/T7 ABRIDGED AND APPLIED COLLEGE PREPARATORY M.AUERBACH C.B.WALSH \ GIFT OF PLANE GEOMETRY LIPPINCOTT'S SCHOOL TEXT SERIES EDITED BY WILLIAM F. R USSELL, PH.D. DEAN, COLLEGE OF EDUCATION, STATE UNIVERSITY OF IOWA PLANE GEOMETRY I. ABRIDGED AND APPLIED II. COLLEGE PREPARATORY BY MATILDA AUERBACH SUPERVISOR OF MATHEMATICS IX THE ETHICAL CULTURE HIGH SCHOOL, NEW YORK CITY AND CHARLES BURTON WALSH PRINCIPAL OF THE FRIENDS' CENTRAL SCHOOL, PHILADELPHIA PHILADELPHIA, LONDON, CHICAGO J. B. LIPPINCOTT COMPANY GA455 A* COPYRIGHT, 1920, BY J. B. LIPPINCOTT COMPANY PRINTED BY J. B. LIPPINCOTT COMPANY AT THE WASHINGTON SQUARE PRESS PHILADELPHIA, U. S. A. PREFACE IN separating this text book on Plane Geometry into two parts, the Authors have followed what appears to them to be a normal and logical requirement essential to a proper presentation of the subject, and the most appropriate reference to these divisions would seem to be a designation of them as a First Study and a Second Study. In the former, the objective is to afford a general view of the subject, with emphasis on applications, the study being intended for the use of all high school pupils, and the material is so presented as to make it available also for use in junior high schools. The Second Study is devoted to a more intensive treat- ment of Plane Geometry, with special emphasis on theoretical work, and is addressed particularly to Regents 7 and college entrance requirements. In the entire text the inductive method is followed as far as practicable, and simplicity is gained rather by the means of scientific accuracy than at its expense. In view of the fact that the purpose and method of the two parts of the book differ somewhat, a separate consideration of each of these divisions seems desirable. PART I The Authors feel confident that the First Study will serve a fourfold purpose. First: That it will contribute to a solution of the question as to how much mathematics shall be required of high school pupils who do not intend to enter college, and with this objective, an effort has been made to plan a course adapted to the needs and interests of pupils meeting minimum requirements in mathematics. The Authors believe that it has been customary in many schools to meet this situation by a study of only a portion three or four books of the geometry in the required course, thus giving students merely an intensive knowledge of a part of the subject, instead of a broadly comprehensive view. The course here outlined covers 459936 vi PREFACE not only all that is really important in the five books although the syllabus contains fewer propositions than are now comprised in the first three books of most available texts but also work in the application of three of the trigonometric functions. Second: That it will suggest a course in elementary geometry so thoroughly adapted to the mental development of pupils in the ninth or tenth school year that it may be profitably used to super- sede the conventional course in formal geometry. The course here outlined will not in any respect restrict the preparation of the student for college; on the contrary, will find him much more ready and willing to proceed to the collegiate preparatory work with his knowledge of the subject enriched by application and vitalized with interest. Third: That it will open the eyes of the pupil to the relation of geometry to the activities and necessities of every-day life, and emphasize the practical application of the science, both in specific professions and trades, and in the affairs of daily life. Fourth: That it will arouse in the pupil a conception of the dignity and power of the subject. To this end the Authors have treated it scientifically, endeavoring to develop gradually in the mind of the pupil a natural assumption of this treatment; and it has been their purpose, by departing from formal methods, to lead the pupil to reason rather than merely to remember. The following means have served in the attainment of the ends just stated: REVISION OF THE SYLLABUS In the First Study the NUMBER OF PROPOSITIONS has been reduced to approximately half of those given in the standard texts. Propositions have been retained or selected on three bases : First: Those that are rich in application. Second: Those of peculiar interest to the young student. Third: Those essential to the sequence of the study. The wording of the propositions retained has departed materially from the traditional phraseology in an effort to avoid the formidable and stilted qualities of the latter, while retaining its scientific correctness. In fact, the language of the students hi the classroom has suggested many of these changes: e.g. "Three PREFACE vii sides determine a triangle " to replace "Two triangles are congruent if three sides of one are respectively equal to three sides of the other." The Authors feel that some of these changes in wording are desirable scientifically as well as from a practical standpoint: e.g., "If the ratio of the sides of one triangle to those of another is constant, the triangles are similar," to replace "Two triangles are similar if the sides of one are respectively proportional to the sides of the other;" and "An angle whose vertex is outside a circle is measured by half the difference of the intercepted arcs," to replace three statements. A transition, more gradual than usual, from the geometry of the grade school to the more scientific work of the secondary school has been secured. The proofs given hi the first section would be wholly convincing but the forms and detail less conventional than those demanded in more rigorous demonstrations, some of which are preceded by explanations. This does not mean that the text is one of Concrete or Observational Geometry. The bare essentials are retained from the outset, and subsequently there is a gradual introduction and demand for all the rigor obtainable in secondary school work. This gradual transition tends to prevent the discouragement so often manifested during the beginning of the study. MINOR DETAILS. First: No authorities are required for auxiliary construction con- sistent reference to them makes a proof unduly formidable no other reference should be permitted. Second: Nothing is introduced in the text until it is required thus avoiding long lists of definitions, axioms, postulates, and the like. Third: Throughout the text emphasis is laid upon the idea of classification; and by means of proper grouping of definitions, postulates, and developed facts, the student is trained to regard the subject, not as a miscellany of isolated facts, but as a frame- work of interrelated sub-topics. A few reference books are men- tioned. The use of these has been found so exceedingly helpful and inspiring in the classes which have been led by the Authors that they feel that the omission of some such lists would be a great calamity. The Authors recall many instances in which new life viii PREFACE has been given to the subject through such references; in fact there are instances where a pupil who might never have discovered him- self mathematically has developed true mathematical enthusiasm and ability by browsing through suggested supplementary reading. In using the text, the Authors earnestly suggest that actual writing of proofs be deferred until the class is quite ready to fall in naturally with a more or less set form of proof. Let the need of a form become apparent to the student who is trying to write a proof unaided by conventions before insisting upon the adoption of one in written work. Indeed the Authors feel that the entire first chapter of this text could well be developed before the necessity arises. for a single written proof from the stu- dent. Sufficient material for written work will be available from the exercises during this period. PART II In addition to a review of the First Study, the Authors desire to direct attention to some details of this Second Division of the work which seem worthy of special mention. FIRST: The size of the syllabus. The number of propositions developed in the First Study has been considerably augmented to include all the demands of College Entrance and Regents' Examina- tions. This has been accomplished by inserting the additional theorems between the theorems of the original syllabus, thus preserving the sequence of the former Study. A separate syllabus of construction problems is given in the chapter entitled " Methods of Attacking Problems " (page 306). SECOND : The grouping of the syllabus. To facilitate the reten- tion of the frame-work of the subject, the propositions are collected in groups by topics, so far as the sequence permits, and such grouping has necessitated certain departures from the traditional arrangement by books. THIRD : The type of exercises. In this part emphasis is placed on theoretical exercises, as contrasted with special reference to prac- tical exercises in the First Study; and a large collection of college entrance papers, together with a still larger selection of isolated exercises of this character, is included. PREFACE ix FOURTH : A chapter on methods of proof. More scientific habits of work are fostered through a discussion and a careful classification of methods of proof with illustrative exercises. FIFTH : A chapter on methods of attacking probkms. This chapter is largely similar in purpose to the chapter just mentioned, except that it deals with construction work. A special syllabus of what might be termed fundamental constructions is given, including those presented hi standard texts, and those propositions are of such character as to form a necessary and sufficient basis for the work required by colleges. This chapter groups many typical constructions and methods employed, and definite reference throughout the book is made to this, as well as to the chapter on methods of proof. SIXTH : Suggestions for club or other additional work. The chap- ter entitled "Suggestions," and exercises preceded by the letter "d" frequent in the book may be omitted (not constituting a requirement for college preparation) without impairing the integ- rity of the course. They are included to give additional interest and breadth to the subject where time and the ability of the student permit. This chapter contains in the suggestions for club work a list of topics suitable for discussion by students or teacher in a mathematics club of high school grade, and appended to this list will be found a bibliography of appropriate references. In summary, then, it may be said that this Second Study is intended to enlarge upon the course outlined in the First Study, not only in that it answers the requirements of college entrance examinations, but in that it also makes possible at the same time a richer and fuller course for those students whose interest and ability warrant it. In closing, the Authors desire to acknowledge three distinct sources of assistance. Realizing the present eclectic tendency of teachers in the matter of exercises, as evidenced by the general use of typewritten lists of problems, the Authors have availed themselves frequently of many of the standard texts in the selection of material of this kind. To Mr. Lewi Tonks, a former pupil of the Authors, who read the proof and criticised the contents of the book, the Authors fee) x PREFACE especially indebted, and wish at the same time to acknowledge the assistance of Mr. H. W. Smith, of the Ethical Culture School, and of Mr. P. S. Clarke, of Pratt Institute, Brooklyn, whose suggestions for revision of the English of the text were of value. Finally, the Authors take pleasure in this opportunity to record their deep appreciation of the endorsement and encouragement they have received from the authorities of the Ethical Culture School Prof. Felix Adler, Rector; Franklin C. Lewis, Superin- tendent; and Dr. Henry A. Kelly, High School Principal. Their approval made possible the experimental work in the School which has developed and justified the course here presented, and the Authors feel that their kindly sympathy and intelligent cooperation in the growth of the experiment have contributed in large measure to its success. THE AUTHORS. OCTOBER, 1919. SYMBOLS AND ABBREVIATIONS = is equal, or equivalent, to ^ is not equal, or equivalent, to * is similar to ^ is congruent to = approaches as a limit oc varies as <*_ is measured by > is greater than > is not greater than < is less than < is not less than + plug; or increased by minus, or diminished by -T- , divided by X,()/ multiplied by a parallel, or is parallel to JT not a parallel, or is not parallel to ||s parallels J_ a perpendicular, or is perpendicu- lar to JL not a perpendicular, or is not per- . pendicular to -Ls perpendiculars arc A B straight line AB O (D circle, circles A, A triangle, triangles O, EU parallelogram, parallelograms " ; * 3 angle, angles since therefore and so on >{}>[]' signs of aggregation V the nth root of TT 3.14159.... adj. adjacent alt. alternate ap. apothem approx. approximately ax. axiom circf. circumference comp. complement, com- plementary con. conclusion cong. congruent const. construction cor. corollary corres. corresponding def. definition diff. difference ex. exercise ext. exterior fig. figure ht. height, or altitude horn. homologous hy. hypotenuse hyp. hypothesis int. interior isos. isosceles lat. lateral peri. perimeter pi. plane pt., pts. point, points n-gon polygon of n sides post. postulate prob. problem proj. projection prop. proposition rect. rectangle reg. regular rt. right sec. sector seg. segment sq. square St. straight subst. substitute sup. supplement, sup- plementary sym. symmetrical th. theorem vert. vertical si CONTENTS PAGE PREFACE v LIST OF SYMBOLS AND ABBREVIATIONS xi PART ONE FIRST STUDY CHAPTER I INTRODUCTION A. A FEW FACTS CONCERNING THE EARLY DEVELOPMENT OF GEOMETRY 3 Summary 10 Bibliography 10 Exercises. Set I. Based Upon Historic Facts 11 B. A FEW ILLUSTRATIONS OF GEOMETRIC FORM 12 Exercises. Set II. Illustrations of Geometric Form. . .* 13 C. MEANING OF GEOMETRIC FORM 14 Exercises. Set III . Meaning of Geometric Form 15 D. SUGGESTIONS OF A FEW USES OF GEOMETRY 16 Exercises. Set IV. Mensuration 17 Exercises. Set V. Constructions. Designing 17 Exercises. Set VI. Some Other Uses of Geometry 24 E. THE BASIC PRINCIPLES OF GEOMETRY 24 Axioms 25 Exercises. Set VII. Illustrations of Postulates 26 Postulates of the Straight Line 27 Definitions 27 Exercises. Set VIII. Sums of Angles in Pairs 28 Exercises. Set IX. Measurement of Angles 29 Postulates of the Angle 31 Exercises. Set X. Relative Position of Angles 31 Exercises. Set XI. Instruments for Measuring Angles 36 F. THE DISCOVERY OF SOME FACTS AND THEIR INFORMAL PROOF .... 36 I. Classification Based upon Sides 37 II. Classification Based upon Angles 37 Experiments. Theorems 37 Exercises. Set XII. Meaning of Congruence and Classi- fication of Triangles 38 Exercises. Set XIII. Application of Congruence of Triangles 39 III. Some Properties of the Isosceles Triangle 42 Experiment. Theorem 42 a. Some Properties of the Equilateral Triangle 43 Exercises. Set XIV. Equilateral Triangles 43 siii xiv CONTENTS PAGE IV. Further Discussion of Congruence of Triangles 44 Experiment. Theorem 44 Exercises. Set XV. Further Applications 44 Summary of Chapter 47 CHAPTER II THE PERPENDICULAR, THE RIGHT TRIANGLE, AND PARALLELS A . THE PERPENDICULAR 48 Postulate. Theorem 48 Exercises. Set XVI. Distance from a Point to a Line 48 Axioms of Inequality 49 Exercises. Set XVII. Numerical Inequality 49 Exercises. Set XVIII. Inequality of Sects 50 B. THE RIGHT TRIANGLE 50 Experiment Theorems 50 C. PARALLELS 51 Exercises. Set XIX. Parallels 52 Postulate of Parallels 52 Exercises. Set XX. Relative Position of Angles . % . 53 Exercises. Set XXI. Construction of Parallels .' . 55 Exercises. Set XXII. Related Statements 56 Exercises. Set XXIII. Applications of Parallelism 58 Summary of Chapter 59 CHAPTER III ANGLES OF POLYGONS AND PROPERTIES OF PARALLELOGRAMS A. ANGLES OF POLYGONS 60 Exercises. Set XXIV. Sum of Angles of a Triangle 60 Exercises. Set XXV. Sums of Angles 01 Polygons 62 Exercises. Set XXVI. Sides and Angles of a Triangle 66 B. PARALLELOGRAMS 67 Exercises. Set XXVII. Parallelograms 68 Exercises. Set XXVIII. Parallels 73 Summary of Chapter 73 CHAPTER IV AREAS A. INTRODUCTION. REVIEW OF FRACTIONS 74 Underlying Principles 74 Exercises. Set XXIX. Fractions 75 B. AREAS. DEVELOPMENT OF FORMULAS 77 Exercises. Set XXX. Comparison of Sects 78 Exercises. Set XXXI. Areas of Rectangles 80 CONTENTS xv PAGE Exercises. Set XXXII. Areas of Parallelograms 81 Exercises. Set XXXIII. Altitudes of Triangles 81 Exercises. Set XXXIV. Areas of Triangles 82 Exercises. Set XXXV. Areas of Trapezoids 83 Summary of Chapter 85 CHAPTER V ALGEBRA AS AN INSTRUMENT FOR USE IN APPLIED MATHEMATICS A. LOGARITHMS 86 I. Introduction 86 Exercises. Set XXXVI. Meaning of Logarithms 86 Exercises. Set XXXVII. Applications of Laws of Expo- nents 88 Exercises. Set XXXVIII. Use of Tables of Powers 90 Historical Note 90 II. Principles of Common Logarithms 91 Exercises. Set XXXIX. Common Logarithms 92 III. Fundamental Theorems 93 IV. Use of the Table of Common Logarithms 94 Exercises. Set XL. Use of Table 95 Exercises. Set XLI. Computation by Logarithms 98 B. RATIO, PROPORTION, VARIATION 105 I. Ratio and Proportion 105 Exercises. Set XLII. Ratio 106 Exercises. Set XLIII. Proportion 106 Exercises. Set XLIV. Applications of Proportion 108 II. Variation 110 Exercises. Set XLV. Applications of Variation Ill Summary of Chapter . . . 114 CHAPTER VI SIMILARITY A. INTRODUCTORY EXPERIMENTS AND THEOREMS 115 Exercises. Set XLVI. Proportional Sects 116 B. IDEA OP SIMILARITY 119 Exercises. Set XL VII. Meaning of Similarity 121 C. SIMILARITY OF TRIANGLES 122 Exercises. Set XL VIII. Similarity of Triangles 124 D. PERIMETERS AND AREAS OF SIMILAR TRIANGLES 132 Exercises. Set XLIX. Areas of Similar Triangles 133 E. APPLICATIONS OF SIMILAR TRIANGLES 133 Exercises. Set L. Projections. Pythagorean Relation 135 Exercises. Set LI. Trigonometric Ratios 140 Summary of Chapter 148 xvi CONTENTS CHAPTER VII THE Locus PAGE A. REVIEW OF ALGEBRAIC IDEA OF Locus 149 Exercises. Set LII. Location of Points 149 Exercises. Set LIII. Applied Problems in Graphic Mathematics 150 Exercises. Set LIV. Graphic Solution of Equations 151 Exercises. Set LV. The Equation as the Statement of a Locus 152 B. PECULIARITY OF THE PROOF OF A Locus PROPOSITION 155 Exercises. Set LVI. Related Statements Direct, Converse, Opposite 155 Theorems 156 Exercises. Set LVII. Applications of Locus 157 Summary of Chapter 159 CHAPTER VIII THE CIRCLE DEFINITIONS 160 A. PRELIMINARY THEOREMS 161 Exercises. Set LVIII. Circle as a Locus 161 B. STRAIGHT LINE AND CIRCLE 162 Exercises. Set LIX. Congruence of Curvilinear Figures 163 Exercises. Set LX. Constructions Based Upon Circles '. . 164 Exercises. Set LXI. Equal Chords 165 Exercises. Set LXII. Tangent and Circle 167 Exercises. Set LXIII. Tangent Circles 168 C. THE ANGLE AND ITS MEASUREMENT 172 Exercises. Set LXIV. Secant and Circle 173 Exercises. Set LXV. Circles 174 Exercises. Set LXVI. Inscribed Angles 175 Exercises. Set LXVII. Measurement of Angles 177 Exercises. Set LXVIII. Tangent and Secant 179 D. MENSURATION OF THE CIRCLE 180 Exercises. Set LXIX. Regular Polygons and Circles 181 Historical Note 183 Postulates of Limits 184 Exercises. Set LXX. Perimeters of Regular Polygons 187 Axioms of Variables 187 Historical Note 188 Exercises. Set LXXI. Value of TT 188 Exercises. Set LXXII. Circumference 189 Exercises. Set LXXIII. Area of Circle 192 Summary of Chapter 195 Miscellaneous Exercises. Set LXXIV 197 PART I FIRST STUDY PLANE GEOMETOY- r j CHAPTER I INTRODUCTION NOTE. For pupils who have had no intuitional geometry section A of the Introduction may be postponed until the work in Areas has been completed p. 77. A. A FEW FACTS CONCERNING THE EARLY DEVELOP- MENT OF GEOMETRY In a cabinet in the British Museum there is a piece of clay some- what over an inch thick and perhaps fifteen inches square which might be referred to as the first book about geometry. Near it, on a roll of papyrus, yellowed by age, is a collection of notes con- taining instructions for finding the contents of areas and solids. When we reflect that this clay tablet and the manuscript are con- siderably over thirty-five hundred years old, we can see that the study of geometry is by no means a modern development. The tablet and the manuscript represent respectively the earliest available records of the geometric knowledge of the Babylonians and the Egyptians. Centuries ago these two races found it neces- sary to devise some means for accomplishing what today seems a very simple undertaking. Perhaps the necessity was forced upon the Egyptians for a reason that does not seem very apparent at first. The River Nile, as we know, rises twice a year and inundates the country bordering on it for many miles. Naturally this flood produces changes in the line of the river banks, and new turns and curves give the adjacent land a very different appearance on each occasion. A farmer whose land bordered the river might therefore find himself one year in possession of a good deal of property, and the next year with much less. This condition, we are told by the historian Herodotus, caused Ranleses II, who was king of Egypt about 1350 B.C., to declare a law: "This king divided the land among all Egyptians so as to give each a quadrangle of equal size, 3 4 PLANE GEOMETRY v . - ; t t ' v and to draw^oln each his revenues by imposing a tax to be levied yearly; but everyone from whose part the river tore away anything hatl to go ta him aiid notify him of what had happened; he then sent the overseers who had to measure out by how much the land had become smaller, in order that the owner might pay on what was left in proportion to the entire tax imposed." As a result of this it became necessary for the Egyptians to employ surveyors who should determine the areas of the land lost or gained, and these surveyors put into practical use such rudimentary knowledge as was then available. Restoration of the Great Hallof Karnak Seven centuries before this Egyptian kings had undertaken large construction operations, the very nature of which showed that their contractors and builders understood the elementary prin- ciples of the science of mensuration. Menes, the first Egyptian king, built a large reservoir and two temples at Phthah and Mem- phis, the ruins of which are still in existence, and under Amenembat III, a later king, the Egyptians designed and constructed a very large irrigating system covering considerable territory and requir- ing a careful calculation of areas, water flow, gradients, etc. Stu- dents of Egyptian art and religion find frequent evidence that this race had a crude knowledge of geometrical principles. The pave- ments of the temples show designs of triangles, squares, five- pointed stars, and rectangles, and the locations of the buildings themselves show geometric knowledge, for their temples were supposed to be constructed with reference to a certain fixed point. The Egyptians were sun-worshippers, and their temples were INTRODUCTION 5 designed to receive sunlight through the doorways at certain times of the day, as a part of the religious ceremonies. It is interesting to note that the movement of the North star has been in a measure demonstrated to our later-day astronomers by the fact that Egyp- tian temples, built three or four thousand years ago and designed to face the North star, are no longer in the perpendicular to it. The Egyptians were astronomers, and in locating their temples used the sun and the North star to establish base lines. The surveyors, called the " harpedonaptse " or " rope-stretchers," fixed the right angle to the north-south line by stretching a rope knotted in three places around pegs. The distances between the knots were in the ratio of 3-4-5, showing that they knew this to be the ratio of sides of a right triangle. Our present day surveyors are still following the same method and have improved upon the method of the Egyptians only by substituting a steel tape for the rope. We mentioned the ancient manuscript of the Egyptians now in the British Museum. The writer of this manuscript was called Aah-mesu (The Moon-born), an Egyptian scribe commonly called Ahmes. The original from which he copied it was probably in existence about 2300 B.C., but has never been discovered. The commercial value of the document is shown by the fact that it contained rules and formulas for finding the capacity of the wheat warehouses constructed in ancient Egypt, as well as a treatise of considerable length on a crude algebraic system. A temple built for the worship of the god Horus on the island of Edfu has on its walls hieroglyphics describing the land which the priests of the temple owned, and the formulas for finding the areas of these plots. Less is known about the Babylonians in these par- ticulars, but so far as we can learn, their geometrical knowledge was used more in the arts than for practical purposes. Their monuments, found in the ruins of Babylon, show geometrical designs, such as a regular hexagon in a circle, and the pictures of their chariots show the wheels divided into sixths. The Babylonians appear to have fol- lowed this division into sixths in their arrangement of the calendar, PLANE GEOMETRY for their year consisted of three hundred sixty days, and they divided the circle into three hundred sixty degrees, on the theory that each degree represented the supposed revolution of the sun round the earth. Although the geometric knowledge of the Egyptians and Baby- lonians may seem to us somewhat crude and simple, we must remember that, as compared with the savage races which sur- rounded them, these people represented the greatest advancement in civilization and scientific knowledge. We see that much of this was due to the very necessities of life; that to build public works, levy taxes, determine boundaries, required a knowledge of the science of mensuration. In the case of the Egyptians, their require- ments, so far as we are able to estimate them, were even broader than those of the Babylonians. The construction of the pyramids shows clearly a geometrical design, executed scientifi- cally, and this work, as well as the erec- tion of wheat ware- houses and storage reservoirs, necessi- tated what was doubtless to them a very advanced conception of the principles of solid geometry. Strangely enough, however, we are indebted to neither of these races for the development of this knowledge into a science, but to a race whose place in history is much later. The Greeks in the ancient world occupied a position in some respects similar to that which America has held in the modern world. They were a people much given to exploitation and expansion, as well as to scientific and philosophical pursuits, and hi addition to this, prided them- selves on their high degree of adaptability. Plato said, "Whatever we Greeks receive we improve and perfect." They did not origin- ate ideas so much as they adopted those of other races and improved upon them to a degree which causes history to associate the Greeks themselves with the original conception. The Greeks were travelers Egyptian Pyramids INTRODUCTION and traders, interested in the arts and sciences, and a distinguishing character- istic of the race was their desire to learn and experiment with new things. Seven hundred years before Christ, Greek mer- chants began sending their ships across the Mediterranean to Egypt. Travelers began to bring back accounts of this other great nation, and the Greeks were imme- diately interested in the reports of what the Egyptians had done and were doing. Thales (640-546 B.C.), a merchant of Miletus, was among those who became acquainted with the Egyptians as a result of commercial intercourse. Thales was at heart a student, and the geometrical theories and practices of the Egyptians interested him. In later life he terminated his business activities and opened a school in his native city, Miletus, where he began teaching the principles of geometrical science as it was then known. The problems with which his pupils concerned themselves would seem elementary to us. They dealt merely with finding the heights of objects or the distances of ships from the shore, but his school, which has come down to us under the name of the Ionic school (so called from the Greek province in which Miletus was situated), was the first intelligent effort to systematize the study of geometry. One of the students in the school of Thales was a noteworthy successor of the first Greek geometrician. Pythagoras (580-501 B.C.) founded a school of mathematics at Crotona in the southern part of Italy. His plan was much more elaborate than that of Thales. Pythagoras felt that the study and character of the school would create a deeper impression if it were organized as a secret society. The watchword was " Silence," and its members were pledged to secrecy as to the nature of the work which was done. The Greek government felt that the secret methods of the school might be used to conceal harmful activities, and finally ordered the institution closed. This circumstance, and the pledge of secrecy 8 PLANE GEOMETRY imposed upon the members of the school, have prevented our learning much about it. The Pythagorean proposition, which states that the square on the hypothenuse of a right triangle is equal to the sum of the squares on the other two sides, bears the name of the school, although the fact was known for a long time before Pythagoras proved it to be true. " Py- thagoras changed the study of geometry into the form of a liberal education, for he examined its principles to the bottom and investigated its pro- Pythagoras positions in an immaterial and intel- lectual manner." Archytas (430-365 B.C.), who followed Pythagoras, was not so much interested in speculative or geometrical subjects as he was in the application of the science to practical uses. He invented several mechanical toys operated on geometrical principles. Very few sailors realize that the ability of one man to move a tremendous weight of sail was made possible by the discovery of this Greek mathematician who lived over twenty centuries ago, for it was Archytas who worked out and applied the principles of the pulley. He is be- lieved to have been the first student to find a solution of the problem called the " duplication of the cube," that is, to find the dimensions of a cube the volume of which shall be twice that of a given cube. Plato (429-348 B.C.) was a contem- porary of Archytas, and his name is associated even more generally with geometrical science than that of his compatriot. Plato called his school the " Academy/' and the underlying prin- pi ato INTRODUCTION 9 ciple of his theory was the abstract and systematic development of geometric science. Plato insisted that the only instruments needed for the study of the subject were the straight edge and the com- passes, and the history of the science has demonstrated the ac- curacy of his conclusions, as they are the only scientific tools needed for all elementary work in the science. Eudoxus (408-355 B.C.) studied under Plato for a time, and subsequently did some independent work in the science. He directed his attention chiefly to the principles of proportion and certain methods of proof, to which we shall make refer- ence later. He was the first scientist to begin to put into book form the mathematical knowledge of his time, and may properly be considered the logical forerunner of the mathemati- cian Euclid. Euclid, who was a teacher in a school of mathematics in Alex- andria, Egypt, about 300 B.C., was the author of what is probably the most famous book on geometry. He collected and arranged all the knowledge of the science down to his own time, and his book still stands today in many respects as a final authority and the background of the entire science. Despite the fact that the Egyp- tians and Babylonians first developed a crude knowledge of the subject, it is perhaps appropriate that the Greeks, whose generations of scientists, culminating in Euclid, gave so much study to it, should have furnished the name by which we call it. The Greek word "ge" meaning the earth, and "metron" to measure, are the roots which formed our name for the science From its first crude beginning in the necessity for measuring the destruction wrought by an ancient river, its instruments, crude pegs and a knotted rope, developed and applied as a science by the mathematicians of five centuries before the Christian era, supple- mented and enlarged by the observations and discoveries of nearly twenty centuries of research, the science by which the surveyors of Egypt located their boundaries is today the method used for deter- mining the power of a battleship or the contents of a mountain range. 10 PLANE GEOMETRY SUMMARY I . GEOMETRY AMONG THE BABYLONIANS AND EGYPTIANS A. DERIVATION OF THE WORD "GEOMETRY." B. EVIDENCES OF KNOWLEDGE OF GEOMETRY. 1. Among the Babylonians. a. Documentary evidence. (I) Clay tablets. (II) Talismans. (Ill) Monuments. 2. Among the Egyptians. a. Evidences in practical life. (I) Surveying. (II) Reservoirs. (III) Irrigation. (IV) Pavements. 6. Evidences in religious life. (I) Orientation of temples. (II) Pyramids, c. Documentary evidence. (I) Ahmes Papyrus. (II) Hieroglyphics II. GEOMETRY AMONG THE GREEKS. A. SOURCE. B. SCHOOLS THAT CONTRIBUTED TO THE DEVELOPMENT OF GEOMETRY. 1. Ionic School, a. Thales (640-546 B.C.). . (a. Pythagoras (580-501 B.C.). 2. Pythagorean School ], A J , IMOA o (6. Archytas (430-365 B.C.). Q Plot Q fc i fa- Plato (429-348 B.C.). 3. Platonic School i . j /^oorr \ ( 6. Eudoxus (408-355 B.C.) n /-, -i f o. Hippocrates (C. 440 B.C.). C. Compilers ] , _ * (b. Euclid (C. 300 B.C.). BIBLIOGRAPHY ALLMAN, G. J.: "Greek Geometry from Thales to Euclid": pp. 2, 3, 5, 7, 15, 16, 22, 29, 41, 139-140, 143 (footnote) and 154. BALL, W. W. R.: "A Primer of the History of Mathematics": pp. 3-6, 8-14, 32, 42-48. BALL, W. W. R.: "A Short History of Mathematics": pp. 5-8, 15-16, 25-27, 40, 54-63, 69-72. FINK, K.: "Brief History of Mathematics": pp. 190-214 (with omissions). Gow, J.: "A History of Greek Mathematics": pp. 176-178. BOYER, J. F.: "HistoiredesMathematiques": Chapters I-V. CANTOR, MORITZ: "Vorlesungen iiber Geschichte der Mathematik": Vol. I, pp 17-52, 90-114, 134-146, 170-187, 201-277. INTRODUCTION 11 EXERCISES. SET I. BASED UPON HISTORIC FACTS 1. From the derivation of the word geometry, can you suggest any studies or professions in which geometry may be applied? 2. What gave rise in the first place to the art and eventually to the science of geometry? 3. From the little told you in the foregoing paragraphs and any references you may have read, what would you judge to be the essential difference between the geometry of the Egyptians and the geometry of the Greeks? 4. Judging from the character of the Roman, would you expect him to do much to advance the science of geometry? 5. Among the formulas given in Ahmes Papyrus for determining areas are the following: I. The area of an isosceles triangle equals half the product of the base and one of the equal sides. II. The area of an isosceles trapezoid equals half the product of the sum of the bases, and one of the equal sides. a. Using 6, 61, for the bases, and s for each of the equal sides write an algebraic formula for each of these areas, A. b. What is the error in each of these formulas? c. Draw figures to show that at times this error would not matter much. d. Draw figures showing cases where the error would make a considerable difference. ,/_ _ T ,1 ... e. In the accompanying dia- gram find j ust what error is made a (correct to tenths) by using the Egyptian formula. k =9 6. Another formula given in Ahmes Papyrus is that for finding the area of a circle. It tells you to multiply the square of the radius by !%. What value must the Egyptian then have assigned to ?r? 7. What is meant by saying that a 3-4-5 triangle is a right triangle? 12 PLANE GEOMETRY 8. Show by knotting a piece of cord so that the parts have the ratio 3 to 4 to 5 how the Egyptian " rope-stretchers " obtained their east-west line. (Stretch the cord around pins on a board and after it is in place test the accuracy of the method with your right triangle.) 9. Plato and his school interested themselves in the so-called Pythagorean numbers. Such numbers are those that would repre- sent the lengths of the sides of a right triangle. In this kind of triangle they must be such that c 2 . The school of Plato found that +1, ft, and (-J^) 2 1 were Pythagorean numbers. a. Verify the statement. b. Find ten sets of Pythagorean numbers. 10. Pythagoras himself found that n, %(n 2 1), and M(ft 2 +l) were numbers such as described in the last exercise. Verify this statement. 11. Bramagupta, a Hindu writer of the seventh century, gave P> J(gfq)> and ^2(3 ~q) as Pythagorean numbers. a. Give various values to p and q to test his statement. b. Verify his statement. 12. In the Culvasutras, a Hindu manuscript, directions for con- structing a right angle are as follows: Divide a rope by a knot into parts 15 and 39 units in length respectively, and fasten the ends to a piece 36 units in length. a. Draw a diagram to show what is meant by this. 6. Check to see whether these are Pythagorean numbers. c. Is it true that all numbers having the ratio of these three are Pythagorean numbers? 13. Archimedes proved that the value of TT lay between 3)^ and How does this compare wiih the value we use to-day? B. A FEW ILLUSTRATIONS OF GEOMETRIC FORM Before we begin a systematic study of geometry, let us see if we can find any illustrations of the kind of forms about which we hope to learn something. Do we not find such forms in nature? We INTRODUCTION 13 recall symmetrical trees and conical mountains; we think of a circular moon, spherical raindrops and crystals of many forms. Symmetrical tree Conical mountain Crystals EXERCISES. SET II. ILLUSTRATIONS OF GEOMETRIC FORMS 14. Make a list of some geometric forms you have found in nature, under the following heads: spherical, conical, cylindrical, prismatic, circular, etc. Aside from natural objects, geometric forms continually appear in the works of man. The building and room in which we are, the furniture, windows, doorways, all are geometric in form. The familiar objects of our daily life coins, boxes, cylindrical tubes, balls all illustrate the application of geometrical principles. 14 PLANE GEOMETRY 15. As in the preceding exercise, make a list of some geometric forms you can find in the works of man. C. MEANING OF GEOMETRIC FORMS If we note these forms carefully, we see that they are various combinations of the simple elements points, lines, surfaces and solids. At the outset, therefore, we should be sure that our ideas about these elements are correct. First let us consider a geometric solid. We have seen cones made of wood, and others of ice cream; we have looked into a well and said it was cylindrical; we have watched a soap bubble and called it spherical. So we see that it is the shape or form, and not the substance of which an object is made, to which we refer in speaking of a geometric solid. When we mention a sphere, we mean the space which it occupies or its shape without reference to its physical properties or the material of which it is made. If, as we have just noted, a solid is a limited portion of space, what limits it? How is a solid separated from the rest of space? The boundaries of a solid are surfaces, and since a solid is identified, not by its material, but only by its shape, so must its boundary be identified by its shape. A chalk box, for instance, is in the form of a prism, i.e., the space it occupies is a geometric prism. The boundaries of this prism are called surfaces, and they divide it from the rest of space. The surfaces meet and form lines. The edges of the chalk box are referred to as the union of its sides. Now if we think of the geometric prism the space occupied by the box the intersection of the surfaces are lines. It is evident, then, that lines crossing form points. A limited portion of space is called a solid, the boundaries of a solid are called surfaces, the intersections of surfaces are called lines, and the places where lines cross are called points. We see thus that a point is a place or position, and can, therefore, have no length, breadth, or thickness. For convenience, we repre- sent a point by a dot of lead, ink, or chalk. Such a dot is obviously not a point, because it has some size, however small it may be, but it marks a location, which is the real point. INTRODUCTION 15 If we could imagine a point to move, we would call its path a line. A line, then, would have no width, but it would have length. If we could now imagine a line to move, not along itself, we would say it generated a surface. Again, the surface would have only length and width, but, of course, no depth. If we consider a surface to move, not along itself, it would form a solid, which would have three dimensions. In our study of geometry we shall have to deal with straight and curved lines. Let us note, then, that a straight line is one which is fixed (or determined) by any two of its points, and a curved line is one no part of which is straight. Throughout the book, it is to be understood that when the word line is used without a qualifying adjective, a straight line is designated. A line is of indefinite length, so that when we wish to refer to a limited portion of a line we shall call it a sect. All the facts with which we shall be concerned for a time will be those relating to a single plane. A plane is a surface such that if any two points in it be connected by a straight line, that line lies wholly within the surface. EXERCISES. SET III. MEANING OF GEOMETRIC FORMS 16. If a series of 600 points were put within an inch would they form a line? 17. A machine has been manufactured which will rule 10,000 dis- tinct lines within the space of one inch. Are these lines geometric? 