths, connected with each other through that one by which they were originally suggested; and often among truths so related, a general principle may be detected, that shall prove of the utmost value in the treatment of entire classes of propositions, which now stand in an uninteresting state of isolation from each other. Moreover, by systematising the propositions of Geometry, we simplify their didactic development; and by contemplating in Connexion with each other, such attempted analyses of single theorems, great benefits may be conferred upon Geometrical Science and its practical applications. & THE ANALysis of PROBLEMs. 291 There is not the slightest difference between analysis and synthesis, as far as the course of consecutive deduction is concerned. Both are direct applications of the Ordinary enthymeme ; and both require the same specific habits of mind, and the same resources as regards truths already known. The only distinction consists, as far as mere reasoning is concerned, in the difference of the starting points of the investigation. In Analysis we start from the enunciated property as a truth tempo- ' rarily admitted; and ultimately arrive at some property which we previously knew to be true of the hypothetical figure. We have only to reverse the Order of the Syllogisms, and of the subject and predicate in each of them, to convert the analysis into the synthesis in one case, or the synthesis into the analysis in the other. They are so connected, in fact, that had the hypothesis of the proposed theorem been already proved by one process; the reversed process, the analysis of which we have spoken, would have become the synthesis of the theorem. It must now be obvious that the synthesis of the theorem can be at once formed from the analysis, by the reversal of the steps already described, and when this has been effected, the analysis may, if desirable, be altogether suppressed. On the other hand, for all the purposes of giving full and legitimate conviction of the truth of atheorem, the analysis is always sufficient, without adding the synthesis. It is, how- ever, desirable, in a course of Geometrical study, to complete the formal draft of the investigation both in the analytic and synthetic form. THE ANALYSIS OF PROBLEMs. IN every Geometrical operation we perform in the construction of a Problem, we have in mind some precedent reason, a knowledge of some properties of the figure, either assumed or proved, which would result from that operation, and a perception of its tendency towards accomplishing the object proposed in the Problem. Our processes for construction are founded on our knowledge of the properties of the figure, supposed to eacist already, subjected to the conditions which are enunciated in the proposition itself. No Problem could be constructed (except by mere trial, and verification by mere instrumental experiments) antecedently to the admission of our knowledge of some properties of the figure which it is proposed to construct. The simple reason for the operations employed, is, that they collectively and ultimately fulfil the prescribed conditions; and their so fulfilling the conditions, is only known by previously reasoning upon the figure supposed already to be so constructed as to embody those conditions. Let any Problem be selected from Euclid, and at each step of the operation, let the question be asked, “Why that step is taken” It will in all cases be found that it is because of some known property of the figure required, either in its complete or intermediate states, of which the inventor of the construction must have been in possession. This antecedency of Theorems to all Geometrical construction in Scientific Geometry is universal and essential to its nature. Let the construction of Euc. IV. 10 be taken in illustration of what has been stated. There are five operations specified in the construction :- f - 1. Take any line A.B. 2. Divide that line in C, so that, &c. 3. Describe the circle BDE with center A and radius A.B. 4. Place BD in that circle, equal to A.C. 5. Join the points A, D. U 2 . 292 THE ANALYSIS OF PROBLEMS. Why should these operations be performed in order rather than any others? And what clue have we to enable us to foresee that the result of them will be such a triangle as was required? The demonstration affixed to the Proposition by Euclid, does undoubtedly prove that these operations must, in conjunction, produce such a triangle : but we are furnished in the Elements with no obvious reason for the adoption of these steps, unless we suppose them accidental. To suppose that ali the constructions, even the simple ones, were the result of accident only, would be supposing more than could be shewn to be admissible. No construction of the problem could have been devised without a previous knowledge of some of the properties of the figure which was to be constituted. In fact, in directing the figure to be constructed, we assume the possibility of its existence; and we study the properties of such a figure on the hypothesis of its actual existence. It is this study of the properties of the figure that constitutes the Analysis of the Problem. - Let then the existence of a triangle BAD (fig. Euc. IV. 10) be admitted which has each of the angles ABD, ADB double of the angle BAD, in order to ascertain any properties it may possess which would assist in the actual construction of such a triangle. Then, since the angle AIDB is double of B.AD, if we draw a line . DC to bisect ADB and meet AB in C, the angle ADC will be equal to OAD; and hence (Euc. I. 6) the side AC, is equal to CD. - Again, there are three points A, C, D, not in the same straight line, lettis examine the effect of describing a circle through them : that is, describe the circle A CD about the triangle A CD (Euc. Iv. 5). Then, since the angle ADB has been bisected by DC, and since ADB is double of DAB, the angle CDB is equal to the angle DAG in the alternate segment of the circle; the line BD therefore coincides with a tangent to the circle at D (converse of Euc. III. 32). - Whence it follows that the rectangle contained by AB, BC, i equal to the square on BD (Euc. III. 36). - But the angle BCD is equal to the two interior opposite angles CAD, CDA ; or since these are equal to each other; BCD is the double of CAD, that is of B.A.D. And since ABD is also double of BAD, by the conditions of the triangle, the angles BCD, CBD are equal, and BD is equal to DC, that is, to A. C. It has been proved that the rectangle AB, BC, is equal to the Square on B D; and hence the point C in AB, found by the intersection of the bisecting line DC, is such, that the rectangle AB, BC is equal to the square on AC (Euc. II. 11). Finally, since the triangle ABD is isosceles, having each of the angles ABD, ADB double of BAD, the sides AB, AD are equal, and hence the points B, D, are in the circumference of the circle described about A with the radius AB. And since the magnitude of the triangle is not specified, the line AB may be of any length whatever. From this “Analysis of the Problem,” which obviously is nothing more than an examination of the properties of such a figure supposed to exist already, it will be at once apparent, why those steps which are prescribed by Euclid for its construction, were adopted. The line AB is taken of any length, because the problem does not prescribe any specific magnitude to any of the sides of the triangle: the circle BDE is described about A with the distance AB, because the triangle is to be isosceles, having AB for one side, and therefore the other extremity of the base is in the circumference of that circle: the THE ANALYSIS OF PROBLEMS. - 293 line AB is divided in C so that the rectangle AB, BC shall be equal to the square on AC, because the base of the triangle must be equal to the segment AC; and the line AD is drawn, because it completes the triangle, two of whose sides AB, BD are already drawn. A careful examination of this process will point out the true character of the method by which the construction of all problems (except perhaps a few simple ones which involve but very few and very obvious steps) have been invented: although the actual analysis itself has been suppressed or concealed, as, amongst the ancient Geometers, appears to have been the general practice. It will be inferred at once, that the use of the Analysis in reference to the construction of problems, is altogether indispensable in its actual form, where the problem requires several steps for its construction; as it has been shewn to be virtually (though the operations may in certain simple problems be carried on mentally and almost unsuspectedly) essential to the construction of all problems whatever. When we have reduced the construction to depend upon problems which have been already constructed, our analysis may be terminated; as was the case when, in the preceding example, we arrived at the division of the line AB in C; this problem having been already con- structed as the eleventh of the second book. From the nature of the subject, it must be at once obvious that no general rules can be prescribed which will be found applicable to all cases, and lead to the solution of every problem. The-conditions of problems must suggest what constructions may be possible; and the consequences which follow from these constructions and the assumed solution, will shew the possibility or impossibility of arriving at some known property consistent with the data of the Problem. In the following exercises, many will be found to be of so simple a character, (being obvious deductions from the Elements) as scarcely to admit of the principle of the Geometrical Analysis being applied, in their solution. - A clear and exact knowledge of first principles must necessarily precede any intelligent application of them. Indistinctness or defec- tiveness of understanding with respect to these, will be a perpetual Source of error and confusion. The learner is therefore recommended to understand the principles of the Science, and their connexion, clearly, before he attempt the application of them. The following directions may assist him in his proceedings. 1. In general, any given problem will be found to depend on Several problems and theorems, and these ultimately on some problem or theorem in Euclid. * 2. Describe the diagram as directed in the enunciation, and sup- pose the solution of the problem effected. g 3. Examine the relations of the lines, angles, triangles, &c. in the diagram, and find the dependence of the assumed solution on some . theorem or problem in the Elements. 4. If such cannot be found, draw other lines parallel or perpen- dicular as the case may require, join given points, or points assumed in the solution, and describe circles if need be: and then proceed to trace the dependence of the assumed solution on some theorem or problem in Euclid. - - 5. Let not the first unsuccessful attempts at the solution of a Pro- blem be considered as of no value; such attempts may lead to the solu- tion of other problems, or to the discovery of new geometrical truths. A GEOMETRICAL EXERCISES ON BOOK I. PROPOSITION I. PROBLEM. To trisect a given straight line. Analysis. Let AB be the given straight line, and suppose it divided into three equal parts in the points D, E, C —w A D 1. B On DE describe an equilateral triangle DEF: then DF is equal to AD, and FE to EB. On AB describe an equilateral triangle ABC, and join AF, F.B. Then because AD is equal to DF, therefore the angle AFD is equal to the angle DAF, and the two angles DAF, DFA are double of one of them DAF. - But the angle FDE is equal to the angles DAF DFA, and the angle FDE is equal to DAC, each being an angle of an equilateral triangle; . - therefore the angle DAC is double the angle DAF; wherefore the angle DAC is bisected by A.F. Also because the angle FA C is equal to the angle FAD, and the angle FAD to DFA; - therefore the angle CAF is equal to the alternate angle AFD: and consequently FD is parallel to A.C. Synthesis. Upon AB describe an equilateral triangle ABC, bisect the angles at A and B by the straight lines AF, BF, meeting in F; through F draw FD parallel to AC, and FE parallel to BC. Then AB is trisected in the points D, E, For since AC is parallel to FD, and FA meets them, therefore the alternate angles FAC, AFD are equal; but the angle FA D is equal to the angle FA C, hence the angle DAF is equal to the angle AFD, and therefore. DF is equal to D.A. But the angle FDE is equal to the angle CAB, and FED to CBA; (I. 29.) - therefore the remaining angle DFE is equal to the remaining angle A CB. * - Hence the three sides of the triangle DFE, are equal to one another, and DF has been shewn to be equal to DA ; therefore AD, DE, EB are equal to one another. - Hence the following theorem:— If the angles at the base of an equilateral triangle be bisected by two lines which meet at a point within the triangle; the two lines drawn from this point parallel to the sides of the triangle, divide the base into three equal parts. BOOK 1. PROP. II, III. 295 PROPOSITION II. THEOREM. If two opposite sides of a parallelogram be bisected, and two lines be drawn from the points of bisection to the opposite angles, these two lines trisect the diagonal. - Let ABCD be a parallelogram of which the diagonal is AC. . Let A B be bisected in E, and DC in F, - - also let DE, FB be joined cutting the diagonal in G, H. * Then AC is trisected in the points G, H. A E B. F Through E draw Eſparallel to AC and meeting FB in K. Then because EB is the half of AB, and DF the half of DC, therefore EB is equal to DF; and these equal and parallel straight lines are joined towards the same parts by DE and FB; - therefore DE and FB are equal and parallel. (I. 33.) And because A.EB meets the parallels EK, AC, therefore the exterior angle B.EK is equal to the interior angle E.A.G. - For a similar reason, the angle EBR is equal to the angle AEG. Hence, in the triangles AEG, EBK, there are the two angles G.A.E, AEG in the one, equal to the two angles KEB, EBK in the other, and one side adjacent to the equal angles in each triangle, namely A.F. equal to EB; - • . therefore AG is equal to EK, (I. 26.) but EK is equal to GH, (1.34.) therefore AG is equal to G.H. By a similar process, it may be shewn that GH is equal to HC. Eſence AG, GH, HC are equal to one another, and therefore AC is trisected in the points G, H. It may also be proved that BF is trisected in H and K. PROPOSITION III. PROBLEM. Draw through a given point, between two straight lines not parallel, a straight line which shall be bisected in that point. Analysis. Let BC, BD be the two lines meeting in B, and let 4 be the given point between them. Suppose the line EAF drawn through A, so that EA is equal to AF; 296 - GEOMETRICAL ExERCISEs through A draw AG parallel to BC, and GH parallel to EF. Then AGHE is a parallelogram, wherefore AE is equal to GB, . . . but EA is equal to AF by hypothesis; therefore GH is equal to 4 F. Hence in the triangles BHG, GAF, the angles HBG, AGF are equal, as also BGH, GPA, (r. 29.) also the side GH is equal to 4 F; whence the other parts of the triangles are equal; (I. 26.) - thereforé BG is equal to G.F. Synthesis. Through the given point A, draw 46 parallel to BC, r on GD, take GF equal to GB; then Fis a second point in the required line : join the points H, A, and produce F4 to meet BC in E; then the line FE is bisected in the point A. Draw GH parallel to A.E. Then in the triangles BGH, GFA, the side BG is equal to GF, and the angles GBH, BGH are respectively equal to FGA, GFA ; wherefore GH is equal to A.F. (I. 26.) \ but GH is equal to AE; (I. 34.) therefore AE is equal to AF, or EF is bisected in 4. PROPOSITION IV. PROBLEM. - Given one angle, a side opposite to it, and the sum of the other two sides, construct the triangle. - Analysis. Suppose BAC the triangle required, having BC equal to the given side, BAC equal to the given angle opposite to BC, also BD equal to the sum of the other two sides. º - Join DC. Then since the two sides BA, AC are equal to BD, by taking BA from these equals, the remainder AC is equal to the remainder A.D. Hence the triangle A CD is isosceles, and therefore the angle ADC is equal to the angle A CD. - - - But the exterior angle BAC of the triangle ADC is equal to the two interior and opposite angles A CD and ADC: - wherefore the angle BAC is double of the angle BDC, and BDC is the half of the angle BAC. - . Hence the synthesis. - * the point D in BD, make the angle BDC equal to half the given angle, - and from B the other extremity of BD, draw BC equal to the given side, and meeting DC in C, --" at C in CD make the angle DCA equal to the angle CDA, so that CA may meet BD in the point A. Then the triangle ABC shall have the required conditions. Book I. PROP. V, VI. 297 PROPOSITION V. THEOREM. If straight lines be drawn bisecting the angles of a triangle, they shall all pass through the same point. - - Let ABC be a triangle, and let BE, CF be drawn bisecting the angles ABC, ACB, and intersecting each other in 6: join A G and produce it to meet BC in D. Then A G D shall bisect the angle B.A. C. A B D. H. C Qe From G draw GH, GK, GL perpendicular to BC, CA, AB respectively. Then in the triangles BGH, BGL, the angles GBH, GHB in the one, are equal to GBL, GLB in the other, and the side GB is common to the two triangles; - therefore GH is equal to GL. (Euc. I. 26.) Similarly, it may be proved that GH is equal to GK. - Wherefore GK is equal to GL. , r And since in the two right-angled triangles AKG, ALG, the squares on the sides A.K, KG, are equal to the squares on AL, LG, (Euc. I. 47, and ax. 1.) - and the square on KG is equal to the square on LG, - therefore the square on AK is equal to the square on AL, (ax. 3.) g and the line AK is equal to the line AL. Whence, in the triangles AKG, ALG, the two sides AK, A G of the one, are respectively equal to AL, AG of the other, and the base GK is equal to the base GL, wherefore the angle KAG is equal to the angle L.A.G., (Euc. I. 8.) and the three straight lines AD, B.E, CF bisecting the angles of the triangle ABC, pass through the same point G. - --- PROPOSITION VI. PROBLEM. To bisect a triangle by a line drawn from a given point in one of the sides. Analysis. Let ABC be the given triangle, and D the given point in the side AB. A. D IB E F C Suppose DF the line drawn from D which bisects the triangle; therefore the triangle DBF is half of the triangle ABC. Bisect BC in E, and join AE, DE, AF, - then the triangle ABE is half of the triangle ABC: hence the triangle ABE is equal to the triangle DBF'; take away from these equals the triangle DBE, 298 GEOMETRICAL ExERCISES therefore the remainder A.D.E is equal to the remainder DEF But ADE, DEF are equal triangles upon the same base DE, and on the same side of it, *, - they are therefore between the same parallels, (I. 39.) that is, AF is parallel to DE, therefore the point Fis determined. Synthesis. Bisect the base BC in E, join DE; from A, draw AF parallel to DE, and join D.F. Then because DE is parallel to A.F. the triangle ADE is equal to the triangle DEF; (I. 37.) to each of these equals, add the triangle BDE; therefore the whole triangle ABE is equal to the whole DBF, but ABE is half of the whole triangle ABC; therefore DBF is also half of the triangle ABC. PROPOSITION VII. THEOREM. If from a point without a parallelogram lines be drawn to the extremi- ties of two adjacent sides, and of the diagonal which they include; of the tri- angles thus formed, that, whose base is the diagonal, is equal to the sum of the other two. Let ABCD be a parallelogram of which A C is one of the diagonals, and let P be any point without it: and let A.P, PC, BP, PD be joined. Then the triangles APD, APB are together equivalent to the tri- angle APC. , Draw PGE parallel to AD or BC, and meeting AB in G, and DC in E; and join DG, G C. Then the triangles CBP, CBG are equal: ' (I. 37.) and taking the common part CBH from each, the remainders PHB, CHG are equal. Again, the triangles DAP, DAG are equal; (I. 37.) also the triangles DAG, AGC are equal, being on the same base AG, and between the same parallels AG, DC: therefore the triangle DAP is equal to the triangle AGC: but the triangle PHB is equal to the triangle CHG, wherefore the triangles PHB, DAP are equal to AGC, CHG, or ACII, add to these equals the triangle APH, therefore the triangles APH, PHB, DAP are equal to APH, ACH, that is, the triangles APB, DAP are together equal to the triangle PA O. * * CoR. If the point P be within the parallelogram, then the differ- ence of the triangles APB, DAP may be proved to be equal to the triangle PAC. ON BOOK I. 299 I. 8. Describe an isosceles triangle upon a given base and having each of the sides double of the base. 9. Draw from the center of the larger circle in the figure Euc. I. 2, a line so that the part of it intercepted between the two circles shall be equal to the given straight line. - - 10. In the fig. Euc. I. 5, if FC and BG intersect in H, then prove that AH bisects the angle BAC. And if the angle FBG be equal to the angle ABC, the angle BHFis equal to twice the angle B.A. C. 11. From the extremities of the base of an isosceles triangle straight lines are drawn perpendicular to the sides, the angles made by them with the base are each equal to half the vertical angle. 12. A line drawn bisecting the angle contained by the two equal sides of an isosceles triangle, bisects the third side at right angles. * 13. If a straight line drawn bisecting the vertical angle of a tri- angle, also bisect the base, the triangle is isosceles. 14. Given two points one on each side of a given straight line; find a point in the line such that the angle contained by two lines drawn to the given points may be bisected by the given line. 15. In the fig. Euc. I. 5, let F and G be the points in the sides AB and AC produced, and let lines FH and GK be drawn perpen- dicular and equal to FC and GB respectively: also if BH, CK, or these lines produced meet in 0; prove that BH is equal to CK, and JB O to CO. | 16. From two given points on the same side of a straight line given in position, draw two lines which shall meet at a point in that line, and make equal angles with it. § 17. From every point of a given straight line, the straight lines drawn to each of two given points on opposite sides of the line are equal: prove that the line joining the given points will cut the given line at right angles. - 18. If A be the vertex of an isosceles triangle ABC, and BA be produced so that AD is equal to BA, and DC be drawn; shew that BCD is a right angle. - , ºr 19. The straight line EDF, drawn at right angles to BC the base of an isosceles triangle ABC, cuts the side AB in D, and CA pro- duced in E; shew that AED is an isosceles triangle. - 20. In the fig. Euc. I. 1, if A B be produced both ways to meet the circles in D and E, and from C, CD and CE be drawn; the figure CDE is an isosceles triangle having each of the angles at the base, equal to one-fourth of the angle at the vertex of the triangle. 21. From a given point, draw two straight lines making equal angles with two given straight lines intersecting one another. 22. From a given point to draw a straight line to a given straight line, that shall be bisected by another given straight line. 23. Place a straight line of given length between two straight lines which meet, so that it shall be equally inclined to each of them. 24. To determine that point in a straight line from which the straight lines drawn to two given points shall be equal, provided the line joining the two points is not perpendicular to the given line. 25. In a given straight line to find a point equally distant from two given straight lines. In what case is this impossible? 26. If a line intercepted between the extremity of the base of an 300 GEOMETRICAL EXERCISES ©. isosceles triangle, and the opposite side (produced if necessary) be equal to a side of the triangle, the angle formed by this line and the base produced, is equal to three times either of the equal angles of the triangle. - - 27. In the base BC of an isosceles triangle ABC, take a point D, and in CA, take CE equal to CD, let ED produced meet AB produced in F; then 3.AEF= 2 right angles + AFE, or = 4 right angles + AFE. 28. If from the base to the opposite sides of an isosceles triangle, three straight lines be drawn, making equal angles with the base, viz. one from its extremity, the other two from any other point in it, these two shall be together equal to the first. 29. 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. 30. In a triangle, if the lines bisecting the angles at the base be equal, the triangle is isosceles, and the angle contained by the bisect- ing lines is equal to an exterior angle at the base of the triangle. 31. No two straight lines drawn from the angles of a triangle and terminated by the opposite sides can bisect each other. - 32. In a triangle, if lines be equal when drawn from the extremi- ties of the base, (1) perpendicular to the sides, (2) bisecting the sides, (3) making equal angles with the sides; the triangle is isosceles: and these lines which respectively join the intersections of the sides, are parallel to the base. 33. If one angle of a triangle be double or triple of another; or if it be equal to the sum, or to one half of the difference of the remain- ing two; the triangle may either be divided into two isosceles triangles, or else an isosceles triangle may be added to it in such a manner as to form together with it a single isosceles triangle. Prove this, and dis- tinguish the different cases. & II. 34. If two straight lines are respectively at right angles to two others which intersect, shew that each pair of lines includes the same angle. . - *. ABC is a triangle right-angled at B, and having the angle A double the angle C; shew that the side BC is less than double the side A.B. 36. If one angle of a triangle be equal to the sum of the other two, the greatest side is double of the distance of its middle point from the opposite angle. 37. If from the right angle of a right-angled triangle, two straight lines be drawn, one perpendicular to the base, and the other bisecting it, they will contain an angle equal to the difference of the two acute angles of the triangle. } •º 38. If the vertical angle CAB of a triangle ABC be bisected by AD, to which the perpendiculars CE, BF are drawn from the remain- ing angles: bisect the base BC in G, join GE, GF, and prove these lines equal to each other. . - 39. The difference of the angles at the base of any triangle, is double of the angle contained by a line drawn from the vertex perpen- dicular to the base, and another bisecting the angle at the vertex. ON BOOK I. 301 40. If one angle at the base of a triangle be double of the other, the less side is equal to the sum or difference of the segments of the base, made by the perpendicular from the vertex, according as the angle is greater or less than a right angle. 41. If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines, a line be drawn to the oppo- site angle of the triangle, it will bisect that angle. 42. From the vertex of a scalene triangle draw a right line to the base, which shall exceed the less side as much as it is exceeded by the greater. -- 43. Divide into three equal angles, (1) a right angle; (2) one- fourth of a right angle. 44. Prove that the sum of the distances of any point within a triangle from the three angles is greater than half the perimeter of the triangle. - 45. If from the angles of a triangle ABC, straight lines ADE, BDF, CDG be drawn through a point D to the opposite sides, prove that the sides of the triangle are greater than the three lines drawn to the point D, and less than twice the same, but greater than two-thirds of the lines drawn through the point to the opposite sides. • 46. If two triangles have the same base as a third triangle, and their vertices upon its two sides, the sum of their sides will be less than twice the sum of the sides of the third triangle, and the sum of their vertical angles will be greater than twice its vertical angle. 47. In a plane triangle an angle is right, acute or obtuse, accord- ing as the line joining the vertex of the angle with the middle point of the opposite side is equal to, greater or less than half of that side. - 48. If the straight line AD bisect the angle A of the triangle ABC, and BDE be drawn perpendicular to AD and meeting AC or AC produced in E, shew that BB = D.E. 49. The side BC of a triangle ABC is produced to a point D. The angle ACB is bisected by a line CE which meets AB in E. A line is drawn through E parallel to BC and meeting AC in F, and the line bisecting the exterior angle A CD, in G. Then EF is equal to FG. 50. The sides AB, AC, of a triangle are bisected in D and E respectively, and B.E, CD, are produced until EF= EB, and GD = DC; shew that the line GF passes through A. - 51. AEB, CED are two straight lines intersecting in E.; lines A C, DB are drawn forming two triangles ACE, BED; if the angles A CE, DBE be bisected by the straight lines CF, BF meeting in F, shew that the angle CFB is equal to half the sum of the angles JEAC, E/).B. - 52. In a triangle ABC, AD being drawn perpendicular to the straight line BD which bisects the angle B, shew that a line drawn from D parallel to BC will bisect A.C. • , , 53. If the sides of a triangle be trisected and lines be drawn through the points of section adjacent to each angle so as to form another triangle, this shall be in all respects equal to the first triangle. 54. Between two given straight lines it is required to draw a straight line which shall be equal to one given straight line, and parallel to another. 55. ABC is a given triangle, it is required to draw from a given point P, in the side AB, or AB produced, a straight line to AC, so that it shall be bisected by BC. 302 GEOMETRICAL EXERCISES 56. If from the vertical angle of a triangle three straight lines be drawn, one bisecting the angle, another bisecting the base, and the third perpendicular to the base, the first is always intermediate in magnitude and position to the other two. 57. In the base of a triangle, find the point from which, lines drawn parallel to the sides of the triangle and limited by them, are equal. Q 58. In the base of a triangle, to find a point from which if two lines be drawn, (1) perpendicular, (2) parallel, to the two sides of the triangle, their sum shall be equal to a given line. III. 59. In the figure of Euc. I. 1, the given line is produced to meet either of the circles in P; shew that P and the points of intersection of the circles, are the angular points of an equilateral triangle. - 60. If each of the equal angles of an isosceles triangle be one- fourth of the third angle, and from one of them a line be drawn at right angles to the base meeting the opposite side produced; then will the part produced, the perpendicular, and the remaining side, form an equilateral triangle. ſº 61. In the figure Euc. I. 1, if the sides CA, CB of the equilateral triangle ABC be produced to meet the circles in F, G, respectively, and if C be the point in which the circles cut one another, on the other side of AB : prove the points F. C., G to be in the same straight line; and the figure CFG to be an equilateral triangle. 62. ABC is a triangle, and the exterior angles at B and C are bisected by lines BD, CD respectively, meeting in D: shew that the angle BDC and half the angle BAC make up a right angle. 63. If the exterior angle of a triangle be bisected, and the angles of the triangle made by the bisectors be bisected, and so on, the tri- angles so formed will tend to become eventually equilateral. 64. If in the three sides AB, BC, CA of an equilateral triangle ABC, distances A.E., B.F. CG be taken, each equal to a third of one of the sides, and the points E, F, G be respectively joined (1) with each other, (2) with the opposite angles: shew that the two triangles so formed, are equilateral triangles. - IV. 65. . Describe a right-angled triangle upon a given base, having given also the perpendicular from the right angle upon the hypotenuse. 66. Given one side of a right-angled triangle, and the difference between the hypotenuse and the sum of the other two sides; to con- struct the triangle. . 67. Construct an isosceles right-angled triangle, having given (1) the sum of the hypotenuse and one side; (2) their difference. 68. Describe a right-angled triangle of which the hypotenuse and the difference between the other two sides are given. - - 69. Given the base of an isosceles triangle, and the sum or dif- ference of a side and the perpendicular from the vertex on the base. Construct the triangle. s * 70. Make an isosceles triangle of given altitude whose sides shall pass through two given points, and whose base shall be on a given straight line. * 71. Given of any triangle the perpendicular let fall from the ver- on Book I. 303 ** tical angle on the base, and the difference between each segment made by the perpendicular and its adjacent side, construct the triangle. 72. Having given the straight lines which bisect the angles at the base of an equilateral triangle, determine a side of the triangle. 73. Having given two sides and an angle of a triangle, construct. the triangle, distinguishing the different cases. 74. Having given the base of a triangle, the difference of the sides, and the difference of the angles at the base; to describe the triangle. 75. Given the perimeter and the angles of a triangle; to construct it. 76. Having given the base of a triangle, and half the sum and half the difference of the angles at the base; to construct the triangle. 77. Having given two lines, which are not parallel, and a point between them; describe a triangle having two of its angles in the re- spective lines, and the third at the given point; and such that the sides shall be equally inclined to the lines which they meet. 78. Construct a triangle, having given the three lines drawn from the angles to bisect the sides opposite. - 79. Given one of the angles at the base of a triangle, the base itself, and the sum of the two remaining sides, to construct the triangle. 80. Given the base, an angle adjacent to the base, and the differ- ence of the sides of a triangle; to construct it. - 81. Given one angle, a side opposite to it, and the difference of the other two sides; to construct the triangle. * - 82. Given the base and the sum of the two other sides of a triangle: construct it so that the line which bisects the vertical angle shall be parallel to a given line. 83. Construct a triangle two of whose sides shall be together, double of the third side. If the triangle be isosceles, prove that it is also equilateral. ; 84. In a right-angled triangle, given the sums of the base and hypotenuse, and of the base and perpendicular, also of the perpendi- cular and hypotenuse; to construct the triangle. 85. Given the base, the difference and the sum of the other two sides; to construct the triangle. . 86. Construct a triangle equiangular to a given triangle, and having its angular points upon three given straight lines which meet in a point. º º o 87. If through the angular points of any triangle straight lines be drawn, making equal angles with the sides taken in order, the triangle formed by these lines shall be equiangular to the original triangle. 88. Given the base of a triangle, one of the angles at the base, and also the angle which the perpendicular drawn upon the base from the opposite angle makes with the side opposite to the given angle of the triangle; construct the triangle. g º 89. With the three altitudes of a triangle ABC as sides, form a triangle; and with the three altitudes of the triangle so formed, form another triangle: shew that this third triangle is similar to the original triangle ABC. Hence determine a triangle having given its three altitudes. W. - 90. From a given point without a given straight line, to draw a line making an angle with the given line equal to a given rectilineal angle. 91. Through a given point A, draw a straight line ABC meeting 304 GEOMETRICAL EXERCISES two given "parallel straight lines in B and C, such that BC may be equal to a given straight line. 92. If the line joining two parallel lines be bisected, all the lines drawn through the point of bisection and terminated by the parallel lines are also bisected in that point. 93. Three given straight lines issue from a point: draw another straight line cutting them, so that the two segments of it intercepted between them may be equal to one another. 94. AB, AC are two straight lines, B and C given points in the same; BD is drawn perpendicular to AC, and DE perpendicular to AB; in like manner CF is drawn perpendicular to AB, and FG to A.C. Shew that EG is parallel to BC. 95. ABC is a right-angled triangle, and the sides AC, AB are produced to D and F; bisect FBC and BCD by the lines B.E, CE, and from E let fall the perpendiculars EF, ED. Prove (without assuming any properties of parallels) that ADEF is a square. 96. AD, BC are two parallel straight lines, cut obliquely by AB and perpendicularly by AC; BED is drawn cutting A C in E, so that J.D is equal to twice B.A.; prove that the angle DBC is equal to one- third of the angle ABC. 97. Having given the angles and diagonals of a parallelogram; construct it. -- 98. The four straight lines which bisect the angles of a parallelo- gram either meet in a point or form a parallelogram. - -- 99. Find a point such that the perpendiculars let fall from it upon two given straight lines shall be respectively equal to two given straight lines. How many such points are there? * - 100. Draw a line parallel to one of the sides of a triangle, such that the portion of it intercepted between the other two sides, shall be equal to the difference between one of those sides, and the side parallel to the line. - ^ 101. If ABC be a triangle in which C is a right angle, draw a straight line parallel to a given straight line so as to be terminated by CA, CB and bisected by A.B. - 102. On the sides AB, BC, CD of a parallelogram are described the equilateral triangles A B.E, CDE without, and BCG within the figure; prove that EG is equal to one, and FG to the other diagonal. 103. Having given one of the diagonals of a parallelogram, the sum of the two adjacent sides and the angle between them, construct the parallelogram. - 104. One of the diagonals of a parallelogram being given, and the angle which it makes with one of the sides, complete the parallelo- gram, so that the other diagonal may be parallel to a given line. 105. ABCD, A'B'C'D' are two parallelograms whose corresponding sides are equal, but the angle A is greater than the angle A'; prove that the diameter AC is less than A'C', but BD greater than B'D. 106. If in the diagonal of a parallelogram any two points equi- distant from its extremities be joined with the opposite angles, a figure will be formed which is also a parallelogram. - 107. From each angle of a parallelogram a line is drawn making the same angle towards the same parts with an adjacent side, taken always in the same order; shew that these lines form another parallelo- gram equiangular to the original one. - 108. Along the sides of a parallelogram taken in order, measure 44' = BB = CC" = DD': the figure A'B'C'D' will be a parallelogram. 'ON BOOK 1. 305 109. Lét ABCD be a parallelogram, bisect AB, DC in E and F, join AF, B.F. CE, DE, cutting the two diagonals in P, Q, R, S. Shew that PQRS is a parallelogram whose area is one-ninth of the parallelogram ABCD. - - 110. From the angular points of a parallelogram ABCD perpen- diculars are drawn on the diagonals meeting them in E, F, G and H respectively; prove that EFG H is a parallelogram equiangular to ABCD. 111. On the sides AB, BC, CD, DA, of a parallelogram, set off AE, BF, CG, DH, equal to each other, and join AF, BG, CH, DE: these lines form a parallelogram, and the difference of the angles, , AFB, BGC, equals the difference of any two proximate angles of the two parallelograms. * 112. OB, OC are two straight lines at right angles to each other, through any point P any two straight lines are drawn intersecting OB, OC, in B, B, C, C", respectively. If D and D′ be the middle points of BB' and CC", shew that the angles BPD, DOD are equal. 113. If straight lines be drawn from the angles of any parallelo- gram, at right angles to any line, the stim of those from one pair of opposite angles is equal to the sum of those from the other pair of opposite angles. - - - 114. OPQ is any triangle, OR bisects PQ in R, PST bisects OR in S and meets 00 in T; then 00 = 3.0T. 115. AD, GH, are two parallel straight lines, K the middle point of GH; AG, DK meet in B; AK, DH in C; shew that BC is parallel to AID. - 116. Upon stretching two chains AC, JBD, across a field ABCD, it is found that BD and A C make equal angles with DC, and that A C makes the same angle with AD that BD does with BC; hence prove that AB is parallel to CD. W . - - - 117. To find a point in the side or side produced of any parallelo- gram, such that the angle it makes with the line joining the point and one extremity of the opposite side, may be bisected by the line joining it with the other extremity. - 118. When the corner of the leaf of a book is turned down a second time, so that the lines of folding are parallel and equidistant, the space in the second fold is equal to three times that in the first. 119. The line joining the points of bisection of any two sides of a triangle is parallel to the third side, and the triangle so formed is one- fourth of the given triangle. - 120. If in the triangle ABC, BC be bisected in D, AD joined and bisected in E, BE joined and bisected in F, and CFjoined and bisected in G; then the triangle EFG will be equal to one-eighth of the triangle ABC. 121. Shew that the areas of the two equilateral triangles in I’rob. 64, p. 302, are respectively, one-third and one-seventh of the area of the original triangle. 122. To describe a triangle equal to a given triangle, (1) when the base, (2) when the altitude of the required triangle is given. . 123. To describe a triangle equal to the sum or difference of two given triangles. . . . . X 306 - GEOMETRICAL ExERCISES 124. Upon a given base describe an isosceles triangle equal to a given triangle. - 125. Describe an equilateral triangle equal to a given triangle. 126. To a given straight line apply a triangle which shall be equal . to a given parallelogram, and have one of its angles equal to a given rectilineal angle. - 127. Transform a given rectilineal figure into a triangle whose vertex shall be in a given angle of the figure, and whose base shall be in one of the sides. ſ 128. Divide a triangle by two straight lines into three parts, which when properly arranged shall form a parallelogram whose angles are of a given magnitude. 129. Shew that a scalene triangle cannot be divided into two parts which will coincide. 130. Divide a triangle into three equal parts, (1) by lines drawn from a point in one of the sides: (2) by lines drawn from the angles to a point within the triangle: (3) by lines drawn from a given point within the triangle. In how many ways can the third case be done 2 131. Bisect a parallelogram, (1) by a line drawn from a point in one of its sides: (2) by a line drawn from a given point within or without it: (3) by a line perpendicular to one of the sides; (4) by a line drawn parallel to a given line. . 132. From a given point in one side produced of a parallelogram, draw a straight line which shall divide the parallelogram into two equal parts. d . 133. To trisect a parallelogram by lines drawn (1) from a given point in one of its sides, (2) from one of its angular points. VII. 134. To describe a rhombus which shall be equal to any given quadrilateral figure. - * 135. Describe a parallelogram which shall be equal in area and perimeter to a given triangle. 136. Find a point in the diagonal of a square produced, from which if a straight line be drawn parallel to any side of the square, and meeting another side produced, it will form together with the produced diagonal and produced side, a triangle equal to the square. 137. If from any point within a parallelogram, straight lines be drawn to the angles, the parallelogram shall be divided into four tri- angles, of which each two opposite, are together equal to one-half of the parallelogram. gi 138. If ABCD be a parallelogram, and E any point in the dia- gonal AC, or AC produced; shew that the triangles EBC, EDC, are equal, as also the triangles EBA and EBD. 139. ABCD is a parallelogram, draw DFG meeting BC in F, and AB produced in G; join AF, CG ; then will the triangles ABF, CFG be equal to one another. 140. ABCD is a parallelogram, E the point of intersection of its diagonals, and K any point in AD. If KB, KC be joined, shew that the figure BKEC is one-fourth of the parallelogram. 141. Let ABCD be a parallelogram, and 0 any point within it, through 0 draw lines parallel to the sides of ABCD, and join OA, OC; prove that the difference of the parallelograms D0, B0 is twice the triangle OA. C. y on Book 1. º 307 142. The diagonals AC, BD of a parallelogram intersect in 0, and P is a point within the triangle A OB; prove that the difference of the triangles APB, CPD is equal to the sum of the triangles APC, BPD. 143. If K be the common angular point of the parallelograms about the diameter AC (fig. Euc. I. 43.) and BD be the other diameter, the difference of these parallelograms is equal to twice the triangle BKD. - VIII. 144. If a quadrilateral figure is bisected by one diagonal, the 'second diagonal is bisected by the first. 145. If two opposite angles of a quadrilateral figure are equal, shew that the angles between opposite sides produced are equal. 146. Prove that the sides of any four-sided rectilineal figure are together greater than the two diagonals. 147. The sum of the diagonals of a trapezium is less than the sum of any four lines which can be drawn to the four angles, from any point within the figure, except their intersection. 148. The longest side of a given quadrilateral is opposite to the shortest; shew that the angles adjacent to the shortest side are together greater than the sum of the angles adjacent to the longest side. 149. From a given isosceles triangle, cut off a trapezium which ‘shall have the same base as the triangle, and shall have its remaining three sides equal to each other. - 150. Given any two points in the opposite sides of a trapezium, in- 'scribe in it a parallelogram having two of its angles at these points. 151. Shew that in every quadrilateral plane figure, two parallelo- grams can be described upon two opposite sides as diagonals, such that the other two diagonals shall be in the same straight line and equal. 152. Describe a quadrilateral figure whose sides shall be equal to four given straight lines. What limitation is necessary 7 153. If the sides of a quadrilateral figure be bisected and the points of bisection joined, the included figure is a parallelogram, and equal in area to half the original figure. 154. If two sides of a trapezium be parallel, the triangle contained by either of the other sides, and the two straight lines drawn from its extremities to the bisection of the opposite sides, is equal to half the trapezium. f 155. If a quadrilateral have two of its sides parallel, shew that the line drawn parallel to either of these sides through the intersec- tion of the diagonals is bisected in that point. 156. Shew that the diagonals of a quadrilateral, two of whose sides only are parallel, one of them being double of the other, cut one another in a point of trisection. - 157. If two opposite sides of a trapezium be parallel to one another, the straight line joining their bisections, bisects the trapezium. 158. If of the four triangles into which the diagonals divide a trapezium, any two opposite ones be equal, the trapezium has two of its opposite sides parallel. - - 159. If two sides of a quadrilateral be parallel but not equal, and the other two sides be equal but not parallel, the opposite angles of the quadrilateral are together equal to two right angles: and conversely. 160. If two sides of a quadrilateral be parallel, and the line join- ing the middle points of the diagonals be produced to meet the other X 2 308 GEOMETRICAL ExeRCISEs sides; the line so produced will be equal to half the sum of the parallel sides, and the line between the points of bisection equal to half their difference. - . 161. To bisect a trapezium, (1) by a line drawn from one of its angular points: (2) by a line drawn from a given point in one side. 162. To divide a parallelogram into four equal portions by lines drawn from any point in one of its sides. 163. It is impossible to divide a quadrilateral figure (except it be a parallelogram) into equal triangles, by lines drawn from a point within it to its four corners. - - 164. 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 that of the quadrilateral. 165. Through a given point P on one side BC of a parallelogram produced, draw a straight line cutting the sides CD, DA on E and F, and AB produced in H, so that the triangle EDF may be equal to the sum of the triangles PCE and AFH. IX. 166. If the greater of the acute angles of a right-angled triangle, be double the other, the square on the greater side is three times the square on the other. 167. Upon a given straight line construct a right-angled triangle such that the square on the other side may be equal to seven times the Square on the given line. tº E 168. If from the vertex of a plane triangle, a perpendicular fall upon the base or the base produced, the difference of the squares on the sides is equal to the difference of the squares on the segments of the base. - 169. If from the middle point of one of the sides of a right-angled triangle, a perpendicular be drawn to the hypotenuse, the difference of the squares on the segments into which it is divided, is equal to the Square on the other side. 170. If a straight line be drawn from one of the acute angles of a right-angled triangle, bisecting the opposite side, the square upon that line is less than the square upon the hypotenuse by three times the square upon half the line bisected. 171. If the sum of the squares on the three sides of a triangle be equal to eight times the square on the line drawn from the vertex to the point of bisection of the base, then the vertical angle is a right angle. 172. If a line be drawn parallel to the hypotenuse of a right- angled triangle, and each of the acute angles be joined with the points where this line intersects the sides respectively opposite to them, the squares on the joining lines are together equal to the squares on the hypotenuse and on the line drawn parallel to it. - 173. Let ACB, ADB be two right-angled triangles having a common hypotenuse AB, join CD, and on CD produced both ways draw perpendiculars AE, B.F. Shew that CE* + CF*= DE” + DF". 174. If perpendiculars AD, B.E, CF, drawn from the angles on the opposite sides of a triangle intersect in G, the difference of the squares on AC, AB, is equal to the difference of the squares on CG, BG. 175. If ABC be a triangle of which the angle A is a right angle; and B.E, CF be drawn bisecting the opposite sides respectively: shew ON BOOK. I. - 309 times the square on B.C. - 176. If ABC be an isosceles triangle, and CD be drawn perpendi- cular to AB; the sum of the squares on the three sides is equal to AD" + 2. BD” + 3. CD". 177. The sum of the squares described upon the sides of a rhombus is equal to the squares described on its diameters. - 178. Upon the sides AB, BC, CA of any triangle ABC, or upon these produced, let fall perpendiculars DE, DF, DG, from the point D within or without the triangle. - Then A. E* + BF's + G Cº - BE; 4 OF8 H A Gº". w * 179. ABC is a triangle, right-angled at C ; D, D, D, ... D, are points in AB, and E, E, E,. . E-1 are points in BC such that B, D, E, D, ... E. D.- are all parallel to CA, and CD, E, D, E, D, ... E., D. are all perpendicular to A.B. Prove that the square on AB is equal to the sum of the squares on AD, D, D, ...D.B, together with twice the sum of the squares on CD, E, D, E, D, ... E. D. 180. In the figure Euc. I. 47, (a) Shew that the diagonals FA, AK of the squares on AB, A C, lie in the same straight line. - - (b) If DF, EK be joined, the sum of the angles at the bases of the triangles BFD, CER is equal to one right angle. (c) If BG and CH be joined, those lines will be parallel. (d) If perpendiculars be let fall from F and K on BC produced, the parts produced will be equal; and the perpendiculars together will be equal to BC. (e) Join GH, KE, FD, and prove that each of the triangles so formed, equals the given triangle. ABC. - (f) The sum of the squares on GH, KE, and FD will be equal to six times the square on the hypotenuse. * (g) The difference of the squares on AB, AC, is equal to th difference of the squares on AD, AE. 181. The area of any two parallelograms described on the two sides of a triangle, is equal to that of a parallelogram on the base, whose side is equal and parallel to the line drawn from the vertex of the triangle, to the intersection of the two sides of the former paral- lelograms produced to meet. - 182. If one angle of a triangle be a right angle, and another equal to two-thirds of a right angle, prove that the equilateral triangle described on the hypotenuse, is equal to the sum of the equilateral triangles described upon the sides which contain the right angle. 2 - s 83. On the sides AC, B C of a triangle right-angled at C, equila- teral triangles A CD, BCE are described externally, and AE, BD are joined; prove that the two triangles A.B.E, ABD are together equal to the whole figure ABECD. & 184. Equiangular triangles ABD, CBE are described on the sides . AB, CB of a given triangle, so that the angles A.BD, B.A.D are respec- tively equal to CBE, BCE; DF’ and EG are drawn parallel to B.E, BD respectively; and meet the sides AB, BC in F and G; shew that FG is parallel to A.C. that four times the sum of the squares on BE and CF is equal to five GEOMETRICAL ExERCISES ON BOOK II. PROPOSITION I. PROBLEM. Divide a given straight line into two parts, such that their rectangle may be equal to a given squaré. Let AB be the given straight line, and let M be the side of the given square. . It is required to divide the line AB into two parts, so that the rectangle contained by them may be equal to the square on M. --- - I) E. - A. F. C. B. Bisect AB in C; with center C, and radius C.A. or CB, describe the semicircle ADB. - - - At the point B draw BE at right angles to AB and equal to M. Through E, draw ED parallel to AB and cutting the semicircle in D; - and draw DF parallel to EB meeting AB in F. - Then AB is divided in F as was required. - : For the rectangle A.F, FB is equal to the square on FD; (II. 14.) t but FD is equal to EB, (r. 34.) . and EB was drawn at right angles to AB and equal to M. Therefore the rectangle A.F, FB is equal to the square on M. PROPOSITION II. THEOREM. The square on the eaccess of one straight line above another is less than, the squares on the two lines by twice their rectangle. - t Let AB, BC be the two straight lines, whose difference is AC. Then the square on A C is less than the squares on AB and BC by twice the rectangle contained by AB and BC. - A C B H G}~k D , E. E. Constructing as in Prop. 4, Book II. Because the complement A6 is equal to GE, - add to each CK, therefore the whole AK is equal to the whole CE; and AK, CE together are double of AK; but A K, CE are the gnomon AKF and CK, and AK is the rectangle contained by AB, BC; therefore the gnomon AKF and CK are equal to twice the rectangle AB, BC, BOOK II. PROP. III, IV. - 311 but AE, CK are equal to the squares on AB, BC; - taking the former equals from these equals, therefore the difference of AE and the gnomon AKF is equal to the difference between the squares on AB, BC, and twice the rectangle AB, BC; but the difference AE and the gnomon AKF is the figure HF which is equal to the square on A. C. Wherefore the square on AC is equal to the difference between the squares on AB, BC, and twice the rectangle AB, BC. PROPOSITION III. THEOREM. In any triangle the squares on the two sides are together double of the squares on half the base and on the straight line joining its bisection with the opposite angle. * Let ABC be a triangle, and AD the line drawn f.om the vertex A to the bisection D of the base B.C. A. B D E C From A draw AE perpendicular to BC. - Then, in the obtuse-angled triangle ABD, (II. 12.) the square on AB exceeds the squares on AD, DB, by twice the rectangle BD, DE: - and in the acute-angled triangle ADC, (II. 13.) the square on AC is less than the squares on AD, DC, by twice the rectangle CD, DE: wherefore, since the rectangle BD, DE is equal to the rectangle CD, JDE; it follows that the squares on AB, A C are double of the squares on AID, D.B. - PROPOSITION IV. THEOREM. If straight lines be drawn from each angle of a triangle bisecting the opposite sides, four times the sum of the squares on these lines is equal to three times the sum of the squares on the sides of the triangle. Let ABC be any triangle, and let AD, B.E, CF be drawn from A, B, C, to D, E, F, the bisections of the opposite sides of the triangle: draw A G perpendicular to BC. A B D - G C Then the square on AB is equal to the squares on BD, DA together with twice the rectangle BD, D6, (II. 12.) 312. GEOMETRICAL ExERCISES - and the square on A.C. is equal to, the squares on CD, DA dimi- nished by twice the rectangle CD, DG ; (II. 13.) therefore the squares on AB, 4.C. are equal to twice the square on BD, and twice the square on AD; for DC is equal to BD: . and twice the squares on AB, A C are equal to the square on B6, and four times the square on AD: for BC is twice B.D. Similarly, twice the squares on AB, BC are equal to the square on A C, and four times the square on B.E.: also twice the squares on BC, CA are equal to the square on AB, and four times the square on FC: hence, by adding these equals, - four times the squares on AB, AC, B C are equal to four times the squares on AD, B.E, CF together with the squares on AB, AC, BC: and taking the squares on AB, A C, BC from these equals, therefore three times the squares on AB, AC, BG are equal to four times the squares on AI), B.E, CF. * PROPOSITION V. THEOREM. The sum of the perpendiculars let fall from any point within an equila- teral triangle upon the sides, will be equal to the perpendicular let fall from one of its angles upon the opposite side. Is this proposition true when the point is in one of the sides of the triangle 3. In what manner must the proposition be enunciated when the point is without the triangle 2 Let ABC be an equilateral triangle, and P any point within it: and from Plet fall PD, PE, PF perpendiculars on the sides AB, BC, CA respectively, also from A let fall AG perpendicular on the base B.C. Then AG is equal to the sum of PD, PE, PF. l- P E G C From P draw PA, PB, PC to the angles A, B, C. . Then the triangle ABC is equal to the three triangles PAB, PBC, PCA. \ But since every rectangle is double of a triangle of the same base and altitude, (I. 41.) therefore the rectangle 46, BC, is equal to the three rectangles AB, PD; A.C., PF and BC, PE. Whence the line AG is equal to the sum of the lines PD, PE, PF. If the point Pfall on one side of the triangle, or coincide with E: then the triangle ABC is equal to the two triangles APC, BPA : whence AG is equal to the sum of the two perpendiculars PD, PF. If the point Pfall without the base BC of the triangle: then the triangle AB6 is equal to the difference between the sum of the two triangles APC, BPA, and the triangle PCB. Whence AG is equal to the difference between the sum of PD, PF, and PE. - A ON BOOK Fr. 313. I. 6. If the straight line AB be divided into two unequal parts in D; and into two unequal parts in E, the rectangle contained by AE, EB, will be greater or less than the rectangle contained by AD, DB; º's as E is nearer to, or further from, the middle point of AB; than D. ... • 7. Produce a given straight line in such a manner that the square. on the whole line thus produced, shall be equal to twice the square on the given line. - 8. If AB be the line so divided in the points Cand D, (fig. Euc. II. 5) shew that A B*= 4. CD +4. A:D. D.B. - 9. If a straight line be bisected and produced to any point, the Square on the whole line when thus produced, is equal to the square on the part produced and four times the rectangle contained by the half line, and the line made up of the half and part produced. 10. Produce a given line so that the rectangle contained by the whole line thus produced, and the part produced may be equal to the Square on half the line. 11. If AM, MG, GA be joined, fig. Euc. II. 5, twice the triangle AMG will be equal to A C*4 CD'; or A.C.A.D – CD. B.D. 12. Divide a line into two parts, such that the sum of their Squares shall be double the square on another line. 13. Shew that the difference between the squares on the two un- equal parts (fig. Euc. II. 9) is equal to twice the rectangle contained by the whole line, and the part between the points of section, 14. Shew how in all the possible cases, a straight line may be geometrically divided into two such parts, that the sum of their squares shall be equal to a given Square. * ... " 15. Divide a given straight line into two parts, such that the squares on the whole line and on one of the parts, shall be equal to twice the Square on the other part. 6. Any rectangle is the half of the rectangle contained by the diameters of the squares on its two sides. 17. If a straight line be divided into two equal and into two un- equal 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. 18. If the points C, D be equidistant from the extremities of the straight line AB, shew that the squares constructed on AD and A C, exceed twice the rectangle AC, AD by the square constructed on CD. 19. If any point be taken in the plane of a parallelogram from which perpendiculars are let fall on the diagonal, and on the sides which include it, the rectangle of the diagonal and the perpendicular on it, is equal to the sum or difference of the rectangles of the sides and the perpendiculars on them. 20. ABCD is a rectangular parallelogram, of which A, C are op- posite angles, E any point in BC, F any point in CD. Prove that twice the area of the triangle AEF together with the rectangle B.E, DF is equal to the parallelogram A.C. - II. 21. Shew how to produce a given line, so that the rectangle con- tained by the whole line thus produced, and the produced part, shall be ‘equal to the square, (1) on the given line, (2) on the part produced. 314 GEOMETRICAL, EXERCISES: 22, If in the figure Euc. II. 11, we join BF and CH, and produce. CH to meet BF in º; is perpendicular to B.F. 23. If a line be divided, as in Euc. II. 11, the squares on the whole line and on one of the parts are together three times the square on the other part. - - * 24. If in the fig. Euc. II. 11, the points F. D be joined cutting. AHB, GHK in f, d respectively; then shall Ff= Dd. 25. Shew that if a line be divided as in Euc. II. 11, the rectangle contained by the sum and difference of the parts; is equal to the rect- angle contained by the parts. - III. 26. If from the three angles of a triangle, lines be drawn to the points of bisection of the opposite sides, the squares on the distances between the angles and the common intersection, are together one-third of the squares on the sides of the triangle. 27. ABC is a triangle of which the angle at C is obtuse, and the angle at B is half a right angle: D is the middle point of AB, and CE" is drawn perpendicular to A.B. Shew that the square on AC is double. of the squares on AD and D.E. - 28. If an angle of a triangle be two-thirds of two right angles, shew that the square on the side subtending that angle, is equal to the Squares on the sides containing it, together with the rectangle con- tained by those sides. - 29. The square described on a straight line drawn from one of the angles at the base of a triangle, to the middle point of the opposite. side, is equal to the sum or difference of the square on half the side. bisected, and the rectangle contained between the base and that part of it, or of it produced, which is intercepted between the same angle and a perpendicular drawn from the vertex. - ſº 30. If the straight lines AD, B.E, CF, drawn from the angles of a triangle to D, E, F, the points of bisection of the opposite sides, inter- sect in G: the squares on the sides AB, BC, and CA, are together equal to three times the squares on the lines AG, BG, and C.G. 31. Produce one side of a scalene triangle, so that the rectangle. under it and the produced part may be equal to the difference of the squares on the other two sides. - 32. Given the base of any triangle, the area, and the line bisect- ing the base, construct the triangle. - 33. If ABC be a triangle in which C is a right angle, and DE be drawn from a point D in AC at right angles to AB; prove that the rectangles AB, AE and AC, AD, will be equal. - 34. In any acute-angled triangle ABC, AD, B.E, CF are drawn respectively perpendicular to the opposite sides; prove that twice the rectangles contained by BD and CD, by CE and A.F., by A.F and BF are together less than the sum of the squares on the sides of the triangle. IV. - 35. Shew that the square on the hypotenuse of a right-angled triangle, is equal to four times the area of the triangle together with the square on the difference of the sides. - 36. In the triangle ABC, if AD be the perpendicular let fall upon the side BC; then the square on AC together with the rectangle con- º: ON BOOK II, 315 tained by BC, BD is equal to the square on AB together with the rectangle CB, CD. - - 37. ABC is a triangle, right-angled at C, and CD is the perpen- dicular let fall from 0 upon AB; if HK be equal to the sum of the sides AC, CB, and LM to the sum of AB, CD; shew that the square on HIſ together with the square on CD is equal to the square on L.M. 38. ABC is a triangle having the angle at B a right angle : it is required to find in AB a point P such that the square on A C may ex- ceed the squares on AP, and PC, by half the square on A.B. - 39. In a right-angled triangle, the square on that side which is the greater of the two sides containing the right angle, is equal to the rectangle by the sum and difference of the other sides. 40. The hypotenuse AB of a right-angled triangle ABC is tri- sected in the points D, E, prove that if CD, CE be joined, the sum of the squares on the sides of the triangle CDE is equal to two-thirds of the square on A.B. ºt 41. From the hypotenuse of a right-angled triangle, portions are cut off equal to the adjacent sides: shew that the square on the middle segment is equivalent to twice the rectangle under the extreme seg- ments. 42. If the hypotenuse AC of a right-angled triangle ABC be pro- duced to D, so that CD = AC, and BD be joined, BD = A B*4 4.BC*. W. 43. Prove that the square on any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square on a side of the triangle, by the rectangle contained by the segments of the base: and conversely. e 44. If from one of the equal angles of an isosceles triangle a per- pendicular be drawn to the opposite side, the rectangle contained by that side and the segment of it intercepted between the perpendicular and base, is equal to the half of the square described upon the base. 45. If in an isosceles triangle a perpendicular be let fall from one. of the equal angles to the opposite side, the square on the perpen- dicular is equal to the square on the line intercepted between the other, equal angle and the perpendicular, together with twice the rectangle. contained by the segments of that side. 46. The square on the base of an isosceles triangle whose vertical angle is a right angle, is equal to four times the area of the triangle. 47. Describe an isosceles obtuse-angled triangle, such that the, square on the side subtending the obtuse angle may be three times the Square on either of the sides containing the obtuse angle. 48. If AB, one of the sides of an isosceles triangle ABC be pro- duced to D, so that BD = AB, shew that CD = A B*4 2. B Cº. 49. If ABC be an isosceles triangle, and DE be drawn parallel to: the base BC, and EB be joined; prove that B.E." – BC & DEF CE*. 50. If ABC be an isosceles triangle of which the angles at B and C are each double of A ; then the square on AC is equal to the square, on BC, together with the rectangle contained by A C and BC. 51. ABC is an isosceles triangle, A B being equal to AC, and JBCD a triangle having a right angle at C. If AD be joined, and JE, F be the points where the perpendiculars from B and C on AD, or, AD produced meet it, then BC*= 2 A.D.E.F. 316 GEOMETRICAL ExERCISEs WI. 52. Shew that in a parallelogram the squares on the diagonals are equal to the sum of the squares on all the sides. A 53. If ABCD be any rectangle, A and C being opposite angles, and 0 any point either within or without the rectangle : OA*4 OC*= OB" + OD". 54. In any quadrilateral figure, the sum of the squares on the diagonals together with four times the square on the line joining their middle points, is equal to the sum of the squares on all the sides. 55. In any trapezium, if the opposite sides be bisected, the sum of the squares on the other two sides, together with the squares on the diagonals, is equal to the sum of the squares on the bisected sides, together with four times the square on the line joining the points of bisection. - 56. The squares on the diagonals of a trapezium are together double the squares on, the two lines joining the bisections of the oppo- site sides. 57. If the diagonals of a quadrilateral intersect each other at right angles, the sum of the squares on the sides of the figure, is equal to twice the sum of the Squares on the four segments of the diagonals. 58. In any trapezium two of whose sides are parallel, the squares on the diagonals are together equal to the squares on its two sides which are not parallel, and twice the rectangle contained by the sides which are parallel. 59. If the two sides of a trapezium be parallel, shew that its area is equal to that offirectangle contained by its altitude and half the sum of the parallel sides. - - 60. If a trapezium have two sides parallel, and the other two equal, shew that the rectangle contained by the two parallel sides, together with the square on one of the other sides, will be equal to the square on the straight line-joining two opposite angles of the trapezium. 61. If squares be described on the sides of any triangle and the angular points of the squares be joined; the sum of the squares on the sides of the hexagonal figure thus formed is equal to four times the sum of the squares on the sides of the triangle. - VII. 62. Find the side of a square equal to a given equilateral triangle. 63. Find a square which shall be equal to the sum of two given rectilineal figures. - -4 64. To divide a given straight line so that the rectangle under its segments may be equal to a given rectangle. 65. Shew how to construct a rectangle which shall be equal to a given square; (1) when the sum, and (2) when the difference of two adjacent sides is given. - . 66. Shew how to describe a rectangle equal to a given square, and having one of its sides equal to a given straight line. * GEOMETRICAL EXERCISES ON BOOK III. PROPOSITION I. THEOREM. If AB, CD be chords of a circle at right angles to each other, prove that the sum of the aros AC, BD is equal to the sum of the arcs AD, BC. Draw the diameter FGH parallel to AB, and cutting CD in H. g - D C gº Then the arcs FDG and FCG are each half the circumference. Also since CD is bisected in the point H, the arc FD is equal to the arc FC, ſº c and the arc FD is equal to the arcs, FA, AD, of which, AF is equal to BG, . therefore the arcs AD, BG are equal to the arc FC; add to each CG, * therefore the arcs AD, BC are equal to the arcs FC, GG, which make up the half circumference. Hence also the arcs AC, DB are equal to half the circumference. Wherefore the arcs AD, BC are equal to the arcs AC, D.B. PROPOSITION II. PROBLEM. The diameter of a circle having been produced to a given point, it is required to find in the part produced a point, from which if a tangent be drawn to the circle, it shall be equal to the segment of the part produced, that ts, between the given point and the point found. - Analysis. Let AEB be a circle whose center is 0 and whose dia- meter AB is produced to the given point D. - Suppose that G is the point required, such that the segment GD is equal to the tangent GE drawn from G to touch the circle in E. • «Pºs- \"y" G Join DE and produce it to meet the circumference again in F; join also CE and CF. . Then in the triangle GDE, because GD is equal to GE, therefore the angle GED is equal to the angle GDE; and because CE is equal to CF, , , the angle CEF is equal to the angle CFE; 318 - GEOMETRICAL EXERCISES therefore the angles, CEF, GED are equal to the angles CFE, G DE: . but since GE is a tangent at E, therefore the angle CEG is a right angle, (III. 18.) hence the angles CEF, GED are equal to a right angle, and consequently, the angles CFE, EDG are also equal to a right angle, wherefore the remaining angle FCD of the triangle CFD is a right angle, - , ºr- and therefore CF is perpendicular to A.D. Synthesis. From the center C, draw CF perpendicular to AD meeting the circumference of the circle in F: - join DF'cutting the circumference in E, join also CE, and at E. draw EG perpendicular to CE and inter- secting BD in G. *. Then G will be the point required. - For in the triangle CFD, since FCD is a right angle, the angles CFD, CDF are together equal to a right angle; also since CEG is a right angle, r therefore the angles CEF, GED are together equal to a right angle; ,- therefore the angles CEF, GED are equal to the angles CFD, iCDF; sº but because CE is equal to CF, the angle CEF is equal to the angle CFD; wherefore the remaining angle GED is equal to the remaining angle CDF, and the side GD is equal to the side GE of the triangle EGD, therefore the point G is determined according to the required conditions. PROPOSITION III. THEOREM. AB, AC and ED are tangents to the circle CFB; at whatever point between C and B the tangent EFD is drawn, the three sides of the triangle AED are equal to twice AB or twice AC: also the angle subtended by the tangent EFD at the center of the circle, is a constant quantity. Take & the center of the circle, and join GB, GE, GF, GD, 60. Then EB is equal to EF, and DC to DF; (III. 37.) ºtherefore ED is equal to EB and DC; to each of these add AF, AID, wherefore AD, AE, ED are equal to AB, AC; ." and AB is equal to AC; therefore AD, A E, ED are equal to twice AB, or twice AC; or the perimeter of the triangle AED is a constant quantity. Again, the angle EGF is half of the angle BGF, ON BOOK III. 319 and the angle DGF is half of the angle CGF; . therefore the angle DGE is half of the angle CGB, or the angle subtended by the tangent ED at G, is half of the angle contained between the two radii which meet the circle at the points where the two tangents AB, AC meet the circle. PROPOSITION IV. PROBLEM. -- Given the base, the vertical angle, and the perpendicular in a plane tri- . angle; to construct it. - |Upon the given base AB describe a segment of a circle containing 'an angle equal to the given angle. (III. 33.) E P c A - B At the point B draw BC perpendicular to AB, and equal to the altitude of the triangle. (I. 11, 3.) Through C, draw CDE parallel to AB, and meeting the circum- ference in D and E. (I. 31.) sº Join DA, DB ; also EA, EB ; - then EAB or DAB is the triangle required. It is also manifest, that if CDE touch the circle, there will be only one triangle which can be constructed on the base AB with the given altitude. PROPOSITION V. THEOREM. If a chord of a circle be produced till the part produced be equal to the zadius, and if from its extremity a line be drawn through the center and meeting the convez and concave circumferences, the convex is one-third of the concave circumference. Let AB any chord be produced to C, so that BC is equal to the radius of the circle : . A 2TSB /T. D \ IE G →0 F and let CE be drawn from C through the center D, and meeting the convex circumference in F, and the concave in E. Then the are BF is one-third of the arc A.E. Draw EG parallel to AB, and join DB, DG. Since the angle DEG is equal to the angle DGE: (T. 5.) and the angle GDF is equal to the angles DEG, DGE"; (I. 32.) therefore the angle GDC is double of the angle DEG. But the angle BDC is equal to the angle BCD, (I. 5.) • .- . .” and the angle CEG is equal to the alternate angle ACE; (I. 29.) therefore the angle GDC is double of the angle CDB, 820 GEOMETRICAL ExERCISEs add to these equals the angle CDB, therefore the whole angle GDB is treble of the angle CDB, but the angles GDB, CDB at the center D, are subtended by the arcs B.F, BG, of which BG is equal to A.E. Wherefore the circumference AE is treble of the circumference BF, and BF is one-third of AE. 4. Hence may be solved the following problem: . AE, BF are two arcs of a circle intercepted between a chord and a given diameter. Determine the position of the chord, so that one arc shall be triple of the other. PROPOSITION VI. THEOREM. If two chords of a circle intersect each other at right angles either within or without the circle, the sum of the squares described upon the four segments, is equal to the square described upon the diameter. tº Let the chords AB, CD intersect at right angles in E. T)raw the diameter AF, and join A C, AD, CF, D13. Then the angle ACF in a semicircle is a right angle, (III. 31.) and equal to the angle AED: also the angle ADC is equal to the angle AFC. (III. 21.) Hence in the triangles ADE, AFC, there are two angles in the one respectively equal to two angles in the other, consequently, the third angle CAF is equal to the third angle LAB; - - - therefore the arc DB is equal to the aro CF, (III. 26.) and therefore also the chord DB is equal to the chord C.F. (III. 29.) - - Decause AEC is a right-angled triangle, -- the squares on AE, EC are equal to the square on AC; (1.47.) similarly, the squares on DE, EB are equal to the square on DB ; therefore the squares on A.E, EC, DE, EB, are equal to the squares on A C, DB ; but DB was proved equal to FC, - and the squares on AC, FC are equal to the square on A.F. wherefore the squares on AE, EC, DE, EB, are equal to the square on AF, the diameter of the circle. . * - When the chords meet without the circle, the property is proved in a similar manner. - - º ** - ON BOOK III. 321 PROPOSITION VII. THEOREM. Two chords of a circle, produced if necessary, meet in a point; prove that the angle between them is equal to half the angle subtended at the center, by the sum or difference of the intermediate ares, according as the point of concourse lies within or without the circle. , First. Let AB, CD be two chords intersecting each other in the point E within the circle ADBC. - Then the angle AEC between the chords, is equal to half the angle at the center of the circle subtended by an arc equal to the sum of the arcs AC, B.D. Q / At the point B draw BF parallel to DC. Because CD, FB are parallel chords in the circle ADBC, therefore the arc BD is equal to the arc FC, - add to these equals the arc AC, - then the whole arc AF is equal to the sum of the arcs AC, BID: and the angle ABF at the circumference, is subtended by the arc AF'; - but the angle at the circumference of a circle is half of the angle at the center subtended by the same arc; (Euc. III. 20.) and the angle AEC is equal to the angle ABF (Euc. I. 29.) therefore the angle AEC between the chords, is equal to half the angle at the center subtended by the arc AF or the sum of the arcs A C, B.D. - Secondly. Let the chords AB, CD intersect each other at the point E without the circle ADBC. - Then the angle AEC is equal to half the angle at the center of the circle, subtended by an arc equal to the difference of the arcs BD, A.C. - Draw CF parallel to A.B. - Then the arc BF is equal to the arc AC, f and the arc DF is the difference of the arcs BD, AC: but the angle FCD at the circumference is subtended by the arc FD, and the angle FCD is equal to the angle AEC. (Euc. I. 29.) Wherefore the angle AEC is equal to half the angle at the center Subtended by an arc equal to the difference of the arcs BD, A. C. If the two chords become tangents to the circle at the points A, B : the angle E contained by the two tangents AE, BE, is equal to half the angle at the center of the circle subtended by FB, the difference of the concave and convex parts of the circumferences AFB, AB. Y 322 GEOMETRICAL EXERCISES - I. 8. To draw that diameter of a given circle which shall pass at a given distance from a given point. - - 9. Draw a chord in a circle, so that it may be double of its per- pendicular distance from the center. 10. AB, CD are two parallel chords of a circle, AB being twice as far from the center 0 as CD. If A 0 be joined, shew that it is bisected by CD. cr - 11. The straight line joining the middle points of two parallel chords in a circle will pass through the center. 12. Two chords of a circle being given in position and magnitude, describe the circle. 13. The two straight lines which join the opposite extremities of two parallel chords, intersect in a point in that diameter which is perpendicular to the chords. 14. The arcs intercepted between any two parallel chords in a circle are equal: and conversely. 15. A, B, C, A, B, C, are points on the circumference of a circle; if the lines AB, AC be respectively parallel to A'B', 4'C', shew that BC is parallel to B. C. f II. 16. If through C the center of a circle two straight lines AB, CD be drawn at any angle; and if through o any other point within the circle there be drawn two other straight lines ab, cd parallel to AB, CD respectively; then the sum of the arcs AD, BC is equal to the sum of the arcs ad, be. 17. Through a given point within or without a circle, it is required to draw a straight line cutting off a segment containing a given angle. 18. In a circle with center 0, any two chords A.B, CD are drawn cutting in E, and 0A, OB, OC, OD are joined; prove that the angles A 00 + BOD = 2.A.EC, and AOD + BOC = 2. A ED. - 19. If two chords in a circle make equal angles with a diameter, or make equal angles at a point in the diameter produced; these lines are equal, and are at equal distances from the center of the circle. 20. Draw a straight line which shall be parallel to a given straight line, and cut off, from a given circle, a segment which shall contain an angle equal to a given rectilineal angle. 21. If on two lines containing an angle, segments of circles be described containing angles equal to it, the lines produced will touch the segments. 22. If an arc of a circle be divided into three equal parts by three straight lines drawn from one extremity of the arc, the angle contained by two of the straight lines is bisected by the third. - 23. If the chord of a given circular segment be produced to a fixed point, describe upon it when so produced a segment of a circle which shall be similar to the given segment, and shew that the two Segments have a common tangent. g 24. A, B, are two given points; find a point P in the circum- ference of a given circle, so that APB may equal a given angle, exhi- ºgº number of possible solutions according to the positions Of 4, 5. - - ON BOOK III. 323 25. Through a given point within a circle, draw a straight line so that the parts of it between the given point and the circumference shall have a given difference. What is the limiting magnitude of this given difference for which the problem is possible? 26. AB, AC are any two chords in a circle; D, E the middle points of the arcs AB, AC. If DE cut AB and AC in the points F and G respectively, shew that AF is equal to AG. - 27. If two equal chords be drawn in a circle, and another chord be drawn through their middle points, the portions of this last chord intercepted between the middle points and the circumferences are equal. Q 28. From a given point without a circle a straight line is drawn cutting a circle. Draw from the same point another line so as to intercept two arcs which together shall subtend an angle equal to a given angle. 29. AB is the diameter of a circle whose center is 0. Through O two radii OP, OQ are drawn at right angles to each other. A Q, BP cut in R, and AP, B9 produced meet in V. Shew that the angle PVQ is half a right angle, and if R, V be joined, the angle POA will be twice the angle PVR. 30. If A, B, C be three points takenin order, in the circumference of a semicircle whose center is E.; then the angle EBC exceeds the angle EAC by the angle ACB. - 31. Let ABC be a semicircle whose diameter is AC, and center 0; from AB, which is greater than the semi-diameter, PB is cut off equal to OB, and through A, AD is drawn meeting OP produced in D, and making the angle ADO equal to the angle AOD; shew that the angle OAD is equal to three times the angle O.A.B. - 32. Given two points in the circumference of a circle on the same side of a given diameter, find a point in the circumference on the other side of the diameter, such that when joined with the given points, the joining lines may intersect the diameter in two points equally distant from the center. 33. Two chords PQ, P'Q of a circle intersect a given diameter at equal distances on opposite sides of the center; shew that PP", QQ also intersect the same diameter at equal distances from the center. 34. If from any point in a circular arc, perpendiculars be let fall on its bounding radii, the distance of their feet is invariable. 35. If AB, AC be any two chords passing through a point A in the circumference of a circle, and if from P, the middle point of the arc BC, the lines Pm, Pn be drawn perpendicular to AB, 40 res- pectively: shew that Bm will equal Ch. III. 36. If both tangents be drawn, (fig. Euc. III. 17.) and the points of contact joined by a straight line which cuts EA in H, and on HA as diameter a circle be described, the lines drawn through E to touch this circle will meet it on the circumference of the given circle. 37. Draw, (1) perpendicular, (2) parallel to a given line, a line touching a given circle. . 