18. Fold over a piece of paper. What will the crease represent? 19. If oil is poured on water, of what material is the surface formed? 20. Put your foot in a heap of snow and quickly withdraw it. Is the impression that is left a physical or geometric solid? 21. If I place a piece of red paper on a blue one, what is the color of the surface between them? 22. If 1000 geometric surfaces were placed one on top of the other would a geometric solid be formed? 23. Is a cake of ice a geometric solid? 24. (a) Make a list of some things in life which are referred to as points. (6) How many of these are geometric points? 25. (a) Make a list of some things in life which are referred to as lines. (6) How many of these are geometric lines? 16 PLANE GEOMETRY D. SUGGESTIONS OF A FEW USES OF GEOMETRY We have reviewed briefly the historical development of geometry and have called to mind illustrations of geometric forms, both in nature and in manufactured articles and have clarified our ideas of these forms. Let us now consider a few of the uses of geometry. The subject grew out of the need of land-measuring. Hence, historically at least, surveying is the first known use of geometry. The following illustrations show some of the things that we ought soon to be able to do. Laying out bounda- ries of property so that the owner shall have his just share, or finding the areas of pieces of ground, Diagram i are problems requiring practical mensuration A B Diagram II Diagram III Transit about which we shall soon study. Finding the distances between inaccessible points, as from X to Y in diagram I, across rivers and INTRODUCTION 17 over swamps, as in Exercises 111-113, and heights of objects as in diagrams II and III, are all geometric problems such as are included in surveying. We have all seen men in our streets with transits on tripods. (A transit is an instrument for measuring angles.) Geometry is necessary to solve the problems for which they are getting the data. EXERCISES. SET IV. MENSURATION 26. Make a list of mensuration formulas with which you are already familiar. 27. Find the area of the following piece of property (three lots). The measurements taken by a surveyor are noted on the diagram. Other surveying problems will be found later in the book. 65' We do not yet know enough geometry to solve many such problems. 50 ' 95 ' Another use of geometry that quickly comes to the mind is designing. We mark the use of geometric design in parquet floors, linoleums, tilings, wall and ceiling papers, grill-work, stained-glass windows, arches, and in similar objects. By learning how to make five fundamental constructions (Exercises 28-32, inc.), we shall be able to combine them into many geometric designs, and thus get a clearer idea of one of the uses of geometry. The reasons why these constructions are correct, and more elaborate work in design, must be postponed until later in the text. EXERCISES. SET V. CONSTRUCTIONS DESIGNING All the constructions in these exercises are to be made with the use of compasses and unmarked straight edge only. The pupil is reminded that unfamiliar technical terms will be found by refer- ring to the index. In general, in geometry, auxiliary lines (those needed only as aids) are indicated by dotted lines, preferably light. 28. From a given point on a given straight line required to draw a perpendicular to the line. Let AB be the given line and P be the given point. It is required to draw from P a line perpendicular to AB. 2 18 PLANE GEOMETRY \X Y / With P as center and any convenient radius strike arcs cutting AB at X and F. With X as center and XFas radius strike an arc, and with F as center and the same radius strike another arc, and call one intersection of the arcs C. With a straight edge draw a line J g through P and C, and this will be the perpendicular required. 29. From a given point outside a given straight line required to let fall a perpendicular to the line. Let AB be the given straight line and P be the given point. It is required to draw from P a line perpendicular to AB. With P as center and any convenient radius describe an arc cutting AB at X andF. With X as center and any convenient radius describe an arc, and with F as center and the same radius describe an- other arc, and call one intersection of the arcs, C. With a straight edge draw a straight line through P and C, and this will be the perpendicular required. It is interesting to test the results in Ex- ercises 28 and 29 by cutting the paper and fitting the angles together. 30. Required to bisect a given sect. Let AB be the given sect. It is required to bisect AB. With A as center and AB as radius de- scribe an arc, and with B as center and the same radius describe another arc. Call the two intersections of the arcs X andF. Draw the straight line XY. Then XY bisects the sect AB at the point of intersection M. INTRODUCTION 19 31. From a given point on a given line required to draw a line making an angle equal to a given angle. Let P be the given point on the given linePQ, and let angle A OB be the given angle. What is now required? With as center and any radius de- scribe an arc cutting A at C and BO at D. With P as center and OC as radius de- scribe an arc cutting PQ at M. With M as center and CD as radius de- scribe an arc cutting the arc just drawn at N, and draw PN. Then angle MPN is the required angle. 32. Required to bisect a given angle. Let A OB be the given angle. It is required to bisect the angle AOB. With as center and any conve- nient radius strike an arc cutting OA at X and OB at Y. With X as center and the sect XY as radius strike an arc, and with Y as center and the same radius strike an arc, and call one point of intersection of the arcs P. Draw the straight line OP. Then OP is the required bisector. 33. Make constructions similar to the following : Suggestion: AB=OB. 20 PLANE GEOMETRY 34. Draw the fol- lowing figure. 35. Make a con- struction similar to a. d36.* Copy b. a b 37. From a study of Exercise 28, suggest how to erect a per- pendicular at the end of a sect. 38. On a given sect construct a square. 39. These figures show a parquet floor design, and one of the units of the design enlarged. Construct figures similar to these. ABCD is a square, and X, Y,Z, and IF are the mid-points of the semidiameters OE, OF, OG, OH, respectively. G H A E 40. Make a construction similar to the ad- joining figure. The vertices of the square are used as centers for four of the arcs. The radius equals one side of the square. 41. Make a construction similar to a. / 42. Make a construction similar to b. a * As here, d will be prefixed to any exercise which the student is likely to find difficult at this stage. INTRODUCTION 21 d43. Make a construction similar to the following: The figure is based on an equilateral triangle, the centers of the interior arcs being the midpoints of radii drawn to the vertices of the equilateral tri- angle inscribed in a circle (i.e., having its ver- tices on the circle) . Note: See Fig. 2, exercise 33. N. B. The design shown in MableSykes, "Source- Hook of Problems for Geometry," page 160, II, 5 (Fig. 138a), shows a good application of exercise 43. 44. Make a construction similar to a. d45. Make an ornamental design similar to b. The circle is divided into how many equal arcs? How many de- grees in each central angle? What kind of triangle is formed by two consecutive radii and the sect j oining their ends? What other method does this suggest of dividing a circle into six equal arcs? 46. Make an ornamental drawing similar to the one in the accompanying figure. Describe the construction. Suggestion: First draw an equilateral poly- gon with six sides in a circle. 47. Construct a six- __j_ pointed star. 48. Bisect each of the four right angles formed by two lines intersecting each other at right angles. 49. Make construc- tions similar to the following : 22 PLANE GEOMETRY 50. Make constructions similar to the following: In such figures artistic patterns may be made by coloring vari- ous portions of the drawings. In this way designs are made for stained-glass windows, oil-cloth, colored tiles, and other dec- orations. 51. Draw a sect of any convenient length, and upon it construct a design similar to the one in the figure. 52. On a line LM take a sect AB. Divide it into 8 equal parts. With your compasses make an ornamental scroll as shown in the diagram. 53. Using the hints given, make a copy of the accompany- L ing outline drawing of a Gothic window. The arc BC is drawn with A as center and AB as radius. The small arches are described with A, D, and B as centers and AD as radius. The center P is found by taking A and B as centers and AE as radius. How may the points D, E, and F be found? 54. In many different machines, such as the sewing machine, printing press, A F D E B etc., there is a wheel called a cam, which is used to modify the motion of the machinery. Cams are con- structed in various shapes and dimensions, depending upon the use for which they are designed. The figure shows the method of draw- ing the pattern of a heart-shaped or " uniform-motion " cam. Let INTRODUCTION 23 Divide A B into eight equal . . . , B, draw circles with the " throw " be AB and the center 0. parts at C, D, etc. Through A, C, D, centers at 0. Draw sects dividing the angular magnitude around into six- teen equal parts. Beginning at A, mark the points where the consecutive circles and consecutive sects intersect, and through these points draw a smooth curve, as in the figure. Draw such a cam with AB equal to a given sect m, and OA equal to a given sect n. (Taken with modifications from Stone-Millis, Elementary Plane Geometry.) 55. Select and copy some geometric design. 56. Make an original design based on the fundamental construc- tions given in exercises 28-32. Pupils particularly interested in this part of the work are referred to: Sykes, Mabel, "Source Book of Problems in Geo- metry." (Pub. Allyn and Bacon.) Geometry is used in architecture. Whether the architect is drawing the plans for an ordinary dwelling-house or a massive cathedral, he is constantly concerned with geometric forms and constructions. Consider for a moment what problems of this character must have confronted the architect of some large building in our community. The list of the direct uses of geometry would be very long if complete. In a few sentences let us, t therefore, simply enumerate a few more miscellaneous suggestions for its uses. Problems scattered throughout this book show more concretely how geometry is used in the cases enumerated. In making all kinds of diagrams, reducing and enlarging maps, the principles of geometry are applied. In engineering, geometry is needed for such matters as laying out railroads, and planning the constructions of machines, bridges, and tunnels; and in astronomy, ascertaining the altitude of stars and similar problems require geometric principles. 24 PLANE GEOMETRY EXERCISES. SET VI. SOME OTHER USES OF GEOMETRY 57. State any other uses of geometry which you know. 58. At the entrance to New York Harbor is a gun having a range of 12 mi. Draw a line inclosing the range of fire, using any convenient scale. 59. Two forts are placed on opposite sides of a harbor entrance, 13 mi. apart. Each has a gun having a range of 10 mi. Draw a plan showing the area exposed to the fire of both guns, using any convenient scale. 60. Make an accurate diagram of a tennis court or a foot-ball field noting all lime lines. d61. Draw to a convenient scale a plan of the ground floor of your school building. E. THE BASIC PRINCIPLES OF GEOMETRY As in all scientific work of an exact nature, the discoveries in geometry rest upon a few basic principles. These may be classified under three heads : definitions, axioms, and postulates. You all can probably recall having heard people argue most heatedly about some question and reach no conclusion at all. This is often the case simply because when two people argue, they very often do so without having clearly in mind the conditions about which they are arguing. In all debates or discussions it is essential that we start with the same premises, and our work in geometry should help us to learn to collect our premises in orderly fashion. The premises upon which the early parts of geometry rest are to a great extent definitions, and it is therefore very necessary that we have a clear image and definition of each new technical term we meet. The wording of our definitions may differ, but the con- tent must be the same. Every good definition should include all that may fall under a particular class, and clearly exclude all that does not fall under that class. Suppose, for instance, we want to define the word botany. We might say, to begin with, that it is a science but we have not differentiated it from the physical sciences, so we say it is a natural science. But, then, so is zoology. Hence it is necessary to differentiate still further, and say it is the natural science which deals with plant life. Now have we fully INTRODUCTION 25 and finally defined it? Can you possibly think of any science now that it is thus defined with which it can be confused? Make all possible tests, and if you find that it cannot be confused with any other science, well and good, then we have found an accept- able definition. Throughout this book no word will be defined until we are ready to make use of it, but then it will be our duty to see the word in its full meaning. From the knowledge we have of algebra, we already know what some of the axioms are, and some uses to which they may be put, though we may not have defined the word " axiom." Some of the axioms for which we shall have immediate use are : 1. The sums of equals added to equals are equal. Example If 5 =5 and a=b then 5+0^5+6. 2. The remainders of equals subtracted from equals are equal. Example If a=c and b=d then ab = cd 3. The products of equals multiplied by equals are equal. Example If a=x and b=y then ab=xy. 4. The quotients of equals divided by equals are equal. Example If x=y and m=p then -^ m p Cases in which the divisor is zero will not be considered in this text. 5. A quantity may be substituted for its equal in a statement of equality or inequality. Example If x = 5 If a = b and x+y=7 and 2a+5>a+2 then by substitution 5+y=7 then 26 + 5 > a + 2 26 PLANE GEOMETRY 6. Two quantities which are equal to equal quantities, are equal to each other. Example If a=b b=c and c=d, then a=d 7. The whole is equal to the sum of Us parts. T. a . a . a Example 3 6^ NOTE: "part" is here used in the sense of a common or vulgar fraction. Other axioms will be stated as we need them. Thus we see that an axiom is the statement of a general mathematical truth which is granted without any proof. A postulate is simply a geometric axiom. That is, it is the state- ment of a geometric truth which is granted without any proof. We shall now attempt to formulate a few such truths. EXERCISES. SET VII. ILLUSTRATIONS OF POSTULATES 62. Why is it shorter to cut across a field than to go around it? g63.* How many pairs of roots are there when two simultaneous linear equations are solved? What is the graphic explanation of this? g64. Why is it that in making the graph of a linear equation, such as x-\-y = 13, we need to plot but two points, and that a third point may be used to check the correctness of our work? 65. Why is it that the Panama Canal is a great advantage over the route formerly used to reach a point on the western coast of South America from the West Indies? 66. Why is it that in putting up a croquet set all one needs to do to get the wickets in line with the stakes is to tie a string tightly to one stake and stretch it and fasten it to the other stake? * As here, g will be prefixed to any exercise in this text which presupposes an acquaintance with the graph. The pupil is here referred to M. Auerbach, An Elementary Course in Graphic Mathematics (Allyn and Bacon), pp. 29-31, for review, and previous pages in the same if the subject is new to him. INTRODUCTION 27 67. What does the bricklayer do to get a row of bricks in a straight line? Why? Does the gardener do anything similar to this? 68. Point out which of the following postulates upholds each of your answers to exercises 62-67. POSTULATES OF THE STRAIGHT LINE NEEDED IN PROOFS 1. Two intersecting straight lines determine a point. 2. Two points determine a straight line. 3. A straight line is the shortest distance between two points. 69. How often can two straight lines intersect? 70. Use your answer to the last question to state postulate 1 in another way. 71. How many straight lines can be drawn between two points? 72. Use your answer to the last question to state postulate 2 in another way. 73. Give at least one good illustration of how each of the three postulates mentioned may be used in practical life. We have used several other postulates in making some of the constructions on pages 13 to 18. They are: (1) A sect may be produced indefinitely. (2) A circle may be described with any point as center and any sect as radius. (3) A point and direction determine a straight line. But these are so exceedingly obvious that we shall not feel obliged to quote them. DEFINITIONS An angle is the opening between two lines. The lines are called the sides of the angle, and the point at which they meet the vertex. An angle may be named in any one of three ways as < A in Fig. 1 where there is no danger of confusion, or as in Fig. 2 *ABC, , and how much longer. Then test your estimate by measuring as in (a). (c) Look at this figure and state whether . D A B and CD are both straight ines. If one is not straight, which one is it? Test your an- swer by using a ruler or the folded edge of a piece of paper. (d) Look at this figure and state A ////////////////////, whether AB and CD are the same dis- tance apart at A and C as at B and D. Then test your answer as in (a) . 3 34 PLANE GEOMETRY (e) Look at this figure and state whether AB will, if prolonged, lie on CD. Also state whether WX will, if prolonged, lie on YZ. Then test your answer by laying a ruler along the lines. A (/) Look at this figure and state which of the three lower lines is AB prolonged. Then test your answer by laying a ruler along AB. (g) In the figures below, are the lines everywhere the same distance apart? Test your answer by using a ruler or a slip of paper. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (h) In the diagrams given below, tell which' sect of each pair is the longer, a or 6, and test your answer by careful measurement. {Z (i) In the figures below, tell which lines are prolongations of other lines. Test your s /f answers. i ^~i 104. (a) Draw two un- X \ / [ equal supplementary-adja- L ~7 cent angles. * (b) Extend the common side of these angles through the vertex, and call the angles thus formed a, & y, d. INTRODUCTION 35 (c) What relation exists between a and 0? (d) What relation exists between and 7? 0) What further relation do you notice that is based upon the relatons stated in (c) and (d)? (/) Do the same, using angles 0, 7, 5 - (0) Do the same, using angles 7, 6, a. (&) What conclusion can you draw? 105. If a plumb-line is fastened to a horizontal wire nail at the vertex of the angle of a quadrant, and the quadrant is turned so that the plumb-line falls along 90 (here indicated by OA), by noting where the shadow of the nail strikes the quadrant the angular alti- tude of the sun may be obtained. Ex- plain why OC in the diagram gives the angular altitude of the sun. INSTRUMENTS FOR MEASURING ANGLES Surveyors and engineers em- ploy for measuring angles costly instruments called theodolites.* An inexpensive substitute for a theodolite is shown in the accom- panying figure 1. It consists of * The identical theodolite with which the historic M ason-Dixon line, be- tween Maryland and Pennsylvania was run, settling a controversy of a century growing out of the overlapping charters of Charles I to Lord Baltimore and Charles II to William Penn, has lately become the possession of the Royal Geo- graphical Society, of London, through Edward Dixon, descendant of Jeremiah Dixon, who used it. Mason and Dixon had observed, for the Royal Society, the transit of Venus at the Cape of Good Hope in 1761, and did their American work two years later. When the line was resurveyed 150 years later, by the Coast and Geodetic Office of Washington, it was proved to be exceptionally accurate, with no errors of latitude of more than two or three seconds cer- tainly a creditable result for the time and the primitive instrument with which the work was done. The Mason-Dixon theodolite has two sights, a large com- pass in the center of its horizontal plate, and is adapted for measuring either horizontal circles or magnetic bearings. The graduated circle is twelve inches in diameter, divided into five minutes, and read by a single vernier. B FIG. 1 36 PLANE GEOMETRY FIG. 2 two pieces of wood shaped like rulers mounted on a vertical axis, by a pin driven through their exact centers. The vertical needles inserted near the end of the rulers are used for sighting. In place of the needle near- est the eye, it is better to employ a thin strip of wood, A, having a fine vertical slit; and in place of the other needle, a vertical wire fixed in a light frame, B. By the help of this instrument, and a protractor, one can measure with considerable accuracy an angle on the ground; for instance, the angle MON (figure 2) . The following is a simple substitute for the theodolite. By means of it angles may be measured in both horizontal and vertical planes. The vertical rod MM' is free to revolve in the socket at M , carrying a horizontal pointer which indicates readings on the horizontal circle divided into degrees. These divisions must be marked. The pointer at M' is provided with sights and is free to move in a vertical circle around M 1 '. By sighting along this pointer, vertical angles may be measured on the quadrant. (This instrument was suggested in Betz and Webb, Plan e Geometry .) EXERCISES. SET XI. INSTRUMENTS FOR MEASURING ANGLES 106. Construct an instrument such as that shown in figure 1 of the precedin g section or an astrolabe or a good substitute by means of which angles may be measured in vertical and horizontal planes. F. THE DISCOVERY OF SOME FACTS AND THEIR INFORMAL PROOF A theorem is the statement of a fact which is to be proved. The fact which you discovered if you worked exercise 104 and applied in 105 is one which was known to Thales about 600 B.C. INTRODUCTION 37 It is very important, so we shall attempt to prove it again. It is the first theorem in our syllabus. Theorem 1. Vertical angles are equal. In the accompanying .diagram, what angle is the supplement of ^BOZl What other angle is the sup- plement of OC? 133. Prove the directions given for bisecting an angle correct. 134. Prove that the directions for erecting a perpendicular to a line at a given point in it are correct. Can you suggest any varia- tion in this construction such as is pointed out in exercise 132? d!35. Prove that the directions for bisecting a sect perpendicu- larly are correct. Compare this construction with the bisection of any angle. 136. Prove that the directions for dropping a perpendicular from a point to a line are correct. d!37. In the sixteenth cen- tury, the distance from A to *^ the inaccessible point B was "~ B found by use of an instrument consisting of a vertical staff AC, to which was attached a horizontal cross bar DE that could be moved up and down on the staff- Sighting from C to B, DE was lowered or raised until C, E and B were in a straight line. Then the whole instrument was re- volved, and the point F at which the line of sight CEi struck the ground again was marked, and FA measured. Show that FA = AB- (This exercise is taken with modifications from Stone-Millis, Plane Geometry.) 46 PLANE GEOMETRY d!38. Cor. 1, theorem 4, is known as the "Pons asinorum," or "Bridge of Asses." Its discovery is attributed to Thales. The proof suggested in this exercise, however, is due to Euclid. He produced BA and BC, the equal sides of the triangle to D and E, so that BD=BE. Then he proved (1) &BCD^ABAE,C2) &ACD ACAE, and so, by subtracting *ACD from ^BCD, and < CAE from < BAE found < BAG = < CA . Give the details of the proof. LIST OF WORDS DEFINED IN CHAPTER I Solid, surface, line, point; straight, curved line, sect; plane. Angle, vertex, sides; straight, right, acute, obtuse angles; complementary, supplementary angles; adjacent, vertical angles; perpendicular. Polygon, vertices, sides, angles; triangle; scalene, isosceles, equilateral triangles; acute, right, obtuse triangles. Congruent, homologous. Theorem, corollary, axiom, postulate. SUMMARY OF AXIOMS IN CHAPTER I 1. The sums of equals added to equals are equal. 2. The remainders of equals subtracted from equals are equal. 3. The products of equals multiplied by equals are equal. 4. The quotients of equals divided by equals are equal. 5. A quantity may be substituted for its equal in a statement of equality or inequality. 6. Two quantities which are equal to equal quantities are equal to each other. 7. The whole is equal to the sum of its parts. SUMMARY OF POSTULATES IN CHAPTER I Straight Line 1. Two intersecting straight lines determine a point. 2. Two points determine a straight line. 3. A straight line is the shortest distance between two points. Angle 4. All straight angles are equal. Cor. 1. All right angles are equal. Cor. 2. Complements of the same angle or equal angles are equal. Cor. 3. Supplements of the same angle or equal angles are equal. Superposition 5. Any geometric figure may be moved about without changing its size or shape. INTRODUCTION 47 SUMMARY OF THEOREMS PROVED IN CHAPTER I 1. Vertical angles are equal. 2. Triangles are determined by two sides and the included angle. 3. Triangles are determined by two angles and the included side. 4. The bisector of the vertex angle of an isosceles triangle divides the triangle into two congruent triangles. Cor. 1. The angles opposite the equal sides of a triangle are equal. Cor. 2. The bisector of the vertex angle of an isosceles triangle bisects the base, and is perpendicular to it. Cor. 3. An equilateral triangle is equiangular. Cor. 4. The bisectors of the angles of an equilateral triangle bisect the opposite sides and are perpendicular to them. Cor. 5. The bisectors of the angles of an equilateral triangle are equal. I 5. Triangles are determined by their sides. CHAPTER II THE PERPENDICULAR, THE RIGHT TRIANGLE AND PARALLELS A. THE PERPENDICULAR We have been studying the congruence of triangles in general, and as a necessary and interesting part of that topic we have considered some properties of the isosceles triangle. Now we are to give our attention to another special kind, the right triangle, but before doing so, we need to know more than we do about perpendiculars. Two facts that we should note at the beginning are sufficiently obvious to permit our accepting them without proof, i.e., postu- lating them. Postulates of Perpendiculars. 1. At a point in a line only one perpendicular can be erected to that line. 2. JZwmapoint outside a line only one perpendicular can be drawn jjine. icause of this property of perpendiculars, by the distance from point to a line is meant the length of a perpendicular from the point to the line. TEeTess we postulate and the more we prove, the more scientific is our work. Hence, later in our study of geometry, we shall prove these postulates of perpendiculars. SET XVI. DISTANCE FROM A POINT TO A LINE 139. Prove the familiar fact that the image of an object in a mirror appears to be as far behind the mirror as the object is in front of it. Hints: (a) It is proved in physics that a ray of light striking a plane surface is reflected from it at the same angle as it strikes it. Assume that fact here, (b) i, M, R lie in a straight line. See the diagram, (c) Prove triangles congruent. Before proving another important property of perpendiculars we must add to our list of axioms. 48 PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 49 Axioms of Inequality. 1. // unequals are operated on by positive equals in the same way, the results are unequal in the same order. In other words, if we add a positive number to each of two unequal numbers the sums will be unequal in just the same way that the original numbers were unequal, i.e., the greater num- ber increased will still be greater than the smaller number increased; e.g., 7>5 and 7+2>5+2. Would the same be true if we started with 7> 5, or -7< 5? Give your reason. If instead of adding equal numbers to the two unequal num- bers, we had subtracted equal numbers from each, or divided or multiplied each by equal numbers, the remainders, quotients, or products would have been unequal in the same order as the original numbers. 2. // unequals are subtracted Jrom equals, the remainders are unequal in the reverse order. Example. 10=10 and 6>3 .'. 4<7 EXERCISES. SET XVII. NUMERICAL INEQUALITY 140. State in algebraic symbols the above axioms of inequality. 141. (a) Why is it that if -2< -1, ( -2) 2n >( -I) 2 "? (6) Why is this not an exception to our axiom? We are now ready to prove another important fact about per- pendiculars which will give us another reason for measuring dis- tance from a point to a line by means of the perpendicular. Theorem 6. The perpendicular is the shortest sect that can be drawn from a point to a line. To show that from P a point outside of ABj the shortest sect that we can draw to AB is the perpendicular PQ, let us draw any other sect, say^P/S to A Q_ AB, and then show that PQ e just discussed might suggest, however, that we can derive the conclusion we wish by showing that 2PQ<2PS. Then, we also recall that we ha\>e listed one fact concerning the inequality of sects and that is the postulate that a straight line is the shortest distance be- tween two points. Thus we are led to make the construction shown in the diagram^ i.e., produce PQ to R so that connect R and S. Why? Finish the proof. Now PS+RS=2PSti PS=RS. Try to prove PS=RS. How do we prove sects equal? EXERCISES. SET XVIII. INEQUALITY OF SECTS 142. Given the point P outside AB and L in the line. Which is shorter, the distance from P to L or the distance from P to AB1 -P 143. Which is the longest side of a right triangle? Give a reason for your - : j -- ^ answer. B. THE RIGHT TRIANGLE We are now ready to go on with the study of the congru- ence of triangles by noting two cases of the congruence of right triangles. EXPERIMENT Construct a right triangle given the side opposite the right angle, called the hypotenuse, and an angle adjacent to it. Let us see if we have data sufficient to determine the triangle. (a) What is the only side of the triangle that is known? (6) Where then will you have to start the construction? (c) Is the direction of a second side fixed? Why? (d) Is the direction of the third side then fixed? Why? We are thus led to expect : *Theorem 7. The hypotenuse and adjacent angle determine a right triangle. PERPENDICULAR, RIGHT TRIANGLE, PARALLELS* 51 Let us convince ourselves of the truth or falsity of this conclusion by actual proof. Given: A ABC and AXYZ with 4^ and 4F each a rt. 4 and 4C = 4Z and AC = XZ. To prove: &ABC = &XYZ Outline of proof : Place AABC on &XYZ so that A(? #_ ~~C coincides with its equal XZ and CB falls along ZY. do this? Y Z What right have we to Then AB will faU along X Y. Why? Finally B will coincide with Y. Why? The other case of congruence of right triangles is: Theorem 8. The hypotenuse and another side determine a right triangle. Given: AABCand A^FZwith AB=XY } AC^XZfmd 4# and 4Feach art. 4. Toprove: AAC AXYZ. A Suppose we place &XYZ next to AABC (as in the fol- lowing diagram) with XY coinciding with AB. (Do we know that we can?) (1) Why will the figure formed be a triangle? ^ (2) What kind of triangle will be formed? (3) Can you now throw this theorem back to the previous one? C. PARALLELS Parallels, or parallel straight lines, are coplanar lines (lines lying in the same plane) which never meet. Do you see any reason for emphasizing the fact that the lines must lie in the same plane? Take two pencils and hold them so that they would neither meet if continued nor be parallel. Draw a line on a piece of paper and erect two perpendiculars to it. Do these perpendiculars appear to be parallel? Since they lie in the same plane they must either be parallel or meet. Can they meet? Give a reason for your answer. This leads us to state another theorem for our syllabus, one which is frequently used in mechanical drawing. 52 PLANE GEOMETRY Theorem 9. Lines perpendicular to the same line are parallel. EXERCISES. SET XIX. PARALLELS 144. What principle is a carpenter using when he lays off parallel lines on a board by moving one arm of his square along a straight edge of the board, and marking along the other arm? 145. What is the principle involved in the use of the T-square for drawing parallels? d!46. The accompanying picture shows a carpenter's plumb-level, the forerunner of the spirit-level. AE and EB are strips of wood of equal length. CE=ED and is the midpoint of CD. A and B rest on the points to be levelled, and they are found to be level when EF passes through 0. Explain. Before proceeding with our study of parallels, we need the Postulate of Parallels. Through a given point only one line can be drawn parallel to a givtn line. Not both lines a and b can be . parallel to c. Why? Cor. 1. Lines parallel to the same line are parallel to one another. Suggestion for proof: If lines a, b, both parallel to c, should meet how would the postulate of parallels be violated? *Theorem 10. A line perpendicular to one of a series of par- allels is perpendicular to the others. Given: AXB \\ CYD\. JXYF _L AXB. To prove: EFA.CD. Suggestions: Draw YKEF. What relation will exist between YK and AB1 Why will YK and CD coincide? What relation exists between CD and EF1 Why? Why is it necessary to consider only two parallels? * When naming a line by more than two of its points it is necessary to use a bar over the letters. In the case of two points it is immaterial. Why? f How do the statements given show that EF cuts AB and CD? C D PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 53 EXERCISE. SET XIX (continued) 147. Representing a series of lines by p, q, r, . . . . construct the following figures, stating in each case all possible relative positions of the first and last line of the series : (a) p || q, p || r. (b) p \\ q, q \\ r. (c) pq, qr. (d) pq, q\\r. (e) p_\_q, q\[r, r_Ls. (/) p \\ q, qr, r || s. ANGLES FURTHER DEFINED ACCORDING TO RELATIVE POSITION If, as in the accompanying diagram, ABC and DEF are cut by GBEH, which is called a transversal (since it cuts across), certain sets of angles are formed to which for brevity we give the following names : ^ABE and ^FEB are called alter- nate-interior angles. and ^HEF are called alternate-exterior angles. and $.FEB are called consecutive-interior angles. and <$.DEH are called consecutive-exterior angles. and ^FEB are called corresponding angles. EXERCISES. SET XX. RELATIVE POSITION OF ANGLES 148. Only one pair of each kind of angles is mentioned in the last paragraph, though there are two pairs of each except the last kind. Name the second pair in each case, and three remaining pairs of corresponding angles. 149. Explain the meaning of alternate as used here. 150. Explain the meaning of consecutive as used here. 151. Explain the meaning of interior as used here. 152. Explain the meaning of exterior as used here. 153. What kind of angles with regard to relative position are formed in the letter Z? In the letter A? In the letter #? HI N1 54 PLANE GEOMETRY 154. In the accompanying diagram select angles under each j class you know (including adjacent and vertical) using: (a) c and d with a as transversal. (b) c and d with b as transversal. (c) a and b with c as transversal. (d) a and 6 with d as transversal. PROPERTIES OF PARALLELS Theorem 11. If, when lines are cut by a transversal, the alter- nate-interior angles are equal, the lines thus cut are parallel. ^ - Given: ABC, XYZ cut by BY. a To prove: ABC \\XYZ. F' # Z Take Q in BY so that QB Draw QP1.AB cutting AB at P. Extend PQ to R in XZ. Then in APQB and ARQY, (1) QB = QY (2) (3) (4) PROOF QY. (5) (6) But PQRAB (7) (8) (9) .'. PQRA.XRZ (10) . (1) Construction. (2) Data. (3) Vertical angles. (4) Two angles and the included side equal each to each. (5) Homologous parts of congruent triangles. (6) Construction. (7) Definition of perpendicular. (8) Quantities equal to the same quan- tity are equal to each other. (9) Definition of perpendicular. (10) Lines perpendicular to the same line are parallel. The student's attention is called to the form and arrangement of this demonstration, as it is the first formal proof given in the text. Note that after the general statement of the theorem following the words "given" and "to prove,"* specific statements are given * In place of the word "given" either "data" or "hypothesis" is frequently used, and in place of "prove" the word "conclusion." PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 55 referring to the particular diagram drawn. These statements should be as brief as possible, and such, that were the diagrams erased, it could be reconstructed. The steps of the proof and the reasons for them are arranged in parallel columns. The con- venience of such an arrangement is at once apparent if it be com- pared with a proof written in essay form. Write the proof that way and draw your own conclusions as to which you would prefer to use, giving your reasons. Show that the following corollaries are true by showing that a pair of alternate-interior angles are equal. Cor. 1. If the alternate-exterior angles or corresponding angles are equal when lines are cut by a transversal, the lines thus cut are parallel. Cor. 2. If either the consecutive-interior angles or the con- secutive-exterior angles are supplementary when lines are cut by a transversal, the lines thus cut are parallel. EXERCISES. SET XXI. CONSTRUCTION OF PARALLELS 155. Parallels may be constructed by using a T-square and a triangle. Explain. 156. Draw three pairs of parallel lines using successively each of the three sides of one of your triangles against a ruler. 157. By using your knowledge of corresponding angles, draw a line through a given point, and parallel to a given line. 158. (a) Practice drawing parallels with compasses and ruler until you can draw them accurately. (6) Test your work by draw- ing any transversal, and measuring a pair of angles that should be equal, (c) Which is more likely to be in error, your drawing or your test? 159. The diagram suggests a method of running one line parallel to another when you are on a field without a transit. Explain and justify the procedure. Presently we shall prove a theorem which is closely related to theorem 11. Before doing so, however, we shall want to see why it need be proved at all. 56 PLANE GEOMETRY EXERCISES. SET XXII. RELATED STATEMENTS 160. (a) Is it true that if a triangle is equilateral, it is also isosceles? (6) Is it true that if a triangle is isosceles, it is also equilateral? 161. (a) Is it true that if two angles are right angles, they are equal? (6) Is it true that if two angles are equal, they are right angles? 162. (a) Is it true that all men are bipeds? (6) Is it true that all bipeds are men? 163. (a) Is it true that if a man lives in Cincinnati, he lives in Ohio? (6) Is it true that if a man lives in Ohio, he lives in Cincinnati? 164. Explain how each of the statements (a) and (6) in each of the four preceding exercises is related to the other: that is, how can (a) in each case be formed from (6), and how can (6) be formed from (a)? 165. Make a statement related to each of the following as (6) is related to (a) in each of exercises 16Q to 163, and tell whether or not your statements are true. (a) If a man lives, he breathes. (6) If a polygon is a triangle, it has three sides. (c) If it rains, the ground is wet. 166. From exercises 160 to 163 and 165 what can you conclude as to the truth or falsity of two statements related in this way? (a) May both of them be true? (6) May one of thfem be true and the other false? 167. If, then, we have proved that a statement is true, is it neces- sary to prove a statement related to it as the second is to the first in each of those exercises, or may we take it for granted that the second will be true without proof? Statements related as those in the last set of exercises, are called converse statements. We saw that each of two converse state- ments could be formed from the other by interchanging the data or hypotheses with the conclusion or conclusions; that is, the two state- ments were so related that what was given in each was what was supposed to follow in the other. From the fact that it is so much PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 57 easier to make a statement whose converse is absurd than one whose converse is true, it appears that we should never claim that the converse of a theorem in geometry is true without having proved it so. EXERCISES. SET XXII (continued) 168. Make a statement of something in life which you know to be true, but whose converse is false. 169. Make a statement of something in life which you know to be true, but whose converse is true. 170. Do the same as you were requested to do in the last two exercises, but take the statements from geometry. 171. Select from the theorems already proved two that are converse statements. 172. The converse of a definition is always true. Test your definitions of Chapter I from this point of view. (See lists at end of Chapter I, page 46.) 173. State what was given us in theorem 11. 174. State what we proved in theorem 11. 175. State what would be given in the converse of theorem 11. 176. State what would have to be proved in the converse of theorem 11. Theorem 12. Parallels cut by a transversal form equal alter- nate-interior angles. \ Fill in all the blank spaces X ~ and answer the questions in the following: Given: p R To prove: PROOF (1) If through the mid-point of YQ we were to draw a line perpendic- ular to XZ what other fact would we know about that line? (2) What parts have we then equal in the triangles thus formed? (3) By what method could we then prove the fact we wish to prove? (1) Why? (2) How do you know each of these pairs are so related? 58 PLANE GEOMETRY R COT. 1. Parallels cut by a transversal form equal corresponding and equal alternate-exterior angles. Cor. 2. Parallels cut by a transversal form supplementary con- secutive-interior ', and supplementary consecutive-ex- terior angles. EXERCISES. SET XXIII. APPLICATIONS OF PARALLELISM 177. The accompanying diagram suggests a convenient method of measuring the distance from B to an inaccessible point F. Explain. Note DE || FB. How can this be done on the ground? 178. The " square network" shown in the figure is used in designing for drawing a great variety of patterns. The patterns A and B drawn upon it are examples of Arabian frets. The best way to rule the square network is to draw a horizontal line MN and mark off equal divisions on it. At each point of division, by use of the tri- angle, draw two lines, such as PQ and PR, each making an angle of 45 with MN. Draw such a network, then upon it construct a pattern, either an original design or a copy of these Arabian frets. (Taken from Stone-Millis, Plane Geometry.) 179. In the annexed dia- gram if XY\\PQzudRS\\ VT how many angles would you need to know in order to find the remaining angles? 180. If the side AC of a triangle ABC is extended, as in the annexed diagram, how could a line be drawn ~~b through C to make an angle Would any other angles be equal? Why? A equal to the PERPENDICULAR, RIGHT TRIANGLE, PARALLELS 59 LIST OF WORDS DEFINED IN CHAPTER II Distance (point to a line). Hypotenuse. Parallels, coplanar, transversal; alternate-interior, alternate-exterior, consecutive-interior, consecutive-ex- terior, corresponding angles. Converse. SUMMARY OF AXIOMS IN CHAPTER H Inequality 1. If unequals are operated upon by positive equals in the same way, the results are unequal in the same order. 2. If unequals are subtracted from equals, the remainders are unequal in the reverse order. SUMMARY OF POSTULATES IN CHAPTER H Perpendiculars 1. At a point in a line only one perpendicular can be erected to that line. 2. From a point outside a line only one perpendicular can be drawn to that line. Parallels 3. Through a given point only one line can be drawn parallel to a given line. Cor. 1. Lines parallel to the same line are parallel to one another. SUMMARY OF THEOREMS IN CHAPTER H 6. A perpendicular is the shortest sect that can be drawn from a point to a line. 7. The hypotenuse and an adjacent angle determine a right triangle. 8. The hypotenuse and another side determine a right triangle. 9. Lines perpendicular to the same line are parallel. 10. A line perpendicular to one of a series of parallels is perpendicular to the others. 11. If when lines are cut by a transversal the alternate-interior angles are equal the lines thus cut are parallel. Cor. 1. If the alternate-exterior angles or corresponding angles are equal when lines are cut by a transversal, the lines thus cut are parallel. Cor. 2. If either the consecutive-interior angles or the consecu- tive-exterior angles are supplementary when lines are cut by a transversal, the lines thus cut are parallel. 12. Parallels cut by a transversal form equal alternate-ulterior angles. Cor. 1. Parallels cut by a transversal form equal corresponding, and equal alternate-exterior angles. Cor. 2. Parallels cut by a transversal form supplementary consecu- tive interior, and supplementary consecutive-exterior angles. CHAPTER III ANGLES OF POLYGONS AND PROPERTIES OF PARALLELOGRAMS A. ANGLES OF POLYGONS In attempting to develop a formula for the sum of the angles of a polygon, it is best for us to begin with the simplest polygon, the triangle. EXERCISE. SET XXIV. SUM OF ANGLES OF A TRIANGLE. 181. By reproducing the angles of a given triangle, place these angles adjacent to one another. What does the sum of the angles appear to be in this case? Theorem 13. The sum of the angles of a triangle is a straight angle. Prove this proposition, using the hints given by the following diagrajn and notes: Produce CA and draw AP || CB. What is the sum of A !, 2, and 3? WTiat is the relation of ^2 to <2i? Of ^3 to $33 Cor. 1. A triangle can have but one right or one obtuse angle. Cor. 2. Triangles having two angles mutually equal are mutu- ally equiangular. Cor. 3. A triangle is determined by a side and any two homolo- gous angles. An exterior angle of a polyyon is one formed by a side of the polygon and the prolongation of an adjacent side. In the preceding diagram < RAB is an exterior angle of the A ABC. Cor. 4. An exterior angle of a triangle is equal to the sum of the non-adjacent interior angles. 60 POLYGONS AND PARALLELOGRAMS 61 EXERCISES. SET XXIV (continued) 182. (a) The theorem that the sum of the angles of a triangle equals a straight angle may be proved by drawing a line through a vertex parallel to the opposite side. Give proof. NOTE. This proof is attributed to the Pythagoreans. 1 ' "* n (b) Prove the same fact by drawing a sect from a vertex parallel to the opposite side. (c) Prove the same theorem by drawing through any point on one ^ ^^ side lines A ^ parallel to the other sides of the triangle. 183. Prove this same fact another A way by erecting perpendiculars to one side at its extremities and dropping a perpendicular to the same side from the opposite vertex. 184. Show how the following procedure may be used to test accu- racy with which you measure angles with an instrument. Select the three positions not in a straight line; call them stations A, B, and C. From station A measure the angle between the directions to B and C; at B, measure the angle between the directions to C and A; at C, the angle between the directions to A and B. Would your measurements be accurate, and if not, what error would there be, if you found the angles to be respectively: 100 27', 23 13', and 56 237 185. If one angle of an isosceles triangle is 60, find the other angles. 186. (a) If the vertex angle of an isosceles triangle is v write an expression for each of the other angles. (b) If a base angle of an isosceles triangle is 6 write an expres- sion for each of the other angles. 187. What angle do the bisectors of the acute angles of a right triangle form? 188. Construct the following angles: 60,30 ,120 ,75 ,150,105 . The pupil is again reminded that the use of instruments is 62 PLANE GEOMETRY restricted to the straight edge and the compasses in scientific geo- metric constructions. d!89. (a) Two mirrors, mi and ra 2 , are set so as to form an acute angle with each other. A ray of light is reflected by mi so as to strike m 2 . The ray is again reflected by m 2 and crosses its first path. Prove that course ABC and notes a lighthouse L when he is at A, he takes the angle A; and if he notices when the angle that the lighthouse makes with his course is just twice the angle noted at A , then BL =AB. He has AB from his log (an instrument that tells how far a ship goes in a given time), so he knows BL. He has "doubled the angle on the bow" to get this distance. B. PARALLELOGRAMS A parallelogram is a quadrilateral whose opposite sides are parallel. Theorem 16. Either diagonal of a parallelogram bisects it. Suggestion: In the proof note that the diagonal is a transversal of the parallel sides. Cor. 1. The parallel sides of a parallelogram are equal, and the opposite angles are equal. The sects of common perpendiculars included by parallels are catted the distances between the parallels. Cor. 2. Parallels are everywhere equidistant. PLANE GEOMETRY EXERCISES. SET XXVII. PARALLELOGRAMS 209. One angle of a parallelogram is 20 more than three times another. Find all the angles. 210. (a) Establish a relation between consecutive angles of a parallelogram. (b) How many angles of a parallelogram must be known in order to determine the others? 211. A stairway inclined 45 to the horizontal leads to a floor 15' above the first. What is the length of the carpet required to cover it if each step is 10" high? If each is 12" high? If each is 9"? Can this problem be solved without knowing the height of the steps? Is it necessary to know that the steps are of the same height? (Taken from Slaught and Lennes, Plane Geometry.) 212. Is the converse of the fact that a diagonal bisects a paral- lelogram true or false? Give a reason for your answer. Theorem 17. A quadrilateral whose opposite sides are equal is a parallelogram. (1) What fact would we need to know about these opposite sides? (2) By what method can we obtain this fact? EXERCISES. SET XXVII (continued) 213. An adjustable bracket such as dentists often use, is out- lined in the figure. It is fastened to the wall at A, and carries a shelf B. Why is it that as the bracket is moved so that B is raised and low- ered, the shelf remains horizontal? (Taken from Stone-Millis, Plane Geometry.) 214. The accompanying figure is a diagram of the " parallel ruler," which is used by designers for drawing parallel lines. POLYGONS AND PARALLELOGRAMS 69 (a) Upon what principle of parallelograms must its construction depend? (b) Make such an instrument. A rectangle is a parallelogram one of whose angles is a right angle. Cor. Each angle of a rectangle is a right angle. Why? EXERCISE. SET XXVII (continued) 215. Prove that a parallelogram whose diagonals are equal is a rectangle. (This fact is used as a check by carpenters and builders. Could it be used in laying out a tennis court?) Theorem 18. A quadrilateral having a pair of sides both equal and parallel is a parallelogram. EXERCISES. SET XXVII (continued) 216. Justify the following method used by surveyors for pro- longing a line beyond an obstacle; that is, show that in the diagram EF is A B prolonged beyond 0. BC is run at right angles to AB; CDBC; DECD and DE=CB; EFA.DE. 217. The accompanying diagrams show another way of extending a line be- yond an obstacle, (a) By reference to dia- gram, state the proce- dure in words, (b) Show that it is correct, (c) Compare this method with that of Ex. 216. Note, for example, why two lines (CB and DE) are used in Ex. 217, and only one (CB) in Ex. 216. Under what con- D H ditions would you use each method? 218. If the vertices of one paral- lelogram are in the sides of another, the diagonals of the two parallelo- grams pass through the same point (called the center of the parallelograms) . Suggestions: Call the intersection of the diagonals AC and BD, 0. Draw OE, OF, OG, OH and prove EOG and FOH straight lines. E 70 PLANE GEOMETRY Pi d219. An interesting outdoor application of the theory of paral- lelograms is the folio whig: Suppose that you are on the side of this stream opposite to XY, and wish to measure the length of XY. Run a line AB along the bank. Then take a carpenter's square, or even a large book, and walk along A B until you reach P, a point from which you can just see X and B along two sides 4 ^ ^^ B of the square. Do the same for F, thus fixing P and Q. Using the tape, bisect PQ at M. Then walk along YM produced until you reach a point FI that is ex- actly in line with M and F, and also with P and X. Then walk along XM produced until you reach a point Xi that is exactly in line with M and X } and also with Q and F. Then measure YiXi and you have the length of XY. For since YXi_LPQ, and XYiPQ, YXi \\XYi. And since PM = MQ, therefore XM=MX l and FjM=MF. Therefore Y^YX is a parallelo- gram. Give the reason for each of these steps. * Theorem 19. A parallelogram is determined by two adjacent sides and an angle, or two parallelograms are congruent if two adjacent sides and an angle are equal each to each. Methods of proof. (1) Congruence of triangles, or (2) Superposition, or (3) Properties of parallelograms. NOTE. If (1) or (3) is used it must be shown that the parts proved equal are arranged in the same order in the two parallelograms. EXERCISES. SET XXVII (continued) 220. Construct a parallelogram, given (a) Two adjacent sides and an included angle. (6) Two adjacent sides and a non-included angle, (c) A side, a diagonal, and the angle between them. 221. In physics it is shown that if two forces (such as a push and a pull) are exerted in different directions upon the same object, they have the same effect as a single force called their resultant. POLYGONS AND PARALLELOGRAMS 71 How is this fact illustrated by the bean-shooter? Referring to the accompanying diagram, if the directions and magnitudes of two forces working on object A are represented by the sects AB and AC, the direction and magnitude of the resultant will be represented by the sect AD, which is the diagonal of the parallelogram, with A B and AC as adjacent sides. In physics, such a diagram is, for obvious reasons, called the Paral- lelogram of Forces. A force of 50 pounds is exerted upon a body pulling it in one direction, and at the same time another force of 100 Ibs. pulls it in a direction at an angle of 45 with the first. Show by the parallelogram of forces the effect on the body. NOTE: (a) Use only compasses and ruler in solving. Represent 50 Ibs. by any given sect as unit; draw the forces to scale, and find the resultant. (6) Use protractor and marked edge in reading result. (c) Could you read the resultant to any degree of accuracy by any other method? 222. Two forces, 250 Ibs. and 400 Ibs. respectively, are exerted upon a body at right angles with each other. Find their resultant as in Ex. 221. Check the result by computation. Why was such a check not available for you in Ex. 221? 223. Find the resultant of two forces exerted upon a body at an angle of 150 with each other, one of 50 Ibs., the other of 60 Ibs. 224. When a train is approaching a station at a velocity of 40 ft. per second, a mail bag is thrown at right angles from the car with a speed of 20 ft. per second. Find the actual direction and speed of the moving bag. (Taken from Stone-Millis, Plane Geometry.) 225. The resultant of three forces may be found by getting the resultant of two of the original forces and then finding the resultant of that and the third original force. This process may be continued to obtain the resultant of several forces. Find the resultant of three coplanar forces of 200 Ibs., 150 Ibe., and 175 Ibs. respectively, acting on the same body at the same 72 PLANE GEOMETRY time. The angle between the first and second force is 45, and that between the second and third is 60. Why note that the forces are coplanar? 226. A force of 100 Ibs. makes an angle of 60 with a second force of 120 Ibs. exerted on the same body, and makes an angle of 90 with a third force of 140 Ibs., and an angle of 120 with a fourth force of 160 Ibs. If the forces are coplanar and act simultaneously, find their resultant. Theorem 20. The diagonals of a parallelogram bisect each other. EXERCISES. SET XXVII (continued) 227. Draw any line through the intersection of the diagonals of a parallelogram. (a) Give a list of the pairs of congruent triangles formed. Give reasons for your asser- tions. (b) Give a list of the pah's of congruent quadrilaterals. Verify your statements. 228. Cut a parallelogram out of cardboard. Placing a pin at the intersection of the diagonals, try to balance the parallelogram. Why would you expect it to balance? The intersection of the diagonals is called the center of gravity of a parallelogram. Why? Theorem 21. A quadrilateral whose diagonals bisect each other is a parallelogram. EXERCISE. Set XXVII (continued) 229. The same principle is often used in the construction of iron gates that was employed in the making of a parallel ruler used in the eighteenth century (see dia- gram). What is the principle? The student is encouraged to make such an instrument. , POLYGONS AND PARALLELOGRAMS 73 EXERCISES. SET XXVIII. PARALLELS 230. Classify quadrilaterals. 231. Summarize ways of proving: (a) Sects equal. (6) Angles equal. (c) Lines parallel. LIST OF WORDS DEFINED IN CHAPTER m Exterior angle and angles of a polygon. Convex polygon, diagonal; quadri- lateral, pentagon, hexagon, heptagon, octagon, nonagon, decagon; regular polygon. Parallelogram, rectangle. Distance between parallels. SUMMARY OF THEOREMS DEVELOPED IN CHAPTER in 13. The sum of the angles of a triangle is a straight angle. Cor. 1. A triangle can have but one right or one obtuse angle. Cor. 2. Triangles having two angles mutually equal are mutually equiangular. Cor. 3. A triangle is determined by a side and any two homolo- gous angles . Cor. 4. An exterior angle of a triangle is equal to the sum of the non- adjacent interior angles. 14. The sum of the angles of a polygon is equal to a straight angle taken as many times less two as the polygon has sides. Cor. 1. Each angle of a regular polygon of n sides equals the ^ 2 th part of a straight angle. Cor. 2. The sum of the exterior angles of a polygon, made by produc- ing each of its sides in succession, is two straight angles. Cor. 3. Each exterior angle of a regular polygon is the |th part of a straight angle. 15. If two angles of a triangle are equal, the sides opposite them are equal. Cor 1. Equiangular triangles are equilateral. 16. Either diagonal of a parallelogram bisects it. Cor. 1. The parallel sides of a parallelogram are equal, and the opposite angles are equal. Cor 2. Parallels are everywhere equidistant. 17. A quadrilateral whose opposite sides are equal is a parallelogram. 18. A quadrilateral having a pair of sides both equal and parallel is a parallelogram. 19. A parallelogram is determined by two adjacent sides and an angle. 20. The diagonals of a parallelogram bisect each other. 21. A quadrilateral whose diagonals bisect each other is a parallelogram. CHAPTER IV AREAS A. INTRODUCTION. REVIEW OF FRACTIONS * In dealing with areas, we are largely concerned with ratios. A ratio is a fraction, and therefore our work in this chapter presup- poses a familiarity with fractions. For those of us who need a review of this topic the following will be helpful. A fraction is an indicated quotient, the dividend oi which is the numerator and the divisor the denominator. Since this is review, let us summarize without discussion the fundamental facts which we need to recall about fractions. PRINCIPLES I. The value of a fraction is not changed if both numerator and denominator (i.e., both terms of the fraction) are multiplied or divided by the same quantity. Why? This statement includes cancellation, for that is the process of dividing both numerator and denominator by a common factor. Illustrations: 1. . = <*? = !^? ^. 64o 3 6 _ a 2 F = 56 = ^ 3. a 2 -4 o+2 2a 2 -8a+8 2(a-2) II. Considering as the signs of a fraction that of the numerator, that of the fraction, and that of the denominator, any two may be changed without changing the value of the fraction. Why? Recall that the numerator and denominator of a fraction are treated as if each were enclosed in a parenthesis. Illustrations: 1. a_ _ a _ __ - - .Zf? 2. a 2_ 2 a_ b~ b ' b~b a5~ 5a~ 3. abc _ a( 6)c _ abc _ def = det^f) = ^def = * Those who have not studied algebraic fractions are advised to take up this topic at this time. Any standard algebra text will furnish sufficient material. 74 AREAS 75 III. Fractions may be added or subtracted by changing them to equivalent fractions having the same denominators and adding or subtracting their numerators, putting the sum or difference over the common denominator. Why? Changing to a same denominator depends for its validity upon preceding principle I. The same or common denominator may be found by getting the Lowest Common Multiple of the denominators. Illustration : m 2n m z 3n 2 3m n m 3 n 3 mn The L. C. M. of ra 2 +mrc+n 2 , m 3 n 3 , and mn is m 3 n 3 . _(m-2n)(m-n) (m 2 -3n 2 ) . (3ra-n)(m 2 +mn+n 2 ) . . IIIG Sliin = - - - - - - - - T - ; - - - m 3 n 3 m 3 n 3 m 3 -n 3 IV. Fractions may be multiplied by multiplying their numerators and denominators separately, obtaining thus the numerator and de- nominator respectively of the product. Why? Illustration: 5c 2 -20d 2 c 2 -2cd+4d z 5(c+2d)(c-2d) ~ (c +2d) (c 2 -2cd+4d 2 ) ' ~ 25cd* _c-2d 5 " 5cd 4 V. Fractions may be divided by inverting the divisor and proceeding as in multiplication. Why? Illustration: a 2 -7a+12 ^a 2 -16 o-l '' 1-a 2 -1 _ (o-4) (o-3) (l-o a-1 "(a-4)(a+4) 4+a EXERCISES. SET XXIX. FRACTIONS 232. Would the value of a fraction be changed if both numerator and denominator were squared? Illustrate and give a reason for your answer. 233. Why in dividing fractions do we invert the divisor and multiply? 76 PLANE GEOMETRY 234. Would the result of cancellation ever be zero? Illustrate and give a reason for your answer. 235. Write in four ways the fraction ^g 236. Find the difference between (a) 1 . 1 1 (b) I 1 1 (c)n+l A n , v ' h-and- v/ --- and- ' -5- and -- + 1. n m n-\-m n m n-m 3 6 237. Are any or all of the expressions in the following groups identical? (a)bh L, 6, , , h (b)b+B b .B ,1,,, ~2 > 2 bh) 2 ' 2 ~2~' 2 + ~2' 2 ( + ^ Justify your answers. 238. State, with reasons, whether or not the following are identities: (a) 1 1 (6) 1 1 (5_ a )2-( a _ 6 )2 (6 -a) 3 (a-6) 3 239. Add: 3a 6 2 Why is it correct to refer to such a a 2 b 2 b a combination as addition? Simplify (which means do whatever is indicated by the symbols) . 241 3___ - 6xy V 2-x - 4a 2 -4a6-36 2 / 96 2 244. 245. 246. A+p 1+p' < v 2 AREAS 77 248 . (0) *+-= (6) a a+b , N , [x+i z-ii (c) x+\ r-jLl I a a+bl 249. a-b-^^ a-b 250. a 2 ra 2 -a 2 ra-56a 2 Qr 2 -49)(:r 2 -16;r+63) ' 252 m+2 ra -3/ \4 -mV m 2 +m - B. AREAS. DEVELOPMENT OF FORMULAS Just as a sect is measured by finding the number of linear units it contains, so a surface is measured by finding the number of square units it contains, or better, itsratio to the unit square, which is known as the area of the surface. A square unit is a square each side of which is a linear unit. For example, if we were to measure the length and the width of this page (taking the inch as a linear unit) a square inch would be the corresponding square unit, and the area would be found in square inches. The selection of the unit in practical measurements is just a matter of convenience. Why, for instance, select an inch instead of a mile in measuring the length of this page? In comparing areas or sects, we compare the abstract numbers which express the ratio of their measures in terms of a common unit. When we say that two sects compare as 5 to 4 we refer to the fact that when measured in terms of the same unit their measures would compare as 5 to 4 say one was 5" and the other 4". Sim- ilarly, when we say two rectangles compare as 6 to 5 we mean that if the area of one were 6 sq. ft. the area of the other would be 5 sq. ft. Such quantities as we have just referred to are said to be com- mensurable because they can be measured in terms of the same unit. In comparing two sects it is sometimes impossible to get a common unit of measure, that is, to select a unit so that it will be contained an exact number of times in both. Such sects are said to be incom- 78 , PLANE GEOMETRY mensurable. It is known from experimental work in mensuration that a circumference and its diameter are incommensurable, for if the diameter were 1 inch, the circumference would be 3. 141 59 4- inches (TT inches). EXERCISES. SET XXX. COMPARISON OF SECTS 253. Assuming the fact that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides, show that the diagonal of a square and a side of the square are incommensurable. 254. When two sects are commensurable a common measure can be found as follows: ^ _ X z v Z z B K *h. e sects are 4^ ' andPQ. Layoff PQ on AB. It will be p y z Q contained once with a remainder XB. _Lay off XB_on PQ. It will be contained twice with a remainder ZQ. Lay off ZQ onXB. It will be contained just 3 times with no remainder. Then Z 2 B, is a common measure (in- deed the greatest common measure) of APand PQ. Any part of Z 2 B would also be a common measure of AB and PQ. Call Z 2 B, u. How many u's does XB contain? How many u's does YZ contain? How many u's does AX contain? Show that u is a common measure of AB and PQ. We shall refer to the consecutive sides of a rectangle as its dimen- sions, calling one the altitude, and the other the base. For the sak^ of brevity the expressions "the ratio of two sects" or "the products of two sects" will be used to indicate the ratio or the product of the abstract numbers expressing their lengths in terms of the samo unit of measure. For the same reason "rec- tangle," "parallelogram," etc., will be used for the "area of rec- tangle," "area of parallelogram," etc. Therefore, the expression "two rectangles compare as the products of their dimensions" is a conventionally abbreviated form of "the areas of two rectangles compare as the products of the lengths of their dimensions ex- pressed in terms of the same linear unit." AREAS 79 Theorem 22.* Rectangles having a dimension of one equal to that of another compare as their remaining dimensions A E D PL ^T \\ * u\ , , B F G C M N S Given: Rectangles ABCD (call it R) and PQST (call it ft) with AB=PQ, and BC and QS commensurable. To prove: R_ = BC Ri~ QS Suggestions for proof: Let u be a common measure of C and QS. Lay uoff on "BC and Q and erect JLs at points of division F, G. . . M,N,.... If w is contained m times in BC and n times in QS what is the ratio ? What kind of figures are ABFE, . . . , PML, . . . ? Q S ' Are they congruent? Why? If R and ft are respectively composed of m and n congruent parallelograms n what is the ratio of ? ft Theorem 23. Any two rectangles compare as the products of their dimensions. Given: Rectangles R and ft with dimensions h, b, and hi, bi respectively. To prove :-~- = r-r- Ki Oi/ll Suggestions for proof: Construct a rectangle (ft) having as dimensions h and 61. S^ ? ft^ ? -''ft^ 7 *Although the proof here suggested applies only to those cases where the sects are commensurable, the theorem is always valid. 80 PLANE GEOMETRY Theorem 24. The area of a rectangle is equal to the product of its base and altitude. Given : Rectangle R with base b and altitude h. To prove: Area of R = bh. Suggestions for proof: b What does area mean? Call the unit of surface u. u will be a rectangle with base and altitude 1. ^. why? For what does stand? Why? EXERCISES. SET XXXI. AREAS OF RECTANGLES 255. What per cent of surface is allowed for joints and waste, if 120 rectangular sheets of tin, 14" by 18", are just sufficient to cover a roof 165 sq. ft.? 256. A map is drawn to a scale of 1" to 1000 miles. What actual area would be represented on the map by a foot square? 257. From a rectangular sheet of paper cut a strip J^ of the sheet in width. What part of the sheet is left? Then from the same sheet cut off J^ of its length. What part of the original sheet is now left? 258. Prove, by letting a and b represent the lengths of two sects, that (a=*=6) 2 =a 2 +6 2 2a6. 259. How long would a rectangular strip of paper 1 sq. ft. in area be if it were .01" wide? 260. Both a square and a rectangle three times as long as it is wide have a perimeter of 64 ft. Compare their areas. 261. Thucydides (430 B.C.), a Greek historian, estimated the size of the Island of Sicily by the time it took him to sail around it, knowing how long it took him to sail around a known area. Was his method correct? Give a reason for your answer. 262. 144 sq. ft. is the area of a square and also of a rectangle four times as long as wide. How do their perimeters compare? // one side of a parallelogram is selected as its base, the distance between it and the opposite side is called its altitude. Would it make any difference at what point its altitude was measured? Why? How many altitudes has a parallelogram? Is this definition consistent with what we have referred to as an altitude in the case of the rectangle? AREAS 81 H Theorem 25. The area of a parallelogram is equal to the product of its base and altitude. A b D Given : EJABCD with base 6 and altitude A. To prove: Area of ElABCD^bh. Suggestions for proof: Draw AH and DH i _L BC. What kind of figure is AHHiD? What is the area of AHHJ). Compare A ABH and DCHi. How do the areas of AHHiD and ABCD compare? What is the base of each, and what the altitude of each? Suppose AH cut BC produced, will the theorem still hold true? EXERCISE. SET XXXII. AREAS OF PARALLELOGRAMS 263. If one side of your parallel ruler is held fixed while the opposite side is raised and lowered to various positions, will the areas of the various parallelograms be changing? If so, what will be the greatest area obtainable, and what the least? Any side oj a triangle may be taken as its base, and the perpen- dicular from the opposite vertex to that side will be its altitude. EXERCISE. SET XXXIII. ALTITUDES OF TRIANGLES 264. (a) How many altitudes has a triangle? Illustrate. (6) Does your answer to (a) hold for a right triangle? Illustrate. (c) Will the altitudes of a triangle always fall within the triangle? (d) What fact have we already proved about the altitudes of an equilateral triangle? Theorem 26. The area of a triangle is equal to half the product of its base and altitude. Given: AABC with b as base and _P, 7 h as altitude. To prove : Area of AABC = Suggestions for proof: DT&wBX || CA and IT meeting at P. CB, What kind of figure is APBCt What is the area of APBC1 How is the area of AABC related to that of APBC? Cor. 1. Any two triangles compare as the products of their bases and altitudes. Cor. 2. Triangles having one dimension equal compare as their remaining dimensions. 6 82 PLANE GEOMETRY EXERCISES. SET XXXIV. AREAS OF TRIANGLES 265. Prove theorem 26, using the suggestion given by the accom- panying diagram. S*^"\ # ~7 266. Calculate the area of the letter Z shown in the figure, the dimensions being indicated in centimeters. 267. With a marked edge draw a triangle, and tak- ing necessary measurements find its area, using in turn its three bases and altitudes. 268. Where do the vertices of all triangles having the same base and the same area lie? Give reasons for your answer. A trapezoid is a quadrilateral with one, and only one, pair of par- allel sides. The parallel sides are called its bases, and the distance between them is called its altitude. Why do the words "and only one" need to be included in the definition? Theorem 27. The area of a trapezoid is equal to half the product of its altitude and the sum of its bases. Given: Trapezoid ABCD with bases 6 and bi and altitude h. To prove: Area of trapezoid equals \h (6+61). Notes on proof: This fact may be proved by making con- structions that will divide the figure into rectangles, parallelograms, or triangles, or combinations of them. Why? The following diagrams show some such constructions. The student may use one of them or suggest another. It is interesting as an exercise to show that the same formula for the area of a trapezoid may be derived from each of the following diagrams. Which gives the simplest deri- if C vation? 1 2. DrawAtf \\XY. B D Y 3. Draw ZfiJLXF and ~CD LXY AREAS 4. Draw PQ and MN AB produced. B 5. Through N,the midpoint of #(7, draw WN || DA cutting ~AB produced in P. B _ _ _^ ~0 6. Extend Aby DC to E and DC 7. Draw CX \\ DA cutting AB produced by AB to F. in F. 8. Extend ~YX to A _ p ^ R so that XR=AB. Draw_RZ || FA. Bi- ,.,./ \P sect AF in M and draw WP 1 1 FX" cut- ting 1Z in Q. Y X~ R EXERCISES. SET XXXV. AREAS OF TRAPEZOIDS 269. The diagram shows how the area of an irregular polygon may be found, if the distance of each vertex from a given base line, as XY, is known. These distances AAi, BBi, etc., are called offsets, and are the bases of trapezoids whose altitudes are A\B\ t Bid, C\D\, etc. The area ABCDEF may now be found by the proper additions and subtractions. On cross-section paper plot the points whose coordinates are given below, join them in order, draw/' 7 A, and find the inclosed area in each case: X Case A B C D # F (a) 5,7 4,5 4,1 7,0 5,3 6,3 (b) 2,7 3,3 6,0 5,2 6,6 8,7 (c) 2,5 3,6 2,4 4, 2 6, 5 4,7 84 PLANE GEOMETRY d d d d d d d 270. In order to determine the flow of water in a certain stream, soundings are taken every 6 ft. on a line AB at right angles to the current. A diagram may then be made to represent a vertical cross-section of the stream. If the area of this cross-section and the speed of the current are known, it is possible to determine the amount of water flowing through the cross-section in a given time. The required area is often found approximately by joining the ex- tremities of the offsets y , y i} y 2 , etc., by straight lines, and finding the sum of the trapezoids thus formed. That is, the strips between successive offsets are replaced by trapezoids. This gives the Trape- zoidal Rule for finding an area. It may be stated as follows: To half the sum of the first and last offsets add the sum of all inter- mediate offsets, and multiply this result by the common distance between the offsets. (a) State this as an algebraic formula. (6) Verify the formula. (c) Find the area of the cross-section of a stream if the soundings taken at inter\ 7 als of 6 ft. are respectively 5 ft., 6.5 ft., 11 ft., 14.5 ft., 16 ft., 9 ft., and 6.5 ft. (d) In the midship section of a vessel the widths taken at intervals of 1 ft. are successively 16, 16.2, 16.3, 16.4, 16.5, 16.7, 16.8, 15, 10, 4, 0, measurements being in feet. Find the area of the section. (Use the line drawn from the keel to the deck as base line.) 6271. A third rule for finding plane areas, known as Simpson's Rule, usually gives a closer result than the Trapezoidal Rule. In proving Simpson's rule two consecutive strips are replaced by a rectangle and two trapezoids as follows: Divide 2d into three equal parts, erect J_s at the points of division, and complete the rec- tangle whose altitude is the middle offset 2/1, as in the figure. Join the extremities of y and y 2 to the nearer upper vertex of this rectangle. AREAS 85 Then if the areas of the trapezoids are T and Ti, and if the area of the rectangle is R, If, now, the number of strips is even, and if the offsets are lettered consecutively y , y\, y%, . . . ., y n , the addition of the areas of suc- cessive double strips, found by the above formula, gives the result : In words : To the sum of the first and last offsets add twice the sum of all the other even offsets and four times the sum of all the odd offsets, and multiply by one-third the common distance between the offsets. (a) Verify this rule. (6) By Simpson's Rule find the area of the stream in Ex. 270 (c) . (c) By Simpson's Rule find the area of the section of the vessel in Ex. 270 (d). LIST OF WORDS DEFINED IN CHAPTER IV Ratio, fraction, numerator, denominator, terms of fraction, cancellation, simplify. Area, commensurable, incommensurable; dimensions, base and altitude of rectangle, parallelogram, triangle, trapezoid. SUMMARY OF THEOREMS PROVED IN CHAPTER IV 22. Rectangles having a dimension of one equal to that of another compare as their remaining dimensions. 23. Any two rectangles compare as the products of their dimensions. 24. The area of a rectangle is equal to the product of its base and altitude. 25. The area of a parallelogram is equal to the product of its base and altitude. 26. The area of a triangle is equal to half the product of its base and altitude. Cor. 1. Any two triangles compare as the products of their bases and altitudes. Cor. 2. Triangles having one dimension equal compare as the re- maining dimensions. 27. The area of a trapezoid is equal to hah" the product of its altitude and the sum of its bases. CHAPTER V ALGEBRA AS AN INSTRUMENT FOR USE IN APPLIED MATHEMATICS A. LOGARITHMS Although logarithms are introduced at this point, as a part of a chapter on Algebra, it is not essential that they be studied at this time. They might well be omitted until there is a feeling of necessity on the part of the pupils in the solution of the more diffi- cult problems based upon similarity and the trigonometric func- tions. In fact, those schools wishing to omit such problems under Ratio, Proportion, Variation and Similarity need not include the topic at all. For this reason, when logarithms are desirable in the solution of problems, this fact has been indicated. The integ- rity of the course will not be injured in the least by the omission of the topic of logarithms or any of these exercises. I. INTRODUCTION If 8 2 = 64, then 2 is called the logarithm of 64 to the base 8. This is written Iog 8 64 = 2. Again 3 4 = 81 or Iog 3 81 = 4. Hence we see that the logarithm of a number is simply the exponent of the power to which another number (called the base) must be raised to obtain that number. Thus, in the last example 4 is the exponent of the power to which the base 3 must be raised to obtain the number 81. EXERCISES. SET XXXVI. MEANING OF LOGARITHMS 272. Write in logarithmic form: (a) 5 3 = 125 (c) 7 2 = 49 (e) 10 1 - 30103 = 20 (g) 103.47712 = 3000 (6) 2 7 =128 (d) 10* = 10 (/) 10 3 =1000 (h) 10 5 - 4 ? 71 = 300,000 (i) 10 6 = 1,000,000 (j) lOO^lOOO 273. Write in exponential form: (a) Iog 6 216 = 3 (d) Iog 10 2-.3010 (g) logw.01 = -2 (6) Iog 9 81=2 (e) lo glo l = (h) Iogi .001= -3 (c) Iognl331 = 3 (/) Iogi .l=-l (i) Iog 10 500 = 2.6990 86 ALGEBRA IN APPLIED MATHEMATICS 87 Before studying logarithms, their principles and applications, it will be well for us to recall the laws concerning exponents in multi- plication, division, involution and evolution. We have learned to multiply, divide, raise to powers and extract roots of simple expressions when the exponents have been positive integers, so that we shall here only summarize what we already know, adding the statement that the laws governing positive integral exponents govern all exponents, both fractional and negative a. PRINCIPLES OF EXPONENTS 1. The exponent of the product of any number of factors of like base is equal to the sum of the exponents of the factors. Illustrations: (1) a n - a p -a q =a n+p+q (2) a*'a*=a* (3) b- n 'b~ b =b- n - p (4) b +H ~ A = 6* ~ A = 6" = 6* 2. The exponent of the quotient of two quantities of like base is equal to the exponent of the dividend diminished by that of the divisor. Illustrations : (1) x<*-x b = x a -b (2) x a + (3) m g -r-m 3 3. The exponent of any power of a quantity is equal to the product of the exponent of the base and the index of the power. Illustrations: (1) (6 p ) fl = 6 M / p\r pr (2) \a*)=a< s ^ 4. The exponent of any root of a quantity is equal to the quotient of the exponent of the base and the index of the root when that index is not zero. Illustrations : (1) _ m _ _* Vb x =b m p _aq (2) Vx~ a =x p 88 PLANE GEOMETRY EXERCISES. SET XXXVII. DRILL IN APPLICATION OF LAWS OF EXPONENTS Perform the operations indicated in the following problems: 274. x*x ~ 7 x~ 8 282. 10 3 - 4362 - 10- 7856 275.|ai|a* 283.10 noA in 97ft kn+2_-_Qk2 6< b. HISTORICAL NOTE 64(1024) 3)(65536X1024) 2 256 2 There is a large amount of computation necessary in the solution of some of the practical applications of mathematics. The labor of making extensive and complicated calculations can be greatly lessened by employing a table of logarithms. About the year 1614 a Scotchman, John Napier (1550-1617), Baron of Merchiston, invented a system by which multiplication can be per- formed by addition, division by subtraction, involution by a single multiplica- tion, and evolution by a single division. From Henry Briggs (1556-1631), who was a professor at Gresham College, London, and later at Oxford, this invention received modifications which made it more convenient for ordinary practical purposes. Laplace, the great French astronomer, said: "The employment of logar- ithms by reducing to a few days the labors of months, doubles, as it were, the life of an astronomer, besides freeing him from the errors and disgust insepar- able from long calculations." ALGEBRA IN APPLIED MATHEMATICS 91 The logarithms now in general u^e are known as common logarithms, or as Briggs' logarithms, in order to distinguish them from another system, also a modified form of Napier's system. The logarithms of this other modified system are frequently employed in higher mathematics, and are known as natural or hyperbolic logarithms. H. PRINCIPLES OF COMMON LOGARITHMS For practical purposes, the exponents of the powers to which 10, the base of our decimal system, must be raised to produce various numbers are put in table form. That is, the logarithms of numbers to the base 10 are tabulated. For the positive integral powers of 10 we would need no tables, for those we can find by inspection. But exponents may be negative and they may be fractional. For the negative integral powers of 10 as we shall see presently, we would need no tables either. But fractional exponents or the frac- tional parts of exponents we cannot readily find, and hence for them we need tables. We all know that Similarly it can be shown that Y10 3 = 1000, .'.*log 1000=3. V 10 - 1 = Jo> .' log .1 = -1 Y10 2 =100, /. log 100 =2. VlO- 2 = 5, .'. log .01 = -2 MO 1 = 10, /. log 10 = 1. YlO- 3 = ^- 3 , .'. log .001= -3 10 1 10 1 V =10i-i= 10, and = 1; .'. 10" =1, /. log 1 = 0. 10-47712 or iQTfrj that is, the one-hundred-thousandth root of 10- 47712 is nearly 3. .'. log 3 = .47712 nearly. Although log 3 can never be expressed exactly as a decimal fraction, it can be found to any required degree of accuracy. In this book logarithms are given to four decimal places. These are sufficient for ordinary computations. * When the base 10 is used the base is not indicated in writing the loga- rithms of numbers. Thus we write log 3 = .47712, not log 10 3 = .47712. 92 PLANE GEOMETRY EXERCISES. SET XXXIX. COMMON LOGARITHMS 305. What are the logarithms of the following to the base 10: (a) 10000? (c) .0001? (/) \/10? (V 1 9 W) 10 9 ? (g) 10*? ' 10000* (6) 10 - 9 ? (h) 10*? 