38. If two straight lines intersect, the centers of all circles that . be inscribed between them, lie in two lines at right angles to each other. Y 2 324. GEOMETRICAL ExFRCISEs 39. Draw two tangents to a given circle, which shall contain an angle equal to a given rectilineal angle. & 40. Determine the distance of a point from the center of a given circle, so that if tangents be drawn from it to the circle, the concave part of the circumference may be double of the convex. - 41. In a chord of a circle produced, it is required to find a point, from which if a straight line be drawn touching the circle, the line so drawn shall be equal to a given straight line. , - 42. Find a point without a given circle, such that the sum of the two lines drawn from it touching the circle, shall be equal to the line drawn from it through the center to meet the circle. - * 43. If from a point without a circle two tangents be drawn; the straight liae which joins the points of contact will be bisected at right angles by a line drawn from the center to the point without the circle. 44. If tangents be drawn at the extremities of any two diameters of a circle, and produced to intersect one another; the straight lines joining the opposite points of intersection will both pass through the center. - . 45. If from any point without a circle two lines be drawn touching the circle, and from the extremities of any diameter, lines be drawn to the points of contact-cutting each other within the circle, the line drawn from the point without the circle to the point of intersection, shall be perpendicular to the diameter. - . 46. If any chord of a circle be produced equally both ways, and tangents to the circle be drawn on opposite sides of it from its extre- mities, the line joining the points of contact bisects the given chord. - 47. AB is a chord, and AD is a tangent to a circle at A. D.PQ any secant parallel to AB meeting the circle in P and Q. Shew that the triangle PAD is equiangular with the triangle QAB. 48. If from any point in the circumference of a circle a chord and tangent be drawn, the perpendiculars dropped upon them from the middle point of the subtended arc, are equal to one another. ſº 49. AB, A'B' are two chords of a circle intersecting in O. Shew that the line joining the intersection of the tangents at A, A’ with the intersection of the tangents at B, B' passes through O. 50. Two diameters AOA', BOB' of a circle are at right angles to each other; P is a point in the circumference, the tangent at P meets BOB' produced in Q, and AP, A'P meet the same line in C, C, res- pectively. Prove that CQ is equal to O'Q. - 51. If a tangent be drawn at any point of a circle, and from the point of contact a perpendicular be let fall on a diameter of the circle, and from the extremities of the diameter perpendiculars be let fall on the tangent, these perpendiculars will be respectively equal to the segments of the diameter, • - IV. . 52. AB, AC are equal arcs of a circle, prove that the chord BC is parallel to the tangent at A. - # 53. ABD is a diameter of a circle, A C a chord, CD a tangent at C, and equal to the chord: the triangles A CD, CBD are equiangular. 54. If two straight lines be drawn touching a circle and meeting * two other straight lines also touching it at points on the same side of the line joining the points of contact of the first two; then the angles, ON BOOK III. - 325 subtended at the center by the two points of intersection on one of the first mentioned lines, will be equal to the angle so subtended by the like points on the other line. - * 55. If from a point two, tangents be drawn to a circle, and also two secants through the CXtremities of any chord which cuts the chord passing through the points of contact; then the intersection of the two chords and the other two points in which the secants meet the circumference are on one straight line. 56. In a given straight line to find a point at which two other straight lines being drawn to two given points, shall contain a right angle. Shew that if the distance between the two given points be greater than the sum of their distances from the given line, there will be two such points; if equal, there may be only one; if less, the pro- blem may be impossible. - - - 57. DF is a straight line touching a circle, and terminated by AD, BF, the tangents at the extremities of the diameter A.B., shew that the angle which DF subtends at the center is a right angle. 58. If tangents An, Bm be drawn at the extremities of the dia- meter of a semicircle, and any line in m Pn crossing them and touching the circle in P; and if Am, Bn be joined intersecting in 0 and cutting the semicircle in E and F; shew that O, P, and the point of intersec- tion of the tangents at E and F, are in the same straight line. 59. If from a point P without a circle, any straight line be drawn cutting the circumference in A and B, shew that the straight lines joining the points A and 8 with the bisection of the chord of contact of the tangents from P, make equal angles with that chord. 60. In the circumference of a given circle, to determine a point to which two straight lines drawn to two given points shall contain an angle equal to a given angle, pointing out the limitations within which the problem is possible. - 61. AB is a diameter of a circle, AC, AD are two chords meeting the tangent at B in E and Frespectively. Shew that the angles FCE and EDF are equal. 62. The angle contained by tangents drawn at the extremities of any chord of a circle is equal to the difference of the angles in the segments which are made by the chord. - - 63. From a given point T in the tangent to a circle at A, draw , a line cutting the circle in B and 6, so that the angle ABC is twice the angle ACB. Shew that the diameter of the circlé must be greater than AT: * . . . 64. If AB touch a circle in the point B, and A601) be drawn through 0 the center cutting the circle in C, D; and BE be drawn perpendicular to CIA; the angles ABC, CBE, OBD are all equal. 65. A.C. is a chord of a given circle, B, B, two given points in the chord, both within or both without the circle; if a circle be described, to pass through B and D, and touch the given circle, shew that AB and CD subtend equal angles at the point of contact. 66. From the extremity of the vertical radius of a circle draw a straightline through a given point A, in the horizontal diameter cutting the circle again in B, and let the tangent at B meet the diameter pro- duced in C, prove that the triangle ABC is isosceles. 67. If AB be the diameter of a circle, and AD be drawn perpen- dicular to the tangent at C, then will AC bisect the angle DAB. 68. If the diameter AB of a circle be produced one-third of its y 326 GEOMETRICAL EXERCISES length to C, and CD be drawn touching the circle, shew that AC is double of CD. - - 69. Two tangents are drawn to a given circle from a given point; it is required to draw a third tangent to the circle so that the part of it intercepted between the former two may be of a given length. 70. Let AB be the diameter of a circle, CD a chord parallel to AB, and equal to one-half of it; join A C and let the tangent at B meet AC produced in E, shew that AE=2. AB. If ED produced meet AB in F, shew that AB = 3. B.F. - º 71. Shew that, if from any point without a circle lines be drawn touching it, the angle contained by the tangents is double the angle contained by the line joining the points of contact and the diameter through one of them. - 72. A line touches a circle, and from the point of contact A, any chord AB is drawn; BC is another chord parallel to the touching line, and BD a chord parallel to A.C. Shew that the lines AB, AC, and CD are equal. g 73. If the chord of a given circular segment be produced to a fixed point, describe upon it when so produced a segment of a circle which shall be similar to the given segment, and shew that the two segments have a common tangent. --- 74. If AB, CD be two equal chords in a circle, and tangents to the circle at B and D be produced to intersect one another in F, and to meet the produced chords in G and H, the triangles A.BG, CDH shall be equal, and shall have the sides and angles of the one respectively , equal to the sides and angles of the other. is 75. Draw two tangents to a given circle from a point in a diameter produced, which shall contain an angle equal to a given angle. - * 76. Two radii of a circle at right angles to each other, when pro- .duced are cut by a line which touches the circle : shew that the tangents drawn from the points of section are parallel to each other. W. 77. Two equal circles intersect, the lines joining the points in which any straight line through one of the points of section, which meets the circles with the other point of section, are equal. - 78. ABB, ACB, are the arcs of two equal circles cutting one another in the straight line AB, draw the chord A CD cutting the inner circumference in C and the outer in D, such that AD and DB together may be double of A C and CB together. 79. Two equal circles are drawn intersecting in the points A and B, a third circle is drawn with center A and any radius not greater than AB intersecting the former circles in D and C. Shew that the three points, B, C, D lie in one and the same straight line. 80. A CB, APB are two equal circles, the center of APB being on the circumference of A CB, A B being the common chord, if any chord A C of ACB be produced to cut APB in P, the triangle PBC is equilateral. - 81. A straight bine cuts two circles in the points E, H; F, G; EF is equal to 6H, shew that the circles are equal, if another straight line be cut in the same way by the circles. ON BOOK III. - - 327 '82. If from the points of intersection of two equal circles, two parallel lines be drawn to meet both circles, two of the lines joining their extremities will be parallel. - 83. " Draw through one of the points in which any two circles cut one another, a straight line which shall be terminated by their circum- ferénces and bisected in their point of section. - 84. Describe two circles with given radii which shall cut each other, and have the line between the points of section equal to a given line. «» • 85. If two unequal circles cut one another, and the straight line, drawn through one of the points of intersection to the extremity of the diameter through the centers, be bisected in that point; then the cir- cumference of the lesser circle will bisect the distance between the centers. 86. ACB, ADB are two segments of circles on the same base AB, take any point C in the segment A CB; join A C, BC, and pro- duce them to meet the segment ADB in D and E respectively: shew that the arc DE is constant. & 87. If two circles intersect, the common chord produced bisects the common tangent. ** - 88. Shew that, if two circles cut each other, and from any point in the straight line produced, which joins their intersections, two tan- . gents be drawn, one to each circle, they shall be equal to one another. 89. Two circles intersect in the points A and B; through A and B any two straight lines CAD, EBF, are drawn cutting the circles in the points C, D, E, F, prove that CE is parallel to D.F. 90. If two circles cut each other, the straight line joining their centers will bisect their common chord at right angles. - 91. Two circles cut one another; if through a point of intersection a straight line is drawn bisecting the angle between the diameters at that point, this line cuts off similar segments in the two circles. 92. A circle is described having its center at a point A in the circumference of a given circle, and cutting this circle and the diameter AB in E and D respectively; prove that if ED be joined and produced to meet the given circle in F, the arc EBF will be trisected in B. - 93. If through one of the points of section of two circles which cut each other any chord be drawn, and its extremities be joined with the other point of section, the lines so drawn divide the two circles into similar segments. - . 94. If two circles cut one another, and diameters be drawn through one of the points of section, the straight line joining the other extremities of these diameters shall pass through the other point of section. - 95. Two circles cut one another, and if any two parallel straight lines be drawn through the points of section, to meet the circles, they shall be equal to one another. - --> 96. Common tangents AB, CD are drawn to.two circles AEC, BFD which intersect. The line joining the points of contact B, C, cuts the circles AEC, BFD respectively in the points E and F. Prove that BE is equal to CF. - 97. On a straight line AB as a base, and on the same side of it are described two segments of circles, AP, A Q are chords of the two, including an angle equal to the angle between the tangents to the two circles at A, shew that PQ produced will pass through B. aft s= * * * * * * * *rs = 328 - GEOMETRICAL-ExERCISES 98. Two circles intersect in A, B, CBD, is drawn through B. perpendicular to AB to meet the eireles, and EAF through. A bisecting . either the internal or external angle between AC, AD, and terminated by the circles: shew that the tangents to the circles at E, F will inter- sect in a point lying on AB produced. 99. Two circles intersect in A, and through A any two straight lines BAC, ...I)AE are drawn meeting the circles in B, C, D, E re- spectively: BD, CE are joined and produced to meet in F. Shew that the angle at Fis the same for all positions of the lines. 100. ABC, ABD are any two circles cutting each other in A and B; AE, AF tangents to the circles at the point A, each of them being terminated by the circumference of the other circle. If AB, B.E., B.F. be drawn, AB will biseet the angle EBF. 101. Two circles intersect; shew how to draw a straight line from one of the points of intersection in such a manner that the segment between the points where it again cuts the circles, shall be equal to a given straight line, this-being of such a length that the problem may be possible. . . 102. If from any point A without a given circle as center, a circle be described cutting the former in B and C, and through any point P, in the arc within the first circle, lines BP, CP be drawn meeting the circle again in D and E, DE shall be a diameter. - 103. From two given points to draw right lines to a point in the circumference of a given circle, so that the chord of the intercepted segment shall be parallel to the right line joining the two points. 104. Let A CB be a quadrant of a circle, of which the center is C, and the terminating radii 6A, 6B; join AB, and on it describe a semicircle : from the point A draw a straight line AED cutting the quadrant in E, and terminated by the semicircle in D; and join D.B.; it is required to prove that DE is always equal to DB, and that the greater only of the two lines AD, DB can cut the quadrant. . 105. Two circles intersect each other in the points A and B, from which are drawn chords to a point 0 on one of the eircles, and these (produced if necessary) cut the other circle in D and E; shew that the line DE makes right angles with a diameter of ABDE drawn through C. ' . . - 106. If two circles (1), (2) be drawn cutting in 6 and H, and at G. a tangent be drawn to (1) cutting the eircumference of (2) in B, and from B a line be drawn through H cutting the circumference of (1) in A; and from any point E on the circumference of (1) a line be drawn through H cutting the circumference of (2) in F, and lastly F6 be joined and produced so as to cut off an are 6 E, from (1); shew that arc GL = arc A.F. - . VI 107. If two circles touch each other externally, and two parallel lines be drawn, so touching the circles in points A and B respectivel that neither circle is cut, then a straight line AB will pass º the point of eontact of the circles. f 108. A common tangent is drawn to two eircles which touch each other externally; if a circle be described on that part of it which lies between the points of contact, as diameter, this circle will pass through the point of contact of the two circles, and will touch the line which joins their centers. - - -- 109. If two circles touch each other externally or internally, and ON BOOK III. 329 *~ parallel diameters be drawn, the straight line joining the extremities of these diameters will pass through the point of contact. f 110. If two circles touch each other internally, and any circle be described touching both, prove that the sum of the distances of its center from the centers of the two given circles will be invariable. 111. If two circles touch each other, any straight line passing through the point of contact, cuts off similar parts of their circumfe- renceS. 112. Two circles touch each other externally, the diameter of one being double of the diameter of the other; through the point of con- tact any line is drawn to meet the circumferences of both; shew that the part of the line which lies in the larger circle is double of that in the smaller. : - 113. Two circles touch one another internally in the point A: ; it is required to draw from A such a chord that the part of it intercepted between the circles may be equal to a given liné, not greater than the difference between the diameters of the circles. 114. Two equal circles touch one another externally, and through the point of contact chords are drawn, one to each circle, at right angles to each; prove that the straight line joining the other extre- mities of these chords is equal and parallel to the straight line joining the centers of the circles. - 115. - Two given circles touch each other internally. Find the semichord drawn perpendicularly to the diameter passing through the point of contact, which shall be bisected by the circumference of the inner circle. < g .g. 116. Two circles whose radii are as 2 to 3 touch each other internally; through the center of the smaller circle a straight line is drawn perpendicular to their common diameter; and from the points where this straight line meets the circumference of the larger circle, tangents are drawn to the smaller circle; shew that these tangents will be perpendicular to each other. -- 117. Two circles tough each other. Any two lines passing through their point of contact will cut off arcs whose chords are parallel. WTT. 118. Two circles are traced on a plane; draw a straight lime cutting them in such a manner that the chords intercepted within the circles shall have given lengths. . 119. From two given points in the circumference of a circle draw two chords to meet in the circumference, which shall intercept between them a portion of a given chord equal to a given straight line. 120. Between two given circles to place a straight line terminated by them, such that it shall be equal a given straight line, and be in- ‘clined at a given angle to the straight hine joining their centers. 12k. To describe a circle through two given points to cut a straight line given in position, so that a diameter of the circle drawn through the point of intersection, shall make a given angle with the line. 122. Through two given, points to describe a circle bisecting the circumference of a given circle. - • 123. Describe a circle which shall pass through two given points and cut a given circle, so that the chord of intersection may be of a given length. J - - - - 330 GEOMETRICAL ExERCISEs 124. Through a given point with a given radius, describe a circle bisecting the circumference of a given circle. - 125. Describe two circles each of which passes through two given points, and cuts out of a given unlimited straight line a portion of given length. - 126. Find a point in a given circle, from which if two tan- gents be drawn to an equal circle, given in position, the chord joining the points of contact is equal to the chord of the first circle formed by joining the points of intersection of the two tangents produced; and determine the limit to the possibility of the problem. 127. Describe a circle which shall pass through two given points, . and cut a given circle at right angles. - 128. Describe a circle which shall pass through a given point and bisect two given circumferences. * 129. A circle described on the common tangent to two circles as diameter will cut them at right angles. * 130. Describe a circle passing through a given point and cutting at right angles each of two given circles. ... " - . 131. Describe a circle which shall cut at right angles three given circles. - 132. The chord A B of a circle is bisected in C, and divided in D, so that the square on CB is equal to twice the square on CD; prove . that the circle described from center D and with radius CD, has its circumference bisected by that of the given circle. VIII. 133. Two circles are drawn, one lying within the other; prove that no chord to the outer circle can be bisected in the point in which it touches the inner, unless the circles are concentric, or the chord be perpendicular to the common diameter. If the circles have the same center, shew that every chord which touches the inner circle is bisected in the point of contact. - - 134. Draw a straightline through two concentric circles, so that the chord terminated by the exterior circumference may be double" that terminated by the interior. What is the least value of the radius of the interior circle for which the problem is possible? 135. . If a straight line be drawn cutting any number of concentric circles, shew that the segments so cut off are not similar. 136. If from any point in the circumference of the exterior of two concentric circles, two straight lines be drawn touching the interior and meeting the exterior; the distance between the points of contact will be half that between the points of intersection. - 137. Through a given point draw a straight line so that the part intercepted by the circumference of a circle, shall be equal to a given straight line not greater than the diameter. º • . 138. Two circles are described about the same center, draw a chord to the outer circle, which shall be divided into three equal parts by the inner one. How is the possibility of the problem limited 7 139. Draw two concentric circles such that those chords of the outer circle which touch the inner, may be equal to its diameter. 140. Find a point in a given straight line from which the tangent drawn to a given circle, is of given length. 141. If any number of chords be drawn in the inner of two con- centric circles, from the same point A in its circumference, and each ON BOOK III. 331, of the chords be then produced beyond A to the circumference of the outer circle, the rectangle contained by the whole line so produced and the part of it produced, shall be constant for all the cases. 142. If two straight lines AEFB, CEGD be drawn touching the inner of two concentric circles in F and G, and having their extremities A, B, and C, D in the circumference of the outer circle; and if the points A, D be joined, and also the points C, B : the two triangles AED, CEB will be equal, and the several sides and angles of the one will be respectively equal to those of the other. * 143. Three concentric circles are drawn in the same plane. Draw a straight line, such that one of its segments between the inner and outer circumference, may be bisected at one of the points in which the line meets the middle circumference. - 144. If a chord be drawn cutting two concentric circles, and from . the points of its intersection with the exterior circle, pairs of tangents be drawn touching the interior, prove that the line joining the inter- Sections of these tangents will pass through the center of the circle. 145. Through a given point to draw a straight line so that the part of it intercepted between two given concentric circles shall be equal to a given straight line. Within what limits is this possible 2 146. If there be two concentric circles, and any chord of the greater circle cut the less in any point, this point will divide the chord into two segments whose rectangle is invariable. 147. In the chord of a circle produced, it is required to find a point, from which if a straight line be drawn touching a circle, the line so drawn shall be equal to a given straight line. 148. With a given point as center describe a circle cutting off similar segments from two given concentric circles. -- IX. 149. The circles which are described upon the sides of a right- angled triangle as diameters, meet the hypotenuse in the same point; and the line drawn from the point of intersection to the center of either . of the circles will be a tangent to the other circle. -- " 150. If on the sides of a triangle circular arcs be described contain- ing angles whose sum is equal to two right angles, the triangle formed by the lines joining their centers, has its angles equal to those in the segments. . t º 151. The perpendiculars let fall from the three angles of any tri- angle upon the opposite sides, intersect each other in the same point. . . 152. If AD, CE be drawn perpendicular to the sides BC, AB of the triangle ABC, and DE be joined, prove that the angles ADE, and ACE are equal to each other. & 153. The lines which bisect the vertical angles of all triangles on the same base and with the same vertical angle, all intersect in one point. 154. A segment of a circle being described on the base of a triangle; describe on the other sides segments similar to that on the base. 155. In the triangle ABC, AD, BE are drawn perpendicular to BC and AC respectively, and BA' is drawn making the angle CBA' equal to the angle CBE and meeting AD produced if necessary in A'; shew that a circle may be described about ACA'B. 156. It is required within an isosceles triangle to find a point such, that its distance from one of the equal, angles may be double its dis- tance from the vertical angle. 332 GEOMETRICAL ExERCISEs . 157. To find within an acute-angled triangle, a point from which, if straight lines be drawn to the three angles of the triangle, they shall make equal angles with each other. 158. The triangle formed by the chord of a circle (produced or not) the tangent at its extremity, and any line perpendicular to the diameter through its other extremity will be isosceles. - 159. AD, BE are perpendiculars from the angles A and B on the opposite sides of a triangle, BF'perpendicular to ED or ED produced; shew that the angle FBD = EBA. w 160. 0 is the center of a circle, Pany point in its eircumference, PM a perpendicular to a fixed diameter; shew that the straight line, which bisects the angle 6 PM, always passes through a fixed point. , 161. Three points being in the same plane, find a fourth, where lines drawn from the former three shall make given angles with one another. - . 162. AB is the diameter of a semicircle, P a point on the circum- ference, PM perpendicular to AB; on AM, BM as diameters, two semicircles are described, and AP, BP meet these latter semieircles in Q, R.; shew that @R will be a common tangent to them. * I63. Let the diameter B.A. of a circle be produced to P, so that At P equals the radius; through. A draw the tangent AED, and from P draw PEC touching the circle in C, and meeting the former tangent in JE; join BC and produce it to meet A.E.D. in D: then shall the triangle DEC be equilateral. . . , 164. On the side. AB of any triangle, as diameter, a circle is de- scribed, EF is a diameter parallel to B 0; prove that the straight line EB and FB bisect the interior and exterior angles at B. z' 165. Through B, E, F, the middle points of the sides of a triangle, a circle is described cutting the sides again in D, E, F. Of the three arcs DD, EE", FF", one is equal to the sum of the other two. 166. On the sides of a triangle, semicircles are deseribed, the points in which they cut two and two lie in the sides, or in the sides produced. Also if these points be joined, the sides of the new triangle so formed . are equally inclined to the sides of the original triangle. gº 167. If in a circle two chords AB, AC be drawn, and if any line parallel to the tangent to the circle at A meet AB, A C or these pro- duced in D and E, shew that B, C, D, E will all lie in the circumference of a circle. - X. - 168. If three equal circles have a common point of intersection, prove that a straight line joining any two of the points of intersection, will be perpendicular to the straight line joining the other two points of intersection. - 169. A number of circles touch each other at the same point, and a straight line is drawn from it cutting them : the straight lines joining each point of intersection with the center of the circle will be all parallel. - 170. If three circles intersect one another, two and two, the three chords joining the points of intersection shall all pass through one point. 171. If two equal circles intersect one another in A and B, and from one of the points of intersection as a center, a circle be described which shall cut both of the equal circles, then will the other point of intersection, and the two points in which the third circle cuts the other two on the same side of AB, be in the same straight line. *. ON BOOK III. 333 172. If three circles touch each other in one point, and if from a point external to all, a pair of tangents be drawn to each circle, the three chords joining the points of contact will all pass through one oint. 173. If on any three chords drawn through the same point in the circumference of a circle, as diameters, three circles be described: the points of intersection of these circles two and two lie in the same straight line. - •º 174. If two circles cut each other; to draw from one of the points of intersection a straight line meeting the circles, so that the part of it intercepted between the circumferences may be equal to a given line. 175. ABCD is a parallelogram, AC, B.D the diagonals. On AB, AC, AD describe circles, then their common chords AE, AF, AG are perpendicular to DC, BD, CB respectively. • * 176. Two equal circles cut one another, and a third circle touches each of these two equal circles externally: the straight line which joins the points of section will, if produced, pass through the center of the third circle. Y- 177. If three circles touch each other externally, and the three common tangents be drawn; these tangents shall intersect in a point equidistant from the points of contact of the circles. 178. Three circles touch one another in A, B, C. If O, O, O, be the centers of the circles in which B and C, C and A, A and B respectively lie, shew that twice the angle BAC is equal to the sum of the angles A0,0, AO, B. * 179. Two circles touch each other in C, and ACB the line drawn through their centers meeting the circumference in A, B, is bisected in D. Shew that if any circle be described with center D, cutting the two circles in E, F, the line joining EF shall pass through C the point of contact. g XI. 180. Given the base, the vertical angle, (1) the sum, and (2) the difference of the sides, to construct the triangle. - * 181. Describe a triangle, having given the vertical angle, and the segments of the base made by a line bisecting the vertical angle. 182. Given the perpendicular height, the vertical angle and the sum of the sides, to construct the triangle. - - 183. Construct a triangle in which the vertical angle and the difference of the two angles at the base shall be respectively equal to two given angles, and whose base shall be equal to a given straight IIlê. • 184. Given the vertical angle, the difference of the two sides con- taining it, and the difference of the segments of the base made by a perpendicular from the vertex: construct the triangle. 185. Given the vertical angle, and the lengths of two lines drawn from the extremities of the base to the points of bisection of the sides, to construct the triangle. - .rº 186. Given the base, the altitude, and the sum of the two re- maining sides; construct the triangle. 187. Describe a triangle of given base, area, and vertical angle. 188. Construct a triangle whose sides are given in magnitude and pass through three given points. ... - 334 GEOMETRICAL ExFRCISES .* -** 189. Given the base, the vertical angle, and the perpendicular from the extremity of the base upon the opposite side, find the triangle. 190. Given the perimeter, the vertical angle and the perpendicular from one extremity of the base upon the opposite side of a triangle, construct the triangle. - 191. On a given straight line describe a triangle which shall have its vertical angle equal to a given angle, and a side equal to a given straight line. t -- 192. On a given base describe a triangle having a given vertical angle, and such that one of the sides cuts off a given segment from a given circle. Within what limits is this possible? * 193. Find a triangle of which the vertical angle, the sum of the squares on the two sides containing it and the area are given. 194. Given the base, the altitude, and the difference of the two remaining sides; construct the triangle. 195. Describe on a given straight line a triangle which shall be equal to a given rectilineal figure, and have its vertical angle equal to a given angle. - . - 196. The base, vertical angle, and rectangle under the sum of the other sides, and one of them are given. Construct the triangle. 197. Describe a right-angled triangle upon a given hypotenuse, so that the hypotenuse and one side shall be together double of the third side. - - XII. 198. Shew that the perpendiculars to the sides of a quadrilateral inscribed in a circle from their middle points intersect in a fixed point. 199. The lines bisecting any angle of a quadrilateral figure in- scribed in a circle, and the opposite exterior angle, meet in the cir- cumference of the circle. • * 200. Divide a circle into two parts so that the angle contained in one segment shall equal twice the angle contained in the other. • 201. Four circles are described so that each may touch internally three of the sides of a quadrilateral figure, or one side and the ad- jacent sides produced; shew that the centers of these four circles will all lie in the circumference of a circle. 202. One side of a trapezium capable of being inscribed in a given circle is given, the sum of the remaining three sides is given; and also one of the angles opposite to the given side: construct it. - 203. If the sides of a quadrilateral figure inscribed in a circle be produced to meet, and from each of the points of intersection a straight line be drawn, touching the circle, the squares on these tan- gents are together equal to the square on the straight line joining the points of intersection. - - 204. A quadrilateral ABCD is inscribed in a circle, BC and DC are produced to meet AD and AB produced in E and F. The angles ABC and ADG are together equal to AFC, AEB, and twice the angle B.A. C. - - 205. If the hypotenuse AB of a right-angled triangle ABC be bisected in D, and EDF drawn perpendicular to AB, and DE, DF cut off each equal to DA, and CE, CF joined, prove that the last two lines will bisect the angle at C and its supplement respectively. - 206. The angle contained by the tangents drawn at the extremi- ties of any chord in a circle is equal to the difference of the angles in --- ON BOOK III. 335. ** the segments made by the chord: and also equal to twice the angle contained by the same chord and a diameter drawn from either of its extremities. - - * 207. If ABCD be a quadrilateral figure, and the lines AB, A C, AD be equal, shew that the angle BAD is double of CBD and CDB together. 208. In a quadrilateral figure ABCD is inscribed a second quadrilateral by joining the middle points of its adjacent sides; a third is similarly inscribed in the second, and so on. Shew that each. of the series of quadrilaterals will be capable of being inscribed in a circle if the first three are so. Shew also that two at least of the opposite sides of ABCD must be equal, and that the two squares upon these sides are together equal to the sum of the squares upon the other two. - 209. Shew that the triangles into which any quadrilateral figure in- scribed in a circleis divided by itsdiagonals, are equiangular, two and two. 210. The opposite sides produced of a quadrilateral inscribed in a circle meet in E, F, and the diagonals in G; shew that the perpendi- culars from E, F, G on FG, GE, EF respectively intersect in the center of the circle. - - * - 211. In an acute-angled triangle ABC, perpendiculars AI), BE are let fall on BC, CA respectively; circles described on AC, BC as diameters meet BE, AD respectively in F, G, and H, K ; shew that JF, G, H, K lie on the circumference of a circle. - 212. ABCD is a quadrilateral inscribed in a circle; AC, BD meet in E; AB, CD produced meet in F; and AD, BC in G ; shew that the lines bisecting the angles AED, AFD are parallel to one another, and also those bisecting the angles AEB, AGB. 213. If any quadrilateral figure be inscribed in the inner of two concentric circles, and its sides be produced to meet the outer; and if the adjacent points of intersection belonging to each pair of opposite sides be joined; shew that the quadrilateral figure formed by pro- ducing these joining lines may have a circle described about it. 214. Shew that the lines bisecting the angles of any quadrilateral, form a quadrilateral which can be inscribed in a circle. - 215. If the opposite sides of a quadrilateral ABCD inscribed in a circle be produced to intersect in E and F, the diagonal AC will be a chord of the circle on EF when the angle at A is a right angle. 216. In a given circle to inscribe a quadrilateral figure which shall be equiangular to a given quadrilateral whose opposite angles are, each pair of them, equal to two right angles; and so that one of the angular points of the inscribed figure shall fall at a given point in the circumference of the circle. - 217. ABCD is a quadrilateral figure having the angles at A and JB right angles; CD is bisected in E, and with center E and distance. JEC a circle is drawn cutting AD and BC, or these lines produced in II, C. AF is a tangent to the circle, and FEG, a diameter. Join AG, and shew that the squares on the four sides of the quadrilateral figure are together equal to twice the square on AG. • . . 218. If E be the middle point of a semicircular arc AEB, and EDC be any chord cutting the diameter in D and the circle in C, prove. that the square on CE is equal to twice the quadrilateral AEBC. 219. In every quadrilateral described round a circle, the middle -points of the diagonals are in a line with the center of the circle. '836 t GEOMETRICAI, ExERCISEs’ C 220. Two sides of a quadrilateral inscribed in a circle are produced to meet each other outside the circle at right angles. Shew that the distance of their point of intersection from the center of their circle is equal to the distance between their middle points. . . . . 221. Let C be the center of a circle, A.A., BB' any two diameters at right angles to one another. Draw. AP, AP, BQ, B'Q' perpendi- culars on any tangent to the circle, then AP. A'P' H B (). B Q = CA*. - 222. Let ABCD be any quadrilateral figure, let the line joining the middle points of the diagonals be bisected in E, and with center E any circle be described. Prove that for all points P in this circle, PA* + PB' | PC". PD is the same, and that it = EA*4 EB"4 ED"14. EP". - 223. If a semicircle be divided into any three parts; the square on the diameter is greater than the sum of the squares on the three chords. - 224. If two opposite sides of a quadrilateral figure inscribed in or about a circle be equal, prove that the other two are parallel. 225. The angles subtended at the center of a circle by any two opposite sides of a quadrilateral figure circumscribed about it, are together equal to two right angles. i 226. If a quadrilateral figure be described about a circle, the sums and the differences of the opposite sides are equal; and each sum equal to half the perimeter of the figure. . 227. ABCD is a quadrilateral figure inscribed in a circle. Through its angular points tangents are drawn so as to form another quadrilateral figure FBLCHDEA circumscribed about the circle. * Find the relation which exists between the angles of the exterior and the angles of the interior figure. XIII. 228. If from any point in the diameter of a semicircle, there be drawn two straight lines to the circumference, one to the bisection of the circumference, the other at right angles to the diameter, the squares upon these two lines are together double of the square upon the semi-diameter. - 229. If from any point in the diameter of a circle, straight lines be drawn to the extremities of a parallel chord, the squares on these lines are together equal to the squares on the segments into which the . diameter is divided. - 230. From a given point without a circle, at a distance from the circumference of the circle not greater than its diameter, draw a straight line to the concave circumference which shall be bisected by the convex circumference. 231. If any two chords be drawn in a circle perpendicular to each other, the sum of their squares is equal to twice the square on the diameter diminished by four times the square on the line joining . the center with their point of intersection. - 232. Two points are taken in the diameter of a circle at any equal distances from the center; through one of these draw any chord, and join its extremities and the other point. The triangle so formed has the sum of the squares on its sides invariable. 233. If chords drawn from any fixed point in the circumference of a circle, be cut by another chord which is parallel to the tangent at that point, the rectangle contained by each chord, and the part of . . . .” - . . . . . .”. ge & ** . on BOOK III. . . 337 it intercepted between the given point and the given chord, is con- stant. - - - - . . . 234. If A B be a chord of a circle inclined by half a right angle to the tangent at A, and AC, AD be any two chords equally inclined to AB, then A C* F AD” = 2. A B*. * → . 235. A chord P00 cuts the diameter of a circle in Q, in an angle equal to half a right angle; P0°4 00 = 2 (rad.)". 236. . Let A CDB be a semicircle whose diameter is AB; and AD, BC any two chords intersecting in P; prove that - y’ - AB* = D.A . A P+ C.B. B.P. 237. If ABDC be any parallelogram, and if a circle be described passing through the point A, and cutting the sides AB, AC, and the diagonal AD, in the points F, G, H respectively, shew that - w A.B. A.F.F.A.C. A G = AD. A.H. . 238. If A be a point within a circle, B0 the diameter, and through A, AD be drawn perpendicular to the diameter, and BAE meeting the circumference in E, then B.A. B.E- BC. B.D. . 239. The diameter A CD of a circle, whose center is C, is produced to P, determine a point F in the line AP such that the rectangle PF. PC may be equal to the rectangle PD. P.A. . . 240. To produce a given straight line, so that the rectangle con- tained by the whole line thus produced, and the part of it produced, shall be equal to a given square. 241. From the vertex of any triangle ABC, draw a straight line meeting the base produced in D, so that the rectangle DB. DC = AD". 242. Two straight lines stand at right angles to each other, one of which passes through the center of a given circle, and from any point in the other, tangents are drawn to the circle. Prove that the chord joining the points of contact cuts the first line in the same point, what- ever be the point in the second from which the tangents are drawn. . . 243. A, B, C, D, are four points in order in a straight line; find a point E between B and C, such that A.E. EB = ED. EC, by a geometrical construction. - 244. If any two circles touch each other in the point 0, and lines be drawn through 0 at right angles to each other, the one line cutting the circles in P, P', the other in Q, Q ; and if the line joining the centers of the circles cut them in A, A'; then PP”4. Q Q = A'A*. . . . .245. If from the vertex A of a triangle ABC, a perpendicular AD be drawn to the base, and from D be' drawn DE, DF, perpendiculars upon AB, AC respectively, shew that the rectangle AB, AE is equal to the rectangle AC, AF . * - 246. From each angular point of a triangle, a perpendicular is let fall on the opposite side; prove that the rectangles, contained by the segments into which each perpendicular is divided by the point of intersection of the three, are equal to each other. - T * 247. If AD, CE be drawn perpendicular to the sides BC, AB of the triangle ABC, prove that the rectangle contained by BC and BD, is equal to the rectangle contained by BA and B.E. . . . 248. In a given straight line ADCB, whose middle point is C, determine the point D, so that the square on CD shall be equal to the rectangle contained by AD and D.B. - 249. Let AP be a tangent to any circle, and AB a diameter. To determine the point P, so that PCB being drawn, cutting the circum- * . 2. 338 - GEOMETRICAL ExERCISEs ference in C, the rectangle contained by PC, CB, shall be equal to a given Square; and shew in what cases this is impossible. * Tx 250. If two chords AB, AC be drawn from any point A of a circle, and be produced to D and E, so that the rectangle A C, AE is equal to the rectangle AB, AD, then, if 0 be the center of the circle, AO is perpendicular to D.E. 251. On 0A the radius of one circle as diameter another is described, and a chord PQ is drawn in it parallel to OA. If through P, Q lines be drawn perpendicular to OA and meeting the larger circle in D, E, and F, G respectively, then PD. PE4 QF. QG = O.A. 252. Two equal circles touch one another at the point A ; from any point P in the one, PQ is drawn to touch the other, P and A are joined, then PQ = 2 PA*. - s 253. Two circles ABC, ADE touch internally in A, and from F (any point in their common tangent A.F.) a circle is described cutting them in points B, Drespectively. Draw the radii FB, FD meeting the circumferences in C and E; and shew that BC = D.E. 254. Two circles cut one another, and from a point A without them the straight lines ABC, ADE are drawn cutting the circles in B, C and D, E respectively, so that the rectangle AB, A C is equal to the rectangle AD, A.E. Shew that the point A is in the chord of inter- section of the two circles. - 255. If in the diameter or diameter produced of a circle, two points be taken equally distant from the center; the sum of the squares of the distances of .any point in the circumference from these two points is constant. 256. ABC is a triangle whose acute vertex is A and base BC, shew that BC* is less than A C*-* AB" by twice the square of a line drawn from A to touch the circle of which BC is the diameter. 257. Through a given point draw a line terminating in two lines given in position, so that the rectangle contained by the two parts may be equal to a given rectangle. - - 258. If P be a point without a circle whose center is 0, and A 0B a diameter perpendicular to P0: draw a line PMEC cutting the circle in M and C and the diameter in E, so that the rectangle PM, PC, may be four times the rectangle AE, EB. 259. If on the radius AC of a circle another semicircle be described and a perpendicular BDE be drawn to the diameter cutting the circles in B and D, and AB, AD be joined, then the square on AB is double of the square on A.D. 260. The circumference of the circle ACE is divided into six equal parts in the points A, B, C, D, E, F, G is the center of ACE; with radius AC, and centers A, D, describe two circles intersecting in H; with radius AC, and centers C, E, describe two circles intersecting in K, - A. K. A G = A Hº. - 261. If there be two circles in the same plane not cutting each other, and two lines be drawn (one of them meeting the line joining the centers) to touch both circles and terminate at the points of contact, prove that the difference of the squares of these lines will be equal to the rectangle of the diameters of the circles. 262. If A 0B be a chord to the outer, and a tangent to the inner of two concentric circles, whose diameters are D and d, and if any two chords EOF COD be drawn through O': and E, D, and C, F be joined, cutting AB in M and N respectively, then . 4. CN. N.F. MN” – D – d'. GEOMETRICAL EXERCISES ON BOOK IV. PROPOSITION I. THEOREM. If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius, or on the side of the regular heavagon inscribed in the same circle. Let ABD be an equilateral triangle inscribed in the circle ABD, of which the center is 0. a Join BC, and produce BC to meet the circumference in E, also join AE. And because ABD is an equilateral triangle inscribed in the circle; *A. therefore AED is one-third of the whole circumference, and therefore AE is one-sixth of the circumference, and consequently, the straight line A.E is the side of a regular hexagon (IV. 15.), and is equal to EC. - And because B.E is double of EC or AE, therefore the square on BE is quadruple of the square on AE, but the square on B.E is equal to the squares on AB, AE; * therefore the squares on AB, AE are quadruple of the square on AB, and taking from these equals the square on A.E, - therefore the square on AB is triple of the square on A.E. PROPOSITION II. PROBLEM. Inscribe a circle in a given sector of a circle. Analysis. Let CAB be the given sector, and let the required circle whose center is 0, touch the radii in P, Q, and the arc of the sector in D. . . .* E 2 D F Join OP, 09, these lines are equal to one another. - - Join also CO. Then in the triangles CPO, COO, the two sides PC, CO, are equal to QC, CO, and the base OP is equal to the base 00; - * Z 2 340 GEOMETRICAL ExERCISEs therefore the angle PCO is equal to the angle Q00; and the angle ACB is bisected by 00: also CO produced will bisect the arc AB in D. (III. 26.) If a tangent EDF be drawn to touch the arc AB in D; and CA, CB be produced to meet it in E, F: the inscription of the circle in the sector is reduced to the inscrip- tion of a circle in a triangle. (IV. 4.) PROPOSITION III. THEOREM, In any triangle inscribed in a circle, if from one extremity of that diameter which bisects the base, a perpendicular be drawn to the longer of the two sides; the segments of this side are respectively equal to half the sum and half the difference of the two sides of the triangle. Let ABC be a triangle inscribed in a circle, of which the diameter D.E. bisects the base BC in F, and also the are BEC in E. Let EG be drawn from E, one extremity of the diameter DE, perpendicular on the longer side A.B. H Then AG is equal to half the sum of the two sides AB, AC, ... and BG equal to half their difference. From E draw EH perpendicular on AC produced in H, and join AE, EB, EC. First. Because the angles B.A.E, CAE are equal (Euc. III. 21.), and the angles EGA, EHA are right angles, also the side AE is common to the two triangles AGE, A.HE: therefore EG is equal to EH, and AG to A.H. (Euc. I. 26.) Again, because EB is equal to EC, and EG to EH in the right-angled triangles EBG, ECH; therefore BG is equal to CH. - ...And AG has been shewn to be equal to AH, therefore AB and A C are equal to AG and AH; but AG and AH are double of A6; - wherefore the double of AG is equal to AB and AC, and AG is equal fo half the sum of AB and A.C. Secondly. It has been shewn that AG is equal to A.H, and BG to (J.H. - - . Because AB is equal to the sum of the lines AG, GB, and AC equal to the difference of AH, CH; therefore the difference of AB and AC is double of BG, and BG is half the difference of AB and A.C. ON BOOK IV, 341 PROPOSITION IV. THEOREM. In any triangle inscribed in a circle, if from the extremities of that diameter which bisects the base there be drawn two lines to the vertea of the triangle; the angle contained between the diameter and the line drawn to that extremity below the base, is equal to half the difference of the angles at the base of the triangle; and the angle contained by the line drawn to the other eactremity and the adjacent side of the triangle, is equal to half their &4%. . Let ABC be a triangle inscribed in a circle whose diameter DE bisects the base BC, also the arc B.EC in E; and let AE, AD be ioined. J Then the angle AED is equal to half the difference of the angles ABC, ACB at the base of the triangle, and the angle BAD is equal to half their Sum. - - * First. Draw AG parallel to BC, and join BG, GE. Because BC, AG are parallel chords in the circle, the arc BG is equal to the arc A.C.; add to these equals the arc GDA, therefore the arc BGA is equal to the arc CAG: but the angle ACB stands on the arc BGA, and the angle GBC on the are CAG, - therefore the angle ACB is equal to the angle GBC; and if APC be taken from these equals, the difference of the angles ACB, ABC is equal to GBA, the difference of the angles GBC, ABC, - but GBA is equal to GEA, (Euc. III.) or twice AED, for ED bisects the angle GEA : therefore twice the angle AED is equal to the difference of the angles ACB, AIBC, and the angle AED is equal to one-half the difference of ACB, ABC, the angles at the base of the triangle. Secondly. Let CA be produced to H. Then because the arcs BE, EC are equal, the angle BAE is equal to the angle EAC, and AE bisects the angle BAC. And because AD being at right angles to AE, (Euc. III. 31.) bisects the adjacent angle BAH, - therefore BAD is half of the angle B.A.H. * - e. But BAH is equal to the two angles ABC, ACB ; (Euc. I. 32.) therefore BAD is equal to half the sum of the angles ACB, ABC. } 342 GEOMETRICAL ExERCISEs f PROPOSITION V. PROBLEM. ABCD is a rectangular parallelogram. Required to draw EG, EG parallel to AD, DC, so that the rectangle EF may be equal to the figure EMD, and EB equal to FD. - Analysis. Let EG, FG be drawn, as required, bisecting the rectangle ABCD, Draw the diagonal BD cutting EG in H and FG in K. Then BD also bisects the rectangle ABCD; and therefore the area of the triangle KGH is equal to that of the two triangles EHB, FKD. A E B H R L F GTM D N C Draw GL perpendicular to BD, and join GB, also produce FG to M, and EG to W. If the triangle LGH be supposed to be equal to the triangle EHB, by adding HGB to each, the triangles LGB, GEB are equal, and they are upon the same base GB, . and on the same side of it; therefore they are between the same parallels, that is, if L, E were joined, LE would be parallel to GB; and if a semicircle were described on GB as a diameter, it would pass through the points E, L; for the angles at E, L are right angles: also LE would be a chord parallel to the diameter GB; therefore the arcs intercepted between the parallels LE, GB are equal, Q1 and consequently the chords EB, LG are also equal; but EB is equal to GM, and GM to GN; wherefore LG, GM, GN, are equal to one another; hence G is the center of the circle inscribed in the triangle BDC. Synthesis. Draw the diagonal B.D. Eind G the center of the circle inscribed in the triangle BDC; through G draw EGN parallel to BC, and FKM parallel to A.B. Then EG and FG bisect the rectangle ABCD. Draw GL perpendicular to the diagonal B.D. . In the triangles GLH, EHB, - the angles GLH, HEB are equal, each being a right angle, and the vertical angles LHG, EH B are equal, also the side LG is equal to the side EB; therefore the triangle LHG is equal to the triangle EHB. . Similarly, it may be proved, that the triangle GLK is equal to the triangle KFD; « - therefore the whole triangle KGH is equal to the two triangles EHB, KFD; - and consequently EG, FG bisect the rectangle ABCD. ON BOOK IV, 343 I. 6. In a given circle, place a straight line equal and parallel to a given straight line not greater than the diameter of the circle. 7. If an equilateral triangle be inscribed in a circle; shew that the radii drawn to its angles trisect the circle. 8. If an equilateral triangle be inscribed in a circle, and a straight line be drawn from the vertical angle to meet the circumference, it will be equal to the sum or difference of the straight lines drawn from the extremities of the base to the point where the line meets the circum- ference, according as the line does or does not cut the base. ... 9. If from the angles of an equilateral triangle, perpendiculars be let fall on any diameter of the circumscribing circle, the sum of the perpendiculars on the same side of the diameter will be equal to the perpendicular on the other side. - - 10. The perpendicular from the vertex on the base of an equila- teral triangle, is equal to the side of an equilateral triangle inscribed in a circle whose diameter is the base. Required proof. 11. If an equilateral triangle be inscribed in a circle, and the adjacent arcs cut off by two of its sides be bisected, the line joining the points of bisection shall be trisected by the sides. * • 12. If an equilateral triangle be inscribed in a circle, any of its sides will cut off one-fourth part of the diameter drawn through the opposite angle. - 13. If a triangle be inscribed in one and described about another of each of two concentric circles, shew that it is equilateral. 14. If an equilateral triangle be turned about its center in its own plane, any two positions of the altitude will always make the same angle as those of the sides; and there will be three positions of coincidence of the triangle. 15. About a given triangle, a circle is described, and another triangle is formed by drawing tangents to the circle through the angular points. If this process be continually repeated, the last of the Series of triangles so successively formed will be equilateral. - 16. If on the sides of any triangle three equilateral triangles be described, and circles inscribed in each of these triangles, the straight lines joining the centers of the circles will form an equilateral triangle. 17. If a circle can be described cutting the sides AB, BC, CA, of a triangle in the points D, E, F, G; H, K; in such a manner that #. dBF= CH, and EB= G0 = KA, shew that the triangle is equi- ateral. 18. ABCP, and A'B'C'P' are two concentric circles, ABC, A'B'C' are any two equilateral triangles inscribed in them. If P, P’ be any two points in the circumference of these circles, shew that A P^* B. P* + C P = AP* + BP" + CP”. II. 19. If ABCDEFbe an equilateral and equiangular hexagon, and AB, DC be produced to meet in 0; B00 is an equilateral triangle. 20. If ABCDEF be a regular hexagon, and AC, BD, CE, DF, BA, FB be joined; another hexagon is formed whose area is one- third of that of the former. - 344 - GEOMETRICAL EXERCISES 21. BC, CD and DE are contiguous sides of an equilateral hexa- gon, and BE is joined; prove that the line BE is inclined to one of the sides of the hexagon at an angle which is equal to half a right angle. - - - - . *. If an equilateral triangle be inscribed in a circle, and tangents be drawn parallel to the sides; the perimeter of the hexagon thus formed will be two-thirds of the perimeter of the triangle. - 23. If two equilateral triangles be described about a circle, they will by their intersections, form a hexagon equilateral, but not gene- rally equiangular. 24. Prove that the area of a regular hexagon is greater than that of an equilateral triangle of the same perimeter. 25. If any six-sided figure be inscribed in a circle, the sum of either three alternate angles is equal to four right angles. 26. If two equilateral triangles be inscribed in a circle so as to Thave the sides of one parallel to the sides of the other, the figure com- mon to both will be a regular hexagon, whose area and perimeter will be equal to the remainder of the area and perimeter of the two triangles. 27. Determine the distance between the opposite sides of an equi- lateral and equiangular hexagon inscribed in a circle. 28. Through the angular points of a given triangle, draw straight lines which shall form an equilateral hexagon, whose area shall equal twice that of the triangle. - vº 29. ABC is a triangle, Aa bisects BC in a, and is produced to a', so that aa' is one-third of Aa, the same construction is made for Bb', 66. Shew that a'b'c' is a triangle similar to ABC and equal to it. Shew that the hexagon common to the two triangles is two-thirds of either triangle. - 30. If any two consecutive sides of a hexagon inscribed in a circle be respectively parallel to their opposite sides, the remaining sides are parallel to each other. - 31. If the alternate sides of a regular hexagon be produced to meet, the figure so formed will be regular, and have an area equal to twice the area of the hexagon. r 32. If the alternate sides of a regular hexagon be produced to meet one another, and the angular points of the triangles thus formed be joined, a regular hexagon will be formed, the area of which is equal to three times the area of the original hexagon. : 33. If a regular hexagon be inscribed in a circle, six circles equal to it may be described, every one touching the original circle, and two of the others. If the centers of these circles be joined successively, a regular hexagon will be formed, whose area is four times the area of the former: and if the outermost points of the six circles be joined, i. hexagon will be formed, whose area is seven times the area of the first. - - 34. If ABCDEF be a hexagon such that AB is equal and parallel to DE, and BC to EF, prove that CD is equal and parallel to FA. Also that AID, B.E, CF all meet in a point, bisecting each other, and that - - AD** BE*4 CF = A B*4 BC*4 CD + A C*4 B.D.'ſ CE". - 85. To inscribe a regular duodecagon in a given circle, and shew that its area is equal to the square on the side of an equilateral triangle inscribed in the circle, - - ON BOOK IV. " - 345 - III. 36. The center of the circle which touches the two semicircles described on the sides of a right-angled triangle is the middle point of the hypotenuse. - - - 37. If a circle be inscribed in a right-angled triangle, the excess of the sides eontaining the right angle above the hypotenuse is equal to the diameter of the inscribed circle. . - 38. Three circles whose radii are as 1, 2, 3, touch each other ex- ternally; the lines joining their centers form a right-angled triangle. 39. Two right-angled triangles ABC, ABD, have a common hypotenuse. If E be the point where the other sides cross, and 0 the center of the circle described about CDE, then OA*4 OB% = A B* + 2. OE*. 40. If two circles be drawn, one within, and the other about, a given right-angled triangle, the sum of their diaméters will equal the sum of the sides containing the right angle. 41. The rectangle of the segments into which the hypotenuse of a right-angled triangle is divided by the point of contact of the in- scribed circle, is equal to the area of the triangle. IV. 42. If the three points be joined in which the inscribed circle touches the sides of the triangle, prove that the resulting triangle is acute-angled. 43. Any number of triangles having the same base and the same vertical angle, will be circumscribed by one circle. - 44. Find a point in a triangle from which two straight lines drawn to the extremities of the base shall contain an angle equal to twice the vertical angle of the triangle. Within what limitations is this possible 2 - - - 45. If any point in the circumference of a circle described about a triangle be joined with the angular points of the triangle, the joining lines shall be in the direction of the sides of a triangle which is equi- angular with the inscribed triangle. - 46. If two isosceles triangles be inscribed in a circle upon a common base; then an isosceles triangle which has the center for its vertex, and a side of either of the former for its base, is equiangular with the other one. 47. On each side of an acute-angled triangle as base, an isosceles triangle is constructed, the sides of each being equal to the radius of the circumscribed circle; if the vertices of these be joined, a triangle will be formed equal and equiangular to the original. 48. If the line joining an angle of a triangle with the center of the circumscribed circle, cuts the other side at right angles, the triangle is isosceles. - - - 49. In a triangle ABC let AD bisecting the angle A meet BC in D: from 0 the center of the inscribed circle draw OE perpendicular to BC; then is the angle BOE equal to the angle DOC. 50. A circle is described round the isosceles triangle ABC in which AB = BC; from B a straight line is drawn meeting the base in D and the circle in E; prove that the circle which passes through A, D, and E, touches A.B. - 51. From A and B the extremities. of a diameter of a circle; two straight lines are drawn meeting the circumference in C and D, and *- 346 GEOMETRICAIA EXERCISES each other in a point E within or without the circle. Shew that the straight lines drawn from C and D to the center of the circle described about the triangle CED are tangents to the first circle, and those drawn to the center of the first, tangents to the second. - 52. From a point O without a circle draw two lines O.A., OB, to touch the circumference, and about OAB describe a circle. Draw any line BCD cutting the circles in C and D, then shall AC = CD, and moreover the circle described about AOB shall pass through the center of the first circle. 53. If in a given triangle a circle be inscribed, and tangents to it be drawn parallel to the sides; the sum of the perimeters of the three small triangles cut off by these tangents, will equal the perimeter of the given triangle. . - 54. If a circle be inscribed in a triangle, and the points of contact be joined by straight lines, shew that the angles of the triangle thus formed will be the halves of the supplements of the corresponding angles of the original triangle. * - 55. From any point B in the radius CA of a given circle whose center is C, a straight line is drawn at right angles to CA meeting the circumference in D; the circle described round the triangle CBD touches the given circle in D. 56. If a circle be described about a triangle ABC, and perpen- diculars be let fall from the angular points A, B, C, on the opposite sides, and produced to meet the circle in D, E, F, respectively, the circumferences E.F, FD, DE, are bisected in the points A, B, C. 57. If from the angles of a triangle, lines be drawn to the points where the inscribed circle touches the sides; these lines shall intersect in the same point. - - 58. Let three perpendiculars from the angles of a triangle ABC on the opposite sides meetin P; a circle described so as to pass through P and any two of the points A, B, C, is equal to the circumscribing circle of the triangle. *, a 59. If perpendiculars Aa, Bb, Co be drawn from the angular. points of a triangle ABC upon the opposite sides, shew that they will bisect the angles of the triangle abo. * , 60. A circle is described about the triangle ABC, the tangent at C meets AB produced in D, the circle whose center is D and radius DC cuts AB at E; shew that EC bisects the angle ACB. 61. The perpendiculars from the angles A and B of a triangle on the opposite sides meet in D; the circles described about ADC and BDC cut AB or AB produced in the points A' and B' respectively: prove that AA = BB". * - 62. If the base BC of a triangle ABC inscribed in a circle pass through the center, and from the extremity B of the base, a circle be described at the distance of the other extremity; the circumference cut off from that circle by the side CA produced, will be bisected by the remaining side produced. 63. B is the middle point of the arc of a segment ABC less than a semicircle. Produce AB to D, and draw DC perpendicular to BC. Then the circle described about BCD will pass through the intersec- tion of the tangents at A and 0. - - - 64. 0 is the center, OA a radius, and P a point in a circle. Another circle is described about the triangle POA, and cuts in Q the line PM perpendicular to OA. In the triangle PAQ the angle Q is double the angle P. - ON BOOK Iv. * 347 65. ff the circle inscribed in the triangle ABC touch the sides AB, AC in the points D, E, and a straight line be drawn from A to * the center of the circle, meeting the circumference in G; shew that G is the center of the circle inscribed in the triangle AIDE. 66. A triangle ABC being inscribed in a circle, and the side AC produced to cut the tangent at B in D, shew that the angles ABD, B0D are equal. 67. If from a point A without a circle, A B be drawn to touch, and A C to meet the circle, and Obe the center of the circle described about ABC; then if a circle described about A 00 cuts AB in D, DC touches the first-mentioned circle. t 68. ABCD is a circle, and AC, BD are any two chords at right angles to each other: from A any chord AP is drawn cutting BD in Q; shew that the center of a circle described about PQD lies in the chord C.D. - & - 69. If a circle be inscribed in a triangle, prove that the part of any side intercepted between one of its angles and the point in which the side touches the circle, is equal to the difference be- tween one-half of the three sides and the side opposite to the angle in question. - • 70. If the arcs, cut off by the sides of a triangle inscribed in a circle be bisected, and from the points of bisection, as centers, circles be described passing through the extremities of the respective sides; shew that they all pass through the center of the circle inscribed in the triangle. * - 71. From a point A in a circle are drawn two chords ABC, AB'C' through the middle points B, B of two equal chords. If a circle can be described round BB CC, BB is parallel to CC” and ABB" is isosceles. - 72. If from the center A of a circle, a straight line ADB be drawn, cutting the circumference in D, and equal to the diameter of the circle, and from the point B a straight line be drawn touching the circle in C, the straight line CD subtends at the center of the circle, an angle equal to one-third of two right angles. - - 73. If from any point in the circumference of a circle perpendi- culars be let fall upon the sides of a triangle inscribed in it, the feet of the perpendiculars will be in the same straight line. - 74. D is any point in AB the base of a triangle CAB; let CD'be joined, and circles be described about the triangles CDA, CDB cutting the sides CA, CB in E, F, and let DE, DF be drawn. Prove that DE, DF’ are equally inclined to the base. 75..., Shew that the perpendiculars drawn from the center of the circle inscribed in a triangle upon the sides, are tangents to three circles which touch one another, and have their centers at the angles of the triangle. - - • * 76. From the angle A of a triangle ABC, AD is drawn to any point D of the opposite side. If O and O' are the centers of the circles inscribed in the triangles ABD, A CD; shew that ODO is a right angle. * 77. A circle is described about a triangle ABC in which the angle BAC is half a right angle. From A perpendiculars are drawn on the lines touching the circle at B and C, and meeting the circumference in E and D respectively. If EC, BD be joined, the angles ACE, ABD will together equal a right angle. - 348 GEOMETRICAL EXERCISES 78. 0 is the center of the circumscribing circle of a triangle ABC; D, E, F the feet of the perpendiculars from A, B, C on the opposite sides: shew that 0A, OB, 00 are respectively perpendicular to EF, FD, DE. -- 79. ABC is a triangle inscribed in a circle whose center is 0. If three circles be described about the triangles A 0B, AOC, B 00: the sum of the angles subtended at their centers by AB, AC, B C respec- tively, is equal to four right angles. 80. 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 its 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. 81. From each angular point of a triangle, a perpendicular is let fall on the opposite side; prove that the distance of the point of inter- section of these perpendiculars from any one of the angular points, is equal to twice the distance of the center of the circumscribed circle from the side opposite to that angular point. 82. 0 is the intersection of the perpendiculars from A, B, C on the opposite sides of the triangle ABC, shew that the circle which passes through the middle points of OA, OB, OC passes also through the feet of the perpendiculars, and through the middle points of the sides of the triangle. - - 83. The line joining the centers of the inscribed and circumscribed circles of a triangle, subtends at any one of the angular points an angle equal to the semi-difference of the other two angles. 84. If from any point P, whether within or without a triangle ABC, perpendiculars Pa, Pb, Po be dropped upon the sides BC, AC, AB; and circles be described about the triangles Pab, Pac, Pbo; the area of the triangle formed by joining the centers of these circles will equal one-fourth of the area of the triangle ABC. 85. A, B, C, 0 are any four points in a plane; D is the center of the circle described about the triangle OBC, E that of the circle de- scribed about the triangle OCA, F that of the circle described about the triangle OAB; prove that the triangle DEF is equiangular to the triangle ABC. : - 86. If four straight lines intersect so as to form four triangles, the circles described round these triangles shall all pass through the same point. W. 87. Shew how to describe a circle touching one side of a given triangle and the other two produced. - 88. Describe a circle touching the inscribed circle and two of the sides of the triangle. 89. The centers of the escribed circles of a triangle must lie with- out the circumscribing circle, and cannot be equidistant from it unless the triangle be equilateral. ' - - - 90. Shew that the lines joining the centers of the circles, each of which touches one side externally and the other two produced, pass through the angular points of the triangle. - - 91. If O be the center of the circle inscribed in the triangle ABC; O' that of the circle touching the base BC and the sides AB, AC pro- º, on Book IV. 349 duced; shew that a circle may be described passing through the points B, C, 0, 0. - - - • . 92. The straight line joining the center of the circle inscribed in the triangle ABC with the middle point of the side BC, is parallel to the straight line joining A with the point of contact of the circle touch- ing. BC externally, and AB, A C produced. . - * 93. A circle is described touching one of the sides of a triangle and also the two others produced, shew that the first mentioned point of contact, and the point at which the inscribed circle touches the same side, are equidistant from the two adjacent angles, and also that the distance between these two points is equal to the difference between the other two sides. . - 94. If a circle be described touching the base of a triangle and the sides produced, and a second circle be inscribed in the triangle; prove that the distances between the points where the two circles touch the sides and the sides produced, are each equal to the base of the triangle. 95. Let a circle be described about a given triangle; bisect the arc of this circle cut off by one side, and with the point of bisection as center describe a circle through the extremities of this side, shew that this circle passes through the centers of the inscribed circle of the triangle, and of the escribed circle touching that side. 96. If there be a series of triangles having the same vertical angle and equal perimeters, and if a circle touch the sides produced and one of the bases, it will also touch all the other bases of the triangles. - 97. The four circles, each of which passes through the centers of three of the four circles touching the three sides of a triangle, are equal to each other. - 98. If O, O, O, be the centers of the escribed circles of a triangle, and C the center of the inscribed, and if OC, 0,0, be bisected in points P, Q, shew that PQ is a diameter of the circumscribing circle of the triangle. - 99. The sum of the radii of the three escribed circles of any triangle, is equal to the sum of the radius of the inscribed, and four times the radius of the circumscribed circle. . - - • 100. If perpendiculars be dropped from the angular points of a triangle upon the sides, the sum of the distances of their point of in- tersection from the angular points, equals the sum of the diameters of the inscribed and circumscribed circles. Also the difference between the sum of any two of the above distances and the third, equals the difference between the diameter of one of the escribed circles and the diameter of the circumscribed circle. - 101. The straight lines joining the center of the inscribed, to the centers of the escribed circles of a triangle are bisected by the circumscribing circle. - - 102. A circle is drawn, touching the hypotenuse of a right- angled triangle and the two sides produced: prove that the diameter of this circle is equal to the perimeter of the triangle. tº 103. Having given the hypotenuse of a right-angled triangle, and the radius of the inscribed circle, to construct the triangle. 104. In a given circle inscribe a triangle equal in area to a given triangle. * - 350 - GEOMETRICAL ExERCISES 105. In a given circle inscribe a triangle, two of whose sides shall be parallel to given lines, and whose third side shall pass through a given point. 106. Inscribe a triangle in a given circle having a given chord for its base, and its sides dividing another given chord, in such a way that the rectangle contained by the extreme segments may be equal to a given Square. 107. Inscribe a triangle in a given circle having its vertex at a given point on the circumference, its vertical angle, and area being IWerl. £ 108. Given the base of a triangle, and the point from which the perpendiculars on the three sides are equal; construct the triangle. To what limitation is the position of this point subject, in order that the triangle may lie on the same side of the base? - 109. Having given the vertical angle of a triangle, and the radii of the inscribed and circumscribed circles, to construct the triangle. 110. Given the base and vertical angle of a triangle and also the radius of the inscribed circle, required to construct it. 111. Given the three angles of a triangle, and the radius of the inscribed circle, to construct the triangle. 112. Construct a plane triangle, having given the three perpendi- culars let fall on the three sides from the center of the circumscribed circle. . 113. Construct the triangle when one side, the circumscribed circle, and the sum of the other two sides are given. - - 114. On the sides of an equilateral triangle three squares are described. Compare the area of the triangle formed by joining the centers of these squares with the area of the equilateral tri- angle. VII. 115. In a given triangle to inscribe a triangle, the sides of which shall be parallel to the sides of a given triangle. 116. D is any point in AB the base of the triangle CAB, DE is drawn parallel to AC cutting BC in E, EF is drawn parallel to BA cutting CA in F, and so on. Shew that after two complete rounds the path returns to D. * 117. In a given triangle inscribe a parallelogram which shall be equal to one-half the triangle. Is there any limit to the number of such parallelograms? 118. If any number of parallelograms be inscribed in a given parallelogram, the diameters of all the figures shall cut one another in the same point. 119. A square is inscribed in another, the difference of the areas is twice the rectangle contained by the segments of the side, which are made at the angular point of the inscribed square. 120. If from the point of intersection of the diagonals of a rhombus, perpendiculars be drawn to the sides, the lines joining the points at which these perpendiculars meet the sides, will form a rectangle. 121. Inscribe a rhombus within a given rhombus, so that one of the angular points of the inscribed figure may bisect a side of the other. - ON BOOK IV. * , 351 122. Inscribe a rhombus in a given parallelogram, so as to have an angle on each side or on each side produced. Shew that the number is unlimited. - 123. In a given quadrilateral figure inscribe a rhombus. 124. Describe a parallelogram similar to a given parallelogram so that its four sides shall pass through the four angular points of a given square respectively. With what limitation is this possible? VIII. 125. In a given square, inscribe a regular octagon ABCDEFGH; and shew that the octagon is twice the rectangle AB, EF. - 126. Inscribe a circle in a rhombus. h 127. In a given circle inscribe a rectangle equal to a given rectilineal figure. - 128. If a circle can be described so as to cut out equal pieces from the four sides of a parallelogram, the parallelogram must be equilateral. Shew how to find the center of all such circles. 129. In a given circle inscribe a quadrilateral figure, Só that two of its adjacent sides may be parallel to two given lines, and that the other two may pass through two given points. w 130. If two squares be described about a circle in any way, shew that the perimeter of the octagon thus formed is always the S8.Iſle. 131. If circles be described having for centers the corners of a Square, and the radius of each equal to the semi-diagonal, they cut the sides in points which are the corners of a regular octagon. 132. The sum of the alternate angles in any octagonal figure in a circle is equal to six right angles. - 133. Shew that the difference between the circumscribed square, and the inscribed duodecagon in a circle, is equal to one-fourth of the circumscribing square. & .* 134. A circle having a square inscribed in it being given, to find a circle in which a regular octagon of a perimeter equal to that of the Square, may be inscribed. 135. On a given straight line describe an equilateral and equi- angular octagon. 136. A regular octagon inscribed in a circle is equal to the rec- tangle contained by the sides of the squares inscribed in, and circum- Scribed about the circle. 187. If from any point in the circumference of a circle, straight lines be drawn to the angular points of the inscribed square, the sum of the squares on these four lines shall be equal to twice the Square on the diameter of the circle. - 138. If a circle be inscribed in a square, an equilateral triangle in the circle, and a circle in the triangle, the lines drawn from the vertices of the triangle to meet in any point of the inner circle, and the radius of that circle, have their squares together equal to the given Square. - 139. If in any circle the side of an inscribed hexagon be produced till it becomes equal to the side of an inscribed square, a tangent drawn from the extremity, without the circle, shall be equal to the side of an inscribed octagon. - - Aº 352 GEOMETRICAL ExERCISES 140. Let two straight lines be drawn from any point within a circle to the circumference: describe a circle, which shall touch them both, and the arc between them. - *gs - 141. In a given triangle having inscribed a circle, inscribe another circle in the space thus intercepted at one of the angles. * - 142. Two equal circles touch each other, and a straight line touches both, describe a circle which shall touch the straight line and circles. º - 143. Let AB, AC, be the bounding radii of a quadrant; complete the square ABDC and draw the diagonal AD; then the part of the diagonal without the quadrant will be equal to the radius of a circle inscribed in the quadrant. - 144. On one radius of a quadrant describe a semicircle, and then draw a circle touching it and the quadrantal arc at the extremity of the other. * - - 145. A circle being inscribed in a quadrant, inscribe a circle in each of the remaining spaces of the quadrant. . 146. Quadrants are described on the sides of a right-angled isosceles triangle so as to touch each other at the vertex. Shew that the radius of the circle which touches the hypotenuse and each of the two quadrants, is equal to one-eighth of the hypotenuse. - 147. Two intersecting circles being given in position, inscribe a square in the space common to both. c 148. If two equal circles intersect, and the space common to them be bisected by the straight line joining their centers, shew how to inscribe a circle on one of the half spaces. . 149. If three equal circles touch each other; to compare the area of the triangle formed by joining their centers, with the area of the triangle formed by joining the points of contact. - 150. In a given segment of a circle inscribe an isosceles triangle, such that its vertex may be in the middle of the chord, and the base and the perpendicular together equal to a given line. 151. Inscribe three circles in an isosceles triangle touching each other, and each of them touching two of the three sides of the triangle. - - - X. - 152. In the fig. Prop. 10, Book IV, shew that the base BD is the side of a regular decagon inscribed in the larger circle, and the side of a regular pentagon inscribed in the Smaller circle. 153. If the points of intersection of the two circles be joined with the vertex of the triangle and with each other, another triangle will be formed similar and equal to the former. - 154. In the figure (Euc. IV. 10.), if the two equal chords of the smaller circle be produced to cut the larger, and these points of section be joined, another triangle will be formed having the property re- quired by the proposition. : º - 155. If A be the vertex, and BD the base of the constructed triangle, D being one of the points of intersection of the two circles . employed in the construction, and E the other, and A.E be drawn meeting BD produced in F, prove that FAB is another isosceles triangle of the same kind. . *r oN BOOK Iv. - - 353 156. The obtuse angle A CD in the figure (Euc. IV. 10.) is three times the angle at the vertex of the isosceles triangle ABD. - 157. If E be the second point of intersection of the two circles, the straight line joining A and E shall be parallel to CD. 158. In the figure (Euc. IV. 10), shew that the angles subtended at the center of the smaller circle by the side of the triangle nearest to that center, is double of each of the angles at the base of the triangle. 159. If from one extremity of the base (fig. Euc. IV. 10.) a per- pendicular be drawn upon the opposite side, this line will divide the angle at the base into two parts, one of which is three times the other. 160. In the fig. Prop. 10, Book IV, produce DC to meet the circle in F, and draw BF; then the angle ABF shall be equal to three times the angle BFD. * 161. Shew how to construct an isosceles triangle, whose vertical angle is three times either of the angles at the base. - 162. Describe an isosceles triangle so that four times either of the base angles is three times the vertical angle. 163. Describe an isosceles triangle having each of the angles at the base equal to one-eighth of the third angle. 164. Divide a right angle into five equal parts. 