306. Between what two consecutive integers does the logarithm of each of the following numbers lie? Why? (a) 600 (c) 13 (e) 46923 (6) 5728 (d) 496,287 (/) 9 307. Between what two consecutive negative integers does the logarithm of each of the following numbers lie? Why? (a) .06 (c) .0008 (e) .00729 (g) 0.5 (6) .007 (d) .0625 (/) .00084 308. What is meant by saying that: (a) log 880 = 2.94448? (6) log 92.12 is 1.96435? (c) log 4.37 is .64048? Since 3585 lies between 1000 and 10,000, its logarithm lies between 3 and 4. It has been calculated as 3.55449. The integral part 3 is called the characteristic, and the decimal part .55449, the mantissa of the logarithm. Y 358.5 = 3585 MO, .'. log 358.5 = log (3585 MO) = log 3585- log 10 = 3.55449 -1 = 2.55449. That is, since log 3585 = 3.55449 log 358.5 = 2.55449 and similarly it can be shown that log 35.85 = 1.55449 log 3.585 = 0.55449 log .3585 = .55449-1* log .03585 = .55449 -2. Thus we see that (a) The characteristic can be found by inspection in all cases. '.'the number 589 lies between 100 and 1000, log 589 lies between 2 and 3 . . ' . log 589 = 2 + some mantissa. *Log 0.3585 = .55449 1 may be written in two other ways as follows: 1.55449 or 9.55449 10. The last method is the most practical, as we shall see as we proceed. ALGEBRA IN APPLIED MATHEMATICS 93 (b) The mantissa is the same for any given succession of digits, wherever the decimal point may be. (See last table of numbers with their logarithms.) (c) Asa result of (a) and (b) only a table of mantissas need be given. EXERCISES. SET XXXIX (continued) 309. What is the characteristic of the logarithm of: (a) 384? (c) .297? (e) A number of n integral places? (6) 5286? (d) Any number of millions? (/) Any decimal fraction whose first significant digit is in the first decimal place? (g) In the second decimal place? (h) In the third decimal place? (i) In the seventh decimal place? (j) In the n ih decimal place? 310. From the last exercise formulate a principle by means of which the characteristic of the logarithm of any positive number may be found. HI. THE FUNDAMENTAL THEOREMS OF LOGARITHMS (a) The logarithm of the product of two numbers equals the sum of their logarithms to the same base. 1. Leta^lO* 1 , then log a=h 2. Let &=ao h , then log b = k 3. .*. ab=W h+l2 , and log o& = Zi+k=log a-f log b. (b) The logarithm of the quotient of two numbers equals the logarithm of the dividend minus the logarithm of the divisor, all to the same base. 1. Leta=10 u then log a=h 2. Let 6 = 10 e , then log b=l 2 3. .'. 5" Jofe=10 Zl ~S and log -^-Z^log a -log b. (c) The logarithm of the n th power of a number equals n times the logarithm of that number. 1. LetaH=10 z , then log a=l 2. .'. a n =10' n , and log a n =nl=n log a. (d) The logarithm of the n th root of a number equals ^th of the logarithm of the number. 1. Leta=10', then log a=l i i i i i 2 : ,", a^10 n , and log a* = n = ~ n log a. 94 PLANE GEOMETRY Th. (c) might have been stated more generally, so as to include - x Th. (d) thus: Log a y =~ log a. The proof would be substantially the same as in Ths. (c) and (d). EXERCISES. SET XXXIX (concluded) Given log 2 = 0.3010, log 3 = 0.4771, log 5 = 0.6990, log 7 = 0.8451, and log 514 = 2.7110, find the following: 311. Log 6. 312. Log 14. 313. Log 7 10 . 314. Log A/2- 315. Log 42. 316. Log 5*. 317. Log 105. 318. Log 1.05. 319. Log V514- 320. Log 514 2 . 321. Log 1542. 322. Log 257^ 323. Logl799[=log(i-514-7)]. 324. Log\/3l 325. Log\/21. 326. Show how to find log 5, given log 2. IV. USE OF THE TABLE (a) Given a number, to find its logarithm. In the table on p. 103 only the mantissas are given. For in- stance, in the row beginning 43, and in columns headed 0, 1, 2, 3, ,9 will be found : N 1 2 3 4 5 6 7 8 9 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 This means that the mantissa of log 430 is .6335, of log 431 is .6345, and so forth, to log 439. Therefore log 431 = 2.6345, log 434 = 2.6375, log 43.7 = 1.6405, log 4.39 = 0.6425, log .438 = .6415 -1, log .0433 = .6365 -2. Now, since 437.8 is .8 of the way from 437 to 438, .'.log 437.8 is about .8 of the way from log 437 to log 438. .'. log 437.8 = log 437+.S of the difference between log 438 and log 437. .'. log 437.8 = 2.6405+.8 of .0010 = 2.6405+.0008 = 2.6413. This process of finding the logarithm of a number lying between two tabulated numbers is called interpolation. This is not wholly accurate, since the numbers do not vary as their logarithms, but it is sufficiently accurate for most practical purposes. If greater accuracy is desired, tables of five or six or even more places are used. The mantissas here given are correct to .0001. This will give a result which is correct to three figures in general, and an approximation to four figures, which will be sufficiently accurate for the computations in this book. ALGEBRA IN APPLIED MATHEMATICS 95 EXERCISES. SET XL. USE OF TABLE Find by using the table: 327. Log 49. 332. Log 14.7. 328. Log 723. 333. Log 14.73. 329. Log 1580. 334. Log 5.93. 330. Log 4285. 335. Log .00432. 331. Log 14.5. 336. Log 1.672. (6) Given a logarithm to find the corresponding number. The number corresponding to a given logarithm is called its anti- logarithm. Example: Y 0.4771 = log 3, .*. antilog 0.4771=3. 337. Log .00002376. 338. Log V29. 339. Log 5.692 3 . 340. Log -v/36.54. 341. Log .0057 6 . N 1 2 3 4 5 6 7 8 9 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 Here we see that antilog 4.6345 = 43100, antilog 0.6395 = 4.36, antilog 3.6405 = .00437. Now suppose we wished to find antilog 2.6417. V 2.6417 is .2 the way from 2.6415 to 2.6425. .'. antilog 2.6417 is about .2 the way from antilog 2.6415 to anti- log 2.6425. .'. antilog 2.6417 is about .2 the way from .0438 to .0439. .*. antilog 2.6417 = .04382. The following form is a good one for use. Required: Antilog 2.7361. Antilog 2.7364 = 545 2.7361 Antilog 2.7356 = 544 2.7356 Tabular diff . = 8 .'. Antilog 2.7361 = 5441 = 544.6. Diff. EXERCISES. Find from the table: 342. Antilog 1.9321. 343. Antilog 2.9049. 344. Antilog 2.7813. 345. Antilog 3.0354. 346. Antilog 1.0354. 347. Antilog 3.1628. 348. Antilog 4.8393. 349. Antilog 10.5843. SET XL. (concluded) 350. Antilog 0.6923 -2. 351. Antilog 8.6923 -2. 352. Antilog 7.5194 -10. 353. Antilog 9.2490 -10. 354. Antilog 10.4687 -10. 355. Antilog 3.5357. 356. Antilog 0.3471. 96 PLANE GEOMETRY (c) Computation by logarithms. Since many errors occur because of failure to arrange work care- fully, the pupil is advised to arrange all work in as compact and neat a form as possible. A few examples worked out in full may be suggestive, therefore we append the following: 1. In how many years will $600 double itself at 3% interest compounded annually? Solution: Let the number of years be n. At the end of one year the amount will be 1.03 of $600, at the end of the second year it will be 1.03 of 1.03 of $600, or 1.03 2 of $600, and so forth. /. at the end of n years it will be 1.03 n of $600. /. 1.03* X600 = 1200 or 1.03" = 2 /. n log 1.03 = log 2 and .'. ft^.j^gj log2 = .3010J. = .3Q10 log 1.03 = .0128j .0128 .'. log n= log .3010 -log .0128 log .3010 = 9.4786 -10 log .0128 = 8. 1072 -10 .'. log n= 1.3714 antilog 1.3714 = 23.5+ *. n = 23.5+, or the sum will double itself in 24 years. ^ . - 5286X\/427 2. Required the value of 3 . \/1754X3292 Solution : log 529 = 2.7235 log 528 = 2.7226 Tab. diff. =9 A 5.4.'. log 528.6_ = 2.7231 log 427 = 2.6304 .'. log \/427= 1.3152 /.log numerator = 4.0383 14.0383 -10 log 1760 = 3.2455 log 1750 = 3.2430 Tab. diff. = 25 _^4 10.0 ' ALGEBRA IN APPLIED MATHEMATICS 97 /.log 1754 = 3.2440 .'. log A/1754 =1.0813; log numerator = 14.0383-10 log 3300 = 3.5185 log 3290 = 3.5172 Tab. diff.= 13 J2, 2.6 or 3. '.log 3292 =3.5175 /. log denominator = 4.5988 ......... 4.5988 antilog 9.4409 - 10 = .276 .'. log fraction (log numerator log denominator) =9. 4395 - 10 antilog 9.4393-10 = .275 9.4393 - 10 Diff. for 1 = 16 2 /. antilog 9.4395 -10 = .275A=.275|. .'. Fraction = .2751 As soon as the pupil is able to interpolate mentally the left column may be omitted. Often in problems involving the process of evolution difficulties may arise owing to the fact that the characteristic may be negative and not a multiple of the divisor, while the mantissa is always positive. For instance, it may be desired to find \/0.03. log = 32- 12 = 384, 5 12 12.160 or by the second method -^ = - .*. v = =384. loU v o (c) How long will a body fall before acquiring a velocity of 520 feet per second? 416. The distance through which a body falls from rest varies as the square of the time during which it falls. (a) State this fact as a formula. (6) If a body falls 576 ft. in 6 sees., how far does it fall in 10 sees.? (c) How far will a body fall in 12 seconds? (d) How far will a body fall during the twelfth second? (e) How long will it take a body to fall a mile? 417. The pressure of wind on a flat surface varies jointly as the area of the surface and the square of the wind's velocity. (a) State this fact as a formula. (b) The pressure of the wind on 1 sq. ft. is 0.9 Ib. when the velocity of the wind is 15 miles per hour. What is the pressure of the wind against the side of a house 120 feet deep and 70 feet high when the wind is blowing 40 miles an hour? (c) What is the pressure on the same house when the wind is blowing 60 miles per hour? 112 PLANE GEOMETRY 418. The heat one derives from a stove varies approximately inversely as the square of one's distance from the stove. If I move my position from 10 feet away from a stove to 35 feet away, what part of the original heat will I then receive? 419. The law of gravitation states that the weight of a body varies inversely as the square of its distance from the center of the earth. (a) If a body weighs 10 Ibs. on the surface of the earth what will it weigh 5 miles above the surface? (Consider the radius of the earth to be 4000 miles.) (6) How high would a body have to be raised above the surface of the earth to lose half its weight? 420. The intensity of light varies inversely as the square of the distance from its source. (a) How much farther from a lamp 20 feet away must a piece of paper be moved to receive half as much light? (6) What is the relation of the intensity of light 15 feet from an electric light and 37 feet from the same light. 1421. Kepler proved that the squares of the times of revolution of the planets about the sun vary as the cubes of their distances from the sun. The earth is 93,000,000 miles from the sun, and makes a revolution in approximately 365 days. How far is Venus from the sun if it makes one revolution in 226 days? (Use logs.) 422. The strings of a musical instrument produce sounds by vibrating. The number of vibrations in any fixed interval of time varies directly as the length of the string, if the strings are alike in other particulars. A C string 42" long vibrates 256 times per second. A G string, like the C string except for length, vibrates 384 times per second. How long must it be? 423. The relation between the time of oscillation of a pendulum and its length is given by the following formula: If two pendulums are of lengths L and I respectively and the number of oscillations per second are T and t respectively, then: T 2 L ^J' (a) A pendulum which makes 1 oscillation per second is 39. 1* long. How often will a pendulum 156.4" long vibrate per second? ALGEBRA IN APPLIED MATHEMATICS 113 (6) How long would a pendulum have to be to oscillate once a minute? 1424. The relation between Q, the quantity of water in cubic feet per second passing over a triangular gauge notch, and H, the height in feet of the surface of the water above the bottom of the notch, is given by Q ac When H is 1, Q is found to be 2.634. What is the value of Q when H is 4? If the area of the reservoir supplying the notch is 80000 sq. ft., find the time in which a volume of water 80000 sq. ft. in area and 3 inches in depth will be drawn off when H remains constant and equal to 4 ft. (The relation between Q and H may be written Q=kH^ where k is a constant.) 425. In steam vessels of the same kind it is found that the relation between H, the horse-power; V, the speed in knots; and D, the displacement in tons, is given by HocVD*. Given H = 35640, 7 = 23, and D = 23000, find the probable numerical value of H when V is 24. 426. Some particulars of steam vessels are given. Assuming in each case the relation H.P. oc V 3 D% to hold, where H.P. denoted the horse-power at a speed of V knots and displacement D in tons, find in each case the probable H. P. necessary to give the indicated speed. Name H. P. V D (i) Paris 20000 20.25 15000 (ii) Teutonic 19.50 13800 (iii) Campania 22.10 19000 (iv) Kaiser 22.62 20000 (v) Oceanic 20.50 28500 (vi) Communipaw 23.00 23000 114 PLANE GEOMETRY g427.* Assuming that the circumference of a circle is 3| times its diameter, make a graph showing that the circumference varies as the diameter. Some of the numerous applications of Ratio, Proportion, and Variation to geometry will be given or suggested in the pages of this book. LIST OF WORDS DEFINED IN CHAPTER V Logarithm, base, characteristic, mantissa, interpolation, antilogarithm. Ratio, proportion, antecedent, consequent, terms, extremes, means, mean proportional, mean proportion, addition or composition, alternation, subtrac- tion or division, inversion. Variation, direct, inverse, joint. SUMMARY OF THEOREMS PROVED IN CHAPTER V 28. Any proportion may be transformed by alternation. 29. In any proportion the terms may be combined by addition. 30. In a series of equal ratios, the ratio of the sum of any number of ante- cedents to the sum of their consequents equals the ratio of any antecedent to its consequent. * As here, "g" preceding the number of an exercise indicates that its solu- tion involves a knowledge of graphs. CHAPTER VI SIMILARITY A. INTRODUCTORY THEOREMS EXPERIMENT I 1. Construct a scalene triangle. (a) Divide one side into 4 equal sects, and through the first point of division construct a line parallel to a second side of the triangle. Compare the lengths of the sects thus cut on the third side of the triangle. (6) Repeat the work of (a) , dividing the first side of the triangle into 5 equal sects. (c) Repeat the work of (a)', dividing the first side of the triangle into 9 equal parts and drawing the line through the fourth point of division. 2. Repeat the work of 1, using an equilateral triangle. Theorem 31. A line parallel to one side of a triangle, and cutting the other sides, divides them proportionally. (External division of the sides will be considered later.) Given: A ABC, D in ~BC and E in AB, so that I)E \\ CA. AE CD Prove: ^ (D ADEA EZ EB' (2) Similarly it can be shown that A EDC = CD "EB: h-DE . , . , h being PROOF (1) Triangles having one dimen- sion equal compare as their remain- ing dimensions. ADBE (3) But if A DCE the distance between DE and CA, (3) Data and equidistant. Is are everywhere 115 116 PLANE GEOMETRY (4) and (5) A PEA _ A EDC A DBE~ ADBE' /f*\ L/ (o; . .-==7= __ (4) Quantities equal to the same quantity equal each other. (5) Quotients of equals divided by equals are equal. (6) See (4). EA = (JD EB~ DB Cor. 1. One side of a triangle is to either of the sects cut off by a line parallel to a second side, as the third side is to its homologous sect. AE CD Suggestion: If = = ^=- why would EB DB ~AE+EB CD + DB \AA,/ /y x/ BI/\B Y/C//\C *-/ ^i/ / \G 2 /z V / w Cor. 3. \ Parallels i EB DB Cor. 2. A series of parallels cuts off proportional sects on all trans- versals. AB BC ,'BC ~CD - =CiA, if AXY and BZW are how drawn? Parallels which intercept equal sects on one trans- versal, do so on all transversals. Cor. 4. A line which bisects one side of triangle, and is parallel to the second, bisects the third. EXERCISES. SET XLVI. PROPORTIONAL SECTS 428. A sheet of ruled paper is useful in dividing a given sect into equal parts, (a) Explain. (6) Make such an instrument by using a sheet of tracing paper and drawing at least twenty parallels. (What must you be sure that these parallels do?) 429. Draughtsmen and designers sometimes divide a given sect into any required number of equal parts by the fol- lowing method: To di- vide AB into 5 equal SIMILARITY 117 parts, draw AC at any convenient angle with A B. Draw BD parallel to AC. Beginning at A, mark off on AC five equal sects a, b, c, etc., of any convenient length. Beginning at B, mark off on BD five sects equal in length to those on AC, a\ t bi, c\, etc. Join their extremities as in diagram. These lines divide A B into 5 equal sects. Prove that this is a correct method. 430. Among the applications of the propositions on parallel lines is an interesting one due to Arab Al-Nairizi (ca. 900 A. D.). The problem is to divide a sect into any number of equal parts. He begins with the case of trisecting a sect AB. Make BQ and AQi perpendicular to AB, and make BP=PQ = Q APi = PiQi. Then A XYZ is con- gruent to A YBP, and also to :APi. Therefore AX=XY = YB. In the same way we might continue to produce BP, until it is made up of n lengths BP, and so for A PI, and by properly join- ing points we could divide AB into n-f-1 equal parts. In par- ticular, if we join P and PI, we bisect^ the sect AB. Prove the truth of these statements. Divide a sect into seven equal parts by this method. 431. Find the cost of fencing the field represented in the diagram. Field is drawn to scale indicated, and the fence costs $2.75 per rod. 50 100 200 rods 432. If DE is parallel to BC in triangle ABC, compute the sects left blank from those given in the following table: AD DB AE EC AB AC 24 14 27 28 56 112 18 12 18 42 418 342 433. Divide a sect 11 units long into parts proportional to 3, 5, 7, and 9. 118 PLANE GEOMETRY First compute the lengths of the required sects, then construct them, and measure the sects so obtained. Compare the results. Which method is more convenient? Which is more accurate? 434. The accompanying diagram suggests a method for finding the distance from a point A to a second point B, visible from but inacces- sible to the first point A . Hints: C is selected, from which both A and B are visible. CB \\ED. 435. If D^is parallel to C in triangle ,4 5C, prove that = = =. AE EC ,ioc TT j +u j-+- 4.1, +AD+AE AD 436. Under the same conditions show that = = - 437. State Exs. 435^36 in words. EXPERIMENT H 1. Construct a scalene triangle. (a) Divide two sides of the triangle into 8 equal sects. Join the corresponding points of division. By comparing certain angles, establish a relation between the lines just drawn and the third side of the triangle. (6) Repeat (a), dividing the sides into 7 equal sects. 2. Repeat 1, using an equilateral triangle. Theorem 32. A line dividing two sides of a triangle proportion- ally is parallel to the third side. B UD Given: A ABC and ==, E , V.A/m JDU, & m a Kt ' EA To prove: DE \\ CA. Proof: Details to be sup- / J^sj^. plied by the student. A T~\ r> "r> 77T Draw CY \\ DE cutting BA in X. Then = ==. Why? Show why X must coincide with A. Cor. 1. A line dividing two sides of a triangle so that those sides bear the same ratio to a pair of homologous sects is parallel to the third side. SIMILARITY B. IDEA OF SIMILARITY 119 Earlier in the book it has been noted that two figures are called similar if they have the same shape. The symbol ( c/3 ), due to Leibnitz, for "is (or are) similar to," it has been pointed out, is an S thrown on its side. The S was doubtless used because it is the initial letter of the word "similis" (Latin for "like"). Before developing the subject we need, however, a more careful definition of similarity, for shape is only a vague notion and not a scien- tifically defined term. Before defining similar figures we must note the meaning of similar sets or systems of points. The points Ai, BI, Ci, .... and Az, B 2) C 2 , .... are said to be similar systems if they can be so placed that all the sects joining corresponding points, AiA*, BiB 2 , CiCz, . . . . , pass through the same point, and are divided by that point into sects having the same ratio. Flo. 1. FIG. 2. In Figs. 1 and 2, points A 1} B i} Ci, DI and A 2 , B 2 , C 2 , D 2 are similar systems. A\A Z , BiB 2 , CiC 2 , and DiD 2 pass through P and An p> ~p r~i "p JT)P = = - - r - In Fig - 2 where p lies on the longation of sects AiA z , BiB 2 , etc., it is said to divide these sects externally. The topic "External Division of a Sect" will be further developed in the "Second Study." The point P is called the center of similitude, and the ratio r is called the ratio of similitude. Similar figures are those which can be placed so as to have a center of similitude. 120 PLANE GEOMETRY The following are illustrations of similar figures : ^ A, (a) Triangles. ^SI B * (c) Circles. , (d) General Curvilinear We are now in a position to prove our right to the use of the double symbol ( ^ ) for congruence. Cor. Congruent figures are similar. If n-gon ABC .... n^AiBiCi . ... HI, they may be made to coincide. Then any point O may be taken as the center of OA OB similitude, and == = -i SIMILARITY 121 EXERCISES. SET XLVII. MEANING OF SIMILARITY 438. In (a) what is the ratio of similitude? What is the center of similitude? If ~OAi = A^A~ 2 and OB* = 0.5", what does ~BJ1 equal? 439. In (a) draw a sect through which will be divided by sides of triangles in the ratio of OAi to OA 2 . How many such lines can be drawn? 440. In (c) if 0#i = 5 cm. and 0# 2 = 1 dm. what is the ratio of OAi to OA Can you mention any other sects in this figure having the same ratio? 441. In (d) if OCi is 2 of dC 2 what is the ratio of similitude? Under the same conditions what is OA 2 if OA i is f "1 Frangois Vieta Shadows furnish familiar illustrations of similar figures, such cases the source of light is the center of similitude. In R6n6 Descartes The lens of the camera gives a figure similar to the object in front of it with the image inverted. 122 PLANE GEOMETRY In reducing or enlarging maps we have another familiar illus- tration of the application of the principle of similarity. EXERCISES. SET XLVII (concluded) 442. State any illustrations or applications of similarity with which you are familiar. 443. If the ratio of similitude is 1, what relation between the figures exists besides similarity. C. SIMILARITY OF TRIANGLES * Theorem 33. The homologous angles of similar \ triangles are equal, and the homologous sides have a constant ratio. Given : A XYZ A ABC. Prove: (I) ^X = %.A, XrY^^B, ^Z^Z-C. .. XY YZ ZX SIMILARITY 123 PROOF (1) .'.AXYZ ~ A ABC, it may be placed in the position of X\Y\Z]_ with O a center of similitude of it and A ABC. (2) :.OXi^OTi = OZi 'OA~~OB'~OC (3) :. CA (4) /. 4. OBC and (1) Data and def. of sim. figs. (2) Def. of center of similitude. (3) Why? (4) Why? (5) Why? (6) Why? (7) Why? (8) Why? (5) .'. What two 3* of A XiY^ or A X YZ are equal to what two 3? of A ABC? (6) Are the third 2 equal? (7) V %-X = ^A how can A XYZ be placed with respect to A ABC? (8) What proportion will this give us? (9) In how many ways will you have to superpose A XYZ in order to prove (II)? Complete the proof. ^Theorem 34. Triangles are similar when two angles of one are equal each to each to two angles of another. Given: ^.P = ^.A, ^.Q = ^.B in &PQS and ABC. Prove: A PQS A ABC. Suggestions for proof: Draw PiQi 1 1 AB and equal to PQ. _ _ Draw APiX and BQiZ. (1) If AJ\X \\BQZ, then ABQ is a O. (2) andAB=PQ (3) .'. APQS * AABC. If APiX intersects ~BQZ at 0, draw PiR \\ AC and intersecting CO at R. Draw QiR. *S (1) Why? (2) Why? (3) Why? 124 PLANE GEOMETRY ... BO AO . (4) Thenand (6) Any sect through O cut- ting PiQi in V and A in T 7 is TO AO divided so that s ~ (4) Why? (5) Why? (6) Why? (7) Why? (8) Why? (9) Why? (10) Def. of congruent figures. (11) Def. of similar figures. (7) :. (8) /. (9) /. (10) .'. APQS may be made to coincide with APiQi-R. (11) .'. A PQS " AABC. Discussion : Consider the instance in which PQ=AB. Note that the figures may be so placed that the center of similitude lies between them. Is a proof necessary for this case? EXERCISES. SET XLV1II. SIMILARITY OF TRIANGLES 444. State a necessary and sufficient condition for the similarity of (a) right triangles, (b) isosceles triangles, and (c) equilateral triangles. 445. If rays of light from a tree (TTi) pass through a hole (H) in a fence (F) and strike a wall, an inverted outline (0\O) of the tree will be seen on the wall. (a) Explain why this should be. (6) If the distance HD is 35 ft. and HP is 9 ft, and the height of the image (O^O) is 8' 8", find the height of the tree. (c) Under what conditions, if ever, will the image be the height of the tree? d446. The location of the image AI of a point A, formed in a photographer's camera, is approximately found by drawing a SIMILARITY 125 D straight line AAi through the center of the lens L. If CE is the position of the photographic plate, then AiBi is the image of AB. How large is AiBi if AB = 6 ft., LD=12 ft., and LF = Q in.? 447. Show that an object which appears of a certain height, will, when moved twice as far away, appear to be comparatively of only one-half the height. Hint: Show that XB = %AB. 448. To measure indirectly from an accessible point A to an inaccessible point B } construct AD perpendicular to the line of sight from A to B, and ED perpendi- cular to AD. Let C be the point on AD which lies in line with E and B. By measurement, ED is 100 ft., CD 90 ft., and CA 210 ft. What is the distance across the river? 449. A man is riding in an automobile at the uniform rate of 30 miles an hour on one side of a road, while on a footpath on the other side a man is walking in the opposite direction. If the dis- tance between the footpath and the auto track is 44 ft., and a tree 4 ft. from the footpath continually hides the chauffeur from the pedestrian, does the pedestrian walk at a uniform rate? If so, at what rate does he walk? 450. A mirror (referred to as a "speculum") has been used for crudely measuring the height of objects, such as trees. In the diagram a mirror is placed hori- G zontally on the ground at M. The obser- ver takes such a position that the top of the tree (C) is visible in the mirror. What distances must he measure to be able to compute the height of the tree (B(?)? NOTE. Light is reflected from the surface of a X M mirror at an angle equal to the angle at which it strikes it. 126 PLANE GEOMETRY 451. The accompanying diagrams show a simple device for measuring heights, using a square (such as ABCD), with a plumb- line (AE) suspended from one corner. BC and DC are divided into equal parts (say 10, 100, or 1000). Study the diagrams, and show how to use the square in each case. 452. Look up a description of the hypsometer, and construct one of wood, stiff pasteboard, or any material that can be used prac- tically. (See D. E. Smith, " Teaching of Geometry." Ginn & Company.) 453. The distance from an accessible point B to an in- accessible one, .4, was meas- ured in the sixteenth century by the use of drumheads. On a drumhead placed at B a sect I was drawn toward A, and another m toward an accessible point C. BC was* measured. The drumhead was then placed at C with m in the direction CB. / YYl The ratio of ~~ was noted. A sect p was drawn in the direction CA. Would any further measurements be necessay to make it possible to compute AB1 Explain. 454. To a convenient scale draw .N" a symmetrical roof, pitch 7 inches to the foot, on MN, which is to repre- sent 30' 6". The figure suggests the construction. SIMILARITY 127 455. Another way of determining the distance from A to an inaccessible point B is to align A, D, and B. Run DE at random. Run^LC parallel to DE. Align C, E, and B. Measure off FA equal to ED. Measure CF y EF, CA. Show how to compute AB and justify the method. 456. The accompanying diagram J- ^ - 4 shows how coordinate (squared) paper may be used to divide a given sect into a number_of equal parts (here 9). If the sect AB is laid off on one of the horizontal lines, the vertical lines will be perpendicular to it (CB-LAB). Draw AC. At the first division on CB from C draw FE \\ BA. (a) Why, then, is c EF = ? (6) What part of ~AB is JVM? y (c) Can you find a sect equal to f (AB) in the diagram? 457. The principle of the diagonal scale is the same as that underlying the division of a sect by the method of Ex. 456. In the accompanying diagram of a diagonal scale the unit is marked u. What is the length ofAB,CD, W, and MF1 u p I 9 I 1 1 8 7 1 6 n 5 A/ P k DJ K 2 B 1 1 1 1 BIO A 10 9876543210 1 2 3 458. Show how by the use of the diagonal scale to measure 0.3w, 0.56w, 0.75w, 1.8w, Z.lu, 0.35^, 0.82^, 2.67u. 459. Draw a triangle and measure the lengths of the sides to hundred ths of an inch by the use of a diagonal scale in which u = 1 inch. In measuring adjust the dividers to the side of the triangle, then apply them to the diagonal scale. 460. Make a diagonal scale on tracing paper, or of stiff card- board or of wood, and with its aid measure correct to .01 in. (a) 128 PLANE GEOMETRY the hypotenuse of a right triangle whose legs are 1 in. and 2 in.; (6) the diagonal of a square of side 1 in.; (c) the altitude of an equi- lateral triangle of side 2 in.; (d) Verify each measurement by com- putation, assuming the Pythagorean Theorem. d461. To find the height of an object AB: Place the rod CD in A. an upright position. Stand at X and sight over D to A. Move the rod any conve- nient distance, so that it takes the position Did, and D ^-- -"""A^"" CE It C X 1 C 1 sight over D l to A. What how may the height of the measurements are necessary, and object be determined from them? Theorem 35. Triangles which have two sides of one propor- tional to two sides of another and their included angles equal are similar. Given: A^BCandA . AC EC To prove: A ABC < A AiBiCi. Suggestions for proof: Place AAiBiCi in position of A 2 B 2 C. How do you know this is possible? Why will AtB2 be parallel to ZB? What foUows about %.B 2 A 2 C and ^.BACl Can you throw this theorem back to the one immediately preceding? EXERCISES. Set XLVIII (continued) 462. Extend your arm and point to a distant object, closing your left eye and sighting across your finger tip with your right eye. Now keep your finger in the same position and sight with your left eye. The finger will then seem to be pointing to an object some distance to the right of the one at which you were pointing. If you can estimate the distance between these two objects, which can often be done with a fair degree of accuracy, especially when there are buildings of which we can judge the width intervening, then you will be able to tell approximately the distance of your finger from the objects by the distance between the objects, for it will be ten times the latter. Find the reason for this. SIMILARITY 129 463. Explain how the accom- panying figure can be used to find the distance from A to B on opposite sides of a hill. CE=\BC, CD=\AC. El) is found by measurement to be 125 ft. What is the distance AB? 464. The accompanying picture shows a pair of proportional compasses. Note that rods A B and CD are of equal length and pivoted together at 0. (a) Prove AAOC&" BOD. /T\ T- x a (6) Prove -=5. (c) How may such an instrument be used to divide a sect? (d) How may such an instrument be used to con- struct a triangle similar to a given triangle? c*465. This picture shows a pair of sector compasses. It can be used in much the same way as the proportional compasses. Show how by means of it to get any part of a given sect. Hint: To bisect a sect, open the compasses so that the distance from 10 to 10 is equal to the given sect. Then the distance from 5 to 5 equals one-half the given sect. What part of the given sect would the distance from 6 to 6 be? c466. Show how by using the sector compasses to divide a given sect into 10 equal parts. c467. The sector compasses may be used to find the fourth proportional to three given sects as follows : From center on OL mark off OA equal to a. Open the a ^ sector until the transverse distance & at A equals 6. Then if OB be marked off on OL equal to c, the transverse distance at B is the required fourth proportional. Why? * c is prefixed to the numbers of exercises better suited to class dis- cussions than to written or home work. 9 130 PLANE GEOMETRY Theorem 36. // the ratio of the sides of one triangle to those of another is constant, the triangles are similar. Given: A ABC and A AiB^d with - = ~ = -. a b c To prove: A ABC < A AiBiCi. Suggestions for proof: AY = A^. DrawjfF. Prove A ABC A AXY. Prove A AXY a b NOTE : = = , and on AC lay off c . .a b - (given) and - = - Ci Z X (Why?) buty=ci. G E EXERCISES. SET XLVIII (concluded) d468. In the figure, E, F, G, and H are the mid-points of the sides of the square ABCD, and the points j) m are joined as shown. Show that the following triangles are similar: (a) EBC, ELB, ELY, and EUN; (b) BLZ,BST,tmdBXO; (c) EYZ and EUC; (d) YLZ, YMB, and BHP; (e) BOY and BHD; (/) BEZ and ATB; (g) BYZ and BHT; (h) BYF and BHC. d469. The following gives a procedure used in surveying for running a line through a given point parallel to a wholly inacces- B sible line. Study the diagram and notes and then justify the method. NOTES : Take C in sect MB. Select D at any convenient place. Run M F \ \ DB. Find E in MF in line with D and C. Run EN || AD, meeting AC at N. Then MN \\AB. Why is it that we can run paral- lels through M and D, whereas we cannot run the one through M directly? SIMILARITY 131 d470. The pantograph, invented in 1603 by Christopher Schemer, is an instrument for drawing a plane figure similar to a given plane figure, and is, hence, useful for enlarging and re- ducing maps and diagrams. The pantograph, shown in two positions, consists of four bars so pivoted at B and E that the opposite bars are parallel. Pencils are carried at D and F, and A turns upon a fixed pivot. BD and DE may A /? J?J? A D be so ad jus ted as to make the ratio of -T-~ (and hence ^-^and-r-^) whatever is desired. So if F traces a given figure, D will trace a AD similar one, the ratio of similitude being the fixed ratio -r-^. C FIG. 1 FIG. 2 (a) Prove that A, D, and F are always in the same straight line, (6) Prove that -j-= is constant and equal to -r-^. FIG. 3 FIG. 4 (c) Make a pantograph. A crude one can be made of stiff card- board and brass brads. NOTE: Figs. 3 and 4 show interesting elaborations of the pantograph. 471. Summarize the conditions necessary and sufficient to make triangles similar. 472. How may four sects be proved proportional? 132 PLANE GEOMETRY D. PERIMETERS AND AREAS OF SIMILAR TRIANGLES Theorem 37. The perimeters of similar triangles are propor- tional to any two homologous sides, or any two homologous altitudes. Cl x l Given: Aa&coo Ai6iCi with hA.b and /YiJ_6i. a+6+c a To prove: Suggestions for proof: a b c h (or or ) = . 61 ci hi = = -. W T hy? 3+6+c a = . Why? h a Prove = by showing A CBX v> ACiBiXi. Cor. 1. Any two homologous altitudes of similar triangles have the same ratio as any two homologous sides. Theorem 38. The areas of similar triangles compare as the squares of any two homologous sides. Given: Aa6c <*> AaifriCj. Aa6c c 2 / 6 2 To prove: Suggestions for proof: A<* *'.iL. Why, /iiOi hi bi - 61 -. Why? a 2 / 6 2 c 2 \ ^r^ or c7j- SIMILARITY 133 EXERCISES. SET XLIX. AREAS OF SIMILAR TRIANGLES 473. Draw a triangle. Construct a second one similar to it, having an area nine times as great. 474. Connect the mid-points of two adjacent sides of a parallelo- gram. What part of the area of the whole figure is the triangle thus formed? 475. Fold a rectangular sheet of paper from one corner as shown. The successive creases are to be equally distant from each other and parallel. Prove that the ratio of the successive areas between creases is 1 to 3 to 5 to 7, etc. 476. When, in forestry, shadows cannot be used, justify the following method of getting the height of a tree. A staff is planted upright in the ground. A man sights from S to the top and foot of the tree. His assistant notes where his line of sight crosses the staff. (a) What measurements does he need to take? (6) Assume a reasonable set of data and calculate AB. E. APPLICATIONS OF SIMILAR TRIANGLES Theorem 39. The altitude upon the hypotenuse of a right triangle divides the triangle into triangles similar each to each and to the original. Given : A abc with a _|_ b and h c- To prove: ABCH ^ ACHA < ABC. Suggestions for proof: %-B is common to the two right triangles BCH and ABC. 4-A is common to the two right triangles CHA and ABC. I. PROJECTIONS The word " projection" has a variety of meanings in general use. We refer to projecting ourselves into a situation; or to projecting a picture on a screen by means of a lantern. In geometry the word 134 PLANE GEOMETRY has a technical significance which we exemplify in the following little experiment. Place in the sunlight a table covered with a white cloth. Hold over the table (parallel to it) a plate of thin glass on which small figures of dark paper have been pasted. The sun will project these diagrams on the cloth. Upon reflection you will note a certain relationship between the point or sect represented on the glass and its manifestation on the cloth. To use scientifically the idea of projection we need exact defini- tions rather than these vague assumptions. Let us express these ideas in geometrical terms. The intersection of a perpendicular to a line with that line is called the foot of the perpendicular. The projection of a point on a line is the foot of the perpendicular from the point to the line. The pro- jection of a sect on a line is the sect cut off by the projections of its extremities on that line. For example: * F B F is projection of P on AB. F B A E EF is projection of CD on AB. a E CF is projection of CD on AB. EF is the projection of CD on AB. Would the projection ever be as long as the original sect? Ever longer? NOTE: These projections are sometimes referred to as orthogonal to dis- tinguish them from other types met with in higher mathematics. The notion of projections was originally obtained from that of shadows. The projection of a circle in one plane on another plane was its shadow. It is evident that a scientific study of shadows becomes very complicated. Consider, for instance, the effect on the shadow caused by the various relative positions of the planes and the positions of the light. Projective geometry, an advanced SIMILARITY 135 study, concerns itself with the more complex phases of the subject. In elementary geometry, we refer only to one very small instance of shadow geometry, reduced, as you note, to geometric definitions. These projections we refer to as right or orthogonal. Cor. 1. Each side of a right triangle is a mean propor- tional between the hypote- nuse and its projection upon the hypotenuse. NOTE: &BCH > ACBA^Fonn a B proportion involving BC, I3A, BH. Cor. 2. The square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. NOTE: a*=c.BH, and b z =c-HA. Why? This fact is very important in geometry, and has an interesting history. The first proof of the theorem is attributed to Pythagoras about 500 B. c., although the fact was known much earlier. Reference has already been made to the history of the theorem in Chapter I. Its later development consists of numerous proofs worked out by later mathematicians. In the following exercises specimens of such will be found, and further interesting proofs are contained in Heath's Monograph, "The Pythagorean Theorem." EXERCISES. SET L. PROJECTIONS. PYTHAGOREAN RELATION 477. What would be the projection of a sect 10" long on a line with which it makes an angle of (a) 30? (6) 45? (c) 60? 478. In the sixteenth century the distance from A to the inac- cessible point B was determined by means of an instrument called the "squadra." The squadra, like a modern carpenter's square, con- sisted of two metallic arms at right angles to each other. To measure AB the squadra was supported, as in the figure, on a vertical staff AC. One arm was pointed toward B, and the point D on the ground, at which the other arm 136 PLANE GEOMETRY pointed, was noted. By measuring AD and AC, show how AB may be computed. d479. Two stakes are set on a hillside whose slope is 20% (i.e., 20 ft. rise in 100 ft. measured along the slope). The distance between the stakes, measured along the slope, is 458 ft. What is the horizontal distance between them? 480. The accompanying drawing represents a plot of land divided as indicated. DE = 22'8", EB = 1W, BF = 50', FG (alley) = 16', GH = 150', HI = 66', and IK = 232'6". Find the length of AB in feet, and the area of triangle ACD in square rods. d481. The figure shows a ground plan of a zigzag or "worm" fence. The rails are 11 feet long, and a lap of 1 foot is allowed at each corner. Stakes, supporting the rider rails, are set along the boundary line. Find the amount of ground wasted by the construc- tion of such a fence 100 rods long. How much more fence is needed in this zigzag fence than in a straight one? d482. Let ABC be any right-angled triangle, right-angled at C, and let the square ABDE be described on the hypotenuse AB, overlapping the triangle. Prove that the perpendicular from E upon AC is of length b, and hence that the area of the triangle ACE is ^6 2 . Similarly, prove that the area of the triangle BCD is %a 2 . Notice that these two triangles have equal bases c and total height c. Hence prove that a 2 -j-6 2 = c 2 . 483. The great Hindu mathematician, Bhaskara (born 1114 A. D.), proceeds in a somewhat similar manner. He draws this figure, but gives no proof. It is evident that he had in mind this relation: Boundary Line Give a proof. SIMILARITY 137 d484. A somewhat similar proof can be based upon the following figure : If the four triangles, 1 + 2+3+4, are taken away, there remains the square on the hypo- tenuse. But if we take away the two shaded rectangles, which equal the four triangles, there remain the squares on the two sides. Therefore the square on the hypotenuse equals the sum of these two squares. Give details of the proof. d485. This exercise makes the Pythagorean Theorem a special case of a proposition due to Pappus (fourth century A. D.), relating to any kind of triangle. Somewhat simplified, this proposition asserts that if ABC is any kind of triangle, and MC, NC are parallelograms on AC, BC, the opposite sides being produced to meet at P; and if PC is produced making QR = PC; and if the paral- lelogram AT is constructed, then AT=MC-\-NC. For MC=AP?