165. C and F being the points in which the smaller circle cuts the side AB and the larger circle, respectively; join AF, CF, and produce AF, BD, to meet in G; then CDGF'is a parallelogram. XI. 166. If a pentagon be equilateral, and three of its angles be equal, shew that it is equiangular. 167. Each diagonal of a regular pentagon is parallel to the side with which it is not conterminous. 168. If an equilateral pentagon have any two of its diagonals parallel to two of its sides, the other three diagonals are also parallel. to the other three sides, and the pentagon is equiangular. 169. If the alternate angles of a regular pentagon be joined, the figure formed by the intersection of the joining lines will itself be a regular pentagon. - - - 170. A watch-ribbon is folded up into a flat knot of five edges, shew that the sides of the knot form an equilateral pentagon. 171. The three points of intersection of the diagonals within the pentagon form an isosceles triangle, such that the base and one of the sides together, are equal to a side of the pentagon. 172. The sum of all the angles subtended by the sides of a pen- tagon at any point of the circle which circumscribes it, is the fifth part of sixteen right angles. Explain the apparent error when the point is at an angle of the pentagon. - • 173. If circles be circumscribed about the two small triangles in the figure (Euc. IV. 11.) which are equiangular with the triangle first constructed, find the magnitude of the angle at which the circles will intersect. ** 174. ABCDE is an equiangular and equilateral pentagon, AC and AD being joined intersect in F; shew that the circle passing through A, F, and D touches the sides AB, CD. § 175. Shew that the area of a regular pentagon is less than four times the area of the triangle formed by joining the extremities of two adjacent sides. - A A 354 GEOMETRICAL EXERCISES 176. Describe an equilateral and equiangular pentagon on a given straight line without assuming Euclid's construction of the pentagon. - 177. If ABCDE be any pentagon inscribed in a circle, and AC, J3D, CE, BA, EB be joined, then are the angles A.B.E, JPCA, CDB, DEC, EAP, together equal to two right angles. 178. The difference between the side and the diagonal of a regular pentagon is equal to the side of another regular pentagon, whose dia- gonal is equal to the side of the first. . 179. If C be the center of the circle, and CB be a perpendicular on one of the sides of the pentagon, shew that CB is equal to half the sum of a side of a decagon and the side of a hexagon in the same circle. 180. If from the extremities of the side of a regular pentagon inscribed in a circle, straight lines be drawn to the middle of the arc subtended by the adjacent side, their difference is equal to the radius: the sum of their squares to three times the square of the radius: and the rectangle contained by them is equal to the square of the radius. 181. Shew that the alternate angles of a decagon inscribed in a circle are equal to eight right angles. - 182. Prove that the sides of the inscribed hexagon and decagon are together equal to the triple chord of the decagon. 183. The rectangle under the sides of a regular inscribed hexagon and decagon, is equal to the difference of the squares inscribed on them. 184. If the sides of an equilateral and equiangular pentagon be . produced to intersect; the stellated figure which results, is equilateral and equiangular, and the straight lines joining the points of intersec- tion will form another equilateral and equiangular pentagon. 185. The square described upon the side of a regular pentagon in a circle, is equal to the square of the side of a regular hexagon, together with the square upon the side of a regular decagon in the same circle. . XII. 186. In a given circle inscribe three equal circles touching each other and the given circle.- 187. AB is the diameter, and 'C the center of a semicircle; shew that O the center of any circle inscribed in the semicircle, is equi- distant from C and from the tangent parallel to AB. .* 188. Three equal circles touch each other; shew that the dia- meter of the circle which circumscribes them, is sixteen times the diameter of the smaller circle which touches them all. - 189. Shew that if two circles be inscribed in a third to touch one another, the tangents at the points of contact will all meet in the same oint. g p 190. Upon the diameter of a semicircle two equal circles are de- scribed, each having the radius of the semicircle as diameter; and another circle is described touching both of these circles and also the semicircle. Shew that the radius of this last circle has its radius one- third of the radius of the semicircle. r - 191. Four equal circles touch each other and a given circle ex- ternally: compare the radius of one of the four circles with the radius of the given circle. -- r - 192. In a given circle inscribe five equal circles touching one another. * ON BOOK IV. 355 * 193. Shew that a circle may be touched by six circles of radius equal to itself which shall also touch one another. -- 194. If there be three concentric circles, whose radii are 1, 2, 3; determine how many circles may be described round the interior one, having their centers in the circumference of the circle, whose radius is 2, and touching the interior and exterior circles, and each other. XIII. 195. Produce the sides of a given heptagon both ways, till they meet, forming seven triangles; required the sum of their vertical angles. - 8. To convert a given regular polygon into another which shall have the same perimeter, but double the number of sides. 197. A point is taken within a polygon, shew that the straight lines joining the point to the angular points of the polygon, are together greater than half the sum of the sides of the polygon. 198. An infinite number of polygons can be inscribed in a circle so that their sides may be parallel to those of a given inscribed polygon with an even number of sides. Why is this not true if the number of sides be odd? - 199. When an equilateral polygon is described about a circle, it must necessarily be equiangular, if the number of sides be odd, but not otherwise. - 200. If straight lines be drawn bisecting the angles of a polygon of n sides, shew that if n – 1 of the bisectors pass through a fixed point, the remaining one will also pass through the same. How must the enunciation be altered when n = 37 201. Shew that a polygon which is inscribed in the larger, and circumscribed about the smaller of two concentric circles, must be a regular polygon. 202. If two circles touch one another externally, and in one of them any polygon be inscribed, shew that straight lines drawn from each of the angles of the polygon to the point of contact of the circles, will if produced, intersect the other circle in a series of points, by join- ing which a polygon will be formed similar to the former. 203. If a point be assumed in a regular polygon of n sides, from which perpendiculars are drawn to each of the sides, or the sides pro- duced, the sum of these perpendiculars is n times the radius of the inscribed circle. ,4 - 204. . In any plane equilateral and equiangular polygon of n sides, when n is an odd number, if lines be drawn from one of the angles to the extremities of the opposite side, the angle contained by these lines is equal to one nth of two right angles. 205. One of the angular points of an equilateral and equiangular polygon is joined by straight lines with all the rest, perpendiculars are let fall from the center of the polygon on these lines and the feet of successive perpendiculars are joined. Shew that the figure thus constructed is similar to the original polygon, and compare the areas of the two. . 206. The sum of the squares on the lines drawn from any point in the circumference of a circle to the angular points of a regular polygon of 2n sides inscribed in it, is equal to n times the square on the diameter. A A 2 GEOMETRICAL EXERCISES ON BOOK WI. PROPOSITION I. PROBLEM. To inscribe a square in a given triangle. Analysis. Let ABC be the given triangle, of which the base BC, and the perpendicular AD are given. A E. G I prº ‘B H - D K C Let FGHK be the required inscribed square. Then BHG, BDA are similar triangles, and GH is to GB, as AD is to AB, º but GF is equal to GH; therefore GF is to GB, as AD is to A.B. Let BF be joined and produced to meet a line drawn from A parallel to the base BC in the point E. Then the triangles B.G.F, BAE are similar, and AE is to AB, as GF is to GB, but GF is to GB, as AD is to AB; wherefore AE is to AB, as AD is to AB; hence AE is equal to A.D. - Synthesis. Through the vertex A, draw AE parallel to BC the base of the triangle, a' make AE equal to AD, . 4 • join EB cutting AC in F, - through F, draw FG parallel to BC, and FK parallel to AD; also through G draw GH parallel to A.D. Then GHKF is the square required. The different cases may be considered when the triangle is equi- lateral, scalene, or isosceles, and when each side is taken as the base. PROPOSITION II. THEOREM. If from the extremities of any diameter of a given circle, perpendiculars be drawn to any chord of the circle, they shall meet the chord, or the chord produced in two points which are equidistant from the center. First, let the chord CD intersect the diameter AB in L, but not at right angles; and from A, B, let AE, BF be drawn perpendicular to CD. Then the points F, E are equidistant from the center of the chord CD. . - Join EB, and from I the center of the circle, draw IG perpendi- cular to CD, and produce it to meet EB in H. - - ON BOOK WI. 357 sº - Then IG bisects CD in G.; (III. 2.) * and IG, A.E being both perpendicular to CD, are parallel. (I. 29.) Therefore BI is to BH, as IA is to HE; (VI. 2.) and BH is to FG, as HE is to GE; therefore BI is to FG, as IA is to 6 F; but BI is equal to IA ; therefore FG is equal to G.E. It is also manifest that DE is equal to CF. When the chord does not intersect the diameter, the perpendiculars intersect the chord produced. r PROPOSITION III. THEOREM. If two diagonals of a regular pentagon be drawn to cut one another, the greater segments will be equal to the side of the pentagon, and the diagonals will cut one another in eactreme and mean ratio. Let the diagonals AC, BE be drawn from the extremities of the side AB of the regular pentagon ABCDE, and intersect each other in the point H. Then B E and AC are cut in extreme and mean ratio in H, and the greater segment of each is equal to the side of the pentagon. Let the circle ABCDE be described about the pentagon. (Iv. 14.) Because EA, AB are equal to AB, BC, º they contain equal angles; J therefore the base EB is equal to the base AC, (1. 4.) and the triangle EAB is equal to the triangle CBA, - and the remaining angles will be equal to the remaining angles, each to each, to which the equal sides are opposite. Therefore the angle BAC is equal to the angle ABE; and the angle A HE is double of the angle BAH, (1. 32.) but the angle EAC is also double of the angle BAC, (VI. 33.) therefore the angle HAE is equal to A.H.E, and consequently HE is equal to EA, (I. 6.) or to AB. And because BA is equal to AE, the angle ABE is equal to the angle AEB; but the angle ABE has been proved equal to BAH: therefore the angle BEA is equal to the angle BAH; and ABE is common to the two triangles A.B.E, ABH; therefore the remaining angle BAE is equal to the remaining angle AHB; * ~ 358. GEOMETRICAL EXERCISEs and consequently the triangles A.B.E, ABH are equiangular; therefore EB is to BA, as AB to BH : but BA is equal to EH, therefore EB is to EH, as EH is to BH, but BE is greater than EH; therefore EH is greater than HB; --- therefore B.E has been cut in extreme and mean ratio in H. Similarly, it may be shewn, that AC has also been cut in extreme and mean ratio in H, and that the greater segment of it CHis equal to the side of the pentagon. . . . T. PROPOSITION IV. PROBLEM. Divide a given aro of a circle into two parts which shall have their chords in a given ratio. - - Analysis. Let A, B be the two given points in the circumference of the circle, and C-the point required to be found, such that when the chords A C and BC are joined, the lines AC and BC shall have to one another the ratio of E to F. Draw CD touching the circle in C; join AB and produce it to meet CD in D. Since the angle BAC is equal to the angle BCD, (IIE, 32.) and the angle CDB is common to the two triangles DBC, DAC; therefore the third angle CBD. in one, is equal to the third angle DCA in the other, and the triangles are similar, - therefore AD is to DC, as DC is to DB ; (VI. 4.) hence also the square on AI) is to the square on DC as AD is to BI), (VI. 20. Cor.) - But AD is to AC, as DC is to CB, (VI. 4.) and AD is to DC, as AC to CB, (v. 16.) also the square on AD is to the square on DC, as the square on A C is to the square on CB; but the square on AD is to the square on DC, as A.D. is to DB ; wherefore the square on AC is to the square on CB, as AD is to BD; * but AC is to CB, as E is to F, (constr.) - therefore AD is to DB as the square on E is to the square on F. • - Henee the ratio of AD to DB is given, - and AB is given in magnitude, because the points A, B in the cir- cumference of the circle are given. - Wherefore also the ratio of AD to AB is given, and also the mag- nitude of AD, - . Synthesis. Join A.B and produce, it to D, so that AD shall be to BD, as the square on E to the square on F. & From D draw DC to touch the circle in C, and join CB, CA. Since AD is to DB, as the square on E is to the square on F, (constr.) and AD is to DB, as the square on AC is to the square on BC; , therefore the square on AC is to the square on BC, as the square on E is to the square on F, - - and AC is to BC, as E is to F. on Book VI. 359 PROPOSITION. W. THEOREM, The diagonal and the side of a square are incommensurable. Let ABCDºbe a square, and AC its diagonal. Then the side AB and diagonal AC are incommensurable; or, there is no line which will exactly measure 4.B and 40. - A, IB F. IE G. F. k 6. D, On AC, take AE equal to the side AB, join BE, and at E. draw EF perpendicular to AC and meeting BC in F. - Then EC the difference between the diagonal A-C and the side AB of the square, is less than AB, or BC (Euc, I, 19.) and CE, EF, FB, may be proved to be equal to one another; also Cº., EF are the adjacent sides of a square whose diagonal is FC. On FC take FG equal to CE, and join E.G. Then, as in the first square, the difference CG between the diagonal JFC and the side E6 or EF, is less than the side EC. . Hence EC the difference between the diagonal and the side of th given square, is contained twice in the side AC or BC with a re- mainder C6 : and CG is the difference between the side CE and the diagonal CF of another square. 6- By proceeding in a similar way, it may be shewn, that GG the difference between the diagonal CF and the side CE, is contained twice in the side CE with a remainder: and the same relations may be shewn to exist between the difference of the diagonal and the side of every square of the series of squares which is so constructed. Hence, therefore, as the difference of the side and diagonal of every square of the series, is contained twice in the side with a re- mainder; - It follows, that there is no line which exactly measures the side and diagonal of a square. - PROPOSITION VI. PROBLEM. . From the obtuse angle of an obtuse-angled triangle, *t is required to draw a line to the base which shall be a mean proportional between the segments of the base. ... , Analysis. Let A BC be an obtuse-angled triangle, having the base BC, and BAC the obtuse angle. - Let AD be drawn from A to meet the base BC in D, so that AD . be a mean proportional between AD and DB the segments of the ase B 0. - Produce AD to E making DE equal to AD. - Then because the rectangle contained by BD, DC is equal to the rectangle contained by AD, DE: a circle can be described passing through the points A, B, E, C. (Euc. III. 35.) 360 - GEOMETRICAL ExeRCISEs Let this circle be described, and let O be its center; join A0, OD: then 0D is at right angles to AE, (Euc. III. 3.) - and ADO is a right-angled triangle. `-- Hence the semicircle described on A0 as a diameter will pass through the point D, (Euc, III. 31.) Therefore the point D is determined to be that point in the base, where the circumference of the semicircle described upon the radius of the circumscribing circle of the triangle, intersects the base. If the circle be completed, it will intersect the base in another point G, so that AG is also a mean proportional between B6 and G.C. Synthesis. Describe a circle about the given triangle ABC, and let O be the center: join AO, and on AO as a diameter describe a semicircle 4D0 intersecting the base BC in D, join AD, then AD is a mean proportional between AD and D.B. Produce AD to meet the circumference in E, and join O.D. Then because OD is at right angles to A.E, and is drawn from the center 0, * . - therefore AD is equal to D.E. (Euc. III. 3.) And because AE, BC intersect each other within the circle; the rectangle BD, DC is equal to the rectangle AD, DE, (Euc. III. 35.) - - but DE is equal to AD, therefore the rectangle BD, DC is equal to the square on AD : and AD is a mean proportional between BD and DC. (Euc. VI. 17.) If A G be joined, then AG can be shewn to be a mean proportional between BG and G.C.; which is a second solution of the Problem. Cor. 1. If the given triangle ABC become right-angled at A, the center O of the circle will bisect the base BC, and the lines AD, AG will coincide at the second point D, where the circle cuts the base BC, and be at right angles to BC, and AD will be a mean proportional between BD and D.C. CoR. 2. If the triangle ABC be acute-angled at A, the problem is impossible; as it is obvious that the circle which circumscribes the triangle ABC, will have its center 0 within the triangle, and the semicircle described upon the radius AO as a diameter cannot intersect the base BC of the triangle. -- - If, however, a tangent AD be drawn to touch the circumscribing circle at the point A, and CB be produced to meet the tangent AD in the point D: the rectangle contained by DC, DB is equal to the square on DA : (Euc, III. 36.) and DA is a mean proportional between DC and DB (Euc. v.I. 17.) Hence a straight line A.D has been drawn from A, an angle of an acute-angled triangle to meet the base BC pro- duced in D, so that AD is a mean proportional between CD and D.B. on Book VI. 361. - - I. - - 7. If A CB, ADB be two triangles upon the same base AB, and 'between the same parallels, and if through the point in which two of the sides (or two of the sides produced) intersect two straight lines, be drawn parallel to the other two sides so as to meet the base AB (or. AB produced) in points E and F. Prove that A F- B.F. 8. In the base AC of a triangle ABC take any point D; bisect AD, DC, AB, BC, in E, F, G, H respectively : shew that EG is equal to HF. 9. If, in similar triangles, from any two equal angles to the opposite sides, two straight lines be drawn making equal angles with the homologous sides, these lines will have the same ratio as the sides on which they fall, and will also divide those sides proportionally. 10. BD, CD are perpendicular to the sides AB, A C of a triangle ABC, and CE is drawn perpendicular to AD, meeting AB in E: shew that the triangles ABC, ACE are similar. - 11. In any triangle, if a perpendicular be let fall upon the base from the vertical angle, the base will be to the sum of the sides, as the difference of the sides to the difference or sum of the segments of the base made by the perpendicular, according as it falls within or with- out the triangle. tº 12. If two triangles on the same base have their vertices joined by a straight line which meets the base or the base produced; the parts of this line between the vertices of the triangles and the base, are in the same ratio to each other as the areas of the triangles. 13. In the triangle ABC there are drawn AD bisecting BC, and EF parallel to BC and cutting AB in E and AC in F. Shew that JBF and CE will intersect in AD. ! -- 14. If the side BC of a triangle ABC be bisected in D, and the angles ADB, ADC be bisected by the straight lines DE, DF, meeting AB, A C in E, F respectively, shew that EF is parallel to BC. 15. APB, CQD are two parallel right lines, and AP is to PB as J)(Q is to QC, prove that the right lines PQ, AD, BC meet in a point. 16. CAB, CEB are two triangles having a common angle CBA, and the sides opposite to it CA, CE, equal; if BAE be pro- duced to D, and ED be taken a third proportional to BA, AC, then shall the triangle BDC be similar to the triangle BAC. w 17. From a point E in the common base of two triangles ACB, ADB, straight lines are drawn parallel to AC, AD meeting BC, BD in F and G, shew that the lines joining F, G and C, D will be parallel. 18. AB is a given straight line, and D a given point on' it; it is required to find a point P, in AB produced, such that AP is to PB as AD is to DB. 19. ABb, AeC are two given straight lines, cut by two others JBC, be, so that the two triangles ABC, Abo may be equal; the lines BC, bo, divide each other proportionally. 20. If AB, A C be two given straight lines, and AD be taken on AB, and A.E on AC, such that AB = m. AD, and AC = m. AE; join B.E, CD to meet in F; shew that DE = (m+ 1). D.F. 21. ABC is a triangle; Ab, Ac are taken on AB, AC res- pectively, such that AB = n. Ab, and AC = n. Ac; Cb, Bo meet in D, and AD produced meets BC in E: shew that 2. AE= (n+1). AD, and E is the middle point of BC. - 22. AB, A C are two straight lines, B and C given points in the 362 GEOMETRICAL EXERCISES same; BD is drawn perpendicular to AC and DE perpendicular to AB; in like manner CF is draw perpendicular to AB, and FG to AC, shew that EG is parallel to BC. 23. Lines drawn from the extremities of the base of a triangle intersecting in the line joining the vertex with the point of bisection of the base, cut the sides proportionally: and conversely. 24. ABC is any triangle, D any point in AB produced; E the point in BC, such that CE: EB :: AD: BD. Prove that DE produced will bisect A. C. - - 25. In a given triangle draw a straight line parallel to one of the sides so that it may be a mean proportional to the segments of the base. 26. Through E, F, two points on the side AB of a triangle ABC, so taken that A.E. is equal to BF, two straight lines EG, BH are drawn respectively parallel to the other sides and meeting them in G and H.; if GH be joined, it will be parallel to AB. 27. If the three sides of a triangle be bisected, the lines which join the points of bisection will divide the triangle into four equal triangles, each of them similar to the whole triangle. - 28. If the side BC of a triangle ABC be bisected in D, and the angles ADB, ADC be bisected by the straight lines DE, DF, meeting AB, AC in E, F respectively, shew that EF is parallel to BC. 29. A straight line drawn through the middle point of one side of a triangle divides the two other sides, the pne internally and the other externally in the same ratio. - 30. If two of the interior angles of a triangle ABC be bisected by the lines COE, BOD intersecting in 0, and meeting the opposite sides in the points E and D, prove that - OD: OB:: AD : AB, and OC: OE:: A C : AB. - 31. If the vertical angle CAB of a triangle ABC be bisected by AD, to which the perpendiculars CE, BF, are drawn from the remaining angles; bisect the base BC in G, join GE, GF, and prove these lines equal to each other. - 32. If a side BC of a triangle ABC be bisected by a straight line which meets the sides AB, A C (produced if necessary) in D and E respectively; the line AE will be to EC, as AD to D.B. 33. The base AB of an isosceles triangle ABC is produced both ways to A' and B, so that AA’. BB = AC*; shew that the triangles . A'A C, BBC are similar to one another. - 34. To find a point P in the base BC of a triangle produced, so that PD being drawn parallel to AC, and meeting AC produced to D, A C : CP:: OP: PD. 35. If from the extremities of the base of a triangle, two straight lines be drawn, each of which is parallel to one of the sides, and equal to the other, the straight lines joining their other extremities with the other extremities of the base, will cut off equal segments from the sides, and each of these will be a mean proportional between the other two segments. • 36. A straight line CD is drawn bisecting the vertical angle C of a triangle ACB, and cutting the base AB in D; also on AB pro- duced a point E is taken equidistant from C and D: prove that A E. BE's DE". '' 37. ABC is an equilateral triangle; E any point in A C; in BC produced take CD = CA, CF= CE, AF, DE intersect in H; then JHC': A C :: A C : A C+ E.C. - on Books vi. -, 363 - 38. ABC is, an isosceles triangle; draw CE perpendicular to the base AB; draw. ADF' intersecting ØB, in D, and CB in F; then I) E: OE:: OA – FF: CA + CF. 39. From an angle of a triangle a line is drawn to the middle point of the opposite side. And through the point of bisection of this line another is drawn from either angle to the side subtending it. Prove that the latter line divides this side into segments which are as 2 to 1. ' * . 40. Determine the point in the produced side of a triangle, from which a straight line being drawn to a given point in the base, shall be intersected by the other side of the triangle in a given ratio. * 41. If from two points P, Q, four perpendiculars be dropped upon the straight lines AB, AC, such that the perpendiculars are pro- portionals; shew that P and Q lie in the same straight line through A. , 42. If one side of a triangle be produced, and the other shortened by equal quantities, the line joining the points of section will be di- vided by the base in the inverse ratio of the sides. 43. If the triangle ABC has the angle, at C a right angle, and from G a perpendicular be dropped on the opposite side intersecting it in D, then AD : DB :: A C*: CB". . . . * 44. 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. : 45. If on the two sides of a right-angled triangle squares be described, the lines joining the acute angles of the triangle and the opposite angles of the squares, will cut off equal segments from the sides; and each of these equal segments will be a mean proportional between the remaining segments. . - 46. ABC is a right-angled triangle, having a right angle at C; find a point P in the hypotenuse, so situated, that PA may be half of the perpendicular dropped from Pupon BC the base. - 47. Determine that point in the base produced of a right-angled triangle from which the line drawn to the angle opposite the base, shall have the same ratio to the base produced, which the perpendicular has to the base itself. 48. 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. 49. From B the right angle of a right-angled triangle ABC, Bp is let fall perpendicular to AC, from p, pg is let fall perpendicular to B.A., &c.; prove that r º - Bp +pg + &c. : AB :: A B+ AC: BC. 50. If the triangle A CB has a right angle C, and AD be drawn bisecting the angle 4, and meeting CB in D, prove that \ A C*: AD :: BC: 2.BD. - 51. In any right-angled triangle ABC, (whose hypotenuse is AB) bisect the angle A by AD meeting CB in D, and prove that g - 2A C*: A C*- CD":: BC: CD. 52. ABC is a right-angled triangle, CD a perpendicular from 364 - GEOMETRICAL ExERCISEs the right angle upon AB; shew that if AC is double of BC, BD is one-fifth of A.B. ** - - 53. If through C the extremity of the hypotenuse BC of a right-angled triangle ABC there be drawn two straight lines bisecting the internal and external angles at C, and meeting BA produced in D º: respectively, AC will be a mean proportional between AD and AB. - ** . 54. If F be a point in the side CB, and CD, FE be perpendi- culars on the hypotenuse AB, then AD. AE4 CD. EF=A C*. III. 55. Triangles and parallelograms of unequal altitudes are to each other in the ratio compounded of the ratios of their bases and altitudes. 56. If triangles AEF, ABC have a common angle A, triangle ABC': triangle AEF':: A.B. A C : A E. A.F. . - / 57. Construct an isosceles triangle equal to a given scalene triangle and having an equal vertical angle with it. - - 58. On two given straight lines similar triangles are described. Bequired to find a third, on which, if a triangle similar to them be described, its area shall equal the difference of their areas. 59. In the triangle ABC, AC = 2. BC. If CD, CE respectively bisect the angle C, and the exterior angle formed by producing AC; prove that the triangles CBD, A CD, ABC, CDE, have their areas as 1, 2, 3, 4. - 60. If two sides of a triangle be bisected and the points joined with the opposite angles, the joining lines shall divide each other proportionally. And the triangle formed by the joining lines and the remaining side shall be equal to a third of the original triangle. 61. A square is inscribed below the base of an isosceles triangle, prove that if the vertex be joined to the corners of the square, the middle segment of the base will be to the outer one in double the ratio of the perpendicular on the base to the base. 62. D is the middle point of the base BC of an isosceles triangle, CF 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 of BD to D.A. . 63. If a straight line be drawn through the points of bisection of any two sides of a triangle, it will divide the triangle into two parts which are to each other as 1 to 3. 64. If perpendiculars be drawn from the extremities of the base to a triangle on a straight line which bisects the angle opposite to the base, the area of the triangle is equal to the rectangle contained by either of the perpendiculars, and the segment of the bisecting line between the angle and the other perpendicular. 65. If the angles at the base of an isosceles triangle (which are double of the vertical angle) be bisected by lines meeting the opposite sides, and the points of intersection with the sides be joined, then the joining line will divide the triangle into two parts, which have the same constant ratio to one another, as one of the sides of the triangle bears to the base. º - - 66. The square on the line bisecting the vertical angle of any triangle is a mean proportional between the differences of the squares on each side containing that angle, and the square on the adjacent Segment of the base. * * ON BOOK WI. 365 & 67. If a line be drawn bisecting the vertical angle BAC of a triangle, and meeting the base BC in P, and if from P any line be drawn meeting B.A, CA (one of them being produced) in Q and R respectively; shew that the rectangle PQ, PB will be to the rectangle PR, PC as BQ is to CR. * . . 68. If a triangle be inscribed in another having one side parallel to a side of the latter, its area shall be to that of the larger triangle as the rectangle contained by the segments of one side to the square of the whole side. . 69. . If perpendiculars be raised upon the middle points of the sides of a triangle, and respectively equal to half of those sides, and the extremities of those perpendiculars be joined; the sum of the Squares on these last lines is equal to the sum of the squares on the sides of the triangle together with six times its area. 70. If the interior angle CAD of a triangle BAC be bisected by a straight line AE which likewise cuts the base BC produced in E; shew that the rectangle B.E, EC is equal to the rectangle BA, AC together with the square on A.E. 71. D, E, are the middle points of the sides CA, CB of a triangle; join D and E, and draw A.E, BD, intersecting in 0; then shall the areas of the triangles DOE, EOB, BOA, be in geometrical progression. 72. Through the point of intersection D of two straight lines drawn from two angles of a triangle to bisect the opposite sides, a straight line EDFG is made to pass, meeting two sides in E and F and the remaining sides produced in G.; prove that DG. DF-DE. DG = D.E. D.F. - - 73. D is any point in the side AC of a triangle ABC, CE drawn . parallel to BD meets AB produced in E, AF and AG are mean pro- portionals between AD, A C and AB, AE respectively; shew that the triangle AGF'is equal to the triangle ABC. - 74. 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: triangle BCD :: A B : BD. - - 75. 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 and F, join EF; shew that the triangle AEF is a mean proportional between the triangles FBD, EDC. 76. If in the figure Euc. IV. 10, the straight lines DC, BA be produced to meet the circle again in E, F, and EF be joined: shew that the triangle ABD is a mean proportional between CEF and BCD. 77. If a straight line A B be produced to a point C, so that AB is a mean proportional between AC and CB; then the square on A.B.: Square on BC :: A B : AB - BC. - 78. If the straight line AB be perpendicular to the two parallel straight lines AD, BE; and if from C the middle point of AB; CD, and CE be drawn at right angles to each other to meet AD, BE in D and E, and if DE be joined; prove that the area of the triangle DCE is equal to the sum of the areas of ADC and CBE. 79. If ABC be a right-angled triangle having the right angle at B, and squares ABDE, CBFG, be described on the sides AB, CB; and DE, FG be produced to meet AC produced in the points H and K, the triangle ABC will be a mean proportional between the triangles A EH, CGK. - f *. 366. * GEOMETRICAL EXERCISES 80. Draw a straight line such that the perpendiculars let fall from any point in it on two given lines may be in a given ratio. 81. It is required to cut off a part of a given line so that the part cut off may be a mean proportional between the remainder and another given line. - 82. It is required to divide a given finite straight line into two parts, the squares of which shall have a given ratio to each other. te 83. Through a given point within a given angle draw a straight line such that the segments intercepted between the given point and the lines, which form the anglès, shall be in a given ratio to ‘One another. - ſº 84. If a straight line be divided in two given points, determine a third point, such that its distances from the extremities, may be proportional to its distances from the given points. 85. Find a point the distances of which from three given points, not in the same straight line, are proportional to p, q, and r respectively, the four points being in the same plane. - 86. Two equal straight lines meet at one extremity; draw a straight line passing through their common extremity so that the perpendiculars upon it from the other extremities may be in a given ratio. - 87. A point and a straight-line being given in position, to draw a line parallel to the given line, so that all the lines drawn through the given point may be divided at it in a given ratio. 88. C is a point in a given line AB, it is required to determine a point P, so that AP and BP together may be equal to a given line, and the angle APB be bisected by PC. 89. From two straight lines cut off two parts having a given ratio, so that the sum of the squares of the remainders may be equal to a given square. - 90. Two straight lines and a point between them being given in Bosition, to draw two lines from the given point to terminate in the lines given in position, so that they shall contain a given angle, and ave a given ratio. l 91. If a perpendicular be drawn from a given point of a given line; find a point in it, such that lines drawn from it to the extre- mities of the given line, may be in a given ratio. - 92. Find a point, such that the perpendiculars let fall from it on three straight lines given in position may be in a given ratio. 93. Let DACEB be points in a straight line such that A C : CB :: A D : DB, and AE = EB; shew that rectangle DCE equals rectangle ACB. 94. If two lines intersecting each other, are each terminated by two unlimited lines given in position, the ratio of the rectangles con- tained by their respective segments will be the same with the ratio of the rectangles made by the segments of any other two intersecting each other which are similarly terminated, and are respectively parallel to the former. * * 95. Four lines AB, CD, EF, GH, drawn in any direction, inter- sect in the same point P; then if from any point in one of these lines, another be drawn parallel to the next in order, cutting the remaining ON BOOK WI. - 367. two in p and q ; the ratio of mp : pg is the same in whichever line the point m is taken. - w 96. In the figure to Proposition 47, Book I. draw through G a right line, so that the sum of the perpendiculars falling on it from B and C may be equal to BC. * - V. 97. AD is drawn bisecting the angle BAC, and AE is drawn at right angles to AD. If BDCE be any straight line cutting AB, AD, AC, AE respectively in the points B, C, D, E, shew that B.E, BC, BD are in harmonical proportion. Af 98. If a straight line be harmonically divided, and the angle con- tained by lines drawn from the extremity and one point, be bisected by a line drawn from the other point; that line shall be at right angles to the line which is drawn from the other extremity. 99. If diverging lines cut a straight line, so that the whole is to one extreme, as the other extreme is to the middle part, they will in- tersect every other intercepted line in the same ratio. . 100. The diameter CD of a circle cuts a chord AB at right angles. If E be any other point on the circle and the lines EA, EB, EC, ED (produced if necessary) meet any other line in the points a, b, c, d; shew that ac : cb :: ad : db. * - 101. If a given straight line A B be divided into any two parts in the point C, it is required to produce it, so that the whole line pro- duced may be harmonically divided in C and B. & 102. To divide a given line in harmonical proportion. 103. To find, by a geometrical construction, an arithmetic, geo- metric, and harmonic mean between two given lines. 104. If through the point of bisection of the base of a triangle any line be drawn, intersecting one side of the triangle, the other pro- duced, and a line drawn parallel to the base from the vertex, this line shall be cut harmonically. - g 105. If through a given point within a triangle lines be drawn from the angles to the opposite sides, and the points of section be joined, the three first drawn lines will be harmonically divided. 106. If from any point four straight lines be drawn, so that a line parallel to one of them falling upon the other three may be bisected by the middle one of these, then will the four lines cut any line in harmonic proportion. - 107. AB, CD are two parallel straight lines, AD and BC are joined meeting in E, and AC and BD meeting in F. Shew that the straight line passing through E and F, bisects AB and CD, and is itself divided harmonically. - 108. BE, AC are parallel lines; F, G, H, &c. a series of equi- distant points in AC; draw Bfh cutting BE in B, and EF, EG, E.H., &c. in f, g, h, &c. Bf, Bg, Bh, &c. are in harmonic progression. VI. 109. It is required to bisect any triangle (1) by a line drawn parallel, (2) by a line drawn perpendicular, to the base. . 110. To divide a given triangle into two parts, having a given ratio to one another, by a straight line drawn-parallel to one of . its sides. … 368 - GEOMETRICAL EXERCISES 111. Bisect a triangle by a straight line drawn through a point within or without the triangle. f 112. From a given point in the side of a triangle, to draw lines to the sides which shall divide the triangle into any number of equal arts. - .* p 113. Any two triangles being given, to draw a straight line parallel to a side of the greater, which shall cut off a triangle equal to the less. - 114. Trisect a triangle by straight lines parallel to one of its sides. 115. Through a given point, either within or without a given triangle, to draw a straight line, which shall cut off from the triangle any part required. * * VII. 116. The rectangle contained by two lines is a mean propor- tional between their squares. - * 117. Describe a rectangular parallelogram which shall be equal to a given Square, and have its sides in a given ratio. - 118. If from any two points within or without a parallelogram, straight lines be drawn perpendicular to each of two adjacent sides and intersecting each other, they form a parallelogram similar to the former. - 119. It is required to cut off from a rectangle a similar rectangle which shall be any required part of it. - - - 120. If from one angle A of a parallelogram a straight line be drawn cutting the diagonal in E and the sides in P, Q, shew that AE?– PE. EQ. - 121. The lines drawn from an angle of a parallelogram perpendi- cular to the sides containing the opposite angle, or to those sides pro- duced, are inversely proportional to these sides. 122. If through any points in the diagonal of a parallelogram a straight line be drawn meeting two opposite sides of the figure, the segments of this straight line will have the same ratio as those of the diagonal. - - 123. If lines be drawn through the angles of a parallelogram so as to form a second parallelogram, the segments which these taken in Order produced or not, cut off from the sides taken in order of the first produced or not, shall be proportionals. - **- 124. If ABCD be a parallelogram; P, Q, points in a line which is parallel to AB; also if PA, QB meet in R, and PD, QC in S; then JES is parallel to A.D. * 125. In the figure Euc. VI. 26, either of the complements is a mean proportional between the parallelograms about the diameter. 126. Shew that if the diameters of the complements of the paral- lelogram, about the diameter of a parallelogram, be drawn, the two which do not meet in the diameter of the parallelogram will, if pro- duced, meet in it produced. VIII. 127. In a given circle place a straight line parallel to a given straight line, and having a given ratio to it; the ratio not being greater than that of the diameter to the given line in the circle. 128. In a given circle place a straight line, cutting two radii which ON BOOK vſ. 369 are perpendicular to each other, in such a manner, that the line itself may be trisected. - - 129. AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary; if from any point C of AB, a perpendicular be drawn to AB meeting AP and BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF. - t 130. If from the extremity of a diameter of a circle tangents be drawn, any other tangent to the circle terminated by them is so divided at its point of contact, that the radius of the circle is a mean proportional between its segments. *- 131. From a given point without a circle, it is required to draw a straight line to the concave circumference, which shall be di- vided in a given ratio at the point where it intersects the convex circumference. º 132. From what point in a circle must a tangent be drawn, so that a perpendicular on it from a given point in the circumference may be cut by the circle in a given ratio 2 - 133. Through a given point within a given circle, to draw a straight line such that the parts of it intercepted between that point and the circumference, may have a given ratio. 134. Let the two diameters AB, CD, of the circle ADBC be at right angles to each other, draw any chord EF, join CE, CF, meeting AB in G and H; prove that the triangles CGH and CEF are similar. 135. A circle, a straight line, and a point being given in position, required a point in the line, such that a line drawn from it to the given point may be equal to a line drawn from it touching the circle, What must be the relation among the data, that the problem may become porismatic, i.e. admit of innumerable solutions? • 136. If from a point without a circle two lines be drawn touching . the circle, and from one of the points of contact a perpendicular be let fall on the diameter passing through the other, it will be bisected by a straight line drawn from the point without the circle to the farthest extremity of the diameter. • . . - 137. If one chord in a circle bisect another, and tangents drawn from the extremities of each be produced to meet, the line joining their points of intersection will be parallel to the bisected chord. 138. If two chords in a circle intersect each other, and the two segments of one chord have the same ratio to each other as the two segments of the other chord; a straight line bisecting the angle con- tained by the two segments which form corresponding terms of the ratios, will pass through the center of the circle. - 139. If AB be any chord of a circle, and from any point C in the circumference a line CD be drawn perpendicular to AB, and be pro-. duced to meet the circle in E, and from A a perpendicular AF be drawn to the tangent at C, the triangles A.EC, FAD are similar. 140. AB is any chord of a circle, P any point in the circumference of the circle, Pm perpendicular to AB is produced to meet the circle in Q, An is drawn perpendicular to the tangent at P; shew that the triangle n Am is similar to PAQ. - * 141. To draw a line from the vertex of a triangle to the base, which shall be a mean proportional between the whole base and one segment. - - B B 370 - GEOMETRICAL ExERCISEs 142. If a straight line touch a circle, and a perpendicular be drawn from the point of contact upon any diameter; and if from the extremities of the same diameter, and from the center, perpendiculars be raised upon the diameter and produced to meet the line which touches the circle, these four perpendiculars are proportional. 143. A point is taken in the diameter of a circle, and another point in the diameter produced, so that the radius is a mean propor- tional between their distances from the center; prove that the lines drawn from these points to any, the same point in the circumference, are always in the same ratio. 144. If to any point C in a circle, a tangent ACB be drawn, and from A, C, B perpendiculars be drawn to the diameter DE, and CF be produced to Y; the rectangle contained by AB, CI is equal to that contained by DE, GH. 145. Through a point 0 outside or inside a circle, two straight lines are drawn making chords AB, CD; join A C, BD, AD, B.C. Prove that OA. 00:: OB. OD :: AC": BD", and OA. O.D. : OB. OC :: A D* : B0°. 146. If tangents be drawn from the extremities of any diameter of a circle, the portion of any other tangent, intercepted between them, is divided at the point of contact so that the radius of the circle is a mean proportional between its segments. 147. If from any point of a circle lines be drawn through the extremities of a chord, intersecting a diameter at right angles to the chord, shew that the radius is a mean proportional between the dis- tances of the points of intersection from the center. 148. The perpendicular drawn from any point of a circle upon a chord of the circle, is a mean proportional between the perpendiculars drawn from the same point upon the tangents at the extremity of the chord. - - 149. If through the middle point of any chord of a circle two chords be drawn, the lines joining their extremities shall intersect the first chord at equal distances from its extremities. 150. Given a circle and two points A, B, exterior to it, find a point Y in the circle such that if XA, XB, be drawn cutting the circle in P, Q, PQ shall be parallel to A.B. 151. In the diameter BB' of a circle produced, a point A is taken such that AB is one-fourth of the radius; from A a tangent AC is drawn to the circle ; find a point D in the circle such that AD shall be a mean proportional between AB and A.C. - 152. Let ABC be the diameter of any circle whose center is C, P any point in it, Q a point so taken that CP: CA :: CA : 00; take Many point in the circumference, join QM, AM, PM. Prove that AM always bisects the angle QMP. - 153. If a circumference be divided into any number of equal parts AB, BC, CD, DE, &c., and the straight lines AB, AC, AD, A.E., &c. be drawn, then AB : A C :: AB+ AD and A C : A D + AB :: AD : A C-H AE. 154. The perpendicular drawn from the center of a circle on the chord of any arc, is a mean proportional between half the radius and the line made up of the radius and the perpendicular drawn from the center on the chord of double the arc. - 155. To describe a circle through two given points to cut a * ... on Book v1. 371 straight line given in position, so that a diameter of the circle drawn fºugh the point of intersection shall make a given angle with the 6. - 156. From a given point without a given circle a line is drawn cutting the circle. It is required to draw from the same point another line also cutting the circle, so that the sum of the arcs intercepted be- tween these two lines shall be equal to a given arc. * 157. ABC is a plane curvilinear triangle formed by arcs of circles which meet in P; shew that the sum of its angles is equal to two right angles. IX. 158. Prove that there may be two, but not more than two, similar triangles in the same segment of a circle. - 159. If as in Euclid VI. 3, the vertical angle BAC of the triangle BAC be bisected by AD, and BA be produced to meet CE drawn parallel to AD in E; shew that AD will be a tangent to the circle described about the triangle EAC. - 160. If a triangle be inscribed in a circle, and from its vertex, lines be drawn parallel to the tangents at the extremities of its base, they will cut off similar triangles. - 161. If the line bisecting the vertical angle of a triangle be divided into parts which are to one another as the base to the sum of the sides, the point of division is the center of the inscribed circle. - 162. The rectangle contained by the sides of any triangle is to the rectangle by the radii of the inscribed and circumscribed circles, as twice the perimeter is to the base. - 163. If from the extremities of the base of a triangle, perpen- diculars be let fall on the opposite sides, and likewise straight lines drawn to bisect the same, the intersection of the perpendiculars, that of the bisecting lines, and the center of the circumscribing circle, will be in the same straight line. 164. ACB is a triangle in a semicircle whose diameter is AB, DEFG a straight line, perpendicular to AB at any point D in AB, cutting one side of the triangle, the circle, and the other side of the triangle produced, in the points E, F, G : then DE: DF:: DF: DG. 165. A CB is an isosceles triangle inscribed in a semicircle, of which the diameter is AB; from any point in AC produced, DE is drawn perpendicular-to AB, and cutting the semicircle in F, and CB in G. Prove that JEF is a mean proportional between EG and ED. . 166. AB is the diameter of a circle, D any point in the circum- ference, and C the middle point of the arc A.D. If AC, AD, BC be joined, and AD cut BC in E, the circle described about the triangle AEB will touch AC, and its diameter will be a third proportional to B0 and AB. - - 167. One circle is inscribed in an equilateral triangle ABC, and another is described about it; through A and B tangents are drawn to the outer circle meeting in P, and from Pa line PES is drawn touch- ing the inner circle in E, and meeting the outer circle in S; find the ratio of PS to AB. = - - 168. A tangent at any point C of a circle, center 0, meets the tangents at the extremities of a diameter AB in I, N. Prove that the B B 2. 372 GEOMETRICAL ExFRCISES radii of the circles inscribed in the triangles ACB, ION, are in the same ratio as the radii of the circumscribed circles. 169. If from the center of the circle circumscribing any triangle, perpendiculars be let fall upon the sides, the sum of these perpen- diculars is equal to the sum of the radii of the inscribed and cir- cumscribed circles. - 170. If a triangle ABC be inscribed in a circle, and a straight line BD be drawn parallel to the tangent at A, meeting AC or AC produced in D, then will AD : AB :: A B : A G. 171. If a circle be inscribed in a triangle, and another be de- scribed about it; the distance between their centers is a mean propor- tional between the radius of the circumscribed circle, and its excess above twice the radius of the inscribed circle. - 172. If from the angles of a triangle, lines be drawn to the points where the inscribed circle touches the sides; these lines shall intersect in the same point. 173. In a plane triangle, if the line joining the center of the cir- cumscribed circle and the point of intersection of the perpendiculars, be divided into parts, one of which is double of the other; (the smaller segment being estimated from the center of the circle) the point so determined is distant from the vertical angle, two-thirds of the line drawn from that angle to the bisection of the base. 174. If a circle be inscribed in a right-angled triangle and another be described touching the side opposite to the right angle and the produced parts of the other sides, shew that the rectangle under the radii is equal to the triangle, and the sum of the radii equal to the sum of the sides which contain the right angle. - 175. If a perpendicular be drawn from the right angle to the hypotenuse of a right-angled triangle, and circles be inscribed with- in the two smaller triangles into which the given triangle is divided, their diameters will be to each other as the sides containing the right angle. 176. If from 0 the center of the circle inscribed in a triangle ABC, OD, OE, OF be drawn perpendicular to the sides AB, AC, BC, and if OP be drawn parallel to the side AB, to any point P, and from P perpendiculars PQ, PR be drawn upon OE and OF, or those lines produced, prove that the rectangle contained by BC, OR is equal to the rectangle contained by AC, OQ. 177. About a triangle ABC describe a circle, and through the points A, B, C draw tangents DAF, F.B.E, ECD, forming a triangle DFE; shew that the rectangle DE, AF, is to the rectangle EC, DF, in the duplicate ratio of AB to CB. 178. In any triangle, the rectangle contained by the excess of half the perimeter above each of the two sides including an angle, is equal to the rectangle contained by the radius of the inscribed circle, and the radius of the circle which touches the base and the two sides produced. - . 179. If a circle be inscribed in any plane triangle, and a right line drawn from the point of the vertex to the point where the circle touches its base: prove that the middle point of that line, the center of the circle, and the middle point of the base of the triangle, are in the same straight line. 180. If the sides of a triangle be in arithmetic progression, and a and a, be the longest and shortest sides, and r and ri the radii of the inscribed and circumscribed circles, 6rri - aa. • ON BOOK WI. . . 373 181. BEF is a circle inscribed in the triangle ABC touching the sides BC, AC, in D, E, F, HIK is a circle touching AB in K and CB, CA, produced in H, I; in CH take CL= CA, and in CI take CM- CB; then FK = AM and 4AF. FB = AL. M.B. 182. A triangle ABC is inscribed in the circle BAC. Through the point B a line BL is drawn parallel to the tangent at A, and meeting AC in L; compare the areas of the triangles ABL and ABC. 183. G is the center of a circle in which an equilateral triangle ABC is inscribed; E is a point in the circumference AC; join EA, EB, EC, and draw EF perpendicular to AC, then EB" is greater than EA*4 EC* by 4. EG. E.F. - X. 184. In any triangle, inscribe a triangle similar to a given triangle. 185. From the vertex of an isosceles triangle two straight lines drawn to the opposite angles of the square described on the base, cut the diagonals of the square in E and F; prove that the line EF is parallel to the base. 186. Inscribe a square in a segment of a circle. - 187. Inscribe a square in a sector of a circle, so that the angular points shall be one on each radius, and the other two in the circumference. 188. Inscribe a square in a given equilateral and equiangular pentagon. - 189. Inscribe a parallelogram in a given triangle similar to a given parallelogram. - - 190. In a given rectangle inscribe another, whose sides shall bear to each other a given ratio. - 191. In a given segment of a circle to inscribe a similar segment. 192. The square inscribed in a circle is to the square inscribed in the semicircle :: 5: .2. - - 193. , If a square be inscribed in a right-angled triangle of which one side coincides with the hypotenuse of the triangle, the extremities of that side divide the base into three segments that are continued proportionals. - - 194. The square inscribed in a semicircle is to the square inscribed in a quadrant of the same circle :: 8: 5. 195. Shew that if a triangle inscribed in a circle be isosceles, having each of its sides double the base, the squares described upon the radius of the circle and one of the sides of the triangle, shall be to each other in the ratio of 4: 15. 196. APB is a quadrant, SPT a straight line touching it at P, PM perpendicular to CA; prove that triangle SCT : triangle ACB :: triangle A CB : triangle CMP. - 197. 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; shew that rº" = R*. 198. If R be the radius of the circle inscribed in a right-angled triangle ABC, right-angled at A ; and a perpendicular be let fall from A on the hypotenuse BC, and if r, r' be the radii of the circles in- scribed in the triangles ADB, A CD; prove that r* + r"= R". 199. If on the diameter of a semicircle, two equal semicircles be described, and in the curvilinear space a circle be inscribed touching the three semicircles, the radius of this circle shall be to the radius of either less-semicircle :: 2: 3. .." * 374 GEOMETRICAL EXERCISES 200. Given the perimeter of a right-angled triangle, it is required - to construct it, (1) If the sides are in arithmetical progression; (2) If the sides are in geometrical progression. 201. Given the vertical angle, the perpendicular drawn from it to the base, and the ratio of the segments of the base made by it, to con- struct the triangle. 202. Construct a triangle, having given the vertical angle, the radius of the inscribed circle, and the rectangle contained by the straight lines drawn from the center of the circle to the angles at the base. 203. Describe a triangle with a given vertical angle, so that the line which bisects the base shall be equal to a given line, and the angle which the bisecting line makes with the base shall be equal to a given angle. 204. Given the base, the ratio of the sides containing the vertical angle, and the distance of the vertex from a given point in the base; to construct the triangle. 205. Given the vertical angle and the base of a triangle, and also a line drawn from either of the angles, cutting the opposite side in a given ratio, to construct the triangle. 206. Upon the given base AB construct a triangle having its sides in a given ratio and its vertex situated in the given indefinite line CD. 207. Given the hypotenuse of a right-angled triangle, and the side of an inscribed square. Required the two sides of the triangle. - 208. To make a triangle, which shall be equal to a given triangle, and have two of its sides equal to two given straight lines; and shew that if the rectangle contained by the two 'straight lines be less than twice the given triangle, the problem is impossible. 209. In a right-angled triangle, given the sums of the base and the hypotenuse, and of the base and the perpendicular; to determine the triangle. -º- 210. In the triangle ABC, let D be the middle point of AB, and JE the foot of the perpendicular let fall from C on AB; then prove that, if CB – CA = DE, the three sides CA, AB, BC are in arithmetical progression. Also shew how such a triangle may be constructed. 211. Having given the base, the sum of the other two sides and the point in which the line bisecting the exterior angle at the vertex meets the base produced. Describe the triangle. 212. Given in any triangle the base, the ratio of the sides, and the distance between the points in which the internal and external bisec- $ors cut the base, construct the triangle. 213. Given the base, the vertical angle, and the ratio of the parts that the line bisecting the base is divided into, by the line joining the centers of the circumscribed and inscribed circles, to construct the plane triangle. - 214. Given the difference of the angles at the base, the ratio of the segments of the base made by a perpendicular from the vertex and the sum of the sides: construct the triangle. r 215. Given the base, the vertical angle, and theratio of the sides which contain that angle, Čonstruct the triangle. XII 216. If through the vertex, and the extremities of the base of a triangle, two circles be described, intersecting one another in the base or its continuation, their diameters are proportional to the sides of the triangle. ON BOOK WI. 375 217. If a tangent to two circles be drawn cutting the straight line which joins their centers, the chords are parallel which join the points of contact, and the points where the line through the centers cuts the circumferences. . . A 218. If from the centers of each of two circles exterior to one another, tangents be drawn to the other circle, so as to cut one another, the rectangles of the segments are equal. . 219. The circumference of one circle passes through the center 0 of another; and through A, one of the points of intersection, a radius AB is drawn to the first circle, meeting the second circle in C. Shew that the rectangle AB, A C, is equal to the square on 00. 220. If two circles cut each other; to draw from one of the points of intersection a straight line meeting the circles, so that the part of it intercepted between the circumference may be equal to a given line. 221. A straight line ABCD is terminated and trisected at the points where it meets the circumferences of two equal circles that cut each other, and it is intersected in E by FG which joins their centers; shew that A. E. : ED :: 2 EF'4 EG : 2 EG + EF. º 222. Two circles cut at right angles. A line is drawn from the center O of the first and cuts the second in A, B. If any radius OC be drawn, and AC, BC joined; the triangles OAC, OBC will be similar. g - - 223. The line A.D is divided into three equal parts in the points B, C; a circle is described with B as a center and BA as radius, and any circle cutting this is described with D as center. Shew that if a chord to both the circles be drawn from it through one of the points of intersection, it will be bisected by this point. - 224. If two circles cut each other and straight lines are drawn from the centers through one of the points of section to the circumfe- rences, they will be divided at that point in inverse proportion. 225. Two equal circles intersect, the center of one being on the circumference of the other; shew that their circumferences are divided in the ratio of two to one. - 226. If two circles cut each other so that the circumference of one may pass through the center of the other, and from that center, lines be drawn cutting both circumferences, the segments of the lines inter- cepted between the circumferences, will be to each other as the per- pendiculars let fall from the intersections with the inner circumference upon the line which joins the points of intersection of the circles. 227. If two circles cut each other, and a common chord ABCDE be drawn, meeting one circle in A and D, the other in B and E, and the line joining these points of intersection in C; then the square on BD shall be to the square on AE as the rectangle BC, CD, is to the rectangle A C, CE. . g 228. If two circles cut each other in any two points, and from either point diameters be drawn, the extremities of these diameters and the other point of intersection are in the same straight line. Also, if on the line joining the points of intersection, as a diameter, a circle be described, and any line be drawn from one extremity of this dia- meter cutting the two circles, the parts of this line intercepted between the circumference of the circle, and the circumferences of the other two circles, shall be proportional to the perpendiculars drawn in these circles from the extremity of the diameter. 229. Two circles intersect in A and B; AD, AD are diameters; 376. GEOMETRICAL ExERCISEs AC, AC" are chords, each of which touches the circle of which it is not a chord; the line AEE" bisects the angle DAD and cuts the -circles in: F and E’; then the common tangent to the circles is a mean propor- tional between the chords DE, DE"; and their common chord A.B is a mean proportional between the chords BC, BC". - 230. If two circles out each other and from one of the points of intersection a straight line be drawn cutting the circles, then the lines joining the new points of intersection with the other point of intersec- tion of the two circles, will be to each other as the radii of the circles. of which they are the chords. * * 231. If two circles touch each other externally and also touch a straight line, the part of the line between the points of contact is a mean proportional between the diameters of the circles. 232. Two circles touch each other internally at the point A, and from two points in the line joining their centers, perpendiculars are drawn intersecting the outer circle in the points B, C, and the inner in D, E: shew that AB : A C :: A D : AIE. 233. To draw a straight line which shall intersect two given circles, so that the chords intercepted by each circle may be equal to a given 6. 234. Let two circles touch one another in the point 0, and through this point let two chords P, OP, Q,00, be drawn meeting the circles in the points P., P., shew that O.P. OQ, = O.P. O.Q. 235. Two circles touch each other externally or internally and through the point of contact a straight line is drawn cutting the two circles, prove that the parts of the line intercepted are proportional to the diameters of the circles in which they may lie. - 236. If two circles touch each other and from the point of contact any two straight lines be drawn cutting the circles, and their extre- mities be joined, the triangles so formed are in the duplicate ratios of the diameters of the circles. - - * . . . 237. From the point T of intersection of the common tangents to two circles whose radii are as 2 to 1, any two lines are drawn cutting. the circles in A, B, and a, b respectively, shew that the rectangles TA. TB, and Ta. Th are as 4 to 1. º i 238. If at any two points A, B, AC, BID be drawn perpendicular to AB on the same side of it, of such magnitude that AB is a mean proportional between A C, BD; the circles whose diameters are AC, BD will touch each other. Af 239. AB is the diameter of a circle, E the middle point of the radius OB. On A.E., 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 smaller circle. . 240. If two circles touch each other internally in a point A, from which a straight line ABC is drawn cutting the circles in B, C; then if another straight line which joins the centers, cut the other circle in D, AD shall be a mean proportional between AB and A. C. 241. Given the magnitude and position of two circles; to draw a chord AB of the greater to touch the less at C, so that CA shall have to CB a given ratio. - 242. Two circles whose centers are A and B, touch each other in the point C ; shew that if D be a point without the circles such that * on Book v. - 377 DA, DB make equal angles with DC, the rectangle contained by the tangents drawn from D is equal to the square on DC. - 243. FEI is a common tangent to two circles, meeting the line which joins the centers produced in the point I; IDCGH is any straightline cutting the circles. Prove that IEx IF=IDxIH =ICx IG. 244. If two given circles touch each other and also touch a given line, the rectangle contained by the diameters of the circles, is equal to the rectangle contained by the sum of the diameters and the perpen- dicular distance of the point of contact from the touching line. . XIV. 245. The diagonals of a trapezium, two of whose sides are parallel, cut one another in the same ratio. - 246. If two sides of a quadrilateral figure be parallel to each other, a straight line drawn parallel to either of them shall cut the other sides or them produced, proportionally. - 247. ABCD is a trapezium of which the opposite sides AD, BC are parallel, shew that * Adº-Bº; AB - Do”: Bc4 AD : BC-AD. 248. Two straight lines AB, CD intersect in E. If when AC, BD are joined, the sides of the triangle ACE taken in order are pro- portional to those of the triangle DBE taken in order, shew that A, C, B, D lie on the circumference of the same circle. 249. On a given line describe a rectangle which shall be equal to the difference of the squares on two given straight lines, any two of the three given lines being greater than the third. 250. Given the sides of a quadrilateral figure inscribed in a circle, to find the ratio of its diagonals. * 251. The diagonals AC, BD, of a trapezium inscribed in a circle, cut each other at right angles in the point E; - the rectangle A.B. BC : the rectangle AD. DC :: BE : ED. 252. If the sides of a quadrilateral figure circumscribing a circle, touch the circle at the angular points of an inscribed quadrilateral figure; all the diagonals will intersect at the same point. . 253. If two circles can be described so that each shall touch the other and three sides of a quadrilateral figure, one-fourth of the diffe- rence of the sums of the opposite sides is a mean proportional between the radii. - *. - * 254. ABCD is a quadrilateral figure inscribed in a circle; BA, CD are produced to meet in P, and AD, BC are produced to meet in Q. Prove that the ratio of PC : PB :: QA : QB. - 255. ABCD is a quadrilateral inscribed in a circle; the straight lines CE, DE, which bisect the angles A CB, AIDB, cut BD and AC in F and G respectively; shew that EF is to EG, as ED to EC. - 256. AB is a diameter of a circle, and CD a chord at right angles to it, E any point in CD. If A E and BE be drawn and produced to cut the circle in F and G, the quadrilateral FCGD has any two of its adjacent sides in the same ratio as the remaining two. 257. ABCD is a quadrilateral inscribed in a circle, and its diago- mals intersect in F; shew that the rectangle AF, FD is to the rectangle BF, FC, as the square on AD is to the square on B.C. 258. Two sides of a quadrilateral described about a circle are 878 GEOMETRICAL ExERCISEs parallel; if the points of contact divide the other two sides propor- tionally, they are equally inclined to the first two. 259. The diagonals of a quadrilateral figure inscribed in a circle, are to one another as the sums of the rectangles contained by the sides which meet their extremities. 260. A quadrilateral is circumscribed about a given circle, and any tangent is drawn to the circle. If perpendiculars be drawn from the angular points of the quadrilateral on the tangent, shew that the rect- angle of those from one pair of opposite angles, is to the rectangle of those from the other pair in a constant ratio. * 261. ABCD is any trapezium inscribed in a circle. Its sides are produced to meet in P and Q. Prove the following proportion; PD. DQ : AD. DC :: PB. BQ : A.B. B.C. * 262. Prove that in order that all the rectangles circumscribed about a quadrilateral figure may be similar to a given rectangle, the diagonals of the quadrilateral must be at right angles to each other, and proportional to the sides of the given rectangle. - XV. 263. If in a given equilateral and equiangular hexagon another be inscribed, to determine its ratio to the given one. - 264. A regular hexagon inscribed in a circle is a mean propor- . tional between an inscribed and circumscribed equilateral triangle. 265. The diameter of a circle is a mean proportional between the sides of an equilateral triangle and hexagon which are described about that circle. - 266. If two equilateral triangles be inscribed in the same circle with their vertices at the opposite extremities of the same diameter, compare the magnitude of either with that of the hexagon formed by the intersection of the sides. 267. If a hexagon be described about a circle, the three lines join- ing the opposite angular points, intersect in a point. - 268. The area of the inscribed pentagon, is to the area of the circumscribing pentagon, as the square on the radius of the circle inscribed within the greater pentagon, is to the square on the radius of the circle circumscribing it. 269. A regular decagon inscribed in a circle is a mean propor- tional between an inscribed and circumscribed regular pentagon. 270. If A1, A2 be the areas of the squares inscribed in, and cir- cumscribed about a given circle, A that of a regular octagon inscribed in the same circle, shew that A1 : A :: A : A2. - - 271. If one circle touch another internally passing through its center, and any number of chords be drawn cutting the two circles, the polygons formed in the two circles by joining the extremities of conseoutive chords will be in the ratio of 1:4. . 272. Any polygons whatsoever described about a circle are to one another as their perimeters. - 273. If two equilateral polygons be drawn, one within, and the other about a circle, and another of half the number of sides be inscribed in the circle, this shall be a mean proportional between the other two. • - - 274. If the sides of any regular figure be bisected, and each two on Book VI. --- 379 *s. º: adjacent points of bisection be joined, a similar figure will be formed, and its area will be to that portion of the original area of the figure which is without it, as the square on the diameter of the circle circum- scribing it is to the square on one of its sides. 275. Two similar rectilineal figures being given, to find a third figure, which shall be similar to them, and also have its area a mean proportional between them. 2. 276. To describe a figure similar to each of two other figures, and equal to their sum or difference. 277. If there be three equal rectangles contained by AB, AG; CD, CH; EF, EK, then if EK-CH=CH-AG, then AB – CD will be to CD - EF, as AB is to EF. Also if this latter proportion hold, * JEK - CH = CH- AG; - prove this, and explain how this is really a geometrical proof of the common propositions of harmonic progression. 278. If a square be inscribed in a right-angled triangle, twice the side of this square is an harmonic mean between the sides which contain the right angle. * -- - 279. If from a point without a circle there be drawn three straight lines, two of which touch the circle, and the other cuts it, the line which cuts the circle will be divided harmonically by the convex circumference, and the chord which joins the points of contact. 280. The straight line ABCD is divided harmonically in the points B, C; prove that if a circle be described on AC as diameter, any circle passing through B and D will cut it at right angles; and conversely. - - 281. If two circles touch one another externally and from the center 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 of the circles. * 282. In any triangle, if the radii of the circles, touching two sides produced and one externally, be in harmonic progression, the mean radius is three times that of the inscribed circle, and the sides are in arithmetical progression. 283. S is a fixed point in OA the radius of a circle, center O'; in 04 produced, OL is taken a third proportional to OS and OA, and from the point L, any line LPQ is drawn cutting the circle in P and Q, and if the line LPQ cut the line drawn from S. at right angles to AO in the point E, prove (1) that LQ is divided harmonicaEly in the points. Jº and P. (2.) If PS be joined, the ratio of LP to PS, is a constant ratio. (3.) If QS be also joined, the rectangle contained by PS, SQ. is a constant rectangle. (4.) If through S a line be drawn at right angles to 40, tangents to the circle at P and Q will cut this line in one and the same point. 284. ABCDE is a regular pentagon whose sides D0 and AB are produced to meet in P, and from BP, Bø is eut off equal to AB; shew that AP, A Q and BP form an harmonic series. - - 285. If A, B, C, D, E, &c. be the successive angles of a regular decagon whose center is 0, and AD cut OB in B, B.E cut OC in C, and so on; shew that OA, O.B., 00', &c. are in harmonic progression. 380 GEOMETRICAL EXERCISES. LOCI. THE STRAIGHT LINE AND THE CIRCLE. PROPOSITION I. THEOREM. Fºnd the locus of the middle points of all straight lines which have one extremity in a given point and the other in a given straight line. Let A be the given point, and B0 the given line, and let any line AD be drawn from A to meet the line BC in any point D, and be bisected in P. - - It is required to determine the locus of P for all positions of the line AD, while the point A and the line BC remain in the same fixed position. B - D P A - Q E C From A draw. A E perpendicular to BC, bisect AE in the point Q, and join PE, PQ. * Then because AED is a right-angled triangle, and EP is drawn from the right angle to the bisection of the hypotenuse, the line EP is equal to AP or PD. - ... And in the triangles APQ, EPQ, - the sides A Q, QP are respectively equal to EQ, QP, . and the base AP is equal to the base EP, therefore the angle AQP is equal to the angle EQP, and these are adjacent angles, therefore each of them is a right angle ; wherefore the line PQ is at right angles to A.E at Q, the middle point of the distance of A from BC. Eſence the point P for all positions of the line AD is situated in PQ or PQ produced. - t Wherefore the locus of P, the middle point of AD is a straight line parallel to BC, drawn through Q the middle point of the distance of the given point A from the given line BC. PROPOSITION II. PROBLEM. Having given the base and vertical angle of a triangle, to find the locus of the intersection of the perpendiculars from the three angles drawn to the . opposite sides. . - On the given base AB describe a segment ACB of a circle contain- ing an angle equal to the given vertical angle of the triangle, and Complete the circle. r t 4 LOCI. - - - 381 Draw. A C, Bc to any point C in the circumference; then ABC is one of the triangles. ſº Let the perpendiculars Aa, Bb, Co be drawn to the sides BC, AC, AB, and intersecting each other in the point 0. Then Cb Oa is a quadrilateral figure having the two opposite angles Cb 0, CaO, right angles, - - therefore also the other opposite angles b Ca, b Oa are equal to two right angles, a constant magnitude, (Euc. III. 22.) but the angle b Ca or ACB is constant, g therefore also the angle boa is constant for all positions of the point 0: and the base AB is given, wherefore the point O is in the arc of a circle described upon AB as a chord. y - PROPOSITION III. THEOREM. The locus of the centers of the circles, which are inscribed in all right- angled triangles on the same hypotenuse, is the quadrant described on the hypotenuse. * - Let ABC be one of the right-angled triangles, C being the right angle and AB the hypotenuse. Let O be the center of the circle inscribed in the triangle ABC, and let the triangle ABC be circumscribed by the circle A CBD. Join C0 and produce it to meet the semicircle A DB in D. D Then since 0 is the center of the circle inscribed in the triangle ABC, COD bisects the right angle ACB, (Euc. Iv. 4.) and hence also the semicircle ADB which subtends that angle. Join A 0, BO, AD, DB. • * ~~ Then the angles DAB, DCB are equal, (Euc. III. 21.) and the angles A CD, BCD are also equal; (Euc. Iv. 4.) therefore the angle DAB is equal to the angle DCA, 382 GEOMETRICAL ExERCISES. and likewise the angle CA0 to OAB, (Euc. Tv. 4.) .* whence DAO is equal to the angles CA 0, A CO; but the angles CA0, AC0 are equal to AOD, (Euc. I. 32.) therefore the angle DAO is equal to AOD, * and the straight line D0 is equal to DA ; and also to DB. The point 0 is therefore in the arc of a circle whose center is D and radius DB or D.A. - And since DB is equal to DA and at right angles to it, the locus of 0 is a quadrantal arc of the circle whose radius is DA or DB upon the given base AB of the right-angled triangle. CoR. If the triangle be an acute angled-triangle, the construc- tion holds good, but the locus of the centers of the inscribed circle is an arc of a circle upon the base, but greater than a quadrant; and if the triangle be an obtuse angled-triangle, the locus of the centers of the inscribed circle is an arc of a circle upon the base, but less than a quadrant. PROPOSITION IV. PROBLEM. Given a circle and a point within it: if a straight line be drawn from the given point to the circumference and divided in a given ratio, to deter- mine the locus of the point of section. Tet A be the given point within the given circle whose center is C, and let the straight line A B be drawn to meet the circumference in JB ; and let A B be divided in D so that the ratio of AD to DB is a given ratio; it is required to determine the locus of the point D. . Draw AC and produce it both ways to meet the circumference in E and F. - Join BC and draw DG parallel to BC meeting the diameter EF'in G. Then since the ratio of AD to DB is given, the ratio of AB to AD is also given. And AB is to AD, as BC to DG. (Euc. VI. 4.) Hence the ratio of BC to DG is given; and because BC the radius is given in magnitude, , therefore DG is also given in magnitude. But AD is to AB, as AG is to AC, (Euc. VI. 4.) Hence AC is divided in the given ratio in G, and therefore G is a fixed point for any position of the point B. Wherefore, since the point G is fixed, and the line GD is given in magnitude, the locus of the point D is the circumference of a circle whose radius is DG and whose center is G. LocI. 383 I. 5. Determine the locus of the vertices of all the equal triangles, which can be described on the same base, and upon the same side of it. 6. Straight lines are drawn from a fixed point to the several points of a straight line given in position, and on each base is de- scribed an equilateral triangle. Determine the locus of the vertices. 7. The base of an isosceles triangle lies in a given infinite straight line, and has one of its extremities at a given point of that line. If the sum of the base of the triangle and its altitude are together equal to a given straight line, find the locus of its vertex. 8. D, E, F, G are the middle points of the sides of a rectangle ABCD. Join DF, EG; then if P be a point such that AP + PC = BP+ PD, shew that P can only lie in DF, EG or these lines produced. Q & © 9. A square is moved so as always to have the two extremities of one of its diagonals upon two fixed lines at right angles to each other in the plane of the square: shew that the extremities of the other diagonai will at the same time move upon two other fixed straight lines at right angles to each other. 10. AB, BÖ, DE, EF are rods joined at B, F, E, and D, capa- ble of angular motion in the same plane, and so placed that FBDE is a parallelogram. If, when the rods are in any given position, points A, E and C be taken in the same line, shew that these points will always bé in the same line, whatever be the angle the rods make with each other. - 11. Two lines of given length slide upon two given lines; shew that the locus of a point, such that the sum of the areas made by join- ing with it the ends of the given lines is constant, is a straight line, and determine its position. Is the property true for all points of this line 2 II. 12. Find the locus of the middle points of any system of parallel chords in a circle. * 13. Shew that all equal straight lines in a circle will be touched by another circle. tº 14. If equal straight lines be placed similarly round a circle just without it, the loci of their extremities will be concentric circles. 15. If from a point in the circumference of a circle any number of chords be drawn, the locus of their points of bisection will be a circle. . 16. A is a fixed point in the circumference of a circle, BC any chord, BP and CP are drawn making angles with AB, A C equal to those which B C makes with these lines. The locus of P is the dia- meter through A. 17. If in a circle two chords of given length be drawn so as not to intersect, and one of them be fixed in position, and if the opposite ex- tremities of the chords be joined by two lines intersecting within the circle; then the locus of the point of intersection will be a portion of a circle passing through the extremities of the fixed cord. - 18. If from two fixed points in the circumference of a circle, straight lines be drawn intercepting a given arc and meeting without the circle, the locus of their intersections is a circle. - 384 GEOMETRICAL EXERCISES. 19. P is any point in a semicircle whose diameter is AB, AP is produced to Q so that PQ is equal to PB; find the locus of Q. 20. A CB is the diameter of a circle, CP, CG) are perpendicular radii, shew that the locus of the intersection of AP, and BQ is a circle whose center is in the given circle, and radius the diagonal of the square on the radius. 21. If through two given points in the circumference of a circle pairs of equal chords be drawn, one set of their intersections will lie in a diameter of a circle and the other in the circumference of a second circle, passing through the given points. 22. If from any external point any number of straight lines be drawn cutting a circle, find the locus of the middle points of the chords thus formed. • 23. Find a point without a given circle from which if two tangents be drawn to it, they shall contain an angle equal to a given angle, and shew that the locus of this point is a circle concentric with the given circle. - 24. Determine the locus of the extremities of any number of straight lines drawn from a given point, so that the rectangle contained by each, and a segment cut off from each by a line given in position, may be equal to a given rectangle. * 25. If from a given point S, a perpendicular Sy be drawn to the tangent Py at any point P of a circle whose center is C, and in the line MP drawn perpendicular to CS, or in MP produced, a point Q be always taken such that MQ = Sy, then the locus of Q is a straight line. III. 26. If one side of a triangle be constant and the difference of the squares described on the other two also constant, the locus of the vertex is a straight line. 27. Given the base and sum of the squares of the sides of any triangle, find the locus of its vertex. 28. . When the vertical angle and the sum of the sides of a plane triangle are given, prove that the locus of the middle of the base is a line given in position. 3. - - 29. Find the locus of the vertex of a triangle, having given the radius of the inscribed circle and the difference of the sides of the triangle. - , 30. The base of a triangle and the radius of its circumscribing circle being given, find the locus of its vertex. 31. If on a given base a triangle be described such that a straight line drawn from the vertex to a given point in the base bisects the vertical angle, shew that the vertex will generally lie in a certain circle. What will be the locus, when the given point is the center of the base? . ... . ſº 32. If a series of triangles be described upon the same base, the perpendiculars drawn to the sides of each triangle from their middle points will intersect in a certain straight line. J 88. Given of any triangle the base, and the point where the line bisecting the exterior vertical angle cuts the base produced, find the locus of the triangle's vertex. 34. Two fixed straight lines AB, A C are cut by any straight line Loci, 385 in B and C: shew that the straight lines bisecting the angles ABC, ACB, will always intersect in a third fixed straight line. 35. If the sides AB, A C of a given triangle be cut proportionally in any two points D and F, and perpendiculars to the sides from those points intersect in L: prove that all the points L. correspónding to different positions of D and F will be in a straight line passing through A. *_ - 36. Shew that the locus of the vertices of all the triangles Con- structed upon a given base, and having their sides in a given ratio, is a circle. & - 37. On the base of a given isosceles triangle and on the same side of it, describe a triangle having the angle at the vertex double the angle at the vertex of the original triangle. Prove that the vertices of all such triangles lie in a certain circle. - - 38. Triangles are constructed on the same base, with equal verti- cal angles; prove that the locus of the centers of the escribed circles, each of which touches one of the sides externally and the other side and base produced, is an arc of a circle, the center of which is on the circumference of the circle circumscribing the triangles, - IV, º 39. If through any point in the circumference of a circle, chords be drawn and divided, so that the parts are in a constant ratio, the points of division shall all lie on the circumference of another circle. 40. Determine a series of points within a given circle, such that the rectangle contained by the parts of all chords drawn through them shall be equal to the square whose side is half the radius of the circle. 41. Let 0, C, be any two fixed points in a circle, and 0A any chord; then if A C be joined and produced to B so that OB may be equal to OA, the locus of B is an equal circle. - 42. BCA, PCP are two straight lines, BCA is fixed, and PCP' revolves about C; shew that the locus of the intersection of BP, P’A is a circle. - . 43. Find the locus of a point such that the distance between the feet of the perpendiculars from it upon two straight lines given in position may be constant. - 44. A ladder is gradually raised against a wall; find the locus of its middle point. - d W. 45. . If a circle roll within another of twice its size, any point in its circumference will trace out a diameter of the first. - 46. Two segments of a circle are upon the same base A.B. P is any point in one of the segments. The lines APD, BPC are drawn, meeting the other segment in D and C. A C and BD are drawn, in- tersecting in Q. Shew that the locus of Q is a circle. 47. P, Q are any points in the circumferences of two segments described on the same line AB; the angles PA Q, PBQ are bisected . by the lines A.R., BR meeting in R; prove that R will always lie on a third segment described on A.B. - 48. If a rod move betwixt two fixed straight lines CP, CO; and the perpendiculars from P, Q upon CP, CQ respectively meet in R, whilst those from P, Q upon CG., CP respectively meet in S: the loci of R and S will be two circles having their common center in C. . C C ... .º 386 GEOMETRICAL ExPRCISEs. 49. If any number of circles be drawn having a common chord AB, and from a point C in AB produced, lines be drawn intersecting the circles; shew that when the locus of the first points of intersection. of the straight lines and circles is a circle passing through B and C, then the locus of the second points of intersection is a straight line. 59. If there be two given circles, shew that the locus of points from which tangents drawn to eachinclude agiven angleis another circle. 51. Find the locus of a point such that the tangents drawn from it to two fixed circles are equal. 52. A series of circles is described having a common chord, and from a point in the chord produced, tangents are drawn to the circles, shew that the points of contact of the tangents with the circles lie in the circumference of another circle. - 53. A series of circles touching each other at one point are cut by a fixed circle; shew that the intersections of the pairs of tangents to the latter, at the points where it is cut by each of the other circles, lie in a straight line. - 54. Construct the locus of a point whose distance from one given point is double of its distance from another. f 55. If a straight line be divided into any two parts, find the locus of the point at which these parts subtend equal angles. 56. Three given straight lines are in the same straight line; find a point from which lines drawn to their extremities shall contain equal angles. - - - . 57. If a circle be given in magnitude and position, and a given point either within or without the circle, and if tangents be drawn at the extremities of all the chords which pass through the given point; to find the locus of the intersections of the tangents. 58. If two chords be drawn, one in each of two circles which intersect one another, through any point in the chord which joins the points of intersection of the circles; the locus of the extremities of the two chords is the circumference of a circle. WI. 59. Two straight lines which intersect and whose point of inter- section is fixed, contain a given angle and are always to each other in the same ratio, and the extremity of one of these lines moves on a given straight line; prove that the extremity of the other moves also on a straight line, and determine its position. 60. Find the locus of a point, such that if straight lines be drawn from it to the four corners of a given square, the sum of the squares shall be invariable. t - - 61. If in two parallel lines two points A and B be taken, and through two other points C, D, in the same lines, any pair of parallels be drawn to meet any pair of parallels through the points A, B, in the points E, F, the line passing through the points E, F, shall pass through a point which shall remain fixed as long as A, B, C, D remain fixed. Shew also that if C and D move in a certain manner along the fixed parallels, the line passing through E and F will still pass through the fixed point. - - 62. If any rectangle be inscribed in a given triangle, required the locus of the point of intersection of its diagonals. 63. From any point in the base of a triangle lines are drawn parallel to the sides; shew that the intersection of the diagonals of every parallelogram so formed lies in a certain straight line. MAXIMA AND MINIMA. PROPOSITION I. PROBLEM. From two given points on the same side of a straight line given in post- tion, draw two straight lines which shall meet in that line, and make equal angles with it; also prove, that the sum of these two lines is less than the sum of any other two lines drawn to any other point in the line. Analysis, Let A, B be the two given points, and CD the given line. Suppose G the required point in the line, such that AG and B G being joined, the angle AGC is equal to the angle B6.D. - E z Draw AF perpendicular to CD and meeting BG produced in E. Then, because the angle BGD is equal to AGF, (hyp.) and also to the vertical angle FGE, (I. 15.) * therefore the angle AGF is equal to the angle EGF; also the right angle AFG is equal to the right angle EFG, and the side FG is common to the two triangles AFG, EFG, therefore AG is equal to EG, and AF to F.E. Hence the point E being known, the point G is determined by the intersection of CD and B.E. - Synthesis. From A draw AF perpendicular to CD, and produce it to E, making FE equal to AF, and join B.E cutting CD in G. ~- Join also AG. - Then AG and BG make equal angles with CD. * For since AF is equal to FE, and FG is common to the two tri- angles AGF, EGF, and the included angles AFG, EFG are equal; therefore the base AG is equal to the base EG, - and the angle AGF to the angle EGF, but the angle EGF is equal to the vertical angle BGD, therefore the angle AGF'is equal to the angle BGD; that is, the straight lines AG and BG make equal angles with the straight line CD. - * , Also the sum of the lines AG, GB is a minimum. For take any other point Hin CD, and join EH, HB, AH. Then since any two sides of a triangle are greater than the third side, therefore EH, HB are greater than EB in the triangle EHB. But FG is equal to AG, and EH to AH; - therefore AH., HB are greater than AG, GB; and this will always . 'be so, how near soever the point H may be to the point G. If any - other point H’ be taken on the other side of G, and AH', BH' be drawn, it may be shewn in the same way that AH', BH' are always greater than AG, GB, how near soever the point H' may be to G. . But when three consecutive values of a variable magnitude are found, such that the first and third are always greater than the second, the second magnitude has then its least value. Hence the sum of AG, GB is less than the sum of any other two lines which can be drawn from A and B to any other point in CD. - - C C 2 388 GEOMETRICAL EXERCISES. PROPOSITION II. PROBLEM. From two given points on different sides of a straight line, draw two straight lines to meet at a point in the line, so that their difference shall be the greatest possible. Let A, B, be the two given points, and CD the given line. It is required to find a point Gin CP, so that the difference of the lines AG, BG drawn from A and B shall be the greatest possible. From B draw B.E perpendicular to CD, and produce B.E. to F, making EF equal to E.B. ſº - * Join AF and produce it to meet CD in G.; then G shall be the point required, such that the difference of 46, and BG is greater than the difference of any other two lines drawn from A and B to any other point in CD. - Take any other point H in CD, and join A.H., HF, HB. Then GF is equal to GB, (Euc. 1. 4.) also HF is equal to HB; and the difference of A.H., HB is equal to the difference of A.H., HF, but in the triangle AHF, the difference of the two sides A.H., HF is less than AF, the third side of the triangle. And AF is the difference of A G and GB ; therefore the difference of AH and HB is always less than AF, the difference of AG and GB, how near soever the point H may be to the point G. If any other point H be taken on the other side of G, it may be shown in the same manner, that the difference of AH' and H'B is always less than the difference of AG and GB, how near soever the point H' may be to the point G. - And when three consecutive values of a variable magnitude are found, such that the first and third are always less than the second magnitude, the second magnitude has then its greatest value. Hence the difference of AG and GB is greater than the difference of any other two lines drawn from A and B to any other point in CD. PROPOSITION III. THEOREM. Of all triangles having "the same base and equal vertical angles, the isosceles triangle is the greatest. Let ABC be an isosceles triangle having the sides AB, A C equal. Describe a circle about the triangle ABC, and take any point in the arc AC, and join DB, DC. Then ABC, DBC are two triangles upon the same base BC, and having equal vertical angles. } - Through A draw EAF parallel to BC, produce BD to meet EA in F, and join FC. Then because BC is parallel to EF, the triangle BAC is equal to the triangle BFC; (Euc. I. 37.) - MAXIMA AND MINIMA w 389 - but the triangle BFC is greater than the triangle BP C, and will be always greaten, so long, as the 99.ht 42 does not coincides with the point A, therefore the triangle BAC is always greater than BDC. E. A F C TZ If any other point D be taken in the arc AB, and DB, D C be drawn, it may be shewn in the same ºanner that the triangle ABC is always greater than the triangle Dº O. - Hence the triangle ABC is greater than any other triangle which has the same base and an equal vertical angle. - PROPOSITION IV. THEOREM, If two circles touch each other internally, and from any two points ºn the circumference of the interior circle, straight lines be drawn to the point of contact, they will contain a greater angle than any other two lines drawn from the same points to any other point in the exterior circumference. . Let the circle PAB touch the circle PDE internally in the point, and let A, B be any two points in the circumferençº of the circle PAB, and let AP, Bpbe drawn to the point of contact P. Then the angle APB shall be greater than the angle contained by any other two lines drawn from A, B to any other point in the circumference of the exterior circle. Join AB and produce it to meet the exterior circle in D, E, a Take any point Q in the circumference of the exterior circle; and draw BQ intersecting the interior circumference in 6, and join &A. QA. Theangles APB, ACB are equal, being in thesame segment, (Euc. mi. 21.) but the angle ACB is greater than the angle A QB, (Euc. I. 32.) - therefore the angle 4 PB is greater than 49B; and APB will be always greater than A QB, how near soever the If point & . *to the point P. - a poln e taken in the arc PE, and QA, Q B be joined, it be shown that the angle APB is always #: º ; angle 'º. 4. *#. angle 4. is a º at P, the point of contact. * : If a common tangent be drawn to the circle - and PD, PE be joined, the angle APD is equal to };" the point P, 390 GEOMETRICAL ExERCISEs. - I. - 5. Two pairs of equal straight lines being given, shew how to construct with them the greatest parallelogram. - 6. Within a square there is no point so distant from the point of intersection of the diagonals as the corners. - !- 7. Determine the shortest path from one given point to another, subject to the condition, that it shall meet two given lines. -8. A point is taken within a square, and straight lines drawn from it to the angular points of the square, and perpendicular to the sides; the squares on the first are double the sum of the squares on the last. Shew that these sums are least when the point is in the center of the square. - 9. To find a point in a given right line such that the sum of the squares on the lines from the two points to that point shall be less than that of the lines from the two given points to any other point in the given line, g - - Aº’ 10. The sum of the squares of two lines is never less than twice their rectangle. - 11. I)ivide a given straight line so that the rectangle under the arts may be equal to a given Square, and point out the limit which the side of the given square must not exceed so that the problem may be possible. - - - 12. If two straight lines of given magnitude cut each other at right angles, the sum of the rectangles contained by the segments into which each divides the other is least when they bisect each other, II. 13. The perimeter of a square is less than that of any other parallelogram of equal area. - 14. Shew that of all quadrilateral figures having the same peri- meter, that which is a square is the greatest. 15. Shew that of all equiangular parallelograms of equal peri- meters, that what is equilateral is the greatest. - 16. Prove that the perimeter of an isosceles triangle is greater than that of an equal right-angled parallelogram of the same altitude. 17. The sum of the diagonals of a trapezium is less than the sum of any four lines which can be drawn to the four angles, from any point within the figure, except their intersection. 18. If two sides of a triangle begiven, the triangle will be greatest when they contain a right angle. 19. Given one of the angles and the perimeter of a plane triangle, to find the sides, when the area is the greatest possible. 20. ABC is a right-angled triangle; find the point P in AC, so that the sum of the distances from A and A C is the least possible. –21. Given the base and one side of a triangle, to find the third side, so that the area may be the greatest possible. 22. If from a point in the base of an isosceles triangle, two straight lines be drawn parallel to and terminated by the sides, the sum of the areas of the new triangles so formed will be least when the point is at the center of the base. 23. Of all triangles having the same vertical angle, and whose bases pass through a given point, the least is that whose base is bisected in the given point. . . *- MAXIMA AND MINIMA. 391 24. Of all the triangles having the same base and the same peri- meter, that is the greatest which has the two undetermined sides equal. - z T ; Q. 25. The perimeter of an isosceles triangle is less than that of any. other equal triangle upon the same base. T e - 26. Of all triangles on the same base, having equal perimeters, the equilateral has the greatest area. g 27. Find the greatest of all triangles having the same vertical angle and equal distances between that angle and the bisection of the opposite sides. e - 28. Of all triangles on the same base and between the same parallels, the isosceles has the greatest vertical angle. - 29. Of all triangles which have the same vertical angle, and whose bases pass through the same point, to determine that in which the rectangle contained by the sides is the least possible. III. 30. Shew that the perimeter of the triangle, formed by joining the feet of the perpendiculars dropped from the angles upon the opposite sides of a triangle, is less than the perimeter of any other triangle, whose angular points are on the sides of the first. - 31. In an acute-angled triangle, to find a point from which if three lines be drawn to the three angles, the sum of these lines shall be the least possible. - 32. Draw through the angles of an equilateral triangle three. straight lines which shall form by their intersection another equilateral triangle; shew that there may be an infinite number of such triangles, and describe the greatest. - 33. The sides of a triangle are divided in the same ratio; shew that the sum of the squares on the lines that join the points of division with each other, is greatest when the sides are bisected. 34. Inscribe a square in a given square. Shew that an infinite number can be inscribed, and find the least. - 35. In a given rectangle ABCD inscribe another rectangle which shall have one angle at a given point P in AB; giving all the solu- tions and the limits of the possibility of the problem. 36. Determine which is the greatest, and which the least, of the three squares which may be inscribed in a given triangle. 37. Through the angular points of a quadrilateral, lines are drawn forming a rectangle; shew that the difference of the areas of the greatest and least rectangles which can be so formed, is equal to twice the area of the quadrilateral. IV. 38. Through a point in a circle which is not the center, to draw the least chord. ' 39. From two given points draw two straight lines to the same point in a given straight line, so as to include a given angle; and find the limit which the given angle must not exceed. - 40. Of all straightlines which can be drawn from two given points $o meet in the convex circumference of a given circle, the sum of those &- 392 GEOMETRICAL EXERCISES. two will be the least, which make equal angles with the tangent at the point of concourse. - 41. If any point P be taken in the plane of a circle, and PA, PB, PC, ... be drawn to any number of points A, B, C, ... situated sym- metrically in the circumference, the sum of PA, PB, ... is least when P is the center of the circle. 42. From a given point within a circle, to draw a line to the circle, so that the angle which it makes with the tangent at the point of con- tact may be the least possible. - - 43. Find a point in the circumference of a given circle, the sum of whose distances from two given lines at right angles to each other is the greatest or least possible. 44. Of all straight lines which can be drawn through a given point within a circle, find that which cuts the circumference in the greatest and least angles. - 45. Let N be any point in the diameter of a circle, whose center is S, PNQ a chord drawn through N, and join SP; shew geometrically that PQ is a minimum, and the angle SPQ a maximum when PQ is perpendicular to the diameter, 46. Determine that point in the arc of a quadrant from which two lines being drawn, one to the center, and the other bisecting the radius, the included angle shall be the greatest possible. - 47. A flag-staff of a given height is erected on a tower whose height is also given: at what point on the horizon will the flag-staff appear under the greatest possible angle 2 48. A and B are two points within a circle; find the point P on the circumference such that, if PAH, PBK be drawn meeting the circle in H and K, the chord HK shall be the greatest possible. 49. Find the point in a given straight line at which the tangents to a given circle will contain the greatest angle. 50. If from a point without a circle two tangents be drawn to the circle, and lines be drawn terminated by these two tangents and them- selves touching the circle, shew that the least of these straight lines is that which falls between the circle and the intersection of the tangents, and makes equal angles with the two tangents, 51. Draw through a given point in the diameter of a circle a chord, which shall form with the lines joining its extremities with either extremity of the diameter, the greatest possible triangle. 52. From a given point A without a circle whose center is 0, draw a straight line cutting the circle in the points B and C, such that the area B00 may be a maximum, r W. 53. The circumference of one circle is wholly within that of another. Find the greatest and the least straight lines that can be drawn touching the former and terminated by the latter. - 54. Let one circle touch another internally, and let straight lines touch the inner circle, and be terminated by the outer; shew that the greatest of these lines is the one parallel to the common tangent at the point of contact. - 55. Find the longest and shortest lines which can be drawn join- ling the circumferences of two circles. 56. Find the points from which two unequal circles subtend equal © Mºxima Asd MINIMA. 393 º, angles. Find also the positions of the points when the equal angles are the greatest and least possible. .* . 57. Through either of the points of intersection of two given in- tersecting circles, draw the greatest possible line terminated both ways by the two circumferences. - - 58. Find the radii of the greatest and least circles which can be de- scribed touching two given circles neither concentric nor intersecting. 59. Two circles can be described, each of which shall touch a given circle, and pass through two given points outside the circle; shew that the angles which the two given points subtend at the two points of contact, are one greater and the other less than that which they subtend at any other point in the given circle. 60. The centers of three circles are equidistant from each other. Describe an equilateral triangle of given magnitude such that the three circles shall touch its three sides respectively on the part external to the triangle. What are the greatest and least magnitudes of the tri- angle that this may be possible * WT. 61. Through a given point between two indefinite straight lines not parallel to each other, to draw a straight line which shall be termi- nated by them, so that the rectangle contained by its segments shall be less than the rectangle contained by the segments of any other straight Hine drawn through the same point and terminated by the same straight lines. 62. A, B, are two given points either within or without a given circle; find in the circumference points P, so that AP*+ BP, may be the greatest and least possible. 63. Find the least triangle which can be circumscribed about a given circle. 64. The perimeter of an equilateral triangle inscribed in a circle is greater than the perimeter of any other isosceles triangle inscribed in the same circle. 65. Prove that the greatest right-angled triangle that can be in- scribed in a circle is an isosceles triangle. © º The square is greater than any rectangle inscribed in the same CITCL6. - - 67. Inscribe the greatest parallelogram in a given semicircle. 68. AB is a fixed chord in a circle. Find the position of the chord A C, such that the diagonal through A of the parallelogram con- structed on A B and A C as sides may be a maximum. 69. Of all quadrilateral figures contained by four given straight lines, the greatest is that which is inscriptible in a circle. 70. If on a straight line there stand two polygons, without re- entrant angles, one of which encloses the other, the exterior polygon has the longest perimeter. & 71. Prove that of the polygons of a given number of sides, which can be inscribed in a given circle, the greatest is that which is equila- teral. Q 72. Of all polygons formed with given sides, the greatest is that which may be inscribed in a circle. 73. Of all polygons having equal perimeters, and the same number of sides, the equilateral polygon has the greatest area. . **, * - * TANGENCIES. - PROPOSITION I. PROBLEM. To describe a circle which shall touch a given circle in a given point, and pass through a given point in a line given in position. Analysis. Let A be the given point in the circumference of the given circle whose center is C, and B the given point in the line D.E. given in position. Suppose the point G. to be the center of the required circle which touches the given circle in the point A, and passes through the point B in the given line D.E. - - Draw AB, GB, G.C.; * then GA is equal GB, G being the center of the circle: and GC passes through the point of contact A ; (Euc. III. 12.) Through C draw CF parallel to AB meeting GB produced in F. Then GC is equal to GF (Euc. I. 6.) and GA is equal to GB, - therefore BF is equal to AC the radius of the given circle. Synthesis. Draw the lines AB, A C ; through C draw CF parallel to AB; * from B draw BF'equal to CA and meeting CF in F, produce CA, FB to meet in G. Then G is the center of the circle which passes through the given point B in the line DE, and touches the given circle whose center is C in the given point A. PROPOSITION II. PROBLEM. To describe two circles with given radii, which shall touch each other and a given straight line, one of the circles touching the given line at a given point. Analysis. Let AB be the given straight line and C the given pointin it. Suppose the circles whose centers are 0, O' described with the given radii, touch each other in the point P, and the circle whose center is O touch the line AB in the given point C, and the circle whose center is 0 in some other point D. , TANGENCIES. - 395. * - Join OO, OC, Ol); then 00 passes through P the point of contact, (Euc. III. 12.) sº and OC, O’D are perpendicular to AB, (Euc. III. 18.) and are therefore parallel to each other. (Euc. I. 29.) Through D draw DE parallel to O'0, and produce 00 to meet DE in E. * Then OO'DE is a parallelogram, of which the side OE is equal to the radius of the larger circle, and the adjacent side ED is equal to the sum of the given radii. g Synthesis. At the given point C in the line AB, draw CO at right angles to AB, and equal to the given radius of the smaller circle; produce CO to E, making OE equal to the radius of the larger circle, with center E, and radius equal to the sum of the radii of the two circles, describe a circle intersecting AB in D; at D draw D'O' at right angles to AB, and equal to the radius of the larger circle. • With center O and radius OC describe a circle, this circle will touch the given line in the given point C. - - With center O and radius O'D describe a circle ; it will touch the line AB in D, and the other circle in the point P. * * Therefore two circles are described touching each other at a point P, and each touching the given line AB, one of them at the given point C in it. & * PROPOSITION III. PROBLEM. To describe a circle which shall touch a straight line given in position, and pass through two given points. Analysis. Let A B be the given straight line, and C, D the two given points. - - - - Suppose the circle required which passes through the points C, D to touch the line AB in the point E. - - Join CD, . and produce DC to meet AB in F. and let the circle be described having the center Z, join also LE, and draw LH perpendicular to CD. Then CD is bisected in H, and LE is perpendicular to A.B. - Also, because from the point F without the circle, are drawn two º lines, one of which FE touches the circle, and the other FDC cuts it ; - - therefore the rectangle contained by FC, FD, is equal to the square on F.E. (III, 36.) º 396 GEOMETRICAL EXERCISES. Synthesis. Join C, D, and produce CB) to meet AB in F, take the point E in FB, such that the square on FE shall be equal to the rectangle FD, F.C. (II. 14.) Bisect CD in H, and draw HK perpendicular to CD ; then HK passes through the center. (III. 1, Cor. 1.) At E. draw EG perpendicular to FB, then EG passes through the center, (III. 19,) consequently L, the point of intersection of these two lines, is the center of the circle. - It is also manifest, that another circle may be described passing through C, D, and touching the line AB on the other side of the point F'; and this circle will be equal to, greater than, or less than the other circle, according as the angle CFB is equal to, greater than, or less than the angle CFA. - PROPOSITION IV. PROBLEM. To describe a circle passing through two given points and touching a given circle, --- - Analysis. Let PAB be the required circle passing through the two given points A, B, and touching the given circle whose center is 0 in the point P. * At Plet PQ the common tangent be drawn to the two circles, and let AB be joined and produced to meet PQ in Q. - Then the rectangle QA, QB is equal to the square on QP. From Q draw any line QCD cutting the given circle in C, D, then the rectangle QC, QD is equal to the square on QP; but the rectangle QA, QB is equal to the square on QP; therefore the rectangle QC, QD is equal to the rectangle QA, QB, wherefore the points A, B, C, D are in the circumference of a circle. Synthesis. Describe a circle passing through the given points A, B and intersecting the given circle in any points C, D Draw AB, CD and produce them to meet in Q. From Q draw QP to touch the given circle in P. : Then the problem is reduced to that of describing a circle which shall pass through two given points A, B, and touch a given straight line QP at a given point P in the line. If therefore a circle ABP be described by the preceding Problem passing through the two points 4, B, and touching the line QP at the point P; then the circle PAB is the circle required. TANGENCIES, - 397 I. - 5. Draw a straight line which shall touch a given circle, and make a given angle with a given straight line. - 6. Draw a tangent to an arc (of given length) of a circle, so that the length of the tangent intercepted between the extreme radii of the arc shall equal a given straight line. - e. 7. In a given straight line find a point such that the straight line drawn from it to touch a given circle, shall be equal to a given straight line. • * -- 8. Draw a tangent to one circle the portion of which intercepted by the circumference of another circle shall equal a given straight line; provided that if the center of that circle lie without the circumference of the other, the straight line be not greater than the diameter; or if within the circumference, than the chord which touches the former circle where the line joining the centers cuts it. 9. If two tangents AB, A C be drawn to a circle, and D be any point in the circumference, the sum of the angles ABD, A CD, is constant. - - - II. 10. Describe a circle which shall pass through a given point and which shall touch a given straight line in a given point. * 11. Describe a circle touching a given straight line and cutting a given circle in such a way that the chord of intersection may pass at a given distance from the center of the circle. . 12. Describe a circle which shall have its center in a given straight line, touch another given line, and pass through a fixed point in the first given line. r - 13. Describe the circles which shall pass through a given point and touch two given straight lines. - -14. Shew how to describe a circle that shall have its center in a given straight line, which shall pass through a given point, and also touch another given straight line. - - * 15. Describe a circle which shall touch one given line in a given point, and shall from another given line intercept a chord of a given length. - 16. Describe a circle through a given point, and touching a given straight line, so that the chord joining the given point and point of contact may cut off a segment containing a given angle. 17. Draw a circle touching a given line in a given point, and such that the angle in the segment cut off by another given line may be equal to a given angle. - - 18. With a given radius describe a circle which shall touch two given straight lines. 19. Describe a circle to touch two lines given in position, and such that a tangent drawn to it from a given point may be equal to a given line. - - r 20. Describe a circle to touch two right lines given in position, so that lines drawn from a given point to the points of contact shall contain a given angle. • 21. Describe a circle touching a given line at a given point, such ... tangents drawn to it from two given points in the line may be parallel. - 398 GEOMETRICAL ExERCISEs. 22. A, B, are two fixed points on the circumference of a given circle, P a moveable point on the circumference, on PB is taken a point Aſ such that PA’: PA in a constant ratio; and on PA a point B, such that PB' : PB in the same ratio; prove that A'B' always touches a fixed circle. - 23. Describe a circle touching three given lines which are not all parallel. How many such circles are there in general? and under what circumstances are there only two such circles? 3 * III. 24. Describe a circle which shall have a given radius and touch a given straight line and a given circle. 25. Describe a circle with a given center, such that the circle so described and a given circle may touch one another internally. 26. To describe a circle which shall touch a given circle in a given point, and also a given straight line. 27. Describe a circle which shall have its center in a given line, and shall touch a circle and a straight line given in position. 28. Describe a circle which shall touch a given circle, have its center in a given straight line, and pass through a given point in the same straight line. .* - & 29. Find the position of a straight liné, such that every two tangents drawn from the same point in this line to two given circles, may be equal. 30. 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 a fixed point in the circumference of the given circle. 31. The common chord of two circles is produced to any point P; PA touches one of the circles in A, PBC is any chord of the other: shew that the circle which passes through A, B, C touches the circle to which PA is a tangent. 32. A CB being the arc of a circle, it is required to find in it a point C such that the circle described with center. C and radius CA, shall touch a given circle whose center is in B. - 33. Describe a circle touching a straight line in a given point, and also touching a given circle. When the line cuts the given circle, shew that your construction will enable you to obtain six circles touch- ing the given circle and the given line, but not necessarily in the given point. IV. 34. Two points are given, one in each of two given circles; de- scribe a circle passing through both points and touching one of the circles. - 35. Draw a circle touching a given circle in a given point, and also touching another given circle. - - 36. To describe a circle touching a given circle in a given point, and passing through a given point not in the circumference of the given circle. In what case is this impossible? - 37. Describe a circle which shall touch a given circle, and each of two given straight lines. - - 38. Two given circles touch each other externally, describe a circle with given radius which shall touch them both. TANGENCIES. - 399 39. Find the semicircle which circumscribes two given circles which touch each other, and find the condition that the problem may be ossible. - P 40. If two circles touch each other externally, describe a circle which shall touch one of them in a given point, and also touch the other. In what case does this become impossible 7 } 41. Describe a circle touching a given straight line, and also two - given circles. - 42. Describe a circle which shall pass through a given point and touch a given straight line and a given circle. 43. With a given radius to describe a circle, touching two given circles. - …” 44. Through a given point draw a circle touching two given circles. 45. If a circle be described touching two given circles, prove that the line joining the points of contact always passes through a given oint. p 46. Draw a circle which shall touch two given circles and have its center upon a given straight line. * 47. One circle cuts another at the extremities of a diameter, to draw a circle touching these two circles, and having its center in the line perpendicular to the diameter at its extremity. , - 48. Describe a circle touching one given circle, and bisecting the circumference of another. 4 49. If two circles touch each other externally, and a straight line which touches both of them intersect another straight line passing through their centers, at a point whose distance from the nearer circle. is equal to its diameter, the radius of one of the circles will be twice as great as that of the other. - - - - 50. The center of a given circle is equidistant from two given straight lines; to describe another circle which shall touch the two straight bines, and shall cut off from the given circle a segment con- . taining an angle equal to a given rectilineal angle. - V. 51. To find a point P, so that tangents from it to the outsides of two equal circles which touch each other, may contain an angle equal to a given angle. 52. Given two circles; it is required to find a point from which tangents may be drawn to each, equal to two given straight lines. 53. A straight line and two circles are given. Find the point in the straight line from which tangents drawn to the circles shall be equal. 54. Find a point without two circles, such, that the tangents drawn therefrom to the circles shall contain equal angles. WI. 55. Describe three equal circles touching each other, and each passing through an angle of a given equilateral triangle. 56. With three given points, not lying in a straight line, describe three circles which shall have three common tangents. - 57. Find a point from which, if straight lines be drawn to touch 400 GEOMETRICAL EXERCISES. * x three given circles, none of which lies within the other, the tangents so drawn shall be equal. 58. Describe three circles touching each other and having their centers at three given points. In how many different ways may this be done P - -. 59. Describe in a given circle three circles which shall touch one another and the given circle. • º 60. Find the center and diameter of a circle that touches three given circles, each of which touches the other two. 61. From three given points as centers, describe three circles each of which touches the other two. In how many ways may this be done P Find also the center of the circle which passes through the points of contact. e } 62. If three circles touch each other in any manner, the tangents at the points of contact pass through the same point. • 63. Given three unequal circles which do not intersect, and let pairs of double tangents be drawn internally to each pair of them, the three intersections will be in one right line. & 64. The centers of three circles (A, B, C,) are in the same straight line, B and C touch each other externally and A internally, if a line be drawn through the point of contact of B and C, making any angle with the common diameter, then the portion of this line intercepted between C and A, is equal to the portion intercepted between B and A. 65. A, B, C, are three given points, find the position of a circle such that all the tangents to it drawn from the points A, B, C shall be equal to one another. What is that circle which is the superior limit to those that satisfy the above condition? - 66. A, B, C, are three given points in the same plane, but not in the same straight line, determine the center and the position of a circle, such that three tangents AP, AQ, AR, drawn from the points A, B, C, shall be respectively equal to three given straight lines. 67. The straight line AB joining A, B, the centers of two circles, whose radii are R, r respectively, is divided in C, so that A C*- B Cº’s Jº — r", and a straight line is drawn from C perpendicular to AB; prove that the tangents drawn to both circles from any point in this line are equal. VII. 68. Draw a straight line which shall touch two given circles; (1) on the same side; (2) on the alternate sides. 69. A common tangent is drawn to two circles which touch each other externally; if a circle be described on that part of it which lies between the points of contact, as diameter, this circle will pass through the point of contact of the two circles, and will touch the line which joins their centers. - * 70. Let C, C' be the centers of two circles, draw two lines touch- ing them on the same side in A, A', and on opposite sides in B, B', then AA*– BB” = 4 OA. O'A'. - 71. Find the point in the line joining the centers of two circles which do not meet, from which the tangents drawn to the two circles are equal. GEOMETRICAL EXERCISES ON BOOK XI. ** THEOREM I. If a straight line be perpendicular to a plane, its projection on any other, plane, produced if necessary, will out the common intersection of the two planes at right angles. Let AB be any plane, . and CEF another plane intersecting the former at any angle in the line EF; and let the line GH be perpendicular to the plane CEF. H c Draw GK, HL perpendicular on the plane AB, - - - and join LK, - - then LK is the projection of the line GH on the plane AB; • * * produce EF, to meet KL in the point L; then EF, the intersection of the two planes, is perpendicular to LK, the projection of the line GH on the plane A.B. Because the line GH is perpendicular to the plane CEF, every plane passing through "GH, and therefore the projecting plane GHKZ is perpendicular to the plane CEF; . - * the º ecting plane GHLK is perpendicular to the plane AB; constr. - - hence the planes CEF, and AB are each perpendicular to the third plane G.HLPſ; -- - therefore EF, the intersection of the planes AB, CEF, is perpen- dicular to that plane; . - and consequently, EF is perpendicular to every straight line which meets it in that plane; but EF meets LK in that plane. s . Wherefore, EF is perpendicular to KL, the projection of GH on the plane A.B. ID D 402. GEOMETRICAL ExPRCISEs THEOREM II. Prove that four times the square described wpon the diagonal of a rectangular parallelopiped, is equal to the sum of the squares described on the diagonals of the parallelograms containing the parallelopiped. Tet AD be any rectangular parallelopiped; and AI), JBG two dia- gonals intersecting one another; also AG, BD, the diagonals of the two opposite faces HF, CE. A F H G. T} i— E C - D Then it may be shewn that the diagonals AD, BG, are equal; as also the diagonals which join CF and HE: and that the four diagonals of the parallelopiped are equal to one another. The diagonals AG, BD of the two opposite faces HF, CE are equal to one another: also the diagonals of the remaining pairs of the oppo- site faces are respectively equal. - - And since AB is perpendicular to the plane CE, it is perpendicular to every straight line which meets it in that plane, - - - - therefore AB is perpendicular to BD, and consequently ABD is a right-angled triangle. Similarly, GDB is a right-angled triangle. And the square on AD is equal to the squares on AB, BD, (I. 47.) also the square on BD is equal to the squares on BC, CD, therefore the square on AD is equal to the squares on AB, BC, CD; similarly the square on BG or on AD is equal to the squares on AB, BC, CD. Wherefore the squares on AD and BG, or twice the square on AID, is equal to the squares on AB, BC, CD, AB, BC, CD; but the squares on BC, CD are equal to the square on BID, the diagonal of the face CE; . - similarly, the squares on AB, BC are equal to the square on the diagonal of the face HB; - - also the squares on AB, CD, are equal to the square on the diagonal of the face BF; for CD is equal to B.E. - Hence, double the square on AD is equal to the sum of the squares on the diagonals of the three faces HF, HB, BC. \ - In a similar manner, it may be shewn, that double the square on the diagonal is equal to the sums of the squares on the diagonals of the three faces opposite to HF, HB, BC. - r Wherefore, four times the square on the diagonal of the parallelopi- ped is equal to the sum of the squares on the diagonals of the six faces. - w . - on Book XI. - 403 - I. * 3. If two straight lines are parallel, the common section of any two planes passing through them is parallel to either. 4. If two straight lines be parallel, and one of them be inclined at any angle to a plane; the other also shall be inclined at the same angle to the same plane. ſo - 5. Parallel planes are cut by parallel straight lines at the same angle. - g --- - - º If two straight lines in space be parallel, their projections on any plane will be parallel. - 7. Shew that if two planes which are not parallel be cut by two other parallel planes, the lines of section of the first by the last two will contain equal angles. | 8. If four straight lines in two parallel planes be drawn, two from one point and two from another, and making equal angles, with another plane perpendicular to these two, then if the first and third be parallel, the second and fourth will be likewise. 9. Draw a plane through a given straight line parallel to another . given straight line. - - 10. Through a given point it is required to draw a plane parallel to both of two straight lines which do not intersect. II, 11. From a point above a plane two straight lines are drawn, the one at right angles to the plane, the other at right angles to a given line in that plane; shew that the straight line joining the feet of the perpendiculars is at right angles to the given line. - 12. AB, AC, AD are three given straight lines at right angles to one another, AE is drawn perpendicular to CD, and BE is joined. Shew that BE is perpendicular to CD. • ' - 13. If perpendiculars AF, AF be drawn to a plane from two points A, A above it, and a plane be drawn through A perpendicular to AA'; its line of intersection with the given plane is perpendicular to FF". - - 14. A, B, C, D are four points in space, AB, CD are at right angles to each other, and also AC, BD; shew that AD, BC will also be at right angles to one other. 15. Two planes intersect each other, and from any point in one of them a line is drawn perpendicular to the other, and also another line perpendicular to the line of intersection of both ; shew that the plane which passes through these two lines is perpendicular to the line of intersection of the plane. . 16. ABC is a triangle, the perpendiculars from A, B on the oppo- site sides meet in D, and through Dis drawn a straightline perpendicular to the plane of the triangle ; if E be any point in the line, shew that . J.A., BC; EB, CA; and EC, AB are respectively perpendicular to each other. - - - - 17. Find the distance of a given point from a given line in space. 18. , Draw a line perpendicular to two lines which are not in the same plane. . - \ 19. Two planes being given perpendicular to each other, draw a third perpendicular to both. . • DD 2 404 GEOMETRICAL ExPRCISEs III. - - 20. Two perpendiculars are let fall from any point on two given planes, shew that the angle between the perpendiculars will be equal to the angle of inclination of the planes to one another. s 21. If through any point two straight lines be drawn equally in- clined, the first to one plane and the second to another, shew that the angle between the lines is equal to the angle between the planes. - 22. Two planes intersect, straight lines are drawn in one of the planes from a point in their common intersection making equal angles with it, shew that they are equally inclined to the other plane. 23. Two planes intersect at right angles in the line AB; at a point C in this plane are drawn CE and CF in one of the planes so that the angle ACE is equal to ACF. CE and CF will make equal angles with any line through C in the other plane. IV. 24. Three straight lines not in the same plane, but parallel to and equidistant from each other, are intersected by a plane, and the points of intersection joined; shew when the triangle thus formed will be equilateral and when isosceles. - 25. Three parallel straight lines are cut by parallel planes, and the points of intersection joined, the figures so formed are all similar and equal. - 26. If a straight line PBob cut two parallel planes in B, b, P and p being equidistant from the planes, and PAa, poC be other lines drawn from P, p, to cut the planes, then the triangles ABC, abo will be equal to one another. - - 27. If two straight lines be cut by four parallel planes, the two segments, intercepted by the first and second planes, have the same ratio to each other as the two segments intercepted by the third and fourth planes. 28. If three straight lines, which do not all lie in one plane, be cut in the same ratio by three planes, two of which are parallel, shew that the third will be parallel to the other two, if its intersections with the three straight lines are not all in one straight line. 29. To describe a circle which shall touch two given planes, and pass through a given point. - 30. Three lines not in the same plane meet in a point; if a plane scut these lines at equal distances from the point of intersection, shew that the perpendicular from that point on the plane will meet it in the center of the circle inscribed in the triangle, formed by the portion of the plane intercepted by the planes passing through the lines. V. 31. A solid angle is contained by the planes BOC, COA, AOB: AD is drawn perpendicular to the plane BOC, and DB, DC are drawn in that plane perpendicular to OB, OC respectively: if AB, A C be joined, shew that they are perpendicular to OB, OC respectively. 32. Three straight lines, not in the same plane, intersectin a point, and through their point of intersection another straight line is drawn within the solid angle formed by them; prove that the angles which ON BOOK XI. 405 this straight line makes with the first three are together less than the sum, but greater than half the sum, of the angles which the first three make with each other. ... we 33. If a solid angle be formed at A by three plane angles BAC, BAD, CAD; the three planes which bisect the three angles contained by the planes ABC, A CD; ACD, ADB ; ADB, ABC, respectively, intersect each other in a straight line passing through A. 34. If two solid angles bounded by any number of plane angles, and having a common vertex, be such that one lies wholly within the other, the sum of the plane angles bounding the latter will be greater than the sum of the plane angles bounding the former. 35. If a polygon having only salient angles lie within another, and these polygons be made the bases of pyramids having a common vertex, the sum of the plane angles at the vertex of the outer pyramid will be greater than the sum of those at the vertex of the inner. - 36. Given the three plane angles which contain a solid angle. Find by a plane construction, the angle between any two of the containing planes. - - 37. Two of the three plane angles which form a solid angle, and also the inclination of their planes being given, to find the third plane angle. WI. 38. If a straight line be divided into two parts, the cube of the whole line is equal to the cubes of the two parts, together with thrice the right parallelopiped contained by their rectangle and the whole 6. . - 39. If planes be drawn through the diagonal and two adjacent edges of a cube, they will be inclined to each other at an angle equal to two-thirds of a right angle. - 40. When a cube is cut by a plane obliquely to any of its sides, the section will be an oblong, always greater than the side, if made by cutting opposite sides. Draw a plane cutting two adjacent sides so that the section shall be equal and similar to the side. 41. A cube is cut by a plane perpendicular to a diagonal plane, and making a given angle with one of the faces of the cube. Find the angle which it makes with the other faces of the cube. 42. Shew that a cube may be cut by a plane, so that the section shall be a square greater in area than the face of the cube in the pro- portion of 9 to 8. , , 43. Shew that if a cube be raised on one of its angles so that the diagonal passing through that angle shall be perpendicular to the . plane which it touches, its projection on that plane will be a regular hexagon. 44. If a four-sided solid be cut off from a given cube, by a plane passing through the three sides which contain any one of its solid angles, the square of the number of standard units in the base of this solid, shall be equal to the sum of the squares of the numbers of simi- lar units contained in each of its sides. * - - 45. If any point be taken within a given cube, the square described on its distance from the summit of any of the solid angles of the cube, is equal to the sum of the squares described on its several perpendi cular distances from the three sides containing that angle. * * * 406 - GEOMETRICAL ExERCISEs •º ºr • 46. Trisect a cube. - 47. A rectangular parallelopiped is bisected by all the planes drawn through the axis of it. * { • , , 48. If three limited straight lines be parallel, and planes pass through each two of them, and the extremities be joined, a prism will be formed, the ends of which will be parallel if the straight lines be equal. g - - Q. 49. Given the lengths and positions of two straight lines which do not meet when produced and are not parallel; form a parallelopiped of which these two lines shall be two of the edges. 50. The content of a rectangular parallelopiped whose length is any multiple of the breadth, and breadth the same multiple of the depth, is the same as that of the cube whose edge is the breadth. 51. If a right-angled triangular prism be cut by a plane, the volume of the truncated part is equal to a prism of the same base and of height equal to one-third of the sum of the three edges. 52. In an oblique parallelopiped the sum of the squares on the four diagonals, equals the sum of the squares on the twelve edges. 53. Construct a rectangular parallelopiped equal to a given cube, and such, that its three edges shall be continued proportionals, WIT. 54. How many triangular pyramids may be formed whose edges are six given straight lines, of which the sum of any three will form a triangle 7 55. Having three points given in a plane, find a point above the plane equidistant from them. - - - 56. A, B are two fixed points in space, and CD a constant length of a given straight line; prove that the pyramid formed by joining the four points A, B, C, D is always of the same magnitude, on whatever part of the given line CD be measured. , 57. Bisect a triangular pyramid by a plane passing through one of its angles, and cutting one of its sides in a given direction. 58. Shew that the six planes passing through one edge of a tri- angular pyramid and bisecting the opposite edge meet in a point. 59. Shew how to find the content of a pyramid, whatever be the figure of its base, the altitude and area of the base being given. 60. Compare the content of a triangular pyramid with the content of the parallelopiped of whose faces the edges are diagonals. 61. ABC, the base of a pyramid whose vertex is 0, is an equila- teral triangle, and the angles BOC, COA, A 0B are right angles; shew that three times the square on the perpendicular from 0 on ABC, is equal to the square on the perpendicular, from any of the other angular points of the pyramid, on the faces respectively opposite to them. * 62. Two triangles have a common base, and their vertices are in a straight line perpendicular to the plane of the one; there are given the vertical angle of the other, the angles made by each of its sides with the plane of the first and the distance of the vertices of the two triangles, to find the common base. - ON BOOK XI. 407 63. ABCDE is a regular pentagon, on AD, A C are described equilateral triangles with a common vertex F"; if a plane through B0 cut AF, DF, in extreme and mean ratio in G, H, shew that ‘G.HCB is a Square. - - - . . 64. If a pyramid with a polygon for its base he cut by a plane parallel to the base, the section will be a polygon similar to the base. WIII. 65. If a straight line be at right angles to a plane, the intersection of the perpendiculars let fall from the several points of that line on another plane, is a straight line which makes right angles with the common section of the two planes. 66. Find the locus of those points which are equidistant from three given planes. i 67. Two planes intersect; shew that the loci of the points, from which perpendiculars on the planes are equal to a given straight line, are straight lines; and that four planes may be drawn, each passing through two of these lines, such that the perpendicular from any point in the line of intersection of the given planes upon any one of the four planes, shall be equal to the given line. 68. If there be two straight lines which are not parallel, but which do not meet, though produced ever so far both ways, shew that two parallel planes may be determined so as to pass, the one through the one line, the other through the other; and that the perpendicular' distance of these planes is the shortest distance of any point that can - be taken in the one line from any point taken in the other. 69. Of all the angles, which a straight line makes with any straight lines drawn in a given plane to meet it, the least is that which mea- sures the inclination of the line to the plane. - 70. Shew that if a straight line meets two others not in the same plane with one another, and is perpendicular to both; the part of it intercepted between them is the shortest line that can be drawn from any point in one of them to any point in the other. - 71. Find a point in a given straight line such that the sum of its distances from two given points (not in the same plane with the given straight line) may be the least possible. - 72. If, round a line which is drawn from a point in the common section of two planes at right angles to one of them, a third plane be made to revolve, shew that the plane angle made by the three planes , is then the greatest, when the revolving plane is perpendicular to each of the two fixed planes. - • 73. Two points are taken on a wall and joined by a line which passes round a corner of the wall. This line is the shortest when its parts make equal angles with the edge at which the parts of the wall meet. - - 74. Prove that among all parallelopipeds of given volume, a cube is that which has the least surface. - . . . . GEOMETRICAL EXERCISES ON BOOK XII. THEOREM I. If semicircles ADB, BEC be described on the sides AB, BC of a right- angled triangle, and on the hypotenuse another semicircle AFBGC be des- cribed, passing through the vertex B ; the lunes AFBD and BGCE are together equal to the triangle ABC. It has been demonstrated (XII. 2.) that the areas of circles are to one another as the squares on their diameters; it follows also that semicircles will be to each other in the same proportion. Therefore the semicircle ADB is to the semicircle ABC, as the square on AB is to the square on AC, and the semicircle CEB is to the semicircle ADC as the square on BC is to the square on AC, hence the semicircles ADB, CEB, are to the semicircle ABC as the squares on AB, BC are to the square on AC; - - but the squares on AB, BC are equal to the square on A C : (I. 47.) therefore the semicircles ADB, CEB are equal to the semicircle ABC. (v. 14.) From these equals take the segments AFB, BGC of the semicircle on AC, and the remainders are equal, - n that is, the lunes AFBD, B.G.C.E are equal to the triangle B.A. C. THEOREM II. If on any two segments of the diameter of a semicircle semicircles be described, all towards the same parts, the area included between the three circumferences (called åp}m}\oc) will be equal to the area of a circle, the diameter of which is a mean proportional between the segments. Let ABC be a semicircle whose diameter is AB, and let AB be divided into any two parts in D, and on AD, DC let two semicircles be described on the same side; also let DB be drawn perpendicular to A. C. ON Book XII. - 409 Then the area contained between the three semicircles, is equal to the area of the circle whose diameter is B.D. - - IB A I} C. - . Since AC is divided into two parts in C, - the square on AC is equal to the squares on AD, DC, and twice the rectangle AD, DC; (II. 4.) and since BD is a mean proportional between AD, DC; ; the rectangle AD, DC is equal to the square on DB, (VI. 17.) therefore the square on AC is equal to the squares on AD, DC, and twice the square on D.B. But circles are to one another as the squares on their diameters or radii, (XII. 2.) - therefore the circle whose diameter is AC, is equal to the circles whose diameters are AD, DC, and double the circle whose diameter is BD ; . . - wherefore the semicircle whose diameter is A C is equal to the circle whose diameter is BD, together with the two semicircles whose diameters are AD and DC: if the two semicircles whose diameters are AD and DC be taken from these equals, - - therefore the figure comprised between the three semi-circumfer- ences is equal to the circle whose diameter is D.B. THEOREM III. Every section of a sphere by a plane is a circle. If the plane pass through the center of the sphere, it is manifest that the ºtion is a circle, having the same diameter as the generating semi- ©ll'Cle. But if the cutting plane does not pass through the center, let AEB be any other section of the sphere made-by a plane not passing through the center of the sphere. - } - Rs. 2. I - . . . K %3 Take the center C, and draw the diameter HCK perpendicular to the section AEB, and meeting it in D; - draw AB passing through D, and join A C; take E, F, any other points in the line AEB, and join CE, DE; CF, DF, * 410 GEOMETRICAL EXERCISEs * Then since CD is perpendicular to the plane AEB, it is perpen- dicular to every straight line which meets it in that plane, . . * therefore the angles CDA, CDE, CDF are right angles, ! and CA, CF, CE, being lines drawn from C, the center of the sphere to points in the surface, are therefore equal to one another. -- - - Hence, in the right-angled triangles CDA, CDF, CDE; the square on DA is equal to the difference of the squares on CA and CD; - . and the square on DF is equal to the difference of the squares on CF and CD; - * also the square on DE is equal to the difference of the squares o CE and CD: - therefore the squares on DA, DF, B.E are equal to one another; and therefore the lines DA, DF, DE are equal to one another. But DA, DF, DE are three equal lines drawn from the same point D, in the same plane, - hence the points A, F, E lie in the circumference of a circle of - which D is the center. THEOREM IV. There can be only five regular solids. If the faces be equilateral triangles. The angle of an equilateral triangle is one-third of two right angles; and six angles, each equal to the angle of an equilateral triangle, are equal to four right angles; and therefore a number of such angles less than Sia, but not less than three are necessary to form a solid angle. Hence there cannot be more º three regular figures whose faces are equal and equilateral tri- angles. • - If the faces be squares. Since four angles, each equal to a right angle, can fill up space round a point in a plane. A solid angle may be formed with three right angles, but not with a number greater or less than three. Hence, there cannot be more than one regular solid figure whose faces are equal squares. If the faces be equal and regular pentagons. Since each angle of a regular pentagon is a right angle and a fifth of a right angle: the magnitude of three such angles being less than four right angles, may form a solid angle, but four, or more than four, cannot form a solid angle. Hence, there cannot be more than one regular figure whose faces are equal and regular pentagons. - If the faces be equal and regular hexagons, heptagons, octagons, or any other regular figures; it may be shewn that no number of them can form a solid angle. - - . f Wherefore there cannot be more that five regular solid figures, of which, there are three, whose faces are equal and equilateral tri- angles; one whose faces are equal squares; and one, whose faces are equal and regular pentagons. & - on Book XII. 411 PROPOSITION V. PROBLEM. - To construct the five regular solids. The regular Tetrahedron. - Each of the angles of an equilateral triangle is one-third of two right angles; a solid angle may therefore beformed by three angles of three equal and equilateral triangles, and the figureformed by the three bases of the triangles is manifestly an equilateral triangle equal in magnitude to each of the three given equilateral triangles. The angles of incli- nation of every two of the four faces are also equal. - The regular Octahedron. .* - - Through any point 0 draw three straight lines perpendicular to each other, take OA, Oa, OB, Ob, OC, Oc equal to one another, and join the extremities of these lines. The faces ABC, A50, &c. are equilateral triangles equal to one another and eight in number; also the inclinations of every two contiguous faces are equal. Thé regular Icosahedron. - A solid angle may be formed with five angles, each equal to the angle of an equilateral triangle. At the point A of an equilateral triangle ABC, let a solid angle be formed with it and four other equal and equilateral triangles ABD, ADE, AEF, AFC each equal to the triangle ABC. Next at the point B, let another solid angle be formed with the triangle ABC and four others BCH, BHK, BKD, BDA, each equal to it. The solid angle at B is equal to the solid angle at A, and the inclinations of every two contiguous faces are equal; also the two solid angles have two faces ABC, ABD common. Next let a third solid angle be formed at C, by placing the two triangles CFG, CGH contiguous to the three CAB, CFA, CHB. The solid angle at C is equal to that at A or B, and the inclinations of the contiguous faces make equal angles. Thus two equal and equilateral triangles are placed contiguous one to another, forming three solid angles at A, B, C, and having every two contiguous faces equally inclined: also the solid angles formed at D, E, F, G, H, K, have alternately three and two angles of the equilateral triangles. In the same manner let another figure equal to this be formed with ten equal and equilateral triangles, each equal to the triangle ABC. If these two figures be connected together, so that the points at which there are two angles of one figure, may coincide with the points which contain three angles with the other, there will be formed at the points D, E, F, G, H, K, six equal solid angles, each contained by five angles of the equilateral triangles, and every two contiguous faces will have the same inclination. - - - Hence a figure of twenty faces is formed, each equal to the equi- lateral triangle ABC, and having the inclinations of every two con- tiguous faces equal. - - . The regular Hexahedron. : p - Since three right angles may form a solid angle, it is therefore obvious that the solid angle formed by three equal squares has every. two of the faces equally inclined to one another; and with three other squares, each equal to the former, a figure is formed, bounded by six equal squares, and having every two contiguous faces at right angles to one another. - R. The regular Dodecahedron. Since three angles each equal to the angle of a regular pentagon may 412 GEOMETRICAL ExERCISEs form a solid angle; let ABCDE be a regular pentagon, and with two others each equal to this, let a solid angle at A be formed; the inclina- tions of every two contiguous faces will be equal. At the points B, C, D, E successively, let solid angles be formed by pentagons equal to 4 BCDE. The solid angles at B, C, D, E, are each equal to the solid angle at A, and the inclination of every two contiguous faces is the same. Thus is formed a figure with six equal and regular pentagons, having the inclination of every two contiguous faces equal, and the angles at the linear boundary of the figure alternately consisting of an angle of a pen- tagon and of two angles of two pentagons equally inclined to each other. Next, let another figure equal to this be constructed with six pen- tagons, each equal to the pentagon ABCDE. If these two figures be so placed that the angular points of the plane angles in the linear boundary of one, may coincide with the points at which there are two angles in the other figure; at each of these points will be formed ten solid angles, each equal to the angle at A, and having the inclination of every two contiguous faces equal to one another. Hence a regular figure is formed having twelve equal faces, and the inclinations of every two contiguous faces equal to one another. I. 6. Find the radius of a circle whose area is eight times as great as that of another whose radius and area are known. - - 7. Construct a circle the area of which shall have a given ratio to that of a given circle. - - 8. Divide a circle into any number of equal parts by means of concentric circles. g - 9. ABC is a circle of given radius, describe another concentric circle abo whose area shall be equal to three times the area of ABC. 10. To divide a circle into any number of qual parts, the perimeters of which shall be equal to the circumference of the circle. ~. 11. If the diameter AB of a circle be divided into an odd number (n) of equal parts, and C and D be the # (n − 1) th and # (n − 1)th divi- sions; and AEC, AFD, CGB, DHB be semicircles; shew that the perimeter of the figure AECGBHDF is equal to that of the circle, and its area an nth part of the area of the circle. 12. Let AB and DC be two diameters of a given circle, at right angles to each other; A.E.B a circular arc described with radius DB or DA; prove that the area of the lune AEBC = area of triangle ADB. 13. Two circles touch each other internally, and the area of the lune cut out of the larger is equal to twice the area of the smaller circle. Required the ratio of the diameters of these circles. - 14. The diameter of a circle is divided into two parts, upon each of which as diameters circles are described; when the remaining area of the great circle is equal to that of one of these two circles, find the ratio which the parts of the diameter bear to one another. 15. The diameter of a semicircle ADB is divided into two parts in C (so that the length of A C is twice that of BC), and upon them are, described the semicircles AEC, CFB. Compare the areas of the circles . which are described on éach side of the common tangent CD So as to touch it and the two semicircles x on Book XII. 4:13 - 16. The centers of three circles A, B, and C are in the same right - line, B and C touch A internally, and each other externally; P, Q be- ing the points where A is touched by B, C respectively; to find a point R on A such that the portion of the lune PR intercepted between I3 and A may be equal to the portion of QR between C and A. 17. On the chord of a quadrant a semicircle is described ; re- quired the area of the crescent thus formed. 18. Semicircles are described upon the radii CA, CB of a quad- rant, and intersect each other in a point D, shew (a) That the area common to both semicircles is equal to the area without them. (b) That the remaining areas of the two semicircles are equal, each is one-fourth of the square on A.C. . 19. . If on one of the radii of a quadrant a semicircle be described; and on the other, another semicircle so described as to touch the former and the quadrantal arc; compare the area of the quadrant with the area of the circle described in the figure bounded by the three curves. 20. Any right-angled triangle BAC is inscribed in a semicircle, A being the right angle, and AD a perpendicular on the base B.C. If circles be described on the sides BA, AC as diameters, prove that the areas of these circles will always be to each other in the same ratio as the segments into which the base is divided by the line AID. 21. If on the two sides of a right-angled triangle, semicircles be described, and a circle be described touching them both, it will include the circle whose diameter is the hypotenuse; and the space between the two circles will be to the outer circle as twice the rectangle of the sides of the triangle to the square on the sum of the sides. II. 22. In different circles the radii which bound equal sectors contain angles reciprocally proportional to their circles. - 23. Prove that the sectors of two different circles are equal, when their angles are inversely as the squares of the radii. 24. If the arc of a semicircle be trisected, and from the points of Section lines be drawn to either extremity of the diameter, the diffe- rence of the two segments thus made will be equal to the sector which stands on either of the arcs. - . 25. If A B be a circular arc, center 0, and AD be drawn perpen- dicular to BO, and the arc AC taken equal to AD, then the sector B00 equals the segment ACB. 26. If two points B, D, be taken at equal distances from the ends of the arc of a quadrant, and perpendiculars BG, DH be drawn to the extreme radius; the space BGHD shall be equal to the sector BOD. 27. If circles be inscribed in the triangles formed by drawing the altitude of a triangle right-angled at the vertex, the circles and the triangles are proportional. 28. If a semicircle be described on the hypotenuse AB of a right- angled triangle ABC, and from the center E, the radius ED be drawn at right angles to AB, shew that the difference of the segments on the two sides equals twice the sector C.E.D. - , 29. If semicircles be described upon the sides of a right-angled triangle on the interior, the difference between the sum of the circular 414 GEOMETRICAL, EXERCISES. segments thus standing upon the exterior of the sides and segments of the base, equals the space intercepted by the circumferences described on the sides. e - * 30. , AB, BD are two radii of a circle at right angles to each other. Produce BD to C, and make BD equal to the arc A.D. Join AC cutting the circumference in E. Then the area. EDC is equal to the area of the segment A.E. * 31. ABC is an isosceles right-angled triangle. On BC is described a semicircle BDEC, and BFC is a circle whose radius is AB and center A. The segment BCF is equal to the segments BAD, A CE. 32. The circle inscribed in a square is equal to four equal circles touching one another and the sides of that square internally. t 33. If the diagonals of a quadrilateral inscribed in a circle cut each other at right angles, and circles be described on the sides; prove that. the sum of two opposite circles will be equal to the sum of the other two. - . . . 34. If two chords of a circle intersect each other either within or without the circle at right angles; and if on these segments as dia- meters, circles be described, the areas of these four circles are together equal to that of the original circle. 35. Shew that the semicircles described on the diagonals of a right-angled parallelogram together equal the sum of the semicircles described on the sides. 36. A quadrilateral inscribed in a circle has a diameter passing through the center ; or has its two diameters at right angles to one another; on the sides of the quadrilateral semicircles are described; the four crescents outside, are together equal to the quadrilateral. III. 37. Two straight lines are inclined to each other at a given angle, find the area of all the circles which can be described touching each other and the two given lines, the position of the center of the last circle being given. - 38. The centers of three circles A, B and C are in the same right line, B and C touch A internally, and each other externally. Shew that the portion of the area of A, which is outside B and C, is equal to the area of the semicircle described on the chord of A, which touches B and C at their point of contact. - - - _ • 39. If three equal circles intersect, so that each of the circumfe- rences pass through the centers of the other two, the spaces bounded by the circumferences intercepted will be all equal. 40. Take any three points A, B, C in the circumference of a circle. Join AB, BC, AC, and draw AD, AE parallel to the tangents at B and C, and meeting BC produced if necessary in D and E; and prove that the segments BD and EC are to each other in the duplicate ratio of AB to A.C. g - . IV. 41. If from a point above the plane of a circle straight lines be drawn to the circumference; there will be only two of them equal in length, and they will be equidistant from the shortest and longest, and on opposite sides. What is the exception to this proposition? As-- ON BOOK XII. - 415. 42. Apply Euc. VI. 19, to shew that the area of a section of a cone parallel to the base varies as the triangle which any diameter of that section makes with the slant sides. - | - * * 43. If through a point without the plane of a circle straight lines be drawn to its circumference and a plane be drawn parallel to the circle on either side of the point, the points of intersection of the lines with the plane will be in a circle, and the area of this and of the first circle will be as the squares on their distances from the given point. 44. If two circles be so placed that their planes are parallel, and the straight line which joins their centers perpendicular to the plane of each ; the straight lines which join the opposite extremities of any pair of parallel diameters will all intersect the straight line which joins the centers in the same point: if the circles be equal, that point. will be the bisection of the aforesaid straight line. - W. 45. Equal straight lines whose extremities are in the surface of a sphere, are equally distant from the center of the sphere. 46. Shew that the angle between the plane of two great circles of a sphere, is measured by the arc of a great circle which joins their poles. , - - . . - 47. If two parallel planes cut a sphere so that the sections are equal, they are equidistant from the center. • ‘ 48. A straight line or a plane can only touch a sphere at one point; and at that point the radius of the sphere will be perpen- dicular to the line or plane. - 49. Shew that all lines drawn from an external point to touch a sphere are equal to one another; and thence prove that if a tetrahe- dron can have a sphere inscribed in it touching its six edges, the sum of every two opposite edges is the same. * - - 50. If two equal circles cut one another in the diameter, and a plane cut them perpendicularly to the same diameter, the points of section of this plane with the circumferences, are in a circle. 51. The squares on the diameter of a sphere and the side of the inscribed cube, are as 3 to 1. - - 52. If three straight lines intersect each other within a sphere at right angles, each at right angles to the plane of the other two; . the sum of the squares on the six segments is equal to the Square on the diameter of the sphere, together with twice the rectangle of the segments of the diameter made at the point of intersection. 53. Having given an irregular fragment, which contains a portion of spherical surface; shew how the radius of the sphere, to which the fragment belongs, may be practically determined. • 54. Let three given spheres be placed on a horizontal plane in mutual contact with each other; find the sides of the triangle formed by joining the points in which the spheres touch the plane. - 55. A pyramid of triangular base is composed of ten spherical balls of given radius; the base is composed of six, the next larger of three, and the remaining one is placed upon them. Find the distance of the upper ball from the ground. * - VI. 56. All the sections of a tetrahedron made by planes parallel to the base are similar to the base. - 416 GEOMETRICAL ExFRCISEs. 57. Find the inclination of two contiguous faces of a tetrahedron to each other. - - - 58. If on the base of a regular tetrahedron lines be drawn from any two angles to bisect the opposite sides; the line joining their point of section with the vertex of the solid is at right angles with the base. 59. ABCD is a regular tetrahedron; from the vertex A, a perpen- dicular is drawn to the plane BCD meeting it in 0. Shew that three times the square on A0 is equal to twice the square on A.B. 60. If the shortest distances between opposite edges of a tetra- hedron be mutually at right angles, they will bisect the edges. 61. Prove that the shortest distance between two opposite edges of a regular tetrahedron is equal to half the diagonal of the square de- scribed on an edge. - * * - 62. If in a tetrahedron the shortest distances between the opposite edges are mutually at right angles, prove that these distances meet in a point, that they bisect each other, and that the opposite edges of the tetrahedron are equal. - - g - 63. If the line joining the bisections of two edges of a tetrahedron which do not meet be bisected, the point so found is distant from the base one-fourth of the perpendicular altitude of the tetrahedron. 64. If the angles of a regular tetrahedron be joined to the centers of the circles inscribed in its faces, the joining lines will form the edges of a new tetrahedron parallel to those of the old. 65. The perpendicular drawn from any angle of a regular tetrahedron upon the opposite face, will meet that face in the center of the circle which circumscribes that face. VII. - 66. The angles of inclination of the faces of a regular tetrahedron and of a regular octahedron are supplementary to each other. 67. Find the dihedral angle contained by two adjacent faces of a regular octahedron; and find its solidity. - 68. If lines be drawn joining the centers of the faces of a cube; these will be the edges and diagonals of a regular octa- hedron, and the Square on the diagonal is double the square on the edge. VIII. 69. Inscribe a sphere within a tetrahedron. - 70. Describe a sphere about a regular tetrahedron. º 71. Given the side of a regular octahedron, find the radius of the inscribed and circumscribed spheres. & . 72. Draw three diameters of a sphere, each at right angles to the other two ; then the six points where the extremities of the diameters meet the surface of the sphere, will be the angles of a regular octahedron, and the lines joining the adjacent points will be the edges, also the three diameters of the sphere its dia- gonals. * . . . " 73. If three planes intersect each other at right angles, so that the planes pass through a fixed point; find the locus of a point, such that the sum of the squares on its distances from the three planes may be constant for all points in the locus. s 74. Describe a sphere about a triangular pyramid, three of whose faces are at right angles to one another. - e * *- GEOMETRICAL EXERCISES ON BOOK I. HINTS, &c. 8. This is a particular case of Euc. I. 22. The triangle however may be de- scribed by means of Euc. 1.1. Let AB be the given base, produce AB both ways to meet the circles in D, E (fig. Euc. I. - 1.9; with center A, and radius AE, describe a circle, and with center B and radius BD, describe another circle cutting the former in G. Join GA, G.B. t 9. In the diagram, Euc. I. 2, with center D and radius DA, describe a circle intersecting the circle CGH in the point M, join DM and produce it to meet the circumference of the circle GRL in N : then MN shall be equal to AL. What is the condition that this construction may be possible * 10. One property is proved by Euc. I. 6, 8: the other by Euc. I. 32, 13, 5. 11. Let fall also a perpendicular from the vertex on the base. 12. Apply Euc I. 4. 13. Let CAB be the triangle (fig. Euc. I. 10.), CD the line bisecting the angle ACD and the base AB. Produce CD, and make DE equal to CD, and join AE. Then CB may be proved equal to AE, also AE to AC. 14. Let AB be the given line, and C, D the given points. From C draw CE perpendicular to AB, and produce it making EF equal to CE, join FD, and pro- duce it to meet the given line in G, which will be the point required. 15. Make the construction as the enunciation directs, then by Euc. 1, 4, BII is proved equal to CK; and by Euc. I. 13, 6, OB is shewn to be equal to OC. 16. Let A, B be the given points, PQ the given line; and let AF, BF be the lines required meeting the line PQ in F, and making the angle AFP equal to the angle BFQ. Draw AC perpendicular to PQ and let it meet BF produced in G. Then AC may be shewn to be equal to CG. *. - This problem has been proposed as a theorem in the following form:—A and B are two given points, AC and BD perpendicular to a given straight line CD ; AD and BD intersect in E, EF is perpendicular to CD ; prove that AF and BF make equal angles with CD. w 17. Let A, B, be the two given points, join AB cutting the given line in E; in the line take any other point F, and join AF, B.F. Then Euc. I. 8, the angles AEF, BEF are equal, and they are adjacent angles. 18. The angle BCD is equal to the sum of the angles ABC, ADC. 19. The angles ADE, AED may be each proved to be equal to the comple- ments of the angles at the base of the triangle. 20. The angles CAB, CBA, being equal, the angles CAD, CBE are equal, Euc. I, 13. Then, by Euc. I.4, CD is proved to be equal to CE. And by Euc. 