=AR, having equal altitudes and bases. Similarly, NC^QT. Adding, MC+NC=AT. If, now, ABC is a right triangle, and if MC and NC are squares, it is easy to show that A T is a square, and the proposition reduces to the Phythagorean Theorem. Show this. d486. The Arab writer, Al-Nairizi (died about 922 A.D.), attri- buted to Thabit ben Korra (826-901 A.D.) a proof substantially as follows: The four triangles T 7 can be proved congruent. Then if we take from the whole figure T and Ti, we have left the squares on the two sides of the right angle. If we take away the other two triangles instead, we have left the square on the hypotenuse. Therefore the former is equivalent to the latter. Give details of proof. d487. A proof attributed to the great artist, Leonardo da Vinci (1452-1519), is as follows: 138 PLANE GEOMETRY Q^ The construction of the following figure is evident. It is easily shown that the four quadrilaterals ABMX, XNCA, SBCP, and SRQP are congruent. /. ABMXNCA equslaSBCPQRS but is not congruent to it, the *M congruent quadrilaterals being dif- ferently arranged. Subtract the congruent triangles MXN, ABC, RAQ, and the pro- position is proved. a b A Give details of proof. d488. A proof attributed to President Garfield is suggested by the accompanying diagram. Work itout. a NOTE: a=a it b=bi, c=ci. ABDC is a trapezoid. What is the altitude of the trapezoid? Its bases? Its area? How else may the area of the trapezoid be found? d489. Show that if AB=a (in Fig. 3), (a) #C^ (6) BLm H E 490. In the middle of a pond 10 ft. square grew a reed. The reed projected 1 ft. above the surface of the water. When blown aside by the wind, its top part reached to the mid-point of a side of the pond. How deep was the pond? (Old Chinese problem.) 491. Show that the following dia- grams illustrate methods of representing FIG. 3. Moorish Design, from Mabel the square roots of integers. Sykes' Source Book of Problems for Geometry. B = BE = \3, etc. C D E F etc. SIMILARITY 139 AC is a, square. D E AG=AE=\/3, etc. NOTE : Such methods would not be practical for large numbers Why? J5 H The first of these methods is used on the "line of squares" on the sector compasses. 492. The Hindus said that triangles having the following sides are right triangles. How is the assumption they apparently made related to Theorem 40, Cor. 2? (a) 5, 12, 13. (6) 15, 36, 39. (c) 8, 15, 17. (d) 12, 35, 37. II. TRIGONOMETRIC RATIOS Let ABC be an acute angle. Drop a series of perpendiculars to BA from any points on BC. It will readily be noted that the right triangles formed, BPiRi, BP^Rz, etc.. A are similar, having the angle B in com- mon. The equality of the following ratios will result : (D -BPi BP^ BP, BP 2 ^5 = ^ and (3) Why? BP '4 BP& BRi JoRz BR$ BR BR$ If we think of $iABC as being generated by sect BC revolving counterclockwise from the position BA to BC we may call BA the initial side and BC the terminal side. These perpendiculars from points on the terminal side may be thought of as projectors form- ing on the initial side the projections of sects of the terminal side. We may then summarize the facts given as ratios in what preceded as follows : For any acute angle if perpendiculars be dropped to the initial side from any points on the terminal side the ratios (1) of the projector to the sect of the terminal side, (2) of the projection to the sect of the terminal side, and (3) of the projector to the projection of a sect of the terminal upon the initial side are constants. These ratios are given the names sine, cosine, and tangent, respectively. 140 PLANE GEOMETRY Initial Side TP Thus (1) the sine of angle ABC is=- BP (2) The cosine of angle ABC is = ~jp -^ (3) The tangent of angle ABC is BJ These are referred to as trigonometric ratios of an angle. If in the fixed A ABC, right angled at C, the side opposite A is called o, the side adjacent, a, and the hypote- nuse, h, fill in the following: EXERCISES. SET LI. TRIGONOMETRIC RATIOS 493. Make a table of the values of the sines, cosines, and tangents of angles of 30, 45, and 60. 494. Showbymeans of similiar trianglesthat the sine of 60 is the same as the cosine of 30. d495. When a wagon stands upon an incline, its weight is resolved into two forces, one the pressure against the incline, the other tending to make it run down the incline. Show that the force along the incline is to the weight of the wagon as the height of the incline is to its length. If the incline makes a 30 angle with the horizontal, with what force does a loaded wagon, weighing three tons, tend to run down the incline, i.e., disre- garding friction, what force must a team exert to pull it up the slope? 496. Fill out the following table: Polygon Dimensions Perimeter Area Parallelogram Base 18 Angle 60 300 300 Rectangle Base 18 300 300 Rhombus Angle 60 300 300 Square 300 300 SIMILARITY 141 D 497. To measure the height of an object AB by drawing to scale: Measure a distance CD towards A. Measure the angle ACB and the angle ADB. Then draw a plan thus: Representing, say, 100 ft. by a sect an inch long, draw EF to represent CD in the plan, and draw the angle HFG equal to the angle ADB, and the angle PEG equal to the angle ACB, and draw GH at right angles to EF prolonged. Measure GH. Show how to compute A B. (a) If CD = 175 ft., the angle ADB = 45, and angle DCB = 30, compute axis are to be found on a line parallel to the z-axis and 2 units above it; likewise we found that all points in this line, no matter how far it may be extended, will be +2 units from the o>axis. Another way of expressing these V-'axis & facts is to say that the path of all points the y-value of which is 2, is the line parallel to the z-axis y and 2 units above it, and next, that the 7/-value of every point in the line parallel to the x-axis and 2 units above it is +2. This is stated algebraically by means of the equation y = 2. EXERCISES. SET LV. THE EQUATION AS THE STATEMENT OF A LOCUS 564. Where are all points -2 units from the z-axis to be found? (Answer in a complete sentence.) 565. What can you say of all points in the line described in your answer to the last question? (Answer in a complete sentence.) 566. State the law which is obeyed by the line described in exercise 564 by means of an algebraic equation. 567. Where are all points + 10 units from the ?/-axis to be found? 568. What can you say of all points in the line described in your answer to the last question? 569. State this law by means of an algebraic equation. 570. Answer the last three questions, inserting the following words in place of "+10 units from the 7/-axis": * In this work, those for whom it is not review will find Auerbach, An Ele- mentary Course in Graphic Mathematics, Chapter I and Chapter III, pp. 22, 23, and 28-31, helpful. THE LOCUS 153 (a) - 15 units from the 2/-axis. (6) +12 units from the o>axis. (c) 18 units from the z-axis. (d) 7 units from the y-axis. (e) +13 units from the 2/-axis, 571. What is the y-value of every point in the line parallel to the z-axis and +6 units from it? 6 units from it? 572. What is the z-value of every point in the line parallel to the 7/-axis and +17 units from it? -17 units from it? 573. What is the path of every point whose t/-value is +18? -18? +20? -3? 574. What is the path of every point whose x-value is 30? 27? +16? -8? 575. Make a list of the equations expressing the facts stated, in order, in the last four questions. 576. (a) What is the ?/-value of every point in the o>axis? (6) What is the path of every point whose 2/-value is this? (c) What, then, is the equation of the z-axis? 577. x = 19 expresses algebraically what two facts? 578. What is the equation of the 2/-axis? Why? 579. What is the equation of the parallel to the 2/-axis through the point ( 5, 7)? The arrangement of points that completely fulfills a given geometric condition is called the locus of that condition. This arrangement usually gives rise to a line or group of lines either straight or curved. For instance, the locus of the condition expressed by the equation x = 7 is the line drawn parallel to the 2/-axis at a distance 7 units to the right of it. This is a brief way of saying that (1) all points in this line are 7 units to the right of the ?/-axis and (2) all points 7 units to the right of the 2/-axis lie in this line. Because of the idea of motion involved, another acceptable definition of the word locus would be : The complete path of a point that moves in accordance with some specified geometric condition. For instance, the complete path of a point that moves so that its distance from the ?/-axis is -7 is the line 7 units to the left of the 2/-axis and parallel to it. Hence this line is called the locus of the point which moves so as to remain constantly 7 units to the left of the i/-axis. 154 PLANE GEOMETRY EXERCISES. SET LV (concluded) 580. What is the locus of points: (a) 3 units from the z-axis? (6) -5 units from the 2/-axis? (c) 6 units from the x-axis? (d) 17 units from the t/-axis? 581. What two facts do you imply in the answer to each of the parts in the last question? 582. Give the equation expressing the condition which deter- mined each of the loci in exercise 580. 583. What is the locus of the condition expressed in each of the following equations? (a) z = 15 (b) y = -9 (c) x= -12 (d) y = 20 584. What locus is represented by the equation x 1 = 25? 585. What locus is represented by the equation (a) x = yl (d) x-y=lW (g) (6) x= -yt (e) x= -3y? (h) (c) x+y = lW (/) x = 3y7 (i) 586. Give the equation of the locus of a point: (a) Just as far from the -axis as from the y-axis. (b) Three times as far from the rr-axis as from the y-axis. (c) Three times as far from the y-axis as from the x-axis. (d) Minus five times as far from the y-axis as from the z-axis. (e) Minus seven times as far from the x-axis as from the y-axis. (/) Such that the sum of its distances from the axes is 11. (g) Such that three times its distance from the z-axis increased by 9 times its distance from the y-axis is 26. (h) Such that five times its distance from the z-axis diminished by twice its distance from the y-axis is 7. (i) Such that the sum of the squares of its distances from the axes is 49. (j) Such that four times the square of its distance from the z-axis increased by the square of its distance from the y-axis is 144. Check the answer to each part of the last two questions by plotting the equation. 587. The theorem of Pythagoras is employed to find the " equa- tion of a circle" about the origin as a center. THE LOCUS 155 Take any point P in a circle about the origin 0. Draw the ordi- nate MP. Let OM=x, and MP=y. Then If the radius OP =r, this becomes x 2 -}-y 2 =r 2 . This equation holds for the co- ordinates of any point on the circle, and is called the equation of the circle, r being any known number. Form the equation of the circle with the origin as center and (a) 7 as radius, (b) 7\/2 as radius. B. THE PECULIARITY OF THE PROOF OF A LOCUS PROPOSITION When we say that the locus of points on this page just one inch from its right edge is a line parallel to that edge, and one inch in from it, we really imply three facts. First, that any point on that line is one inch from that edge; second, that any point which is one inch from that edge and on the page is on that line; and third, that any point not on that line and on the page is not one inch from that edge. If the first of these three facts be called the direct statement, we already know that the second is its converse. The third is known as its opposite. In symbols, three theorems so related might be stated as follows : Direct theorem. If a = b, then c = d. Converse theorem If c = d, then a = b. Opposite theorem. If a^b, then c^d. Hence we see that while the converse of a fact simply inter- changes its data or what is given with its conclusion, the opposite of a fact negates both the data and conclusion. Now let us discover what we can as to the truth or falsity of converse and opposite theorems when the direct theorem is true. EXERCISES. SET LVI. DIRECT, CONVERSE, OPPOSITE The following exercises will help us in this task: 588. Form (1) the converse, and (2) the opposite of each of the following facts : (a) If a man lives in Boston, he lives in Massachusetts. 156 PLANE GEOMETRY (6) If it rains, the ground is wet. (c) If two lines meet at right angles they are perpendicular to each other. (d) All vertical angles are equal. (e) The supplements of equal angles are equal, or, If two angles are equal, their supplements are equal. (/) All men are bipeds. 589. (a) Of which of the six facts mentioned in the last exercise are the converse facts true? (b) Of which of them are the opposite facts true? (c) Of which of them are the converse facts false or at least not necessarily true? (d) Of which of them are the opposite facts either false or not necessarily true? 590. (a) Can you draw any conclusion as to the truth or falsity of converse and opposite theorems? (b) Test this conclusion with several more instances. Though a statement is true: (1) Its converse may or may not be true. (2) If its converse is true, its opposite is also true. (3) If its converse is false, its opposite is also false. The proof of this fact follows: Given: That when a = b, c=d, and when c =d, a = b. Prove : That when a ^ b, c ^ d. Proof : Suppose c = d. Then what follows? Why? What conclusion can you draw? Let us see how these conclusions help us to decide just how much must be proved in order to establish the truth of a locus proposition. Since the converse of a fact is not necessarily true, in order to prove a line a required locus, not only must we prove (1) that any point in it fulfills the required conditions, but also (2) that any point that fulfills the required conditions is in the line. It is, however, unnecessary to prove the opposite, since if the converse has been proved true, we know the opposite is also true. In short, if we know (1) and (2) are true, we know without further proof that any point not in the line does not fulfill the required conditions. Suppose we wished to prove Theorem 40. The locus of points equidistant from the ends of a sect isjhe perpendicular bisector of the sect. P, THE LOCUS 157 Given: XOY AOB so that AO=OB. Prove: XY is the locus of points equidistant from A and B. If we proved first that any point P in -^ XY is equidistant from A and B, and second that point PI which is equidistant y from A and B lies in line XY, what would x / x we know about any point not in X F? X X .'. (1) Prove: PA=PB, given: AO^OB /' and P in XY _\_AB at 0; and (2)_prove: A^- OPlLAB, given: PA =PiB and AO=OB. To prove (1): What parts of A AOP and A BOP do you know are equal? Are the A congruent? To prove (2): What parts of A AOPi and A BOPi do you know are equal? Are the A congruent? What fact must you prove to show that OP\.ABl Is there any other converse which we might have proved in place of the one here proved? Why are there two converses in this case? Cor. 1. Two points each equidistant from the ends of a sect fix its perpendicular bisector. How many points determine a straight line? EXERCISES. SET LVII. APPLICATIONS OF LOCUS 591. Show why a circle may be defined as the locus of points at a fixed distance from a given point. Describe without proof: 592. The locus of the tip of the hand of a watch. 593. The locus of a point on this page and just 3" from the upper right corner. 594. The locus of the center of a hoop as it rolls along the floor in a straight line. 595. The locus of the edges of the pages of a book as it is opened. 596. The locus of the handle of a door as it is opened. 597. The locus of the end of a swinging pendulum. 598. The locus of places described as 1 mile from where you are standing. 599* The locus of points 1' above a given shelf. d00. The locus of points 1' from a given pbelf. 158 PLANE GEOMETRY 601. The locus of the center of a circle as it rolls around another circle, its circumference just touching that of the other circle. 602. The locus of the center of a ball as it rolls around another ball, its surface just touching that of the other ball. 603. The locus of one side of a rectangle as it revolves about the opposite side as axis. 604. The locus of the entire rectangle in the last exercise. 605. The locus of a point at 3" from a fixed point P. 606. The locus of a point 3" from a given line. 607. The locus of a point equidistant from two parallel lines. 608. The locus of a point equidistant from two given points. 609. The locus of a point equidistant from two intersecting lines. 610. The locus of a point the distance <-d > from a given line I. 611. The locus of a point the distance <-d -> from a given point P. 612. The locus of a point the same distance from the center and the circumference of the circle c. We shall now prove one more veryimportant fact concerning loci. Theorem 41. The locus of points equi- distant from the sides of an angle is the bisector of the angle. I. Given: %.CBA, BXsp that ^CBX = %.XBA. P any point in BX. PR _\_AB cutting AB at R. TSA.BC cutting_BC at S. To prove: PR=PS. (Proof left to the student). II. Given: PI a point within the %.CBA, so that P\R (the _|_ to AB) =/V? (the _L to BC). To prove: PiB bisects %.CBA. ~S~ ~~G (Proof left to the student.) Cor. 1. The locus of a point equidistant from two intersecting lines is a pair of lines bisecting the angles. Hint: At what angle do the bisectors of any two adjacent angles formed by the pair of intersecting lines meet each other? At what angle, then, do the bisectors of the vertical angles meet each other? EXERCISES. SET LVII (concluded) 613. If a gardener is told to plant a bush 10' from the north fence does he know exactly where to plant i t ? If not, state another direction which might be given him that he may know just where to plant it. THE LOCUS 159 614. Is the second direction given the gardener the only one you could give him to have the bush definitely located? If not, state other directions which might have answered. 615. How many loci are needed to locate a point on the floor of the room? On any one of its walls? On the ceiling? 616. (a) Where would all points that are two feet from the floor of a room lie? (6) Where would all points 3' from the front wall lie? (c) Where would all points that are both 2' from the floor and 3' from the front wall lie? (d) Suppose a point was described as being 2' from the floor, 3' from the front wall, and 4' from a side wall. On how many loci is it? Exactly where is it? (e) How many conditions are needed to fix a point in a room? 617. A man wants to build his home at the same distance from two railroad stations, (a) Is the location of his home fixed? (6) If at the same time he wishes to build a half mile from the bank of a river which runs parallel to and four miles from the road connecting the stations, is the location of his home fixed? Make an accurate construction showing how many locations answer the description. 618. Prove theorems 40 and 41 by means of direct and opposite. 619. What is the locus of the vertices of triangles which have a common base and equal areas? 620. What is the locus of points dividing sects which connect a given point and a given line in the ratio of 5 to 7? 621. What is the locus of the vertices of triangles resting on a common base and having fixed areas in the ratio of 5 to 7? LIST OF WORDS DEFINED IN CHAPTER VII Locus, opposite. SUMMARY OF THEOREMS PROVED IN CHAPTER VH 40. The locus of points equidistant from the ends of a sect is the perpen- dicular bisector of the sect. Cor. 1. Two points each equidistant from the ends of a sect fix its perpendicular bisector. 41. The locus of points equidistant from the sides of an angle is the bisector of the angle. Cor. 1. The locus of points equidistant from two intersecting lines is a pair of lines bisecting the angles. CHAPTER VIII THE CIRCLE We have not only used the word "circle" very freely throughout the text, but have also used our compasses for the construction of the circle, or any part of it, whenever necessity arose. This has been due to the fact that the idea of a circle seems to be one with which all of us have grown up. But we have now reached the time to consider the idea scientifically, and add to our stock of facts concerning it. Up to the present we have thought of the circle as a portion of a plane, and the curve bounding it as its circumference. This is not the sense in which the word circle is used as we advance in mathematics, so we shall have to revise our notion of it. The word circle is used to refer to both the portion of a plane and the curve which bounds it the one to which it refers being determined by the context, but the definition covers only the boundary. When there is any danger of ambiguity the word circumference will be used in this text. A circle is a plane curve which contains all points at a given distance from a fixed point in the plane, and no other points* Thus we see that a locus definition may be given as a corollary to this one; namely, a circle is the locus of points at a given distance from a fixed point* The fixed point is called the center, and the given dis- tance (the distance from the center to any point on the circle) is called its radius (plural, radii). The sect through the center and terminated by the circle is called its diameter. At this point we may state several corollaries to these definitions. It will be left to the student to verify them. ^ Cor. 1. All radii of equal circles are equal. - Cor. 2. Circles of equal radii are equal. . 3. All diameters of equal circles are equal. * Refer to Exs. 585 (i), 586 (i), 588, 591, and 611 for previous illustrations of this definition. 160 THE CIRCLE 161 Cor. 4. A point is inside, on, or outside a circle, according as its distance from the center is less than, equal to, or greater than the radius. Cor. 5. A point is at a distance less than, equal to, or greater than the radius from the center according as it is inside, on, or outside the circle. A. PRELIMINARY THEOREMS ^ Theorem 42. Three points not in a straight line fix a circle. Given: Points A, B, D, not in a straight line. To prove: (1) A circle can be passed through A, B,D. (2) Only one circle can be passed through A, B,D. Suggestions for proof: What is the locus of the -D centers of all circles passing through A and D? Why? Through B and D? By means of the transversal XY in the diagram, show that PY and QX are not parallel, and that hence point C exists Why can no second point such as C\ exist? EXERCISE. SET LVIII. THE CIRCLE AS A LOCUS 622. An amusement park is to be located at the same distance from each of three villages V\, V 2 , and V 3 . V 2 is 5 miles from Vi, Vs is 6 miles from Vi, and V 3 is 8 miles from V 2 . Show by an accur- ate construction the actual location of the park. Can you think of any locations of V\, V 2 , and at the same time F 3 which would make it impossible to have a park so located? (Use sect V 2 V 3 io represent 8 mi.) V 2 -F 3 In the accompanying diagram, ^BOA is known as a central angle. Define such an angle. Any portion of a circle, such as BA is known as an arc. When referring to any definite arc, such as BA or CD, we write it thus: BA, CD. |^ Any two points on a circle (unless they are the ends of a diameter) are the ends of two arcs knownjis minor and major arcs.^For instance, CXD is the minor CD, and DYC is the major DC. The shorter of two arcs cut off by any two points on 11 162 PLANE GEOMETRY a circle is known as the minor arc, and the longer is known as the major. When not otherwise stated , the minor arc is referred to. The sect CD is known as a chord . Define the word chord. The diameters of a circle are simply chords. Why? Angles in a circle are said to intercept arcs, and chords are said to subtend arcs. Intercept comes from the Latin "inter," meaning "between," and "capio," meaning "to take," hence the angle intercepts or "takes the arc between its sides." Subtend comes from the Latin "sub," meaning "under," and "tendere," meaning "to stretch." Thus, as we see, the chord is the straight line which "stretches under the arc." Arcs are considered passive, and are referred to as being "intercepted by" angles, and "subtended by" chords, although in engineering the expression "the central angle subtended by an arc of n " is very common. If the expression is* used in this text it will only be when referring to engineering problems. Theorem 43. In equal circles equal central angles intercept equal arcs, and conversely. I. Direct. Given: OC = Od Method of proof, superposition, make coincide in proving the direct? Prove: AB^XY. II. Converse. Given: 0C= 0d, AB=XY. Prove: X,.ACB = %.XCiY. In superposing, which parts will you Which in proving the converse? B. THE STRAIGHT LINE AND THE CIRCLE Theorem 44. In equal circles, equal arcs are subtended by equal chords, and conversely. Suggestion: Prove A ACB^ What means have you of proving the triangles con- gruent in the direct? What in the converse? THE CIRCLE 163 EXERCISES. SET LIX. CONGRUENCE OF CURVILINEAR FIGURES 623. Prove that the curved figures * G BDEF and CFGH are congruent. The figure is based on a network of equi- lateral triangles. The vertices are the centers, and the sides the radii for the arcs. 624. In the accompanying figure prove that the curved triangles ABC, CUD, etc., are congruent. Also BCDEFG and DHML. H Y Theorem 45. A diameter perpendicular to a chord bisects it and its subtended arcs. Given: QO, diameter^ EF^L chord AB at D. Prove: AD=DB. AE=EB. BF=fA. Proof: AE=EB if what angles are equal? These angles are equal if what triangles are con- gruent? Write a complete proof. ^ By means of what angles can you prove BF Cor. 1. A radius which bisects a chord is perpendicular to it. Which of the methods of proving lines perpendicular can be applied here? Cor. 2. The perpendicular bisector of a chord passes through the center of the circle. Hints: Draw the radius which bisects the chord and prove the given bisector coincident with it, or treat as a locus. 164 PLANE GEOMETRY EXERCISES. SET LX. CONSTRUCTIONS BASED UPON CIRCLES 625. Give the con- struction for Fig. 1. From Carlisle Cathedral. (Figs. 1 and 2 applied.) 626. Inscribe Fig. 1 in a given circle. (See Fig. 2.) 627. Give the construction for the design shown in Fig. 3. Suggestion : Construct the equilateral AABC. Divide each side into three equal parts and join the points as indicated. The intersections are the centers, and AM is the radius for the arcs drawn as indicated. From Exeter Cathedral. (Fig. 3 applied.) 628. Construct Fig. IV. Suggestion : Construct the equilateral AABC. A,J1, and C are the centers for CB, CA, and AH. The semi- circles are constructed on the sides of the linear triangle as diameters. THE CIRCLE 165 629. A civil engineer wishes to continue the circular track AB for some distance. Suggest how he can do it. 630. From the measurements of a A piece of broken wheel a new wheel is to be cast of the same size. Show how to find the radius of the new wheel. V Theorem 46. In equal circles, equal chords are equidistant from the center, and conversely. I. In the direct what parts are known to be equal in AOBX and prove . II. State and the converse. EXERCISE. SET LXI. EQUAL CHORDS 631. The following method of locating points on an arc of a circle that is too large to be described by a tape is used by engineers. If part of the curve APB is known, take P as the mid-point. Then stretch the tape from A to B and draw PM perpendicular to it. Then swing the length AM about P, and PM about B, until they meet at L, and stretch the length AB along PL to Q. This fixes the point Q. In the same way fix the point C. Points on a curve can thus be fixed as near to- gether as we wish. Why is this method correct? A straight line is said to be tangent to a circle when it touches it once, and only once. Thus XY is tangent to QC if it touches at pt. P, and nowhere else. 166 PLANE GEOMETRY V ^Theorem 47. A line perpendicular to a radius at its outer extremity is tangent to the circle. Given: OO, AB J_ radius OC at C. Prove: AB tangent to 00. PROOF (1) Given. (2) Authorities left to the student to insert. (1) AB J. radius at C. (2) .-. AB touches OO at C. (3) It remains to prove that no other point such as P in AB touches _OO._/. Draw sect OP. (4) OC^L AB^_ (4) (5) .'.OPjLAB. (5) (6) .*. OP > OC. (6) (7) /. P lies outside of 00. (7) (8) /. CVand only C in AB is in OO. (8) (9) .'. AB is tangent to OO. (9) The first three corollaries following are partial converses of this theorem. We are not yet ready to prove the converse of Theorem 6. It will be proved later under the topic of inequalities. For the present, then, we shall add it as a postulate. Postulate, the third postulate of perpendiculars: The shortest distance from a point to a line is the perpendicular to that line. COT. 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. Given: Xf tangent to OO at P, radius UP. Prove: XY OP. Suggestions f or proof : Draw OQ to any point QmXY. Where doesQ lie? How does OQ compare with OP? What conclusion can be drawn with respect to OP? Cor. 2. The perpendicular to a tangent at the point of contact passes through the center of the circle. Hint: Show that KP coincides with the radius CP, and .'. C lies on KP. - - THE CIRCLE 167 ** Cor. 3. A radius perpendicular to a tangent passes through the point of contact. Hint: Draw the radius to the point of con- tact P, and show that CR coincides with CP. Cor. 4. Only one tangent can be drawn to a circle at a given point on it. Hint: How many s can be erected to CR atjR? EXERCISES. SET LXII. THE TANGENT AND THE CIRCLE 632. When a ray of light strikes a spherical mirror (represented in cross section by the arc of a circle), the angle of incidence is found by drawing a tangent to the circle at the point of incidence, and erecting a perpendicular to the tangent at that point. In this case the perpen- dicular (called the normal) is a diameter. Why? d633. The line of propagation of a sound wave also follows the law of reflection of a ray of light, namely, that the angle of incidence is equal to the angle of reflection. The circular gallery in the dome of St. Paul's in London is known as a whispering gallery, for the reason that a faint sound produced at a point near the wall can be heard around the gallery near the wall, but not elsewhere. The sound is reflected along the circular wall in a series of equal chords. Explain why these chords are equal. 634. What is the locus of the centers of a number of hoops of different sizes (one inside the other) tied together at one point? 635. What is the locus of the centers of all circles tangent to a given line at a given point? 636. What is the locus of the center of a wheel as it rolls straight ahead along level ground? Prove this fact. 637. What is the locus of the centers of all circles tangent to both sides of an angle? 638. Two straight roads of different width meet at right angles. A is the nar- rower, B the wider. It is desired to join them by a road the sides of which are arcs of circles tangent to the sides of the straight roads. What construction lines are necessary? Draw a figure. 168 PLANE GEOMETRY 639. How would you construct a tangent to a given circle at a given point P on the circle? Theorem 48. Sects of tan- gents from the same point to a circle are equal. Why are A OPX \nd OPY congruent? Circles are said to be tangent to each other when they are tangent to the same line at the same point. Why is OCi tangent to Why to QC 3 ? Why is QC 2 tangent to OC 3 ? Circles are externally tangent when their centers lie on opposite sides of the common tangent. Name two pairs of circles that are externally tangent in the diagram. Circles are internally tangent when their centers lie on the same side of the common tangent. Which circles in the diagram are internally tangent? EXERCISES. SET LXIIL TANGENT CIRCLES 640. (a) How are the hoops mentioned in Ex. 634 related to one another? (6) How is the line in which their centers lie related to their common tangent? WTiy? (c) Is this fact true of the centers of two cog-wheels when they A mesh? Why? The line of centers of two circles is the sect con- necting their centers. Theorem 49. The line of centers of two tangent circles passes through their point of contact. (For suggestions see Ex. 634, p. 167, and Ex. 640, p. 168.) THE CIRCLE 169 The common chord of two intersecting circles is the sect connecting their points of intersection. EXERCISES. SET LXIII (concluded) 641. How is the line of centers of two intersecting circles related to their common chord ? Prove your answer. 642. If two cog-wheels mesh, show that the point where they mesh is in a straight line with the centers of the wheels. 643. (a) Show how to construct an equilateral Gothic arch. (See the accom- panying diagram.) Suggestion : Construct the equilateral tri- angle ABC. With A and B as centers and AB as radius, construct the arcs BC and AC. (6) If E, F, and D are the mid-points of the lines AC, CB, and AB, respec- tively, prove that equal equilateral tri- angles are formed. (c) Construct the equilateral arches ADE and DBF and the curved triangle EFC. Suggestion: Points D, E, C, F, B, and A are the centers and AD is the radius for the arcs drawn as indicated. d644. An application of geometry to engineering is seen in cases where two parallel streets or lines of track are to be connected by a "reversed curve." If the lines are AB and CD, and the con- nection is to be mad^ from B to C, as shown, we may pro- ceed as follows: Draw BC and bisect it at M . Erect PO the perpendicular bisector of BM, and BO perpendicular to AB. Then is one center of curvature. In the same way fix 0\. The curves may now be drawn, and they will be tangent to AB, to CD, and to each other. Prove that the curve BMC is a reversed curve tangent to AB and CD; i.e., prove (a) BM tangent to AB at B^CM tangent to CD at C; (b) BM tangent to CM at M] (c) BM = CM. 170 PLANE GEOMETRY 645. The figure represents a Persian arch. Triangles ABC and DEF are congruent and equilateral. The centers of the upper arcs MC and NC are respectively D and F\ while the lower arcs are drawn with the cen- ter E. Prove that the area of the arch equals the area of triangle ABC. 646. The trefoil ADBF, etc., is constructed from circles described on the semisides of A ABC. The points D, E, and F are the centers for the arcs which are tangent to the sides of AABC, and which form the trefoil HYKZGX. If PD and RF in Fig. 1 are radii for the arcs ZK and YK, prove that PD = FR. FIG. 1. FIG. 1. applied. d647. The semicircle AGHB in Fig. 2 is constructed on AB as diameter, and CD is perpendicular to AB at its mid-point. (a) Construct arcs CH and CG tangent to the line CD at point C, and to the semicircle. Suggestion : Make KC = AD. Draw NE, the perpendicular bisector of KD, meeting CK extended at E. E is the center for the arc CH. What general problem in construction of circles is involved here? (6) If the arcs CH, drawn with E as a center, and HB drawn with D as a center, are tangent at H, prove that the points D, H, and E are collinear. A D B FIG. 2. Three-centered ogee archea. THE CIRCLE 171 (c) Prove that C, H, and B are collinear. Suggestion: Join C and H, and B and C, and prove that each is parallel to DK. (d) If AB = 8, and CD = 8, find the length of CE. Suggestion : KD = 4 A/5. Compare the sides of the similar triangles KCD and KNE. Find KE and hence C#. (e) IfAB = s and CZ) = /i, find the length of CK. 648. What must be the relation between the length and width of the rectangle A BCD in order that the tangent circles may be inscribed as shown. Stair Railing. 649. ABCD is a parallelogram with four tangent circles inscribed. (a) If the lines AC and BD are supposed to be indefinite in extent, show how to construct circle tangent to the lines A C, AB, and BD, and circle X tan- gent to lines AC and BD and to circle 0. (b) If E is the mid-point of AB, and and X are the centers of the circles, prove that the points E, 0, and X are collinear. 650. Fig. 1 shows a trefoil formed of the three circles X, Y, and Z tangent to each other at the points T, S, and R. It is E inscribed in the circle as shown. ^~~ /^T^\ 13t (d) Show how to construct the figure. Solution : Circumscribe an equilateral triangle about the circle. Connect each vertex with the center. Inscribe a circle in each of the triangles FOG, GOE, and EFO. (b) Prove that the small circles are tangent to the large circle and tO each Other. FIG. I. Trefoil fcrmed cf tangent circlea- 172 PLANE GEOMETRY 651. (a) Construct the quadrifoil of tangent circles inscribed in a square so that each circle is tangent to two sides of the square and to two other circles. E - Inlaid tile design. FIG. 2. (Fie. 2 applied.) (6) Prove that the lines joining the centers of the circles form a square. C. THE ANGLE AND ITS MEASUREMENT v Theorem 50. In equal circles central angles have the same ratio as their intercepted arcs. Given: OC=JDCi, %.ACB&nd 4 XCiY, AB and XY com- mensurable. Prove: Suggestions for proof: Select a unit of measure for AB and XY Divide these arcs into such units and connect the points of division with the centers C and C\. What can you say of all the central angles thus formed? How do %.s_BCA and YCiX compare? How does BA compare with YX* Cor. 1. A central angle is measured by its intercepted arc. Suppose 4. a were the angular unit of measure, and a the circular unit of measure in a circle of radius r. What would be the numerical measure o What would be the numerical measure o How do these numerical measures compare? THE CIRCLE 173 Up to the present time we have emphasized the fact that a magnitude can be measured by a unit of the same kind only. We must then justify a statement such as that given in Cor. 1, Theorem 50. This corollary should be stated as follows: The ratio of a central angle of a circle to the angular unit is equal to the ratio of its intercepted arc to the circular unit ; or : The numerical measure of a central angle of a circle is equal to the numerical measure of its intercepted arc. But since this fact is one frequently referred to, and the correct statement of it is so lengthy, mathematicians have agreed to the abbreviated statement given in Cor. 1. The symbol " 2 " is used for "is measured by." It suggests the ideas of both equality and variation. A secant is a straight line which intersects the circle. EXERCISES. SET LXIV. SECANT AND CIRCLE = 49, 652. (a) Show by a graphic solution of the equations x 2 and x = 3, that a secant cuts a circle in two points. (b) What numbers would have to replace 3 in the second equa- tion to change the equation to one of a tangent? )f Theorem 51. Parallels intercept equal arcs on a circle. p ^T^ Q x _ T Q Case I. When the parallels are a secant and a tangent. (Fig. 1.) Given: PQ tangent to circle C at T, secant RS \\ PQ, cutting C at R and S. Prove: ST = TR. Suggestions for proof : Draw diameter through T, cutting QC in M. What relation exists between TCM and PQ? Then what relation^exists between TCM and ~RSt What follows as to ST and TRt 174 PLANE GEOMETRY Case II. When both parallels are secants. (Fig. 2.) (Proof left to the student.) Suggestion : Draw the diameter perpendicular to one secant. Case III. When both parallels are tangents. (Fig. 3.) Draw the diameter through T, and give the proof in full. EXERCISES. SET LXV. CIRCLES 653. Does it make any difference in what order the cases under Theorem 51 are proved, if the proofs are given as suggested? 654. Can you suggest methods of proof for Cases II and III which depend upon Case I? Which methods do you prefer? Why? 655. Construct a diagram consisting of: (a) Two concentric circles whose radii are in the ratio of 1 to 3. (6) Six circles lying between them, tangent to them, and each tangent to two others. 656. Make a diagram 01 the mariner s compass, putting in six- teen points of the compass. Angles such as ABC in Figs. 1, 2, 3, Theorem 52, are called inscribed angles. Their vertices are not only in the circle, but their sides are chords. An angle is said to be inscribed in a semicircle when its sides intercept a semicircle, in less than a semicircle when its sides intercept a major arc, and in more than a semicircle when its sides intercept a minor arc. Since measurement is but a numeric relation, axioms may be converted into authorities for statements concerning measure- ment by simple changes such as those illustrated by the following: The measure of the sum of two magnitudes of the same kind is If X OC ft ) equal to the sum of their measures. E.g. : , ~ > x -\-y^a-\-b. The measure of any multiple of a magnitude is equal to that multiple of its measure. E.g.: If xB 1 and 2 draw CX\\AB,and reduce case 3 to case 2 by drawing any secant BSR. ^ Theorem 55. A tangent* is the mean proportional between any secant and its external sect, when drawn from the same point. Given: 0C, AB tangent at A and secant BP cutting GC at Q and P. VP _AB V ^rr~ = 'rnzz Jr Suggestion: Prove * By "tangent" in such cases is meant the sect from the point to the point of tangency. THE CIRCLE 179 EXERCISES. SET LXVIII. TANGENT AND SECANT 670. (a) Assuming that the diameter of the earth is 8000 mi., how far can a man see from the top of a building 200 ft. high? (The height of the building is measured along the prolongation of the diameter.) (6) How far can one see from the top of a mountain 1000 ft. high? 2000ft. high? 3000 ft. high? 4000 ft. high? (c) How far can one see from a balloon 1500 ft. above the sea? d671. Galileo (1564-1642) measured the heights of the moun- tains on the moon, some of which are as much A M as 6 mi. high, as follows: ACS was the illu- minated half of the moon just as the peak of the mountain M caught the beam SM of the c\ rising or setting sun. He measured the dis- tance AM from the half -moon's straight edge A B to the mountain peak M. Then by using the known diameter of the moon, show how he was able to com- pute the height of the mountain. d672. Since the earth is smaller than the sun, it casts a conical shadow in space (umbra), from within which one can see no por- tion of the sun's disk. If S is the center of the sun, E the center of the earth, and V the end or vertex of the shadow, prove that the length of Approximately, ES = 92,900,000 the shadow, VE= ESXEB SD -EB miles, SD = 433,000 miles, and # = 4,000 miles. Compute VE. 673. Look up the terms "umbra" and " penumbra, "and answer the following questions: (a) How are the umbra and penumbra affected by a change in the distance apart of the lumi- nous and opaque bodies? (6) If a golf ball is held be- tween the eye and the sun, is there any penumbra? Explain. 