㺠the angle at the vertex is shewn to be four times either of the angles at the base. - 21. Let AB, CD be two straight lines intersecting each other in E, and let P be the given point, within the angle AEC. Draw EF bisecting the angle AEC, and through P. draw PGH parallel to EF, and cutting ED, EB in G, H. Then EG is equal to EH. And by bisecting the angle DEB and drawing through P a line parallel to this line, another solution is obtained. It will be found that the two lines are at right angles to each other. 22. Let the two given straight lines meet in A, and let P be the given point. Iet PQR be the line required, meeting the lines AQ, AR in Q and R, so that PQ is equal to Q.R. Through P. draw PS parallel to AR and join RS. Then APSR is a parallelogram, and AS, PR the diagonals. Hence the construction. 23. Let the two straight lines AB, AC meet in A. In AB take any point D, and from AC cut off AE equal to AD, and join DE. On DE, or DE produced, take DF equal to the given line, and through F draw FG parallel to AB meeting AC in G, and, through G draw GH parallel to DE meeting AB in H. Then GH is the line required. - E E 418 GEOMETRICAL EXERCISES * . 24. The two given points may be both on the same side, or one point may be on each side of the line. If the point required in the line be supposed to be found, and lines be drawn joining this point and the given points, an isosceles triangle is formed, and if a perpendicular be drawn on the base from the point in the line; the construction is obvious. - 25. The problem is simply this—to find a point in one side of a triangle from which the perpendiculars drawn to the other two sides shall be equal. If all the positions of these lines be considered, it will readily be seen in what case the problem is impossible. - 26. It may be shewn that the angle contained between the side produced and the line drawn from the other extremity of the base to meet it, is twice either of the angles at the base, if the triangle be obtuse-angled; but equal to the angle at the vertex, if the triangle be acute-angled. 27. Construct the figure and apply Euc. I. 5, 32, 15. If the isosceles triangle have its vertical angle less than two-thirds of a right angle, the line ED produced, meets AB produced towards the base, and then 3. AEF = 4 right angles + AFE. If the vertical angle be greater than two-thirds of a right angle, ED produced meets AB produced towards the vertex, then 3. AEF = 2 right angles + AFE. 28. Let ABC be an isosceles triangle, and from any point D in the base BC, and the extremity B, let three lines DE, DF, BG be drawn to the sides and making equal angles with the base. Produce ED and make DH equal to DF and ioin BH. - - - J 29. In the isosceles triangle ABC, let the line DFE which meets the side AC in D and AB produced in E, be bisected by the base in the point E. Then DC may be shewn to be equal to B.E. 30. If two equal straight lines be drawn terminated by two lines, which meet in a point, they will cut off triangles of equal area. Hence the two triangles have s a common vertical angle and their areas and bases equal. By Euc. I. 32 it is shewn that the angle contained by the bisecting lines is equal to the exterior angle at the base. 31. This may be proved indirectly. - 32. (1), (3) Apply Euc. I. 26, 4. (2) The equal lines which bisect the sides may be shewn to make equal angles with the sides. - 33. The different cases offer no occasion for remark. Taking one of them, that in which one angle ABC is triple of another angle ACB of the triangle. At B make the angle CBD equal to BCA, then BD is equal to CD : and the angles CDB, CBD are each double of the angle ACB. 34. Let AB, AC contain any angle at A, and from any point P draw PB, PC respectively perpendicular to AB, AC ; the angle BPC is equal to BAC. 35. At C make the angle BCD equal to the angle ACB, and produce AB to meet CD in D. 36. Let BC be bisected in D, join AD, then AD is equal to BD or DC. Through D draw DE parallel to CA, and DF to BA. Then in the triangles BED, DFC, by Euc. 1. 26, BE is equal to DF; and DF is equal to E.A. Euc. 1. 34, whence in the triangles BED, AED, Euc. I. 4, BD is equal to D.A. 37. Let ABC be a triangle, having the right angle at A, and the angle at C greater than the angle at B, also let AD be perpendicular to the base, and A.E be the line drawn to E the bisection of the base. Then AE may be proved equal to BE or EC independently of Euc. III. 31. 38. Produce EG, FG to meet the perpendiculars CE, BF, produced if necessary. The demonstration is obvious. - 39. If the given triangle have both of the angles at the base, acute angles; the difference of the angles at the base is at once obvious from Euc. I. 32. If one of the angles at the base be obtuse, does the property hold good? * 40. Let ABC be a triangle having the angle ACB double of the angle ABC, and let the perpendicular AD be drawn to the base BC. Take DE equal to DC and join AE. Then AE may be proved to be equal to EB. tw If ACB be an obtuse angle, then AC is equal to the sum of the segments of the base, made by the perpendicular from the vertex A, - 41. Let the sides AB, AC of any triangle ABC be produced, the exterior angles bisected by two lines which meet in D, and let AD be joined, then AD bisects the angle BAC. For draw DE perpendicular on BC, also § DG perpen- diculars on AB, AC produced, if necessary. Then DF may be proved equal to DG, and the squares on DF, DA are equal to the squares on DG, GA, of which the square on FD is equal to the square on DG ; hence AF is equal to AG, and Euc, 1, 8, the angle BAC is bisected by AD. ON BOOK I. HINTs, &c. 419 42. The line required will be found to be equal to half the sum of the two sides of the triangle. - - - 43. Apply Euc. I. 1,9. for the trisection of a right angle. To trisect one- fourth of a right angle. If an equilateral triangle be described on one of the sides of a triangle which contains the given angle, and a line be drawn to bisect that angle of the equilateral triangle which is at the given angle, the angle con- tained between this line and the other side of the triangle will be one-twelfth of a right angle, or equal to one-third of the given angle. It may be remarked, generally, that any angle which is the half, fourth, eighth, &c. part of a right angle, may be trisected by Plane Geometry. 44. Apply Euc. 1. 20. It may also be shewn that the greatest of the three sides of a scalene triangle lies between one-half and one-third of the perimeter of the triangle. - 45. Apply Euc. r. 20. 46. Apply Euc. r. 20, 16. - - 47. For the first case, see Theo. 36, p. 300 : for the other cases, apply Euc. I. 19. 48. This is obvious from Euc I. 26. 49. By Euc 1, 29, 6, FC may be shewn equal to each of the lines E.F, FG. 50. Join GA and AF, and prove GA and AF to be in the same straight line. 51. Apply Euc. I. 32. 52. Let the straight line drawn through D parallel to BC meet the side AB in E, and AC in F. Then in the triangle EBD, EB is equal to ED, by Euc. 1 29, 6. Also, in the triangle EAD, the angle EAD may be shewn equal to the angle EDA, whence EA is equal to ED, and therefore AB is bisected in E. In a similar way it may be shewn, by bisecting the angle C, that AC is bisected in F. Or the bisection of AC in F may be proved when AB is shewn to be bisected in E. 53. The triangle formed will be found to have its sides respectively parallel to the sides of the original triangle. 54. If a line equal to the given line be drawn from the point where the two lines meet, and parallel to the other given line; a parallelogram may be formed, and the construction effected, . - 55. Let P be the given point in the side AB produced, of the triangle ABC. Supposing PDQ the line required cutting the base BC in D and meeting the side AC in Q, so that PD is equal to DQ. Through Q draw QE parallel to BC and DF parallel to AB, Then BE may be shewn to be equal to BP, whence the . If Q be supposed to be given, then the line QDP meets the side AB produced. - 56. Let ABC be the triangle; AD perpendicular to BC, AE drawn to the bisection of BC, and AF bisecting the angle BAC. Produce AD and make DA' equal to AD : join FAſ, EA'. 57. If the point in the base be supposed to be determined, and lines drawn from it parallel to the sides, it will be found to be in the line which bisects the vertical angle of the triangle. - 58. Let ABC be the triangle, at C draw CD perpendicular to CB and equal to the sum of the required lines, through D draw DE parallel to CB meeting AC in E, and draw EF parallel to DC, meeting BC in F. Then EF is equal to DC. Next produce CB, making CG equal to CE, and join EG cutting AB in H. From H draw HK perpendicular to EAC, and HL perpendicular to BC. Then HK and HL together are equal to DC. The proof depends on Theorem 28, p. 300. 59. Let C’ be the intersection of the circles on the other side of the base, and join AC", BC'. Then the angles CBA, CBA being equal, the angles CBP, C'BP are also equal, Euc. 1. 13: next by Euc. I, 4, CP, PC' are proved equal; lastly prove CC' to be equal to CP or PC'. . - 60. In the fig. Euc. I. 1, produce AB both ways to meet the circles in D and E, join CD, CE, then CDE is an isosceles triangle, having each of the angles at the base one-fourth of the angle at the vertex. At E draw EG perpendicular to DB and meeting DC produced in G. Then CEG is an equilateral triangle. - 61. Join CC', and shew that the angles CCF, CCG are equal to two right angles; also that the line FC'G is equal to the diameter. - . 62. Construct the figure and by Euc. I. 32. If the angle BAC be a right angle, then the angle BDC is half a right angle. - * 63. Let the lines which bisect the three exterior angles of the triangle ABC form a new triangle A'B'C'. Then each of the angles at A', B', C may be shewn to be equal to half of the angles at A and B, B and C, C and A respectively. And it will be found that half the sums of every two of three unequal numbers whose sum is constant, have less differences than the three numbers themselves. F. E 2 420 GEOMETRICAL ExERCISEs 64. The first case may be shewn by Euc. 1. 4: and the second by Euc. 1. 32, 6, 15. - , * 65. At D any point in a line EF, draw DC perpendicular to EF and equal to fhe given perpendicular on the hypotenuse. With center C and radius equal to the given base, describe a circle cutting EF in B. At C draw. CA perpendicular to CB and meeting EF in A. Then ABC is the triangle required. 66. Let ABC be the required triangle having the angle ACB a right angle. In BC produced, take CE equal to AC, and with center B and radius BA describe a circular arc cutting CE in D, and join AD. Then DE is the difference between the sum of the two sides AC, CB and the hypotenuse AB; also one side AC the perpendicular is given. Hence the construction. On any line EB take EC equal to the given side, ED equal to the given difference. At C, draw CA perpendicular to CB, and equal to EC, join AD, at A in AD make the angle DAB equal to ADB, and let AB meet EB in B. Then ABC is the triangle required. 67. (1) Let ABC be the triangle required, having ACB the right angle. Produce AB to D making AD equal to AC or CB : then BD is the sum of the sides. Join DC: then the angle ADC is one-fourth of a right angle, and DBC is one-half of a right angle. Hence to construct: at B in BD make the angle DBM equal to half a right angle, and at D the angle BDC equal to one-fourth of a right angle, and let DC meet BM in C. At C draw CA at right angles to BC meeting BD in A : and ABC is the triangle required. - This Problem has been proposed in the following form :—Divide a given straight line into two parts, so that the square on one part may be dºuble of the square on the other part. - (2) Let ABC be the triangle, C the right angle: from AB cut off AD equal to AC; then BD is the difference of the hypotenuse and one side. Join CD ; then the angles ACD, ADC are equal, and each is half the supplement of DAC, which is half a right angle. Hence the construction. - 68. Take any straight line terminated at A. Make AB equal to the difference of the sides, and AC equal to the hypotenuse. At B make the angle CBD equal to half a right angle, and with center A and radius AC describe a circle cutting BD in D : join AD, and draw DE perpendicular to AC. Then ADE is the required triangle. - 69. Let BC the given base be bisected in D. At D draw DE at right angles to BC and equal to the sum of one side of the triangle and the perpendicular from the vertex on the base: join DB, and at B in BE make the angle EBA equal to the angle BED, and let BA meet DE in A : join AC, and ABC is the isosceles triangle. n - - 70. This construction may be effected by means of Prob. 16, p. 299. 71. This problem is reduced to that of constructing two right-angled triangles of the same altitude, one on each side of the perpendicular, when the altitudes i. the difference between the hypotenuse and the base of each triangle is In OWIſle 72. Let ABC be the equilateral triangle of which a side is required to be found, having given BD, CD the lines bisecting the angles at B, C. Since the angles DBC, DCB are equal, each being one-third of a right angle, the sides BD, DC are equal, and BDC is an isosceles triangle having the angle at the vertex the supplement of a third of two right angles. Hence the side BC may be found. - 73. Let the given angle be taken, (1) as the included angle between the given sides; and (2) as the opposite angle to one of the given sides. In the latter case, an ambiguity will arise if the angle be an acute angle, and opposite to the less of the two given sides. 74. Let ABC be the required triangle, BC the given base, CD the given diffe- rence of the sides AB, AC : join BD, then DBC by Euc. I. 18, can be shewn to be half the difference of the angles at the base, and AB is equal to AD. Hence at B in the given base BC, make the angle CBD equal to half the difference of the angles at the base. On CB take CE equal to the difference of the sides, and with center C and radius CE, describe a circle cutting BD in D : join CD and produce it to A, making DA equal to DB. Then ABC is the triangle required. 75 On the line which is equal to the perimeter of the required triangle describe a triangle having its angles equal to the given angles. Then bisect the angles at the base; and from the point where these lines meet, draw lines parallel to the sides and meeting the base. . l 76. Let ABC be the required triangle, BC the given base, and the side AB greater than AC. Make AD equal to AC, and draw CD. Then the angle BCD may be shewn to be equal to half the difference, and the angle DCA equal to half ON BOOK I. HINTs, &c. - .421 the sum of the angles at the base. Hence ABC, ACB the angles at the base of the triangle are known. - - 77. Let the two given lines meet in A, and let B be the given point. If BC, BD be supposed to be drawn making equal angles with AC, and if AD : and DC be joined, BCD is the triangle required, and the figure ACBD may be shewn to be a parallelogram. Whence the construction. 78. It can be shewn that lines drawn from the angles of a triangle to biseet the opposite sides, intersect each other at a point which is two-thirds of their lengths from the angular points from which they are drawn. Let ABC be the triangle required, AD, B.E, CF the given lines from the angles drawn to the bisections of the opposite sides and intersecting in G. Produce GD, making DH equal to DG, and join BH, CH: the figure GBHC is a parallelogram. Hence the con- struction. 79. Let ABC (fig. to Euc. 1. 20) be the required triangle, having the base BC equal to the given base, the angle ABC equal to the given angle, and the two sides BA, AC together equal to the given line B.D. Join DC, then since AD is equal to AC, the triangle ACD is isosceles, and therefore the angle ADC is equal to the angle ACD. Hence the construction. - 80. Let ABC be the required triangle (fig. to Euc. 1, 18), having the angle ACB equal to the given angle, and the base BC equal to the given line, also CD equal to the difference of the two sides AB, AC. If BID be joined, then ABD is an isosceles triangle. Hence the synthesis. Does this construction hold good in all cases 2 81. Let ABC be the required triangle (fig. Euc. I. 18), of which the side BC is given and the angle BAC, also CD the difference between the sides AB, A.C. Join BD ; then AB is equal to AD, because CD is their difference, and the triangle ABD is isosceles, whence the angle ABD is equal to the angle ADB ; and since BAD and twice the angle ABD are equal to two right angles, it follows that AJ3D is half the supplement of the given angle BAC. Hence the construction of the triangle. § Let AB be the given base: at A draw the line AD to which the line bisecting the vertical angle is to be parallel. At Bdraw BE parallel to AE); from A draw AE equal to the given sum of the two sides to meet BE in E. At B make the angle EBC equal to the angle BEA, and draw CF parallel to AD. Then ACB is the triangle required. 83. An indefinite number of triangles may be constructed with the given data; the only essential condition required is, that "the sum of the two sides, be divided into two parts, such that the sum of any two of the three lines shall be greater than the third. 84. By adding the three given lines together is obtained a line equal to double the perimeter of the triangle : whence the three sides may be found and the tri- angle constructed. . - 85. From the sum and difference of the two sides of the triangle, the sides can be found, and the base is given. Hence the triangle can be constructed. , 86. Let ABC be the given triangle, PQ, PR, PS the three given lines meet- ing in P. Bisect the vertical angle BAC of the triangle by AD. Through P draw a line EPF making with PR, the same angle that AD makes with BC. At any point G in PC, make the angle PGF equal to BAD and PGE equal to DAC : the triangle GFE is equiangular to ABC. Draw Fö parallel to PG meet- ing PS in b, ba to FG meeting PQ in a, and ac to GE meeting PQ in c; join be, then abo is a triangle fulfilling the required conditions. - 87. Apply Euc. I. 13, 32. 88. At the point A in AB the given base, make the angle BAC equal to the given angle, and at B, make the angle ABC equal to the complement of the angle given between the side BC and the perpendicular from C upon the base AB : from C draw CD perpendicular on AB; then the triangle ACB has the required conditions. . 89. If the three perpendiculars be equal, the triangle is equilateral, and the problem becomes that of constructing a right-angled triangle of which the base is given and the hypotenuse double of the perpendicular. Next consider the isosceles triangle, and lastly the scalene triangle. 90. Take any point in the given line, and apply Euc. I. 23, 31. 91. On one of the parallel lines take EF equal to the given line, and with center E and radius EF describe a circle cutting the other in G. Join EG, and through. A draw ABC parallel to EG. 92. This will appear from Euc. 1, 29, 15, 26. 422 GEOMETRICAL ExFRCISES 93. Let AB, AC, AD, be the three lines. Take any point E in AC, and on EC make EF equal to EA, through F draw FG parallel to AB, join GE and produce it to meet AB in H. Then GE is equal to G.H. 94. Apply Euc. I. 32, 29. $ 95. From E draw EG perpendicular on the base of the triangle, then ED and EF may each be proved equal to EG, and the figure shewn to be equilateral, Three of the angles of the figure are right angles. 96. ſet ED be bisected in F, and AF be joined. Then AF, FE, FD are equal to one another and to AB. Hence Euc. i. 5, 15, 29, the angle ABE is double of EBC. - 97. Draw the lines AB, AC making an angle CAB equal to one of the angles of the parallelogram. If BAC be less than a right angle, place DE between AB, AC, and equal to the shorter diagonal: bisect DE in F, join AF and produce it making FG equal to AF, then AG is the other diagonal, join GE, GD, then GEAD is the parallelogram. 98. If the parallelogram be equilateral, it will be found that the four lines meet in a point; if not equilateral, the four lines by their intersections form a parallelogram. 99. Let AB, AC be the two given straight lines meeting at A and inclined to each other at an angle BAC. At the point A draw AD at right angles to AB and equal to one of the given lines, and AE at right angles to AC and equal to the other given line. Draw DP, EP, parallel to AB, AC respectively and meeting in the point P. P is the point required. For draw PM, PN perpendicular to AB, AC respectively : then PM is equal to AD and PN to BE. This gives one solution of the question. 100. Let ABC be the triangle, with center B and radius BA describe a circle cutting the base BC in D. Through D draw DE parallel to CA, and through E draw EF parallel to BC and meeting AC in F. Then EF shall be parallel to BC and equal to the difference between the sides AB, BC. * 101. At the point C draw a line parallel to the given line, and let DEF be the required line, intersected by AB and terminated by AC and CB produced, so that DE is equal to EF. Draw DG parallel to AB, then GB is equal to B.F. Whence the method of drawing the line, 102, Let the lines EG and FG be drawn, as also the diagonals AC, B.D. Then the angle ABE is equal to the angle CBG, and when the angle ABG is added to each of these, the angles EBG, ABC are equal in the triangles EGB, ACB ; whence by Euc. I. 4, EG is proved equal to the diagonal AC. And similarly FG is proved equal to BD. 103. This problem is the same as the following:—Having given the base of a triangle, the vertical angle and the sum of the sides, to construct the triangle. This triangle is one-half of the required parallelogram. 104. Draw a line AB equal to the given diagonal, and at the point A make an angle BAC equal to the given angle. Bisect AB in D, and through D draw a line parallel to the given line and meeting AC in C. This will be the position of the other diagonal. Through B draw BE parallel to CA, meeting CD produced in E: join AE, and BC. Then ACBE is the parallelogram required. 105. Construct the figures and by Euc. I. 24. 106. By Euc. I. 4, the opposite sides may be proved to be equal. 107. Let ABCD be the given parallelogram; construct the other parallelogram A'B'C'D' by drawing the lines required, also the diagonals AC, A'C', and shew that the triangles ABC, A'B'C' are equiangular. . If the parallelogram be a square, and the lines drawn from the angles, bisect the sides of the square, the area of the smaller square is one-fifth of the given Square. - - 108. AT)' and B'C' may be proved to be parallel. 109. It may be shewn that AP, CR are each one-third of the diagonal AC ; as also BS, DQ each one-third of the diagonal BD. The diagonals PR, QS of the quadrilateral PQRS, mutually bisect one another. 110. Prove that the diagonals of EFGH bisect one another. 111. Apply Euc. I. 29, 32. 112. The points D, D', are the intersections of the diagonals of two rectangles; if the rectangles be completed, and the lines OD, OD’ be produced, they will be the other two diagonals. 113. Let ABCD be the parallelogram, and first let the straight line PQ be drawn through A one of the angular points and fall without the parallelogram : let 18B, CC, DD’ be the perpendiculars on PQ from B, C, D. From B draw BE ON BOOK I. HINTS, &c. 423, arallel to AC", and the truth is manifest: next let the line PQ be so drawn as to #. within the parallelogram. Thirdly. If the line PQ do not pass through any angle of the figure, but intersect AB, AD two adjacent sides or AB, CD, two op- posite sides, and let AA’ be the perpendicular from A on PQ. Through A, C, draw lines parallel to PQ and produce BB", DD’ to meet these parallels in a, b, c, d. If the perpendiculars drawn from opposite angles be not in the same direc- tion, the difference instead of the sum must be read. - 114. The triangles OSP, SPR are equal, Euc. I. 38, as also OST, SRT : whence the triangle OPT is equal to TPR; but TPR is equal to TRQ, therefore OPT º, fºul to RTG); wherefore the triangle TPQ is twice OPT, and QT is twice OT, 115. On constructing the figure, it will be found to differ from that of Euc. I. 37, only by having the line GKH drawn through the intersection of AC, BD and parallel to AD. The proof offers no difficulty. - Let the converse be tried—If through K the intersection of AC, BD, GKH be drawn parallel to AD or BC, shew that GK is equal to K.H. * - 116. Let the diagonals intersect in E. In the triangles DCB, CDA, two angles in each are respectively equal and one side DE: wherefore the diagonals DB, AC are equal: also since DE, EC are equal, it follows that EA, EB are equal. Hence DEC, AEB are two isosceles triangles having their vertical angles equal, wherefore the angles at their bases are equal respectively, and therefore the angle CDB is equal to DBA. - 117. (1) By supposing the point P found in the side AB of the parallelogram ABCD, such that the angle contained by AP, PC may be bisected by the line PD; CP may be proved equal to CD ; hence the solution. , , (2) By supposing the point P found in the side AB produced, so that PD may bisect the angle contained by ABP and PC ; it may be shewn that the side AB must be produced, so that BP is equal to BD. - - 118. This may be shewn by Euc. I. 35. 119. Let D, E, F be the bisections of the sides AB, BC, CA of the triangle ABC : draw DE, EF, FD; the triangle DEF is one-fourth of the triangle ABC. The triangles DBE, FBE are equal, each being one-fourth of the triangle ABC : DF is therefore parallel to BE, and DBEF is a parallelogram of which DE is a diagonal. #. This may be proved by applying Euc. I. 38. 121. Apply Euc. I. 37, 38. 122. On any side BC of the given triangle ABC, take BD equal to the given base; join AD, through C draw CE parallel to AD, meeting BA produced if necessary in E, join ; then BDE is the triangle required. By a process some- what similar the triangle may be formed when the altitude is given. 123. Apply the preceding problem (122) to make a triangle equal to one of the given triangles and of the same altitude as the other given triangle. Then the sum or difference can be readily found. - 124. First construct a triangle on the given base equal to the given triangle; next form an isosceles triangle on the same base equal to this triangle. 125. Make an isosceles triangle equal to the given triangle, and then this isos- celes triangle into an equal equilateral triangle. - 126. Make a triangle equal to the given parallelogram upon the given line, and then a triangle equal to this triangle, having an angle equal to the given angle. 127. If the figure ABCD be one of four sides; join the opposite angles A, C of the figure, through D draw DE parallel to AC meeting BC produced, in E, join AE:—the triangle ABE is equal to the four-sided figure ABCD. If the figure ABCDE be one of five sides, produce the base both ways, and the figure may be transformed into a triangle, by two constructions similar to that employed for a figure of four sides. If the figure consists of six, seven, or any number of sides, the same process must be repeated. . 128. Draw two lines from the bisection of the base parallel to the two sides of the triangle. + 129. This may be shewn ex absurdo. - 130. (1) DF bisects the triangle ABC (fig. Prop. 6, p. 297). On each side of the point F in the line BC, take FG, FH, each equal to one-third of BF, the lines DG, DH shall trisect the triangle. Or, . Let ABC be any triangle, D the given point in BC. Trisect BC in E, F. Join AD, and draw EG, FH parallel to AD. Join DG, DH ; these lines trisect the tri- angle. Draw AE, AF and the proof is manifest. (2) Let ABC be any triangle; trisect the base BC in D, E, and join AD, AE. * 424 GEOMETRICAL ExeRCISES * From D, E, draw DP, EP parallel to AB, AC and meeting in P. Join AP, BP, CP; these three lines trisect the triangle. - * (3) Let P be the given point within the triangle ABC. Trisect the base BC in D, E. From the vertex A draw AD, AE, AP. Join PD, draw AG parallel to PD and join PG. Then BGPA is one-third of the triangle. The problem may be solved by trisecting either of the other two sides and making a similar construction. - - 131. It is proved, Euc. I. 34, that each of the diagonals of a parallelogram bisects the figure, and it may be shewn that they also bisect each other. It is hence manifest that any straight line, whatever may be its position, which bisects a parallelogram, must pass through the intersection of the diagonals. - The following form of the problem may be tried by the Student:—Bisect a parallelogram by a straight line such that the part intercepted by one pair of op- posite sides, may be half of that intercepted between the other two produced. This may be put more generally in the following form :—Draw a line bisecting a parallelogram, so that the part of it between two of the opposite sides may be equal to a given line. Shew that there are four cases in which only one such line can be drawn. - 132. See the remark on the preceding problem (131.) 133. Trisect the side AB in E, F, and draw EG, FH parallel to AD or BC, meeting DC in G and H. If the given point P be in EF, the two lines drawn from P through the bisections of EG and FH will trisect the parallelogram. If P be in FB, a line from P through the bisection of FH will cut off one-third of the parallelogram, and the remaining trapezium is to be bisected by a line from P, one of its angles. If P coincide with E or F, the solution is obvious. 134. Construct a right-angled parallelogram by Euc. I, 44, equal to the given quadrilateral figure, and from one of the angles, drawea line to meet the opposite side and equal to the base of the rectangle, and a line from the adjacent angle parallel to this line will complete the rhombus. * 135. Bisect BC in D, and through the vertex A, draw AE parallel to BC; with center D and radius equal to half the sum of AB, AC, describe a circle cutting AE in E. - - - g 136. Produce one side of the square till it becomes equal to the diagonal, the line drawn from the extremity of this produced side and parallel to the ad4 jacent º of the square, and meeting the diagonal produeed, determines the point TeC Ullred, - º: Let fall upon the diagonal perpendiculars from the opposite angles of the parallelogram. These perpendiculars are equal, and each pair of triangles is situated on different sides of the same base and has equal altitudes. If the point be not on the diagonal, draw through the point a line parallel to a side of the parallelogram. - 138. One case is included in Theo. 137. The other case, when the point is in the diagonal produced, is obvious from the same principle. 139. The triangles DCF, ABF may be proved to be equal to half of the parallelogram by Euc. I. 41. - 140. Apply Euc. I. 41, 38. 141. If a line be drawn parallel to AD through the point of intersection of the diagonal, and the line drawn through O parallel to AB ; then by Euc. 1. 43, 41, the truth of the theorem is manifest. - - 142. It may be remarked that parallelograms are divided into pairs of equal triangles by the diagonals, and therefore by taking the triangle ABD equal to the triangle ABC, the property may be easily shewn. 143. The triangle ABD is one-half of the parallelogram ABCD, Euc. I. 34. And the triangle DKC is one-half of the parallelogram CDHG, Euc. 1. 41, also for the same reason the triangle AKB is one-half of the parallelogram AHGB: there- fore the two triangles T)KC, AKB are together one-half of the whole parallelogram ABCD. Hence the two triangles DKC, AKB are equal to the triangle ABD : take from these equals the equal parts which are common, therefore the triangle CKF is equal to the triangles AHK, KBID : wherefore also taking AHK from these equals, then the difference of the triangles CKF, AHK. is equal to the triangle KBC; and the doubles of these are equal, or the difference of the parallelo- grams CFKG, AHKE is equal to twice the triangle KBD, 144. . Let the diagonal AC bisect the quadrilateral figure ABCD. Bisect AC in à join BE, ED, and prove BE, ED in the same straight line and equal to one another, - 145. Apply Euc, I. 15. ON BOOK I. HINTS, &c. : 425, 146. Apply Euc. I. 20. - 147. This may be shewn by Euc. 1. 20. - - 148. Let AB be the longest and CD the shortest side of the rectangular figure. Produce AD, BC to meet in E. Then by Euc. 1. 32. - 149. If the angles at the base of the isosceles triangle be bisected, the line joining the points where the bisecting lines meet the opposite sides of the triangle, will cut off the trapezium required. - 150. Let ABCD be the quadrilateral figure, and E, F, two points in the opposite sides AB, CD, join EF and bisect it in G ; and through G draw a straight line HGK terminated by the sides AD, BC; and bisected in the point G. Then EF, HK are the diagonals of the required parallelogram. 151. After constructing the figure, the proof offers no difficulty. 152. If any line be assumed as a diagonal, if the four given lines taken two and two be always greater than this diagonal, a four-sided figure may be con- structed having the assumed line as one of its diagonals: and it may be shewn that when the quadrilateral is possible, the sum of every three given sides is greater than the fourth. o - 153. Draw the two diagonals, then four triangles are formed, two on one side of each diagonal. Then two of the lines drawn through the points of bisection of two sides may be proved parallel to one diagonal, and two parallel to the other diagonal, in the same way as Theo. 118, supra. The other property is manifest from the relation of the areas of the triangles made by the lines drawn through the bisections of the sides. * By means of this proporty a quadrilateral figure may be constructed, if the four sides be given and the line which bisects either two of the opposite sides. 154. Let ABCD be a trapezium having the side AB parallel to CD, let E be the middle point of BC, and let AE, DE be joined, then the triangle AED is one- half of the trapezium. Through E draw GEF parallel to DA and meeting AB produced in F. j 155. Let ABCD be a quadrilateral of which the side AB is equal to CD, and let the diagonals AC, BD intersect in O, the line POQ drawn through O parallel to AB or CD, is bisected in the point O. Bisect AB in E, join EO and produce it to meet CD in F. Then CD is bisected in F, whence it may be shewn that PQ is bisected in O. 156. Let ABCD be the quadrilateral whose sides AB, DC are parallel, and DC double of AB; the diagonals AC, BD intersect in a point E, such that AE is one-third of AC, and BE one-third of BC, Draw CF parallel to DA meeting AB produced in F. See Theo. 2, p. 295. - g . - 157. Let the side AB be parallel to CD, and let AB be bisected in E, and CD in F, and let EF be drawn. Join AF, BF, then Euc. I. 38. 158. Let BCED be a trapezium of which DC, BE are the diagonals intersecting each other in G. If the triangle DBG be equal to the triangle EGC, the side DE may be proved parallel to the side BC, by Euc. I. 39. - 159. Let ABCD be the quadrilateral figure having the sides AB, CD, parallel to one another, and AD, BC equal. Through B draw BE parallel to AD, then ABED is a parallelogram. - 160. Let ABCD be the quadrilateral having the side AB parallel to CD. Let E, F be the points of bisection of the diagonals BD, AC, and join EF and produce it to meet the sides AD, BC in G and H. Through H draw LHK parallel to DA jº DC in L and AB produced in K. Then BK is half the difference of DC 8 Il g e 161. (1) Reduce the trapezium ABCD to a triangle BAE by Prob. 127, supra, and bisect the triangle BAE by a line AF from the vertex. If F fall without BC, through F draw FG parallel to AC or DE, and join AG. - Or thus. Draw the diagonals AC, BD : bisect BD in E, and join AE, EC. Draw FEG parallel to AC the other diagonal, meeting AD in F, and DC in G. AG being joined, bisects the trapezium. ., (2) . Let E be the given point in the side AD. Join EB. Bisect the quad- rilateral EBCD by E.F. Make the triangle EFG equal to the triangle EAB, on the same side of EF as the triangle EAB. Bisect the triangle EFG by EH. EH bisects the figure. : - • 162. If a straight line be drawn from the given point through the intersection of the diagonals and meeting the opposite side of the square : the problem is then reduced to the bisection of a trapezium by a line drawn from one of its angles. 163. If the four sides of the figure be of different lengths, the truth of the theorem may be shewn. If, however, two adjacent sides of the figure be equal to 426. - GEOMETRICAL EXERCISES one another, as also the other two, the lines drawn from the angles to the bi- section of the longer diagonal, will be found to divide the trapezium into four triangles which are equal in area to one another. Eug. I. 38. - 164. Through the angles A, C of the quadrilateral, draw EAF, HCG, parallel to the diagonal DB, and through the angles B, D, draw FBG, EDH, parallel to *: dºgonal AC, join HF; then the triangle EHF is equal to the quadrilateral D. - 165. Produce BP to K making CK equal to C.B. Join AK, then the triangle AKB is equal to the parallelogram ABCD. Next draw PEFC cutting CD, D.A. in E, F, and meeting BA produced in H, so that the triangle HPB is equal to the triangle AKB. .- - 166. Apply Euc. 1. 47, observing that the shortest side is one-half of the longest. - d -- • 167. Find by Euc 1.47, a line the square on which shall be seven times the square on the given line. Then the triangle which has these two lines containing the right angle shall be the triangle required. - 168: Apply Euc. I.47. - 169. Let the base BC be bisected in D, and DE be drawn perpendicular to the hypotenuse AC. Join AD ; then Euc. I, 47. & 170. Construct the figure, and the truth is obvious from Euc. 1, 47. 171. See Theo. 36, p. 300, and apply Euc I. 47. * 172. Draw the lines required, and apply Euc. 1. 47. 173. Apply Euc. I.47. 174. Apply Euc. I.47. - 175. Apply Euc. I, 47, observing that the square on any line is four times the square on half the line. - 176. Apply Euc. 1. 47, to express the squares of the three sides in terms of the squares on the perpendiculars and on the segments of AIB. 177. By Euc. I, 47, bearing in mind that the square described on any line is four times the square described upon half the line. - 178. This is obvious from Euc. I. 47. 179. Apply Euc. I.47. 180. (a) This is obvious from Euc. I. 13. b) Apply Euc. I. 32, 29. c) Apply Euc. I. 5, 29. d) Let AL meet the base BC in P, and let the perpendiculars from F, K meet BC produced in M and N respectively; then the triangles APB, FMB may be proved to be equal in all respects, as also APC, CKN. (e) Let fall DQ perpendicular on FB produced. Then the triangle DQB may be proved equal to each of the triangles ABC, DBF, whence the triangle DBF is equal to the triangle ABC. & Perhaps however the better method is to prove at once that the triangles ABC, FBD are equal, by shewing that they have two sides equal in each triangle, and the included angles, one the supplement of the other. - (f) . If DQ be drawn perpendicular on FB produced, FQ may be proved to be bisected in the point B, and DQ equal to AC. Then the square on FD is found by the right-angled triangle FQD. Similarly, the square on KE is found, and the sum of the squares on FD, EK, G H will be found to be six times the Square on the hypotenuse. g - (g) Through. A draw PAQ parallel to BC and meeting DB, EC produced in P,Q. Then by the right-angled triangles. In Leybourn's Mathematical Repository, (New Series, 1814,) Vol. III, Part II. pp. 71–80, will be found a collection of more than fifty properties of a right-angled triangle, and of the lines drawn and figures described in and about the diagram which accompanies the demonstration of Euc. I, 47. • 181. Let any parallelograms be described on any two sides AB, AC of a triangle ABC, and the sides parallel to AB, AC be produced to meet in a point P. Join PA. Then on either side of the base BC, let a parallelogram be described having two sides equal and parallel to A.P. Produce AP and it will divide the parallelogram on BC into two parts respectively equal to the parallelograms on the sides. Euc. 1, 35, 36. 182, Let the equilateral triangles ABD, BCE, CAF be described on AB, BC, CA, the sides of the triangle ABC having the right angle at A. Join DC, AK : then the triangles DBC, ABE are equal. Next draw DG perpendicular to AB and join CG : then the triangles BDG, DAG, DGC are equal to one another. Also draw AH, EK perpendicular to BC; the triangles ON BOOK II. HINTS, &c. 427 EKH, EKA are equal. Whence may be shewn that the triangle ABD is equal to the triangle BHE, and in a similar way may be shewn that CAF is equal to CHE. The restriction is unnecessary: it only brings AD, AE into the same line. 183. On AB describe the equilateral triangle ABF ; then the quadrilateral figure ABECD is equal to the figure ACBF. Shew that the triangles ABD, ABE are together also equal to ACB.F. - 184. Shew that the exterior angle FGB is equal to the interior angle ACB, or that GFB is equal to CAB. GEOMETRICAL EXERCISES ON BOOK II. HINTS, &c. 5. This property may also be shewn by drawing through the point P, a line QPR parallel to BC, and drawing QS perpendicular to BC. Then by Prob. 28, p. 300, P.D., PF together are equal to QS. 6. See the figure Euc. II. 5. • 7. This problem is equivalent to the following:—Construct an isosceles ºneled triangle, having given one of the sides which contains the right angle. 8. Construct the square on AB, and the property is obvious. 9. Let AB be the given line bisected in C and produced to the point D. On AD describe a square ADEF, join DF and through B, C, draw BGH, CK parallel to AF or D.E. 10. It may be shewn that the given line must be produced to a length equal § the difference between the diagonal and the side of a square described on half the line. 11. Produce AK, FE to meet in N. If the sum of the triangles ABM, MPG, GNA be taken from the figure ABFN, the difference may be shewn to be equal : ºf the sum of the squares on AC and CD: the other property offers no ifficulty. . 12. %. line may be found the square on which is double the square on the given line. The problem is then reduced to :—Having given the hypotenuse and the sum of the sides of a right-angled triangle, construct the triangle. 13. This follows from Euc. II. 5, Cor. 14. This problem is, in other words?—Given the sum of two lines and the sum of their squares, to find the lines. Let AB be the given straight line, at B draw BC at right angles to AB, bisect the angle ABC by BD. On AB take AE equal to the side of the given square, and with center A and radius AE describe a circle cutting BD in D, from D draw DF perpendieular to AB, the line AB is divided in F as was required. 15. Let AB be the given line. Produce AB to C making BC equal to three times the square on AB. From BA cut off BID equal to BC ; then D is the point5. section such that the squares on AB and BD are double of the square OIl g º 16. In the fig. Euc. 11.7. Join BF, and draw FL perpendicular on G.D. Half the rectangle DB, BG, may be proved equal to the rectangle AB, BC. Or, join KA, CD, KD, CK. Then CK is perpendicular to BD. And the tri- angles CBD, KBID are each equal to the triangle ABK. Hence, twice the tri- angle ABK is equal to the figure CBKD; but twice the triangle ABK is equal to. the rectangle AB, BC; and the figure CBKD is equal to half the rectangle- DB, CK, the diagonals of the squares on AB, BC. Wherefore, &c. 17. The difference between the two unequal parts may be shewn to be equal: to twice the line between the points of section. 18. This proposition is only another form of stating Euc. II. 7. 19. In the figure, Theo. 7, p. 298, draw PQ, PR, PS perpendiculars on AB, . AD, AC respectively: then since the triangle PAC is equal to the two triangles PAB, PAD, it follows that the rectangle contained by PS, AC, is equal to the sum of the rectangles PQ, AB, and PR, AD. When is the rectangle PS, AC equal to the difference of the other two rectangles: - 20. Through E draw EG parallel to AB, and through F, draw FHK parallel 428 GEOMETRICAL ExERCISEs to BC and cutting EG in H. Then the area of the rectangle is made up of the areas of four triangles; whence it may be readily shewn that twice the area of the triangle AFE, and the figure AGHK is equal to the area ABCD. 21. Apply Euc. II. 11. * ..., 22. The vertical angles at L are equal, and each of them is a right angle. 23. . Apply Euc, II. 4, 11 ; I, 47. 24. Produce FG, DB to meet in L, and draw the other diagonal LHC, which passes through H, because the complements AG, BK are equal. Then LH may be shewn to be equal to Ff, and to Dd. 25. This property follows directly from Euc. II. 11. and Euc. II. 5. Cor. 26. The common intersection of the three lines divides each into two parts, one of which is double of the other, and this point is the vertex of three triangles which have lines drawn from it to the bisection of the bases. Euc. II. 12, 13. 27. Apply Theorem 3, p. 311, and Euc I. 47. 28. This theorem is that particular case of Euc. II. 12, in which the distance of the obtuse angle from the foot of the perpendicular, is half of the side sub- tended by the right angle made by the perpendicular and the base produced. 29. (1) Let the triangle be acute-angled, (Euc. II, 13, fig. 1). Let AC be bisected in E, and BE be joined; also EF be drawn perpendicular to BC. EF is equal to FC. Then the square on BE may be proved to be equal to the square on BC and the rectangle BD, BC. (2) If the triangle be obtuse-angled, the perpendicular EF falls within or with- out the base according as the bisecting line is drawn from the obtuse or the acute angle at the base. 30. This may be shewn from Theorem 3. p. 311, AG being double of GD; &c. 31. Let the perpendicular AD be drawn from A on the base BC. It may be shewn that the base BC must be produced to a point E, such that CE is equal to the difference of the segments of the base made by the perpendicular. 32. Since the base and area are given, the altitude of the triangle is known. Hence the problem is reduced to :—Given the base and altitude, and the line drawn from the vertex to the bisection of the base, construct the triangle. 33. Join BD, and by the first case of Euc. II. 13 and Euc. II. 7, is proved that twice the rectangle contained by AB, AE, is equal to twice the rectangle con- tained by AC, AD. 34. Apply Euc. II, 12, 13. 35. This follows immediately from Euc. 1. 47. 36. Apply Euc. 11. 13. 37. The truth of this property depends on the fact, that the rectangle contained by AC, CB is equal to that contained by AB, CD. - 38. Let P the required point in the base AB be supposed to be known. Join CP. It may then be shewn that the property stated in the Problem is contained in Theorem 3. p. 311. > 39. This may be shewn from Euc. I.47; II. 5. Cor. 40. From C let fall CF perpendicular on AB. Then ACE is an obtuse-angled, * and BEC an acute-angled triangle. Apply Euc. II. 12, 13; and by Euc. 1. 47, the Squares on AC and CB are equal to the square on AB. 41. Apply Euc. 1. 47, II. 4; and the note p. 84, on Euc. II. 4. 42. Ely Theo. 3. p. 311 and Euc. I.47, the property is proved. 43. Draw a perpendicular from the vertex to the base, and apply Euc. I.47; II. 5. Cor. Enunciate and prove the proposition when the straightline drawn from the vertex meets the base produced. vºy 44. This follows directly from Euc. II. 13, Case 1. 45. The truth of this proposition may be shewn from Euc. I.47; 11. 4. * 46. Let the square on the base of the isosceles triangle be described. Draw the diagonals of the square, and the proof is obvious. . . . . - 47. Let ABC be the triangle required, such that the squaré on AB is three times the square on AC or BC. Produce BC and draw AD perpendicular to BC. Then by Euc II, 12, CD may be shewn to be equal to-one half of BC. 48. Apply Euc. II. 12, and Theorem 44, p. 315:” ‘’’ - * 49. Draw EF parallel to AB and meeting the base in F; draw also EG per- pendicular to the base. Then by Euc. I.47; II. 5, Cor. 50. Bisect the angle at B by BD meeting the opposite side in D, and draw BE perpendicular to AC. Then by Euc. 1. 47; 11 5, Cor. 51. Construct the figure, and apply Euc. ii. 12, 13, observing that AB is equal to AC in the triangles ABD, ACD. 52. This follows directly from Theorem 3, p. 311. ON BOOK III. HINTS, &c. 429 53. Draw the diagonals intersecting in P, and join OP. By Theo. 3, p. 311. 54. Draw from any two opposite angles, straight lines to meet in the bisection of the diagonal joining the other angles. Then by Euc. II. 12, 13. By means of this property of a quadrilateral, it may be proved that, In every figure of five sides, if the diagonals be drawn and be bisected, and the points of bisection be successively joined, three times the sum of the squares on the sides of the figure is equal to the sum of the squares on the diagonals together with four times the sum of the squares on the lines joining the points of bisection of the diagonals. The student may try whether any analogous property exists for irregular figures of six, seven, &c. sides. - - 55. Draw two lines from the point of bisection of either of the bisected sides to the extremities of the opposite side; and three triangles will be formed, two on one of the bisected sides and one on the other, in each of which is a line drawn from the vertex to the bisection of the base. Then by Theo. 3, p. 311. 56. If the extremities of the two lines which bisect the opposite sides of the trapezium be joined, the figure formed is a parallelogram which has its sides re- spectively parallel to, and equal to, half the diagonals of the trapezium. The sum of the squares on the two diagonals of the trapezium may be easily shewn to be equal to the sum of the squares on the four sides of the parallelogram. 57. This is obvious from Euc. I. 47. - 58 Draw perpendiculars from the extremities of one of the parallel sides, meeting the other side produced, if necessary. Then from the four right-angled triangles thus formed, may be shewn the truth of the proposition. 59. Let ABCD be any trapezium having the side AD parallel to BC. Draw the diagonal AC, then the sum of the triangles ABC, ADC may be shewn to be equal to the rectangle contained by the altitude and half the sum of AD and BC. 60. Let ABCD be the trapezium, having the sides AB, CD parallel, and AD, BC equal. Join AC and draw AE perpendicular to DC. Then by Euc. II. 13. 61. Let ABC be any triangle; AHKB, AGFC, BDEC, the squares upon their sides ; EF, GH, KL the lines joining the angles of the squares. Produce GA, RB, EC, and draw HN, DQ, FR perpendiculars upon them respectively: also draw AP, BM, CS perpendiculars on the sides of the triangle. Then AN may be proved to be equal to AM; CR to CP; and BQ to BS; and by Euc. II. 12, 13. 62. Convert the triangle into a rectangle, then Euc. II. 14. 63. Find a rectangle equal to the two figures, and apply Euc. II. 14. 64. Find the side of a square equal to the rectangle. See Prob. 1. p. 310. 65. On any line PQ take AB equal to the given difference of the sides of the rectangle, at A draw AC at right angles to AB, and equal to the side of the given square; bisect AB in O and join OC; with center O and radius OC describe a semicircle meeting PQ in D and E. Then the lines AD, AE, have AB for their difference, and the rectangle contained by AD, AE is equal to the square on AC. 66. Apply Euc. II, 14. GEOMETRICAL EXERCISES ON BOOK III. HINTS, &c. 8. The given point may be either within or without the circle. Find the Center of the circle, and join the given point and the center, and upon this line describe a semicircle, a line equal to the given distance may be drawn from the given point to meet the arc of the semicircle. When the point is without the circle, the given distance may meet the diameter produced. 9. On any radius construct an isosceles right angled triangle, and produce the side which meets the circumference. . . 10. Let AQ intersect C.D. in E, and draw EF and OG perpendicular to AB. ... 11. Shew that the line joining the middle points of the chords when produced, is perpendicular to the chords. 12. The point determined by the lines drawn from the bisections of the chords and at right angles to them respectively, will be the center of the required circle. 38. The two chords form by their intersections the sides of two isosceles tri- 430 GEOMETRICAſ, EXERCISES angles, of which the parallel chords in the circle are the bases. If the chords be equal, the intersection of the lines joining their opposite extremities takes place at the center of the circle. - • 14. Join the extremities of the chords, then Euc. I. 27; III. 28. 2— --> 15. Construct the figure and the arc BC may be proved equal to the arc B'C'. 16, This is obvious from Theorem 7, p. 321. 17. This is the same as Euc. III. 34, with the condition, that the line must pass through a given point, - 18. Join AD, and the first equality follows directly from Euc, III. 20, 1, 32. Also by joining AC, the second equality may be proved in a similar way. If how- ever the line AD do not fall on the same side of the center O as E, it will be found that the difference, not the sum of the two angles, is equal to 2. AED. See note to Euc. III. 20, p. 155. - 19. First, Let the chords intersect the diameter at equal angles: then by Euc. III. 7. Secondly, let them intersect the diameter produced in the same point, then Euc. III. 8. - * 20. Let C be the center of the given circle, and AB the given line to which the required line must be parallel. Draw a line in the circle cutting off a segment containing an angle equal to the given angle Euc. III. 34: describe a concentric circle touching this chord, and lastly parallel to the given line draw a line touch- ing this last circle ; this line will cut off the segment required. 21. Let the segments AHB, AKC be externally described on the given lines AB, AC, to contain angles equal to BAC. Then by the converse to Euc. III. 32, AB touches the circle AKC, and AC the circle AHB. - 22. This is obvious from the note to Euc. III. 26, p. 156. 23. The segment must be described on the opposite side of the produced chord. By converse of Euc. III. 32. - 24. Join AB, and upon AB describe a segment of a circle containing an angle equal to the given angle, and meeting the circumference of the given circle. 49 25. Let É be the given point within the circle. Take the center C, and join PC. On PC describe a semicircle, and place in it a line PQ equal to half the given difference, and produce PQ to meet the circumference of the circle, 26. From Euc. I. 32; III. 21, the angle AFG is shewn to be equal to the angle AGF. . * 27. Let the diagram be drawn, and from the center of the circle draw a per- pendicular on the chord which passes through the middle point of the two equal chords. Then Euc. III. 3. - 28. Let P be the given point, and PBA the given line cutting the circle ABC in the points B, A. Let PCD be the line required : join OC, OD, O being the center. Then the arc AB being given, and the sum of the arcs BC, AD; the arc CD is also given in magnitude, and the angle COD which it subtends at the Center. * Whence the construction. Take the arc RS equal to the defect of the sum of the three arcs AB, DA, BC from the whole circumference: join RS, and with center C describe a circle touching RS, and from P draw PCD to touch this circle. 29. Construct the figure, and the sum of the angles at A and B which stand upon the arcs AP, Bº may be shewn to be equal to the angle PVQ. Also PQA is twice PBO or PWR. 30. Let A, B, C be any three points taken in order in the circumference of a semicircle whose diameter is PQ and center E; and let EA, EB, AC be joined; then the angle EBC exceeds EAC by the angle ACB. For join CE and produce AE, BE to meet the circumference of the circle in D, F. T. angle FBC stands on the arc FC, and the arc FD is equal to the arc A.B. 31. Construct the figure, observing that BO, BP, OA, AD, are equal to one another. . 32. Let A, B, be the two given points in the circumference of the circle whose center is C. Bisect AB in D, join DC and produce it to meet the circumference in E on the other side of the given diameter. E is the point required. * 33. . Let the chords PQ, P'Q' produced intersect the diameter AB produced, on opposite sides of the center C in the points T, T', so that CT is equal to CT". Join PP', QQ' intersecting AB in D, E ; then CD is equal to DE. - 34. Let AB, AC be the bounding radii, and D any point in the arc BC, and DE, DF, perpendiculars from D on AB, A.C. The circle described on AD will always be of the same magnitude, and the angle EAF in it, is constant: whence the arc EDF is constant, and therefore its chord E.F. . 35. Join PB, PC : then in the triangles PBm, PCn, PC is equal to PB, Pan ON BOOK III. HINTs, &c. & 431 to Pn, and the angle PCn equal to the angle PBm. Euc. 111. 22, note p. 130 whence Bm is equal to Cn. • * * - - 36. Construct the figure, and let the eircle with center O, described on AH as a diameter, intersect the given circle in P, Q, join OP, PE, and prove EP at right angles to OP. - - - 37. If the tangent be required to be perpendicular to a given line: draw the diameter Fºl. to this line, and the tangent drawn at the extremity of this dia- meter will be perpendicular to the given line. 38. The straight line which joins the center and passes through the intersec- tion of two tangents to a circle, bisects the angle contained by the tangents. 39. Draw two radii containing an angle equal to the supplement of the given angle; the tangents drawn at the extremities of these radii will contain the given angle. - - - 40. It is sufficient to suggest that the angle between a chord and a tangent is equal to the angle in the alternate segment of the circle. Euc. III. 32. 41. Let AB be the given chord of the circle whose center is O. Draw DE touching the circle at any point E and equal to the given line; join DO, and with center O and radius DO describe a circle: produce the chord AB to meet the circumference of this circle in F : then F is the point required. - - 42. Let D be the point required in the diameter BA produced, such that the tangent DP is half of DB. Join CP, C being the center. Then CPD is a right- angled triangle, having the sum of the base PC and hypotenuse CD double of the perpendicular PD. 43. If BE intersect DF in K (fig. Euc. III. 37). Join FB, FE, then by means of the triangles, BE is shewn to be bisected in K at right angles. 44. Let AB, CD be any two diameters of a circle, O the center, and let the tangents at their extremities form the quadrilateral figure EFGH. Join EO, OF, #3. EO and OF may be proved to be in the same straight line, and similarly , OK. , NotE.—This Proposition is equally true if AB, CD be any two chords whatever. It then becomes equivalent to the following proposition:—The diagonals of the circumscribed and inscribed quadrilaterals, intersect in the same point, the points of contact of the former being the angles of the latter figure. 45. Let C be the point without the circle from which the tangents CA, CB are drawn, and let DE be any diameter, also let AE, BD be joined, intersecting in P, then if CP be joined and produced to meet DE in G : CG is perpendicular to D.E. Join DA, EB, and produce them to meet in F. Then the angles DAE, EBD being angles in a semicircle, are right angles; or DB, EA are drawn perpendicular to the sides of the triangle DEF: whence the line drawn from F through P is perpendicular to the third side DE. 46. . Let the chord AB, of which P is its middle point, be produced both ways to C, D, so that AC is equal to BD. From C, D, draw the tangents to the circle forming the tangential quadrilateral CKDR, the points of contact of the sides, be- ing E, #. F, G. Let O be the center of the circle. Join EH, GF, CO, GO, FO, DO. Then EH and GF may be proved each parallel to CD, they are there- fore parallel to one another. Whence is proved that both EF and D.G. bisect AB. 47. This is obvious from Euc. 1. 29, and the note to III. 22. p. 156. 48. From any point A in the circumference, let any chord AB and tangent AC be drawn. Bisect the arc AB in D, and from D draw DE, DC perpendiculars on the chord AB and tangent AC. Join AD, the triangles ADE, ADC may be shewn to be equal. - 49. Let the tangents at A, A' intersect in T; and the tangents at B, B' inter- sect in T', join TO, TO, and shew that TOT is a straight line. - 50. Constructing the figure and producing the tangent QP, the triangle CPQ may be shewn to be an isosceles triangle, as also the triangle C'QP: also CQ and CQ may be each shewn equal to QP. - 51. Let the tangent AB touch the circle in C, and let CD be drawn from C perpendicular on any diameter EF, and let the perpendiculars from E and F meet the tangent in B and A respectively Join CE, CF. Then the angles BCE, DCE, may be shewn each equal to the angle CFE; and by means of the triangles BCE, DCE, BE may be shewn equal to ED, and in a similar way FA may be shewn equal to FD. *** 52. By Euc. III. 32, I. 29, the property is proved. 53. The angle CDA is equal to CAD, Euc. 1. 5. and the angle DCB equal to CAD, Euc. III. 32. - - 54. Let the tangents TP, TG, be drawn from the point T to meet the circle 432 2 GEOMETRICAL EXERCISES whose center is C at P and Q, and from another point t, on the same side of PQ as T, lettp, tº be drawn to touch the circle in p, q; and let to, tº, produced meet TP in r, r'; and TQ in s, s': it is required to shew that the angle rCr' is equal to the angle scs'. - T55. Let AP, AQ be two tangents, PQ the chord joining the points of contact, ROS a chord passing through O any point in the chord PQ ; ART, AVS two secants passing through the extremities of the chord R.S, and meeting the circum- ference in T, V. Join TO, WO and prove TOW a straight line. 56. Let A, B, be the given points. Join AB, and upon it describe a segment of a circle which shall contain an angle equal to the given angle. If the circle cut the given line, there will be two points; if it only touch the line, there will be one ; and if it neither cut nor touch the line, the problem is impossible. 57. Let C be the center of the circle, and E the point of contact of D.F with the circle. Join DC, CE, CF 58. Let the tangents at E, F meet in a point R. Produce RE, RF to meet the diameter AB produced in S, T. Then RST is a triangle, and the quadrilateral RFOE may be circumscribed by a circle, and RPO may be proved to be one of the diagonals. - 59. Let C be the middle point of the chord of contact: produce AC, BC to meet the circumference in B", A', and join AA’, BB'. 60. If a segment of a circle containing an angle equal to the given angle be described upon the line which joins the two given points, the circumference of the segment will meet the circumference of the given circle in two points, which will in general give two solutions of the problem. 61. The angles FCE, EDF may be shewn by Euc. III. 31, to be respectively the complements respectively of a right angle. . - 62. Let the tangents at the extremities of the chord PQ meet in T. Draw TACB passing through the center C, and meeting the circumference in A and B. ... Join also PA, PB. Then the angle APT may be shewn to be equal to PBA, Whence PTA is equal to the difference of the angles PAB and PBA. - , 63. Draw the chord AC equal to the line AT; and join TC cutting the circle in B. Join AB, and the angle ABC is double the angle ACB. ... 64. Apply Euc. III. 32: 1. 5, remarking that the angle EBO is equal to the difference of the angles OCB and ODB. Prob. 37, p. 300. 65. Draw a tangent PEQ at the point E where the circles touch one another. Then the angle PEA is equal to PCA, and PEB to PDB, Euc. III, 32. Whence PEB is equal to CED, DCE (Euc. I. 32,) and AEB is equal to CED. 66. It may be shewn from the right-angled triangles that the angle BAC is equal to the angle BCA. - 67. Take O the center, join CO and produce it to meet the circumference in E, and join BE. Then BE is parallel to AC. : - 68. Take the center O, and join OD: Euc. I.47. * 69. Let the tangents drawn from A touch the circle in the points P, Q ; and let BRC be the line of given length touching the circle in R and meeting AP, AQ in B and C. Then ABC is a triangle of which the sum of the sides AB, BC, CA is equal to the sum of the tangents AP, AQ. Theo. 7. p. 321. Hence the Problem is the same as the following:—To describe a triangle whose base, vertical angle, and sum of the three sides, are given. 70. Shew that the angle BAE is double of the angle AEB in the right-angled triangle ABE. * - * 71. Let TP, TCA touch a circle in P, Q, and let PCD be a diameter passing through P, also let the chord PQ intersect TC in B. Then the triangles PCB, PCT are equiangular. - 72. Let the figure be constructed. Then AB is equal to CD, because AC, BD are parallel chords in a circle. And by Euc. I. 29, III. 32, AB is shewn to be equal to AC. y 73. Let AB the chord of a given circle whose center is C, be produced to a fixed point D. Join CB and produce it, this line passes through the center of the required circle. Bisect B.D in E, at E draw EC’ perpendicular to BD and meet- ing CB produced in C". Then C' is the center of the required circle, and the line drawn through B at right angles to CC' is a common tangent to the two segments. 74. The triangles DBG, BD H are equal in all respects, Euc. I. 26, whence ; is equal to DH. The triangles ABG, CDH, are also equal in all respects (Euc. I, 4. - 75. #! the center draw a radius making an angle with the given diameter equal to half the supplement of the given angle, and at the circumference draw a line at right angles to the radius and meeting the given diameter produced. ON BOOK III. HINTS, &c. 433 76. Let the radii CA, CB be at right angles to each other, and let a tangent at P intersect them when produced in D and E. From D, E draw DR, EQ, to touch the circle in R, Q; then EQ is parallel to DR. 77. Apply Euc. III. 27. I. 6. 78. From A suppose ACD drawn, so that when BD, BC are joined, AD and DB shall together be double of AC and CB together. Then the angles ACD, ADB are supplementary, and hence the angles BCD, BDC are equal, and the triangle BCD is isosceles. Also the angles BCD, BDC are given, hence the triangle BDC is given in species. - Again AD + DB = 2. AC + 2. BC, or CD = AC + BC. Whence, make the triangle bale having its angles at d, c equal to that in the segment BDA ; and make ca = ca. – eb, and join ab. At A make the angle BAD equal to bad, and AD is the line required. 79. By constructing the figure and joining AC and AD, by Euc. III. 27, it may be proved that the line BC falls on B.D. 80, Apply Euc, III, 27; 1, 32, 6. 81. If the two circles intersect or do not, the other line must be drawn parallel to the given line, and intersecting the circles at the same distance from these centers, in order that the chords may be respectively equal to EF or G.H. 82. Let the lines be drawn; also the line joining the points of intersection of the circles. Then shew that the opposite angles of the figure are equal. 83. Let two unequal circles cut one another, and let the line ABC drawn through B, one of the points of intersection, be the line required, such that AB is equal to BC. Join O, O' the centers of the circles, and draw OP, OP' perpen- diculars on ABC, then PB is equal to BP'; through O' draw O'D parallel to PP"; then ODO is a right-angled triangle, and a semicircle described on OO as a diameter will pass through the point D. Hence the synthesis. If the line ABC be supposed to move round the point B and its extremities A, C to be in the ex- tremities of the two circles, it is manifest that ABC admits of a maximum. 84. Suppose the thing done, then it will appear that the line joining the points of intersection of the two circles is bisected at right angles by the line joining the centers of the circles. Since the radii are known, the centers of the two circles may be determined. 85. Let two unequal circles whose centers are C, C' cut one another in A and B, and let CC produced meet the circumference of the smaller circle in D, so that the line DAE drawn is bisected in A : then the point F where the circumference of the smaller circle cuts CC, is equidistant from C and C. Draw CG, CH per- pendicular to DE and join FA: also draw CIC, FL parallel to DG. Then it may be readily shewn that CF is equal to FC. 86. Complete the circle whose segment is ADB ; AHB being the other part. Then since the angle ACB is constant, being in a given segment, the sum of the arcs DE and AHB is constant. But AHB is given, hence ED is also given and therefore constant. 87. This is at once obvious from Euc, III. 36. 88. This follows directly from Euc. III. 36. If instead of the two circles in- tersecting each other, they touch each other, and from the point of contact a tangent be drawn to the circles, the tangents drawn to the two circles from any point on this line will also be equal to one another. 89. Each of the lines CE, DF may be proved parallel to the common chord A.B. 90. By constructing the figure and applying Euc. 1.8, 4, the truth is manifest. 91. The bisecting line is a common chord to the two circles; join the other extremities of the chord and the diameter in each circle, and the angles in the two Segments may be proved to be equal. 92. Let the circles again intersect in E, and let EID produced meet the circle in F. Join EE, FF, B.E, BE : then BE, BE are tangents to the circle whose . is A. The arcs E.F, FB, BF, FE may be proved to be equal to one another. 93. Shew that the angles in the respective similar segments are equal. 94, Let A, B be the points of intersection of the two circles, and let AD, AE be diameters drawn from A one of the points of intersection, let EB, DB be joined and proved to be in the same straight line. 95. Let the circles intersect in A, B ; and let CAD, EBF be any parallels Passing through A, B, and intercepted by the circles. Join CE, AB, D.F. Then º CEFD may be proved to be a parallelogram. Whence CAD is equal 0. - - 434 GEOMETRICAL EXERCISES 96. The tangent AB is equal to CD, and by Euc. III. 36, it may be shewn that BE is equal to CF. - 97. Construct the figure, observing that the lines AQ, AP may fall on the same or on opposite sides of A.B. Join PB, QB, and prove PBQ to be a straight line. 98. Let the line AEF which bisects the angle CAD meet the circumference of the circle ABC in E, and the circumference of ABF in F. Let the tangent at E * AB produced in G : join FG and prove FG to touch the circle ABD IIl iſ . 99. The angle at F may be shewn to be the supplement of the angles FBC, FCB of the triangle CBF, and also the supplement of the angles FDE, FED of the triangle DFE. 100. If the two circles cut each other at right angles, the lines AE, AF are diameters of the circles, and the angles ABE, ABF are obviously right angles. If the circles intersect at any other angle, apply Euc, III, 32. I. 32. 101. Let the two circles intersect in the point A : it is required to draw through A, a line BC which shall be terminated by the circumferences of the circles, and be equal to a given line M. Suppose the thing done, and that BAC is the line drawn through A and equal to M. Take C, C' the centers of the circles, and draw CD, CD' perpendiculars on BAC, then DD is one-half of BC. Through C draw CE parallel to BC and meeting CD in E, then CEC is a right- angled triangle. Whence the synthesis. 102. Let any other point P be taken in the arc within the first circle, and let BP, CP be joined and produced to meet the circumference in D, E, and let D'E' be joined. It may be proved that DE and D'E' bisect each other. 103. Tescribe a circle passing through the two given points and touching the given circle, and join the given points with the point of contact; these lines produced will meet the given circle in the points required. 104. Let the radius BC produced meet the circumference of the quadrantal arc when continued in F, and join FE, CD, B.E. Then FE is parallel to CD, and the angles DEB, EBD may be each shewn to be equal to half a right angle. 105. If the chords AC, BC produced meet the second circle in the points D, E on the same side of the line AB which joins the points of intersection of the circles: then DE is at right angles to that diameter of the circle ABED which passes through C. But if the chords CA, CB meet the second circle on different sides of AB ; then DE produced is at right angles to that diameter of the circle ACB which passes through C. 106. Construct the figure according to the directions given, taking the point E between A and H. Then the angles GFH, GBH are equal, both stand- ing on the same are GH in circle (2); and FG being produced to L, BEI to A and FH to E, points in the circumference of the circle (1), it may be shewn that the arc GLA is equal to the arc LAE. 107. Draw a common tangent at C the point of contact of the circles, and prove AC and CB to be in the same straight line. 108. Let A, B, be the centers, and C the point of contact of the two circles: D, E the points of contact of the circles with the common tangent DE, and CF a tangent common to the two circles at C, meeting DF in E. Join DC, CE. º DF, FC, FE may be shewn to be equal, and FC to be at right angles to - 109. The line must be drawn to the two extremities of the diameters which are on opposite sides of the line joining the centers, 110. The sum of the distances of the center of the third circle from the centers of the two given circles, is equal to the sum of the radii of the given circles, which is constant. - 111. Let the circles touch at C either externally or internally, and their dia- meters AC, BC through the point of contact will either coincide or be in the same straight line. CDE any line through C will cut off similar segments from the twº i. For joining AT), BE, the angles in the segments DAC, EBC are proved to be equal. The remaining segments are also similar, since they contain angles which are supplementary to the angles DAC, EBC. i12. Let the line which joins the centers of the two circles be produced tº meet the circumferences, and let the extremities of this line and any other line from the point of contact be joined. From the center of the larger circle draw perpendiculars on the sides of the right-angled triangle inscribed within it. ON BOOK III. HINTS, &c. 435 113. Let AB, AC be the diameters of the larger and smaller circles. On CB de- scribe a circle, and on CB take CD equal to the given line, with center C and radius CI), describe a circle intersecting the circumference of the circle on CB in E, join CE, and through A draw the chord AFG parallel to CE meeting the circles in F, G. If the diameter AC be one-half of AB, then all the chords drawn from A to cut the inner and to meet the outer circle, are bisected by the circumference of the inner circle. 114. Apply Euc, III, 31. - 115. Let the two circles whose diameters are AD, AD", touch each other at A: and let PQM be the semichord perpendicular to the diameter. If PQ be supposed equal to QM, the line MD may be shewn to be equal to four times M.D.'. 116. I.et ACB be the common diameter of the two circles which touch each other at the point A, and through C the center of the smaller circle, let PP be drawn perpendicular to AB, and meeting the inner circle in Q, Q also let the tangents from P, P touch the inner circle in T, T. Join CT, CT. Then PT, PT may be proved each equal to CT, CT. 117. Through the point of contact draw a common tangent to the two circles. Conversely. If the chords be parallel, the lines joining the extremities of the chords pass through the point of contact. 118. In each circle draw a chord of the given length, describe circles con- centric with the given circles touching these chords, and then draw a straight line touching these circles. - 119. Let A, B, be the two given points in the circumference, and CD the given chord, and let C be the point required such that when ADE, BEC are drawn, DE shall be equal to the given line. Since the arc AB is known, the angle ACB is known. Hence the problem is reduced to that of describing a circle touching the given circle in the point C and cutting off a chord DE from BC which subtends an angle equal to C. 120. Let C, C' be the centers of the given circles, join CC', and let DD’ be the line required, making the given angle with CC. Through C draw CE making with CC. an angle equal to the given angle, and equal to the given line, join CD, ED". 121. Let A, B, be the given points, and CD the given line. From E the middle of the line AB, draw EM perpendicular to AB, meeting CD in M, and draw MA. In EM take any point F; draw FH to make the given angle with CD ; and draw FG equal to FH, and meeting MA produced in G. Through A draw AP parallel to FG, and CPK parallel to F.H. Then P is the center, and C the third defining point of the circle required: and AP may be proved equal to CP by means of the triangles GMF, AMP ; and HMF, CMP, Euc. VI. 2. Also CPK the diameter makes with CD the angle KCD equal to FHD, that is, to the given angle. - 122. Let A, B, be the two given points, and C the center of the given circle. Join AC, and at C draw the diameter DCE perpendicular to AC, and through the points A, D, E describe a circle, and produce AC to meet the circumference in F. Bisect AF in G, and AB in H, and draw GK, HK, perpendiculars to AF, AB respectively and intersecting in K. Then K is the center of the circle which passes through the points A, B, and bisects the circumference of the circle whose center is C. 123. Let A, B be the two given points, join AB and bisect AB in C, and draw CD perpendicular to AB, then the center of the required circle will be in CD. From Ö the center of the given circle draw CFG parallel to CD, and meeting the circle in F and AB produced in G. At F draw a chord FF equal to the given choid. Then the circle which passes through the points at B and F, passes also through F. - - - 124, Let P be the given point and O the center of the given circle, with center P and a radius equal to the given radius, describe a circle. The center of the required circle is in this circumference and is a point equidistant from P and the extremities of a diameter of the given circle. . . 125. Let AB be the given indefinite line, P, Q the two points through which one circle is to pass, P, Q, the two points through which the second circle is to pass, both of which are to cut off a given length from AB. If two points M, N, can be found in AB such that MN is equal to the given length, and the quadrila- terals PQ MN, PQ. NM have each their opposite angles equal to two right angles, the circles which circumscribe PQN M, PQ'NM are the circles required. - 126. The point required Q in the circumference of the first circle will be F F 2 436 GEOMETRICAL EXERCISES - found to be the point of bisection of the line PQR touching the second circle in P and drawn through Q and terminated by the circumference of the second circle in R. 127. Let P, Q be the two given points and O the center of the given circle. Draw OS to meet the circumference in any point S, and at S draw ST at right angles to OS, the center of the required circle is in the line ST. The problem is reduced to describe a circle passing through two given points, having its center in a given line, and touching a given straight line in a given point. 128. Let A be the given point, and suppose the circle ABCDE the required circle which bisects the circumferences of the two given circles whose centers are O, O' in B, C ; D, E, respectively, join AO and produce it to meet the required circle in E: then OF is a third proportional to AO, O.C.: therefore the point E is known. Similarly join AO and produce it to meet the circumference of the re- quired circle in G ; then OG is a third proportional to AO, OD; therefore the point G is known. Hence A, F, G are three points in the required circle, and the circle described through the points A, F, G, bisects the circumferences of the two circles. 129. Shew that at the points of intersection, the tangents pass through the centers of the two given circles. 130. The given point must be so situated that the required circle may cut the two given circles in points where the lines drawn through the centers shall be tangents to the required circle. 131. This problem will be solved by finding the point from which three equal tangents can be drawn to the three circles. 132. Construct the figure, and by Euc. ITT. 35. 133. Let C be the center of the inner, C of the outer circle, join CC and produce this line to meet the circumference of the inner circle in B, A, and the outer in B, A. Prove that any chord DPF, touching the inner circle at P cannot be bisected at P, unless the point P coincide with A' or B : and that if C coincide with C, then all the chords of the outer, which touch the inner circle are bisected at the points of contact. - 134. Let A be the common center of two circles, and BCDE the chord such that BE is double of CD. From A, B draw AF, BG perpendicular to B.E. Join AC, and produce it to meet BG in G. Then AC may be shewn to be equal to CG, and the angle CBG being a right angle, is the angle in the semicircle described on CG as its diameter. 135. The lines joining the common center and the extremities of the chords of the circles, may be shewn to contain unequal angles, and the angles at the centers of the circles are double the angles at the circumferences, it follows that the segments containing these unequal angles are not similar, 136. Let AB, AC be the straight lines drawn from A, a point in the outer circle to touch the inner circle in the points D, E, and meet the outer circle again at B, C. Join BC, DE. Prove BC double of D.E. Let O be the center, and draw the common diameter AOG intersecting BC in F, and join EF. Then the figure DBFE may be proved to be a parallelogram. 137. The given point may be either within or without the circle. Draw a chord in the circle equal to the given chord, and describe a concentric circle touching the chord, and through the givenpoint draw a line touching this latter circle. 138. The diameter of the inner circle must not be less than one-third of the diameter of the exterior circle. 139. Let C be the center of the inner circle; draw any radius CD, at D draw a tangent DE equal to CD, join CE, and with center C and radius CE de- scribe a circle, and produce ED to meet the circle again in F. 140. Take C the center of the given circle, and draw any radius CD, at D draw DE perpendicular to DC and equal to the length of the required tangent, with center C and radius CE describe a circle. 141. This is manifest from Euc. III. 36. - 142. The lines AEFB, CEGD are equal, and therefore AC and DB are paral- lel, whence AD is equal to C.B. Then Euc. III. 21, I. 26. 143. Let O be the common center of the three concentric circles, and let the circumferences of the circles cut the required line in the points A, B, C, so that the segment intercepted between the first and second circumferences is equal to that intercepted between the second and third. Join OC, OB, O.A. Then since O.A., OB, OC are given, AB or BC is found from Prob. 3, p. 311. 144. Take O the center of the circle. Let the tangents drawn from A and B intersect in P and Q : join PO, Q0, and prove POQ to be a straight line. 145. The sum of the two parts of the line intercepted between the circulº ON BOOK III. HINTS, &c. - 437 ferences of the concentric circles must be less than that chord of the outer which touches the inner circle. 146, This is manifest from Euc, III. 35. 147. At any point P in the circumference of the given circle, draw a line APB touching the circle at P, the parts AP, BP being equal to the given line. With center C and radius CA or CB, describe a circle, produce the given chord DE to meet the circumference of this circle in F. From F draw FG to touch the given circle in G, then FG is the line required. 148. If the given point P be without the two circles, and C their common center. Draw the radius CADB meeting the inner and outer circles in A, B, and so that the perpendicular drawn from P on AB is bisected in D: join PA, PB and the circle described with center P and radius PA or PB will cut from the two circles similar segments. 149. Let A be the right angle of the triangle ABC, draw AD perpendicular on BC; the circles on AB, AC, pass through D. Euc, III. 31. Let DE, DE be drawn to E, F the centers of the circles on AB, AC and join EF. Then ED may be proved to be perpendicular to the radius IDF of the circle on AC at the oint D. p 150. Let ABC be a triangle, and let the arcs be described on the sides externally containing angles, whose sum is equal to two right angles. It is obvious that the sum of the angles in the remaining segments is equal to four right angles. These arcs may be shewn to intersect each other in one point D. Let a, b, c be the centers of the circles on BC, AC, AB. Join ab, be, ca; Ab, b0, Ca.; ab, Bc, cA; blj, cD, alſ). Then the angle cba, may be proved equal to one-half of the angle Ab{}. Similarly, the other two angles of abe. 151. It may be remarked that generally, the mode of proof by which, in pure geometry, three lines must, under specified conditions, pass through the same point, is that by reductio ad absurdum. This will for the most part require the converse theorem to be first proved or taken for granted. The converse theorem in this instance is, “If two perpendiculars drawn from two angles of a triangle upon the opposite sides, intersect in a point, the line drawn from the third angle through this point will be perpendicular to the third side.” - The proof will be formally thus: Let EHD be the triangle, AC, BD two perpendiculars intersecting in F. If the third perpendicular EG do not pass through F, let it take some other position as E.H. ; and through F draw EFG to meet AD in G. Then it has been proved that EG is perpendicular to AD : whence the two angles EHG, E.G.H. of the triangle EGH are equal to two right angles:—which is absurd. 152. If a circle be described upon the side AC as a diameter, the circum- ference will pass through the points D, E. Then Euc. III. 21. 153. Since all the triangles are on the same base and have equal vertical angles, these angles are in the same segment of a given circle. The lines bisecting the vertical angles may be shewn to pass through the extremity of that diameter which bisects the base. 154. Let ABC be a triangle of which the base or longest side is BC, and let a segment of a circle be described on BC. Produce BA, CA to meet the are of the segment in D, E, and join BID, CE. If circles be described passing through the points A, B, D ; and through A, C, E, the lines AB, AC shall cut off segments similar to the segment described upon the base BC. - 155. Shew that any two opposite angles of the figure ACAB are equal to two right angles. 156 Let ABC be the given isosceles triangle having the vertical angle at C, and let FG be any given line. Required to find a point P in FG such that the distance PA shall be double of PC. Divide AC in D so that AD is double of DC, produce AC to E and make AE double of A.C. On DE describe a circle cutting FG in P, then PA is double of PC. This is found by shewing that AP* = 4. PC2. - 157. On any two sides of the triangle, describe segments of circles each containing an angle equal to two-thirds of a right angle, the point of inter- section of the arcs within the triangle will be the point required, such that three lines drawn from it to the angles of the triangle shall contain equal angles. Euc, III, 22. 158. The line drawn perpendicular to the diameter from the other ex- . of the tangent is parallel to the tangent drawn at the extremity of the lameter. - 4.38 GEOMETRICAL EXERCISES 159. Apply Euc, III, 21. - 160. The fixed point is the bisection of the arc of the semicircle on the other side of the fixed diameter. 161. Let A, B, C be the three given points. Join A, B, and on AB describe a segment of a circle ADB containing an angle equal to that which the lines drawn to the points A, B are to contain, and complete the circle. On the other side of AB at the point B make the angle ABE equal to the angle which the lines drawn to the points A, C are to contain, and let the line BE meet the circumference in B. Join EC and produce it to meet the circumference again in D. D is the point required. - 162. Take O, O' the centers of the semicircles on AM, MB, and join 00, QT, and prove Q0, RO to be at right angles to QR. 163. Take O the center of the circle and join CO; then in the triangle PCO, PCO is a right angle, and PO is double of OC. The angle COP is equal to one of the angles of an equilateral triangle. Also COP is equal to PEA, and PEA to DEC. 164. The lines EB, FB are at right angles to each other forming the angle EBF in a semicircle; and these lines divide the interior angle of the triangle into two angles, as also the exterior angle into two angles. 165. Join DE, E.F, FD, and these lines may be shewn to be parallel to AC, AB, BC the sides of the triangle. And parallel chords in a circle cut off equal arcs. 166. The lines drawn from the vertices of the triangle to the points on the opposite sides intersected by the circumferences of the circles, bisect the angles contained by the lines drawn from that point to the other two points where the circles intersect on the other sides. 167. If BC be joined, it may be shewn that the angles BDE and BCE are together equal to two right angles. - 168. Let A, B, C, be the centers of the three equal circles, and let them intersect one another in the point D : and let the circles whose centers are A, B intersect each other again in E.; the circles whose centers are B, C in F; and the circles whose centers are C, A in G. Then FG is perpendicular to DE; DG to FC; and DF to GE. Since the circles are equal, and all pass through the same point D, the centers A, B, C are in a circle about D whose radius is the same as the radius of the given circles. Join AB, BC, CA; then these will be perpendicular to the chords DE, DF, DG. Again, the figures DAGC, DBFC, are equilateral, and hence FG is parallel to AB; that is, perpendicular to D.E. Similar for the other two cases. 169. Tet three circles touch each other at the point A, and from A let a line ABCD be drawn cutting the circumferences in B, C, D. Let O, O, O" be the centers of the circles, join BO, CO, DO", these lines are parallel to one another. Euc. I. 5, 28. 170. Proceed as in Theorem 154, supra. 171. With center A and any radius less than the radius of either of the equal circles, describe the third circle intersecting them in C and D. Join BC, CD, and prove BC and CD to be in the same straight line. - 172. This property is true not only for three circles, but for all circles. Let A be the point of contact, C the center of any circle in the line AP, and B the given point from which the tangents are to be drawn. Join BA and make the angle ABf equal to BAF, and produce BF till FG be equal to FB. The chord of contact DE will always pass through G. For join BC cutting DE in H: then BHG is a right angle. Also BF, FA, FG are equal to one another, and A is in a semicircle on BG, and AG being joined, BAG is a right angle. The angles BHG, BAG, therefore are right angles in the same semicircle; and hence HE always passes through G, the extremity of the hypotenuse of the triangle BAG, 173. Let the three chords be AB, AC, AT), the middle points of which are b, c, d; let E, F, G be the intersections of the circles on AB and AC, on AB and AD, and on AC and AD respectively; join BC, CD, DB, be, cd, db, A.E, AF, AG; and let m be the intersection of be, A.E. The proof may be made to depend on the following theorem; If a circle be inscribed in a triangle, and perpendiculars bº drawn upon the sides from any point in the circumference; the three points of intersection are in the same straight line. 174, Let the two circles cut one another in A and B. Join AB, and suppose ACD to be the line required meeting the circumferences in C, D, such that tº part DC is equal to the given line. Join also BC, B.D. Then the angles at C. D may be shewn to be given, and the point D. ON BOOK III. HINTS, &c. 439. 175. It may be shewn that AEC, AFB, AGB are semicircles, having AC, and AB for their diameters. Euc. III. 31. 176. Let E be the center of the circle which touches the two equal circles whose centers are A, B. Join AE, BE which pass through the points of contact F, G. Whence AE is equal to EB. Also CD the common chord bisects AB at right angles, and therefore the perpendicular from E on AB coincides with 177. The three tangents will be found to be perpendicular to the sides of the triangle formed by joining the centers of the three circles. 178. The lines joining the centers O1, O2, Os of the circles pass through the points of contact of the circles. Apply Eug. I. 5, 13, 32. 179. Let O, O' be the centers of the circles, and E, F, the points of intersec- tion of the third circle with the circles whose centers are O, O. Join EC, FC, and prove ECF to be a straight line. At C draw a common tangent to the two given circles. 180. Let ABC be the triangle required; BC the given base, BD the given difference of the sides, and BAC the given vertical angle. Join CD and draw AM perpendicular to CD. Then MAD is half the vertical angle and AMD a right angle: the angle BDC is therefore given, and hence D is a point in the are of a given segment on BC. Also since BD is given, the point D is given, and therefore the sides BA, AC are given. Hence the synthesis. 181. Let ABC be the required triangle, AD the line bisecting the vertical angle and dividing the base BC into the segments BD, DC. About the triangle ABC describe a circle and produce AD to meet the circumference in E, then the arcs BE, EC are equal. 182. Analysis, Let ABC be the triangle, and let the circle ABC be de- scribed about it: draw AF to bisect the vertical angle BAC and meet the circle in F, make AV equal to AC, and draw CV to meet the circle in T; join TB and TF, cutting AB in D; draw the diameter FS cutting BC in R, DR cutting AF in E; join AS, and draw AK, AHI perpendicular to FS and BC. Then shew that AD is half the sum, and DB half the difference of the sides AB, A.C. Next, that the point F in which AF meets the circumscribing circle is given, also the point E where DE meets AF is given. The points A, K, R., E are in a circle, Euc. III. 22. Hence, KF. FR = AF.F.E, a given rectangle; and the segment KR, which is equal to the perpendicular AH, being given, RF itself is given. Whence the construction. -- 183. On AB the given base describe a circle such that the segment AEB shall contain an angle equal to the given vertical angle of the triangle. Draw the diameter EMD cutting AB in M at right angles. At D in ED, make the angle EDC equal to half the given difference of the angles at the base, and let DC meet the circumference of the circle in C. Join CA, CB; ABC is the triangle re- quired. For, make CF equal to CB, and join FB cutting CD in G. 184. Let ABC be the triangle, AD the perpendicular on BC. With center A, and AC the less side as radius, describe a circle cutting the base BC in E, and the longer side AB in G, and BA produced in F, and join AE, EG, FC. Then the angle GFC being half the given angle, BAC is given, and the angle BEG equal to GFC is also given. Likewise B.E the difference of the segments of the base, and BG the difference of the sides, are given by the problem. Wherefore the triangle BEG is given (with two solutions). Again, the angle EGB being given, the angle AGE, and hence its equal AEG is given; and hence the vertex A is given, and likewise the line AE equal to AC the shortest side is given. Hence the construction. 185. Let ABC be the triangle, D, E the bisections of the sides AC, AB. Join CE, BD intersecting in F. Bisect BD in G and join EG. Then EF, one- third of EC is given, and BG one-half of BD is also given. Now EG is parallel to AC ; and the angle BAC being given, its equal opposite angle BEG is also given Whence the segment of the circle containing the angle BEG is also given. Hence F is a given point, and FE a given line, whence E is in the circumference of the given circle about F whose radius is FE. Wherefore E being in two given circles, it is itself their given intersection. - 186. Let AB be the given base and ABC the sum of the other two sides ; at B draw BD at right angles to AB and equal to the given altitude, produce BD to E making DE equal to BD. With center A and radius AC describe the circle CFG, draw FO at right angles to BE and find in it the center O of the circle which passes through B and E and touches the former circle in the point F. The 440 GEOMETRICAL EXERCISES - centers A, O, being joined with the line produced, will pass through F. Join OB. Then AOB is the triangle required. - - 187. Since the area and base of the triangle are given, the altitude is given. Hence the problem is—given the base, the vertical angle and the altitude, de. scribe the triangle. 188. Let P, Q, R., be the three given points; join PQ, QR, RP. Form a triangle whose sides AB, AC, BC, shall be equal to the three given sides. On PQ describe a segment of a circle, containing an angle equal to the angle ABC; and on PR a segment containing an angle equal to ACB. Through the point P, draw bBe equal to BC and terminated by the arcs on PQ, P.R. Join bq, clº, and pro- duce them to meet in a. The triangle abc has its sides and angles equal to the triangle ABC, and its sides pass through the three points P, Q, R. 189. On BC the given base describe a segment of a circle containing an angle equal to the vertical angle of the triangle. Also on BC describe a semicircle, and from C draw CD to meet the arc of the semicircle in D, and so that CD shall be equal to the given perpendicular. Join BD and produce it to meet the circumfe- rence of the segment in A, and join A.C. ABC is the triangle required. 190. Suppose ABC the triangle required, BAC being the given vertical angle and CD the given perpendicular on the side AB. Let PEQ the diameter of the circumscribing circle be drawn through E the bisection of the base, and let QS be drawn perpendicular on AB. Then AS is half the sum and BS half the difference of the sides AB, AC. Hence the sides are known, and the perimeter is known: whence the base is known, and the triangle may be constructed. 191. Describe on the given base BC a segment of a circle containing an angle equal to the given angle, and from B or C draw BA or CA equal to the given side to meet the circumference in the point A ; join AC or AB, and ABC is the triangle required. - 192. The simplest case of this problem, is that in which the side AB of the triangle coincides with the chord AB of the given circle which cuts off the given segment. 193. When the vertical angle and area are given, the rectangle under the sides is also given. Likewise the sum of the squares and the rectangle of the two lines which constitute the sides are given. These lines may hence be found and the triangle constructed. 194. Let CD be the given difference of the sides AB, AC (fig. Euc. I. 18), and BC the given base. Let ABC be the triangle required; at B draw BF per- pendicular to BC, through A draw AE parallel to BC meeting BF in E, and produce CA to meet BF in F. Join BD, then AD, AB, AF are equal to one another, and the triangle may be constructed on the base BC with altitude B.E, and having the difference of the sides equal to CD. For the point D is determined by two circles which touch one another, one described with center C and radius CD, and the other described passing through the points F, B, and touching the circle whose radius is CD in D. - 195. Make a triangle ABC equal to the given figure, by Prob. 127, p. 306. Produce the base BC, if necessary, making BD equal to the given base. On BD make he triangle BDE equal to ABC; through E. draw EG parallel to BD, and upon BD lescribe a segment of a circle containing an angle equal to the given angle, Euc. III. 33, and cutting EG in H ; join HB, HD. H.BD is the required tri- angle. *. Let AB be the base, and AHB the segment containing the vertical angle. At any point H draw the tangent HP, making the square on HP equal to the given rectangle under the sum of the sides of one of them. Find O the center of the circle AHB and join PO, with which as radius describe a circle PDQ. Also from R the middle of the arc AHB, with radius RA or RB describe the circle ADB cutting PDQ in D. Draw DA cutting AHB in C, and join CB, ACB is the triangle required. For, draw the tangent DE, and join HQ, E0: DO. Then HP = DEA = AD. DC = (AC + CB). CB ; for the triangle DCB is isosceles. 197. Let ABC be the required triangle having the right angle ABC, and such that the sum of AC and AB is double of BC. Now since AC + AB = 2BC, therefore AC – BC = BC – AB. On AC take AD equal to this difference. Then AC = BC + AD, AB = BC – AD, and sing: AC* = AB* + BC”, it follows that BC* = 4BC. AD, and BC = 4AD. Hence AD the difference between the hypotenuse and one side BC is known, and therefore the hypotenuse AC = 5A.D. Hence the construction of the triangle depends On the division of the hypotentise into five equal parts. ON BOOK III. HINTS, &c. 441 198. The fixed point may be proved to be the center of the circle. 199. Let the line which bisects any angle BAD of the quadrilateral, meet the circumference in E, join EC, and prove that the angle made by producing DC is bisected by EC. 200. Draw a tangent to the circle, and at the point of contact draw a chord making an angle equal to one-third of a right angle. 201. The centers of the four circles are determined by the intersections of the lines which bisect the four angles of the given quadrilateral. Join these four points, and the opposite angles of the quadrilateral so formed are respectively equal to two right angles. 202. Let ABCD be the required trapezium inscribed in the given circle (fig. Euc. III. 22.) of which AB is given, also the sum of the remaining three sides and the angle ADC. Since the angle ADC is given, the opposite angle ABC is known, and therefore the point C and the side BC. Produce AD and make T).E. equal to DC and join EC. Since the sum of AD, DC, CB is given, and DC is known, therefore the sum of AD, DC is given, and likewise AC, and the angle ADC. Also the angle DEC being half of the angle ADC is given. Whence the segment of the circle which contains AEC is given, also AE is given, and hence the point E, and consequently the point D. Whence the construction. 203. Let ADBC be the inscribed quadrilateral; let AC, BD produced meet in O, and AB, CD produced meet in P, also let the tangents from O, P meet the circles in K, H respectively. Join OP, and about the triangle PAC describe a circle cutting PO in G and join AG. Then A, B, G., O may be shewn to be points in the circumference of a circle. Whence the sum of the squares on Q.H. and PK may be found by Euc, III. 36, and shewn to be equal to the square on OP. 204. Apply Euc. III, 22. 205. A circle can be described about the figure AECBF. 206. Apply Euc. III. 21, 22, 32. 207. Apply Euc. III. 20, and the angle BAD will be found to be double of the angles CBD and CDB together. 208. Apply Euc. III. 22. 209. This is shewn from Euc. III. 21. 210. GE, GF produced to meet the circle again form two chords in the circle, and the two perpendiculars drawn from E, F upon these chords pass through the center of the circle C. If CG be joined and produced to meet EF in H, then CGH may be proved to be perpendicular to EF, 211. The points G, K, coincide with E, D respectively, Euc. III. 31. Join FH, HG, GK, KF, and shew that the opposite angles of FHGK are respectively equal to two right angles. 212. The lines which bisect the adjacent angles AEB, AED may be shewn to be at right angles to one another. Next let the lines which bisect the angles at F, G intersect each other in O, and meet the circumference in the points H, K; L., M. Then the ares DM, MC : as also AL, LB are equal to each other. Similarly the arcs C.K, KB are equal, and DH, HA : from which it may be shewn that the sum of the arcs H.M., KL is equal to the sum of HL, KM: whence by the converse of Theo. 1, p. 317. HK, LM intersect each other at right angles. Whence since the lines bisecting the angles AEF and G being in the same direc- tion, as also the lines bisecting the angles AED and F : it follows that these four lines at their intersections form a rectangular parallelogram. 213. This is shewn by proving that the sums of the opposite angles of the *ew quadrilateral are respectively equal to two right angles. - 214. Let the angles be bisected and let the bisecting lines meet each other in the points a, b, c, d. Shew that the sum of the angles at a and c, as also the sum of the angles at 5 and d, are equal to two right angles. 215. Since the angle A of the quadrilateral is a right angle, the opposite angle C is also a right angle. If FE be joined the semicircle described on EF as a diameter will pass through the points A, C, and the line joining the points A, C, will be a chord of the circle whose diameter is E.F. 216. Let abed be the given quadrilateral whose opposite angles a, c.; b, d, are equal to two right angles, and let A be the given point on the circumference of the given circle. Join bai, and at B draw a line PQ to touch the circle at B. At the point B in BQ make the angle QBD equal to bad, DBC equal to die and DBA equal to dba : join AD, DC : then ABCD is the quadrilateral required. 217. Let the figure be constructed, and join DB, DG, then DABG is a rect- ºngle, and FG is equal to CD, both being diameters of the circle, and by Eug. 1. 47; II. 12, 3 ; III 36, the property may be proved. - 442 GEOMETRICAI, EXERCISES 218. Draw the diameter EF, join CF, and draw CG perpendicular to EF. Then EC* = EF" – CF* = EF" – EF. EG = EF. EG = EF, EO + EF, OG = 2. ABB + 2. ACB = 2 fig, AEBC. 219. Let the diagonals AC, BD be bisected in the points P, Q, and let O be the center of the circle. Join PO, OQ, and prove POQ to be a straight line. 220. Let ABCD be a quadrilateral inscribed in a circle so that the two sides AB, DC when produced meet in a point E at right angles: let F, G, the middle points of AB, DC, and O the center of the circle, join EO and FG : these lines may be shewn to be the diagonals of a square. 221. Let the tangent meet the circle in a point T between the points A and B. Join TC and produce it to meet the circumference again in T', draw a tangent to the circle at T. Produce PA to meet this tangent in R, then AR is equal to A. P. Draw AM perpendicular on TT. Then A.P. A.P = AP, AR = AM”. It may be shewn also that BQ. B'Q' = CM*. Whence by adding these equals the property is shewn to be true. 222. Let ABCD be any quadrilateral figure, AC, BD, the diagonals, F, G, their points of bisection, E the point of bisection of FG. Let P be any point in the circumference of a circle described from center E. Join PF, PE, PG, PA, PB, PC, PD, EA, EB, EC, ED. Then by Theo. 3. p. 311. 223. Let AC, CD, DB, be any three chords of a semicircle whose diameter is AB. From B let fall BE perpendicular on CD produced. Then by Euc. II, 12, III. 31, I. 48. 224. Draw the diagonals of the quadrilateral, and by Euc. III. 21, I. 29. 225. From the center draw lines to the angles: then Euc. III. 27. 226. This will be manifest from the equality of the two tangents drawn to a circle from the same point. 227. Apply Euc, III. 22, 32. 228. Join the center of the circle with the other extremity of the line per- pendicular to the diameter. 229. Let AB be a chord parallel to the diameter FG of the circle, fig. Theo. 1, p. 317, and H any point in the diameter. Let HA and HB be joined. Bisect FG in O, draw OL perpendicular to FG cutting AB in K, and join HK, HL, OA. Then the square on HA and HF may be proved equal to the squares on FH, HG, by Theo. 3, p. 311; Euc. I, 47; Euc. II. 9. - 230. Let A be the given point (fig. Euc. III. 36, Cor.) and suppose AFC meeting the circle in F, C, to be bisected in F, and let AD be a tangent drawn from A. Then 2. AF4 = AF. AC = AD*, but AD is given, hence also AF is given. To construct. Draw the tangent A.D. On AD describe a semicircle AGD, bisect it in G ; with center A and radius AG, describe a circle cutting the given circle in F. Join AF and produce it to meet the circumference again in C. 231. Let the chords AB, CD intersect each other in E at right angles. Find F the center, and draw the diameters HEFG, AFK and join AC, CK, B.D. Then by Euc, II, 4, 5; III. 35. 232, Let E, F be the points in the diameter AB equidistant from the center O; CED any chord; draw OG perpendicular to CED, and join FG, OC. The sum of the squares on DF and FC may be shewn to be equal to twice the square on FE and the rectangle contained by AE, EB, by Euc. I.47; II, 5; III. 35. 233. Let the chords AB, AC be drawn from the point A, and let a chord FG parallel to the tangent at A be drawn intersecting the chords AB, AC in D and E. and join BC. Then the opposite angles of the quadrilateral BDEC are equal º two right angles, and a circle would circumscribe the figure. Hence by uo. T. 36. - 234. Let the lines be drawn as directed in the enunciation. Draw the diameter AE and join CE, DE, BE; then AC*-i- AD" and 2. A B* may be each shewn to be equal to the square on the diameter. - 235. Let QOP cut the diameter AB in O. From C the center draw CH perpendicular to QP. Then CH is equal to OH, and by Euc. II, 9, the squares on PO, OQ are readily shewn to be equal to twice the square on CP. 236. From P draw PQ perpendicular on AB meeting it in Q. Join AC, CD, DB. Then circles would circumscribe the quadrilaterals ACPQ and BDPQ, and then by Euc. III. 36. - - 237. Describe the figure according to the enunciation; draw AE the diameter of the circle, and let P be the intersection of the diagonals of the parallelogram. Draw EB, EP, EC, EF, EG, E.H. Since AE is a diameter of the circle, the ON B00K III. HINTS, &c. 443 angles at F, G, H, are right angles, and EF, EG, EH are perpendiculars from the vertex upon the bases of the triangles EAB, EAC, EAP. Whence by Euc. II, 13, and theorem 3, p. 311, the truth of the property may be shewn. 238. Let AD meet the circle in G, H, and join BG, GC. Then BGC is a right-angled triangle and GD is perpendicular to the hypotenuse, and the rect- angles may each be shewn to be equal to the square on BG, Euc. III. 35 ; II. 5; I. 47. Or, if EC be joined, the quadrilateral figure ADCE may be circumscribed by a circle, Euc. III. 31, 22, 36, Cor. - 239. On PC describe a semicircle cutting the given one in E, and draw EF perpendicular to AD ; then F is the point required. 240. Let AB be the given straight line. Bisect AB in C and on AB as a diameter describe a circle; and at any point D in the circumference, draw a tan- gent DE equal to a side of the given square; join DC, EC, and with center C and radius CE describe a circle cutting AB produced in F. From F draw FG to touch the circle whose center is C in the point G. 241. Produce the longer side BA at the vertex of the triangle to any point E. bisect the angle CAE by AD meeting BC produced in D. Then the square on DA is equal to the rectangle DB, DC. For describe a circle passing through the points B, C, and touching AD at A. Then Euc. III. 36. 242. Let AD, DE be two lines at right angles to each other, O the center of the circle BFQ ; A any point in AD from which tangents AB, AC are drawn; then the chord BC shall always cut FD in the same point P, wherever the point A is taken in AD. Join AP; then BAC is an isosceles triangle, and FD. DE + AD* = AB* = BP. PC + AP* = BP, PC + AD* + DP”, wherefore BP. PC = FID. IDE — IDP”. The point P, therefore, is independent of the position of the point A ; and is consequently the same for all positions of A in the line A.D. 243. The point E. will be found to be that point in BC, from which two tan- gents to the circles described on AB and CD as diameters, are equal, Euc. III, 36. - 244. If AQ, ATP be produced to meet, these lines with AAſ form a right- angled triangle, then Euc. I, 47. 245. If a semicircle be described upon BD as a diameter, it will pass through the point B and touch the line AD in the point D : then the rectangle AE, AB is equal to the square on AD. 246. The semicircles described upon the sides of the triangles will pass through the feet of the three perpendiculars drawn upon the sides. Euc. III. 35. 247. The circle described on AB as a diameter will pass through E and D. Then Euc. III. 36. 248. On AB describe a semicircle, draw CE at right angles to AB meeting the circumference in E, and bisect the angle ACE by CF meeting the circumference in F. From F draw FD perpendicular to AB meeting AB in D. 249. Draw the chord AC in the circle which shall be equal to the side of the given square, join BC and produce it to meet the tangent AP in the point P. B #". By Euc. III. 36, Cor. a circle can be described passing through the points 2 - 2 E, D. 251. Through A, draw the chords APS, AQR ; then AP is equal to PS, and AQ to Q.R. Join Q0. By Euc. III. 35, 31, 1.47, the property is shewn to be true. 252. If the line PA be produced to meet the other circle in A', then PA is equal to AA, by Euc. III. 36. 253. The rectangle FB, FC is equal to the rectangle FD, FE, each being equal to the square on FA. And since FB is equal to FD, it follows that FC is equal to FE: whence BC is equal to D.E. 254. Let the circles intersect in the points P, Q, draw the chord PQ, and join AP, and prove APQ to be a straight line. Euc. III. 36, 37. 255. The construction of the figure suggests a reference to Theorem 3, p. 311. 256. Let the line drawn from A touch the circumference in P, and from B, C, let lines be drawn to the points where the circle intersects the two sides of the triangle. Then by Euc. III, 31; II, 13; III. 36. 257. Let AB, AC be the two given lines meeting at the point A, and D the given point. From D draw DE perpendicular to AB and produce ED to F, so that the rectangle contained by ED, DF may be equal to the given rectangle. On DF as a diameter describe a circle cutting AC in G, join GE and produce it to meet AB in H. Then a circle may be described through the points E, G, F. H. 258. Let the point F in AO be supposed to be found subject to the conditions 444 GEOMETRICAL EXERCISES of the problem. Produce PO to meet the circumference in Q. Then by Euc, i. 5, 6; III, 36, Cor.combined with the given conditions, the squares on OE and of may be shewn to be equal to five times the square on the radius AO. And the triangle PEQ is right-angled. Hence the square on PE is five times the square on AQ, and AO is known, therefore the line PE is known, also the point P is given. Whence the point E is determined subject to the given conditions. - 259. Here A is the extremity of the diameter and C the center of the larger circle, the perpendicular BDE meets AC in E, the smaller circle in D and the larger in B. From the right-angled triangles the truth of the property may be shewn. 260. Draw AG, DK, AE, AC, A.H., E.K, KC, CG, CE, and let M be the in- tersection of AK, EC. Then AEKC is a rhombus, and MK = MA: also GM-MD. Whence GK = AD = 2. AG, and AK = 3. AG : wherefore A.K.AG = 3.AG2. Also AH" = AC* = AG” + GC* + 2.A.G. GM = 3. GM*. Hence A.K.AG – AH2. 261. Let A, B, be the centers, EF the tangent intersecting the line AB joining the centers, and CD the other tangent. Draw the radii AC, AE, BD, BF to the points of contact; and draw. BG parallel to DC meeting AC in G ; and BH parallel to FE meeting AE produced in H. Then BG = CD, BH = EF, AG = AC – BD, and AH = AE + BF = AC + BD: and CD’ – EF" may be proved to be equal to 4. AC, BD, or the rectangle of the diameters of the circles. 262. Let Q be the center of the circles. Join QO, QB. Then QOB is a right-angled triangle. Also ON may be proved equal to OM. And 4. CN. NF + MO* may be shewn to be equal to the difference of the squares on the radii of the circles by Euc, III. 35, II. 6, 1.47. GEOMETRICAL EXERCISES ON BOOK IV. HINTS, &c. 6. LET AB be the given line. Draw through C the center of the given circle the diameter DCE. Bisect AB in F and join FC. Through A, B draw AG, BH parallel to FC and meeting the diameter in G, H. : at G, H draw GK, HL perpendicular to DE and meeting the circumference in the points K, L ; join KL; then KL is equal and parallel to AB. 7. The three radii divide the circle into three equal sectors. 8. Let the line AD drawn from the vertex A of the equilateral triangle, cut the base BC, and meet the circumference of the circle in D. Let DB, DC be joined: AD is equal to DB and DC. If on DA, DE be taken equal to DB, and BE be joined; BDE may be proved to be an equilateral triangle, also the tri- angle ABE may be proved equal to the triangle CBD. The other case is when the line does not cut the base. This theorem has been proposed in the following form :—If three of the angular points of a quadrilateral figure in a circle, be equally distant from each other, then the distance of the fourth from the middle one, equals the sum of the distances from the other two. 9. Let ABC be an equilateral triangle inscribed in a circle, and let AP, BQ, CR be perpendiculars drawn from A, B, C on any diameter DE of the circle. Produce AP to meet the circumference in F, through F draw a line parallel to DE and produce QB, RC to meet this line in G, H, Then BG is equal to CR, and CH to QB. 10. Let a circle be described upon the base of the equilateral triangle, and let an equilateral triangle be inscribed in the circle. Draw a diameter from one of the vertices of the inscribed triangle, and join the other extremity of the di. ameter with one of the other extremities of the sides of the inscribed triangle. The side of the inscribed triangle may then be proved to be equal to the perpendicular in the other triangle. - 11. The line joining the points of bisection is parallel to the base of the triangle and therefore cuts off an equilateral triangle from the given triangle. By Euc. III, 21; 1, 6, the truth of the theorem may be shewn. - - 12. Leta diameter be drawn from any angle of an equilateral triangle inscribed in a circle to meet the circumference. It may be proved that the radius is bisected by the opposite side of the triangle. - 13. The sides are chords of the exterior and tangents to the interior concº tric circle. From the center draw perpendiculars on the sides, Euc, III, 14. ON BOOK IV. HINTS, &c. 445 14 Let the equilateral triangle ABC whose altitude is AD, be turned round its center O, till it assume the position abe, and let the base be of the new position cut BC in E. Produce ad to meet BC in F. Then from the right-angled triangles QDF, dEF, the angle between the two positions of the altitude is proved to be equal to the angle between the bases BC, bo. - 15. Apply Euc. Iv. 5: III. 22, 32. The construction fails, if the triangle be a right-angled triangle. - 16. Let three equilateral triangles be described upon AB, AC, BC, the sides of any triangle, and let D, E, F be the centers of the circles inscribed in the equilateral triangles on AB, AC, BC respectively. Let DE, EF, FD be drawn; then EFD is an equilateral triangle. Join DA, DB, EA, EC, FB, FC. At E in AE make the angle AEG equal to FEC, and take EG equal to ED, and join GA. Then the angles of the triangles GDE, DEF may be proved to be respectively equal, and each equal to two-thirds of two right angles. Shew that the triangle formed by joining the centers of squares described on the sides of any triangle is equilateral. 17. Prove the three chords cut off from the three sides of the triangle to be equal to one another. 18. Let a diameter be drawn from P through the center Q, and join PAſ, PB, PC also from A, B, C, draw lines perpendicular to the diameter through P. Then PAQ, PB'Q, PC Q are three triangles, of which one will be obtuse- angled and the other two acute-angled. Then by Euc II. 12, 13, PA.” -- PB" -- PC” may be found to be equal to 3. PQ.” -- 3. A Q”, Again, by joining PA, PB, PC, &c. in a similar way, PA* + PB” + PC* may be found to be equal to the same quantity. 19. Shew that each of the angles of the triangle BOC, is equal to half the interior angle contained by two sides of the hexagon. 20. Let the lines intersect each other in a, b, c, d, e, f, and form the interior hexagon, the line AC is trisected by the two lines BD, B.F.; and the rest in the Same manner. 21. The line BE is a diameter of the circumscribing circle. - 22. Two tangents can be drawn parallel to each side of the inscribed triangle, which will form a hexagon and an equilateral triangle; the perimeter of the hexagon is two-thirds of the perimeter of this latter equilateral triangle. 23. The hexagon will be equilateral and equiangular, when the lines drawn from the vertices to the bisections of the bases of the two triangles are coincident: in all other positions the hexagons will be equilateral, but not equiangular. 24. It may be shewn that four equal and equilateral triangles will form an equilateral triangle of the same perimeter as the hexagon, which is formed by six equal and equilateral triangles. - 25. This appears from Euc. III. 21. 26. Let the figure be constructed. By drawing the diagonals of the hexagon, the proof is obvious. 27. By Euc. I, 47, the perpendicular distance from the center of the circle upon the side of the inscribed hexagon may be found. 28. Describe a circle about the given triangle ABC, and join the center O with the angular points A, B, C. If from A, B, C lines be drawn parallel and equal to O.A., OB, OC, an equilateral but not equiangular hexagon will be formed equal to twice the area of the triangle. - 29. If O be the point where the lines Aa, Bó, Co meet each other, AO is double of Oa, and so of the rest. The lines ab', be', c'a" are respectively parallel to BA, CB, AC. - - 30. The angles contained in the two segments of the circle, may be shewn to be equal, then by joining the extremities of the arcs, the two remaining sides may be shewn to be parallel. 31. Let the alternate sides of the figure, Euc. Tv. 15, be produced to meet; each of the triangles so formed exterior to the hexagon, may be proved equal in all respects to each of the six triangles into which the hexagon is divided by the diagonals. 32. This appears directly from Euc. I. 38, and Iv. 15. 33. The circles will be the escribed circles of the six triangles formed by º the center of the circle and the angular points of the circumscribed exagon. 3. Join AE, BD ; AEDB is a parallelogram and its diagonals AD, BE bisect each other, Join BF, EC, BFEC is also a parallelogram and the diagonals B.E, CF bisect each other, Join AC, DF, these lines are equal by the triangles ABC, 446 GEOMETRICAL EXERCISES DEF, and the lines FC, AD joining their opposite extremities are also equal therefore FA CD is a parallelogram, apply Theo. 3, p. 311. Shew whether aciº can circumscribe the hexagon. 35. A regular duodecagon may be inscribed in a circle by means of the equilateral triangle and square, or by means of the hexagon. The area of the duodecagon is three times the square on the radius of the circle, which is the square on the side of an equilateral triangle inscribed in the same circle Theorem 1, p. 339. - 36. Let ABC be a triangle, having C a right angle, and upon AC, BC, let semicircles be described : bisect the hypotenuse in D, and let fall DE, DF perſºn. diculars on AC, BC respectively, and produce them to meet the circumferences of the semicircles in P, Q ; then DP may be proved to be equal to DQ. 37. Let the angle BAC be a right angle, fig. Euc. Iv. 4, Join AD. Then Euc. III. 17, note p. 130. 38. The sides of the required right-angled triangle will be 3, 4, and 5 units. 39. AB is the common hypotenuse. Let AD, BC be produced to meet in F. then the circle which circumscribes the triangle CDE, also circumscribes the quadrilateral figure CEDF, and the center O is the bisection of F.E. 40. The excess of the two sides containing the right angle above the hypo- tenuse is equal to the diameter of the inscribed circle. In this theorem the hypotenuse is equal to the diameter of the circumscribed circle. 41. The rectangle contained by the segments of the hypotenuse may be easily shewn to be equal to one-half of the rectangle contained by the sides which contain the right angle. 42. Each line joining the points of contact divides the circle into two segments, one greater and one less than a semicircle. Euc. III. 31. 43. This is manifest from Euc. III, 21. 44. The point required is the center of the circle which circumscribes the tri- angle. See the notes on Euc, III. 20, p. 130. 45. Construct the figure. If of three lines, two be parallel and one coincident, or one parallel and two coincident with three other lines, the three former lines are said to be in the same directions as the three latter. 46. This is shewn from Euc, III. 20, 21. 47. Construct the figure, the proof offers no difficulty. 48. This appears from Euc. III. 3. 49. This property of the figure, Euc. Iv. 4, exhibited by joining AD and producing it to meet the base, is shewn to follow from Euc. 1, 32. 50. By Euc. III. 35. The rectangle BD, DE is equal to the rectangle AD, DC. Also Prob. 43, p. 315, AB = AD* + AD. D.C. Whence it may be shewn that the rectangle B.E, BD is equal to the square on B.A. 51. Produce A.D., BC to meet in F. The circle described about the triangle DEC passes through F. Shew that the circumferences of the two circles intersect at right angles. 53. Join AB, AE, BE; E being the center of the circle. The circle described through the points O, A, B, will pass through E. Euc. III. 22. Prove the angle CAD equal to the angle CDA. 53. Let the figure be constructed; the proof depends on Theorem 4, p. 319. 54. Apply Euc. III. 32. 55. This is manifest from Euc. III. 11, 18. - 56. The base BC is intersected by the perpendicular AD, and the side AC is intersected by the perpendicular B.E. From Theorem I. p. 317; the arc AF is proved equal to Af, or the are FE is bisected in A. In the same manner the arcs FD, DE, may be shewn to be bisected in BC. - If the perpendiculars intersect in P, and meet the sides of the triangle in the points a, b, c, respectively; it may be easily shewn that PD, PE, PF, are bisected by the sides of the triangle in the points a, b, c, respectively. . . 57. Let ABC be a triangle, and let D, E be the points where the inscribed circle touches the sides AB, AC. Draw BE, CD intersecting each other in 0. Join AO, and produce it to meet BC in F. Then F is the point where the inscribed circle touches the third side BC. If F be not the point of contact, let some othº point G be the point of contact. Through D draw DH parallel to AC, and DK parallel to BC. By the similar triangles, CG may be proved equal to CF, or G. the point of contact coincides with F, the point where the line drawn from A. through O meets BC. - - 58. Let ABC be any triangle inscribed in a circle, and let the perpendicular AD, BF, CF intersect in G. Produce AD to meet the circumference in H, and ON BOOK IV. HINTS, &e. 447 join BH, CH. Then the triangle BHC may be shewn to be equal in all respects to the triangle BGC, and the circle which circumscribes one of the triangles will also circumscribe the other. Similarly, it may be shewn by producing BE and CF, &c. 59. First. Prove that the perpendiculars Aa, Bb, Ce pass through the same point O, as Theo. 153, p. 331. Secondly. That the triangles Act, Bca, Cab are equiangular to ABC. Euc. III. 21. Thirdly. That the angles of the triangle abc are bisected by the perpendiculars. 60 Produce CE to meet the circumference in F. Take the center C, join FC, and shew that FC bisects AB at right angles, whence the arc AF is equal to F.B. 61. Produce CD to meet AB in Q, then CQ is perpendicular to AB, Euc. III. 36. 62. Let CA, BA be produced to meet the larger circle in A, B respectively: then CAA' is a chord and BA from the center is perpendicular to CA. 63. Construct the figure, and let E be the intersection of the tangents at B and C, join DE, BE, and shew that CBED is a quadrilateral about which a circle can be described. 64. Let PM produced meet the circumference of the larger circle in P'. Because OMA is drawn from the center O perpendicular upon the chord PQP", therefore PP' is bisected in M ; also the arc PAP' is bisected in A; the angles POA, PQA are equal, being both at the circumference and standing on the arc AP of the smaller circle. And POA is double of APP. Whence the angle Q is double of P. 65. Join DG, EG, and prove that these lines bisect the angles at D and E. Euc. IV. 4. 66. Apply Euc. III. 32: 1. 32. 67. Produce AC to meet the circumference of the first circle in E, and join BE; then the angles DBC, DCB may be proved each equal to BCE. ADB was drawn to touch the circle in B, and DC is drawn to meet it in C, and makes the angle OCB equal to OEB, therefore by converse of Euc, III. 32, DC touches the first circle. 68. At Q draw QE at right angles to BD meeting CD in R and join PR, PD; if RP can be shewn to be at right angles to PD, the circle described about PQD will pass through R, and have its center in the line R.D. 69. This is obvious from the fact, that two tangents drawn from any point to a circle are equal. 70. Let the arcs AB, BC, CA be bisected in the points D, E, F, respectively; and let the circles described, with centers D, F intersect in a, with center D, E intersect in b, and with centers E, F, intersect in c : join Aa, Bb, Ce, and prove these lines to meet in a point which will be the center of the circle in which the triangle is inscribed. 71. If a circle can be described about BBC'C, it may be shewn that BB' and CC are both parallel to the line which joins the extremities of the equal chords. 72. Construct the figure, and shew that ACD is an equilateral triangle. 73. In the arc AB (fig. Euc. Iv. 2.), let any point K be taken, and from K. let KL, KM, KN be drawn perpendicular to AB, AC, BC, respectively, produced if necessary, also let LM, LN be joined : then MLN may be shewn to be a straight line. Draw AK, BK, CK, and by Euc, III. 31, 22, 21; Eue. I. 14. 74. BDEC is a quadrilateral described in the circle CBD, and the side CE is produced, the exterior angle DEA is equal to the interior opposite angle DBC, and the angle CAB is common to the two triangles EDA, BCA : when the angle EDA is equal to the angle ACB. Similarly ADFC is also a quadrilateral des- cribed in the circle ADC; whence also the angle FDB is equal to ACB. 75. Apply Euc. III. 18. - 76. Let the circles touch the base in the points E, F: then the angles ODA, QDE are equal to one another, as also the angles O'DA, O'D.F. Whence the angle ODO is equal to the angles ODE, O'DF. 77. The arc BC being the arc of a quadrant, the tangents at B and C intersect at right angles, and the lines drawn from A perpendicular on these lines also con- tain a right angle. Join ED, then ED is a diameter (Euc. III. 31). And the angle ABD, is equal to ACD, Euc, III. 21. Therefore the angles ACE, ABD are equal to the angle ECD, the angle in a semicircle. 78. Let AO meet FE in s and FC in g : describe a circle touching Fg in F and passing through the point A. Then the angle gFs is equal to FAg: Euc. III. 448 GEOMETRICAL, EXERCISES 32, and the angles gr’s, Fqs are equal to the angles FAQ, Fq.A.: but the two angles FA.g., Fq.A., are equal to a right angle, Euc. I, 32; therefore gFs, Fºs are also equal to a right angle; but the exterior angle, FSA is equal to the angles gFs, Fas: wherefore the angle FSA is a right angle, and EF intersects AO at right angles. #. Let Ol. O, O, be the centers of the circles, join OA, O, B, &c., also join AO, BO, CO. Prove that in the quadrilateral AOBOL, the angle AOB is the supplement of AOB; and similarly for the angles at O2, Os. 80. Let P, Q, R be the points of contact of theinscribed circle with BC, CA, AB, the sides of the triangle ABC : join PQ, QR, RP forming the triangle PQR. From P, Q, R, draw Pa, Qb, Ro perpendiculars on the opposite sides, and join ab, be, ca; then these lines are respectively parallel to AB, BC, CA the sides of the original triangle. The angle BRP is equal to RQP, Euc, III. 32. The quadrilateral Rh20, can be circumscribed by a circle, and the side Rò is produced to P, the angle PöC. is equal to PQR, (Euc, III. 22, note p. 130), therefore the angle Pbc is equal to BRP : but Q5 bisects the angle abo, Theo. 53, p. 346; whence the angle abR. is equal to Pbc. Wherefore the angle abłł is equal to PRB or 5BB, and these are alternate angles, therefore ab is parallel to A.B. 81. Let ABC be a triangle, and let the perpendiculars Aa, Bb, Co from the angles upon the opposite sides intersect in O. Let P be the center of the circle which circumscribes the triangle. Draw PD, PE, PF perpendicular to AB, BC, CA respectively; the sides of the triangle are bisected at the points D, E, F. Draw DG perpendicular to BC intersecting Bb in H; join EH and produce it to meet AB in K; then EHK may be proved to be perpendicular to A.B. But PD is perpendicular to AB; therefore AD, E H are parallel. For a similar reason DG, PE are parallel; wherefore PEHD is a parallelogram, and PE is equal to DH. Through D draw Dm parallel to Bb meeting Aa in m; then the triangles ADm, DBH are equal in all respects and DH is equal to Am, also to mo, whence AO is double of P.E. 82. I.et ABC be a triangle, and let Aa, Bö, Co be drawn from the angles perpen- dicular to the sides, intersecting each other in O. Join ab, be, ca, and let a circle be described about the triangle abe, and let it cut the sides of the triangle ABC, AB in D, BC in E, and CA in F. Then prove that the perpendiculars drawn from D, E, F at right angles to AB, BC, CA respectively, meet in a point which is the center of the circle which circumscribes the triangle ABC. Lastly, shew that the circle which circumscribes abe, passes through the points of bisection of AO, BO, CO. Or, a circle may be described through the three points D, E, F where the perpendiculars drawn from the center of the circumscribing circle of the triangle ABC, meet the three sides, and then prove that the circumference passes through the points a, b, c where the perpendiculars from the three angles, meet the opposite sides of the triangle ABC. 83. Let F, G, (figure, Euc, Iv. 5.) be the centers of the circumscribed and in- scribed circles; join GF, GA, then the angle GAF which is equal to the difference of the angles GAD, FAD, may be shewn to be equal to half the difference of the angles ABC and ACB. - 84. First, let P be taken within the triangle, and let the figure be drawn; Oi, O2, Oa being the centers of the circles, and a, b, c, the points where Pa, Pb, Po intersect OO, O,Os, O,Os. It is obvious that the sides of the triangle OIO,03 are respectively parallel to the sides of the triangle ABC; and Pa, Pb, Pe, are bisected in a', b, c'; also the figures Abpe, Oºb'P6' are equiangular, also the line POAA is a common diagonal. Next let the point be taken without the triangle. 85. From the centers of the three circles draw perpendiculars to the sides of the triangles, and quadrilaterals are formed having the sums of three opposite angles equal to two right angles. 86. Let AB, CD intersect in E, and let AD, CB be joined and produced to meet in F, forming four triangles ADE, BCE, ABF, DC.F. Let the circles described about the triangles ADE, BCE intersect each other in P. The circles which cir- cumscribe the triangles ABF, DCF shall also intersect in P. Join PA, PB, PC, PD, PE. Then the angle PAD is equal to PEC the exterior angle of the in- scribed quadrilateral ADEP. But the angle PEC is equal to PBC (Euc. III. 21.); therefore PBC the exterior angle of the quadrilateral AFBP is equal to the oppo; site angle PAF. Hence the angles PBF, PAF are equal to two right angles, and therefore the circle which is described about the triangle ABF passes through the point P. In a similar way it may be shewn that the circle which is described about the triangle DCF also passes through the point P. The student may readily shew the truth of the three following theorems:- *ON BOOK IV. HINTs, &c. 449 ... 1. If about each of the four triangles into which any quadrilateral figure is divided by its diagonals a circle be described, the centers of these circles are the four angular points of a parallelogram. - 2. If the sides of a quadrilateral inscribed in a circle be bisected and the middle points of adjacent sides joined, the circles described about the triangles thus formed are all equal and all touch the original circle. * 3. Prove that the four circles, each of which passes through the middle points of the sides of one of the four triangles, formed by two adjacent sides and a diagonal of any quadrilateral, all intersect in a point. 87. Bisect the angles at B and C formed by producing the two sides AB, AC of the triangle ABC; the point O where they intersect is the center of the escribed circle. The center also may be found by joining BD, AD, D being the center of the inscribed circle, and at B drawing BO at right angles to DB, and producing AD to meet BO in O. 88. Draw a line from any angle to the center of the circle, and at the point of intersection draw a tangent to the circle. Describe a circle in the triangle so formed. • 89. The centers are the bisections of the exterior angles of the triangle, which points cannot fall within the triangle. 90. Of any triangle ABC, let AO, BO bisecting the exterior angles at A, B meet in O ; and AO’, CO' bisecting the exterior angles at A and C meet in O', prove OA, AO' to be a straight line. 91. This may be shewn from Euc. III, 22. - 92. Let O be the center of the inscribed circle, O' of the escribed circle which touches BC in E and AB, AC produced. Let D be the bisection of BC, and let OD, AE be drawn, prove the angle FDO equal to the angle FEA. 93. If a, b, c be the points where the inseribed circle touches the sides of the triangle ABC ; and E, F, G, the points where the escribed circle touches the base BC and AB, AC produced. By means of the equality of the two tangents drawn from any point to meet the circle, it may be proved that Be is equal to CG, and BE equal to Ca, whence Ea is shewn to be equal to the difference of AB, A.C. 94. This property is proved by means of the equality of the two tangents drawn from a point without a circle. - 95. Let a circle whose center is O be described about the triangle ABC, and let the arceut off by BC be bisected in D. Let the exterior angles formed by roducing the sides AB, AC be bisected by the lines BF, CF meeting in F. }. DC, DO, DB, D F and prove these lines to be equal to one another. 96. See Theorem 3, p. 318. The triangle AED is one of the triangles. - 97. Let G be the center of the circle inscribed in a triangle ABC, and Ol, O2, Os be the centers of the three escribed circles. It will be seen by constructing the figure that the Hines drawn from Qi, O2, Os through O, are respectively per- pendicular to the sides of the triangle O,O2O3. Then by Theorem 58, p. 346, the truth of the property is proved. - 98. It may be shewn that the line PQ passes through the center of the cir- cumscribing circle, and that P, Q are two points in its circumference. 99. Let O be the center of the inscribed circle of the triangle ABC; Oh, O2, Oa, the centers of the escribed circles which touch the sides AB, AC, BC, and P the center of the circumscribing circle. Let Q.M., OIN1, be drawn from Oil per- pendiculars to AB and CA produced, then Oi!M1 is equal to OIN1, each being equal to the radius of the escribed circle which touches AB. Let O.M., O.N. ; OAMs, OANs, be similarly drawn. Let also OQ, PR be drawn Pºpº. to BC, and let PR be produced both ways to meet the circumference of the circum- scribing circle below the base in K and above the base in L. Then the arc BC is bisected in K ; and the line bisecting BAC, the vertical angle of the triangle, passes through the center O and the point K. Since NiN2 is bisected in R, and Oi!N1, RL, Q2N2 are parallel to one another; the sum of Oi Ni, O2N2 two of the escribed radii is equal to twice R.L.; and the difference of Osms and OQ the third escribed radius and the inscribed radius, is equal to twice RK; whence the sum of OINI, O.N., Os Ma, is equal to the sum of OQ and twice the sum of RL and RK, but twice the sum of RL, RK is twice the diameter of the circumscribing circle, and OQ is the radius of the inscribed circle. Wherefore twice the diameter of the circumscribing, and the radius of the inscribed circle is equal to the sum of ON1, O2N2, Oamſs the radii of the three escribed circles. w _ 100. Let the perpendiculars from A, B, C on the opposite sides intersect in G. Let O be the center of the circumscribing circle, Oi, O, Os the centers of the escribed circles which touch the sides AB, BC, CA, respectively; also let P be G G 450 GEOMETRICAL ExERCISEs the eenter of the inscribed circle. Let the perpendiculars from P, Oi, Os, Oa, be drawn meeting the sides or sides produced of £he triangle. Let the perpendiculars OD, OE, OF from O on the sides BC, CA, AB, when produced both ways meet ‘the circumference of the circumscribing circle in K, L ; Q, R.; S, T; and let R, D, r, d, denote the radii and diameters of the circumscribed and inscribed circles. Then Theo. 81; p. 348. AG = 2. OD; BG = 2. OE; CG = 2. OF, and from But DK = OD + R.; EQ = OE + R.; FS = OF + R, wherefore DK -- EQ + FS = 3R, + OD + OE + OF; whence OD + O.E + OF = R + r ; and 2 (OD + O.E + OF) = 2R + 2r; or AG + BG + CG = D + d. Next, Let ri, r2, rs, denote the radii of the escribed circles. Then 2. OD = ri + r. – 2B ; 2. OE = rs + ra – 2B ; 20 F = ri-H rs—2B ; whence 2. OD+2. OE – 2. OF = 2ri – 2R, -or AG + BG – CG = 2r. -2R, - - - 101. If P the center of the inscribed circle be joined to P' the center of the circle which passes through the centers of the escribed circles, and PP’ be bisected in O ; then O is the center of the circumscribed circle, And the radius of the circle whose center is P’ may be shewn to be double the radius of the circle whose center is O. - 102. Let ABC be the triangle, right-angled at C. Let O be the center of the escribed circle touching the hypotenuse. Draw OD, OE perpendicular on CA, CB produced. Then the figure CDOE is a rectangular parallelogram. See Theorem 3, p. 318. - } -- * 103. Suppose the triangle constructed, then it may be shewn that the differ- ence between the hypotenuse and the sum of the two sides, is equal to the diameter of the inscribed circle. º - 104: If the base be not greater than the diameter, a triangle may be inscribed ‘equal in all respects; but if the base be greater than the diameter, a triangle of less base †. be constructed (Prob. 122. p. 305), and an equal triangle inscribed. 105. Produce the given lines to meet, the angle they contain is equal to, the vertical angle of the triangle. And since the circle is given, its diameter is known ; through the given point draw a chord cutting off a segment from the circle con- taining an angle equal to the angle formed by the two given lines. From the extremities of the chord draw two chords parallel to the two given lines. 106. Let BC be the position of the given base and DE that of the other given chord in the given circle, and let ABC be the triangle required, intersecting DE in the points F, G, so that the rectangle DF, GE is equal to EG take n equal to the side of the given square. Hence DE is given and EG is known, the diffe- rence of DE and EG may be divided so that the rectangle DF, GE may be equal to the square on FG. Whence the points F, G are found, and therefore the point A. - - º- : 107. Since the vertical angle of the triangle is given and the circle is given, . of the triangle is given. And the area is given, whence the altitude can € IOUlDiCl, > . 108. If the perpendiculars meet the three sides of the triangle, the point is within the triangle, Euc. Iv. 4. If the perpendiculars meet the base and the two sides produced, the point is the center of the escribed circle. • 109. With the given radius of the circumscribed circle, describe a circle. T)raw BC cutting off the segment BAC containing an angle equal to the given vertical angle. Bisect BC in D, and draw the diameter EDF: join FB, and with center F and radius FB describe a circle : this will be the locus of the centers of the inscribed circle (see Theorem 3, p. 381.) On DE take DG equal to the given ‘radius of the inscribed circle, and through G draw GH parallel to BC, and meeting the locus of the centers in H. H is the center of the inscribed circle. - 110. This may readily be effected in almost a similar way to the preceding Problem. * 111. With the given radius describe a circle, then by Euc. III. 34. 112. If the radius of the circumscribed circle be known and the perpendicu- lars drawn from the center of the circle; the sides of the triangle can be found. 113. Let ABC be the given circle. Place in it the chord BC equal to the given base of the triangle. Then since the circle is given and the chord BC is given, the segment BAC cut off by BC is given, and therefore the angle in that segment is known. Hence the problem is the same as to construct a triangle having given the base, the vertical angle and the sum of the two sides. . . 114. Let ABC be the equilateral triangle, and let a, b, c be the centers of the squares described upon the sides opposite to the angles A, B, C respectively. The ...triangle formed by drawing ab, be, ca may be proved to be an equilateral triangle, 2. ON BOOK IV. HINTs, &c. 451. thesides of which are respectively equal to the line drawn from any angle of the given triangle to the center of the square on the opposite side. If the numerical value of the side of the given triangle.be given, the areas of the two triangles may be expressed in terms of the given side. * r 115. A Let ABC be the given triangle, and A'B'C' the other triangle, to the sides of which the inscribed triangle is required to be parallel. Through any point a in AB draw ab parallel to A'B' one side of the given triangle, and through a, b draw ac, be respectively parallel to AC, BC. Join Ac and produce it to meet BC in D; through D draw DE, DF, parallel to ca, cb, respectively, and join EF. Then DEF is the triangle required. 116. Let the lines be drawn; the proof offers no difficulty. 117. Suppose the parallelogram to be rectangular and inscribed in the given triangle and to be equal in area to half the triangle : it may be shewn that the parallelogram is equal to half the altitude of the triangle, and that there is a restriction to the magnitude of the angle which two adjacent sides of the parallelo- gram make with one another. . º ' -- Il 8. This point will be found to be the intersection of the diagonals of the given parallelogram. - 119. The differenee of the two squares is obviously the sum of the four tri- angles at the corners of the exterior square. . 120. Construct the figure and prove its angles right angles. - 121. Take any point in one side, and similar points in the other sides, join these points and the figure will be a rhombus, except in one case. 122. If two lines be drawn meeting the opposite sides or sides produced, and be bisected at right angles at their intersection, the lines joining their extremities will form a rhombus. - 123. Draw in the quadrilateral two lines bisecting each other at right angles, and terminated by the opposite sides. 124. The lines which form the required parallelogram may intersect each. other outside or inside the square. - 125. Draw the diagonals ac, bd of the square intersecting in e. With centers, a, b, c, d and radius ae, describe circles cutting the four sides of the square in A, B, C, D, E, F, G, H. Join AB, CD, EF, GH, next join BE, AF. 126. Let the diagonals of the rhombus be drawn; the center of the inscribed circle may be shewn to be the point of their intersection. . 127. On the diameters AB describe a rectangle equal to the given rectilineal figure, and let the side parallel to AB meet the circumference in E. Join AE, EB, through A, draw AF parallel to BE and join BF. ... • 128. The bisection of the diagonals determines the center of the circle. 129. “Since the positions of the two lines to which two adjacent sides of the quadrilateral are to be parallel are given, the angle between them is given. Let P, Q be the two given points outside the given circle through which the other two sides are required to pass. Join PQ and on PQ describe a segment of a circle containing an angle equal to the supplement of the angle contained by the two given lines, and touching the circumference of the given circle in C. Join PC, QC and produee them to meet the circumference in D, B: through B, D draw BA, parallel to one of the given lines and join AD; AD shall be parallel to the other. 130. Apply Theo. 3, p. 318. *. 131. ABCD is a square, E the intersection of the diagonals, with radius AE, and centers A, B, D describe circles cutting AB in F, G and AB in H: join GH, and prove GH equal to FG. • . - 132. The sum of the arcs on which stand the first, third, fifth and seventh angles, is equal to the sum of the arcs on which stand the second, fourth, sixth and eighth angles. 133. Construct the figure, and the truth is obvious. - 134. Take half of the side of the square inscribed in the given circle, this will be equal to a side of the required octagon. At the extremities on the same side of this line make two angles each equal to three-fourths of two right angles, bisect these angles by two straight lines, the point at which they meet will be the center of the circle which circumscribes the octagon, and either of the bisecting lines is the radius of the circle. •' 135. Each of the interior angles of a regular octagon may be shewn to be equal to three-fourths of two right angles, and the exterior angles made by producing the sides, are each equal to one-fourth of two right angles, or one-half of a right angle. - - 136. By constructing the figures and drawing lines from the center of the G G 2 452 - GEOMETRICAL ExERCISEs circle to the angles of the octagon, the areas of the eight triangles may be easily shewn to be equal to eight times the rectangle contained by the radius of the circle, and half the side of the inscribed square. , - 137. Let the given point in the circumference be joined with the center of the circle, then by Theo. 3. p. 311. - - 138. Let a circle inscribed touch one of the sides in the point A, let ABC be an equilateral triangle inscribed in the circle, also let a circle inscribed in the tri- angle, touch BC, CA, AB in the points D, E, F respectively. AD being joined passes through O the common center of the two circles. If any point P be-taken in the circumference of the inner circle, and PA, PB, PC be drawn, then PA* + PB* + PC* + OD* = 4. AO”, or the area of the given square. By Theorem 22, #: 309, after joining PF, PC, PD; PA*--PB” + #&#; +PE* + PF*-i- 3. BF". f perpendiculars PL, PM, PN be drawn to the sides of the triangle BC, CA, AB; since the square of the chord PD drawn from any point P to the point of contact of the inscribed circle, is equal to the rectangle contained by the perpendicular PL and the diameter of the circle, PD* + PE* + PF* = (PL + PM + PN). 2DO ; also P.L. -- PM + PN = AD (Theorem 5, p. 312.) = 3. OD (Euc. Iv. 4.) Whence may be deduced the required property. Is the property true when the point P is not on the circumference, but any point within the circle g - 139. Let AB, AC, AD, be the sides of a square, a regular hexagon and an octagon respectively inscribed in the circle whose center is O. Produce AC to E making AE equal to AB; from E draw EF touching the circle in F, and prove EF to be equal to AD. - 140. Bisect the angle contained by the two lines at the point where the bi- secting line meets the circumference, draw a tangent to the circle and produce the two straight lines to meet it. In this triangle inscribe a circle. * 141. From the given angle draw a line through the center of the circle, and at the point where the line intersects the circumference, draw a tangent to the circle, meeting the two sides of the triangle. The circle inscribed within this triangle will be the circle required. 142. The problem may be reduced; to describe a circle in a right-angled isosceles triangle. m 143. Let the diagonal AD cut the arc in P, and let O be the center of the inscribed circle. Draw OQ perpendicular to AB. Draw PE a tangent at P meeting AB produced in E: then BE is equal to PD. Join PQ, P.B. Then AB may be proved equal to QE. Hence AQ is equal to BE or DP. - 144. This problem is a particular case of describing a circle to touch two given circles and a straight line. * - 145. This problem may be reduced to that of inscribing a circle in a triangle. 146. Construct the figure, and by Euc. 1. 47, it may be shewn that the hypo- tenuse of the given triangle is equal to twice the radius of the quadrant; and the radius of the circle inscribed is one-fourth of that radius. - 147. First, If the two intersecting circles be equal, bisect the line joining their centers intercepted by the circles, and at the point of bisection make an angle equal to half a right angle ; the point where the line cuts the circumference, will bé one of the angular points of the square. Secondly. If the intersecting circles be not equal. - º -148. Let the lines joining the intersections of the circles and the centers inter- sect in G. On each side of O take OA equal to OB on the line joining the centers. Gn. AB describe a square ABCD, join OC and produce it to meet the circle in E, then E is one of the angular points of the required circle. / 149. The area of the triangle formed by joining the centers, may be shewn to be four times the area of the triangle formed by joining the points of contact of the circles. © * ^ 150. Let AB be the base of the given segment, C its middle point. Let DCE be the required triangle having the sum of the base DE and perpendicular CF equal to the given line. Produce. CF to H making FH, equal to DE. Join HD and produce it, if necessary, to meet AB produced in K. Then CK is double of DF. Draw DL perpendicular to CK. - 151. From the vertex of the isosceles triangle let fall a perpendicular on the base. Then, in each of the triangles so formed, inscribe a circle, Euc. Iv. 4; next inscribe a circle so as to touch the two circles and the two equal sides of the triangle. This gives one solution: the problem is indeterminate. Next, let the triangle be *ś and lastly, let the triangle be a scalene triangle. Also let. the three circles be equal or unequal. - 152. If BI) be shewn to subtend an arc of the larger circle equal to one-tenth ON BOOK Iv. HINTs, &c. 453 of the whole circumference:—then BD is a side of the decagon in the larger circle. And if the triangle ABD can be shewn to be inscriptible in the smaller circle, BD will be the side of the inscribed pentagon. - 153. Shew that the angles and sides of the two triangles are respectively equal. Q 154. Let DC, CA produced meet the circumference of the larger circle in E, F respectively, join DF; prove CEF to be equiangular to the triangle ABD. 155. This is proved by showing that BAD is bisected by AD, and that BF is divided in D so that the rectangle #. BD is equal to the square on D.F. 156. Apply Euc. I. 32. - - 157. Apply Euc. III. 22, and shew that the angle DCB is equal to EAC. 158. The angle at the center of the smaller circle is equal to double the sup- plement of the angle ACD. t - 159. Apply Euc. I. 32. 160. It may be shewn that the angles ABF, BFD stand on two arcs, one of which is three times as large as the other. . 161. In the fig. Euc. Iv. 10, the triangle ACD has the angle ACD at the vertex three times either of the angles at the base. - 162. The angle at the vertex of the triangle is double the vertical angle, fig. Euc, xv. 10, and the angles at the base are each three halves of that angle. 163. The angles at the base of the triangle are each equal to half the vertical angle BAC of the triangle ABD in the figure, Euc, Iv. 10. - 164. The angle BAC at the vertex of the triangle (fig. Euc. Iv. 10.) is one- fifth of two right angles; if this angle be bisected, each is the fifth of a right angle. A right angle can be divided into 2, 3, or 5 equal parts, and any angle can be divided into two equal parts; but no method is known by which any angle can be divided into 3 equal parts by means of the straight line and circle. ... The condition, how- ever, for the trisection of an angle can be found. ... Let AOC be one-third of any angle AOB ; with centre O and radius OA describe a circle, intersecting OB, QC in B, C. Draw BEF parallel to CO meeting the circle in E and the diameter AOD produced in F, and join EO. . Then the angle BFO = AOC = one-third of AOB, and OBF = BOC = two-thirds of AOB = twice B.F.O. But OBE = OEB = BFO + EOF: therefore twice BFO = BFO + EOF; hence BFO = EOF, and EF = EO. To trisect the angle AOB, the line BEF is required to be drawn in such a manner that the point EF shall be equal to the radius of the circle; which cannot be effected by the straight line and circle. 165. Shew that CF is parallel to BDG and that CD is equal to FG. 166. Let ABCDE be an equilateral pentagon having the angles ABC, BCD, CDE equal. Join AC, CE, then Euc. I.4, the angle BAC is equal to DEC and CA is equal to CE; therefore Euc. I. 5, the angle CAE is equal to CEA ; hence the angle BAE is equal to AED, and these angles may be shewn to be equal to each .#. three given angles. . - 167. This may be shewn from the equality of the alternate angles. 168. By means of the parallels and the triangles, the properties may be proved. 169. It may be proved that the diagonals bisect the angles of the pentagon, and the five-sided figure formed by their intersection, may be shewn to be both equiangular and equilateral. * - 170. If a side CD (figure, Euc. Iv. 11) of a regular pentagon be produced to K, the exterior angle ADK of the inscribed quadrilateral figure, ABCD is equal to the angle ABC, one of the interior angles of the pentagon. From this a construc- tion may be made for the method of folding the ribbon. 171. The diagonals intersect in five points, the triangle is formed by joining the first, third and fourth points. The proof is obvious. - 172. Apply Euc. III. 20, 22. - - 173. The angle at which two circles intersect, is the angle between two lines drawn through a point of intersection and at right angles to two radii of the circles drawn from that point. - - 174. Prove that the two radii of the circle circumscribing the triangle AFD, which are drawn to the points A, B are perpendicular to AB, DC respectively. 175. The triangle formed by joining the first, third and fourth angles of the pentagon, may be shown to be greater than double one of the triangles formed by joining the extremities of two adjacent sides. 176, There are different constructions for the regular pentagon which can be effected without describing the triangle as in Euclid; but all depend on the same principle. ... The following are two methods. (1) At A and B the extremities of the given line with radius AB describe two circles, and divide AB in C, so that sº 454 . GEOMETRICAL ExERCISEs the rectangle AB, AC is equal to the square on BC. From A to D along the cir- cumferenee of the circle whose center is A, set off three chords each equal to BC; and similarly let E be the point in the circumference, where the same chords are set off from B. With D, E as centers and radius AB describe two circles intersect- ing in F; then A, B, D, F, E, are the angular points of the pentagon. (2) Draw BC at right angles to A.B. the given line, and equal to half of AB. With center B and radius BC describe a circle, join AC and produce it to meet the circumference again in D. With A, B as centers and radius AD, describe two circles cutting each other in E; with centers A, E, C and radius AB describe circles cutting each other in F and G. Then A, B, G, E, F, are the angular points of the pentagon. 177. The figure ABCDE is an irregular pentagon inscribed in a circle; it may be shewn that the five angles at the circumference stand upon arcs whose sum is equal to the whole circumference of the circle : Eurc. III. 20. 178. Let the diagonals BD, A.C (fig. Euc Iv. 11.) intersect in K. ; then DK is. equal to DC; and BK the difference of BD and DC, is to be shewn to be equal to the side of a pentagon. - - 179. By means of Euc. 1, 47. III. 35. II. 7, and Theorem 185, p. 354, it may be shewn that the square on the side of the decagon is equal to the square on the difference of twice CB and a side of the hexagon. - 180. In the figure, Euc. Iv. 10, let DC be produced to meet the circumference in F, and join FB. Then FB is the side of a regular pentagon inscribed in the larger circle, D is the middle of the arc subtended by the adjacent side of the pen- tagon. Then the difference of FD, and BD is equal to the radius AB. Next, it may be shewn, that FD is divided in the same manner in C as AB, and by Euc. II, 4, 11, the squares on FD and DB are three times the square on AB, and the rectangle of FD and DB is equal to the square on AB. - - 181. The alternate angles of a decagon stand upon arcs of the circle which are equal to four complete circumferences. 182. Let AB, BC, CD be three consecutive sides of a decagon inscribed in the circle whose center is O. Join AD, AO ; then AD is equal to the sum of AB and AO. Draw OB meeting A.D in E and join OD: then-AB is parallel to OD, and the triangles BAE, EDO are isosceles; whence AE is equal to AB, and ED to DO or AO, therefore AE and ED or AD is equal to the sum of AB and AO. 183. Let AB, BC, be two consecutive sides of a decagon inscribed in the circle whose center is O. Join OB, OC, and produce OC, AB to meet in D. Then BD is equal to BO or AO ; and the rectangle AD, DB is equal to the square on BD or on AO. But the rectangle AD, DB is equal to the rectangle AB, BD and the square on AB, therefore the rectangle AB, BD and the square on AB is equal to the square on AO: whence the rectangle AB, BD is equal to the difference of the squares on AO, A.B. - - 6. - 184. The figure may be proved to be both equilateral and equiangular by means of the isosceles triangles formed by producing the sides of the pentagon. Next prove the five lines joining the points of intersection to be equal, also the angles contained by every two lines which are adjacent to one another. 185. The angle at A the center of the circle (fig. Euc. Ev. 10.) is one-tenth of four right angles, the arc BD is therefore one-tenth of the circumference, and the chord BD is the side of a regular decagon inscribed in the larger circle. Produce DC to meet the circumference in F and join BF, then BF is the side of the in- scribed pentagon, and AB is the side of the inscribed hexagon. Join FA. Then FCA may be proved to be an isosceles triangle and FB is a line drawn from the vertex meeting the base produced. If a perpendicular be drawn from F on BC, the difference of the squares on FB, FC may be shewn to be equal to the rectangle AB, BC, (Euc. I. 47; 11. 5. Cor.) ; or the square on A.C. * 186. Divide the circle into three equal sectors, and draw tangents to the middle points of the arcs, the problem is then reduced to the inscription...of a circle in a triangle. - - - .* 187. Let DF be the tangent at D the extremity of the radius CD perpendicu- lar to AB. Join DO, CO, and prove these lines to be equal to one another. 188. The three lines which join the centers of the three equal circles form an Jquilateral triangle. . tº 189. Let the inscribed circles whose centers are A, B touch each other in G, and the circle whose center is C, in the points D, E ; join A, D ; A, E; at D, draw DF perpendicular to DA, and EF to EB, meeting in F. Let F, G be joined, and FG be proved to touch the two circles in G whose centers are A and B. 190. Draw the figure and the necessary lines. Then by Euc, I. 47. on BOOK IV. HINTs, &c. 455 191. If the radius of the given circle be unity, the radius of each of the four equal circles which touch it externally and each other, may be found numerically. 192. Describe five circles in the sectors formed by drawing lines from the center of the circle to the angular points of the inscribed pentagon. 193. Each of the circles is the escribed circle of the triangle formed by draw- ing lines from the center of the inscribed circle of the hexagon, to the angular oints of it. , * - & * & p 194. The problem is the same as to find how many equal circles may be placed round a circle of the same radius, touching this circle and each other. The number IS Slix, 195. Each of the vertical angles of the triangles so formed, may be proved to be equal to the difference between the exterior and interior angle of the heptagon. . - º 196. Every regular polygon can be divided into equal isosceles triangles by drawing lines from the center of the inscribed or circumscribed circle to the angular points of the figure, and the number of the triangles will be equal to the number of the sides of the polygon. If a perpendicular FG be let fall from F the center, (figure, Euc. Iv. 14) on the base CD of FCD, one of these triangles, and if GF be produced to H. till FH be equal to FG, and HC, HD be joined, an isosceles triangle is formed, such that the angle at H is half the angle at F. Bisect HC, HID in K, L, and join KL ; then the triangle HKL may be placed round the vertex H, twice as many times as the triangle CFD round the vertex F. * , 197. Apply Euc. I. 20. 198. If the number of sides be odd, and AB, BC, CD, DE, EA be sides of any polygon (suppose of five sides) inscribed in a circle, whose sides are respectively parallel to the sides ab, be, cd, de, ea of the polygon inscribed in another circle; if A'B' be drawn parallel to AB, and the polygon A'B'C'D'F' be drawn whose sides shall be respectively parallel to the sides of the given polygon, the angular points do not all fall in the circumference of the circle. -- 199. Ilet any equilateral polygons of an odd and of an even number of sides be described about two circles; each of the sides is bisected at the points of contact. Let the diagonals in the two polygons be drawn. It will appear in the polygon of the odd number of sides, that the triangles formed by one side and two diago- nals cannot have any angles increased or diminished without increasing or di- minishing the sides: but in the polygon of an even number of sides, the parallelo- grams formed by two opposite sides of the figure and two diagonals, may have their angles increased or diminished without increasing or diminishing the sides of the parallelograms. * 200. The mode of proof is analogous to that of theorem 5, p. 297. A polygon regular or irregular may be divided into as many triangles wanting two, as the polygon has sides, drawing lines from one of the angles to the other angles of the figure. Hence to construct any irregular polygon of 3, 4, 5, 6, &c. sides and angles; 3, 5, 7, 9, &c. independent conditions must be given. And generally, if the irregular polygon consist of n sides and n angles, to construct the figure, of these 2n conditions, 3n - 2 must be independent. - 201. The sides of the polygon are chords of the outer, and tangents to the inner-circle. ** - - 202. Let any polygon (suppose of five sides) ABCDE be inscribed in the circle whose center is C, and which touches at the point P, the circle whose center is C’. Draw from the angles of the figure to P the point of contact AP, BP, CP, DP, EP, and produce these lines to meet the circumference of the circle whose center is C’ º # c, d, e : join ab, be, cd, de, ea, and prove the polygon abcde similar to E 203. From the given point draw lines to the angles of the polygon ; then the area of the polygon is equal to the areas of the triangles thus formed, namely, the rect- angle contained by the sum of the perpendiculars and half one of the equal sides. But the area of the polygon is also equal to n times the area of the equal triangles of which the polygon is composed, this is n times the rectangle contained by the radius of the inscribed circle and half one side. Hence, &c. - 204. Let two lines also be drawn from the center of the circumscribed circle to the extremities of the same side. , - - 205. Any side of the polygon formed by joining the feet of the perpendiculars is equal to one-half of a side of the given polygon. - - º: Construct the figure drawing the necessary lines, and apply Theorem 8. p. 311. . . . - - & GEOMETRICAL EXERCISES ON BOOK WI. HINTs, &c. - 7. LET two sides intersect in O, through Odraw POQ parallel to the base A.B. Then by similar triangles, PO may be proved equal to OQ : and POFA, Q O.E.B., are parallelograms: whence AE is equal to F.B. *. 8. Apply Euc. vi. 4, v. 7. - 9. . The lines drawn making equal angles with homologous sides, divide the triangles into two corresponding pairs of equiangular triangles; by Euč, vi. 4, the proportions are evident. - - - 10. A circle may be described about the four-sided figure ABDC. By Euc. I. 13; Eue. III. 21, 22. The triangles ABC, ACE may be shewn to be equiangular. - 11. Apply Euc. 1. 48; II. 5. Cor., v1. 16. 12. Let ABC, DBC be two triangles on the same base BC, and let their ver- tices A, D, bejoined, and produced to meet the base BC produced, if necessary, in E. Draw AF, DG, perpendicular to: B.E, then Euc. vi. 1, 4. 13. Apply Euc. vi. 4. 4 14. By Euc. vſ. 3, BE is proved to have the same ratio to EA, as CF to FA. Then by Euc. vi. 2, EF is parallel to BC. - 15. Draw AD, BC, intersecting in O, join PO, O.Q., and shew that PO, O.Q. are in the same straight Iine: 16. Since CE is equal to CA, the triangles CAB, CED are similar, Euc. v.I. 6, and by Euc. I. 5, 32, the triangles C.A.B., DCB may be shewn to be similar. 17. Apply Euc. v.I. 2. - 18. This may be done by means of Euc, VI, 3, A. See note; p. 244. 19. Let BC, be intersect each other in O. From each triangle take the figure AbOC, and the remaining triangles O.Bb, OCC are equal, and OCb is another tri- angle. Euc. v. 7. VI. I. - 20. Join BC, DC. Draw AF intersecting DC in G, and let AF produced meet BC in H. Then DC is parallel to BC, and from the similar triangles FDG, FHC ; FDE, FBC; ADE, ABC : FC may be shewn to be equal to m. FD; and DC = FC + D.F. 21. Join ba cutting AE in F. Then by the similar triangles DöF, DFC; Dba, DBC ; Aba, ABC ; n. FE = (n + 1). DE, and DE = AE – AD. Also from the similar triangles Aba, ABC; n. FE = (n − 1). AE. Whence is shewn 2. AE = (n + 1). AD. - - 22. The angle between two linesis equal to the angle between the perpendi- culars on these lines. g 23. Apply Euc wr. 1. 24, Join BF; then CE: EB : ; triangle CED : triangle EBD, and CE : EB :: triangle CEF : triangle EBF, therefore CE : EB :: triangle CFD : triangle FBD. Also AD : DB :: triangle ADF : triangle BDF. But CE : EB :: AD : DB, whence triangle CFD : triangle FBD :: triangle ADF : triangle FBI); in which the second term is equal to the fourth, wherefore the first is equal to the third, or triangle CFD is equal to triangle ADF. Whence CF is equal to FA, and AC is bisected in the point F. - 25. Let ABC be the triangle, it is required to draw a line parallel to the side AC, so that it may be a mean proportional between the segments of the base BC. Divide BC in D so that BD may be to DC in the duplicate ratio of BC to CA, and through D, draw DP parallel to AC meeting AB in the point P. Then BD is to DP as BC is to CA (Euc. vi. 4.); and BD is to DC in the duplicate ratio of BC to CA (constr.); therefore BD is to DC in the duplicate ratio of BD to DP, and consequently BD is to DP as DP is to DC; or DP is a mean proportional be- tween BD, DC, the segments of the base B.C. 26. Apply Euc. vi. 2, to prove EG equal to BH; then Eue. I. 33. 27. The lines joining the bisections of every two sides may be proved parallel to the remaining side of the triangle, and the equality of the triangles may be in- ... ferred from Euc. 1. 38. 28. Apply Euc. III. 2. The triangles EBD, FDC are equal; and ABD, ADC are equal; whence AED, ADF are equal, ... EBD : EEA :: CD F : ADF, but EBD : EDA :: BE: EA, and CIDF: ADF :: CF: FA, ... BE: EA :: CF : FA, ... EF is parallel to BC. 29. Let the straight line cha bisect the side AB of the triangle ABC in c, cut ON BOOK VI. HINTs, &c. 457 AC in b, and meet the side BC produced in a. Then Ab is to bO as a B is to a C. º A draw AD parallel to BC, and produce cba to meet it in D. Then uc. vi. 4. s * 30. Join AO and produce it to meet the base BC in F; then AOF bisects the angle BAC. The proportions follow directly from Euc. vi. 3. 31. Produce EG, FG to meet the perpendiculars CE, BF produced, if neces- sary. The demonstration is obvious. - 32. Let DE bisect the base BC in F, draw CG parallel to AB, then CG is equal to DB, Euc. I. 26; and the triangles ADE, CGE are similar. Euc. vi. 4. 33. This follows directly from Euc. vi. 4. 34. In BC produced take CE a third proportional to BC and AC ; on CE describe a circle, the center being O ; draw the tangent EF at E equal to AC ; draw FO cutting the circle in T and T’; and lastly draw tangents at T, Tº meeting BC in P and P'. These points fulfil the conditions of the problem. By combining the proportion in the construction with that from the similar triangles ABC, DBP, and Euc. III. 36, 37; it may be proved that CA. PD=CP*. The demonstration is similar for P’D'. - 35. Let ABC be any triangle, let BD be drawn parallel to AC and º to AB, and CE parallel to AB and equal to A.C. Join DC, BE intersecting AB, AC in F, G respectively. Then by means of the similar triangles, two proportions may be found from which it may be proved that AF is equal to AG, and that either is a mean proportional between BF and CG. 36. The triangles. AEC, CBE may be shewn to be equiangular. Euc. vi. 4. 