180 PLANE GEOMETRY (c) What may be said of the umbra if the luminous body and the opaque body are of the same size? (d) What is the shape of the umbra if the luminous body is larger than the opaque body, as in the case of the sun and the earth? (Exs. 672-673 are taken from Betz and Webb, Plane Geometry.) D. MENSURATION OF THE CIRCLE We now come to a very important section of our plane geometry the section which develops the formula for the length of the circle and for the area enclosed by the circle. The latter is known as the area of the circle. A polygon is said to be cir- cumscribed about a circle when r inscribed each of its sides is tangent to the fl circle circle, and to be inscribed in a circle when each of its sides is a chord of the circle. In the first case, the circle is said to be inscribed in the polygon, and in the second case, the circle is said to be circumscribed about the polygon. Theorem 56. A circle may be circumscribed about, and a circle may be inscribed in, any regular polygon. Given: The regular n-gon PQRST . . . To prove: 1. A circle may be circumscribed about PQRST . . . 2. A circle may be inscribed in PQRST . . . Hints for proof 1 : Three non-collinear points determine a circle. R Draw OP, OQ, OR, OS. By means of A OPQ and ORS prove OS=OP. To what triangle can AOST be proved congruent? Would it make any differ- ence how many sides the original polygon had? Give the details of the proof. Let be the center of the circle determined by P, Q, and R. B THE CIRCLE 181 Hint for proof 2 : If is the center of the circumscribed circle, what are PQ, QR, RS, etc.? Cor. 1. An equilateral polygon inscribed in a circle is regular. Hint: In Fig. 1 what is the measure of each of the angles P, Q, R, etc.? Cor. 2. An equiangular polygon circumscribed about a circle is regular. Suggestion: In Fig. 2 connect consecutive points of contact B, C, D, etc. Prove ABRC, ACSD, etc., congruent isosceles triangles. What can then be said of the sums of any two of the equal sides of the triangles? (Such as RC+CS, SD+DT, etc.) An angle such as $.POQ is called a central angle of the regular polygon. EXERCISES. SET LXIX. REGULAR POLYGONS AND CIRCLES 674. Make a design for tiling or linoleum pat- terns based uponinscribed equilateral hexagt 675. Make a copy of the accompanying design of arose window of six lobes. (Fig. 1.) 676. Make an accurate construction of the accompanying design representing a Gothic win- dow. (Fig. 2.) d677. Fig. 3 shows a star in- Fig. i. scribed in a given square with all of its vertices on the sides of the square. (a) Construct a figure in which points K,L,M,N, etc., shall be the mid-points B C- Fig. 2. D E of AE, EB, BF, FC, etc. (6) Show that a circle can be circum- scribed about the star constructed as in (a) and find its radius, if AB = a. 678. If a series of equal chords are laid off in succession on a circle, what relation exists between: (a) the arcs subtended by the chords? (6) the central angles intercepting the arcs? Fig. 3. (c) the inscribed angles formed by any two successive chords? 182 PLANE GEOMETRY 679. If the series of equal chords mentioned in the last exercise were such as finally to form an inscribed polygon, what kind of polygon would it be? Why? 680. How are the central angles of a regular polygon related? 681. Make a table showing the number of degrees in the central angle of a regular inscribed polygon of 3, 4, 20 sides. 682. Write a formula by means of which we can obtain the number of degrees in the central angle of any regular polygon. 683. Which of the angles found for the table in Ex. 681 can we construct by means of compasses and straight edge? 684. Inscribe in a given circle each of the regular polygons you can by the means mentioned in the last exercise. (Do not go beyond the sixteen-sided polygon.) 685. Express as a formula the number of sides of the regular polygons you can inscribe in a circle up to this point. Theorem 57. // a circle is divided into any number of equal arcs, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the points of division form a regular circumscribed polygon. Given: QO with AB = BC = = FA. (1) Chords AB, BC FA. (2) GH, HK, etc., tangent to QO at A, B, .... F respectively. To prove: (1) ABC . . . .F a regular inscribed polygon. (2) GHK. . . .N a regular circumscribed polygon. Hints: To prove (1) use Cor. 1, Theorem 56. To prove (2) use Cor. 2, Theorem 56. Cor. 1. Tangents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides. How do the vertices of the regular inscribed polygon divide the circle? Cor. 2. Lines drawn from each vertex of a regular polygon to the mid-points of the adjacent arcs subtended by the sides of the polygon form a regular inscribed polygon of double the number of sides. Hint : Show that the polygon is equilateral. Why ? Or throw the corol- lary back to the proposition. THE CIRCLE 183 Cor. 3. Tangents at the mid-points of the arcs between con- secutive points of contact of the sides of a regular circumscribed polygon form a regular circumscribed polygon of double the number of sides. How does the corollary rest on the theorem? Cor. 4. The perimeter of a regular inscribed polygon is less than that of a regular inscribed polygon of double the number of sides; and the perimeter of a regular cir- cumscribed polygon is greater than that of a regular circumscribed polygon of double the number of sides. (Proof left to the student.) HISTORICAL NOTE. This theorem presupposes the possibility of dividing the circle into any number of equal arcs. In the table found in answer to Exercise ]681 it was probably seen that the number of equal arcs into which we are at this time able to divide the circle is very limited, but we have not yet learned to divide the circle exactly in as many ways as possible by means of elementary geometry. Some other methods will be discussed later. As early as Euclid's time it was known that the angular magnitude about a point (and hence a circle) could be divided into 2 n , 2 n -3, 2-5, and 2 n -15 ' equal parts. In 1796 it was discovered by Karl Friedrich Gauss, then only sixteen years of age, that 2-17 equal parts of the circle could be found by the use of only the straight edge and compasses. Gauss also showed that in general it is possible to construct all regular polygons having (2+l) sides, when n is an integer and (2 n +l) is a prime number. He went still further and proved that regular polygons, having a number of sides equal to the product of two or more different numbers of this series, can be constructed. EXERCISES. SET LXIX (continued) 686. Show that according to Gauss's formula, regular polygons of 3, 5, 17, and 257 sides can be constructed. 687. Inscribe a square in a circle, and by means of it a regular octagon, a regular 16-sided, and a regular 32-sided polygon. 688. What is the perimeter of the square in terms of the diameter of the circle in the last exercise? How do the perimeters of the octagon, hexadecagon, and 32-sided polygons compare with it? With the circle? 689. Repeat exercises 687 and 688 with respect to the inscribed equilateral triangle. 690. Repeat exercises 687 and 680 with respect to the regular circumscribed square and triangle. 184 PLANE GEOMETRY 691. Between what two values will the length of the circle always lie? 692. Between what two values will the area of the circle always lie? From Cor. 4, Theorem 57, and Exercises 688-692, it is seen that though the perimeter of the inscribed polygons increases as the number of its sides increases, it is always less than the length of the circle; and that while the perimeter of the circumscribed polygon decreases under the same conditions, it is always greater than the length of the circle. In the first case the perimeter and likewise the area of the regular inscribed polygon are increasing variables. They are always less than the perimeter and the area of the circle, which are fixed or constant. In the second case, the perimeter and the area of the regular circumscribed polygon are decreasing variables which are always greater than the perimeter and the area of the circle. In the first case we say that the perimeter and the area approach as superior limits the circle and its area, while in the second we say that the perimeter and the area approach as inferior limits the circle and its area. Thus, if p n represents the perimeter of a regular inscribed polygon of n sides, a n) its area, P n the perimeter of a regular cir- cumscribed polygon of the same number of sides, A n , its area, and c the circle in and about which they are inscribed and cir- cumscribed, and C its area, we say that as the number n H in- creased, p n approaches c as its limit, P n approaches c as its limit, a n approaches C as its limit, and A n approaches C as its limit. These statements are briefly written as follows: p n =c } P n =c, a n = C, A n = C. POSTULATES OF LIMITS 1. The circle is the limit which the perimeters of regular inscribed and circumscribed polygons approach if the number of sides of the polygon is indefinitely increased. 2. The area of the circle is the limit which the areas of regular inscribed and circumscribed polygons approach as the number of sides of the polygon is increased. THE CIRCLE 185 Theorem 68. A regular polygon the number of whose sides is 3-2" may be inscribed in a circle. (Proof left to the student.) EXERCISE. SET LXIX (continued) 693. Inscribe a regular polygon of 3-2 4 sides. Theorem 59. If i n represent the side of a regular inscribed polygon of n sides, i 2n the side of one of 2n sides, and r the radius of the circle, i' 2ft = V2r 2 - r\/4r* - ft. Given: QO of radius r, AB (i n ) the side of a regular in- ~ scribed n-gon, AC (i 2n ) the side of the regular inscribed ^^ -\ 2n-gon. j ir^* | ^\g Prove: i 2n = \/2r 2 -r\/4r 2 i n 2 . PROOF Draw radius OC cutting AB at D. Draw AO- (1) Then OC A.AB and AD= % (1) 'Why? (2) .'. AC 2 ^(i 2n ) 2 ^( jjrl +CD (2) Why? (3) But CD =r-OD. (3) Why? (Note AOAD.) (5) . ' . t 2n = \/2r - - r Vr 2 - i n z EXERCISES. SET LXIX (continued) 694. Given a circle of radius 1 unit, compute: (a) The length of a side of the inscribed square. (6) The length of a side of the inscribed regular octagon. (c) The length of a side of the inscribed regular 16-sided polygon. (d) The perimeters of each of these polygons. 695. Do the same as you did in the last exercise for the regular hexagon, dodecagon, and 24-sided polygon. 696. Do the same as you did in Ex. 694, using a diameter of 1 unit, 186 PLANE GEOMETRY 697. Do the same as you did in Ex. 695, using a diameter of 1 unit. 698. (a) Which computation is simpler that using a radius of 1 unit or a diameter of 1 unit? (6) If the radius is one unit what must be done to the perimeter found for each of the polygons in order that it be expressed in terms of the diameter? Theorem 60. If i n represent the side of a regular inscribed polygon of n sides, c n that of a regular circumscribed polygon of n sides, and r the radius of the circle, c n = /^.^ 2 Given: AB, a side c n of a regular circum- scribed polygon of n sides, tangent to 00 of radius r at point C; i n the side of a regular inscribed polygon of n sides. A'* O ' Suggestions for proof: Draw AO, BO, CO. Let AO cut QO at P, and BO cut it at R, and CO cut PR at Q. Show that P/2=i n . Note that OC is an altitude of AAOB and OQ of APOfl. Why is AAOB <* APOfl? ' (1) Why? (2) In AOPQ, OQ m J r' - f | ) (2) Why? Substitute (2) in (1) and complete the proof. EXERCISE. SET LXIX (concluded) 699. Repeat the computations made in Exs. 694 and 695, or 696 and 697 for the sides and perimeters of regular circumscribed polygons. The radius of its circumscribed circle is called the radius of a regular polygon, and the radius of its inscribed circle is called the apothem of a regular polygon. Cor. The apothem of a regular polygon is perpendicular to its side. THE CIRCLE 187 Theorem 61. The perimeters of regular polygons of the same number of sides compare as their radii, and also as their apothems. Given: Polygons ABCD Ni, regular, and each of n sides, and with perimeters p and p\ t centers and Oi, radii r and n, and apothems a and a\. To prove: (1) -- s - and (2) 2- m - Pi ri pi a\ Suggestions: Proof of (1) Show that AAOB *> AB If 21 why is P n PI Proof of (2) What are a and in &AOB and EXERCISE. SET LXX. PERIMETERS OF REGULAR POLYGONS 700. Construct a regular hexagon whose perimeter is f of the perimeter of a given regular hexagon. In the proofs that follow, two more assumptions are made, namely : AXIOMS OF VARIABLES 1. // any variable approaches a limit, any part of that variable approaches the same part of its limit. That is, if v\ = li, then / J= 2. // two variables are always equal, the limits which they approach are equal. That is, if v\= l\, and v 2 = lz } and v\ = v 2 then li = l*. Theorem 62. Cir- cumferences have the same ratio as their radii. Given: Circles c and ea of radii r and r t . Prove: e -JL. 188 PLANE GEOMETRY PROOF Inscribe in each circle a reg- ular polygon of n sides, and let p and pi be their perimeters. (1) Then^- s-. Pi ri Let n increase indefinitely. (3) Thenp = candpi = ci and (5) (1) The perimeters of regular polygons of the same number of sides compare as their radii. (2) By alternation. (3) Postulate of limits for inscribed poly- gons. (4) If any variable approaches a limit, any part of that variable approaches the same part of its limit as limit. (5) If two variables are always equal, the limits which they approach are equal. Cl Tl | (6) By alternation. Cor. 1. The ratio of any circle to its diameter is constant. -=^-. Why? .'. ~=-. Why? Ci 2ri 2r 2n Cor. 2. Since the constant ratio ^ is denoted by the Greek letter TT* (which is the initial letter of the Greek word "periphery") in any circle, c=2irr. EXERCISE. SET LXXI. VALUE OF it 701. Show that the second value given to TT by Brahmagupta is another form of that given by Ptolemy. (See historical note.) Theorem 63. The value of TT is approximately 3.14159. If p n stands for the perimeter of the regular inscribed polygon, and P n for the perimeter of the regular circumscribed polygon in * HISTORICAL NOTE. " Although this is a Greek letter, it was not used by the Greeks to represent this ratio. Indeed, it was not until 1706 that an English writer, William Jones, used it in this way." Professor D. E. Smith further tells us that "probably the earliest approxi- mation of the value of TT was 3." In I Kings, vii, 23, we read : " And he made a molten sea, ten cubits from one brim to the other; it was round all about THE CIRCLE 189 a circle of diameter 1, the following table can be derived by using the formulas of the last two theorems: (1) i 2n ^2r 2 -r\4r 2 -i n 2 and (2) c n = /JT in wbich ^ i, V *-T ^", in, and c n retain their original meanings. (The student might verify a few of the values in the table by using logarithms wherever possible, and for convenience basing the calculations on a circle of diameter 2.) No. of sides. p n < c < P n 6 3.0000000 3.4641016 12 3.1058285 3.2153903 24 3.1326286 3.1596599 48 3.1393502 3.1460862 96 3.1410319 3.1427146 192 3.1414524 3.1418730 384 3.1415576 3.1416627 768 3.1415838 3.1416101 1536 3.1415904 3.1415970 EXERCISES. SET LXXII. CIRCUMFERENCE g702. Construct a graph by means of which the circumference of a circle of any given diameter may be obtained. 703. Find the size of the largest square timber which can be cut from a log 24 in. in diameter. and a line of thirty cubits did compass it round about." Again in ii Chronicles iv, 2, we read a similar sentence. And again in the Talmud is found the sen- tence: "What is three hand-breadth around is one hand-breadth through." The following list of various other values given to TT may be of interest to some of us. Value Attributed to 3.106 ............................. Ahmes (c. 1700 B.C.) Archimedes (287-212 B.C.) Ptolemy of Alexandria (87-165 A.D.) -8Jj- or 3.1416. ... . ................ Aryabhatta (c. 500 A.D.) f and J!$ or \/10 ............... Brahmagupta (c. 600 A.D.) m ............................ Metius of Holland (1571-1635) Computed to the equivalent of over 30 decimal places .............. Ludolph von Ceulen (1540-1610) To 140 decimal places .............. Vega (1756-1802) To 200 decimal places .............. Base (1824-1861) To 500 decimal places .............. Richter (1854) To 707 decimal places ............. Shanks (1854) 190 PLANE GEOMETRY G 704. The following problems illustrate the type of problem that is suggested by books on carpentry. To lay off an octagon on the end of a square piece of timber A BCD, draw diagonals AC and BD. With radius EF (the apothem of the square) draw arc cutting BD at G. Square out from G. Make a similar construction at each of the other corners of the square. Justify the rule. 705. How much belting does it require to A C D B E make a belt to run over two pulleys, each 30 in. in diameter, the distance between their centers being 18 ft.? 706. The annexed figure repre- sents a small wire fence used to protect flower beds. How many feet of wire are needed per run- ning foot of fence if AB = 1ft., BD = CD = 9in., and DE = 3 in.? d707. Using 4000 miles as the radius of the earth, find the num- ber of feet in the length of one minute at the equator. Use loga- rithms. (This distance is commonly called a "knot.") 708. (a) If a cable were laid around the earth at the equator, how many feet would have to be added if the cable were raised 10 ft. above the surface of the earth? (6) If the same were done around a gas-tank whose diameter is 100 ft.? (c) In which case is the increase proportionally larger? d709. Carpenters and other tradesmen frequently wish to know the circumference of a circle of given radius. The accompany- ing graphic method is given in some of the self -education books as a substitute for com- putation : Draw radii AO and BO at right angles. Draw chord AB and line OE perpendicular to AB, meeting circle at E and chord at D. THE CIRCLE 191 To 6 times the radius add the sect DE. The resulting sect is approximately the length of the circumference. Compute the approximate per cent of error, using 7r = 3.14159. 710. The central angle whose arc is equal to the radius is often used as the unit of measure of angles. It is called a radian. Find the number of degrees in a radian. 711. Many attempts have been made to construct a sect equal in length to a circle. The following approximate con- struction is one of the simplest. It is due to Kochansky (1685). At the ex- tremity A of the diameter A B of a given G circle of radius r draw a tangent CD, making and r as radius, describe circles. Find the perimeter and the area of the trefoil and of the shaded part of the resulting symmetric pattern. (Using S for the area of the shaded portion, C for that of the large circle, A for that of the small circle, T for that of the triangle, give the formula for S in terms of r, 730. In the papyrus of Ahmes, an Egyptian, the area of a circle was found by subtracting from the diameter one- ninth of its length and squaring the remainder This was equi- valent to using what value of TT? 731. In the Sulvasutras, early semi-theological writings of the Hindus, it is said: " Divide the diameter into 15 parts and take away 2; the remainder is approximately the side of the square, equal to the circle." From this compute their value of TT. d732. The proposition of the so-called lunes of Hippocrates (ca. 470 B.C.) proved a theorem that asserts in somewhat more general form, that if three semicircles be described on the sides of a right triangle as THE CIRCLE 195 diameter, the lunes L and LI as shown in the diagram are together equivalent to the triangle T. Prove it. d733. A problem of interest is one that Napoleon is said to have suggested to his staff on his voyage to Egypt: To divide a circle into four equal parts by the use of circles alone. LIST OF WORDS DEFINED IN CHAPTER VDI Circle, center, radius, diameter; central angle, arc (minor, major); chord, intercept, subtend; tangent to a circle, secant, tangent circles (internally, externally); line of centers,- common chord, inscribed angle; inscribed, circum- scribed regular polygons; center, central angles, radius, apothem of regular polygon; sector. Constant, variables (increasing, decreasing); limits (inferior, superior). AXIOMS OF VARIABLES IN CHAPTER Vm 1. If any variable approaches a limit, any part of that variable approaches the same part of its limit, as limit. 2. If two variables are always equal, the limits which they approach are equal. POSTULATE OF PERPENDICULARS (third) IN CHAPTER VHI 1. The shortest distance from a point to a line is the perpendicular to that line. SUMMARY OF THEOREMS PROVED IN CHAPTER Vm 42. Three points not in a straight line fix a circle. 43. In equal circles equal central angles intercept equal arcs, and conversely. 44. In equal circles, equal arcs are subtended by equal chords, and conversely. 45. A diameter perpendicular to a chord bisects it and its subtended arcs. Cor. 1. A radius which bisects a chord is perpendicular to it Cor. 2. The perpendicular bisector of a chord passes through the center. 46. In equal circles, equal chords are equidistant from the center, and conversely. 47. A line perpendicular to a radius at its outer extremity is tangent to the circle. Cor. 1. A tangent to a circle is perpendicular to a radius drawn to the point of contact. Cor. 2. The perpendicular to a tangent at the point of contact passes through the center of the circle. Cor. 3. A radius perpendicular to a tangent passes through the point of contact. Cor. 4. Only one tangent can be drawn to a circle at a given point on it. 196 PLANE GEOMETRY 48. Sects of tangents from the same point to a circle are equal. 49. The line of centers of two tangent circles passes through their point of contact. 50. In equal circles central angles have the same ratio as their intercepted arcs. Cor. 1. A central angle is measured by its intercepted arc. 51. Parallels intercept equal arcs on a circle. 52. An inscribed angle or one formed by a tangent and a chord is measured by one-half its intercepted arc. Cor. 1. An angle inscribed in a semicircle is a right angle. 53. An angle whose vertex is inside the circle is measured by half the sum of the arc intercepted by it and by its vertical angle. 54. An angle whose vertex is outside the circle is measured by half the difference of the intercepted arcs. 55. A tangent is the mean proportional between any secant and its external sect, when drawn from the same point. 56. A circle may be circumscribed about, and a circle may be inscribed in, any regular polygon. Cor. 1. An equilateral polygon inscribed in a circle is regular. Cor. 2. An equiangular polygon circumscribed about a circle is regular. 57. If a circle is divided into any number of equal arcs, the chords joining the successive points of division form a regular inscribed polygon, and the tan- gents drawn at the points of division form a regular circumscribed polygon. Cor. 1. Tangents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides. Cor. 2. Lines drawn from each vertex of a regular polygon to the mid-points of the adjacent arcs subtended by the sides of the polygon form a regular inscribed polygon of double the number of sides. Cor. 3. Tangents at the mid-points of the arcs between consecutive points of contact of the sides of a regular circumscribed polygon form a regular circumscribed polygon of double the number of sides. Cor. 4. The perimeter of a regular inscribed polygon is less than that of a regular inscribed polygon of double the num- ber of sides, and the perimeter of a regular circumscribed polygon is greater than that of a regular circumscribed polygon of double the number of sides. 58. A regular polygon the number of whose sides is 3*2 W may be inscribed in a circle. 59. If i n represent the side of a regular inscribed polygon of n sides, and i 2re the side of one of 2n sides, and r the radius of the circle, i 2n = V2r 2 r\/4r 2 in THE CIRCLE 197 60. If i ra represent the side of a regular inscribed polygon of n sides, c n that of a regular circumscribed polygon of n sides, and r the radius of the circle, 2ri ra C7l = VV-i re 2 ' 61 . The perimeters of regular polygons of the same number of sides com- pare as their radii, and also as their apothems. 62. Circumferences have the same ratio as their radii. Cor. 1. The ratio of any circle to its diameter is constant. Cor. 2. In any circle c =2-jrr. 63. The value of IT is approximately 3.14159. 64. The area of a circle is equal to half the product of its radius and its circumference. Cor. 1. The area of a circle is equal to IT times the square of its radius. Cor. 2. The areas of circles compare as the squares of their radii. 21 ft. EXERCISES. SET LXXIV. MISCELLANEOUS The problems in this set are miscellaneous in that their solution may depend upon any part of the text; but they are arranged, in general, in the order of difficulty. Those problems requiring the application of trigonometric ratios are preceded by "t." 734. Find the num- 73 ft. ber of feet of lime-line 1 4 - 5 ft> of a tennis-court, as represented. 735. Find the num- ber of yards of lime- line for a football field, which is 300 ft. by 160 ft., including all the ten-yard lines. How long would it take a run- ner to cover the total distance if he can make 40 feet in 12 seconds? 736. Construct an accurate diagram of a rectangular garden with a border inside it one-fourth the width of the garden. 737. A designer, in making a pennant, must make one in B the same proportion as a given one, but larger. Find AB, if D CD = 36", A& = 43", and 198 PLANE GEOMETRY E 738. By use of the steel .square lay out an angle of 45. 739. By use of steel squares lay out an angle of 60. 740. A straight railroad AB strikes a mountain at C, and a tunnel is to be driven at C in the direction ABC. It is de- sired to commence the work at the other side of the mountain at the same time as the work is progressing from C. -&A BD ismade 150, BD = 3 miles, and 60. How far from D in DE must the tunnel be driven, and in what direction? 741. An instrument for leveling A ^ consists of a rectangular frame ABCD. E and F are the mid-points of A B and DC, respectively. A plumb-line is suspended from E. Show that when the plumb-line coincides with the mark F } DC is horizontal. This instrument is shown in French books. B 742. The accompanying diagrams represent an in- strument for locating the center of circular discs. Three pieces of metal are so joined that AB bisects the angle formed by BD and BE, which two sects are equal. Prove that AB passes through the center of the circle. 743. A carpenter bisects an angle by the following rule: Lay off AB=AC. Place a steel square so that BD = CD as shown in the diagram. Draw the line AD. Is this method correct? Give proof. Would this method be correct if the steel instru- ment did not have a right angle at D? 744. Answer the following questions without proof: THE CIRCLE 199 (a) Are all equilateral polygons equiangular? (6) Are the diagonals of a parallelogram equal? (c) A diagonal of a parallelogram divides the figure into two congruent triangles. Is this proposition conversely true? (d) Do the diagonals of a parallelogram bisect its angles? (e) Under what circumstances do two chords bisect each other? (/) An arc of 30 is subtended by a chord of 5". In the same circle will an arc of 60 be subtended by a chord equal to, .greater than, or less than 5"? (g) Is it possible to inscribe a parallelogram in a circle? 745. In an ideal honeycomb the cells are 6 sided. Why is this? In what other regular form might they be built and yet fit snugly? 746. The wheel of an automobile makes 110 revolutions per minute. If it measures 2 ft. 6 in. in diameter, find the speed of the machine. 747. Walking along a straight road a traveler noticed at one milestone that a house was 30 off to the right. At the next mile- stone the house was 45 off to the right. How far was the house from the road? Is there more than one solution? 748. Calculate the diameter of the circle of water visible to an observer at sea, (a) when seated in a small boat, his eyes being 4 ft. above the surface of the sea, (b) when on the bridge of a steamer 25 ft. above the surface, (c) when at the masthead 60 ft. above the surface, (d) when on the top of a mountain 3000 ft. above sea level. (e) How far above sea level does the elevation of the observer begin to make a perceptible difference? a749. In finding the diameter of a wrought-iron shaft that will transmit 90 horse-power when the number of revolutions is 100 per minute, using a factor of safety of 8, we have to find the 3 / 90 diameter d from the formula d=68.5 A / 50000. Find d. o 750. Draw any quadrilateral ABCD. Take such measurements of your figure as you consider necessary and sufficient, and from your measurements construct the quadrilateral a second time. State what measurements you make, and how you draw the second figure. Cut out the two figures and fit one upon the other. 200 PLANE GEOMETRY (6) Discuss a number of other sets of measurements that you could use to reproduce the quadrilateral A BCD. How many meas- urements must there be in every set? (c) Construct a quadrilateral similar to ABC 'D, but 50% greater in area. 751. Find the number of square feet in the floor of the room shown in the accompanying plan. 752. Find the area of the ac- companying polygon by filling out the following table, assuming reasonable values for necessary dimensions. (Fig. 1). Parts of Polygon Factors Products Bases, or sums of parallel sides Altitudes AABB! BBi = 6.8 4i=5-G 38.08 2) ' Polygon ABCDEFGH = B 753. Show how to find the area of polygon ABCDEFGHK, assuming the shaded portion to be inaccessible. rs. FIG. 2. 754. Using goods 20 in. wide, how many strips will it take, cut on true bias, to put a band 12 in. wide around a skirt 3 yds. wide? THE CIRCLE 201 755. In the accompanying diagram how do (a) the perimeters and (6) the areas of the circle and the curvi- linear figures ADCFBX compare? (c) Use this figure as a suggestion and show how, by means of arcs, to divide a circle into any number of equal areas. 756. This is the cross- section of a foot-stool, in which the width of the top is to be 12 in., d 8 in., e 12 in., and the lengths of the legs 8 in. In making the stool, angle a. and angle /? are first laid out on paper. Show how, from the required di- mensions, to lay out these angles on paper. 757. To find the diagonal of a square, multiply the side by 10, take away 1% of this product, and divide the remainder by 7. Test the accuracy of this rule of thumb used by some carpenters. 758. Construct a perpendicular at the end of a sect without producing the sect. Hints: Let AB be the given sect. With any point C, between A and B, but outside the sect, as center, and radius CB, describe a major arc inter- secting AB at E. Draw the diameter EK. KB is the required perpendicular. Prove that this construction is correct. 759. The resultant of two forces acting upon a body is 400 Ibs. One of the forces is 250 Ibs. What are the limiting values for the other force? 760. A man decided to buy some numerals and make the face of a grandfather's clock, but when he came to divide the face into minutes he found that he was not able to do it without gues- sing. How could it be done accurately? t761. A kite string is 250 ft. long, and makes an angle of 40 with the level ground. Find (approximately) the height of the kite above the ground, disregarding the sag in the string. 202 PLANE GEOMETRY 762. Fill out the blank spaces in the table by referring to the diagram. (Fig. 1.) 43 41 3-4 4-5 EA DB 4NZ> AF BE ^1G FH 41 35 41 40 80 30 2L3 4-3 30 50 190 105 FIG. 2 763. The accompanying drawing is one of the earliest of Gothic O tracery windows. The arch^.C# is based on an equilateral trian- gle. A is the center and AB theradius for the arc CB, and.B the cen- A\ ter and AB the radius for arc AC. D is the mid-point of span AB. The arches AED and DFB are drawn on the half span by similar construction. Find the center and the radius of the circle with center that shall be tangent to the four arcs, DE, DF, AC, CB. Suggestions: With A as center and AH as radius cut the altitude CD at 0- H is the mid-point of DB. 764. How many degrees are there in each of the angles of the Pythagorean badge? 765. From a strip of metal 2^" wide it is desired to cut off a rectangle from which two circular disks 2" in diameter can be cut. What length AB must be cut off? 766. In taking soundings to make a chart of a harbor it is necessary at each sounding to determine the position of the boat. This is sometimes done by measuring with a sextant the angles between lines from the observer to three range poles on the shore. When a chart is made, the position on the chart of each sounding is sometimes found as follows: The points A, B, and C represent the positions of the three range poles. Suppose the angles read THE CIRCLE 203 by the sextant were 50, and 35 (50 between lines to A and B, and 35 between lines to B and C) . On the chart lay off at A and B, angles BAM and ABM, each equal to 40, and find the point M. At B and C, lay off angles CBN and BCN each equal to 55 and find N. With M and N as centers, draw circles passing through B. The other common point of these circles is the position of the sounding. Prove the correctness of this construction. Make the construction to scale. 767. The resistance offered by the air to the passage of a bullet through it varies jointly as the square of its diameter and the square of its velocity. If the resistance to a bullet whose diameter is .32 in., and whose velocity is 1562.5 ft. per second, is 67.5 oz., what wiJl be the resistance to a bullet whose diameter is .5 in., and whose velocity is 1300 ft. per second? 768. Roman surveyors, called agrimensores, are said to have used the following method of measuring the width of a stream: A and B were points on opposite sides of the stream, in plain view from each other. The distance AD was then taken at right angles to AB, and bisected at E. Then the distance DF was taken at right angles to AD, such that the points B, E, and F were in a straight line. Make a drawing illustrating the above method, show what measurement affords a solution, and prove that this is so. 769. A student lamp and a gas jet illuminate a screen equally when it is placed 12 ft. from the former and 20ft. from the latter. Compare the relative intensities of the two lights. 770. An endless knife runs on pulleys 48" in diameter at the rate of 180 revolutions per minute. If the pulleys are decreased 18" in diameter, how many revolutions per minute will they have to make to keep the knife traveling at the same speed? 771. In surveying, to determine a line from the inaccessible point P perpendicular to AB, lay off FE perpendicular to AB at an arbit- rary point and of any length. Make EH = EF. Obtain point D in lines PF andAB; next, point N in HP and AB; next, K in DH and FN. PK is perpendicular to AB. Why? 204 PLANE GEOMETRY 772. Make the construction shown in the diagram, and AB will be approximately the quadrant of the circle. Find the per cent by which it differs from the correct value. 773. The steel square may be tested by measuring across from the 9-inch point of the tongue to the 12-inch point of the blade. If this distance is exactly 15 inches, the square is true. Why? 774. Four of the largest possible equal sized pipes are enclosed in a box of square cross-section 18 in. on an edge. What part of the space do the pipes occupy? dt775. A boy pulls a sled with a force of twenty-five pounds by means of a rope ten feet long, and with his hands three feet from the ground. Find the component of his force effective in pulling the sled forward. 776. A camp kettle weighing 20 Ibs. is suspended on a wire from two trees 10 feet apart. The wire is 20 feet long, and the kettle is suspended on it at a point midway between the trees. Find the tension on each strand of the wire. 777. A running track having two parallel sides and two semi- circular ends, each equal to one of the parallel sides, measures exactly a mile at the inner curb. Two athletes run, one at the inner curb, and the other 10 ft. from this curb. By how much is the second man handicapped? 778. The last row of seats in a circular tent is 30 ft. away from the central pole, which is 20 feet high, and which is to be fastened by ropes from its top to stakes driven in the ground. How long must these ropes be in order that they may be 6 feet above the ground over the last row of seats, and at what distance from the center must the stakes be driven? 779. A wheelwright is given a part of a broken wheel to make a duplicate. To do this he needs the diameter. He measures the chord of the arc given him, 24"; the height of the segment is 4". How large is the wheel? THE CIRCLE 205 dt780. The width of the gable of a house is 35 ft. The height of the house above the , eaves is 15 ft. Find the length of the rafters and the angle of inclination of the roof. 781. An aeroplane travels 1000 ft. upwards, 3^ miles due west, and 2-^ miles due north. Find its distance from the starting-point. 782. The sides of a floor are 10 ft. and 6 ft. A man wishes to tile it with tiles in the shape of a hexagon whose sides are 6 in. How many tiles will it take? 783. A flat circular sheet of metal is to be stamped into the form of a spherical segment with a flange. The figure shows a cross-section of the resulting piece of metal, a being the width of the spherical segment, b the depth or altitude of the segment, and c the outer diameter of the flange. The problem is to determine the size to cut the sheet metal in order that when stamped the piece may have these dimensions. Show that the required radius of the circular sheet equals e in the figure. 784. To lay off the length of a brace with the steel square: Suppose that the post is 4 ft. and the beam 3 ft. Apply 3 times to the timber from which the brace is to be cut, the distance across the square from the 12 in. point of the tongue to the 16 in. point of the blade. Why will this give the required length of the brace? t785. In railroad construction and mining the material is sometimes hauled in a tram pulled by a horse. If the pull of the tram in the direction of the track is, say, 200 Ibs., and if the horse walks at the side of track so that its pull is exerted at an angle of 25 with the track, what pull must the horse exert? 786. The force that the wind exerts normally (per- pendicularly) on the sail of a boat is resolved into two components: one useless, in pushing the boat sidewise in the water in spite of the keels; and one useful com- ponent driving the boat directly forward. If, as in the following diagram, the sail is at an angle of 30 to the keel, and a force of wind of 100 Ibs. acts on the sail (considered as applied at one point), find the effective value of the two components mentioned above. NOTE. A similar problem can be applied to the aeroplane. 206 PLANE GEOMETRY 787. An angle inscribed in a circle varies directly as the inter- cepted arc. Show that in this case k = ^. 788. A street-car track is 12' from the curb (GF = BC=12 f ). In passing the corner of two streets which deflect through an angle of 60 the rail must be 5' (DE = &) from the corner, (a) Find the radius of the curve. (6) Find the length of the tangents from G and C to their point of intersection. (c) Find the length of arc GC; also the length of the outer arc. The width of the track is 4' &|-". (All these curves are arcs of circles.) 789- Where two straight streets intersect, each corner is usually "rounded off." Show that the problem of laying out the corner arc is a simple one where the streets made an angle of 90. Show by a plan how to lay out a corner where the angle is 45, and the radius for the curbing is to be 20 ft. Find the total length of curved curbing needed. dt790. A given regular polygon has n sides. How many meas- urements of the figure are both necessary and sufficient to deter- mine it in size and shape? dt791. A flagstaff is seen in a direction due north of a station A at an elevation of 17, and from a station B 120 ft.- due east of A the flagstaff bears 23 west of north. The two stations and the foot of the flagstaff being at the same level, determine the height of the flagstaff. 792. A straight street intersects one which is curved (with a large radius, such as 200 ft.), and the corner is to have a small radius, say 15 ft. Show in a plan how to find the center for the corner arc by means of the intersection of two loci. (Fig. 1.) How would you get, in actual field work, the curved locus? (Note its large radius.) (Fig. 2.) 793. Solve in another plan the corresponding problem where both streets are curved. Fig. 1. Fig. 2. THE CIRCLE 207 4'lK"- D 794. The "gear" of a bicycle is the diameter (in inches) of a wheel whose circumference would equal the distance gone forward with one revolution of the pedals. Find the gear if the diameter of the rear wheel is 28" and the front and rear sprockets have 22 and 8 teeth, respectively. 795. A belt runs over two pulleys, one of which is 4 ft. in dia- meter, and driven by an engine at the rate of 100 revolutions a minute. What must be the diameter of the other pulley if it is to turn a fan at the rate of 400 revolutions a minute? 