37. The triangles HCF, ABF may be shewn to be equiangular, - 38. Draw FG perpendicular to BA, and FH perpendicular to CE. From the similar triangles AED, AGF, BFG, BCE; DE: CE :: BG : AG. But BG = AE – EG = AE – FHE, and AG = AE + EG = AE -- FH. And from the similar triangles CFH, CEA ; AC : CF :: AE : FH. - Whence may be deduced the proportion DE : EC :: AC – CF : AC + CF. 39. If the base BC of the triangle ABC be produced to E, so that CE is equal to BC, and AC be bisected in ID, then if BD be joined and produced to meet AE in F, AF is one half of F.E. r 40. Let ABC be any triangle, and D the given point in the base BC. Divide DC in E so that CE may be to ED in the given ratio. Join AD, and from E draw EF to meet AC in F, and making the angle CFE less than CAD. Through F draw FG parallel to CB meeting AB in G, join DG and produce it to meet CA, produced in H. Then DH is divided in G in the given ratio. What are the limits to the position of the point F * . . . . 41. First. Let P, Q be on the same side of the lines AC, AB which issue from the point A, and let PM, QN be perpendiculars on AB, and Prm, Qn on AC, such that PM : QN :: Prm : Qn: join PQ, QA, and prove PQ and QA to be in the same straight line. If the points P, Q are on opposite sides of the point A, two of the perpendiculars will fall on the line CA, CB produced. e 42. Let ABC be the given triangle, and let the line EGF cut the base BC in G. Join AG. Then it may be proved that AC is to AB as GE is to GF. . 43. This property may be immediately deduced from Euc. vi. 8, Cor. 44. Let ABC be the triangle, right-angled at C, and let AE on AB be equal to AC, also let the line bisecting the angle A, meet BC in D. Join DE. Then the triangles ACD, AED are equal, and the triangles ACB, DEB equiangular. 45. The segments cut off from the sides are to be measured from the right angle, and by similar triangles are proved to be equal; also by similar triangles, either of them is proved to be a mean proportional between the remaining segments of the two sides. & is º gº º 46. The angles made by the four lines at the point of their divergence remain constant. See Note on Euc. vi. A. p. 244. 47. For “the base produced,” read “the part of the base produced.” Let ABC be the given right-angled triangle, C the right angle, and BC the base. At the vertex A in AB, make the angle BAD equal to the angle BAC, and let AD meet the base CB produced in D. Then D is the point required. Euc, VI, 3. 48. Apply Euc, wi. 8, Cor. ; VI. 17. Q • * * 49. The triangles formed by drawing the successive perpendiculars may be shewn to be equiangular, and each equiangular to the original triangle. 50. From D draw DE perpendicular to AB, then DE is equal to DC. And by Euc. vi. 3, 4, DC : AC : : BE : BC. Whence may be shewn, AC* : AD” :: BC : BE" -- BC" : also BE2 = BD”—DE* = BD” – DC* = (BD+DC). (BD–DC) = BC.(BD–DC). Whence it follows that AC*: AD*:: BC”: BC. (BD-DC + BC) :: BC : 2. B.D. * 458- GEOMETRICAL EXERCISES - 51. First prove AC": AD”:: BC: 2.BD ; then 2. AC*:AD":: BC; BD, whence 2. AC” – AD*:AD”:: BC – BD: BD, - - . e . . and since 2. AC” – AD* = 2. AC” – (AC” + DC*) = AC – CD", the property is immediately deduced. - - 52. BC is a mean proportional between AB and BD, Euc. vi. 8. Cor. and the square on AB is five times the square on BC. 53. The angle DCE is a right angle, and AC is at right angles to DE, whence. Euc. VI. 8, Cor. AC is a mean proportional between AD and AC. - 54. By Euc. vi. 8, Cor. AD. AB = AC*: and AD. (AE + EB) = AC*; also ADC, FEB are similar triangles, whence AD. EB = CD. FE, whence AD . AE + CD . FE = AC”. * S - $ 55. In the figure Euc. v.I. 23, let the parallelograms be supposed to be rectan- gular. Then the rectangle AC : the rectangle DG :: BC : CG, Euc. vi. 1; and the rectangle DG : the regtangle CF :: CD : EC, whence the rectangle AC : the rectangle CF :: BC. CD : CG. E.C. In a similar way it may be shewn that the ratio of any two parallelograms is as the ratio compounded of the ratios of their bases and altitudes. * 56. This property follows as a corollary to Euc. v.I. 23: for the two triangles are respectively the halves of the parallelograms, and are therefore in the ratio compounded of the ratios of the sides which contain the same or equal angles: and this ratio is the same as the ratio of the rectangles by the sides. - 57. Let ABC be a scalene triangle, having the vertical angle A, and suppose ADE an equivalent isosceles triangle, of which the side AD is equal to AE. Then Euc. VI. 15, 16, AC. AB = AD . AE, or AD". Hence AD is a mean propor- tional between AC, AB. Euc. vi. 8. º - 58. The construction is suggested by Euc. 1. 47, and Euc. vi. 31. º, 59. See Note Euc. v.I. A. p. 244. The bases of the triangles CBD, ACD, ABC, CDE may be shewn to be respectively equal to DB, 2.BD, 3.BD, 4.B.D. 60. Apply Euc, VI. 2. 61. Let ABC be the isosceles triangle, BCDE the square upon the base; AFD, AGE, the lines drawn from the vertex A to the corners of the square, and AH the perpendicular on the base. Then the triangles AFH, FCD are similar, and FG the middle segment of the base is double of FH. 62. By compounding the proportions obtained from the similar triangles ABD, DGH, I).EG, GAH; EG may be proved to have to GA, the triplicate ratio of BD to DA. * . 63. The triangle cut off may be proved to be one-fourth of the given triangle. 64. Let the perpendiculars from B, C the angles at the base, meet the line bisecting the vertical angle A in E, F, and let the line bisecting the vertical angle, meet the base in D. Then twice the area of the triangle ABC is the sum of the rectangles contained by AD, BE and AD, CF. The triangles AFC, AEB are equiangular, as also the triangles CFD, BED. - 65. Let the lines DC, BF bisect the angles at the base of the triangle ABD and meet the opposite sides in C, F. (fig. Euc. Iv. 10.) Then by Theo. 56, p. 364. Triangle ABD : triangle ACF :: A.B. BD: AC. CF :: AB : CF; therefore figure BDFC: triangle ACF :: AB – CF : CF :: AC : CF :: AB : B.D. 66. This property may be deduced directly from Euc. v.I. B, 3. 67. Apply Euc. v.I. 1, 3, and Prop. 56, p. 364. . 68. Let ABC be any triangle, and abo be an inscribed triangle, having ch parallel to BC and the vertex a in BC. Draw Bö, then the triangles Bcb, gbo are equal. Euc. I. 37. Then Euc. vi. 1, 19. - - 69. Let DG, EH, FK, be drawn perpendicular to the middle of the sides BC, CA, AB of the triangle ABC, and each equal to half the side from the middle of which it is drawn; join GH, HK, KG, AK, KB, BG, GC, CH, and HA; produce GC, and draw HQ perpendicular to it, and AP perpendicular to BC. Then the two right-angled triangles APC, CQH, are similar; whence AP* = 2. CQ*; also GC* = 2. CD*: and Euc. II. 12, GH’ = GC*-- CH* + 2. GC. CQ, which may be shewn to be equal to #. CB’ + #. CA* + 2. (triangle ABC). Similarly for HK*, KG*. Whence GHP + KG* + HK* is found. - 70. This theorem bears the same relation to Euc. v.I. B, as Euc. v1. A does to Eue. VI. 3. Describe a circle about the triangle ABC, produce EA to meet the circumference again in F and join FC. Then by the similar triangles BEA, FCA; the rectangle BA, AC is equal to the rectangle AE, AF. By Euc. 111. 36, Cor. the rectangle FE, EA is equal to the rectangle B}, EC. And since FE is divided in A, Euc. II. 3, the rectangle FE, EA is equal to the rectangle EA, AF together with the square on AE, Hence, &c. * - sº on BOOK VI. HINTs, &c. 459 71, The triangles DOE, EOB are readily proved to have the same ratio as the triangles EOB, BOA by Euc. vſ. 1. . - } * - 72. Let ABC be the triangle, Mthe middle point of BC, and AD = 2. DM ; draw AK, PL parallel to the base, the former meeting EF produced in K, and the latter through D meeting AB and AC in P and L. Then PD = PL, and by the similar triangles KE: FD = KF. ED, or DF. (KD – DE) = (KD + DF). DE; whence is deduced the equality required. * -- 73. From the theorem itself and Euc. vi. 4, it may be proved that the sides of the triangles ABC, AFG are reciprocally proportional. Euc. v., 15. 74. The triangles have the angles ABC, DBC equal. Apply Theo. 56, p. 364. 75. Having constructed the figure. By Theo. 56, p, 364..the triangle EDC: triangle AFE :: ED. EC : AE . AF :: EC : AE, and triangle AFE: triangle BFD :: AE . AF : FD . BF :: AF : BF. But EC : AE :: ED : BF, Euc. vi. 4, there- fore triangle EDC : triangle AFE :: triangle AFE : triangle BFD. 76. Draw AF, and the triangle AFC is equal to the triangle ABD ; therefore the ratio of the triangle FCE to ABD is known. The numerical value of the ratio. may be found from the note on Euc; II. 11, p. 89. 77. Apply Euc. vi. 30. *. - 78. The triangles ACD, EBC, CDE are equiangular. Draw CF perpendicular to EF and apply Euc. vi. 31. 79. Let the figure be constructed, then from the three similar right-angled tri- angles, Euc. v.I. 19. & g 80. Let the two given lines meet when produced in A. At A draw AD per- pendicular to AB, and AE to AC, and such that AD is to AE in the given ratio. Through D, E, draw DF, EF, respectively parallel to AB, AC and meeting each other in F. Join A.F and produce it, and the perpendiculars drawn from any point of this line on the two given lines will always be in the given ratio. g 81. Let AB be the given line from which it is required to cut off a part BC, such that BC shall be a mean proportional between the remainder AC and another given line. Produce AB to D, making BD equal to the other given line. On AD describe a semicircle, at B draw BE perpendicular to AD. . Bisect BD in O, and with center O and radius OB describe a semicircle, join OE cutting the semi- circle on BD in F, at F draw FC perpendicular to OE and meeting AB in C. ' C 5 the point of division, such that BC is a mean proportional between AC and 82. Find two squares in the given ratio, and if BF be the given line (figure, Euc. vi. 4), draw BE at right angles to BF, and take BC, CE respectively equal to the sides of the squares which are in the given ratio. Join EF, and draw CA parallel to EF: then BF is divided in A as required. 83. Let AB, AC be the two lines meeting in A, and P the given point between them. Through P draw PD meeting AB in D and parallel to AC, and on AB take AD to DQ in the given ratio, join QP and produce it to meet AC in R; then QP is to PR in the given ratio. If the point P be without the angle BAC, draw PD parallel to AC and divide AD in Q, so that AD is to DQ in the given ratio. Join PQ and produce it to meet AC in R, then PR is to PQ in the given ratio. * - 84. Let the given line AB be divided in C, D. On AD describe a semicircle, and on CB describe another semicircle intersecting the former in P; draw PE perpendicular to AB; then E is the point required. ~ 85. Let A, B, C be the three given points, and join AB, BC, CA. On AB take AD, AE equal to the lines p, r respectively. Make the angles ADF, AEG, each equal to ACB. From center D and radius equal to q, describe a circle; and from center F and radius equal to AG describe another circle cutting the former in H, and join HA, HD. At B draw BP making the angle ABP equal to the angle AHD, and meeting AH produced in P. Join CP, FH. Then by means of the equiangular triangles, it may be proved that AP, BP, CP are as AD, DH, AE, or as p, q, r. 2 - 86. Let AB, AC be the two equal lines meeting in A. Join BC and divide it in J) so that BD is to DC in the given ratio. Join AD, and draw BM, CN perpen- diculars on AD, producing if necessary. 87. Let P be the given point and AB the given straight line. I\raw PQ per- pendicular to AB, and produce QP to R making QP to PR in the given ratio; through R. draw CD parallel to AB : then any line drawn through P and termi- nated by the parallels will be divided at P in the given ratio. - 88. With the centers O, O' and diameter greater than the sum of AP, BP, de- scribe two equal circles having AB for their common chord. Draw APQ equal to 460 GEOMETRICAL EXERCISES the sum of AP, PB meeting the inner circumference in P and the outer in Q. Join OO' intersecting AB .#5 : join also PB, PC; PC produced passes through O. Then PB is equal to PQ, and the angle APB is bisected by PCO. & 89. Tiet AB, AC, be the two given lines placed at right angles at A. Take AC to AD in the given ratio, and join CD ; with center B and radius BP equal to the side of the given square, describe a circle cutting CD in P; draw PE parallel to AC, and EF parallel to CD : then AB, AC are divided in E and F as required. 90. Let AB, A.C be the two given straight lines meeting at A, and P the given point between them. At A draw AD, AE so that AD is to AE in the given ratio, and containing the angle ADE equal to the given angle. Join DE, PA, through P draw. PF parallel to AD meeting AB in F, and PG parallel to AE meeting AC in G. PF, PG, are the lines required. 91. Let AB be the given straight line, and let a perpendicular be drawn to AB from the point C. Divide AB in D, so that AD is to DB in the given ratio: then if from D a line DE be drawn to meet the perpendicular in E so that when AE, EB are joined, the angle AEB shall be bisected by ED, E will be the point re- quired. Euc. vſ. 3. \ 92. Let the three given lines meet in A, B, C and form a triangle, and let the ratios of the three perpendiculars be as three lines m, n, p. On AC, BC, take AD, BE each equal to m, draw DF, EG each equal to m and parallel to AB. Join AF, BG and produce them to meet in O, the perpendiculars OP, OQ, OR drawn from O to AB, AC, BC, shall have the same ratios as m, n, p. From F draw FH perpen- dicular to AB, FK to AC, and FL parallel to,AC. Then by the similar triangles. 93. By converting the proportion by Euc. v.I. 16, and observing that DB = DC + CB, CB = AC + 2. CE, and AD = DC – AC. 94. Let AB, CD intersect each other in E, and be terminated by two unlimited lines given in position: and let ab, cd be drawn parallel to AB, CD respectively, intersecting each other in e, and also terminated by the two given lines. Then by the similar triangles and the composition of the ratios. -- * 95. Let a point m be taken in CD, and mpg be drawn parallel to AB, and intersecting FE, HG, in p, q : through q, let rqs be drawn parallel to CD, and in- tersecting EF, AB in r, s ; then mp : pg :: sq : qr, may be proved from the similar triangles. In the same manner, if through s, tsv be drawn parallel to EF, and intersecting GH, CD in t, v, then gs : gr:: ts : vs, and similarly if a line be drawn through v parallel to the line next in order, &c. 96. This problem may be constructed in the same way under more general circumstances than those in which it is enunciated ; namely, when A, B, G, are ; points, and any line is substituted for BK, compatible with the construction ollowing. '• Bisect AB in V and join VG, produce VG to P, make GP = 2. WG ; on PG. describe a circle, in which place the chords PQ, PQ' equal to the given sum of the perpendiculars: then the line QG is that required, and Q’G is that upon which §§ difference (instead of the sum) of the perpendiculars shall be equal to '97. This is only another form of stating the general property of three lines in harmonical proportion. It º be deduced from the note to Euc. v.I.A., page 244. 98. Produce AC to G. Then the angle BCG is bisected by CF, Euc. v.I.A., and the angle ACB by CE. Euc. vſ. 3. - t 99. The angles made by the four lines at the point of their divergence, remain constant. See Note on Euc. VI. A., p. 244. - * 100. The angle AEC is equal to the angle CEB, and if AE be produced, it will be seen that ED bisects the exterior angle of the triangle able. The theorem thus is reduced to the same as 97 above. - 101. If a triangle be constructed on AB so that the vertical angle is bisected by the line drawn to the point C. By Euc. v.I. A. the point required may be de- termined. T - 102. This may be effected in different ways, one of which is the following. At one extremity A of the given line AB draw AC making any acute angle with AB and join BC; at any point D in BC draw DEF parallel to AC cutting AB in E and such that EF is equal to ED, draw FC cutting AB in G. Then AB is harmonically divided in E, G. 103. In the figure Euc. v1.13. If E be the middle point of AC; then A.E or EC is the arithmetic mean, and DB is the geometric mean, between AB and BC. If DE be joined and BF be drawn perpendicular on DE; then DF may be proved to be the harmonic mean between AB and BC. - - 104. Let the line DF drawn from D the bisection of the base of the triangle ON BOOK v1. HINTs, &c. 461 ABC, meet AB in E, and CA produced in F. Also let AG drawn parallel to BC from the vertex A, meet DF in G. Then by means of the similar triangles; DF, FE, FG, may be shewn to be in harmonic progression. 105. Let ABC be any triangle, and let Aa, Bb, Ca be drawn from the angles to the opposite sides intersecting each other in O ; and let ab, ac, be be joined in- tersecting Aa, Bb, Co in a', b, c respectively; then Aa, Bb, Ce are divided harmo- nically. Produce cab to meet BC produced in D, and prove BD to be cut harmo- nically by the four limes cB, ca, c0, cD. These lines also cut harmonically Bb, in the points b', O, by Theo. 99, p. 367. g 106. Let PA, PB, PC, PD be four straight lines drawn from P, and let mpg be drawn parallel to PA and meeting PB, PC, PD in m, p, q, so as to be bisected by PC in p. Through p draw any line EFpG meeting the other three lines in E, F, G. Then EG is divided harmonically in F, p. 107. Let FE meet AB in G, and CD in H. Then by means of the triangles, it may be shewn that FG is to GE, as FH is to HE. w - - - 108. Suppose Bfh to intersect EF, EG, EH in f, g, h, and to meet EK in K. Through h draw hkl parallel to EB or AC, and km parallel to EF. Then by means of the similar triangles it may be proved, that Bf is to fſ, as fg is to 2mg. Whence Bf is to Bh as fg is to gh, since 2mg is the difference between fy and gh. . Again draw Hn parallel to Gg, and by a similar process it is proved, that Bg is to BK as gh to hk. The line Bfh might be drawn so as to meet any one of the equi- distant points in the given line A.C. 109. (1) Let ABC be the triangle which is to be bisected by a line drawn parallel to the base B.C. Describe a semicircle on AB, from the center D draw DE perpendicular to AB meeting the circumference in E, join EA, and with center A and radius AE describe a circle cutting AB in F, the line drawn from F parallel to BC, bisects the triangle. The proof depends on Euc. v.I. 19 ; 20, Cor. 2. (2) Let ABC be the triangle, BC being the base. Draw AD at right angles to BA meeting the base produced in D. Bisect BC in E, and on ED describe a semicircle, from B draw BP to touch the semicircle in P. From BA cut off BF equal to BP, and from F draw FG perpendicular to BC. The line FG bisects the triangle. Then it may be proved that BFG : BAD :: BE : BD, and that BAD : BAC :: BD : BC; whence it follows that BFG : BAC :: BE : BC or as 1: 2. 110. Let ABC be the given triangle which is to be divided into two parts having a given ratio, by a line parallel to BC. Describe a semicircle on AB and divide AB in D in the given ratio; at D draw DE perpendicular to AB and meeting the circumference in E ; with center A and radius AE describe a circle cutting AB in F: the line drawn through F parallel to BC is the line required. In the same manner a triangle may be divided into three or more parts having any given ratio to one another by lines drawn parallel to one of the sides of the triangle. - 111. This problem is a particular case of 115, p. 368. 112. Let ABC be any triangle and D be the given point in BC, from which lines are to be drawn which shall divide the triangle into any number of (suppose five) equal parts. Divide BC into five equal parts in E, F, G, H, and draw AE, AF, AG, AH, AD, and through E, F, G, H draw EL, FM, GN, HO parallel to AD, and join DL, DM, DN, DO; these lines divide the triangle into five equal parts. - 113. Let ABC be the larger, abc..the smaller triangle, it is required to draw a line DE parallel to AC cutting off the triangle DBE equal to the triangle abc. On BC take BG equal to be, and on BG, describe the triangle BGH equal to the triangle abc. Draw HK parallel to BC, and join KG; then the triangle BGK is equal to the triangle abc. On BA, BC take BD to BE in the ratio of BA to BC, and such that the rectangle contained by BD, BE shall be equal to the rectangle con- | . by BK, BG. Join DE, then DE is parallel to AC, and the triangle BDE is equal to abe. - 114. Let ABC be the given triangle. Divide the base BC into three equal parts in the points D, E. On BC describe a semicircle, and draw DF, EG, per- pendicular to BC and meeting the circumference in F, G, and join BF, BG." With center B and radii BF, BG describe circles cutting BC in H, K, and through H, K draw HL, KM parallel to the side BC: HL, KM trisect the triangle ABC. For triangle BHL : triangle BKM :: BH" : BK*:: BF*: BG*::BD . BC : BE. BC :: BD : BE:: 1 : 2. And similarly, triangle BKM : triangle BCA :: 2: 3. Whence triangle BHL : triangle BCA :: 1 : 3; and triangle BHL is one-third of triangle BCA. And LHKM, MKCA may be each shewn to be equal to the triangle BHL. Give the Analysis of this Problem. * 462 GEOMETRICAL ExERCISEs 115. Let D be the given point within the triangle ABC. In the base BC take BX to BC as the part to be cut off is to the whole triangle. Join BX, and describe the parallelogram BEFG equal to the triangle ABX, having the sides BE, BG on BA, BC respectively, and the side EF passing through the point D. Draw DH parallel to AB, and on HG describe a semicircle, place HK in this semicircle, equal to HB, join GK, and with center G and radius GK describe a circle cutting BC in L, M ; the line drawn from L or M through D, cuts off the required part from the triangle ABC. Give the analysis. . 116. Let ABCD be any rectangle contained by AB, BC, -- then AB3 : AB . BC :: AB : BC, - and AB . BC : BC*:: AB : BC, \ “ \ whence AB* : AB . BC :: A.B. BC : BC”; or the rectangle contained by two adjacent sides of a rectangle, is a mean propor- tional between their squares. . 117. In a straight line at any point A, make Ac equal to Ad in the given ratio. At A draw AB perpendicular to cAd, and equal to a side of the given square. On cd describe a semicircle cutting AB in 5 ; and join be, bd; from B draw BC parallel to be, and BD parallel to bd: then AC, AD are the adjacent sides of the rectangle. For, CA is to AD as câ to Ad, Euc. vi. 2; and CA. AD = AB", CBD being a right-angled triangle. -*-*- 118. From one of the given points two straight lines are to be drawn perpen- dicular, one to each of any two adjacent sides of the parallelogram; and from the other- point, two lines perpendicular in the same manner to each of the two remaining sides. When these four lines are drawn to intersect one another, the figure so formed may be shewn to be equiangular to the given parallelogram. - 119. It is manifest that this is the general case of Prop. 5; p. 342. If the rectangle to be cut off be two-thirds of the given rectangle ABCD. Produce BC to E so that BE may be equal to a side of that square which is equal to the rectangle required to be cut off; in this case, equal to two-thirds of the rectangle ABCD. On AB take AF equal to AD or BC; bisect FB in G, and with center G and radius GE, describe a semicircle meeting AB, and AB produced, in H and K. On CB take CL equal to AH and draw HM, LM parallel to the sides, and HBLM is two-thirds of the rectangle ABCD. - 120. Let ABCD be the parallelogram, and CD be cut in P and BC produced in Q By means of the similar triangles formed, the property may be proved. 121. The triangles formed by the perpendiculars on the sides, or the sides pro- duced containing the opposite angle are similar. k ... * 122. Apply Euc. vi. 4. - 123. The two pairs of parallel lines drawn through the angles of the given parallelogram form with the segments of the sides, similar triangles. - 124. Apply Euc. v.I. 2. - 125. Apply Euc. vi. 1, 4. * 126. Let FH, GE (fig. Euc. 1. 43.) be joined and produced, they will meet the diagonal CA produced in the same point L. The lines CA, GE, FH, may be proved by similar triangles to eonverge, and when produced, to meet in the point L. 127. Let AB be the given straight line, and C the center of the given circle; through C draw the diameter DCE perpendicular to A.B. Place in the circle a line FG which has to AB the given ratio; bisect FG in H, join CH, and on the diameter DCE, take CK, CL each equal to CH; either of the lines drawn through K, L, and parallel to AB is the line required. - 128. Let C be the center of the circle, CA, CB two radii at right angles to each other; and let DEFG be the line required which is trisected in the points E, F. Draw CG perpendicular to DH and produce it to meet the circumference in K; draw a tangent to the circle at K : draw CG, and produce CB, CG to meet the tangent in L, M, then MK may be shewn to be treble of LK. - 129. The triangles ACD, BCE are similar, and CF is a mean proportional between AC and C.B. - . - - 130. Let any tangent to the circle at E be terminated by AD, BC tangents at the extremity of the diameter AB. Take O the center of the circle and join OC, OD, OE; then ODC is a right-angled triangle and OE is the perpendicular from the right angle upon the hypotenuse. - - 131. This problem only differs from problem 133, p. 169, in having the given point without the given circle. * 132. Let A be the given point in the circumference of the circle, C its center. Draw the diameter ACB, and produce AB to D, taking AB to BD in the given on Book VI. HINTs, &c. - 463 ratio : from D draw a line to touch the circle in E, which is the point required From A draw AF perpendicular to DE, and cutting the circle in G. 133. Let A be the given point within the circle whose center is C, and let BAD be the line required, so that BA is to AD in the given ratio. Join AC and . produce it to meet the circumference in E, F. Then EF is a diameter. Draw BG, DH perpendicular on EF: then the triangles BGA, DHA are equiangular. Hence the construction. - 134. Through E one extremity of the chord EF, let a line be drawn parallel to one diameter, and intersecting the other. Then the three angles of the two triangles may be shewn to be respectively equal to one another. 135. Let AB be that diameter of the given circle which when produced is perpendicular to the given line CD, and let it meet that line in C; and let P be the given point: it is required to find D in CD, so that DB may be equal to the tangent DF. Make BC : CQ :: CQ : CA, and join PQ ; bisect PQ in E, and draw ED perpendicular to PQ meeting CD in D; then D is the point required. Let O be the center of the circle, draw the tangent DF; and join OF, OD, QD, PD. Then QD may be shewn to be equal to DF and to DP. When P coincides with Q, any point D in CD fulfils the comâitions of the problem; that is, there are innumerable solutions. 136. Let AB, AC be drawn from A and touch the circle in B, C ; let AB be perpendicular to the diameter BD, and CE perpendicular to BD, also let AD in- tersect CE in F; then CE is bisected in F. Join DC. and produce it to meet BA produced in G. DG may be shewn to be equal to AD, and EF to FB by means of similar triangles. Euc. vi. 4. * T 137. Let the chord AB be bisected in E by the chord CD, Let the tangents at A, B meet in P, and the tangents at C, D meet in Q, join PQ, and PQ is parallel to AB. Join PE and produce it to the center O, also join OQ cutting CD in F. Draw the radii OC, O.A., Then the triangles OFE, OPQ are equiangular and right-angled, also the right angle B.EP is equal to the alternate angle EPQ. 138. Let the chords AB, CD intersect each other in E, so that AE is to EB as CE to ED. Then it may be shewn that the lines joining DB, AC are parallel, and that the line bisecting the angle at E bisects these parallels. - 139. Let the figure be constructed: then the quadrilateral AFCD may be cir- cumscribed by a circle; whence the angle FCA is equal to FDA, (Euc. III. 21.) and FCA is equal to CEA (Euc. III. 32.) therefore the angle FDA is equal to CEA. Again, the angles FAC, DAE are equal, each being the complement of a right angle, and adding the angle CAD to each of these equals, the angle FAID is equal to CAE. Hence two angles in each triangle are respectively equal, the third angles are equal, and the two triangles are similar. sº 140. A circle may be described about the quadrilateral PrºAm, (Euc. III. 22.) - the angle Anm is equal to APm. (Euc. III. 21.) And the angle APn is equal to AQP, (Euc. III. 32) Whence the complements na P, QAm of these angles are equal; add to each the angle map, then the angle man is equal to QAP. Where- fore the angles Amn, AQP are equal, and the triangles Anm, APQ are equian- gular, and hence the sides about their equal angles are proportional. 141. This Problem is analogous:to Prob. 6, p. 359. - - 142. Let a tangent be drawn to touch the circle at P, and let PM be drawn perpendicular to the diameter ACB, C being the center of the circle. At A, C, B, draw lines perpendicular to the radius meeting the tangent at P in A', C., B'. Then AA’, MP, CC', BB' are proportional. Draw A'R, PQ parallel to AB and meeting PM, BB' in R, Q respectively. Then by the similar triangles PAR, B'PQ, the required proportion may be deduced, observing that AA is equal to A'P; B'P to B'B, and CC' an arithmetic mean between AA’ and BB'. - 143. The lines so drawn may be proved by Euc. v.I. 3, to be proportional to the segments of the base of the triangle. - 144. Let the figure be constructed and let the perpendicular from AG on the diameter be greater than the perpendicular BH. Take O the center of the circle, join CO, and draw BK perpendicular to AG. Then the triangles ABK, OCF be- ing equiangular; AB is to BK or GH as OC is to CF. But DE is equal to twice . OC, and CT is twice CF. Hence. AB is to GH as DE is to CF; and therefore the rectangle contained by AB, CF is equal to that contained by GH, DE. 145. These properties follow directly from compounding the proportions which arise from the similar triangles in the figure. * 146. Let C be the center of the circle, and let the third tangent at P meet the tangents at the extremities of the diameter AB in D, E. Join CP,-CD, CE; then DCE is a right-angled triangle and CP is perpendicular to D.E. 464 GEOMETRICAL ExERCISES f 147; Let P be any point in the circumference of a circle, and let the chord AB intersect the diameter CD at right angles, and let O be the center of the circle; also let the lines drawn from P to A, B, the extremities of the chord intersect the diameter in E and the diameter produced in G. Join PO, and shew that the tri- angles OEP, OPG are similar. 148. Let AB be any chord of a circle, and P any point in the circumference; let PM be drawn perpendicular to AB, also PN, PQ perpendiculars on the tan- . gents at A, B. Join PH, PB ; the triangles PAN, PBM are similar, as also the triangles PAM, PBQ. - *** 149. Let AB a chord in a circle, be bisected in C, and DE, FG two chords drawn through C; also let their extremities DG, FE be joined intersecting CB in H, and AC in K; then AK is equal to HB. Through H draw MHL parallel to EF meeting FG in M, and DE produced in L. Then by means of the equiangular triangles, HC may be proved to be equal to CK, and hence AK is equal to H.B. 150. Join AB, and from A draw AT to touch the circle in T. Divide AB in C so that the rectangle contained by AB, AC shall be equal to the square on AT. Through C draw CP to touch the circle in P, join AP and produce it to meet the circumference in X. Draw XB intersecting the circumference in Q and join PQ. PQ is parallel to AB. The proof requires Euc. v.I. 6. How is the construction to be effected if the two points A, B be given within the circle? . . - we 151. The tangent AC by Euc. III. 36, may be shewn to be equal to 3. AB : the problem is reduced to finding a mean proportional between AB and 3. AB. 152. Join CM, then CM is equal to CA ; and Euc. v.I. 6, the triangles CMP, $º are equiangular. Whence it may be shewn that the angle QMP is bisected. AM . 153. Let the figure be drawn, and BC, CD, BD be joined. Then ABCD is a quadrilateral figure inscribed in a circle, and BD, AC are the diagonals. By Euc. v1. D, 17, the first proportion is deduced; and the other in a similar way. 154. Let the arc AE be double the arc AB of the circle whose center is C. Let CD, CF, be the perpendiculars on the chords of the arcs AB, AE. Produce CF to meet the circumference in B and G, join GA and draw CH perpendicular to GA. The proportion is deduced from the similar triangles CBD, GFA. 155. Let A, B be the given points, and CD the given line. From E the middle of the line AB, draw EMS perpendicular to AB, meeting CD in M, and draw MA. In EM take any point F; draw FH to make the given angle with CD ; and draw FG equal to FH, and meeting MA produced in G. Through A draw AP parallel to FG ; and CPK parallel to FH. Then P is the center, and C the third defining point of the circle required ; and AP may be proved equal to CP by means of the triangles GMF, AMP; and HMF, CMP, Euc, v1. 2. Also CPK the diameter makes with CD the angle KCD equal to FHD, that is, to the given angle. •’ 156. Let P be the given point and PBA the given line cutting the circle ABC in the points B, A. Let PCD be the line required; join OC, OD, O being the center of the circle. Then the arc AB being given, and the sum of the arcs BC, AD; the arc CD is also given in magnitude, and the angle COD which it subtends at the center. Whence the Construction. Take the arc RS equal to the defect of the sum of the three arcs AB, DA, BC from the whole circumference; join RS, and with center O describe a circle touching RS, and draw PCD to touch the circle. - 157. The angles of the curvilinear triangle are measured by the angles be- tween the tangents to the circles at the three angular points of the triangles. 158. It may be proved that the vertices of the two triangles which are similar in the same segment of a circle, are in the extremities of a chord parallel to the chord of the given segment. * 159. For let the circle be described about the triangle EAC, then by the con- verse to Euc. III. 32; the truth of the proposition is manifest. 160., Let the figure be constructed, and the similarity of the two triangles will be at once obvious from Euc. III. 32; Euc. I. 29. 161. Let ABC be a triangle, and let the line AD bisecting the vertical angle A be divided in E, so that BC : BA + AC :: AE : E.D. By Buc. vi. 3, may be . deduced BC: BA + AC :: AC : AD. Whence may be proved that CE biseets the angle ACD, and by Euc. Iv. 4, that E is the center of the inscribed circle. 162. By means of Euc. Iv. 4, and Euc VI. C. this theorem may be shewn to be true. - - - - 163. Let ABC be any triangle, and from A, B let the perpendiculars AD, BE on the opposite sides intersect in P: and let AF, BG drawn to F, G the bisections of the opposite sides, intersect in Q. Also let FR, GR be drawn perpendicular to V ON BOOK WI. HINTs, &c. 465 BC, AC, and meet in R; then R is the center of the circumscribed circle. Join PQ, QR ; these are in the same line. - * * . . . Join FG, and by the equiangular triangles GRF, APB, AP is proved double of FR. And AQ is double of QF, and the alternate angles PAQ, QFR are equal. Hence the triangles APQ, RFQ are equiangular. - - ". . . . 164. By Euc..vi. 8, Cor. AD : DF: DF: DB, and DE: AD ::DB : DG, by similar triangles ADE, GDB, whence DE: DF:: DF: DG. 165. Let the figure be constructed, then from the isosceles triangles, ED is shewn to be equal to EA, and EG to E.B. Then Euc. vi. 13. -: 166. Let O be the center of the circle which circumscribes the triangle AEB. Join OA and produce it to meet the circumference in E, join also BE. Then AO is at right angles to AC, and the triangles ACB, ABE are similar. - : 167. By means of Euc. I.47. III. 36. Iv. 3, 4. AB and PS may be expressed in terms of the radius of the inscribed circle; whence the ratio of PS to AB may be found. - --- - 168. The two triangles are right-angled triangles, - 169. Let ABC be a triangle, F the center of the circumscribed circle (figure, Euc. Iv. 5.) FD, FE, FG, the perpendiculars from F on AB, AC, BC respectively. Draw DE, DG, GE. Then each of the quadrilaterals ADFE, BDFG, GFEC may be circumscribed by a circle, Euc. III. 22. Then by Euc. v.I. E, and observing that twice the area of the triangle ABC is equal to the sum of the rectangles con- tained by the perpendiculars FD, FE, FG and the sides on which they respec- tively fall, and also to the rectangle by the sum of the sides and the radius of the inscribed circle, we may shew that the rectangle contained by the sum of the perpendiculars and the sum of the sides of the triangle, is equal to the rectangle contained by the sum of the sides, and the sum of the radii of the inscribed and ºcircumscribed circles. 170. The triangles ABC, ADB may be shewn to be equiangular. - .” 171. Let ABC be any triangle, and let D, E be the centers of the circumscribed and inscribed circles respectively. Join AD, and through D draw the diameter FDG and join AE; AE produced meets the diameter in F. Draw EH perpendi- cular to AC and join DE, EC, FC, CG. Then FC is equal to FE, and by Theorem 43, p. 315, DE* = DA* – AE. EF = DA* – AE. FC ; also the triangles AEH, sº being similar, AE. FC = GF. E.H. Whence the truth of the theorem may e shewn. - - - - 172. Let ABC be a triangle, and let D, E, be the points where the inscribed circle touches the sides AB, A.C.. Draw BE, CD intersecting each other in O. Join AO, and produce it to meet BC in F. Then F is the point where the inscribed circle touches the third side BC. If F be not the point of contact, let some other point G be the point of contact. Through D draw DH parallel to AC, and DK parallel to BC. By the similar triangles, CG may be proved equal to CF, or G the point of contact coincides with F, the point where the line drawn from A through Q meets BC. - 173. Let ABC be the triangle; Q the center of the circumscribing circle; P. the intersection of the perpendiculars BG, CH : D, E the middle points of BA, CA; divide PQ in R, so that PR = 2. QR ; and join BR, RE: also draw DQ, QE, #. º triangles BPC, EQD may be shewn to be equiangular, and hence Again, PQ meeting the parallels QE, BP, the angles RQE, RPB are equal; and by hypothesis, RP = 2. RQ : whence the sides about the equal angles are pro- portional, that is, EQ : QR : : BP: PR, and the angles QRE, PRB are equal. The points B, R., E are therefore in one line. The same triangles give PR : RQ :: BR : RE, and hence BR = 2. RE ; or the point R is distant from B, two-thirds of the line BE drawn to the middle of the opposite side AC. ~ 174, Let the figure be constructed as in Theorem 3, p. 318, the triangle EAD being right angled at A, and let the circle inscribed in the triangle ADE touch AD, AE, DE in the points K, L, M respectively. Then AK is equal to AL, each. being equal to the radius of the inscribed circle. Also AB is equal to GC, and AB is half the perimeter of the triangle AED. - - Also if GA be joined, the triangle ADE is obviously equal to the difference of AGDE and the triangle GDE, and this difference may be proved equal to the rectangle contained by the radii of the other two circles. - , - . 175. . From the centers of the two circles let straight lines be drawn to the extremities of the sides which are opposite to the right. angles in each triangle, and to the points where the circles touch these sides. Euc. vi. 4. - . 176. The triangles OQR, ABC may be proved to be similar. Euc. vi. 4. . . H. H. 466 GEOMETRICAL ExPRCISEs 177. Draw FG to bisect the angle DFE, and draw DK EH perpendicnlar on FG, and let FK meet A.B. in G._ _ _ _ _ _ _ _ _ _ _ " " . . . Then, 2.6Bſ: Bf'; ; 2.H.E.; FE, and 2.AG : AF :: 2.DK : FD; by similar triangles ; and by compounding these proportions, observing that AF = FB, AG = GB, and 4.H.E.DK = 4.D.C.C.E, there results . . . . . . . . AB*: AF"; ; 4.E.C.Cſ) : FD.FE; similarly BC”; CE* :: 4.F.A.AD : FEED ; . . . . . whence AB* : BC*:: DEFA : EC.DF. . . . . . . 178; Let O, O' be the centers of the inscribed and escribed circles of the tri- *. ABC, and let the former touch AB, AC in D1, E1, and the latter touch AB, ÁC produced in D, E,. Then BD, is the excess of half the perimeter above the side A6, join Öſſ, ÖB, dºB, op. Then the triangles obiş, o'BD, may be shewn to be similar ; whence may be shewn BDI. BD2 = R. r. And by joining OE, OC, O'C1, O'Es, in a similar way may be shewn, R. r = CE,. CE,. . . I79, Let ABC be: ..º the center of the inscribed circle, draw DE erpendicular to the base BC, E is the point where the inscribed circle touches the ase. Join CE and bisect it in F, bisect also BC in G. Then the points G, D, F may be proved to be in a straightline. , - – . Draw. GH pèrpendicular to BC, and DK perpendicular to G.H. Join CK and produce it to meet BC in L, join also GD, DF. Then LG is equal to GE, and by the similar triangles, CE may be proved to be bisected in F. Hence G, D, must be in the same straight line. * … . . . . . 180. By Euc. Iv. 4, twice the area of the triangle is equal to the rectangle contained by the sum of the sides and radius of the inscribed circle. By Euc. v1. E, the area may be expressed in terms of the sides and the radius of the cir- cumscribed circle. Whence the property required may be deduced, observing that one of the sides of the triangle is half the sum of the other two sides. a 181. Let O be the center of the inscribed circle DEF, and P that of the escribed circle HIK; these are in the line bisecting the angle C. Join M.B, LA cutting COP in N and R. ; draw the several radii to the points of contact ; and join OA, OB, PA, PB. Then prove that FK is equal to the difference of the sides AC, CB"; and therefore to AM. Next, the lines BM, AL are perpendicular to CP, which bisects the common vertical angle, and CNB, CRL are right angles, as are also the angles made by OF, AB. Describe semicircles about ONFB and OFRA, and join NF, RF. Then the angle AFR = AOR = BOF = BNF ; and the alternate angles FAR, FBN are equal. The triangles AFR, BNF are therefore equiangular and AR : AF : ; FB : BN ; also 4. AF:FB = 4.A.R.BN = AL.B.N. - 182. The triangles ABL, ABC may be shewn to be similar, and they have the same altitude; by Euc. v.I. 1, ABL, ABC are as their bases, AL, AC; and by Eug. vi. 19, they may be shewn to be in the duplicate ratio of AB to AC. . . . 188, Apply Prop. 8, p. 343 and Euc. vi. G. T. - 184. Let ABC be any triangle, and DEF the given triangle to which the in- scribed triangle is required to be similar. Draw any line de terminated by AB, AC, and on de towards AC describe the triangle def similar to DEF, # Bf, and produce it to meet AC in F". Through F.' draw FD' parallel to fa, FE parallel to fe, and join D'E', then the triangle D'E'F' is similar to DEF. . . . . . .185. Let BCDE be the square on the side BC of the isosceles triangle ABC. Then by Euc. vi. 2, FG is proved parallel to ED or BC. . . . 186. Let AB be the base of the segment ABD, fig. Euc, III. 30. Bisect AB in C, take any point E in AC and make CF equal to CE: upon EF describe a square EFGH: from C draw CG and produce it to meet the arc of the segment. II]. JK. - 187. Take two points on the radii equidistant from the center, and on the line joining, these points, describe a square; the lines drawn from the center through the opposite angles of the square to meet the circular arc, will determine two points of the square inscribed in the sector. P . 188. Let ABCDE be the given pentagon. On AB, AE take equal distances AF, AG, join FG, and on FG describe a square FG.K.H. Join AH and produce." it to meet a side of the pentagon in L. Draw LM parallel to FH meeting AdB in M. Then LM is a side of the inscribed square. . . . . . . . . . . . 189. Let ABC be the given triangle, Draw. AD making with the base BC an angle equal to one of the given angles of the parallelogram. Draw AE parallel ...; take AD to AE in the given ratio of the sides. Join BE cutting AC in F. . t . º - . . . . ; igo, Analysis. Let ABCD be the given rectangle, and EFGH that to be constructed. en the diagonals of EFGH are equal and bisect each other in P the center of the given rectangle. About EPF describe a circle meeting BD in K, ON BOOK v1. HINTs, &c. 467 and join KE, K.F. Then since the rectangle EFGH is given in species, the angle EPF formed by its diagonals is given; and hence also the opposite angle EKF of the inscribed quadrilateral PEKF is given. Also since KP bisects that angle, the angle PKE is given, and its supplement BKE is given. And in the same way, KF is parallel to another gived line; and hence EF is parallel to a third given line. Again, the angle EPF of the isosceles:triangle EPF is given; and hence the quadrilateral EPFK is given in species. . . . * . . . s 191. In the figure Euc. III. 80; from C draw CE, CF making with CD, the angles DCE, DCF each equal to the angle CDA or CDB, and meeting the arc *ś in E and F. Join EF, the segment of the circle, described upon EF and which passes through C, will be similar to ADB. g . . . . . . w . 192. The square inscribed in the circle may be shewn to be equal to twice the square on the radius; and five times the square inscribed in the semicircle to four times the square on the radius. . . . . - - 193. The three triangles formed by the three sides of the square with segments of the sides of the given triangle, may be proved to be similar. Whence by Euc. v1.4, the truth of the property is proved. . . . . . . . 194. By constructing the figure, it may be shewn that twice the square in- scribed in the quadrant is equal to the square on the radius, and that five times the square inscribed in the semicircle is equal to four times the square on the radius. Whence it follows: that, &c. . . . ; - * - - - 195. By Euc. 1. 47, and Euc. v1.4, it may be shewn, that four times the square on the radius is equal to fifteen times the square on one of the equal sides of the triangle. * - . - * , - - - 196. Constructing the figure, the right-angled triangles SCT, ACB may be proved to have a certain ratio, and the triangles ACB, CPM, in the same way, may be proved to have the same ratio. . . . \ ; 197. Let BA, AC be the bounding radii, and D a point in the arc of a quadrant. Bisect BAC by AE, and draw through D, the line HDGP perpen- dicular to AE at G, and meeting AB, AC, produced in H, P. From H draw HM to touch the circle of which BC is a quadrantal arc ; produce AH, making HL equal to HM, also on HA, take HK equal to HM. Then K, L, are the points of contact of two circles through D which touch the bounding radii AB, AC. Join DA. Then, since BAC is a right angle, AK is equal to the radius of the circle which touches B.A., BC in K, K’; and similarly, AL is the radius of the circle which touches them in L, L'. Also, HAP being an isosceles triangle, and AD drawn to the base, AD* is shewn to be equal to AK. KL. Euc. III. 36; II. 5, Cor. - * - - - : - - 198. Let E, F, G be the centers of the circles inscribed in the triangles ABC, ADB, ACD. Draw EH, FK, GL perpendiculars on BC, BA, AC respectively, and join CE, EB ; BF, FA ; CG, GA. Then the relation between R, r, r" or EH, FK, §: may be found from the similar triangles, and the property of right-angled triangles. - - 199. Let O, O’ be the centers of the semicircles on the radii AC, CB. At C draw CD perpendicular to AB meeting the circumference in D. Produce CD and make DE equal to the radius of either of the smaller semicircles. Join EO, EO', and let F be the center of the circle described about the triangle EOO', join FO, FO' meeting the circumferences in G, H. Then FD, FG, FH may be proved to be equal to one another; and by Euc. III, 36, twice OG may be shewn to be equal to three times G.F. . . . . . . * , = 200. (1) . In every right-angled triangle when its three sides are in arithmetical progression, they may be shewn to be as the numbers. 5, 4, 3. On the given line AC describe a triangle having its sides AC, AD, DC in this proportion, bisect the - angles at A, C by AE, CE meeting in E, and through E draw EF, EG, parallel to AD, DC meeting in F and G. - - - - (2) Let AC be the sum of the sides of the triangle, fig. Eue. wr. 13. Upon AC describe a triangle ADC whose sides shall be in continued proportion. Bisect the angles at A and C by two lines meeting in E. From E draw EF, EG parallel to DA, DC respectively. - 3. - .." . 201. Describe a circle with any radius, and draw within it the straight line MN cutting off a segment containing an angle equal to the given angle, Euc. III. 34. Divide MN in the given ratio in P, and at P draw PA perpendicular to MN and meeting the circumference in A. Join AM, AN, and on AP or AP produced, take AD equal to the given perpendicular, and through D draw BC parallel to MN jºins AM, AN, or these lines produced. Then ABC shall be the triangle ) sº SE bisects the interior angle PSQ, and SL the exterior angle PSF, of the triangle PSQ, PE : EQ :: PS : SQ :: PL : LQ ; or LQ is harmonically divided in P, E. This is (1). (h) Produce AP, BQ to meet in G, and let AQ, BP meet in H ; then G, H. will be in the line CD. For in the triangle PSQ, the three lines QA, BP, SC have been shewn to bisect the angles; wherefore these lines meet in a point. Also, SC bisects PSQ, and AP, BQ bisect the exterior angles at P and Q; there- fore they also meet in a point. Whence G, H are in the line CD., - (i) The lines BG, AG, are bisected in K and I by the circle which passes through the points P, S, O, Q. AE) Let the circle PSOQ cut SG in M ; and draw MO, MQ, MP. Then since SM bisects the angle QSP, it bisects the circumference QMP on which QSP stands ; and hence MQ = MP. Also, since QO = OP, it follows that MO is perpendicular to QK, and is a diameter of the circle PSOQ. Whence OQM, OPM are right angles. But OQ, OP are radii of the circle BQ PA, and hence QM, PM are tangents at Q and P : and they meet at M in SC produced. Where- fore tangents at P and Q always meet in the line SC produced. This is (4). (l). If any chord QSF be drawn through the pole S, and QL, FL be drawn; then the angle SLF = SLQ. . For, join SP : then QS. SP = AS. SB = QS. SF: whence SF = SP, and the triangles PLS, FLS are equal in all respects, and hence the angle SLF = SLQ. (m) Conversely. If QF be drawn through S, and lines be drawn from Q, F to make equal angles with LV drawn through L; the line which bisects the angle FLQ is a diameter passing through S. Note. The line SG is called the polar, and L the pole ; as are also the line LV and the point S, so called respectively. Taken together, either point and its re- spective line are called reciprocal polars;–as for instance SG and L. .A. The characteristic property of the pole and polar to which it is most convenient to reter, is, that if the diameter of a circle AB be produced and be harmonically divided in S and L; then a perpendicular to AB through S is the polar of L, and a perpcndicular to AB through L is the polar of S. 284. It may be proved that of the four lines drawn from C which meet AP, in A, B, Q, P, two of them AC, CQ are at right angles to each other. 285. Draw the figure as the enunciation directs, and let AD cut OC in P : then prove OC to be divided harmonically in the points P, C'; and that OC is equal to OA and OP to OB'. 474 GEOMETRICAL EXERCISES. LOCI. HINTS, &c. 5. Let ABC, DBC, EBC, be three equal triangles on the same base BC and on the same side of it. Join AD, DE. Then AD is parallel to BC, and DE is parallel to BC. - 6. From the given point A, let fall AB on the given straight line, and upon AB describe an equilateral triangle PAB. Produce AP to meet the given line in D, and bisect the angle BAD, by AC meeting the given line in C. On AC, BID, describe the equilateral triangles QAC, RAD; if P, Q, R can be proved to be in a straight line the locus of the vertices of the triangles on the same side of AB will be a straight line. 7. This problem is the same as to find the locus of the vertex of a right-angled triangle which has the sum of the base and perpendicular equal to a given line. 8. Whether P be in the line DF or DF produced, shew that AP must always be equal to PD. 9. Let &Aa', yAy’ be two straight lines at right angles to each other, and let a square abcd have the point a coincident with A, and ab on Aa, and ad on Ay, so that b, d the extremities of the diagonal of the square are upon Aa, and Ay. If the square be moved so that the points b, d always move on Air, Ay, it will be found that the points a, c move on two lines at right angles to each other which bisect the angles acAy, a 'Ay. 10. The points A, C are two points in the adjacent sides BF, BD produced of the parallelogram. It may be shewn that so long as the figure BFED is a parallelogram, the angles made by FE, DE at the point E, with AE and CE, are together equal to two right angles, and therefore by Euc. I. 14, the line AE is in the same straight line with EC. * 11. First let the more simple case be taken when the two sliding lines are equal and when the lines on which they slide are at right angles to one another. Next take the two lines unequal, and lastly, let the lines on which they slide not be at right angles to one another. 12. This may be easily shewn to be a straight line passing through the center of the circle. - 13. This appears from Euc. III. 14. 14. Let lines be drawn to the center of the circle from the extremities of the lines and the points of contact, and the loci of the extremities of the lines may be shewn to be in the circumferences of two concentric circles, unless the parts of the lines on each side of the points of contact be equal. - 15. Let AB be the diameter of the given circle of which the center is C, and E the bisection of any chord A.D. Join EC, then the angle AEC may be proved to be always a right angle in whatever position the chord AD may be situated. 16. Join PA and prove that PA produced if necessary passes through the center of the circle. 17. The center of the circle will be found to be a point in that diameter of the circle which bisects the fixed chord. - - 18. The line drawn from the point of intersection of the two lines to the center of the given circle, may be shewn to be constant, and the center of the given circle is a fixed point. - - 19. Join QB, the angle AQB is constant for all positions of the point P in the circumference of the semicircle, being half a right angle. 20. Let AP, BQ be produced to meet each other in the point O. Shew that the angle POQ is constant. - 21. Let P, Q be the given points, and let PA, QB be one pair of equal chords intersecting each other in M, and PA’, QB' another pair of equal chords inter- secting each other in M’ and the former pair in N, N'. The locus of the intersec- tions at M, M’ are in that diameter which bisects the arc PQ, and the intersections at N, N' lie in the circumference of a circle. 22. Let T be the given point and O the center of the given circle, and let TAB be one of the lines from T, and cutting the given circle in A, B, and let P be the bisection of AB. Join OP; then P is a point in the locus. e 23. Suppose AD, DB to be the tangents to the circle AEB containing the given angle." Draw DC to the center C and join CA, CB. Then the triangles ACD, BCD are always equal; DC bisects the given angle at D and the angle LOGI. HINTS, &c. 475 ACB. The angles CAB, CBD, being right angles, are constant, and the angles ADC, BDC are constant, as also the angles ACD, BCD ; also AC, CB the radii of the given circle. Hence the locus of D is a circle whose center is C and radius D - 24. Let A be the given point and let AB be a line divided in C so that the rectangle AB, BC is equal to the given rectangle; then B is one point in the locus. On AC describe a circle and from A draw any line AQP meeting the cir- cumference in Q, and a perpendicular from B to AB in P. Then P is a second point in the locus. For join QC, the quadrilateral PQCB can be circumscribed by a circle, and AP.PQ = AB.BC, and P is a point in the perpendicular PB ; the locus is the line BP perpendicular to AB. The line BP may also cut the circle. , 25. Let the diagram be described as in the enunciation, and let CS meet the circle at A (between C and S) draw the tangent at A, and in it take AE = AE’. = AS; on CD describe a circle cutting the given one in B, B'; join BB' meeting CS in D; draw DE, TXE'. Then these lines will be the loci of the point Q. } Analysis, Suppose it true that QM = Sy, Q being taken in the line DE. Then draw CP and produce it to meet the circle on SC in N, and join NS. g Since Sy and NP are perpendicular to Py, they are parallel, and since CNS is in a semicircle, it is a right angle. Whence Ny is a rectangle, and NP = Sy. Again, by the similar triangles EAD, QMD, ... . . . { QM.AD = EA. MD, or QM. (AC - CD) = (CS – CA). (CM — CD). Also by the similar triangles CMP, CNS, CM. GS = CP. CN = CP” + CP, PN = CA* + CA. Sy, \ - - º or CA. Sy = CM. CS – CA*. - - Now since the proposition is assumed to be true, namely, that QM = Sy, a comparison of their values gives • 2 : * . . CA: CA – CD :: CM, CS – CA* : (CS – CA) (CM – CD), * - . or CD : CA :: CM. CS – CA* – C.S. CM + CS. CD + CA. CM – CA. CD: . CM. CS – CA*; - - - ... . . ‘. . and since CD : CA :: CA. : CS, and C.S. CD = CA*; . . . . . . . this becomes CA : CS :: CA. DM : CM. CS – CA*; whence CS.DM = CM. CS – CA*, or CA* = Cs (CM - DM) = CS. SD, & - a known truth. Whence the Theorem is true. 26. This is obvious from Euc. I. 47. -- : 27. Let ABC be the required triangle on the given base AB, having the sum of the squares on its sides AC, CB equal to the given square. If the base AB be bisected in D and CD be joined; then by Theorem 3, p. 311, the difference be- tween the sum of the squares on the sides and on the square of half the base may be shewn to be equal to double the square on CD. Hence CD is constant, and therefore the locus is a circle whose center is D and radius DC. 28. Let ABC be the isosceles triangle which fulfils the conditions; take any point F in the base BC, from which draw FH and FG parallel to AB and AC; make GD equal to GA and draw DFE. Then since the side AG of the triangle ADE is bisected in G, and GF is parallel to AE, the base DE is bisected in F. Again, since GF is equal to AH, and FH to AG; and, since BGF, HFC, are isosceles triangles: it follows that AG + AH = AG + GB = AB. Wherefore AD + AE = 2. AG + 2. AH = 2. AB = BA + AC. The variable triangle ADE therefore has its vertical angle and the sum of its sides constant, and the middle of its base is in the line BC. . . . 29. Let ABC be the triangle of which the vertex is A, and the base BC. Let the inscribed circle touch the sides AB, AC in D, E, and the base in F, and the escribed circle touch AB, AC produced in G, H and the base in K. Then CF may be proved to be equal to BK, and FK equal to the difference of the sides AB, AC. Also, the line which bisects the vertical angle A passes through O.Q. the centers of the inscribed and escribed circles; and if AK, O'G, OK', OD, OF be drawn, and FO produced to meet AK in L: then by means of the similar tri- angles ADO, AGO’; ALO, AKO’; OL is proved equal to OD; and the points K, L being known, the position of KLA is known. 30. Let AB be the given base, upon AB as base describe an isosceles triangle ABC having the equal sides AC, BC each equal to the radius of the circumscribing circle; bisect AC, BC by lines meeting in O. O will be the center of the circle which circumscribes the triangle, and therefore the locus of its vertex. - 31. Let ABC be a triangle described on the given base BC and let the line drawn from the vertex A to the point D in BC bisect the vertical angle. Describe * sºle about the triangle ABC and produce AD to meet the circumference BC Ill Jºs n 476 : GEOMETRICAL EXERCISES. Ap , 32. The straight line may be shewn to be that diameter of the circumscribing circle which bisects the base of the triangle. - 33. See Euc. v.I. 3. A. - 34. The third fixed line is that line which bisects the angle BAC. - 35. Join AL, and produce it to meet the base BC in G. Join also DF intersect- ing AG in M. Then every line DF drawn parallel to the base BC is divided in the same proportion as the segments of the base made by a line drawn from the vertex of the triangle. Then conversely, if this proportion hold good, the line joining the points L, G must pass through A. - 36. Divide the given base BC in D, so that BD may be to DC in the ratio of the sides. At B, D draw BB', DD’ perpendicular to BC and equal to BD, DC respectively. Join B'D' and produce it to meet BC produced in O. With center O and radius OD, describe a circle. From A any point in the circumference join AB, AC, A.O. Prove that AB is to AC as BD to DC. Or thus. If ABC be one of the triangles; divide the base BC in D, so that BA is to AC as BD to DC. .* BC and take DO to OC as BA to AC : then O is the center of the CIICIe, 37. Let ABC be the given triangle, and let BDC be a triangle described on BC having the vertical angle BDC double of BAC. Since the base BC is con- stant, and the vertical angle is also constant, the locus of D is a circle. 38. Let ABC one of the triangles on the given base BC be circumscribed by a circle; and let the exterior angles made by producing the sides AB, AC be bisected by BO, CO meeting in O. Then O is the center of one of the escribed circles. Draw OE, OF perpendicular to AB, AC produced, and OD perpendicu- lar to BC. The angle BOC may be proved to be constant for all positions of O, also the line BC is constant. 39. Let AB be the diameter of a circle whose center is C. The simplest case is when the chords drawn from A, are divided in a ratio of equality; in that case the locus is the circumference of the circle described on the radius AC. 40. The series of points will be found to be on the circumference of a circle, which is described on the radius of the given circle as a diameter. 41. The triangles OAC, OBC have the sides OA, OB equal, OC common, and the angles opposite to the equal sides OB, OA, one the supplement of the other. 42. Let BP, AP' intersect in Q, prove that the angle PQP" is constant. 43. The locus is a circle. See Theorem 34, p. 323. 4 44. . If the ladder be supposed to be a straight line, the locus of the middle point of this line, while the ends move along two lines at right angles to each other, may be shewn to be the quadrantal arc of which the radius is half the length of the line. If a semicircle be described on the line as a diameter, any point in the circumference may be shewn to move in a straight line, while the line is moved with its extremities along the two lines. 45. Let the point P in the circumference of the smaller circle coincide with A the extremity of a diameter AOB of the larger circle, O being the center. When the smaller circle has rolled on the quadrantal arc AD, the point P will then be at the center O of the larger circle, and at any intermediate point of con- ; the smaller circle may be shewn to intersect the radius AO of the larger CHTCie, 46. The angle APB is constant, and the chord AB is constant; it may be shewn that the angle at Q is constant. - 47. Shew the angle ARP to be constant for all the various positions of P and Q. - - 48. First, Since the angle PCQ and base PQ of the triangle PCQ are constant, the circle about PCQ is constant in magnitude, and consequently in diameter. Also since the angles PCQ, PRQ are supplementary, R is in the circumference of the circle PCQ. But RQC, RPC being right angles, CR is a diameter; and it has been proved to be of constant magnitude. Wherefore R is always at the same distance from C. Secondly, Draw RS, SC; then PQ is bisected in L, since PSQR is a parallelogram. Also bisect RC in K and join KL. Then since RC is the dia- meter of a circle given in magnitude, and PQ a given chord in it, the line KL is of constant magnitude. Moreover, since RS, RC are bisected in L and K, CS is equal to twice LK a given line; and the locus of S is a circle. - 49. The straight line which is the locus of the second points of intersection, will be at right angles to AB the common chord of the circles, and the produced part BC will be the diameter of the locus of the first points of intersection. 50. Let the tangents drawn at T, T' two points in the circles whose centers are O, O', intersect each other in the point P, so that the angle TPT" is constant tº-º- LOCI. HINTS, &c. 477 for all positions of P. Join PO, PO", TO, T'O', and shew that the point P moves in an arc of that circle of which OO is a chord and OPO, the angle subtended at the circumference by that chord. If the tangents be at right angles to each other, the locus of their intersection is a concentric circle whose diameter is equal to the diagonal of the circumscribing square. 51. Let O, O' be the centers of the two fixed circles, and let P be a point in the required locus, so that the tangents PT, PT' drawn to the two circles are equal. Join TO, OP, T'O', O'P', O O". Then the difference of the squares on PO, P'O' is equal to the difference of the squares on the radii OT, O'T', a eonstant quantity. And O, O' are fixed points. The locus is a straight line at right angles to OO'. . 52. The tangents drawn from the fixed point in the chord produced, are equal to one another. Euc. III. 36. 53. Let the fixed circle CDE cut one of the series of circles touching one. another at P, in the points C, D. Draw PT a common tangent to the series of circles at P; join CD and produce it to meet the tangent in T. The rectangle TU, TD is equal to the square on TP, which is constant for the points where the fixed circle cuts each one of the series of circles. - 54. The locus may be proved to be a circle. 55. Let AB be divided internally in C into two parts, and let P be a point in the locus, so that when PA, PC, PB are drawn, the angle APC may be equal to CPB. Then AP is to PB as AC to CB, Euc. vi. 3. Again, let AB be divided externally in C", and let AP be produced to Q and PC' be drawn so that the angle BPC may be equal to CPQ. Then AP is to PB as AC’ to BC, Euc. v.I. A. Whence CPC is a right angle, and the point P lies in the circumference of a circle. 56. Let AB, BC, CD be the three given straight lines in the same straight line. On AC as a base describe the locus of the vertex of the triangle whose sides are as AB to BC. On the base B D describe the locus of the vertex of the triangle whose sides shall be as BC to CD. Let P be the intersection of these loci. Join PA, PB, PC, PD; the angles ABP, BPC, CPD are equal angles. See prob. 36, , 385. - - P 57. Through A the given point either within or without the given circle whose center is C, let any chord BAD be drawn, and let tangents be drawn at B and D meeting in the point P; then P is one point in the locus. Draw CP cutting the chord BD in E and the circumference in F, and PQ perpendieular to CA pro- duced in Q. Then the rectangle CP, CE is equal to the square on the radius CF; also the rectangle CP, CE is equal to CQ, CA; whence the rectangle CQ, CA is equal to the square on CF. Hence Q is also a point in the locus, which is a straight line drawn from the point G ; and G is known, AG being a third propor- tional to CA and CF. & - 58. Let two circles whose centers are O, O' intersect each other in A, B ; and through C any point in the chord AB, let chords DCE, FCG be drawn through C in the circles whose centers are O, O'. Then the rectangle DC, CE is equal to the rectangle FC, CG, each being equal to the rectangle AC, CB. Hence by the * of Euc, III. 35, the four points D, E, F, G, lie in the circumference of a Clf Cle. - 59. Let AB be a line given in position, and P the given point. At P make the angle APD equal to the given angle, taking the point D, so that AP is to PD in the given ratio. Draw a line through the point D making with PD an angle equal to the angle PAB. Then this line is the locus of the extremity D of the line PD. This may be proved by taking another point A’ on the given line, and making the angle APD' equal to the angle APD. The triangles A'PA, D'PD may be shewn to be similar. - - 60. Let ABCD be the given square, and let the diagonals intersect each other in E. Let F be any point in the locus, and join FA, FB, FD, FC, FE. Then by Theorem 3, p. 311, the sum of the squares on the lines from F to the angles of the square is shewn to be equal to twice the area of the square and four times the square on EF. And E is a fixed point; if EF be constant, the locus of F is a circle whose center is E and radius EF. b - - 61. Let the figure be drawn, it will be seen that whatever inclination the pa- rallels AE, BF ; and CC, DD’ have to the lines AC, BD, the line EF is a diago- nal of the parallelogram AEBF. 62. The locus of the intersections of the diagonals of all the rectangles inscribed in a scalene triangle, is a straight line drawn from the bisection of the base to the bisection of the shorter side of the triangle. - * The straight line may be proved to be a line parallel to the base of the triangle. • e . er 478 GEOMETRICAI, EXERCISES. MAXIMA AND MINIMA. HINTS, &c. 5. THE greatest parallelogram constructed with given sides can be proved to be rectangular. s = - 6. This is shewn by Euc. I. 18. - 7. The construction of this problem may be effected from Prob. r. p. 293. 8. The former part is at once manifest by Euc. 1. 47. , Let the diagonals of the square be drawn, and the given point be supposed to coincide with the inter- i. of the diagonals, the minimum is obvious. Find its value in terms of the side. - . . . 9. Join the two given points, and the perpendicular drāwn from the point of bisection of the distance, will determine the point in the given line. ' 10. The sum of the squares on the two parts of any line is least when the two parts are equal. -" - - 11. See Prob. 1, p. 310. The greatest square is determined by the line DE meeting, but not cutting the circle. - 12. The two rectangles are two squares, whose sum may be shewn to be the least possible. •. - 13. First prove that the perimeter of a square is less than the perimeter of an Hequal rectangle : next, that the perimeter of the rectangle is less than the perimeter of any other equal parallelogram. e . 14. This differs from the preceding in that the quadrilateral figure is not a parallelogram. s * ſº 15. This may be proved by shewing that the area of the isosceles triangle is greater than the area of any other triangle which has the same vertical angle, and the sum of the sides containing that angle, equal to the sum of the equal sides of the isosceles triangle. - 16. Let ABC be an isosceles triangle (fig. Euc. 1, 42), AE perpendicular to the base BC, and AECG the equivalent rectangle. Then AC is greater than EA, &c. 17. This may be shewn by Euc. I 20. • , - 18. On the same base AB, and on the same side of it, let two triangles ABC, ABD be constructed, having the side BD equal to BC, the angle ABC a right angle, but the angle ABD not a right angle; then the triangle ABC is greater than ABD, whether the angle ABD be acute or obtuse. - 19. Make an isosceles triangle, having its vertical angle equal to the given angle. Describe a triangle similar to this isosceles triangle, and having its perimeter equal to the given perimeter. Then the area of this triangle may be shewn to be greater than the area of any other triangle which has the same vertical angle and the same perimeter. 20. Let ABC be a right-angled triangle, having the right angle at B, and the base BC greater than the perpendicular A.B. Let P be the required point in BC so that when AP is joined, and PD drawn perpendicular to AC, AP and PD shall be a minimum. Produce DP to meet AB produced in E. Then ED or AP and PD may be proved to be less than the sum of two lines drawn to any other point of BC. - 21. See Theorem 69, p. 304. 22. Let ABC be an isosceles triangle, and from any point D in the base let DE, DF be drawn parallel to BA, CA respectively: and from D the bisection of the base, let D'E', D'F' be similarly drawn ; it is required to prove that the tri- angles D'BF", D'CE’ are less than the triangles DBF, DCE. - f 23. Let ABC be a triangle whose vertical angle is A, and whose base BC is bisected in D; let any line EDG be drawn through D, meeting AC the greater side in G and AB produced in E, and forming a triangle AEG having the same vertical angle A. Draw BH parallel to AC, and the triangles BDH, GDC are equal. Euc. I. 26. 24. Let two triangles be constructed on the same base with equal perimeters of which one is isosceles. Through the vertex of that which is not isosceles draw a line parallel to the base, and intersecting the perpendicular drawn from the ver- tex of the isosceles triangle upon the common base. Join this point of intersection and the extremities of the base. 25, Let ABC, DBC be two equal triangles on the same base, of which ABC MAXIMA AND MINIMA. HINTs, &c. 479 is isosceles, fig. Euc. I. 37. By producing AB and making AG equal to AB or AC, and joining GD, the perimeter of the triangle ABC may be shewn to be less than the peringeter of the triangle DBC. 26. It can be shewn that of all triangles on the same base and between the same parallels, the isosceles triangle has the least perimeter. The equilateral tri-. angle, being also isosceles, may be shewn to be greater than any other isosceles triangle of the same perimeter, and hence of all triangles of equal perimeters, the -equilateral has the greatest area. * 27. It may be proved that of all triangles having equal vertical angles, and their distances from their vertices to the bisections of their bases, equal to one another, the greatest is that which is isosceles. - 28. Let AC be the common base of the triangles, ABC the isosceles triangle, and ADC any other triangle on the same base AC and between the same parallels AC, B.D. Describe a circle about ABC, and let it cut AD in E and join EC. Then, Euc. I. 17, III. 21. - 29. I.et ABC be any triangle, and D any point in the base, and let AEF be another triangle having the same vertical angle, whose base EF is bisected in the point D. It may be shewn that the rectangle of the sides AE, AF is always less than the rectangle AB, AC, and that the triangle AEF is also always less than the triangle ABC. - 30. By means of Prob. I. p. 387, the sum of ab, be, ca, may be proved to be a minimum. See Theo. 59, p. 346. - 31. On any two sides AC, CB describe internally two segments of circles each containing an angle equal to one-third of four right angles, and let the segments intersect each other in D. Join AD, BD, CD. Let a circle be described with center A and radius AD and a tangent be drawn to it at D. The sum of BD, DC is a minimum when these lines make equal angles with the tangent, and therefore when they make equal angles with the radius AD. Similarly it may be shewn that the sum of AD, DC is a minimum. But the lines AD, BD, CD make equal angles with each other, D therefore is the point required. . - 32. If three lines be drawn within the triangle from the angular points, making equal angles, with the respective sides of the triangle, an equilateral tri- angle is formed by the intersection of these lines, except when each line makes equal angles with two sides of the triangle; and the smaller the angles are which are contained between the lines and the sides of the triangle, the greater will be the area of the new triangle, yhich cannot exceed the area of the given triangle. If however the lines are not required to be drawn within the triangle, the greatest triangle will be that whose sides are parallel to the sides of the given triangle. 33. When the sides are bisected, the lines joining the points of bisection are each respectively equal to half the opposite sides of the triangle. - 34. If four points successively be taken in the sides at equal distances from the angles, the lines joining these points will form a square. When the four points coincide with the bisections of the sides of the given square, the area of the in- scribed square is a minimum, - - 35. The diagonals of all the inscribed rectangles intersect each other in the same point as the diagonals of the given rectangle. Take CR equal to AP and join PR, and upon PR as a diameter describe a semicircle cutting AD in Q, take S equal to DQ, then PQRS is the inscribed rectangle, and the least value may be readily determined. 36. The sides of the three squares inscribed in the triangle may be shewn to be inversely as the three sides of the triangle respectively assumed as the base. Hence that square is the greatest which is contiguous to the smallest of the three sides of the triangle. 37. Describe semicircles on the exterior sides of the quadrilateral, the angles of any rectangle which circumscribes the quadrilateral will lie on the circumfe- rences of these semicircles. w 38. The least chord drawn through a given point, is the line perpendicular to that diameter, which passes through the given point. - 39. The point D may be shewn to be that point in the circumference of the circle which passes through the points A, B and touches the line CD in the point D. 40. Let two lines AP, BP be drawn from the given points A, B, making equal angles with the tangent to the circle at the point of contact P, take any other point Q in the convex circumference and join QA, QB ; then by Prob. 1, p. 387, and Euc. I. 21. - 41. Take the center O, and join AP, AO, &c. and apply Euc. I. 20. 42. Let P be the given point within the circle whose center is O. Let PO 480 GEOMETRICAL EXERCISES. be joined and produced to meet the circumference in A, B. From P draw PQ perpendicular to meet the circumference in Q, and at Q draw a tangent QT meeting the diameter produced in T. * - - 43. From the angular point draw a line through the center of the circle, and from the points where this line cuts the circumference, draw lines perpendicular to the two given lines. - - . . 44. The angle at which any straight line cuts the circumference of a circle is the angle contained between the line and a tangent to the circle at the point of intersection. - 45. The chord PQ is proved greater than any other chord TR passing through the same point N, by Euc. I. 19 ; III. 15. If a circle be described about the tri- angle PSQ, it will touch the circumference of the given circle in P, and the angle SPQ may be shewn to be greater than the angle STR. 46. The point required will be found to be that point at which the line drawn loisecting the radius is perpendicular to it. 47. Let A be the base of the tower, AB its altitude, BC the height of the flag- staff, AD a horizontal line drawn from A, If a circle be described passing through the points B, C, and tºuching the line AD in the point E.: E will be the point required. Give the analysis. 48. The chord HK will be the greatest possible when the arc HK is the greatest possible, that is, when the angle HPK is the greatest possible. 49. It may be shewn that the point required is determined by a perpendicular drawn from the center of the circle on the given line. - 50. See fig. Theo. 3, p. 318; DE will be least when F is the middle point of the arc BFC. - 51. Let A be the given point in the diameter BC; through A draw DAE perpendicular to BC, and join DB, BE. Through A draw any other chord FAG ; and join BIF, BG ; draw FH, GK parallel to BC meeting DE in H, K respectively; join BH, BK, and draw FM, GN perpendicular to BC. Then HK may be proved to be less than DE, and the triangle FBG less than the triangle DBA. gº - 52. First find the chord DE so that the rectangle OF, DE may be the greatest possible. With center O and radius OF describe a circle, DE touches this circle in F. From A draw a line touching the smaller circle and intersecting the given circle in A and B. Join BO, CO, then BOC is the triangle required. - 53. Let the straight line joining the centers of the two circles be produced both ways to meet the circumference of the exterior circle. 54. Let perpendiculars from the center of the larger circle be drawn on those straight lines, then Euc. III, 15. 55. The longest and shortest lines may be proved to be in the line joining the centers of the two circles. 56. Let two common tangents be drawn to the circles, first, intersecting each other, next, meeting each other. - - 57. The line drawn through the point of intersection of the two circles parallel to the line which joins their centers, may be shewn to be double of the line which joins their centers, and greater than any other straight line drawn through the same point and terminated by the circumferences. The greatest line therefore depends on the distance between the centers of the two circles. 58. One circle may be within the other, or the circles may be without one another. Let the centers of the circles be joined and the line produced to meet the circumferences of both circles. r 59. Apply Euc. III. 21. - 60. Suppose the equilateral triangle described, and let the three circles be the escribed circles of the triangle. i - 61. The line drawn through the given point and making equal angles with the two given lines is the line required. If a circle be described touching the two given lines at the points where the required line meets them, the rectangle contained by the segments of any other line drawn through the given point, is greater than the rectangle by the segments of that line which makes equal angles with the iven line. - - I - gl 62. The two points required are at the extremities of a certain diameter. 63. The equilateral triangle can be proved to be the least triangle which can be circumscribed about a circle. * * tº 64. Let ABC be an equilateral triangle inscribed in a circle, and let ABC be an isosceles triangle inscribed in the same circle, having the same vertex A: Draw the diameter AD intersecting BC in E, and BC in E', and let B'C' fall below BC. Then AB, BE, and AB", B'E', are respectively the semi-perimeters TANGENCIES. HINTs, &c. - 481. ef the triangles. Draw B'F. perpendicular to BC, and cut off AH equal to AB, and join #. If BE can be proved to be greater than B'H, the perimeter of ABC is greater than the perimeter of AB'C', Next let B'C' fall above BC. • 65. This may be shewn to be that isosceles triangle whose base is the diameter. . - - - d 66. Let one of the diagonals of the square be drawn, then the isosceles right- angled triangle which is half the square, may be proved greater than any other right-angled triangle upon the same hypotenuse, - 67. #: ABC be the given semicircle, O the center, and OG perpendicular to AB. If EFGH be the greatest rectangle inscribed in the semicircle; then half of this rectangle OGHK will be the greatest rectangle inscribed in the quadrant OBC. The tangent ST to the quadrant at H, and terminated by OB, OC produced in S, T, is bisected at the point H. If instead of a semicircle, any segment of a circle be given, it will be found that the tangent SHT is bisected in H. 68. The magnitude of the diagonal AD of the parallelogram of which AB is constant and AC variable, but terminated by the circumference of the circle, depends on the magnitude of AC and the angle CAB. e 69. Let ABCD, A'B'C'D' be two quadrilateral figures, having their eorres- ponding sides respectively equal, and let ABCD be circumscribed by a circle, and A'B'C'D' not capable of being so circumscribed, the quadrilateral ABC13 is greater than the quadrilateral A'B'C'D'. Draw the diameter AE, intersecting-the side CD, and join CE, E.D. On C'D' describe the triangle C'ED' having its sides Fººd * depends on shewing that of the quadrilaterals ABCE, A'B'C'E', which have their sides respectively equal, the greater is that which has its fourth side equal to the diameter of the circumscribing circle. 20. This may be easily shewn by Euc. 1, 20. - 71. Let ABCDE (figure Iv. º: be a regular polygon inscribed in a circle, and in the arc AB take any point M, may be shewn to be greater than the triangle AMC. 72. This Theorem is the general form of theorem 69, p. 393, and may be proved in a similar manner for a polygon of 5, 6, &c, sides. 73. The proof depends on the property, that an isoscelestriangle has a greater area than any scalene triangle of the same perimeter. TANGENCIES. HINTS, &c. IT may be remarked that in most of the Problems on tangencies, numerous cases arise, depending on the magnitude and position of the given circles, and on other circumstances. In Prob. 43, p. 399, no fewer than fifteen different cases have been distinguished. - - . 5. Let Q be the center of the given circle, Draw OA perpendicular to the given straight line ; at O in OA make the angle AOP equal to the given angle, produce PO to meet the circumference again in Q. Then P, Q are two points from which tangents may be drawn fulfilling the required condition. 6. Let AB be the given arc. Find the center C of the circle, and draw the radii CA, CB and produce them. . The tangent TPT" touching the are AP and terminated by the radii CA, CB produced in T, T' has a minimum value when the point P is at the center of the arc A.B. Af * 7. Let AB be the given straight line and C the center of the given circle whose . radius is CD. At the point D draw DE equal to the given tangent and join CE; with center C and radius CE describe a circle touching or intersecting the given straight line; the point or points so determined, are the point or points required. 8. First. If the circles be without each other. In the second circle draw a chord equal to the given portion of the tangent to the first circle. Describe a concentric circle touching this chord, and then draw a common tangent to the first -circle and the concentric circle. Secondly. If the circles be one within the other, the concentric circle must be drawn to the first circle. 9. Take any point E in the arc BC, and join BE, EC. Then Euc. III. 22, 32. 10. Let A be the given point, and B the given point in the given line CD. and join AM, CM. Then the triangle ABC y equal to the sides of the triangle CED, and join A'E'. The proof * * - I I 482 GEOMETRICAL ExERCISES. At B draw BE at right angles to CD, join AB and bisect it in F, and from F draw FE perpendicular to AB and meeting BE in E. : E is the center of the required Clf Cle, t r 11. Let AB be the given line and C the center of the given circle. Since the radius is given and the perpendicular from the center of the chord is given; hence the length of the chord is known. The problem is the same as, to de scribe a circle passing through two given points and touching a given straight line. 12. Let AB be the given line and D the given point in it, through which the circle is required to pass, and AC the line which the circle is to touch. From D draw DE perpendicular to AB and meeting AC in C. Suppose O a point in AD to be the center of the required circle. Draw OE perpendicular to AC, and join OC, then it may be shown that CO bisects the angle ACD. . 13. Let AB, AC be the two given lines which meet at A, and let D be the given point. Bisect the angle BAC by AE, the center of the circle is in A.E. Through D draw DF perpendicular to AE, and produce DF to G, making FG equal to FD. Then DG is a chord of the circle, and the circle which passes through D and touches AB, will also pass through G and touch. AC. 14. Let A be the given point, BC the given straight line which the circle is to touch, and DE the line in which the center is to be situated. Let DE be produced to meet BC in C. Join AC, and through A draw AB perpendicular to BC, and produce it to meet DEC in D. With center D and radius DB describe a circle cutting CA, produced if necessary, in two points F, G, or touching it in one. Join FD, and draw AH parallel to FD meeting DE in H. The circle described with center H and radius HA is the circle required. Draw HK perpendicular to BC, then by similar triangles, HG is proved to be equal to H.A. 15. Let AT, ABC be the two given lines meeting when produced in A, and let T be the point in the line where the circle touches A, and BC the chord of given length. Since AT* = AC . AB, Euc. III. 36, the positions of B, C the ex- tremities of the given chord BC are indeterminate. ~. 16. Let D be the given point and EF the given straight line. (fig. Euc. III.32) Draw DB to make the angle DBF equal to that contained in the alternate segment. Draw BA at right angles to EF, and DA at right angles to DB and meeting BA in A. Then AB is the diameter of the circle. - I 17. Let AP be the given line, and P the given point where the circle is to touch it, and let AQ be the position of the other given line intersecting the former line in A. At P draw PB, PC intersecting AQ in B, C, and making the angle PBC equal to the supplement of the angle in the segment to be cut off by AQ. The circle passing through P, B, C, is the circle required. -- 18. Let AB, AC be the two lines given in position and when produced meetin in A. Bisect the angle BAC by AF, at A draw AG perpendicular to AB and equal to the given radius. Through G draw GH parallel to AB, meeting AF in H. Then H is the center of the required circle. 19. Let AB, AC be the given lines and P the given point. Then if O be the center of the required circle touching AB, AC, in R, S, the line AO will bisect the given angle BAC. Let the tangent from P meet the circle in Q, and draw OQ, OS, OP, AP. Then there are given AP and the angle OAP. Also since OQP is a right angle, we have OP” – QO* = OP” – OS* = PQ” a given magnitude. More- over the right-angled triangle AOS is given in species, or OS to OA is a given ratio. Whence in the triangle AOP there is given, the angle AOP, the side AP, and the excess of OP” above the square on a line having a given ratio to OA, to determine OA. Whence the construction is obvious. 20. Draw any line AB so that AC is equal to BC; on this describe a segment to contain the given angle; through the given point P draw CQ, meeting the seg- ment in Q, and join QA, QB ; draw PD, PE parallel to QA, QB respectively; and DO perpendicular to AC, and EO to BC. Then O is the center. *. 21. Let P be the given point in the line where the circle is to touch it. , Q, R, the other two given points through which the two tangents drawn to the circle are to be parallel. On QR describe a semicircle, and at P draw PC at right angles to QR to meet the circumference in C. Then C is the center of the circle which touches AB in P and two lines drawn from Q, R parallel to one another. 22. Join AA', BB' and shew A'R' to be constant. - 23. These circles are the inscribed circle and escribed circles of a triangle. If two of the given lines be parallel and intersected by the third, only two circles can be described touching the three lines. ſe 24. Let A B be the given straight line and C the center of the given circle; draw CD perpendicular to AB, and in DC or DC produced, take DE equal to the TANGENCIES. HINTs, &c. 483 radius of the required circle, at E draw EF parallel to AB, and with center C and radius equal to the sum of the radii of the given and required circles, describe a circle intersecting EF in F. Then F is the center of the requiréd circle. 25. Euc. III. 11, suggests the construction. 26. Lêt the circle required touch the given circle in P, and the given line in Q. Let C be the center of the given circle, and C that of the required circle. Join CC’, C'Q, QP; and let QP produced meet the given circle in R, join RC and produce it to meet the given line in W. Then RCW is perpendicular to WQ. Hence the construction. - 27. Let the two given lines AB, BD meet in B, and let C be the center of the given circle, and let the required circle touch the line AB, and have its center in BD. Draw CFE perpendicular to HB intersecting the circumference of the given circle in F, and produce CE, making EF equal to the radius CF. Through G draw GK parallel to AB, and meeting DB in K. Join CK, and through B, draw BL parallel to KC, meeting the circumference of the circle whose center is C in L.; join CL and produce CL to meet BD in O. Then O is the center of the circle re- quired. Draw OM perpendicular to AB, and produce EC to meet BD in N. Then by the similar triangles, OL may be proved equal to OM. 28. Let A be the given point in the given line AB through which the re- quired circle is to pass, and to touch the given circle whose center is C. Through C draw CD parallel to AB meeting the circumference in D; join AD intersecting the circle in E, draw CE and produce it to meet AB in O. d is the center of the required circle. - 29. The position of the line may be determined by means of Theorem 67, p. 400. - 30. The fixed point is the extremity of that diameter of the given circle which when produced is perpendicular to the given line. 31. Let the two circles intersect each other in RS, and let RS be produced to P, and from P let the tangent PA be drawn to the circle whose center is O, and PBC be drawn making BC a chord in the other circle whose center is O'. The center of the circle which passes through the points ABC is in the line AO or AO produced. ' - 32. With center B and radius equal to the radius of the given circle, describe a circle cutting the given arc in D. The problem differs from 28, in having the given point and the required center in a circular arc instead of a straight line. 33. Let C be the given point in the given straight line AB, and D the center of the given circle. Through C draw a line CE perpendicular to AB; on the other side of AB, take CE equal to the radius of the given circle. Draw ED, and at D make the angle EDF equal to the angle DEC, and produce EC to meet DF. This gives the construction for one case, when the given line does not cut or touch the other circle. - s 34, Let the circle ACF having the center G, be the required circle touching the given circle whose center is B, in the point A, and cutting the other given circle in the point C. Join BG, and through A draw a line perpendicular to BG ; then this line is a common tangent to the circles whose centers are B, G. Join AC, GC. Hence the construction. ** * 35. Let C, C' be the centers of the given circles and P the given point in one of them. Draw CP and produce it, the center of the required circle is in CP pro- duced. Join CC' cutting the circles in D, D', on PC take PE equal to C'D' and join EC'; at C’ make the angle EC'O equal to CEP, and produce CO to meet CP produced in O. O is the center of the circle required. 36. Let C be the center of the given circle, B the given point in the circum- ference, and A the other given point through which the required circle is to be made to pass. Join CB, the center of the circle is a point in CB produced. The center itself may be found in three ways. 37. Let AB, CD be the two lines given in position and E the center of the given circle. Draw two lines FG, HI parallel to AB, CD respectively and external to them. Describe a circle passing through E and touching FG, H.I. Join the centers E, O, and with center O and radius equal to the difference of the radii o these circles describe a circle; this will be the circle required. * -º 38. The lines joining the centers of three circles which touch one another, pass through the points of contact of the circles. 39. Inscribe two unequal circles in a semicircle, touching each other and the diameter and circumference of the semicircle, and reverse the problem. 40. Let A, B, be the centers of the given circles, which touch externallyin E; and let C be the given point in that whose center is B. Make CD equal to II 2 484 - GEOMETRICAL EXERCISES. & *** AE and draw AD; make the angle DAG equal to the angle ADG : then G is the center of the circle required, and GC its radius. . 41. Let A, B be the centers of the given circles and CD the given straight line. On the Šide of CD opposite to that on which the circles are situated, draw a line EF parallei to CD at a distance equal to the radius of the smaller circle. From A the center of the larger circle describe a concentric circle GH with radius equal to the difference of the radii of the two circles. Then the center of the circle touching the circle GH, the line EF, and passing through the center of the smaller circle B, may be shewn to be the center of the circle which touches the circles whose centers are A, B, and the line CD. , 42. Let A be the given point, BC the given straight line, and D the center of the given circle. Through D draw CD perpendicular to BC, meeting the cir- cumference in E, F. Join AF, and take FG to the diameter FE, as FC is to F.A. The circle described passing through the two points A, G and touching the line BC in B is the circle required. Let H be the center of this circle; join HB, and BF cutting the circumference of the given circle in K, and join EK. Then the triangles FBC, FKE being equiangular, by Euc. vi. 4, 16, and the construction, R is proved to be a point in the circumference of the circle passing through the points A, G, B. And if DK, KH be joined, DKH may be proved to be a straight line,—the straight line which joins the centers of the two circles, and passes through a common point in their circumferences. ſ 43. Ilet the two given circles be without one another, and let A, B be their centers. Join AB cutting the circumferences in C, D ; take CE, DF each equal to the radius of the required circle: the two circles described with centers A, B, and radii AE, BF respectively, will intersect one another, and the point of intersection will be the center of the required circle. Distinguish the different cases. 44. Let A be the given point, B, C the centers of the two given circles. Let a line drawn through B, C meet the circumferences of the circles in G, F; E, D, re- spectively. In GD produced, take the point H, so that BH is to CH as the radius of the circle whose center is B to the radius of the circle whose center is C. Join AH, and take KH to DH as GH to AH. Through A, K describe a circle ALK touching the circle whose center is B, in L. Then M may be proved to be a point in the circumference of the circle whose center is C. For by joining HL and pro- ducing it to meet the circumference of the circle whose center is B in N ; and joining BN, BL, and drawing CO parallel to BI, and CM parallel to BN, the line HN is proved to cut the circumference of the circle whose center is B in M, O ; and CO, CM are radii. By joining GL, DM, M may be proved to be a point in the circumference of the j. ALK. And by producing BL, CM to meet in P, P is proved to be the center of ALK, and BPjoining the centers of the two circles, passes through L the point of contact. Hence also is shewn that PMC passes º M, the point where the circles whose centers are P and C touch each OUIleſ, sº NotE. If the given point be in the circumference of one of the circles, the con- struction may be more simply effected thus: Let A be in the circumference of the circle whose center is B. Join BA, and in AB produced, if necessary, take AD equal to the radius of the circle whose center is C; join DC, and at C make the angle DCE equal to the angle CDE, the Fº ºtermined by the intersection of DA produced and CE, is the center of the circle. 45. The line joining the points of contact, when produced, intersects the line joining the centers, when produced in a fixed point. - - 46. Let O be the center of the required circle in the given line AB, and touching the given circles whose centers are C, C*-in P and Q. Join CP, C'Q, these lines produced pass through O. Euc, III. 12. Join also CC', PQ : POQ is an isosceles triangle. If PC be greater than QC'; on PC take PE equal to QC. ;iºn EC'; then EC'O is also an isosceles triangle having the base EC’ parallel to o . . - 47. This problem is a particular case of the preceding prob. 46. 48. Let C be the center of the first and C’ the center of the second circle. Join the centers CC', meeting the first circle in D. Through C draw a diameter ECF at right angles to C'C'; then the circle described passing through the points D, E, F, will touch the first circle in D and bisect the circumference of the second circle. Next, take the point D not in the line joining the centers of the circles. 49. This may be shewn to follow directly from Euc. III. 36, 1.47. ... , 50. Let the given circle be described. Draw a line through the center and intersection of the two lines. Next draw a chord perpendicular to this line, cutting TANGENCIES. HINTs, &c. -- 485s off a segment containing the given angle. The circles described passing through one extremity of the chord and touching one of the straight lines, shall also pass through the other extremity of the chord and touch the other line. - – 51, Join the centers A, B ; at C the point of contact draw a tangent, and at A draw AF cutting the tangent in F, and making with CF an angle equal to one fourth of the given angle. From F draw tangents to the circles. • 52. At any points P, R, in the circumferences of the circles, whose centers are A, B, draw PQ, RS, tangents equal to the given lines, and join AQ, BS. These being made the sides of a triangle of which AB is the base, the vertex of the tri- angle is the point required. 53. Let AB be the given straight line, and C, C' the centers of the given circles. Let O be a point in AB such that the tangents OP, OQ drawn from O to the two circles are equal, Join CC', OC, OC, PC, QC, and draw OM perpendicular to CC’. Then M is a point in CC" such that CM* – C'M* = CP” – C'Q*. 54. Let C, C' be the centers of the given circles, and let P be the point at º, which the tangents PQ, PR ; PQ', PR contain equal angles. Join QC, CR, Q'C', C.R., CC'. Then PQC, PC Q are similar triangles, and CQ, C'Q' are con- stant ; whence the ratio of CP to C'P is constant, also CC' is constant. The point P lies on the circumference of a circle. Determine its center, - 55. Bisect the sides AB, BC, CA, in D, E, F, and join DE, EF, FD; then the circles described about the triangles ADF, BIDE, CEF shall pass through the angles A, B, C of the equilateral triangle and shall touch each other. * 56. If each of the three circles is to be touched by each of three straight lines: the circles required will be found to be the inscribed circle and two of the escribed circles of a triangle. - - 57. Let A, B, C be the centers of the three given circles. On the same side draw the tangent DE to the circles whose centers are A, B ; and FG to the circles whose centers are B, C ; bisect DE in H, and FG in K : and draw HL, KM perpendicular to AB. A.C. (the lines joining the centers of the circles) to meet in O. - O is the point required. For, join DA, HA, FA, K.A., EB, HB, KC, CG, AO, BO, CO; draw the tangents OP, OQ, OR to the circles whose centers are A, B, C respectively, and join AP, BQ, CR. Then the difference of the squares on OP and OQ may be shewn to be equal to the difference of the squares on DH and H.E. But DH is equal to HE, by construction, it follows that OP is equal to OQ. In the same way it may be shewn, that OP is equal to OR. NoTE. The line HO is called the radical axis of the circles, whose centers are A and B; and similarly, KO of the circles whose centers are B and C. - The point O is the radical center of the circles whose centers are A, B, C. 58. If the three points be such as when joined by straight lines a triangle is "formed; the points at which the inscribed circle touches the sides of the triangle, are the points at which the three circles touch one another. Euc. Iv. 4. Different cases arise from the relative position of the three points. - 59. First, let the three circles be equal. Divide the given circle into three equal sectors, and draw tangents to the middle of the arcs, the problem is then reduced to the inscription of a circle in a triangle. Secondly, let two of the circles be equal; and thirdly, let the three circles be unequal. - 60. The general case of this problem is when the given circles do not touch or intersect one another. Let A, B, C be the centers of the given circles. With center B describe a circle with a radius equal to the difference or sum (as the case may require) of the radii of the circles whose centers are A and B: with center C describe another circle with a radius equal to the difference-or sum of the radii of the circles whose centers are A and C. Then the circle described touching these two circles and passing through the point A (Prob. 59, p. 351) will have its center coincident with the center of the required circle. Give the analysis of the problem. • ** 61. Let A, B, C be the three given points, join AB, BC, CA. Then Euc. Tv. 4. suggests the method of finding the points on AB, BC, CA, through which the three circles pass. - - 62. The three tangents will be found to be perpendicular to the sides of the triangle formed by joining the centers of the three circles. º 63. Let C, C, C" be the centers of the three circles; C is the center of the largest, C" of the smallest. Let the tangents to the circles whose centers are C, C’; C, C"; C, C" meet in A, B, C respectively. ... Join the points A, B, C; then AB shall be in the same straight line as BC. Join C, C", "C" and produce CC, CC", C'C'"; these lines meet the tangents in A, B, C respectively. "Through Cº draw CE parallel to AB, then BC may be proved also parallel to CE. *. 486 GEOMETRICAL EXERCISES 64. Let POQ be the common diameter, O being the point of contact of the circles B, C. Let DEOFG be any line drawn through O and meeting the circum- ferences of the circles. Join PE, QF, then DE is equal to FG. 65. Describe a circle through the three given points; from A any one of them, draw any chord, and from the center D draw the perpendicular DE upon it. With the same center and radius DE describe a circle. Then from B, C draw lines BF, CG touching this circle; then AE, BF, CG are equal to one another. The circle, which is the superior limit, is the circle passing through the points A, B, C. 66. Let A, B, C be the three given points. Suppose O to be the center of the circle required, such that the tangents AP, BQ, CR drawn from A, B, C are equal to three given straight lines. Draw OP, OQ, OR, O.A., OB, OC; then AO” – OP” = AP*, BO" – OO" = BQ”, OC* – OR* = CR*: whence AO” – BO* = AP* – BQ" = a constant square; and the locus of O can be drawn. In a similar way BO* – CO* = BQ* – CR* = some constant square; and the locus of O in this case also can be drawn. . The intersection of the two loci will determine the center of the circle. It may be remarked that the three circles described with centers A, B, C with radii AP, IBQ, CR respectively, intersect the required circle at right angles. 67. Let the tangents TP, TC) be drawn from anyFº T in the perpendicular CT"to meet the circles in P, Q respectively. Join AP, BQ, AT, PT ; then by Buc. I. 47, the square on PT may be shewn to be equal to the square on QT. 68. (1) When the tangent is on the same side of the two circles. Join C, C their centers, and on CC' describe a semicircle. . With center C and radius equal to the difference of the radii of the two circles, describe another circle cutting the semicircle in D ; join DC and produce it to meet the circumference of the given circle in B. Through C draw CA parallel to DB and join BA ; this line touches the two circles. - * (2) When the tangent is on the alternate sides. Having joined C, C*; on CC’ describe a semicircle ; with center C, and radius equal to the sum of the radii of the two circles describe another circle cutting the semicircle in D, join CD cutting the circumference in A, through C draw CB parallel to CA and join A.B. 69. Let A, B, be the centers, and C the point of contact of the two circles; D, E the points of contact of the circles with the common tangent DE, and CF a tangent common to the two circles at C, meeting DF in E. Join DC, CE. Then DF, FC, FE may be shewn to be equal, and FC to be at right angles to AB. 70. Let C, C' be the centers, BB' the tangent intersecting the line CC’ joining the centers, and AA’ the other tangent. Draw the radii AC, BC, A'C', §§ to the points of contact; and draw C'G parallel to AA’ meeting AC in G ; and BH parallel to BB' meeting CB produced in H. Then CG = AA, C"H =BB', CG = AC — A'C', and AH = CB + BH = AC + A'C'; and A'A* – B'B' may be proved to be equal to 4. AC. A'C', or the rectangle of the diameters of the circles. 71. Let O be the point required in the line CC' which joins the centers of the two circles, such that tangents OP, OQ drawn from O to touch the two circles are equal. The lines CC, CP, CQ are known, and the position of the point O may be determined from CO” – C'O* = CP” – C'Q*. GEOMETRICAL EXERCISES ON BOOK XI. HINTS, &c. 3. LET AD, BE be two parallelstraightlines, andlet two planes ADFC, BEFC pass through AD, BE, and let CF be their common intersection, fig. Euc. x1. 10. Then CF may be proved parallel to BE and AD. - 4. This theorem is analogous to Euc, x1. 8. Let two parallel lines AC, BD meet a plane in the points A, B. Take AC equal to BD and draw CE, DF, per- pendiculars on the plane, and join AE, BF. Then the angles CAE, DBF, are the inclinations of AC, BID to the plane, Euc. x.E. def. 5, and these angles may be proved to be equal. t - - 5. Let lines be drawn in each plane through the points where the lines cut the planes, then by Euc. I. 29. - ON BOOK XI. HINTS, &c.; 487 6. Let AB, CD be parallel straight lines, and let perpendiculars be drawn from the extremities of AB, CD on any plane, and meet it in the points A, B, C, D'. Draw A'B', CD', these are the projections of AB, CD on the plane, and may he proved to be parallel. - - - 7. Draw the figure, the proof offers no difficulty. 8. Let AB, AC drawn from the point A, and A'B', A'C' drawn from the point A', in two parallel planes, make equal angles with a plane EF passing through AA', and perpendicular to the planes BAC, B'A'C'. Let AB in the plane ABC be rºle to A'B' in the plane A'B'C': then AC may be proved to be parallel to A'C'. 9. The plane must be drawn through the given line so that the plane and the other given line may be equally inclined to a third plane. 10. The required plane must be drawn through the given point so as to have the same inclination to a third plane, as the plane which passes through the two given lines. ** 11. From the point A let AB be drawn perpendicular to a plane, and AC perpendicular to a given line CD in a plane: join BC, then BC is at right angles to CD. For AB, BC, CD may be considered as three consecutive edges of a rectan- gular parallelopiped, and AC the diagonal of one face. 12. In the triangle BCD in which BE is drawn from the vertex to a point E in the base CD ; it may be proved that the difference of the squares on the sides RC, BD is equal to the difference of the squares on the segments CE, ED of the base. By the converse of Theo, 168, p. 308. 13. The lines AA’, FF" produced in the planes, meet in the line of their inter- section. See Theorem 11. p. 403. 14. Construct the figure by joining the points, and it will be found to be a rectangular parallelopiped of which A, B, C, D are four of the corners. 15. Let BC be the common intersection of the two planes ABCD, EFG H which are inclined to each other at any angle. From K at any point in the plane ABCD, let KL be drawn perpendicular to the plane EFGH, and KM perpendi- cular to BC, the line of intersection of the two planes. Join LM, and prove that the plane which passes through KL, KM is perpendicular to the line BC. 16. Let a line be drawn in the plane of the triangle through the point A. parallel to BC the base. * 17. About the given line let a plane be made to revolve, till it passes through the given point. The perpendicular drawn in this plane from the given point upon the given line is the distance required. - 18. Through any point in the first line draw a line parallel to the second; the plane through these is parallel to the second line. Through the second line draw a plane perpendicular to the fore-named plane cutting the first line in a point. Through this point draw a perpendicular in the second plane to the first plane, and it will be perpendicular to both lines. • 19. Through any point draw perpendiculars to both planes; the plane passing through these two lines will fulfil the conditions required. ... • 20. From the points where the lines meet the planes, draw two lines perpen- dicular to the intersection of the planes. 21. This is only a more general form of the preceding theorem. 22. Let AB, AC in one of the planes make equal angles with DE the line of the intersection of the planes. Let AB be equal to AC. Draw BF, CG perpen- diculars on the other plane, and draw FA, GA in that plane, and prove the angle BAF equal to the angle CAG. * * 23. This differs from the preceding only in the two planes being at right angles to each other. 24. If the intersecting plane be perpendicular to the three straight lines; by joining the points of their intersection with the plane, the figure formed will be an equilateral triangle. If the plane be not perpendicular, the triangle will be isosceles. 25. Let the three parallel straight lines AD, BE, CF be cut by the parallel planes ABC, DEF, and A, B, C, the points of intersection of the lines, be joined, as also D, E, F ; then the figure ABC may be proved to be equal and similar to the figure DEF. *. 26. This is shewn by means of the similar triangles. 27. This may be readily proved by Euc. XI. 17. 28. Apply Euc. XI. 6. 29. This is an indeterminate Problem. If however, the circle be in that plane which passes through the given point, and be perpendicular to the two given planes, the problem is reduced to that of describing a circle which shall pass through a given point, and touch two given straight lines, 488 • GEOMETRICAL ExERCISEs 30. Let the three lines meet in the point A, and let a plane intersect them in the points B, C, D, so that AB, AC, AD are equal to one another. Describe a circle about the triangle BCD, and let O be the center; the line AO is per- pendicular to the plane BCD. ... I - 31. See Theorem 16, p. 403. 32. , Let the straight lines intersect in A, and let a plane be drawn cutting the three given lines in the points B, C, D, and the fourth in E. 33. This theorem is analogous to Theorem 5, p. 297. 34. This will appear from Euc. 1, 19. 35. Apply Euc. I. 19. ^* 36. Let S be the proposed solid angle, in which the three plane angles ASB, ASC, BSC are known, it is required to find the angle contained by two of these planes; such as ASB, ASC. On a plane make the angles B'SA, ASC, B"SC equal to the angles BSA, ASC, BSC in the solid figure; take BS and B"S each equal to BS in the solid figure ; from the points B" and B" at right angles to SA and SC draw B'A and B"C., which will intersect each other at the point O. From O as a center with radius AB" describe the semicircle Bºb E.; at the point O, erect Ob perpendicular to B'E and meeting the circumference in 5; join A5: the angle EAb will be the required inclination of the two planes ASC, ASB in the solid angle. Legendre's Geometry, translated by Sir David Brewster, pp. 125, &c. 37 Let ASC, ASB (same figure as in 36) be the two given plane angles; and suppose for a moment that CSB" is the third angle required; then employing the same construction as in the foregoing problem, the angle included between the planes of the two first, the inclination of these planes would be EAb. Now as BAb can be determined by means of CSB", the other two being given, so likewise may, CSB" be determined by means of EAb, which is just what the problem requires. . Having taken SB' at pleasure, upon SA let fall the indefinite perpendicular B'E; make the angle EAb equal to the inclination of the two given planes; from the point b, where the side Ab meets the circle described from the center A, with the radius AB", draw bo perpendicular to AE; from the point O, at right angles to SC draw the indefinite line OCB"; make SB" equal to SB’; the angle CSB." will be the third plane angle required. Legendre's Geometry, translated by Sir David Brewster, pp. 127, &c. 38. This theorem is analogous to Euc. II. 4. - 39. Construct the figure, and it will be found that the angle between the diagonal and one side of the cube measures the inclination of the two planes. 40. The section will be in all cases a parallelogram, but not necessarily rect- angular. Any plane drawn perpendicular to a face through a line equal to the edge of the cube, will fulfil the condition. - *- 41. The diagonal plane of a cube is at right angles to two of the faces of the cube, and makes angles each equal to half a right angle with the other four faces. 42. Let a rectangular parallelogram ABCD be formed by four squares, each equal to a face of the given cube, and let EF, GH, KL, be the lines of division of the four squares. Let BD the diagonal of ABCD, cut EF in M ; the square on BM to the square on AB is as 17 to 16. Let BG the diagonal of ABHG cut EF in N ; the square on BN is to the square on AB, as 20 is to 16; heaee there is some square between that on BM and BN which bears to the square on AB, the ratio of 18 to 16, or of 9 to 8. , The following addition may be easily proved. If six edges of a cube taken in order round the figure, be bisected, and the points of bisection be joined in suc- cession, these six lines will form a regular hexagon. * 43. From the six points out of the perpendicular, draw perpendiculars to the plane, and join the points where the perpendiculars meet the plane. 44. Let O be one of the solid angles of a cube whose three adjacent edges are OA, OB, OG ; OBEC being the base of the cube. On OA, OB, OC, let any three points A, B, C be taken, and join A'B', B'C', C.A'. Then the square of the area of the base A'B'C' of the solid OA'B'C' is equal to the sum of the squares of the areas of the faces, OA'B', OA'C', OC'B'. Join OE intersecting B'C' in E', and join A'B'. Then A'E' may be shewn to be perpendicular to the base C'B' of the triangle A'B'C', and by Euc. I.47, and note page 82, the truth of the property is shewn. 45. This is to shew that the square on the diagonal of a rectangular parallelo- piped is equal to the sum of the squares on its three edges. * 46. This may be effected in several ways, the most ‘simple is by drawing a plane through a diagonal of the cube. . - - * ON BOOK XI. HINTS, &c. 489 47. This theorem is analogous to the corresponding theorem respecting a rect- angular parallelogram. • - \ The axis of a parallelopiped must not be confounded with its diagonal. 48. See the figure Euc. XI, 10. If the lines be unequal, the figure may be shewn not to agree with the definition of a prism. Euc. x.I. def. 13. 49. Through each line draw a plane parallel to the other ; these planes will be parallel, and obviously form two of the faces of the parallelopiped. Through each line and one extremity of the other, draw a plane; and a second plane E. to it through the remaining extremity. This will complete the figure; ut there will be four varieties of cases according as the extremities are situated. 50. Let the depth be taken as the fixed unit, and let the breadth be double the depth, and the length double the breadth. Let the parallelopiped and the cube be constructed. Then the cube may be shewn to consist of the same number of cubie units as the parallelopiped. See note on def. A. p. 272. Similarly, if the breadth be treble, or any other multiple of the depth, and the length treble, or the same multiple of the breadth, the equality of the two figures may be shewn to exist, - - 51. Let ABCDEF be a truncated prism whose base ABC is a triangle, right- angled at B, and which is cut off by a plane which intersects the edges in D, E, F. Through D the extremity of the shortest edge, draw a plane parallel to the base ABC cutting the edges BF, CE in b, c respectively. Then the sum of the prism ABCabſ), and the pyramid DcEFb may be shewn to be equal to the volume of the prism stated in the question. - -- 52. Let the figure be described in a similar manner to that of Theorem 2, page 402: by employing Euc. II. 12, 13, instead of Euc. I.47, the truth of the theorem may be proved. - - - 53. See Theorem 50, p. 406. 54. Every possible combination of the lines taken three at a time will form the pyramids, since the respective faces may be so formed. These are fifteen in number, - - - 55. Describe a circle passing through the three given points, and from the eenter draw a line perpendicular to its plane. Then every point in this perpendi- cular fulfils the conditions required. - - 56. Let BCD be the base of the pyramid. Take CT)' equal to CD in the same line, and join AC", BC, AD", B.D.'. Then the triangular base BCT) is equal to BCD, Euc. I, 38. And since A is a fixed point, the altitude of the pyramids ABCD, ABC'D' is the same, and pyramids of the same altitude on equal bases are equal. & Ql 57. Bisect the base by a line drawn in the given direction, whether parallel to a given line, or tending to a given point. The plane drawn through the bisecting line and the vertex of the pyramid, gives the solution of the problem. 58. This Theorem is analogous to Theorem 151, p. 331. . 59. Every oblique pyramid is equal to a right pyramid of the same base and altitude. Every right pyramid whose base is not triangular may be divided into triangular pyramids of the same altitude. Every pyramid on a triangular base is equal to one-third of a prism of the same base and altitude. Hence, any pyramid may be proved to be one-third of a prism of the same base and altitude. 60. See the preceding problem, and note on Def. A., p. 272. 61. Draw the necessary lines and by Euc. 1.47. 62. Let ACB, ADB, be the triangles, CD being perpendicular to the plane of ACB. Then the angles DAC, DBC, and the line DC being given, the lines DA, DB, can be constructed. Draw ab through C perpendicular to CD, and make the angles CT)a, C'Db, the complements of the given angles; then Ca, Cb will be of the same length as CA, CB. Whence the angle which they are to form being given, the triangle may be constructed, and its base is the line required. 63. The triangle FCD is equal in all respects to ACD, and CH is obviously. equal to CD or CB ; and it will be sufficient to shew that the angle BCH is a right angle, and that HG and GB are each equal to BC. 64. From the vertex A draw a line to any point B in the base of the pyramid, and meeting the given section in B'. From the angular points of the base draw ines to the point Bº, also from the angular points of the given section to the point B'. Then any triangle in the section, may be shewn to be similar to the corres- ponding triangle in the base. Euc. vi. 20. 65. Let AB be at right angles to the plane BCED, and let the perpendiculars. .. from AB intersect the plane GHKL in the line MN, and let HNK be the common intersection of the planes CBDE, GHKL. Join AM, BN, and prove MN to be a straight line perpendicular to HK. * 490 GEOMETRICAL EXERCISES *-*. 66. The locus may be shewn to be a straight line inclined at half a right angle to each of the planes. • 67. Let the line EF be the intersection of the two planes AB, CD. Take the point P between the planes, such that the perpendiculars PQ, PR on the two planes respectively may be equal to the given straight line; then the locus of P, one of the points may be shewn to be one of the lines, which will be parallel to EF. tº zººlº 68. Let AB, A'B' be any portions of the two straight lines. At B' draw B'C parallel to AB, and B'C perpendicular to the plane passing through A'B'C'. Let the plane passing through A'B'C intersect the line AB in the point A. In the plane A'B'C from A draw AA’ perpendicular to A'B', and AC perpendicular to AA’. Then the plane CAB passing through the line AB may be shewn to be parallel to the plane A'B'C' passing through the line A'B', and that no other parallel planes can be drawn through AB, A'B'. Also AA’ is the perpendicular distance between the two planes, and that AA’ is less than any other line which can be drawn be- tween the two planes. 69. Let AE meet the straight lines BE, DE in the plane BED, fig. Euc. x1.6, and let the angle AEB measure the inclination of AE to the plane BDE; then the angle AEB is less than the angle AED. Draw AB perpendicular to the plane, make ED equal to EB and join BD, AD. Euc. I. 18, 19. s 70. Let AB, CD be two straight lines not in the same plane, and let EF be the portion of the line perpendicular to both lines which is intercepted between them. Through the lines AB, EF draw a plane which will meet the line CD in F, and through the line CFD draw a plane parallel to the line AEB, and intersect- ing the plane through AB, EF in the line GFH ; also draw a plane through the lines, CD, FE; then EF may be shewn to be less than any other line EG drawn in the plane through CD, FE to meet CD. 71. Let A, B, be the given points, and GH the given straight line ; draw AC, BD perpendicular on GH, and in the plane AGH produced, draw DB' perpendi- cular to GH, and equal to DB ; join AB", meeting GH in E, and draw EB. Then AE + EB is the minimum. For the triangles EDB, EB'D are equal, being right- angled at D, and having one side common, and the others equal. Whence the angle BEH is equal to GEA, each being equal to B'EH. The conclusion follows from the demonstration of Theorem 73, p. 407. - 72. Let HM be the common section of the two planes MN, MQ; and let AB be drawn from a point A in HM perpendicular to the plane MN: then, if planes be drawn through AB to cut the planes MN, MQ in lines which make the angles CAD, EAF with each other, and that the plane BACD is perpendicular both to MN and MQ; the angle CAD will be greater than EAF. Shew that the angle BAD is less than the angle BAF, and it follows that CAD is greater than EAF. 73. Let GH be the edge of the wall, A, B the two points, and let the line joining A, B, meet the edge of the wall GH in E. If the points AE, BE make equal angles with GH, then AE, EB may be proved to be less than any other two lines drawn from A, B, to meet GH in any other point E. : . 74. This is analogous to the theorem :--The square has its perimeter less than that of any right-angled parallelogram of the same area. GEOMETRICAL EXERCISES ON BOOK XII. HINTS, &c. - 6. FIND a square which is eight times the given square. Euc. xII. 2. 7. Apply Euc, XII. 2. - 8. First, to bisect a circle by a concentric circle. Let C be its center, AC any radius. On AC describe a semicircle, bisect AC in B, draw BD perpendicular to AC, and meeting the semicircle in D; join CD, and with center C, and radius CD, describe a circle; its circumference shall bisect the given circle. Join AD. Then by Euc. v.I. 20, Cor. 2, the square on AC is to the square on CD as AC is to CB; and Euc XII. 2. In the same way, if the radius AC be trisected, and perpendicu- lars be drawn from the points of trisection to meet the semicircle in D, E, the two circles described from C with radii CD, CE shall trisect the circle. And generally, a circle may be divided into any number of equal parts. NoTE. ... By a similar process a circle may be divided into any number of parts which shall have to each other any given ratios. * ON BOOK XII. HINTs, &c. 491 9. The radius of a circle whose area is three times the area of a given circle, is equal to the side of a square whose area is three times that of the square on the radius of the given circle. - -- - * 10. To divide the circle into two equal parts. Let any diameter ACB be drawn, and two semicircles be described, one on each side of the two radii AC, CB: these semicircles divide the circle into two equal parts which have their perimeters equal. In a similar way a circle may be divided into three parts, by dividing the diameter into three equal parts, AB, BC, CD, and describing semicircles upon AB, AC on one side of the diameter, and then semicircles upon DC, DB on the other side of the diameter. 11. Let the diameter AB be divided into five equal parts, in C, D, E: then C, D are the second and third points of division. The semicircles AEC, AFD are described on one side of the diameter, and BGC, BHD on the other. Then since the semicircumferences of circles are proportional to their diameters, the perimeter of the figure AECGBHDF is shewn to be equal to the perimeter of the original circle. By Euc. xII. 2, the area of the figure AECGBHDF may be shewn to be one-fifth part of the area of the circle. The general case, when the diameter is divided into n equal parts is proved in the same way. - 12. By Eue. xII. 2, the area of the quadrant ADBEA is equal to the area of the semicircle ABCA. 13. By Euc. XII. 2. The squares on the radii of the two circles may be shewn to be in the ratio of 3 to 1. 14. By reference to Theorem 2, p. 408 and Euc. xII. 2, the parts of the dia- meter may be proved to bear to each other the ratio of 1 to 2. 15. Apply Euc. xII. 2. 16. If the circles whose centers are B and C touch each other in S, the prob- lem may mean:—to find the point R, so that the figure between the three circles (see fig. Theo. 2, p. 408) may be bisected by the line RS; or it may mean, if two chords be drawn from P, Q, to R, the portions of the lunes bounded by parts of these chords and portions of the circles may be equal. 17. This will be found by Theorem 1, p. 408. ' 18. (a) Produce CD to meet the arc of the quadrant in E. Then the sector ACE is half of the quadrant: also the semicircle CDA may be shewn to be equal to half the quadrant. (b) The segments on CD and DA are similar and equal, if the figure bounded by DA, AC, and the arc CD be added to each, the remaining part of the semicircle on AC is equal to the triangle ACD which is a right-angled isosceles triangle. - - 19. The area of the circle of which the quadrant is given, is to the area of the circle which touches the three circles, as 36 is to 1. And the quadrant is one-fourth of the area of the circle. Hence the quadrant is to the circle as 9 to 1. I 20. The circles on BA, AC are as the squares on BA, AC ; Euc, xI . 2. and the square on BA is equal to the rectangle BC, BD, also the square on ACis equal to the rectangle CB, CD; whence it follows that the circles are as BD, CD. 21. Let ABC be the right-angled triangle, BC being the hypotenuse, and let semicircles be described on AB, AC as diameters. Bisect AB, AC, BC, in E, F, G; from G draw perpendiculars on AB, AC, meeting the semicircles in H, K, and shew that GH is equal to GK. By Euc. XII. 2. the difference is found. 22. Let AB, A'B' be arcs of concentric circles whose center is C and radii CA, CA', and such that the sector ACB is equal to the sector A'CB'. Assuming that the area of a sector is equal to half the rectangle contained by the radius and the included arc: the arc AB is to the arc A'B' as the radius A/C is to the radius AC. Let the radii AC, BC be cut by the interior circle in A', D. Then the arc A'D is to the arc AB, as A'C is to AC ; because the sectors A/CD, ACB are similar : and the arc AB' is to the arc AD, as the angle ACB' is to the angle ACD, or the angle ACB, Euc. vi. 33. . From these proportions may be deduced the propor- tion :-as the angle ACB is to the angle A'CB', so is the square on the radius A/C to the square on the radius AC. And by Euc. XII. 2, the property is manifest. 23. Let AB, A'B' be arcs of two concentric circles, whose center is C. ACB, A!CB' two sectors such that the angle ACB is to the angle A'CB', as A'C' is to AC”. If AC, BC be cut by the interior circle in A', D ; then the arc A'B' is to the arc Aſ D, as the angle A/CB' is to the angle A"CD, or ACB. Euc. VI. 33. And the arc A'D is to the arc AB, as the radius A'C is to the radius AC, by similar sectors. By means of these two proportions and the given proportion, the rectangle contained by the arc AB and the radius, AC, may be proved equal to the rectangle contained by the arc A'B' and the radius A.C. 24. Let the arc of a semicircle on the diameter AB be trisected in the points ^*. 492 - GEOMETRICAL, EXERCISES D, E ; C being the center; join AD, AE, CD, CE; then the difference of the seg- ments on AD and AE, may be proved to be equal to the sector ACD or DCE. 25. Assuming that the area of a sector of a circle is equal to half the rectangle contained by the radius and the arc, the sector AOC is shewn to be equal to AOB. 26. Let POQ be any quadrant, O being the center of the circle, and let BG, DH be drawn perpendicular to the radius PO, and OB, OD be joined. The tri- angle GBO is equal to DHO. -- 27. The radii of the circles may be proved to be proportional to the two sides of the original triangle. Then by Euc. XII. 2; VI. 19. - 28. The triangles CEA, CEB are equal, and the difference of the two segments is equal to the difference of the parts of the semicircle made by CE. The difference of the same parts may also be shewn to be equal to double the sector DEC. 29. Let AB be the hypotenuse of the right-angled triangle ABC, and let the semicircles described upon the sides AC, BC, intersect the hypotenuse in D. Join AD. AD is perpendicular to AB. The segments on AC, AD, and on one side of CD are similar ; and the segments on AC may be shewn to be equal to the segments on AD, CD. Also the segment on BC may be shewn to be equal to the segments on BD, and the other side of CD. If Euc. vi. 31 be true for all similar figures, the conclusions above stated follow at once. - 30. The area of the triangle ABC is equal to the quadrant ABD. From these equals take the figure AEDB. 31. The segments on BC, BA, AC may be shewn to be similar. And similar segments of circles may be proved to be proportional to the squares on their radii, Euc. XII. 2, and to the squares on the chords on which they stand, Euc. v.I. 6. If Euc. vi. 31 be extended to any similar figures, the equality follows directly. 32. This is shewn from Euc. xII. 2; 1. 47; v. 18. . 33. The sum of the squares on the segments of the diagonals, is equal to the sum of the squares on each pair of opposite sides of the quadrilateral figure. Hence by Euc. xII. 2; 1.47; v. 18, the property is proved. 34. The squares on the four segments, are together equal to the square on the diameter. Theorem 6, p. 320. Then by Euc. XII. 2. - 35. This is shewn by Euc. I.47; XII. 2; v. 18. 36. Apply Theorem 1, p. 408. - - 37. Let the two lines intersect each other in A, and let C be the center of the last circle. Join CA, and draw CB perpendicular to AB one of the lines. Then CA and CB are given. Let C', C.", &c., be the centers of the circles which successively touch one another. Draw C'B', C"B", &c., perpendicular to AB and CD, C"D' parallel to AB meeting CB in D, C'B' in D', &c. Then by means of the similar triangles, the radii C'B', C"B", &c., may be expressed in terms of CB and CA. Hence the sum of the series of the circles may be expressed in terms of the area of the last circle, 38. The proof of this property depends on the demonstration of Theorem 2, p. 408, and the relation between the area of two circles described upon two lines as diameters, one of which is double of the other. 39. Let the figure be described, and the demonstration will be obvious from the consideration of the parts. * 40. The triangles ABD, ABC have the angle ABD common, and ACB may be proved equal to BAID, by Euc. 1. 29; III. 32. And therefore the angle CAB is equal to the angle ADB. Also the triangles CAB, CEA may be shewn to be equi- angular, and AD equal to AE. Then by Euc. v.I. 4. 41. The figure formed by lines drawn from a point above the plane of a circle to every point in its circumference is a cone. If the point be in the perpendicular to the plane drawn from the center of the circle, the cone is a right cone, and all lines from the point to the circumference are equal; if the point be not in the per- pendicular, the cone is an oblique cone and has only the two lines equal. 42. Let A be the vertex, BC the diameter of the base of the cone, AB, AC lines in the surface joining the extremities of BC; and let any other section of which be is the diameter be made parallel to the base, then the triangles Abe, ABC are similar triangles. Euc. v.I. 19 ; XII. 2. 43. Let A be any point above the plane of the circle whose center is C and diameter BCD. Join CA, and let a plane pass through any point c in AC or AC produced. Through c in this plane draw bed parallel to BCD. Join BA, DA and produce them to meet bed in d’and b. Then b, d may be proved to be two points in the circumference of the circle whose diameter is bcd, and by means of the similar triangles Acd, ACD, the areas of the two circles may be shewn to be proportional to the squares on AC and Ac. ON BOOK XII. HINTs, &c. 493 44. . Let Ce be the line joining the centers of the two circles whose planes are parallel, and let ACB, act be parallel diameters drawn in each. Join A.C., Bb, then ABba is a quadrilateral figure having two of its sides AB, ab parallel. It is then required to shew that the lines joining Ab, Ba intersect at the same point D in the line Co.. If ab be equal to AB, the figure ABba is a rectangular parallelogram. 45. This Theorem is analogous to Euc. 111. 14. 46. The arc of a circle being considered as the measure of an angle which the arc subtends ; the angle between the planes of two great circles, can be shewn to be equal to the angle between the two radii of that great circle which bisects the two planes at right angles. Aº . 47. This theorem is analogous to Euc. III. 14. - 48. This may be proved indirectly as in Euc. III. 18. 49, Let D be the given point, and from D let DA be drawn through the center E, and meeting the surface in C, A. Let DB be a line from D touching the sphere at B. Join BE. Then the triangle DBE (fig. Euc. III, 36) is in a plane passing through D, and E the center of the sphere, and the distances DE, EB are always the same. Hence it follows that BD is always of the same length. Euc. 1. 47. The sphere which touches the six edges of any tetrahedron, has four circular sectionstouching the sides of the four triangles which form the surface of it. 50. Let the circle ADB cut the circle AEB in the diameter AB at any angle, C being their common center. Next let the plane perpendicular to AB cut the cir- cumference of the circle ADB in D, F, and the circumference of AEB in E, G. Then E, D, G, F may be proved to be in the circumference of a circle. 51. The diameter of the sphere is equal to the diagonal of the cube. - 52. Let AB, CD, EF be three lines meeting the surface and intersecting each otherat right angles in the point G within a sphere whose center is O. Join OG and produce it to meet the surface of the sphere in H, K ; then HK is a diameter. Erom O draw OL, OM, ON perpendicular on AB, CD, EF respectively, then these three lines are bisected in L, M, N. Next draw OP perpendicular to the plane of AB, EF, and join PL; PL is perpendicular to the line AB; also in the same plane join PN ; PN is also perpendicular to EF. Join also OA, OC, OF. Then Euc. II. 9, the squares on AG, BG, are equal to double the squares on AL, LG. Simi- larly for the lines CD and EF; and by Euc. 1. 48, II, 5. Cor. it may be proved that the squares on AG, GB, CG, GD, EG, GF are together equal to the square on HK and twice the rectangle HG, GK. - - 53. Take a point A on the spherical surface of the fragment as a center, and with any radius AB describe a circle upon it. Take two other points C, D in the circumference of this circle, and describe a plane triangle A'B'C' having its sides equal to the distances AB, BC, CA, respectively. Describe a circle about the triangle A'B'C', and-draw the diameter A'D'; with centers A', D' and the radius ºqual to AB, describe circles intersecting each other in E' and through the points A’, D', E' describe a circle ; the diameter of this circle will be equal to that of the sphere of which the fragment is given. * 54. By Euc. 1. 47, expressions for the squares on the sides of the triangle may be found, from which it will appear that the three sides of the triangle are mean proportionals between every two of the three diameters of the spheres. 55. If the centers of the upper sphere, and the three upon which it rests, be joined, the figure is a tetrahedron : aud the same remark may be made with re- spect to each of the three, and the spheres upon which they severally rest. 56. All the sections may be proved to be equilateral triangles. 57. From the vertex A draw the line AE perpendicular on BCD the base of the tetrahedron, and from E draw the line EF perpendicular on the plane ABC; the angle between the perpendiculars is equal to the inclination of two planes of the tetrahedron. It will be found that in the triangle AEF, the side AE is three times E.F. The inclination may also be found as in Prob. 37, p. 405. 58. The two lines drawn from two angles to bisect the opposite sides of the base of the tetrahedron, are at right angles to the sides of the triangular base. 59. Draw BO and produce it to meet DC in E. Then Euc. I. 47. 60. First, let ABCD be a tetrahedron ; bisect the opposite edges, AB in E, and CD in F; join EF, and prove EF perpendicular to AB, CD. Then conversely. 61. If FE be the shortest distance of the opposite sides AB, CD ; join CE, DE, and shew that the square on EF is one-fourth of the square on CD. 62. First prove the direct proposition, then the converse of it. 63. Let ABCD be a tetrahedron and let the line EF joining the bisections E, F of the two opposite sides AB, CD, be bisected in G ; the line AO drawn from the vertex A to the plane of the base BCD passes through G. Draw the necessary. lines, Euc. vi. 4. 494 GEOMETRICAL ExERCISEs. 64. . The joining lines in the theorem, are the lines joining the centers of the circles inscribed in the four faces of the given tetrahedron. - 65. . From the vertex A of a tetrahedron draw AO to the point O, the center of the circle which circumscribes the face BCD, and prove AO perpendicular to the plane BCD ; then conversely. - 66. Let ABCD be a regular tetrahedron. From A in the plane ABC draw AE perpendicular to BC, and join DE in the plane BCD, also from A draw AG perpendicular to the line D.E. Then the angle AEG is the inclination of the two faces ABC, DBC of the tetrahedron, and the base EG is one-third of the hypote- nuse AE in the right-angled triangle AGE. - Let abodef be a regular octahedron whose faces are equal to those of the tetra- hedron. Join a, f, two opposite vertices. Draw ag in the plane abc perpendicular to be, and ge perpendicular to af. Draw fg in the plane föc, and from f draw fº perpendicular to ag produced. ^. Then agfis the inclination of two faces of the octahedron. Also in the right- angled triangle fºg, gh may be proved to be one-third of fg, and fg is equal to AE. Hence the triangles fgh, AEF are equal in all respects. Therefore the angle fgh is equal to the angle AEB. Hence the angle AEF is the supplement of the angle agf, or the inclination of two contiguous faces of a tetrahedron, is the supplement of the inclination of two contiguous faces of an octahedron. j 67. For the inclination of any two contiguous faces of the octahedron, see Theorem 66, page 416. The octahedron is divisible into two pyramids whose bases are squares, and the four slant sides equilateral triangles. 68. The sides of the equilateral triangles which bound the octahedron so fººd, will be equal, each to one-half of the diagonal of a face of the given *Cll De. - 69. Let A be the vertex, and BCD the triangle forming the base of the tetrahe- dron. Bisect each of the dihedral angles at the base by three planes which mutually intersect each other in the point F, which will be the center. Then a perpendicular from F, drawn upon any one of the faces will be the radius of the inscribed sphere. 70. If planes be drawn through the bisections of three of the edges of the tetra- hedron at right angles to the edges, the point of their mutual intersection, is the center of the sphere which circumscribes the tetrahedron. 71. It may be shewn that the diameter of the sphere which circumscribes a regular octahedron will be to an edge as the diagonal is to the side of a square. 72. Let AB, CD, EF be three diameters of a sphere each at right angles to the other two, and intersecting each other in O the center of the sphere, the extremi- ties of the lines meeting the surface of the sphere. Join AC, CB, BD, DA, then these four edges of the figure may be proved equal to - one another by the right- angled triangles. In the same way the other edges may be proved equal. Having proved all the edges equal, the faces of the figure are equilateral triangles. Lastly prove the inclinations of every two faces to be equal. 73. Let three planes at right angles intersect each other in the fixed point O, and let OA, OB, OC be the intersections of the planes. Let P be any point in the locus such that OP is equal to the line, the square on which is constant, and equal to the sum of the squares on PM, PN, PR the perpendiculars from P on the three planes AOB, AOC, BOC respectively. Complete the rectangular parallelopiped of which PO is the diagonal and PM, PN, PR the edges. The sum of the squares on the three edges of a parallelopiped is always equal to the square on the diagonal. But the edges PM, PN, PR are variable in magnitude according to the position of the point P, while the diagonal is constant in magnitude, and moveable in any direction whatever round the fixed point O. The locus of P is therefore the surface of a sphere whose radius is PO and center O. 74. Let ABCD be a triangular pyramid having the three faces at the vertex A at right angles to each other. At the bisections E, F, G of AB, AC, AD, draw EO, FO perpendicular to AB, AC, and meeting in O; and EO, GO’ perpendi- cular to AB, AID and meeting in O'; then O, O' are the centers of the circles which circumscribe the triangles ABC, ABD. In the plane of OEO' draw OP, O'P perpendicular to OE, O'E meeting in P; join PA, PB, PC, PD, and prove these lines equal to one another. IND EX TO THE PROBLEMS AND THEOR EMS IN THE GEOMETRICAL EXERCISES. Senate House Examination for Degress. S. H. Smith’s Mathematical Prizes. Bell’s University Scholarships. St. Peter's College. Pet. . Clare College. Cla. Pembroke College. Pem. Gonville and Caius College. Trinity Hall. T. H. Corpus Christi College. C. C. King’s College. Ki. Queens' College. Qu. 1 Emm. .22 ,35. ,46. Sid. , 30. Trin. ,37. 2 Trin. , 40, Cai, ,57. , Joh. ,51. Chr. ,58. 3 Trin. , 32. , 37. , 50. T. H. ,52. Joh. ,54. S. H. ,54. 4 Sid. , 30. , 43. Jes. , 50. ,58. Qu.,34. Trin. ,40. Cla. ,47. Emm. , 56. 5 Cai. , 50. 6 S. H. , 17. Trin. , 24. ,37. Qu. ,25. Emm. ,27. ,48. Cath. , 9.,48. Pem.,39. ,47. Sid, ,40. Chr ,45. Cla. , 56. 7 S. H. , 19. Trin. ,29. Qu. , 35. Pem. ,44. Jes. , 49. B. S.,55. 8 Qu. , 26. ,28. S. H. ,49. , 50. Pet. , 56. Lmm. , 50. C. C.,57. Cai. , 55. 9 Trin. , . B. S. ,58. 10 Emm. ,34. ,58. Mag. ,38. Joh. ,58. 11 Cai. ,40. Joh. , 50. T. H.,59. 12 S. H.,40. ,54. 13 Cath. , 31. S. H. ,50. 14 15 Pet. ,57. S. P. B. S. Cai. ABBREWIATIONS. Jesus College. Christ's College. St. John’s College. Magdalene College. Trinity College. Trin. Emmanuel College. Sidney Sussex College. Downing College. St. Catharine’s College. Cath. Jes. Chr. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 + 35 36 37 38 Cath. , 22., 33. Trin. ,37. Cai. ,57. Chr, , 56. Emm. , 56. T. H. ,51. Jes. ,58. S. H. , 50. Qu. , 19. T. H. ,51. Emm. ,51. ,60. Pem. ,57. - Jes. ,58. S. H.,14. Cla.,55.,60. Cai, ;41. Chr.,26. ,41. ,52. Jes. ,52. Joh. , 31. Pet. ,38. Trin. ,39. , 50. Mag. , 51. & S. H.,58. ,60. C. C.,53. S. H. ,59. Pet. ,55. C. C.,53, Qu.,54. Chr. , 56. Joh. ,46. Trin. ,53. Joh. , 56. Cath. ,59. Trin. , 31. S. H.,36. ,48. Mag. ,47. Chr. ,54. ,59. Emm. ,25. Joh. , 19. Qu. ,25. 39 | 40 41 42 43 Joh. Mag. Emm. Sid. Down. - In the years the centuries are omitted, and the places are supplied by a comma prefixed, thus ,45 means l845, - EXERCISES ON BOOK I., p. 294, &c. Chr.,28. Pem. , 42. Jes. , 51. Trin. , 26. Sid. ,43 C. C.,57 Pem. , 29. B. S.,48. Qu. ,52 Qu. , 50. . - Qu. , 31. Cath. ,35. Emm. ,35. ,60. Sid. ,38. B. S.,40. Trin. ,27. ,34. S. H. ,55. Joh. ,59. Trin. ,60, T. H. ,59. Trin. ,54. Emm.,54. Pem. ,59. T. H.,58. Trin. ,58. Cai. ,55. Pet. ,58. Chr. ,55. Cai. ,59. Cai. , 49. Jes. , 54. S. H. ,53. Chr. ,59. Trin. , 39. ,51. ,51. Trin. , 43. Joh. , 26. Pem. ,47. Chr. ,52. ,53. Cai. ,46. Qu. ,48, Jes. , 56. Pem, 496 INDEX, 60 Cai. , 31. Joh. , 30. 61 62 Jes. ,52. 63 Jes. , 55. 64 Pet, ,51. 65 Chr. , 39. 66 Pet. ,36. 67 Trin. ,52. ,54. T. H. ,52. Cla.,58. Chr.,58. 68 Pet. ,51. 69 Trin. ,51. 70 Jes. ,54. 7l Qu. ,55. 72. S. H.,48. 74 T. H.,54. 75 Trin. ,40. "77 Cai. , 33. 78 Trin. , 49. 79 Qu. , 31. Sid. ,36. 80 Qu. , 19. 81 Qu. , 24. 82 Cla. ,51. 83 Joh. , 56. Cai. , 56. Cai. ,60. Qu. , 33. Chr. , 56. Pet. ,53. 86 Pet. ,58. Trin. ,51. 87 Mag.,59. 88 Pet. ,53. 89 Trin. ,48. 90 Qu. ,32. Jes. ,36. S. H. 949. , 50. 91 S. H.,49. Mag.,52. 92 Qu. ,37. 93 Trin. ,48. 94 Chr. ,58. 95 Chr. ,52. 96 Cath. , 31. 97 98 Trin. ,59. 99 Cai. ,60. 100 Jes ,59. 101 S. H.,61. 102 Cath. , 49. 103 Cla. ,57. 104 Pet. ,46. 105 C. C. , 50. Cai. ,53. 106 Mag.,51. ,58. 107 Jes. ,54. T. H. ,51. Trin. ,46. ,57. 1I 1 * 108 Cath. , 49. 109 Jes. ,53. 110 S. H. ,61. Qu. ,55. Jes. , 55. 112 Cai. ,46. 113 Qu. ,55. 114 C. C.,55. 115 Chr. ,59. 116 Chr. , 43. 117 Joh. , 31. 118 119 Emm. , 30. S. H.,56. Gath. ,57. Chr. , 55. ,56. ,57. T. H. ,55. Mag.,53. 120 Trin. ,59. 121 Pet. ,51. 122 Qu. , 29. ,35. ,37. b. S.,39. - 123 124 Mag.,52. 125 Chr.,47. Cla. ,48. 126 Pet. ,51. 127 Sid. ,45. Chr. ,47. IEmm. ,47. 128 S. H. ,52. 129 Emm. ,57. 130 Qu. , 26. ,29. ,37. Trin.,27.,33.,36.,40. ,49. , 50. Chr. , 44. Pem. ,45. Emm, ,54. Jes. ,52. S. H.,50. C. C.,58. Qu. , 19. ,37. Emm. S. H.,54. 131 ,25.,53. Mag.,29.,32. Cai.,34. Trin.,37.,38. Pet. ,44, ,52. ,54. 132 S. H. ,48. 133 Qu. , 39. Mag. ,54. S. H. , 134 Trin. ,29. 135 Emm. , 22. C. C.,58. 136 Pet. ,45. 137 S. H. ,35. ,48. Joh. ,37. Chr.,60. Pet., 50. 138 Pet. ,38. Chr. , 39. S. H. ,59. 139 S. H. ,53. . 140 Joh. ,58. 141 C. C. ,46. 142 Pem. ,46. 143 S. H. 54. Cai. ,36. Cath. ,55. Cath. ,58. | 176 Joh. ,20. 144 Pem. ,47. 145 Cla. . 146 Ki. ,48. S. H. ,53. Chr. , 55.,57. Joh ,59. 147 Qu. , 30. Chr, ,46. 148 Pet. , 58. ,” 149 Emm. , 33. Mag.,45. 150 Jes. ,57. 151 Mag, ,57. - 152 Cai, ,52. - 153 Trin. ,37. , 50. , 51.,52. Joh. , 47. Emm. ,52. ,53. ,56. Chr. , 50. T. H. ,52. Qu. , 50. 154 Emm. , 21. Trin. ,44. 155 Pet. , 55. 156 Qu. ,54. 157 Mag.,49. 158 Cla. ,36. Pet. ,60. 159 Joh. ,58. Chr. ,58. 160 Trin. ,53. ,54. 161 Joh. , 16. - Qu. , 30. Bem.,33.,49. Jes.,46. Trin. ,47. ,58. C. C.,58. 162 Pet. , 27. 163 S. H. ,36. 164 Joh. , 56. 165 Pet. ,55. 166 Chr. ,54. 167 Cla. , 56. 168 Jes.,20. Qu.,32. ,48. Cath. ,35. S. H. ,59. 169 Trin. ,40. 170 Pet. ,32. , 35. 171 Pet. , 49. *~ 172 S. H.,55. Trin. , 56. 173 Jes. ,53. ‘C. Ç. ,60. 174 Chr. , 56. 175 T. H. ,52. Emma.,26. 177 Sid. ,46. Mag. ,58. 178 Joh. , 13. Pet. ,48. 179 Ki. ,59. 180 Emm. ,32. Qu. ,35. ,59. C. C. ,36. ,59. Mag.,39. B. S.,47. 181 Trin ,21. , 50. EXERCISES ON BOOK II., p. 310, &c. 1 S. H., 14.,50. Joh.,18. Trin. ,35. Chr. ,55. 2 Joh. , 17. 3 S. H. , 16. 59. Trin. ,27. , 30. ,37. ,47. ,48. Mag. , 31. ,43. ,29. ,38. ,59. Sid.,34. Emm.,21. , 27.,37.,44. Pet. - Cai. ,43. Qu. ,37. T. H. ,55. 4 Emm. ,34. Mag. ,51. 5 S. H. ,03. Joh. , 18. Qu. , 21. Trin. ,37. Sid. , 42. Chr. ,45. ,46. ,48. Pem. ,57. *. Pem. ,46. 182 Jes. , 35. 183 Pet. ,54. 184 Cai, ,47. 6 Pet. ,52. 7 Pet. ,58. 8 Jes. ,54. 9 - 10 11 Jes. ,59. 12 Cai. ,58. 13 Jes. ,53. v. INDEX. 497 14 Pet, ,25. 15 Chr.,40. 16 Trin. , 42. 17 Pet. ,37. 18 T. H.,40. ,54, 19 Qu. ,37. 28 S. H. ,04. , 10. Trin. , 29. 29 Pet. , 43. 30 Chr, , 49. 31 Qu. ,55. 32 Qu. ,57. 33 34 Joh. ,59. 85 Qu. , 51. | 36 Mag.,57. 37 Cai.,59. 38 Chr.,57. 39 Cai. ,44. 40 Joh. ,44. 41 Trin. , 49. Cai, ,57. 42 T. H.,59. , 43 Joh. , 13. Emm. ,25. ,36. Trin. ,32. ,59. Mag. , 33.,40. Pet., 52. S. H. ,53 44 joh.3i. S. H.,50. Pet. ,54. ' 45 Joh. ,25. 46 Cai. , 42. 47 S. H.,53. 48 T. H. ,58. 49 Joh. , 26. Mag.,42. 50 Pet, ,44. 51 Jes. ,59. Jes. ,37. 52 Emm.,23. , 26.,28.,43. ,51. ,49. , 50. , 61. Pet. , 30. ,59. Trin. , 27. ,44. Qu.,60. | 53 Emm. ,28. Mag.,33.,43. ,46.,52. B. S.,38. C. C.,51. Chr. ,41. ,47. , 50. Sid. , 33. C. C.,39. Chr. ,59. | 54 joi...g. qu.23.30. ,48. Chr, ,59. 55 Chr., 30. Emm. ,36. S. H.,45. Cath. ,52. 56 S. H.,07.,59. T. H. ,44. Pem. , 52. Joh. ,41. Trin. ,53. Emm. ,52. - | 57 | 58 Emm. , 28. ,46. Trin. ,32. 59 Chr. ,51. 60 Pet. ,53. 61 Cai. ,58. Pem. ,47. | 62 63 64 Emm. ,59. 65 Jes. ,58. T. H. ,60. Mag. ,58. 66 S. H. ,59. GEOMETRICAL IXERCISES, Book III. p. 317. 1 Chr. ,28. S. H. ,36. ,59. Cai. ,44. 2 Qu. , 23. T. H. ,54. Mag.,53. 3 Trin. , 27. 4 Mag.,53. 5 S. H. ,04. Sid. ,41. 6 Trin. , 19. , 23. Qu.,21. ,22. Pem. , 30. , 39. Sid. , 36. Pet. , 31. Emm. ,34. ,42, ,44. T. H.,54. 7 Trin. , 19. Sid. , 33. Cai. ,34. Emm. , 24. ,34. Qu. ,36. , 56. Chr.,49. , 56. Cla.,46. Joh. ,57.,58. G.H.,53. B. S. ,51. 8 Pet. , 29. Cla. ,46. 9 Joh. , 21. ,51. 10 Cath. ,60. 11 Cla. ,58. 12 S. H.,48. 13 Mag.,46.,53. 14 Cath. , 49. Chr. ,54. Mag.,48. Ki. ,60. 15 S. H. , 43. 16 Jes. ,46. 17 Joh., 17. Pem. ,47.,60. 18 C. C. , 42. 19 Cai. ,39. ,40. 20 Pet. ,52. 21 Joh. , 21. 22 S. H. ,48. 23 Pet. ,47. 24 C. C. ,52. 25 Trin. ,57. 26 Trin. , 56. 27 Pet. ,44. 28 Joh. , 17. 29 Jes. ,59. 30 Jes. , 56. 31 Sid. ,59. 32 Pet. ,51. 33 Cai. ,53. 34 Trin. ,39. 35 Pet. , 50. 36 Emm. ,54. 37 S. H. ,53. 38 S. H.,59. 39 Qu. ,57. 40 Trin. 30. , 39. C. C.,35. ,45. Emm. ,37. Chr. , 39. Pem. ,40. 41 Sid. ,35. Chr. , 49. 42 Joh. , 30. 43 Joh. ,20. Emm. ,26. 44 Cath. , 31. 45 Qu. , 36. 46 Joh. ,28. 47 Emm. , 56. 48 Trin. ,57. 49 Cath. ,58. 50 Cai. ,45. 51 Trin. ,41. 52 C. C.,47. 53 Cai. ,46. Qu. ,35. | 60 Pet. ,39. 54 Jes. ,57. 55 Sid. ,48. 56 57 S. H. , 14. Qu.,20.,32. Joh.,25. Emm. ,32. Chr. ,45. Cai. ,44. Cla. ,48. Pem. ,51. j 58 Pet. , 56. 59 Trin. ,57. Qu. ,39. T. H.,55. Ki. ,59. 61 Emm. , 56. 62 Jes. ,51. 63 Cai. ,46. 64 Chr, ,51. 65 Cai. ,53. 66 Joh. ,51. 67 Chr. , 50. 68 Qu. , 50. 69 Chr. ,59. 70 Jes. ,53. 71 Qu. , 50. 72 Qu. , 50. 73 Pet. ,47. 74 Pet. ,53. 75 76 Chr.,58. 77 Emm. ,48. 78 Joh. , 22. Chr, ,26. . 80 Chr.,48. Sid.,52. 81 Jes. ,58. 82 T. H.,59. R FC 498 INDEx. 134 Sid. ,35. 83 Trin. , 11. 84 Qu. ,36. 85 Jes. , 56. 87 Joh. , 42. Chr, ,53. 88 S. H.,25. Emm. ,52. S. H.,53. 89 Cai. , 43. Emm. ,44. 90 Trin. ,29. Sid. ,45. - S. H. , 50. Jes. ,60. 91 Jes. ,58. 92 Pet. ,54. 93 Chr. ,52. . 94 Cla. ,47. 95 Pet. ,59. 96 T. H.,52. Cai. ,55. 97 Joh. ,58. º 98 Chr, ,59. 99 Chr ,53. 100 Jes. ,51. 101 Trin. ,59. 102 Pet. , 56. 103 Trin. ,48. 104 Pem., 31. 105 Chr. , 60. 106 C. C. ,59. 107 Trin. , 39. 108 S. H.,36. Jes. ,57. Chr. ,58. . - 109 S. H. ,02. Pem. ,32. T. H. ,44. Chr. ,54. Joh. ,59. - 110 Trin. ,45. 111 S. H. ,04. ,59. Pem, ,55. 112 Pet. ,39. S. H. ,59. 113 Chr. ,58. 114 S. H. ,59. 115 Trin. ,34. 116 Mag.,35. 117 Qu.,50. T.H.,58.,59. 118 Joh. ,47. 119 Trin. , 50. 120 Joh. ,17. 121 Trin. , 24, 122 Qu. , 39. Pem. , 43. 123 Qu. ,54. Trin. ,46. Pet. ,53. 124 Pet. , 51. 125 C. C. ,46. 126 Cai. ,51. 127 Qu. ,52. C. C. ,54. 128 Cai.,57. Joh. ,60. 129 Trin. ,48. 130 C. C.,53. 131 C. C.,52. 132 Pet. ,54. 133 Cath. , 53. Emm. , 56. Pem. , 42. 135 Qu. ,38. 136 Cai. , 31. 137 S. H. , 50. Qu. ,54. Pem. , 50. T. H.,58. 151 | 153 | 174 | 175 183 38 Cai. ,47. Cath., 57. 139 Emm. ,56. ,57. 140 T. H. ,58. 141 Pet. ,52. 142 Pet. ,52. 143 S. H. ,60. 144 Trin. ,52. 145 146 Joh. ,34. 147 Pet. ,25. 148 Pem. ,60. 149 Pet. 39. Pem. ,45. 150 Chr. ,40. , Trin. , 29. ,32..,38. S. H. ,08. Pet. , 19. ,20. , 21. Qu.,20r;28. Mag.,30. JB. S.,39. Chr.,51. S. H.,49, Joh. , 31. Cai. ,29. Emm.,53., 56. Chr. , 27. Cai. ,60. Joh. ,25. S. H. ,20. Qu..,39. Trin. , 22. , 25, Mag.,37. Jes. ,54. Cai. ,51. Cai. ,59. Qu. ,34. ,44. Chr. ,54: Pet. , 49. Cai. , 50. C. C.,52. - Cai. ,39. Jes. ,26. C. C.,38. ,47. Pem. ,60. 167 Pet. , 50. 152 154 155 156 157 158 159 160 16.1 162 163 164 165 166 | 168 Cai. ,37. 169 Cai.,36. ,59. # 170 Pet. Ag."joh.,58. 171 Pet. ,52. 172 Qu. ,37. 173 Chr. ,32. ,33. ,38. Emm. ,32. Trin. , 28. Joh. ,34. Pet. ,37. - C. C. ,51. ,55. Cai. , 42. - Qu. ,54. Pem. ,58. Cath. ,60. Jes. ,60. Trin, ,42. ,59. Chr. ,51. Pem, ,54. B. S.,58. Cla. ,57. S. H. ,04. Qu. ,20. ,29. ,35. Trin. , 22. ,23. 176 177 178 179 180 181 182 Pet. , 43. Trin , 33. Pem. ,44. Sid. ,35. T. H. ,58. Trin. ,46. , 50. | 200 C. C.,49. - 214 Joh. , 16. Pet.,31. B. S., 30. ,34. # 189 Qu.,48. : 190 Jes. ,52. 191 S. H. ,59. 192 Trin. ,54. 193 Pet. , 37. 194 Sid. ,35. - 195 Jes. 28. Sid. ,36. 196 Joh. , 26. 197 Chr. , 39. 198 Qu. ,54. Cath. ,59. 199 Joh. ,58. IEmm. ,60. Joh. ,41. ,42. , 49. Qu. ,55. Qu. , 33. Joh. ,20. ,48. Trin. ,46. 204 T. H. ,58. 205 S. H. ,48. 206 T. H. ,54. 207 Cla. 208 Jes. 209 Joh. ,56. Mag. ,59. 210 Chr. ,59. 211 Chr. ,59. 212 Chr.,59. 213 Pet, 201 202 203 215 216 Fei. 32. 184 185 186 187 188 217 Trin. ;42. 218 Joh. ,60. 219 T. H. ,60. 220 T. H.,60. 221 C. C.,52: 222 C. C. , 42. 223 Joh. ,52. 224 Emm. ,47. 225 Jes. , 49. Pet. ,52. 226 Joh. ,25. T. H.,51. ,55. Trin. , 51. Emm. 460. 227 Cath. ,58. 228 Mag.,49. 229 Joh. ,l 8, ,19. Qu. ,26.,39. Mag, ,29. Emm.,80. Trin. ,46. * Pem. ,44. ,48. Mag. , 35. Cath. , 30, Joh. ,34. Trin. , 26. Pem. ,34. 230 231 232 233 234 C. C. ,46. 235 Joh. ,35. Ki. ,60. 236 Trin. ,38. ,53. Joh. , 19. Chr. , 39. Jes. , 43. S. H. ,42. Pet. ,52. Cath. ,47. T. H. ,53. 237 Jes. ,38. C. C.,38. 238 C. C.,33. 239 Pet. ,32. 240 C. C.,25. B. S. ,28, Mag. ,45. 241 Pet. ,28. ,35. INDEX. 499 242 Cath. ,80. 243 Joh. ,36. 244 C. C.,39. 245 Jes, , 56. 246 Cai. ,58. 247 S. H. , 49. 248 Pet. ,52. 249 Pet. ,25. 250 S. H. ,60. 251 Jes. ,59. 252 Jes. ,59. 253 C. C. ,51. 254 Trin. ,53. 255 Trin. ,23. Chr.,44. 256 Joh. , 19. Trin. ,34. 257 Trin. ,44. Jes. ,34. Qu. ,26. 258 Pem. ,37. 259 C. C.,25. 260 Joh. ,29. 261 Qu. ,42. 262 Trin. , 43. GEOMETRICAL EXERCISES, Book IV. p. 339. I S. H. ,02. , 12. Chr. , 33.JPet. ,34. ,38. Trin. , 49. ,60. 2 Qu. ,20. , 30. ,34. Trin. ,29. ,60. Emm.,30. C. C.,35. B. S. ,36. Pem. ,40. ,48. ,52. 3 Jes. ,52. 4 Qu. ,20. 5 Joh. ,l6. 6 Trin. , 31. 7 Emm. ,53. 8 Trin. ,23. Sid.,39. ,47. Qu.,41. , 50. C. C.,45. Emm. ,46. ,48. Pem. ,52. T. H.,54. 9 Trin. ,54. 10 Chr. , 27. 11 S. H., 16. Qu. ,20.,27. C. C.,28. Joh. ,39. 12 Joh. ,29. Chr.,48. Cath. ,51. Pet.,52. Emm.,52. Pet. ,36. 14 Cai. , 39. 15 Trin, ,59. 16 Emm. ,37. 17 Jes. ,53. 18 Pet. ,45. 19 S. H. ,54. 20 Chr. , 56. 21 Cath. ,53. Cla. ,58. 23 Emm. ,58. 24 Chr. ,45. 25 Chr. ,55. Jes, ,56. Emm. ,60. 26 Cai. ,38. 27 Cai. ,35. 28 Sid. ,59. 29 Jes. , 52. 30 S. H. , 30. 31 Cai. , 42. 32 Trin. ,45. 33 Emm. , 35. 34 C. C.,47. Chr.,58. 35 Trin., 21. Chr., 30. ,34. 36 Pem. , 29. ,35. 37 S. H. , 13. Mag. ,54. Qu. , 19. Emm. , 21. , 23. ,48. B. S. , 26. Cai, ,35. Cla. , 49. Pem, ,36. 38 Chr. ,52. 39 Cath, ,57. 40 Pet. , 26. Emm. ,44. 41 Trin. ,60. 42 S. H.,59. 43 Cai. , 42. 44 Cai. ,32. 45 Mag.,51. 46 Chr. ,54. 47 Joh. ,48. 48 Cla. , 56. 49 Trin. , 43. 50 Cai. , 49. 51 Sid. , 52. 52 Chr. ,54. 53 Trin. ,38. 54 Pem. , 49. 55 Joh. , 22. 56 Trin. , 30. 57 Pem. , 29. 58 Joh. ,42. 59 Jes. , 33. 60 Cai. ,55. 61 Cai. , 50. 62 Pet. ,52. 63 Jes. ,57. 64 Pet, ,52. 65 Joh. ,59. 66 Chr. ,60. 67 Chr. ,52. 68 Chr.,60. 69 Cath. ,55. 70 Sid. ,59. 71 T. H. ,60. 72 Trin. ,58. 73. Pem., 30. Emm. , 56. Chr. ,38. Joh. ,60. Joh. ,60. S. H.,61. IB. S. ,55, S. H.,36. C. C.,41. Pet. , 56. Emm. ,58. Mag. , 60. S. P. ,42, Trin. ,48. ,50. ,53. Chr. ,47. , 56. Qu. ,53. ,55. 74 C. C. , 50. 75 C. C. ,55. 76 Cath. ,59. 77 Jes. ,60. H09 Joh. , 18. 78 Chr. ,58. 79 Jes. ,58. 80 Chr. , 56. 81 S. H. ,59. 82 Chr. ,59. 83 Joh. , 30. 84 Joh. , 50. 85 Cai. ,59. 86 Qu. ,47. ~~~ 87 T. H.,59. Emm. ,54. . S. H.,58. 88 S. H. ,58. 89 S. H. ,60. 90 Qu. ,59. 91 Cai. ,48. 92 Cai. ,56. 93 Cla. ,53. Jes. ,53. S. H.,61. * 94 Pet. , 27. 95 Qu. ,52. Pet. ,59. 96 Ki. , 49. 97 Cai. , 56. 98 Joh. ,59. 99 Chr. ,60. . 100 Joh. , 50. 101 Chr. ,57. 102 Trin. ,54. 103 Jes, , 31. 104 Mag. , 50. 105 Pet. ,55. Ki. ,60. 106 Chr, ,52. 107 Pet. , 50. 108 Mag.,35. Trin. ,36. 110 Joh. , 17. 111 Pet. ,25. 112 S. P.,55. 113 Cla. , 56. 114 Joh. ,44. 115 Joh, ,25. Trin. ,52. 116 C. C. , 50. 117 Pet. , 43. 118 Trin. ,29. 119 Cai. ,37. 120 T. H.,60. 121 Chr.,60. 122 C. C.,56. 123 Trin, ,51. 124 Trin, ,53. 125 C. C.,53. 500 INDEX, 126 Emm. , 21. Trin. ,36. Pem. , 42. 127 Sid. ,40. ,51. C. C.,41. Cla. ,60. 128 C. C. , 56. 129 Pet. , 56. 130 T. H.,55. 131 T. H. ,59. 132 Cla. ,55. 133 Jes. ,46. 134 Trin. , 21. 135 Trin. ,23. Emm. , 23. 136 Joh, , 18. 137 138 Trin. ,41. 139 Pet. ,25. T. H. ,46. 140 Sid. ,29. 141 Trin. , 31. 142 Pet. ,46. 143 C. C. , 38. 144 Cla. ,46. 145 Trin. ,46. 146 Cai. ,47. 147 Chr.,26.,42. Qu.,54. 148 Trin. ,57. 149 Mag.,41. 150 C. C.,44. 151 Qu. ,44. 152 Cath. , 30. .,37. 158 Trin. 53, Cai.,60. ,32. ,36. Jes. , 49. Pem. ,46. Qu. , 43. 158 Jes. , 56. 154 Chr. ,60. 155 S. H. ,60. 156 Chr. ,57. 157 Mag, ,58. Joh. ,58. 159 Pem. ,58. 160 Cai. ,40. 161 Cath. ,59. Jes. ,57. Pet. ,52. 162 Jes. ,59. 163 Pet. ,52. Jes. ,53. 164 Chr. 50. Trin. ,48. 165 Pet. , 49. - 166 C. C. ,48. 167 Chr, ,46. Qu. , 49. Pet. ,52. Cla. ,52. 168 C. C. ,46. 169 Sid. ,38. Trin. , 39. Pet. ,52. Cai. ,54. Qu. , 56. Cath. ,57. 170 Trin. , 33. T. H. ,60. 171 Trin. ,58. Pet. ,59. 172 C. C. ,55. 173 B. S. ,51. 174 Sid. , 50. 175 Pem. ,56, 176 Cla. , 50. 177 Cai. ,41. Emm. , 50. 178 Chr. , 39. 179 C. C.,49. S. P. ,60. 180 C. C. , 24. 181 182 Trin. ,46, 183 C. C.,60, 184 Trin. ,37. Pet ,36. ,52. 185 Trin., 22.,25.,27. , 36. , Sid. , 49. Jes. , 19. Qu.,35. Pem, ,37. Mag. ,45. * 186 Jes. , 19. Trin. , 22, ,25. , 27. , Qu. ,35. Pem. ,37. Mag.,45. Cai, ,50. 187 Chr. ,58. 188 Chr.,57. 189 Qu. , 31. ,40. Trin. , 42. - 190 Trin. , 11. Pem. , 31. Chr. ,36. ,57. S. H. ,36. Jes, ,60. 191 Joh. , 13. - 192 Jes. ,54. 193 T. H. ,58. Cla. ,47, 194 Jes, ,38. 195 Cai. ,38. T. H. ,54. 196 Trin. , 19. 197 Chr. ,57. ,59. 198 Pet. ,59. 199 Joh. , 50. 200 Jes. ,52. 201 Trin. ,46. 202 Trin. ,54. 203 Joh. , 18. 204 Qu. , 32. Pem. ,35. 205 Cai. ,47. - 206 Pet. , 56. GEOMETRICAL EXERCISES, Book VI., p. 356, &c. Qu. ,38. Jes. ,46. C. C.,31. Jes. , 19. Trin.,32. ,44. Qu. ,23. Sid. ,34. C. C. ,40. 5 Joh. , 14. Trin. , 27. ,28.,32. ,34. ,41. ,44. Cath. ,34. Chr. , 44. 6 Trin. , 11. ,28. , 43. , 46. Jes. , 19. C. C.,26. Qu. , 21. , 23. , 26. ,60. Pem. ,32. ,34. , 43. Cai. , 33. Emm. , 21. T. H.,54. Pet. ,28.,35. 7 Cath. , 51. 8 S. H. , 50. 9 Joh. , 23. 10 Trin. ,23. Mag.,37. 11 Trin. , 30. Mag.,44. 12 Trin. ,57. 13 Trin. ,59. 14 Cai. , 56. 15 Pem. ,47. I 2 3 4 Cai. ,35. S. H. ,04. Pem. ,58. 16 Trin , 33. 17 Emm. ,59. 18 Chr, ,59. 19 Chr. ,59. 20 Jes. ,37. 21 Jes. , 39. 22 Joh. ,58. . 23 Chr. , 56. Pet. , 56. 24 Emm. , 56. 25 Chr. ,54. 26 Mag.,57. 27 Pem. ,29. Cath. ,34. Iºmm.,34. Pet. ,35. Chr.,39. 28 Cai. , 56. 29 S. P.,55. 30 Cath. , 53. 31 Joh. , 19. 32 Trin. , 50. 33 Sid ,60. 34 Joh. ,19. & * * 35 Joh. , 18. - 36 Cath. , 33. 37 Joh. ,29. 38 Joh. ,29. 39 Joh. , 23. Qu. ,25. Emm. 29. Chr. , 30. ,32. 40 Pet. , 27. 41 Mag.,57. 42 Joh. , 26. 43 Cai, ,36. 44 Joh. , 26. 45 Joh. , 15. 46 Pem. , 35. 47 T. H.,44. 48 Qu. ,48. 49 Joh. ,25. 50 Qu, ,32. 51 Trin. ,25. 52 Chr, ,58. 53 T. H. ,59. 54 Jes. ,55. 55 Pet. ,38. 56 Qu. ,20. , 26. ,32. ,51. 57 Pem. .46. T. H. ,46. 58 Joh. , 17. 59 Joh. , 42. 60 Cla. ,57. 61 Pet. ,60. 62 Cai. , 50. 63 Cai. ,40. \ 64 Trin. ,28. C. C.,87. INDEX. 50). 65 Pet. ,48. 66 Trin. ,25. 67 Pet, ,50. 68 Emm. ,58. 69 Cai. ,29. 70 Joh. , 19. 71 Sid. ,44. 72 Qu, ,29. Cai. ,55. Cla. ,58. Pet. ,58. 76 Joh. , 39. Joh. ,60. Cla. ,57. Cath. ,51. 79 Sid. , 39. 80 Joh. ,28. 81 Qu. ,34. Qu. , 24. 3 Cla. ,59. Joh. , 19. S. H. ,60. 86 Sid. , 50. Joh. , 23. Trin. , 30. Joh. , 21. Cai.,52. 91 Joh. ,28. 2 Joh. , 29. Trin. ,36. Trin. , 24. 95 Joh. , 14. Joh. , 15. Qu. ,36. Cai. ,45. Trin. , 22. Chr. ,35. Cath. ,35. S H. , 15. Jes. ,44. Trin, ,32. ,35. , 50. Emm. , 26. , 50. Qu. ,28. ,60. Chr.,39. Cai. ,35. B. S.,55. Pet. ,59. 98 Joh. ,20. ,34. 99 Qu. ,38. 100 Qu. ,40. 101 Joh. ,20. 102 Trin. 103 Qu. ,38. Trin. ,33, Chr. ,36. Qu. , 43. Joh. , 21. Sid. ,48. Joh. , 29. Chr. , 43. 44. Qu. ,46. 104 105 106 107 108 109 Jes. ,51. Pet. ,25. Cai. , 31. Joh. , 21. Pet. , 32. 114 Cai. ,55. 115 Qu. , 24. * 116 Joh. ,20. 1 17 Joh. , 14. 118 Qu. ,36. 119 Qu. ,25. 120 Pet. ,54. Ri. , 56. 110 111 112 113 Cai. , 49. Joh.,20. , 21. Trin. ,45. S. H.,43. Emm. ,47. 172 121 Cla. ,46. 122 S. H.,59. 123 Jes. , 56. 124 Jes. ,57. 125 Pem. , 56. 126 C. C. ,45. ,59 Qu. , 47. 127 Joh. , 15. 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 Chr. ,41. S. H. , 50. Mag. ,41. Pet. ,25. Joh. , 17. Qu. , 22. Qu. ,21. Trin. , 26. Joh. , 18. Cath. , 36. Pet. ,45. Trin. , 56. Cai. ,55. Joh. , 26. Pet, ,39. Emm. Joh. , 21. C. C. ,53. Jes. , 49. Joh. ,57. IPem. ,37. Joh. , 42. Cai. ,41. Pet. , 24. Chr.,29. Trin. , 43. Trin. , 24. Trin. , 26. C. C.,51. Pet. , 35. Joh. , 19. Sid, , 30. Joh. ,35. Pet. , 26. Joh. , 18. Jes. , 50. Pem. ,32. Jes. ,58. Cath. , 53. Jes. , 55. Cath. , 30. 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 Joh. ,20. 173 174 175 176 177 Trin. , 43. 178 Joh. , 23. 179 Qu. ,40. 180 Qu.,29. 181 Joh. ,29, 182 Cath. , 56. 183 C. C. ,48. Emm. ,52 Jes. ,57. Chr.,58. o Emm.,58. Qu. ,35.,36. Emm. , 49. Cath. , 31. Mag. ,80. Ki.,44. C. C.,41. Pem. , 43. 191 227 I 184 Pem.,34. C. C. ,30. 185 Cath. , 31. 186 Emm. ,46. 187 Joh. , 13. , 21. Trin. , 29. ,34. Qu. , 38. , 43. C. C.,28. Pem. , 42. C. C.,35. S. H., 11. . Pem. ,46. T. H.,46. S. H. , 36, Sid. , 29. Pet. 36. Cai, ,39. Trin. , 11.,20.,32. , 33. Chr. ,35. - Pet. ,37. Cai. , 31. Joh. , 31. - Qu. ,44. C. C. , 30. Pem. , 31. Chr, ,37. Joh. , 13. Trin. ,20. Emm. , 24. Chr. ,37; ,45. Qu. , 22. .36. ,44. Trin. ,44. 202 Trin. , 32. 203 Qu. , 37. 204 Joh. ,29. 205 Joh. , 18. 206 Qu. , 21. 207 S. H. ,25. 208 Pet. , 33. 209 Joh. , 15. 210 Pem. , 43. 21 1 Cla. ,57. 212 Qu. ,58. 213 Qu..,31. 214 C. C. ,57. 215 Ki. ,47. Emm. , 50. 216 Trin, ,35. 217 Cai. ,45. 218 Chr. ,48. 219 Jes, ,57. 220 Qu. ,25. 221 Pet. , 33. 222 Pet. ,52. 223 T. H. ,55. 224 Cai. ,46. 225 Qu. ,60. 226 Joh. , 21. Pet. ,44. Joh. , 15. S. H.,40. Pem. , 53. Pem. ,3i. , 43. Qu. , 19. ,25. ,48. Trin. , 22. ,37. Cai. , 43. Mag. ,32. Joh. ,51. S. H. , 50. Jes. , 53. S. H. ,50. Sid. ,46. Cai. ,46. Chr. ,57. Chr. ,59. 188 189 190 192 193 194 195 196 197 198 199 200 201 Qu. , 43. Jes. ,19. Cla. ,52. 228 229 230 231 232 233 234 235 236 237 238 239 Joh. ,51. 502 INDEX. 240 Pet, ,35. 241 Qu. ,41. 242 Jes. ,44. 243 Joh. ,25. 244 Mag.,50. 245 Cai. ,44. 246 S. H. ,59. 247 Joh. ,25. 248 Cath. ,60. 249 S. H. ,59. 250 Joh. , 30. 251 Trin. , 24. 252. Joh. ,20. 253, S. H. ,45. 254 Cath. , 55. 255 Cai. ,60. 256 Chr.,58. ; 257 Chr. ,60. # 258 T. H.,60. 259 C. C. ,57, . 260 Pet; ,51. 60. 261 Joh. , 16. 262 Trin. ,58. 263 Joh. ,20, 264 Emm. ,37. 265 Cath. ,48. 266 Emm. , 56. 267 Joh. , 42. 268 Trin. ,20,. 269 Sid. 48. 270 Pet. , 24. 271 Chr.,54. 272 Jes. , 56. 273 Chr.,36. ,38. 274 Cai. ,39. ; 275 Qu. , 24. 276 Chr. , 38. 277 Qu. , 33. 278 279 Joh. , 15. 280 Trin, ,53. 281 Jes. ,58. 282 Jes. ,51. 283 Joh. , 14. 284 Sid. ,52. 285 Cai. ,51. Cai. ,56. 1 Pet, ,47. Chr.,48. 2 Pet. , 29. Qu. , 49. 3 Joh.,31. Emm. , 24. Qu. , 31. Pet. ,51. T. H. ,58. 4 & 5 Trin. , 33. 6 S. H. , 33. 7 Trin. ,47. 8 Pet. ,58. 9 S. P.,47. 10 Qu, ,38. 11 Trin. , 52. 12 Qu. , 56. 13 S. H.,48. Qu. , 50. ,57. Jes. , 50. Cath. ,59. 14 Cai. ,34. ,39. 15. Cath. ,35. Af 16 Pet. , 55. 17 Pet. , 49. Pem. , 49. C. C. , 49. Chr.,54. Cla. ,55. Trin. ,59. 18 S. H. ,29. Chr, ,54. 19 Cath. ,57. 20 Cai. ,46. Jes. , 50. LOCI, p. 380,. &c. 21 Jes. ,55. 22 Emm.,37.,56. Cla.,60. 23 Cai. ,40. Cla. ,52. 24 Chr. , 44. 25 Cai. , 39. 26 Qu. ,52. Emm. ,54. Trin. ,59. - 27 Trin. , 49. Qu, ,58. Pet. ,59. 28 Joh. , 28. 29 | 30 Qu, ,53. 31 Trin. ,54. 32 Qu. ,54. Qu. , 50. Mag. ,57. 33 Qu. ,55. 34 Cai. , 52. 35 Qu. , 31. 36 S. H., 18. Qu.,20.,52. Trin. , 49. Cla. ,51. B. S. ,55. T. H. ,55. 37 Cla. , 56. 38 S. H.,59. 39 Mag. ,58. 40 Mag.,59. 41 Pet. ,53. 42 Jes. ,57. 43 Cai, ,55. S. H. ,58. 44 Trin. ,20. - 45 Joh. , 30. Cai, ,36. Cla. ,46. 46 Pet. ,58. 47 Joh. ,54. 48 S. H. , 39. 49 Pet. , 50. 50 Pet. ,57. 51 Emm. , 56. 52 Qu. , 50. 53 Trin. , 43. 54 Joh. , 50. 55 Trin. , 21. 56 Emm. , 24. 57 Trin. ,60. Joh. ,59. Qu. ,44. 59 Joh.,44. Pem. ,53. ,57. Sid. ,60. 60 Joh. ,34. 61 Qu. ,52. 62 Qu. ,41. ,42. Cai. ,54. 63 Cai. ,54. Qu. ,55. MAXIMA AND MINIMA, p. 387, &c. 1 Sid. , 30. , 43. Jes. ,50. ,58. Qu. ,34. Trin. ,40. Cla. ,47. Emm. , 56. 2 3 Trin. ,20. Cath. ,58. Pet. ,60. 4 Pem. ,32. 5 Trin. ,52. 6 S. H. ,59. 7 Trin. ,36. 8 Cai. , 37. 9 Trin. ,56. -10 Trin. , 31. Emm. ,54. Qu, ,59. Qu. , 56. 11 Qu. , 19. C. C.,26. Cai. , 56. 12 T. H. ,60. 13 Emm. , 31. Chr. ,38. 14 T. H. ,53. Cai. ,54. Sid. ,60. 15 Cath. ,46. Trin. ,48. 16 17 Qu. , 30. Chr. ,46. 18 S. H. ,04. Cai, ,34. Emm. , 39. w 19 S. H. , 26. 20 C. C. , 26. 21 Qu. , 19. 22 T. H. ,60. 23 Qu. ,25. Trin, ,25. ,38. Pet. ,39. Jes. ,52. Pem. , 42. 24 S. H. ,03. , 18. Trin. ,25. ,44. Cla. , 31. ,36. Pet. ,59. 25 S. H. , 14. Cla. , 55. 26 Trin. , 30. Pem. ,35.,40. 27 S. H. , 13. 28 Emm. ,28. 29 30 Jes. , 33. Trin. ,48. 31 Trin. , 21. ,57. ,60. Joh. ,52. 32 Pet. ,40. TNTEX. -503 33 Cai. ,46. 34 Cai. , 43. Pet. , 42. Mag. ,48. Trin. ,48. 35 C. C.,52. 36 Qu. , 24. Joh. ,38. Pem. ,53. 37 S. H.,52. 38 Trin. , 42. Cla. ,54. 39 Trin. ,20. ,47. ,51. Joh. , 49, Ki. ,47. Qu. , 50. Jes. ,51. Pem. ,54. Emm. ,59. 49 Pet. ,58. - 41 Joh. ,57. 42 Trin. ,58. 43 Trin. ,46. 44 Qu. , 49. 45 Sid. ,29. Chr.,44. :| 51 Cai. ,28. 46 S. H. ,02. Qu. ,46. 47 S. H. ,03. 48 Cai.,59. 49 Qu. ,58. 50 Trin. , 56. 52 Cai. , 50. 53 Joh. , 17. Cla. , 56. 54 Sid. ,29. Trin. ,58. 55 C. C.,53. 56 Cla. ,57. 57 Trin. ,43, Qu. ,55. 58 Sid. ,60. 59 Ki. , 50. 60 Trin. , 53. 61 S. H. ,02. Trin. ,20. , 22. , 33. 62 C. C. ,52. Pem. ,56, Trin. ,58. : 66 Cai. ,41. 63 Trin. ,41. - 64 S. H., 13. Trin. ,22. 65 Emm. , 56. iſes. ,51. Sid. , 51. 67 S. H. ,09. B. S. , 30. 31. Trin. , 56. 68 Cla. ,55. ;60. 69 Trin. ,20. Joh. , 33. Sid. ,51. 70 C. C.,52. | 71 Trin. ,34. Joh. ,84. | 72 Joh. , 28. Trin. ,52. 73 S. H. ,03. Trin. , 24. , 30. Qu, ,31...,85. Cai. 35. 1 2 Chr. ,29. 3 Qu. ,20. Cath. ,34. Trin. ,44. , 49. ,59. Jes. ,46. Ki. , 37. Joh. ,57. Cla. ,59. 4 Trin. , 22. , 50. Qu.,39. Chr. , 42. Cai, , 49. Jes. ,58. Cla. ,47, ,51. ,58. Mag. ,52. Cai. ,32. ,41. Qu. ,47. S. H. ,53. Chr, ,59. Cla. ,55. C. C. ,59. Chr. , 55. - 10 Trin. ,34. 11 T. H. ,60. 12 Cla. ,56, Chr, ,54. 13 Cla. , 56. 14 15 Chr. ,55. 16 Joh. , 30. 17 T. H. ,60. 18 Cla. , 50. 19 Qu. ,40. 20 Qu. , 43. 21 Pet. ,55. Mag.,59. Emm. , 27. i Trin, 49. TANGENCIES, p. 394, &c. 22 Cai. ,60. 23 Cla. ,57. 24 Cai. ,54. 25 S. H.,53. 26 Trin. ,37. 27 Qu. ,23. ,36. ,38. 28 Cai. ,60. 29 30 Cai. ,52. Joh. ,60. 31 Cai. ,55. 32 Chr. , 49. 33 Emm. ,25. Pem. ,46. 34 35 Trin. ,48. , 49. 36 Emm. , 21. Pem. ,32. Cla. ,36. Cai. ,45. Trin. ,45. 37 Qu. , 21. , 26.,36. - Trin. ,52. Cla. , 56. 38 Pet. ,46. 39 Joh. ,57. 40 Cai. , 33. 41 Trin. , 23. 42 Qu. , 22. ,38. Trin. , 42. ,44. ,48. , 50. 43 Trin. , 35. Mag. , 42. Joh. , 56. *- 47 Trin. ,58. 48 49 Pet. ,45. 50 Emm. ,57. 51 Joh. , 19. 52 Qu, , 33. 53 T. H. ,59. 54 55 Qu. , 33. 56 Jes. ,57. 57 Qu. ,32. 58 Joh. , 14. , 16. ,37. S. H. ,44. Ki. ,52. 59 Qu. , 31. - 60 S. H. ,04. 61 Cla. ,59. 62 Jes. ,57. 63 Trin. ,19. ,41. Sid.,45. C. C.,44. 64 Pet. ,44. 65 C. C. ,34. 66 Trin. , 11. S. H. ,36. 67 - 68 Trin. , 19. , 56. ,60. Sid. ,59. Qu ,60. 69 S. H. ,36. Jes. ,57 Chr.,58. 70 Qu. , 42. C. C. ,52. Qu. ,26. Chr. ,36. | 71 Cla. ,60. GEOMETRICAL EXERCISES on Book XI. p. 401, &c. | 44 Trin. ,36. 45 Pet. ,61. | 46 Cath. , 49. 1 Trim.,29. Chr.,43.,46. 2 Chr. ,44. 3 S. H. , 12. 4 Pet. ,25. 5 Cai. ,37. 6 S. H.,20. 7 Emm.,58. 8 Emm. ,34. 9 Joh. ,51. 10 B. S. ,51. 11 Trin. , 31. ,36. ,46. Mag.,58. Cla. ,59. 12 Cla. ,57. 13 Emm. ,54. 14 Chr. ,57. 15 Pet. ,40. 16 Chr. ,57. Joh. ,58. Pet. ,61. 17 Trin. ,32. 18 S. H.,25. Qu. , 30. Trin. ,25 0. ,85. { 504. INDEx. 19 Emm. ,32. 20 Trin. , 24, Cla. ,58. 21 Cai. ,53. 22 Cla. ,55. 23 T. H. ,60. 24 Cai. , 43. 25 Cai. ,34. 26 Chr. 27 Trin. , 43. 28 S. H.,60. 29 Trin. 30 Pet. ,60. 31 Cai. , 56. 32 S. H. ,59. 33 34 B. S. ,55. 35 B. S.,58. 36 Trin. , 24. ,47. 37 Trini. , 30. ,41. 38 Joh. , 17. Cai. ,46. Cai. ,36. Trin. ,57. S. H. , 33. Chr. ,51. Trin. Trin. 46 C. C.,61. 47 Cai.,44. 48 Cai. ,38. 49 Trin. ,29. 50 Cai. ,45. 51 Trin. , 42. 52 Trin., 22.,25. Mag.,29. Joh. ,34. Chr. ,45. S. H. , 26. ,37. 55 Cai. ,42, Mag.,59. 56 Cath. , 33. 57 Qu. , 30. | 39 , 40 41 42 43 44 45 53 54 59 | 65 68 58. S. P.,60, S. H. ,01 Trin. , 49. Cai. ,61. Trin. Chr. ,54. Joh. , 22. Trin. , 31. Cai. ,48. S. H. ,58 Cai.,57. Chr, ,58. Joh. , 31. 70 Qu. ,46. 71 Joh. , 29. 72 Joh. , 14. 73 Trin. , 28. 74 Trin. ,53. 60 61 62 63 64 66 67 69 Cai. ,53. Cla.,49. Trin.,57. C. C.,51. GEOMETRICAL EXERCISES on Book XII. p. 408, &c. 1 Trin. , 23. , 27. ,46. Pem. , 30. Sid. ,31...,44. Cai. , 31. ,34. ,41. Emm. ,36. ;40. Chr. , 42. Joh. ,47. Qu. ,54. 2 Jes. ,19. Pem. ,32. Qu, ,48. Chr. ,51. Sid. ,52. 3 Chr.,32. B. S.,35.,37. Cai, ,51. 4 Cath.,30. Trin.,32.,57. S.H. ,03. , 43. Chr. , 35. , 50. Cla.,57. 5 Trin. ,32. ,34. 6 C. C. 7 Chr. , 49. 8 Qu. ,25. , 29. Trin. , 33. ,35. ,44. Chr. ,34. ,41. ,45. IEmm. , 39. Cai. ,52. 9 C. C.,34. Pem. ,52. 10 Trin. , 21. Qu. ,37. 11 Pem. ,37. 12 S. H. ,01. 13 S. H. , 18. Chr. ,83. 14 Pem. ,45. 15 C. C. 16 S. H. ,44. 17 Trin. , 50. | 26 | 18 Joh. , 22. Cai. , 38. S. H. , 43. Emm.,53. 19 Joh. ,21. 20 Cath. , 56. 21 Trin. ,54. 22 Emm. , 33. 23 S. H. , 16. 24 Qu. , 24. 25 Chr. ,37. Joh. , 15. Cai. ,35. Joh. , 31. Cai. ,37. Jes, ,55. Joh. , 17. Joh. , 17. Cai. ,32. Trin. , 21. Qu. ,36. 36 C. C. , 50. 37 S. H.,28. 38 S. H. ,44. 39 Cai. ,39. 40 Pet. ,37. 41 Cai. ,37. 42 Cath. ,57. 43 Cai. ,38 44 Cai. 45 Cai. 46 Cai. 27 28 29 30 31 32 33 34 35 Joh. , 15. CAM Br II, G E Cai. ,45. Joh. , 37. Cai. ,44. 47 48 49 S. H.,37. Sid. ,33. Qu. , 33. 56 Qu. , 50. 57 T. H. ,59. 58 Chr. ,56. ,59. 59 Joh. ,58. 60 Chr. ,59. 61 S. H.,61: 62 S. H.,62. 63 Trin. , 49. 64 T. H. ;60. 65 66 S. P.,39. 67 Trin. Mag.,29 68 - 69 Mag.,29. 70 Trin. , 41. 71 Trin. , 21. 72 73 74 Cla.,56. Printed and Stereotyped by W. Metcalfe, Green Street. /*- ,29.,37. S. H.,20. Trn. ,32. B 1,640,386 | lifiſi. - t 3 9015 O6537 11:17