796. The figure is the diagram of a part of the side of a bridge. The point C must be located on AB and on DE y where the holes must be bored to fasten the brace AB to the upright DE. Required to find the lengths of AB, x, y, and 2, in order to locate C. Suggestion: After finding AB, compare triangles ACF and ABG. 797. It is desired to construct a subway under a river. The bank on one side has a 30% slope from the edge of the river to the river bed. The maximum effective grade of a subway is 5%. On the surface it is five hundred feet from the bank of the river to the bed of the river. Deter- E K- 2 !&--;*: 2' "I mine the necessary length of the subway from the bank to the river bed. 798. WW is a wall with a round cor- ner of dimensions as given in the figure from A to B, on which a molding, gutter, or cornice is to be placed. Find the radius of the circle of which the arc A NB is a part. 799. A, B are two beacons on a coast- line; S is a shoal off the shore, and the angle ASB is known to be 120. Show that a vessel V sailing along the coast will keep outside the shoal if the angle AVB is always less than 110. 208 PLANE GEOMETRY Prove a property of the circle that you used in proving that the ship will clear the shoal. . 800. The method given by Galileo for finding the strongest rectangular beam that can be cut from a round log is as follows: Let the circle A BCD represent the end of the log, and let AC be a, diameter. Divide the diameter into three equal parts at the points M and N, and from these points erect perpen- diculars intersecting the circumference at points D and B. Draw AD, DC, CB } and BA . The rectangle thus formed is the cross-section of the strongest rectangular beam. Show that the dimensions of the rectangle are in the ratio of 1 to \/2. dtSOl. The resultant of two forces is 300 Ibs. One of the forces acting at an angle of 37 with the resultant is 100 Ibs. Find the other force, if the forces act at an angle of 65 with each other. dt802. The radius of a circle is 7 ft. What angle will a chord of the circle 10 ft. long subtend at the center? 803. To prolong a line through an obstacle and to measure the distance along the line through the obstacle. A and B are two points on the given line. Take -X" any point. Measure AX. Take XC=AX and in line with AX. Mark Y on the mid-point of XC. Make DX = BX and in line with BX. MakeZY=YD. Make ZE = ZD and in line with ZD. Make ZF = ZC and in line with ZC. Prove FE in line with AB. What line of the figure equals BFl 804. The cross-section of the train-shed of a railroad station is to have the form of a pointed arch, made of two circular arcs, the centers of which are on the ground. The radius of each arc equals the width of the shed, or 210 ft. How long must the supporting posts be made which are to reach from the ground to the dome of the roof? THE CIRCLE 209 g805. Show by a graph that the area of a triangle having a fixed altitude varies as the base, and that one having a fixed base varies as the altitude. g806. Construct a graph showing the relation between the areas and the sides of equilateral triangles. g807. Construct a graph showing the relation between a side and the area of a regular hexagon. By means of it find the area of a regular hexagon whose sides are 6. Compare with the computed area. g808. If a side of a given regular polygon is a and its perimeter p, graph the relation of the perimeters and the corresponding sides of polygons similar to the given polygon. g809. Given a polygon with area A and a side a. Construct a graph showing the relation between the areas and the sides corre- sponding to a in polygons similar to the one given. dt810. The figure shows a method for determining the horizontal distance PR, and the difference of level QR be- tween two points P and Q. A rod with fixed marks A and B upon it is held vertical at Q, and the elevation of these points ACD ( = a) and BCD (=/3) are read by a telescope and divided circle at C, the axis of the telescope being at a distance CP ( = a) above the ground at P. If .QA = b and AB = s, write down expressions for PR (=x) and QR (=y). Find x and y when a = 6 10' and = 7 36', the values of a, 6, and s being 5 ft., 2^- ft., and 5 ft. respectively., dtSll. Find the radius of a parallel of lati- tude passing through Portland, Me. (43 40' N. lat.), if the radius of the earth is taken as 4000 mi. (Note that in the figure <&x equals . SUMMARY OF METHODS OF PROOF 338 Exercises. Set XCVI. Congruence of Triangles 339 Exercises. Set XCVII. Equality of Sects 340 Exercises. Set XCVIII. Equality of Angles 340 Exercises. Set XCIX. Parallelism of Lines 341 Exercises. Set C . Perpendicularity of Lines 342 Exercises. Set CI. Inequality of Sects 343 Exercises. Set C1I. Inequality of Angles ' 344 Exercises. Set CHI. Similarity of Triangles 345 Exercises. Set CIV. Proportionality of Sects 346 Exercises. Set CV. Equality of Products of Sects . . . 346 Exercises. Set CV1. Miscellaneous Exercises . . 347 CONTENTS 217 CHAPTER IX COLLEGE ENTRANCE EXAMINATIONS PAGE CHICAGO 362 HARVARD 363 CHAPTER X SUGGESTIONS A. LIST OP TOPICS SUITABLE FOR STUDENTS' DISCUSSION: General 374 Arithmetic 375 Algebraic 375 Geometric 375 B. TOPICS WITH DEFINITE REFERENCES: Geometric Fallacies 376 Number Curiosities 376 Pythagorean Proposition 376 C. LIST OF BOOKS SUITABLE FOR STUDENTS' READING: History 376 Recreations 377 Practical 377 General 378 Index of Definitions . 379 SUGGESTIONS FOR A REVIEW OF THE FIRST STUDY OF PLANE GEOMETRY Before beginning our Second Study of Plane Geometry, it might be well for us to review the First Study. The following material furnishes a brief, suggestive outline for such a review. A. CONGRUENCE THEOREMS 1-8 I. Define: * Sect, polygon, axiom, postulate, corollary, adjacent angles, congruent, homologous, perpendicular. II. Summarize: a. Conditions under which triangles are congruent in general; b. Special conditions under which right triangles are congruent; c. Facts about perpendiculars. III. Family Trees of Propositions: To trace a proposition back to its sources, that is, back to the definitions, postulates, and axioms upon which it rests, will be found an interesting and profitable form of review. A convenient arrange- ment is to make a "family tree" of a theorem, the branches of which are the authorities quoted in proving the proposition. Each branch should be followed down as in the main proposition until it ends in a postulate, an axiom, or a definition. Such a tree of Theorem 6 is given as an illustration (Plate 1). The student is advised to make a tree of Theorem 5. B. PARALLELS THEOREMS 9-12 I. Define: Parallels, transversal, alternate-ulterior angles. II. Classify angles according to: a. Individual size; 6. Relative size; c. Relative position. III. Summarize: a. Conditions under which lines are parallel; b. Methods of proving sects equal; c. Methods of proving angles equal. IV. Family Tree: A family tree of Theorem 11, Cor. 2, is appended (Plate 2). The student is advised to study it and make one of Theorem 12. * Consult the First Study only when necessary. 210 220 PLANE GEOMETRY it J Is I cS J2 *- a& a 1! tr* a es are are eq II n ^ ft '& lologous parts of c at n-gons are equ; ular can m a poiu j* a O r9 0-S * M|l H'53 ^3 II i! 3S II 11 a; rt 5 ** o 1 to 5S rl - o a REVIEW OF FIRST STUDY 221 ill II ! ^ I? 2 so* s I I '5 , -2 ha" II $11 LID s g v; g -01555 222 PLANE GEOMETRY C. SUMS OF ANGLES THEOREMS 13-15 I. Define: Exterior angle. II. Summarize: The facts about the sums of the angles, interior and exterior, of an n-gon. III. Make a family tree of Proposition 14. D. PARALLELOGRAMS THEOREMS 16-21 I. Define: Parallelogram, diagonal. II. Classify quadrilaterals. Is it possible to classify them in more than one way? If so, is there any preference? III. Summarize: a. Conditions under which a quadrilateral is a parallelogram; b. Properties of a parallelogram. IV. Make a family tree of Proposition 17. E. AREAS THEOREMS 22-27 I. Define: Commensurable, measurement, area. II. Summarize the facts about areas by giving formulas for the areas of figures studied. III. Complete the family tree of Theorem 27 as appended (Plate 3). F. SIMILARITY THEOREMS 28-39 I. Define. Ratio, proportion, center of similitude, similar figures. II. Summarize: a. Properties arising from similarity of triangles; 6. Conditions under which triangles are similar. III. A family tree of Theorem 39, Cor. 2, is begun (Plate 4). The student is advised to make a tree of Proposition 36. G. Locus THEOREMS 40-41 I. Define: Locus of a point. II. State the facts of which proofs are necessary and sufficient to establish a locus theorem. III. State the two locus theorems proved in the syllabus. S I a a REVIEW OF FIRST STUDY PLATE 4 C mps ' f equal * are equah 223 The altitude upon the hypotenuse I of a right triangle divides it into . ., , ., triangles similar to each other ,,,,.., and to the original. n Triangles are similar if two angles e ,\ of one are equal to those of] another. The homologous sides of similar triangles have a constant ratio. [ The homologous angles of similar | triangles are equal. \ Post, of superposition. If corresponding angles are equal,! lines cut by a transversal are] parallel. A line parallel to the base of a triangle cuts the remaining sides so that they are proportional to either pair of homologous sects.! Quantities equal to the same quantity are equal to each other. The products of equals multiplied by equals are equal. The sums of equals added to equals are equal. The whole is equal to the sum of its parts. (To be completed by the student.) H. THE CIRCLE AND STRAIGHT LINE THEOREMS 42-49 I. Define: Circle, chord, secant, tangent. II. Summarize the facts proved in the syllabus about: a. Chords; 6. Tangents; c. Arcs. III. Make a family tree of Proposition 48. I. THE CIRCLE AND ANGLE MEASUREMENT THEOREMS 50-55 I. Define: Central angle, inscribed angle. 224 PLANE GEOMETRY II. State the method of measuring: a. Central angles; 6. Inscribed angles; c. Angles formed by a tangent and a chord; d. Angles with vertices outside the circle; e. Angles with vertices inside the circle. III. A family tree of Proposition 55 is begun (Plate 5). The student should make one of Theorem 53. J. MENSURATION OF THE CIRCLE THEOREMS 56-64 I. Define: Sector, apothem. II. State formulas for: a. Area of a regular polygon; b. Circumference of a circle; c. Area of a circle. III. Summarize the methods now known to you of proving: a. Sects equal; 6. Angles equal; c. Lines perpendicular; d. Lines parallel; e. Triangles congruent; /. Triangles similar; g. Arcs equal; i. Chords equal. PLATE 5 Th. 55. A tangent from a point to a circle is the mean proportional between any secant and its external sect, from the same point to the circle. 1 A. An inscribed 1 Any two quanti- I l B. Triangles are C. Homologous angle, or one ties of the same similar when two sides of similar formed by a tan- kind compare as angles of one are triangles have a gent and a chord is their numeric equal each to each constant ratio. measured by one- measures. to two angles of | half its intercepted another. arc. 1 NOTE. If the student is in need of further review, he is advised to trace branches A, B, and C of this family tree back to their sources. CHAPTER I FUNDAMENTALS. RECTILINEAR FIGURES Theorem 1. Vertical angles are equal. The pupil is earnestly advised not to refer to the First Study* for suggestions as to proofs of the theorems there taken up if it is possible for him to work them out without any help in the review. For this reason those theorems are stated in this part. Theorem 2. Two sides and the included angle determine a triangle. Given: AABC and ADEF] AB =DE] %.B = %.E; and BC =EF. Prove: AABC&ADEF. E (1) Place ADEF on AABC so that %.E coincides with ^.B and ED falls along BA. (2) Then EF falls along BC. '.' 4#=3_. V ED = BA, D will fall on A. '.' EF = BC, F will fall on C. (3) .'. FD coincides with CA. (4) .'. ADEF AABC. PROOF (1) Superposition post. (2) Data. (3) Two points fix a straight line. (4) Def. of congruence. * Throughout "A Second Study of Plane Geometry" reference will be made to "A First Study of Plane Geometry" as "First Study," and no proofs will be given in this part of propositions contained in the former except in the case of the congruence of triangles. In this instance, one proof by the method of superposition will be given in full, since cutting and pasting were used in the First Study owing to the difficulty of the usual proof for the beginner. 15 225 226 PLANE GEOMETRY Theorem 3. Two angles and the included side determine a triangle. Apply the method of proof used in Theorem 2. What parts of the triangles will you first make coincident in superposing in this case? What postulate is needed in order to clinch this proof? Theorem 4. The bisector of the vertex angle of an isosceles triangle divides it into two congruent triangles. Cor. 1. The angles opposite the equal sides of an isosceles triangle are equal. Cor. 2. The bisector of the vertex angle of an isosceles tri- angle bisects the base and is perpendicular to it. Cor. 3. An equilateral triangle is equiangular. Cor. 4. The bisectors of the angles of an equilateral triangle bisect the opposite sides and are perpendicular to them. Cor. 5. The bisectors of the angles of an equilateral triangle are equal. Theorem 5. A triangle is determined by its sides. Theorem 5a. Only one perpendicular can be drawn through a given point to a given line. Given: (I) Point P in 'AB; (II) Point P outside AB. Prove: Only one line through Pis perpendicular to AB. Suggestion: (I) When P is in AB, in how many positions of PD will the st. ^.APB be di- vided into two right 2$_s? PROOF (II) (1) Suppose PD meets ~AB at rt. 4s at D, and PC is a line drawn to any other point C in AB. (2) Extend PD to PI so that PiD D \ (3) PCPi is not a straight line. (4) /. 4PCPiisnotast.2$_. (5) APCD^APiCD. (6) :. ^PCD^^PiCD. (7) .'. 4PCD is not a rt. %.. (8) .'. PCAB. (3) Why? (4) Why? (5) Why? (6) Why? (7) Why? (8) Why? FUNDAMENTALS. RECTILINEAR FIGURES 227 Theorem 5fr. Two sects drawn from a point in a perpendicular to a given line, cutting off on the given line equal sects from the foot of the perpendicular, are equal and make equal angles with the perpendicular. Given: PQA.AB at Q; P any point in PQ', QC=QD; C and D in AB. Prove: PC=PD; ^CPQ^^DPQ. __^^_*^_ A __*___ Jk _ (Proof left to the student.) AC Q D B In our first study, propositions dealing with inequalities were with one exception omitted. The following group of propositions supplies this omission, and before proving them it will be necessary to study a set of axioms dealing with inequalities. 1. If unequals are operated on in the same way by positive equals, the results are unequal in the same order. e.g., If a>b and c=d, where c and d are positive quantities, a+c> b+d, a-c>b -d, ac>bd } -> -; while a e >b d and Va> %/b under c a certain conditions which do not affect work in elementary geometry. Test these statements, substituting for the symbols the following values : (1) a = 5, 6 = 3, c = d = 4. (2) a=, &=, c = d=4. - (3) a = Z, 6= -Z, c = d = 4. (4) a=|-, 6= --J-, c = d = 3. (5) a= -sV,6= -i,c = d = 3. What conclusions can you draw? What meaning is attached to the phrase "in the same order"? 2. The sums of unequals added to unequals in the same order (or the same sense) are unequal in the same order. e.g., If a>b, and c>d, then a+c>b+d. Show why this statement could not be made for subtraction of unequals. 3. The differences of unequals subtracted from equals are unequal in the reverse order. Illustrate this fact. 4. // the first of three quantities is greater than the second, which in turn is greater than the third, all the more then is the first greater than the third. Illustrate this fact. 228 PLANE GEOMETRY Theorem 5c. The sum of two sects drawn from any point inside a triangle to the ends of one of its sides is less than the sum of its remaining sides. Given: AABC; D any point inside. Prove: DA+DC#+DCCQ. Prove: PR>PQ. Suggestions : Where will Qi lie with res- pect to C and R if CQi = CQt Why? Why is it that whatever you prove true of PQi is true of PQ? Why is it that whatever is true of PQi+PiQi and PR+P t R is true of PQi and Pfl? Write a complete proof of this theorem. II. Converse. Given: PC JLAQCRB; PR>PQ. Prove: CR>CQ. Suggestions: Why can CR not be less than CQ? Why can CR not equal CQ? What remains for the relation of CR to CQ? For convenience PQi+QiPi is at times referred to as PQiPi and is called a broken line. Such a line is always composed of two or more sects. The method of proof here outlined is known as the method of exclusion or elimination. Any two quantities of the same kind (a and b) must bear one of the following relations to each other: FUNDAMENTALS. RECTILINEAR FIGURES 229 a > 6, a = b, or a < b. If any two of these relations can be shown to be false, the remaining relation must therefore be true. For further discussion and illustration see p. 299. Cor. 1. All possible obliques from a point to a line are equal in pairs, and each pair cuts off equal sects from the foot of the perpendicular from that point to the line. (I) Under what condition will obliques from a point to a line be equal ? Why, then will all possible obliques from a point to a line be equal in pairs? (II) Use the method of exclusion to prove the second part of the corollary. Theorem 6. The perpendicular is the shortest sect from a point to a line. Suggestion: Use Theorem 5d to give a much simpler proof than was possible in the First Study. Theorem 6a. The shortest sect from a point to a line is perpen- dicular to it. Hint: Show that this is a special case of Theorem 5d (converse) Theorem 7. The hypotenuse and adjacent angle determine a right triangle. Theorem 8. The hypotenuse and another side determine a right triangle. EXERCISES. SET LXXV. TRIANGLES Numeric 820. The perimeter of an isosceles triangle is 13, and the ratio of one of the equal sides to the base is 1%. Find the three sides. Theoretic 821. In proving triangles congruent, two methods have been used. (a) When what elements are given equal can superposition be used? (b) When must juxtaposition be used? Answer in a single brief sentence. 822. The sects of any bisector of a given sect cut off by per- pendiculars erected at the ends of the given sect are equal. 230 PLANE GEOMETRY 823. The sects of a perpendicular to the bisector of an angle at any point in it and limited by the sides of the angle are equal. 824. Sects joining any point in the bisector of an angle to points on the sides equidistant from the vertex are equal. 825. If in the accompanying diagram, $.BAC = $BCA, and $BAY= $BCX, sect ZC = sect YA. 826. If the sides of an equilateral triangle are prolonged in turn by equal lengths, and the extremities of these sects are joined, another equilateral triangle is formed. 827. Two isosceles triangles are congru- ent if one of the equal sides and the altitude upon that side are equal each to each. 828. Triangles are congruent if two sides and the altitudes upon one of them are equal each to each. 829. A triangle is determined by a side and the median* and the altitude to that side. (X d830. If in the accompanying diagram %A is a right angle, YX=AC, CP = PY, show that PB+PX>CB+CA. 831. The sum of the distances from any point inside a triangle to the vertices A X B is greater than its semiperimeter, but less than its perimeter. 832. The sum of the diagonals of a quadrilateral is greater than the sum of either pair of opposite sides. 833. The sum of the diagonals of a quadrilateral is less than its perimeter, but greater than its semiperimeter. 834. The sum of the medians of a triangle is less than one and a half times its perimeter. (Prove this and the following exercise without assuming the concurrence of the medians). 835. The sum of the medians of a triangle is greater than its semiperimeter. 836. Each altitude of a triangle is less than half the sum of the adjacent sides. *A median is the sect between any vertex of a A and the midpoint of the side opposite that vertex. FUNDAMENTALS. RECTILINEAR FIGURES 231 837. The sum of the altitudes of a triangle is less than the perimeter. 838. Show that the bisector of the vertex angle of an isosceles triangle is coincident with the altitude to its base. Construction * 839. Bisect a reflex angle. d840. To construct a triangle, having given a side, the median and the altitude to a second side. d841. To construct a triangle having given one side, the corre- sponding median, and the altitude to another side. Theorem 9. Lines perpendicular to the same line are parallel. Theorem 10. A line perpendicular to one of a series of parallels is perpendicular to the others. Theorem 11. // when lines are cut by a transversal the alter- nate-interior angles are equal, the lines thus cut are parallel. Cor. 1. // the alternate-exterior angles or corresponding angles are equal when lines are cut by a transversal, the lines thus cut are parallel. Cor. 2. // either the consecutive-interior angles or the consecu- tive-exterior angles are supplementary when lines are cut by a transversal, the lines thus cut are parallel. Theorem 12. Parallels cut by a transversal form equal alter- nate-interior angles. Cor. 1. Parallels cut by a transversal form equal corresponding angles and equal alternate-exterior angles. Cor. 2. Parallels cut by a transversal form supplementary consecutive-interior angles and supplementary con- secutive-exterior angles. * For a discussion of methods of attacking problems in construction see Chapter VII, p. 306, where additional exercises will also be found. At this point, for instance, the topics, "The Synthetic Method of Attacking a Problem" (p. 309) and "The Formal Analysis of a Problem" (p. 318) should be studied. 232 PLANE GEOMETRY Theorem 12a. Two angles whose sides are parallel each to each or perpendicular each to each are either equal or supplementary. ,C Given: (I) X^YX \\ BA', Z^YZ /<* \\BC. (ID A. _^__ BC $ Z/ Prove: (I) X, fr X L> / AI 180- Suggestions : Using the con- struction lines prove (I). (II) If YiCi || EC and YiAi\\BA, what relation exists between %.ABC and 2$f.AiYiCit How are YiCi and YiZi related? How YiAi and Y\Xi! Then how are ^ AiYiCi and ZiYiAi related? Then how are %? XiYiZi and Z\Y\A\ related? Draw conclusions. Theorem 13. The sum of the angles of a triangle is a straight angle. Cor. 1. A triangle can have but one right or one obtuse angle. Cor. 2. Triangles having two angles mutually equal are mutu- ally equiangular. Cor. 3. A triangle is determined by a side and any two homolo- gous angles. Cor. 4. An exterior angle of a triangle is equal to the sum of the non-adjacent interior angles. Theorem 14. The sum of the angles of a polygon is equal to a straight angle taken as many times less two as the polygon has sides. Cor. 1. Each angle of an equiangular polygon of n sides equals the ^th part of a straight angle, n Cor. 2. The sum of the exterior angles of a polygon is two straight angles. Cor. 3. Each exterior angle of an equiangular polygon of n sides is equal to the -th part of a straight angle. FUNDAMENTALS. RECTILINEAR FIGURES 233 Theorem 15. If two angles of a triangle are equal, the sides opposite them are equal. Cor. 1. Equiangular triangles are equilateral. EXERCISES. SET LXXVI. PERPENDICULARS. PARALLELS. SUMS OF ANGLES OF POLYGONS Numeric 842. Find the angles of an isosceles triangle if a base angle is double the vertex angle. 843. If the vertex angle of the isosceles triangle is 30, find the angle formed by the bisectors of the base angles. If the vertex angle is J5? 844. The bisector of the base angle of an isosceles triangle makes with the opposite leg an angle of 53 17'. Find the angles of the triangle. 845. Find the angles of an isosceles triangle if the altitude is one-half the base. 846. If the angle at the vertex of an isosceles triangle is 36, the bisector of a base angle divides the triangle into two isosceles triangles. Find the lengths of all the sects in the diagram if a leg of the given triangle is a and the base is b. 847. What is the sum of the angles of (a) a hexagon, (6) a hepta- gon, (c) an octagon, (d) a nonagon, (e) a decagon, (/) a polygon of 18 sides, (g) a polygon of 24 sides, (h) of 30 sides? 848. Find each angle of an equiangular (a) hexagon, (6) hepta- gon, (c) octagon, (d) nonagon, (e) decagon. 849. In what polygon is the sum of the angles three times as great as in a pentagon. 850. How many sides has a polygon if (a) the sum of the interior angles equals 4 rt.<&? 3 st. ? 6 rt. BCi. Then our problem is to so locate P that PCi=PC. What kind of triangle, therefore, is How, then, shall we draw CiP? FUNDAMENTALS. RECTILINEAR FIGURES 237 Theorem 21e. If two triangles have two sides equal each to each, but the third sides unequal, the A angles opposite those sides are unequal in the same order. Suggestions: Use method of exclusion. Theorem 21/. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter leg, and conversely. Suggestions: (a) Direct. What part of a straight angle is 2a? What kind of triangle is ACCJ How are the sects CB, CCi, and AC related? (6) Converse. How is ACi related to AC? What kind of triangle is ACCi? How are the angles CiAC, BAG, and ACCi related? EXERCISES. SET LXXVIII. INEQUALITIES Numeric 864. Two sides of a triangle are 10" and 13". Between what limits must the third side lie? 865. If two angles of a triangle are respectively 55 and 65, which is the longest, and which the shortest side of the triangle? 866. If one angle of a triangle is one-third of a straight angle, and a non-adjacent exterior angle of the triangle is five-eighths of a straight angle, which side of the triangle is the longest and which is the shortest? Theoretic 867. Either leg of an isosceles triangle is greater than a sect connecting the vertex with any point in the base. 868. The sect joining the vertex of an isosceles triangle to any point in the prolongation of its base is greater than either leg. 869. If one leg AB of an isosceles triangle ABC is pro- duced beyond the base BC to a point D, then ER. 871. If no median of a triangle is perpendicular to the side to which it is drawn, the triangle is not isosceles. 872. If any point in the prolongation of a leg produced through the vertex of an isosceles triangle whose base is shorter r than a leg is joined to the extremity of the base, a scalene triangle will be formed. 873. If a vertex of an equilateral triangle is joined to any point in the prolongation of the opposite side, a scalene triangle is formed. Which is the longest and which the shortest side of each of the triangles thus formed? d874. Prove that the sect joining an extremity of the base of an isosceles triangle to any point in the opposite leg is greater than one sect cut off on that leg. Is it ever greater than either sect? Under what condition is it less than one of the sects? 875. The diagonals of a rhomboid intersect obliquely, and the greater angle formed by them lies opposite the greater side of the parallelogram. (A rhomboid is an oblique O in which the ad- jacent sides are unequal.) 876. If a triangle is not isosceles the median to any side is not perpendicular to it, and the larger angle which it forms with that side lies opposite the greater of the remaining two sides. d877. Any point not on the perpendicular bisector of a sect is unequally distant from the ends of the sect. LIST OF WORDS DEFINED IN CHAPTER I Exclusion, elimination, broken line. SUMMARY OF AXIOMS IN CHAPTER I 1. If unequals are operated on in the same way by positive equals, the results are unequal in the same order. 2. The sums of unequals added to unequals in the same order (or the same sense) are unequal in the same order. 3. The differences of unequals subtracted from equals are unequal in the reverse order. 4. If the first of three quantities is greater than the second, which in turn is greater than the third, all the more then is the first greater than the third. (For a summary of theorems see Chapter VIII, p. 324.) CHAPTER II AREAS OF RECTILINEAR FIGURES Theorem 22. Rectangles having a dimension of one equal to that of another compare as their remaining dimensions. Theorem 23. Any two rectangles compare as the products of their dimensions. Theorem 24. The area of a rectangle is equal to the product of its base and altitude. Theorem 25. The area of a parallelogram is equal to the product of its base and altitude. Cor. 1. Any two parallelograms compare as the products of their bases and altitudes. In proving Cors. 1, 2, 3 of this theorem and the next, express the areas as algebraic formulas and apply axioms. Why is such a procedure both convenient and natural? Cor. 2. Parallelograms having one dimension equal compare as the remaining dimensions. Cor. 3. Parallelograms having equal bases and equal altitudes are equal. Theorem 26. The area of a triangle is equal to half the product of its base and its altitude. Cor. 1. Any two triangles compare as the products of their bases and altitudes. Cor. 2. Triangles having one dimension equal compare as their remaining dimensions. Cor. 3. Triangles having equal bases and equal altitudes are equal. Theorem 26a. The square on the hypotenuse of a right tri- angle equals the sum of the squares on the two legs. NOTE. In this text the conventional phraseology will be followed, and the usual distinction between such expressions as "square on" and "square of" will be observed. Square on will mean the area of the square constructed on the given sect, and square of, the square of the numeric measure of the given sect. 239 240 PLANE GEOMETRY Given: AABC; 2^BCA=Ti. 4; squares CBFG, ABDE, and CAKH. Prove: Sq. ABDE =Sq. CBFG + Sq. CAKH. Proof: (Some steps which the student can readily supply have been pur- posely omitted.) Draw CM .LEWD. Draw CD and AF. (1) .'. CM ||Dand AE. (2) ACG is a st. line-_ (3) AABF = y 2 (BF-CB). (4) AABF=y 2 (Sq. CB/^G). (5) LMDB is a rect. _ (6) ACDB=^ (BD-LB}. (7) ACDBs^ (Rect. LMDB}. (8) BD=AB,CB=BF. (10) /. AABF^ACBD. (11) /.Sq. CF^Rect. LD. (1) Why? (2) Why? (3) Why? (4) Why? (5) Why? (6) Why? (7) Why? (8) Why? (9) Why? (10) Why? (11) Why? \ /" \ x '-*\\ yl 1 \ \ \ \ B M D Draw the necessary construction lines and show that sq. AH equals rect. AM. Finish the proof. Theorem 26fr. The areas of two triangles having an angle of one equal to an angle of the other are to each other as the products of the sides including those angles. R Given: AABC and &AiXY with %.A = AABC AB-AC ToProve: Suggestions: Proof being left to the student. Place AABC in the position of AAiQR AA 1 RQ_ A 1 X' Draw QY. A 1 R , A l Y' a a AAiYQ~A l Y' AA 1 XY Could the case arise in which QR intersects XY between X and F? If so, show that this proof still holds or give one that does hold. AREAS OF RECTILINEAR FIGURES 241 Theorem 27. The area of a trapezoid is equal to half the produc t of its altitude and the sum of its bases. EXERCISES. SET LXXIX. AREAS Numeric 878. Find the expense of paving a path 4' wide inside a square piece of ground the side of which is 50' if the price is 18 cents per square yard. What would be the cost if the path were outside the piece of ground? a879.* Find the dimensions of a rectangle given: (a) Area 216 sq. ft., perimeter 60 ft. (6) Area 600 sq. ft., difference of the sides 10 ft. (c) Area 756, sides in the ratio -J. (d) Area 340 sq. ft., sum of squares of two consecutive sides 689 sq. ft. 880. Find the change in the area of a triangle of base a and altitude h in the following cases : (a) If a and h are increased by m and n, respectively. (6) If a and h are diminished by m and n, respectively. (c) If a is increased by m, and h diminished by n. (d) If a is diminished by m, and h increased by n. 881. A line of division is drawn between two sides of a triangle, dividing it into a triangle and a quadrilateral. What parts are these two figures, respectively, of the entire triangle if the line of division cuts off the following parts of the two sides, reckoned from the intersection of the sides? (a) | and |. (6) f and f . (c) 1 and J. (d) | and \. (e) and (/) \ and i. 882. Find the area of a rhombus, given: (a) The diagonals 18 and 12 units. (6) The sum of the diagonals 12, and their ratio |-. * As here, "a" precedes a problem which calls for the solution of an affected quadratic equation. 242 PLANE GEOMETRY Theoretic 883. (a) Prove, geometrically, the algebraic formula, a(b+c)=ab+ac. (b) Prove, geometrically, the algebraic formula, a(b c)=ab ac. (c) Prove, geometrically, the algebraic formula, 884. The area of a rhombus is equal to half the product of its diagonals. 885. If lines are drawn from any point inside a parallelogram to the four vertices, the sum of either pair of triangles with parallel bases is equal to the sum of the other pair. 886. The accompanying figures show easy methods of trans- forming (a) a triangle into a parallelogram, (b) a parallelogram into a triangle, (c) a trapezoid into a parallelogram. Explain. Can you give more than one explanation? If so, upon what does your explanation depend? A F D For problems in construction based upon this chapter see Chapters VII and VIII. TERMS DEFINED IN CHAPTER H Square on (a sect), square of (a quantity), medians of a triangle, rhomboid. CHAPTER III SIMILARITY The following are terms with which the student should now be familiar: Antecedent, consequent, extremes, means, mean proportion, mean proportional, continued proportion, inversion, composition, division, composition and division.* Theorem 28. Any proportion may be transformed by alternation. Theorem 29. In any proportion the terms may be combined by composition or division. Theorem 30. In a series of equal ratios, the ratio of the sum of any number of antecedents to the sum of their consequents equals the ratio of any antecedent to its consequent. Theorem 31. A line parallel to one side of a triangle divides the other sides proportionally. Cor. 1. One side of a triangle is to either of the sects cut off by a line parallel to a second side as the third side is to its homologous sect. COT. 2. Parallels cut off proportional sects on all transversals. Cor. 3. Parallels which intercept equal sects on one transversal do so on all transversals. Cor. 4. A line which bisects one side of a triangle, and is parallel to the second, bisects the third. Cor. 6. A sect which bisects two sides of a triangle is par- allel to the third side and equal to half of it. Suggestions: What means have you for proving lines parallel? How may one sect be proved half of another? What kind of figure would you like to have? * The student will find an alphabetical index of definitions on p. 379. 243 244 PLANE GEOMETRY The sect joining the mid-points of the non-parallel sides of a trape- zoid is called its median. f Cor. 6. The median of a trape- zoid is parallel to the bases and equal to one- half their sum. Suggestions: Why draw a diagonal? Show that a line through M \\Tffi will bisect AC and therefore DC, and hence coincide with MMi. Cor. 7. The area of a trapezoid equals the product of its median and altitude. What numerical relation exists between the median and the bases of a trapezoid? Theorem 32. A line dividing two sides of a triangle proportion- ally is parallel to the third side. Cor. 1. A line dividing two sides of a triangle so that these sides bear the same ratio to a pair of homologous sects is parallel to the third side. DIVISION OF A SECT A point in a sect is said to divide it internally, and a point in the prolongation of a sect is said to divide it externally, and in both cases the divisions of the sect are reckoned from one extremity to the point of division and from that point to the other extremity of the original sect. A _7 B E I divides sect AB internally in the ratio =. IB AE E divides sect AB externally in the ratio . EB NOTE. When neither "internal" nor "external" is used to qualify the division, internal is understood. Is there any ratio into which a sect cannot be divided externally? A sect is said to be divided harmonically when it is divided intern- ally and externally in the same ratio. SIMILARITY 245 In the foregoing illustration AB is divided harmonically if Tl_AE IB~EB' Is there any ratio into which a sect cannot be divided harmonically? Discuss Theorems 31 and 32 from the point of view of external division of a sect. Theorem 32cz. The bisector of an angle of a triangle divides the opposite side into sects which are proportional to the adjacent sides. Given: AABC; D on AB and 4. A CD = %.DCB. AD AC To prove: 55-53. Suggestions : Compare A A DC and ABDC in two ways. A "~*B Theorem 32b. The bisector of an exterior angle of a triangle divides the opposite side externally into sects which are propor- tional to the adjacent sides. 'TJ Given: AABC with exterior $. BCD; EmAB produced; %.BCE= 4.ECD. AB AC To prove: - = - Suggestions: Com- pare AAEC first with ABEC, and second with ABiEC How should Bi be taken so that ABiEC may be substituted for ABECt Consider special cases where a=b, a>b. Cor. 1. The bisectors of an adjacent interior and exterior angle of a triangle divide the opposite side harmonically. (Proof left to the student. Discuss special cases.) EXERCISES. SET LXXX. RATIO. PROPORTION. PARALLELS Numeric 887. Find the value of n if (a) - = |, (6) - = -. no n c 216 246 PLANE GEOMETRY 889. If find? 890. If =, find? o O 91. Form all possible proportions involving a, b, p, and q if ab=pq. 892. Find the mean proportionals between: (a) 2 and 50. (c) 3 and 21. (e) a and 6. (b) 2 and 75. (d) 3 and 19. (/) a+6 and a -b. 893. Find the third proportional to (a) 5 and 6, (6) a and b. a c 894. Transform the proportion T=-, so that b becomes the third u term. 895. If -r- = ?, what is the ratio of a to &? o o fi AABC Suggestions : Show that : (1) Three angles of A one triangle cannot be supplementary to three angles of the other. (2) Two angles of one triangle cannot be supplementary to two of the other. (3) What, then, is the fact? AB] CA. 250 PLANE GEOMETRY Theorem 35. Triangles which have two sides of one propor- tional to two sides of another and the included angles equal are similar. Theorem 36. If the ratio of the sides of one triangle to those of another is constant, the triangles are similar. EXERCISES. SET LXXXI. SIMILARITY OF TRIANGLES Numeric 935. If the sides of a triangle are 3, 7, and 8, find the sides of a similar triangle in which the side homologous to 7 is 9. 936. If the sides of a triangle are a, 6, and c, find the sides of a similar triangle in which the side homologous to a is p. Construction 937. The sides of a triangle are 5, 6, and 7. Construct a triangle similar to the original, having the ratio of similitude 3 to 2. Q 938. Construct a triangle similar to the accompanying triangle with the ratio of J2 similitude equal to that of the two given sects a and b. Theoretic 939. Two isosceles triangles are similar if an angle of one is equal to the homologous angle of the other. 940. Prove that the altitudes of a triangle are inversely propor- tional to the sides to which they are drawn. 941. If the altitudes AD and BE in A ABC are drawn, prove that AC DC -===-;.. Are the altitudes directly or inversely proportional to DC and EC1 942. If a spider, in making its web, makes A& \\AB, B^Ci \\ BC, CiA || CD, >!#! || DE, and E^ || EF, and then runs a line from F\ \\ FA, will it strike the point AI? Prove your answer. 943. If D is taken in the leg A B of an isosceles triangle ABC, so that CD=AC (the base), then SIMILARITY 251 944. Isosceles or right triangles ABC and PQR are similar if h P PQ 945. The diagonals of a trapezoid divide each other pro- portionally. 946. If in triangle ABC altitudes AD and BE meet at 0, then: (a) BD DC=DO AD; (b) BD AC=BO -'A5. 947. If in a parallelogram PQRS, a sect Q7 7 is drawn cutting the diagonal PR in V, the side #/S in L and the prolongation of PS 948. In similar triangles homologous angle bisectors are directly proportional to the sides of the triangle. 949. In a quadrilateral A BCD, right-angled at B and D, per- pendiculars PE and PF from any point P in AC to the sides BC P'E PF and AD are such that =-+=-= 1. d950. Every straight line cutting the sides of a triangle (pro- duced when necessary) determines upon the sides six sects, such that the product of three non-consecutive sects is equal to the product of the other three. The line XYZ must cut either (a) two sides of the triangle and the third side produced (Fig. 1), -or (b) all three sides pro- duced (Fig. 2). The proof in both instances is the same. Fig. 1 Fig. 2 Draw CD \\AB. From the similar triangles AX AZ . BY BX = and ===== CD CZ , CY CD . AX-BY AZ-BX therefore : == = . CY -CD CZ CD whence AZ BY CZ=AZ -BX-CY. 252 PLANE GEOMETRY This theorem was discovered by Menelaus of Alexandria about 80 B.C. d951. Prove the converse of the last theorem. Let XY produced cut AC produced in a point P. Then AX BY; CP=AP -BX-CY. But, by hypothesis, A ~X BY CZ=AZ - BX CF; CP AP whence ======; whence or AC AC AZ' A ~ B that is, P coincides with Z. d952. Lines drawn through the ver- tices of a triangle, and passing through a common point, determine upon the sides six sects, the product of three non-consecutive sects being equal to the product of the other three. Fig. 2 The common point may lie either inside or outside the triangle (Figs. 1 and 2). In both cases apply Ex. 950 to the AACD and sect BOF and to the ABCD and sect AOE, then multiply the results. Fig. 1 This theorem was first discovered by Ceva of Milan, in 1678. d953. Conversely, if three lines drawn through the vertices of a triangle determine upon the sides six sects, such that the product SIMILARITY 253 of three non-consecutive sects is equal to the product of the other three, the lines pass through the same point. The proof is similar to that of Ex. 951. Theorem 36a. The homologous angles of similar polygons are equal ', and their homologous sides have a constant ratio. Suggestions: Why is it that the n-gons may be placed as in the accompanying diagram? To prove the second part use the C similar triangles thus formed and show that the ratio of the homo- logous sides is equal to the ratio of similitude of the polygons. Give the complete proof. Cor. 1. If the ratio of similitude of polygons is unity, they are congruent. Cor. 2. The homologous diagonals drawn from a single vertex of similar polygons divide the polygons into triangles similar each to each. A Why is A AiBid ^ A ABC? Having proved this can you prove A A \C\D\ v> A A CD, and so forth? Why is it that only these two sets of A need be proved similar? Write the proof in full. Theorem 366. Polygons whose homologous angles are equal and whose homologous sides have a constant ratio are similar. Given: Polygons ABC. N and winch JV . AB BC = Prove: A 1 B 1 C 1 ABC. . .N. 254 PLANE GEOMETRY PROOF Place A iBiCi . . . Ni as in the diagram so that A^B z \\AB t and draw AA^X and BB^Z. (2) If AB=A 1 B 1 then (3) If AB>Aj3 lt then etc. and .ABB^a is not a (4) and AA 2 X and at some point O. (5) AAOB (6) (7) intersect ^ AB ~~BO' (8) and (9) (10) (11) Draw (12) Then (13) (14) (15) C^" cutting BBjZ in 0] coin- (17) cides with 0. (18) .'.BOi through 0. (19) Any O^L cutting sides of the polygons in R and L respectively is BO and CC 2 passes ,. . , , tn divided so that = (1) Data. (2) Congruent polygons are similar. (3) The opposite sides of a paral- lelogram are equal, and conversely. (4) Non-parallel coplanar lines in- tersect. (5) Two angles equal each to each, or sides respectively parallel. (6) Homologous sides of similar triangles have a constant ratio. (7) Corresponding angles of par- allels. (8) Data. (9) The differences of equals less equals are equal. (10) Corresponding 4? equal. (11) See (1), (2), (3), and (4). (12) See (5). (13) See (6). (14) See (7). (15) Data. (16) Quantities equal to equal quan- tities are equal to each other. (17) By division and the products of equals multiplied by equals are equal. (18) Two points determine a straight line. (19) See (6). In the same way it may be proved that DD 2 , . . . NN 2 , pass through and therefore AiBiCi. . .Ni^ABC. . .N by the definition of similar figures. SIMILARITY 255 Cor. 1. If homologous diagonals drawn from a single ver- tex of two polygons divide them into triangles similar each to each and similarly arranged, the polygons are similar. (Proof is left to the student.) Theorem 37. The perimeters of similar triangles are propor- tional to any two homologous sides, or any two homologous altitudes. Cor. 1. Homologous altitudes of similar triangles have the same ratio as homologous sides. Cor. 2. The perimeters of similar polygons have the same ratio as any pair of homologous sides or diagonals. Apply propositions 3690, DA = p c b, the projection of c upon 6. Prove: a 2 = 6 2 -}~c 2 -{-2& /)<.& (Proof left to the student.) Theorem 39c. 7. The sum of the squares of two sides of a triangle is equal to twice the square of half the third side, increased by twice the square of the median to it. This theorem is attributed to Appollonius of Perga (c. 225 B.C.). 77. The difference between the squares of two sides of a tri- angle is equal to twice the product of the third side and the projection of the median upon it. 260 PLANE GEOMETRY Given: AABC in which a>c, mb is the median to b and XD =p is its projec- p tion upon 6. Prove: I. D Outline of proof, which the student is asked to write. (1) Show that $.BDC is obtuse V* and &.BDA is acute by com- paring A ABD and BDC. (2) Use propositions 39a and 39& to give the values of a 2 and c 2 . (3) Combine these values as indicated in I and II. Consider this proposition when a=c. Cor. 1. If m b represents the length of the median to side b of the triangle whose sides are a, b, c, then Solve proposition 39c I for m b . Theorem 39d. // h a represents the altitude upon side a of a triangle whose sides are a, b, c, and s represents its semiperimeter, i.e., s = ' then h a =- n Vs(s-a)(s-b)(s-c). Fig. 1. PROOF (1) At least one of the angles B or C is acute. Suppose C is acute (FiguresJ_and 2). (2) (4) All authorities to be given by the student. or 2a 2a 2 -c 2 \ /c 2 -(a 2 -2ab+b 2 ) \ A 2a ; 4a 2 FIG. 2. SIMILARITY 261 3 x /s(s-a)(s-6)(s-c). a Show that this proof holds for Figures 3 and 4, i.e., when C is obtuse or right. Cor. 1. If A stands for the area of a triangle whose sides are a, b,c, and whose semiperimeter is s, then A = Vs(s-a)(s-b)(s-c) By what must the value of h a be multiplied to obtain the value of Af This formula is known as Heron of Alexandria's, to which atten- tion was called on p. 98, Ex. 359 of the First Study. The Principle of Continuity. By considering both positive and negative properties of quantities, a theorem may frequently be stated so as to include several theorems. For instance, Theorems 39, Cor. 2, 39a and 396 may be stated as a single theorem if we take into consideration the direction of the projection of c upon 6. If AD, the projec- tion of c upon 6 in Fig. 1 be considered positive, AD in Fig. 3 will be negative. Therefore we may say in general that a 2 =6 2 +c 2 -26.p c6 , for in Fig. 1 that means a 2 =6 2 +c 2 -26 (+35), or a 2 ^6 2 +c 2 -26.p c6 , while in Fig. 2 it means a 2 = FlG - 3 " 62+c 2 _- 26(0), or a 2 = b* + c 2 , and in Fig. 3 it meansa 2 =6 2 +c 2 -2b(-DA), or a 2 =6 2 +c 2 +26-pet. 262 PLANE GEOMETRY EXERCISES. SET LXXXIII. METRIC RELATIONS Theorems 39a through 39d enable us to calculate the altitudes and medians of a triangle in terms of its sides, and the length of the projection of any side upon any other, as well as to determine whether a triangle is acute, right or obtuse. 973. If in solving the identity a 2 =6 2 +c 2 -2b-p cb for p c b for any particular values of a, b, and c, we find (a) p c b = 0, then 0, then 90. Explain. 974. Give the formulas for b and c corresponding to those given in propositions 39a and 396 for a. 975. From the three formulas obtained in Ex. 974 derive formulas for p cb , p ab , p ca or for p bc , p ba , p ac - 976. Give a formula for m a , for m c , for h b , for h e . 977. In a right or an obtuse triangle, the greatest side is opposite the right or the obtuse angle. Hence if a is the greatest side of a triangle, show that if a 2 >& 2 -f-c 2 the triangle is acute, if a 2 ==6 2 +c 2 the triangle is right, and if a 2 <& 2 -f c 2 the triangle is obtuse. dt978. Show that p c b=c cos A and hence that the general formula might be a 2 =6 2 +c 2 2bc cos A. Theorem 39e. If similar polygons are constructed on the sides of a right triangle f as homologous sides f the polygon on the hypot- enuse is equal to the sum of the polygons on the other two sides. If PI, PZ, and PS be similar polygons constructed on a, 6, and c respect- r> ively, as homologous sides, when < C is a right angle in A ABC, -= ?, PS _ Give the complete proof. EXERCISES. SET LXXXIII (concluded) Numeric 979. The base of an isosceles triangle is 48 in. Find the altitude if each arm equals 50 in. 980. Let ABC be a right triangle. The two sides about the right angle C are respectively 455 and 1,092 feet. The hypotenuse AB is divided into two sects AE and BE by the perpendicular upon it from C. Compute the lengths of AE, BE, and CE. SIMILARITY 263 981. (a) If two sides of a triangle equal 15 and 25, respectively, and the projection of 15 upon 25 equals 9, what is the value of the third side? (6) Is the triangle right, acute, or obtuse? 982. The altitude of a triangle is 20 in. A line parallel to the base and 12 in. from the base cuts off a triangle that is what part of the given triangle? 983. If the side of an equilateral triangle equals 10 in., what is the length of the projection of one side upon another? 984. Find the projection of AB upon a line XY, if AB and XY include an angle of 45, and AB=2. 985. Find the side a of the square equal to an equilateral tri- angle whose side is s. Solve the equation for s in terms of a. 986. Two sides of a triangle are 5 and 8, respectively, and include an angle of 30. Find the area. 987. Find the area of an equilateral triangle of which (a) The side is 30, (6) The altitude is 34, (c) The side is a, (d) The altitude is h. 988. Find the area of a trapezoid, given: (a) The median ra, and altitude h; (b) The median ra, one leg I, and the angle between this and the base 30; (c) Bases 61 and 6 2 , and legs each I. 989. In a trapezoid, given the two bases a, b, and the altitude h. The legs are divided into three equal parts by lines parallel to the bases. Find in terms of a, 6, and h, the areas of the three parts into which the trapezoid is divided. 990. Find the side of a rhombus composed of two equilateral triangles and equal to another rhombus whose diagonals are 12 and 18. 991. ABC is a triangle and AD the altitude upon BC. If AD = 13, and the length of the perpendiculars from D to AB and AC are 5 and lOf , respectively, find the area of the triangle. 992. Find the area of a square in terms of (a) its perimeter p, (b) its diagonal d. 264 PLANE GEOMETRY 993. Find the dimensions of a rectangle given: (a) Its perimeter p and area a : (b) Its length I and diagonal d; (c) Its diagonal d and the ratio of its length to its width r. 994. Find the projection of AB upon XY, if AB = m, and the two lines include an angle (a) of 60, (6) of 30. 995. In triangle abc, a = 8, 6 = 15, and the angle opposite c equals 60. Findc. 996. In triangle abc, a=3, 6 = 5, and the angle opposite c equals 120. Findc. 997. In triangle abc, a = 7, 6 = 8, and the angle opposite c equals 120. Find c. 998. Two side.s of a triangle are 20 and 30, respectively, and include an angle of 45. Find the third side. 999. Two sides of a triangle are 16 and 12 in., respectively, and include an angle of 60. Find the third side. 1000. Find the area of a rectangle in terms of its length I, and diagpnal d. 1001. In triangle abc } *a = 20, 6 = 15, and c = 7. Find the projec- tion of 6 upon c. Is the triangle obtuse or acute? 1002. In a quadrilateral ABCD, AB = 10, BC = 17, CD = 13, DA =20, and AC = 21. Find the diagonal BD. 1003. Two sides of a triangle are 17 and 10; the altitude upon the third side is 8. What is the length of the third side? 1004. The sides of a triangle are 7, 8, and 9, respectively. Find the length of the median to 8. 1005. The sides of a triangle are 10, 5, and 9, respectively. Find the length of the median to 9. 1006. The sides of a triangle are 22, 20, and 18, respectively. Find the length of the median to 18. 1007. The sides of a triangle are 9, 10, and 17, respectively. Find the three altitudes. 1008. The sides of a triangle are 11, 25, and 30, respectively. Find the three altitudes. 1009. The sides of a triangle are 12, 14, and 15 respectively. Find the three altitudes. 1010. (a) Find the altitude of an equilateral triangle with side s. (6) Find the side of an equilateral triangle with altitude h. SIMILARITY 265 1011. Find the area of a triangle whose sides are respectively (a) 13, 14, 15, (6) 9, 10, 17, (c) 11, 25, 30. d!012. The sides of a triangle are as 8 to 15 to 17. Find the altitudes if the area is 480 sq. ft. 1013. Find one diagonal of a parallelogram, given the sides a, 6, and the other diagonal g. Construction 1014. Given any sect as unit, construct a sect which is -\/2, \/3> \/5 units. PA 1 1015. In a given sect AB find a point P such that (a) -^-5 = ^, "& V2 PA 1 1016. Construct an equilateral triangle equal to (a) the sum of two given equilateral triangles, (6) their difference. 1017. Construct a polygon similar and equal to (a) the sum of two given similar polygons, (&) their difference. 1018. Construct a square equal to the sum of 3, 4, 5 given squares. (For other construction problems based on this character, see Chapter VII.) Theoretic 1019. The median drawn from the extremities of the hypot- enuse of the right triangle ABC are BE, CF\ prove that d!020. In a certain triangle ABC, AC 2 -BC 2 =^AB*\ show that a perpendicular dropped from C upon AB will divide the latter into sects which are to each other as 3 to 1. 1021. If ABC is a right triangle, C the vertex of the right angle, D any point in AC, then BD 2 +AC 2 ^AB 2 +DC 2 . 1022. The sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of its diagonals. _1023 1 _If Jn the parallelogram ABCD 4=60, ZC 2 =ZB 2 + BC 2 +AB-BC. 1024. The sum of the squares of the medians of a triangle is three-fourths the sum of the squares of its sides. 266 PLANE GEOMETRY dl025. The sum of the square on the difference of the legs of a right triangle, and twice the rectangle whose sides are the legs of the triangle, is equal to the square on the hypotenuse. State this exercise in algebraic form and prove it. 1026. One-half the sum of the squares on the sum and difference of the legs of a right triangle is equal to the square on the hypotenuse . State and prove this exercise algebraically. dl027. Two similar parallelograms are to each other as the pro- ducts of their diagonals. 1028. If in triangle ABC, AB=BC and altitudes AD and BE A i s\ ji .oC intersect at 0, then TT= 1029. If in a triangle the ratio of the squares of two sides is equal to the ratio of their projections upon the third side, the triangle is a right triangle. dlOSO. The sum of the squares of the sides of a quadrilateral is equal to the sum of the squares of the diagonals increased by four times the square of the sect joining their mid-points. dlOSl. If perpendiculars are drawn to the sides of a triangle from any point within it, the sum of the squares of three alternate sects cut off on the sides is equal to the sum of the squares of the three remaining sects. 1032. If ABCD is a sect_such that AB=BC==C^D, and P is any other point, prove that PA 2 +3PC 2 =PD 2 -\-3PB*. CHAPTER IV LOCUS Theorem 40. The locus of points equidistant from the ends of a sect is the perpendicular bisector of the sect. Cor. 1. Two points equidistant from the ends of a sect fix its perpendicular bisector. Theorem 41. The locus of points equidistant from the sides of an angle is the bisector of the angle. Cor. 1. The locus of points equidistant from two intersecting lines is a pair of lines bisecting the angles. (Concurrent lines are those which pass through a common point.) Theorem 41a. The bisectors of the angles of a triangle are concurrent in a point equidistant from the sides of the triangle. Given: &ABC; ^BAZ^^ZAC, %_ACY=4.YCB, To prove: AZ, CY, and BX are concurrent in a point equidistant from the sides. Suggestions for proof: If BX were parallel to CY what relation would exist between ^YCB and ^.CBX? If is a point in YC, how is it located with regard to BC and ACT If O is a point in BX, how is it located with regard to BC and ABf Why, then, must O be on AZ? 287 PLANE GEOMETRY X X V/2 M c / / 2 A Theorem 416. The perpendicular bisectors of the sides of a triangle are concurrent in a point equidistant from the vertices. Giveni^AABC; MM^AMB, RR, ARC, PPiA.CPB; AM = MB,AR=RC ; CP=PB. To prove: MMi, PPi, RRi are concurrent in a point equi- distant from A, B, and C. Suggestions for proof: If MMi were parallel to PP\ what relation would exist between MMi and CB1 Therefore, what relation would exist between CB and BA? If is a point in MMi how is it located with regard to A and 5? If O is a point in PPi how is it located with regard to C and 5? Why, then, is a point in RRJ Theorem 41c. The altitudes of a triangle are concurrent. Given: AAJ5C; CH A. AHB, BTA. CTA, ALA. CLB. To prove: AL, BT, CH are concurrent. Analysis of proof: If AL, BT, CH were the perpendicular bisec- A ^-JI tors of the sides of AAiBiCi they would be concurrent. If they are to be such, how should the sides of AAiBiCi be drawn through A, B, and C to make AL_\_BiCi, BTAid t and CH A.A 1 B 1 1 If ACi =AB lf BCi=BAi, and CA l =CBi, then this construction would make possible a proof. Give a synthetic proof.* NOTE. Where an obtuse triangle is involved, show that no separate proof is necessary. H * For notes on various types of proof see Chapter VI, p. 297. LOCUS 269 Theorem 41d. The medians of a triangle are concurrent in a point of trisection of each. Given : A ABC; BM = MC, B M in BC, MI in BA, M 2 in AC. . To prove: AM, BM 2 , CM! meet in a point O such that Ad=2MO, 2M 1 O. Notes on proof: Why can- not CM\ be parallel Bisect AO at X and CO at F. How is JTF related to AC? (Consider AAOC.) How is MMi related to AC? (Consider AABC.) Therefore, how is MMi related to XYt Prove AXOY^ AMOM 1} and hence XO=OM and MiO=OY. Therefore, CO=2MiO and AO=2MO. Consider, now, medians AM and BMi. Can they be parallel, and, if not, why would their point of intersection Oi coincide with 0? EXERCISES. SET LXXXIV. LOCUS * 1033. Find the locus of the mid-points of sects drawn from a common point to a given line. 1034. Find the locus of the points in which the sects mentioned in Ex. 1033 are divided in the ratio 5 to 8. 1035. Find the locus of the points in which the sects mentioned in Ex. 1033 are divided in the ratio of two given sects a and b. 1036. Find the locus of the mid-points of sects connecting points on two parallels. 1037. Lines are drawn parallel to one side of a triangle and are terminated by the other two sides. What is the locus of their mid-points? * Various terms are used in stating locus exercises. We shall follow the most usual interpretations, which are: (1) No locus exercise need be proved unless a proof is definitely called for, indicated by "prove"; (2) an accurate construction is called for when the terms "plot" or "construct" are used; (3) the terms "describe" or "find" are used in calling for a statement of what the locus is. 270 PLANE GEOMETRY 1038. Parallel sects are drawn with their extremities in the sides of an angle. Find the locus of their mid-points. 1039. What is the locus of the vertices of triangles having (a) a given base and a given altitude? (6) a given base and a given area? d!040. Find a point within a triangle such that the lines joining it to the vertices shall divide the triangle into three equal parts. 1041. If AB be a fixed sect, find the locus of a point which moves so that its distance from the nearest point in AB is always equal to a given sect c. How does this locus differ from the one obtained if for the word " sect " we substitute " line "? 1042. If PQRS be a rhombus, such that Q and S lie on two fixed lines through P, find the locus of R. 1043. If PQRS be a parallelogram of constant area and given base PS, find the loci of R and Q.__ 1044. If A be a fixed point, BC a fixed line, n any integral number, P any point in BC, and Q a point in AP or PA produced so that AQ=n-AP, find the locus of Q. 1045. Find the locus in the last exercise if AP=n-AQ. 1046. If in the APQR a sect QS be drawn to any point in the QT base, find the locus of a point T on this sect such that the ratio ; 1 o is constant. Justify the two expressions " the locus of points " and " the locus of a point." 1047. If from the intersection of the diagonals of a parallelo- gram sects are drawn to the perimeter, find the locus of the point in these sects such that the ratio of the parts into which the sect fn is divided is (a) constant, (6) equal to a given ratio -- or (c) equal ft/ to the ratio of two given sects a and 6. 1048. Given a square with side 3 in. Construct the locus of a point P such that the distance from P to the nearest point of the square is 1 in. 1049. Upon a given base is constructed a triangle, one of whose base angles is double the other. The bisector of the larger base angle meets the opposite side at the point P. Find the locus of P. LOCUS - 271 dl050. What is the locus of points, the distances of which from two intersecting lines are to each other as m to ft? dl051. Find the locus of points the sum of whose distances from two given parallel lines is equal to a given length. Discuss all possible cases. d!052. Find the locus of points the difference of whose distances from two given parallel lines is equal to a given length. Discuss. dl053. Find the locus of points the sum of whose distances from two given intersecting lines is equal to a given length. d!054. Find the locus of points the difference of whose distances from two given intersecting lines is equal to a given length. 1055. The vertex A of a rectangle ABCD is fixed, and the direc- tion of the sides AB and AD also are fixed. Plot the locus of the vertex C if the area of the rectangle is constant. d!056. Plot the locus of a point if the product of its distances from two perpendicular lines is constant. d!057. Plot the locus of a point P such that the sum of the squares of its distances from two fixed points is constant. d!058. Plot the locus of a point such that the difference of the squares of its distances from two fixed points is constant. dl059. Given the base of a triangle in magnitude and position and the difference of the squares of the other two sides, plot the locus of the vertex. 1060. Given a square ABCD. Let E be the mid-point of CD, and draw BE. A line is drawn parallel to BE and cutting the square. Let P be the mid-point of the sect of this line within the square. Construct the locus of P as the line moves, always remain- ing parallel to BE. Other locus exercises will be found in the chapter on "Circles," pp. 273, 275, 276, 283, 284, 288, as well as in the chapter on " Methods of Attacking Problems," p. 306, et seq. CHAPTER V THE CIRCLE Theorem 42. Three points not in a straight line fix. a circle. Theorem 43. In equal circles, equal central angles intercept equal arcs,* and conversely. Theorem 43a. In equal circles, the greater of two central angles intercepts the greater arc, and conversely. Suggestion: Lay off the smaller central angle on the greater to prove the direct. What may be done in the case of the converse? Theorem 44. In equal circles, equal arcs are subtended by equal chords, and conversely. Theorem 44a. In equal circles, unequal arcs are subtended by chords unequal in the same order, and conversely. Suggestions: If radii are drawn, what do we know of the triangles formed? Then what method of proving sects unequal may be used in the proof of the direct? In the proof of the converse, what is the only method you are ready to use in order to prove arcs unequal? Write the proofs of both parts of this theorem. Theorem 45. A diameter perpendicular to a chord bisects it and its subtended arcs. Cor. 1. A radius which bisects a chord is perpendicular to it. Cor. 2. The perpendicular bisector of a chord passes through the center of the circle. Theorem 46. In equal circles, equal chords are equidistant from the center, and conversely. Theorem 46a. In equal circles the distances of unequal chords from the center are unequal in the opposite order, and conversely. * Such arcs are actually congruent, but we are following custom in using the word "equal." 272 THE CIRCLE 273 Axioms of Inequal- ity (continued). 5. Squares of positive un- equals are unequal in the same order. Illus- trate. Given: OCaQd; chord AB> chord DE; CY LAYB, CiX DXE. Prove: CYDE, .'. YB>XE; %aadv. Can the same method be used for the proof of the converse? Give the proof in full. NOTE. Why is it better to use a direct method rather than the method of exclusion in the proof of the converse? Theorem 47. A line perpendicular to a radius at its outer ex- tremity is tangent to the circle. Cor. 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. Cor. 2. The perpendicular to a tangent at the point of contact passes through the center of the circle. Cor. 3. A radius perpendicular to a tangent passes through the point of contact. Cor. 4. Only one tangent can be drawn to a circle at a given point on the circle. Theorem 48. Sects of tangents from the same point to a circle are equal. Theorem 49. The line of centers of two tangent circles passes through their point of contact. Theorem 49a. The line of centers of two intersecting circles is the perpendicular bisector of their common chord. What is the locus of points equally distant from the ends of a sect? Where, then, do the centers of these circles lie? 18 274 PLANE GEOMETRY EXERCISES. SET LXXXV. THE STRAIGHT LINE AND THE CIRCLE 1061. What methods can you now add to those known before this chapter of showing: (a) Sects equal? (6) Angles equal? (c) Sects unequal? (d) Angles unequal? (e) Sects perpendicular? 1062. Can you now mention certain relations of a new kind of element? If so, what are they? Numeric 1*063. Two parallel chords of a circle are 4 and 8 units in length, and their distance apart is 3 units. What is the radius? 1064. Two parallel chords of a circle are d and k in length, and their distance apart is /. What is the radius? 1065. Find the length of a tangent from a point 15" from the center of a circle whose radius is 5". 1066. Find the radius of a circle if the length of a tangent from a point 23" from the center is 16". 1067. Find the length of the longest chord and of the shortest chord that can be drawn through a point 1' from the center of a circle whose radius is 20". 1068. The radius of a circle is 5". Through a point 3" from the center a diameter is drawn, and also a chord perpendicular to the diameter. Find the length of this chord, and the distance (to two decimal places) from one end of the chord to the ends of the diameter. 1069. The span (chord) of a bridge in the form of a circular arc is 120', and the highest point of the arch is 15' above the piers. Find the radius of the arc. 1070. The line of centers of two circles is 30. Find the length of the common chord if the radii are 8 and 26 respectively. 1071. Two circles touch each other, and their centers are 8 /r apart. The radius of one of the circles is 5". What is the radius of the other? (Two solutions.) 1072. If the radii of two concentric circles are denoted by a and b, respectively, find the radius of a third circle which shall touch both given circles and contain the smaller. THE CIRCLE 275 Locus 1073. Find the locus of the center of a circle which has a given radius and is tangent to a given circle. 1074. Find the locus of the extremity of a tangent of given length drawn to a given circle. 1075. Two equal circles are tangent to each other externally. Find the locus of the centers of all circles tangent to both. 1076. A sect so moves that it re- mains parallel to a given line, and so that one end lies on a given cir- cle. Find the locus of the other end. Does the accompanying diagram give the complete locus? 1077. What is the locus of the mid-points of parallel chords of a circle? Prove the correctness of your statement. 1078. From a point outside how many tangents are there to a circle? Prove. 1079. Find the locus of the mid-point of a sect that is drawn from a given external point to a given circle. 1080. A straight line 3 in. long moves with its extremities on the perimeter of a square whose sides are 4 in. long. Construct the locus of the mid-point of the moving line. 1081. A circular basin 16 in. in diameter is full of water, and upon the surface there floats a thin straight stick 1 ft. long. Shade that region of the surface which is inaccessible to the mid-point of the stick, and describe accurately its boundary. 1082. The image of a point in a mirror is apparently as far behind the mirror as the point itself is in front. If a mirror revolves about a vertical axis, what will be the locus of the apparent image of a fixed point 1 ft. from the axis? 1083. In the rectangle A BCD the side AB is twice as long as the side BC. A point E is taken on the side AB, and a circle is drawn 276 PLANE GEOMETRY through the points C, D, and E. Plot the path of the center of the circle as E moves from A to B. 1084. Find the locus of a point P such that the ratio of its dis- tances from two fixed points is equal to the constant ratio m to n. Construction * 1085. Find the center of a given circle. 1086. Inscribe a circle in a given triangle. 1087. Circumscribe a circle about a given triangle. 1088. Escribe circles about a given triangle. (See p. 359.) 1089. Through a given point in a circle draw the shortest pos- sible chord. 1090. Inscribe a circle in a given sector. VII. 1091. With its center in a given line construct a circle which shall be: (a) Tangent to another given line at a given point. (6) Tangent to two other given lines.f (c) Tangent at a given point to a given circle. VII. 1092. Construct a circle of given radius r, which shall : (a) Pass through a given point and be tangent to a given line; (b) Pass through a given point and be tangent to a given circle; (c) Be tangent to a given line and a given circle; (d) Be tangent to two given circles. VII. 1093. Construct a circle tangent to two given lines and having its center on a given circle. VII. 1094. An equilateral triangle ABC is 2 in. on a side. Construct a circle which shall be tangent to A B at the point A and shall pass through the point C. VII. 1095. To a given circle draw a tangent that shall be parallel to a given line. 1096. Draw two lines making an angle of 60, and construct all the circles of J^ in. radius that are tangent to both lines. * While some more or less difficult construction problems have been inserted at this point, they have been primarily inserted for the benefit of those pupils who wish to test their power, and when found too difficult may well be omitted until Chapter VII has been studied. Such problems will be followed by the Roman number "VII." t See p. 311 for the section dealing with "The Discussion of a Problem." THE CIRCLE 277 dl097. Construct an equilateral triangle, having given the radius of the circumscribed circle. VII. 1098. Construct a circle, touching a given circle at a given point, and touching a given line. VII. 1099. In a given square inscribe four equal circles, so that each shall be tangent to two of the others, and also tangent to two sides of the square. dllOO. In a given square inscribe four equal circles, so that each shall be tangent to two of the others, and also tangent to one and only one side of the square. VII. dllOl. In a given equilateral triangle inscribe three equal circles tangent each to the other two, each circle being tangent to two sides of the triangle. 1102. Draw a tangent to a given circle such that the sect inter- cepted between the point of contact and a given line shall have a given length. VII. Theoretic 1103. If two chords intersect and make equal angles with the diameter through their point of intersection, they are equal. 1104. The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle. 1105. If two common external tangents or two common internal tangents are drawn to two circles, the sects intercepted between the points of contact are equal. 1106. If two circles are tangent externally, the common internal tangent bisects the two common external tangents. 1107. A line tangent to two equal circles is either parallel to the sect joining their centers or bisects it. 1108. A coin is placed on the table. How many coins of the same denomination can be placed around it, each tangent to it and tangent to two of the others? Prove your answer. 1109. If through any point in the convex arc included between two tangents a third tangent is drawn, a triangle will be formed, the peri- 278 PLANE GEOMETRY meter of which is constant and equal to the sum of the two tangents. 1110. If a triangle is inscribed in a triangle ABC, whose semi- perimeter is s, the sects of its sides from the vertices to the points of contact are equal to s a, s b, and s c. 1111. The perimeter of an inscribed equilateral triangle is equal to half the perimeter of the circumscribed equilateral triangle. 1112. The radius of the circle inscribed in an equilateral tri- angle is equal to one-third of the altitude of the triangle. dl!13. In a circumscribed quadrilateral the sum of two opposite sides is equal to the sum of the other two sides, and a circle can be inscribed in a quadrilateral if the sum of two opposite sides is equal to the sum of the other two sides. dl!14. In what kinds of parallelograms can a circle be inscribed? Prove. 1115. The diameter of the circle inscribed in a right triangle is equal to the difference between the sum of the legs and the hypotenuse. 1116. All chords of a circle which touch an interior concentric circle are equal, and are bisected at the points of contact. Theorem 50. In equal circles central angles have the same ratio as their intercepted arcs. COT. 1. A central angle is measured by its intercepted arc. Theorem 51. Parallels intercept equal arcs on a circle. Theorem 52. An inscribed angle, or one formed by a tangent and a chord is measured by one-half its intercepted arc. Theorem 52a. The mid-point of the hypotenuse of a right triangle is equidistant from the three vertices. What point in the circumscribed circle is the mid-point of the hypotenuse? Theorem 53. An angle whose vertex is inside the circle is mea- sured by half the sum of the arcs intercepted by it and its vertical. Theorem 54. An angle whose vertex is outside the circle is measured by half the difference of its intercepted arcs. Theorem 54a. The opposite angles of a quadrilateral inscribed in a circle are supplementary. (Proof left to the pupil.) THE CIRCLE 279 Cor. 1. A quadrilateral is inscriptible if its opposite angles are supplementary. Suggestion: Show, by the method of exclusion, that the fourth vertex of the quadrilateral lies on the circle passing through three of its vertices. EXERCISES. SET LXXXVI. MEASUREMENT OF ANGLES Numeric 1117. Find the value of an angle which (a) is inscribed, and in- tercepfcs an arc of 160, (6) is inscribed in a segment of 250. 1118. If the tangents from a point to a circle make an angle of 60, what are the values of the arcs they intercept? What if the angle is a right angle? 1119. Find the angle whose sides are tangents drawn from a point whose distance from the center of the circle is the diameter of that circle 1120. An angle between two chords intersecting inside a circle is 35, its intercepted arc is 25 18'; find the arc intercepted by its vertical. 1121. A triangle is inscribed in a circle, and another triangle is circumscribed by drawing tangents at the vertices of the inscribed triangle. The angles of the inscribed triangle are 40, 60, and 80. Find all the other angles of the figure. 1122. The arcs subtended by three consecutive sides of a quad- rilateral are 87, 95, 115; find the angles of the quadrilateral; the angles made by the intersection of the diagonals, and the angles made by the opposite sides of the quadrilateral when produced. 1123. Three consecutive angles of an inscribed quadrilateral are 140 30', 80 30', and 39 30'. Find the numbers of degrees in the arcs subtended by the four sides if the arc intercepted by the largest angle is divided into parts in the ratio of 4 to 5. 1124. Three consecutive angles of a circumscribed quadrilateral are 85, 122, 111. Find the number of degrees in each angle of the inscribed quadrilateral made by joining the points of contact of the sides of the circumscribed quadrilateral. 1125. The points of tangency of a quadrilateral, circumscribed 280 PLANE GEOMETRY about a circle, divide the circumference into arcs, which are to each other as 4, 6, 10, and 16. Find the angles of the quadrilateral. 1126. If the sides AB and BC of an inscribed quadrilateral A BCD subtend arcs of 60 and 130, respectively, and the diagonals form $.AED = 70, find the number of degrees in (a) 15, (b) (c) each angle of the quadrilateral. 0. ^ 1127. In this figure = 41, = 65, and $BCD = 97. Find the number of degrees in each of the other angles, and determine whether or not CD is a diameter. 1128. In this figure P of two circles passes through the direct center of similitude if the radii have the same direction, and through the inverse center if the radii have opposite directions. THE CIRCLE 283 dl!52. Taking any point as a center of similitude of two circles, the two radii of one of them, drawn to its points of intersection with any other line passing through that center of similitude, are parallel, respectively, to the two radii of the other, drawn to its intersections with the same line. Hint: Use an indirect proof depending upon Ex. 1151. dl!53. All secants, drawn through a direct center of similitude P of two circles, cut the circles in points whose distances from P, taken in order, form a proportion. dl!54. If in the last exercise, the line of centers cuts the circles in points A, B, C, D, and any other_secant through P cuts the circle in points M, N, R, S, prove that PNPR is constant and equal to PB-PC. dl!55. The common external tangents to two circles pass through the direct center of similitude, and the common interior tangents pass through the inverse center of similitude. What method of drawing the common tangents to two circles may be derived from this fact? 1156. ABC is an isosceles triangle inscribed in a circle, BD a chord drawn from its vertex cutting the base in any point E. _. 'BD AB Prove === AB BE 1157. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided pro- portionally by the circumference of the smaller. 1158. If two circles touch at M, and through M three lines are drawn meeting one circle in A, B, C, and the other in D, E, F, respectively, the triangles ABC and DEF are similar. Locus 1159. An angle of 60 moves so that both of its sides touch a fixed circle of radius 5 ft. What is the locus of the vertex? 1160. Find the locus of the mid-point of a chord drawn through a given point within a given circle. 1161. Through a point A on a circle chords are drawn. On each one of these chords a point is taken one-third the distance from A to the end of the chord. Find the locus of these points. 284 PLANE GEOMETRY 1162. The locus of the vertex of a triangle, having a given base and a given angle at the vertex, is the arc which forms, with the base, a segment capable of containing the given angle. 1163. Find the locus of the points of contact of tangents drawn from a given point to a given set of concentric circles. 1164. A variable chord passes, when prolonged, through a fixed point outside a given circle. What is the locus of the mid- point of the chord? 1166. Upon a sect AB a segment of a circle containing 240 is constructed, and in the segment any chord CD subtending an arc of 60 is drawn. Find the locus of the intersection of AC and BD, and also of the intersection of AD and BC. 1166. The locus of the centers of circles inscribed in triangles having a given base and a given angle at the vertex is the arc which forms with the base a segment capable of containing a right angle plus half the given angle at the vertex. 1167. The locus of the intersections of the altitudes of triangles having a given base and a given angle at the vertex is the arc forming with the base a segment capable of containing an angle equal to the supplement of the given angle at the vertex. dl!68. Find the locus of a point from which two circles subtend* the same angle. 1169. If A and B are two fixed points on a given circle, and P and Q are the extremities of a variable diameter of the same circle, find the locus of the point of intersection of the lines AP and BQ. dllTO. The lines li and k meet at right angles in a point A . is any fixed point on k> Through draw a line meeting li in B. P is a varying point on this line such that OB'OP is constant. Plot the locus of P as the line swings about as a pivot. Theorem 66. A tangent is the mean proportional between any secant and its external sect, when drawn from the same points to a circle. * If tangents from the same point to two circles form equal angles, the circles are said to subtend equal angles from that point. THE CIRCLE 285 Cor. 1. The product of a sec- ant and its external sect from a fixed point out- side a circle is constant. Hint: Draw tangent PIT. What is the constant in this corollary? State the corollary in another way. Theorem 55a. If chords intersect inside a circle, the product of their sects is constant, Prove by means of similar triangles. Applying the principle of continuity, Theorem 55, Cor. 1, and Theorem 55a can be stated as a single theorem. State them so. Theorem 556. The square of the bi- sector of an angle of a triangle is equal to the product of the sides of this angle, diminished by the product of the sects made by that bisector on the third side. Given: AABC with bisector fo cutting AC at D into sects q and r. Prove: tb z = ac qr. PROOF Circumscribe QO about ABC, and extend BD to E in 00 and draw chord CE. LetDE^s. (1) &BDA c/3 ABCE (1) Why? ^-toTi-l (2> Why? (3) .-.ac=fe 2 +fcs (3) Why? (4) But t b s=qr (4) Why? (5) /. ac = tb*+qr (5) Why? (6) .-. tb*=ac-qr (6) Why? Theorem 55c. In any triangle the product of two sides is equal to the product of the diameter of the circumscribed circle and the altitude on the third side. Hint: Prove AABX ~ AEBC. Consider the special case where 3- B is a right angle and evolve a formula for hb. 286 PLANE GEOMETRY Cor. 1. If R denote the radius of the circle circumscribed about a triangle whose sides are a, b, c, and semiperimeter s, then R = abc 4\A(s a)(s b)(s c)* In the figure for Theorem 55c, ac=hbD. 2 But hb=7\fs(s a)(a 6)(s c) ac a&c s -a)(s -6)(s -c) 4VsO -a)(s -b)(s -c) b EXERCISES. SET LXXXVII METRIC RELATIONS Numeric 1171. A point P is 10 in. from the center of a circle whose radius is 6 in. Find the length of the tangent from P to the circle. 1172. The length of a tangent from P to a circle is 7 in., and the external sect of a secant is 4 in. Find the length of the whole secant. 1173. A point P is 8 in. from the center of a circle whose radius is 4. Any secant is drawn from P, cutting the circle. Find the product of the whole secant and its external sect. 1174. From the same point outside a circle two secants are drawn. If one secant and its external sect are 24 and 15, respec- tively, and the external sect of the other is 7, find that secant. 1175. Two chords intersect within a circle. The sects of one are m and n and one sect of the other is p. Find the remaining sect. 1176. If a tangent and a secant drawn from the same point to a circle measure 6 in. and 18 in., respectively, how long is the ex- ternal sect of the secant? 1177. Two secants are drawn from a common point to a circle. If their external sects are 12 and 9, and the internal sect of the first is 8, what is the length of the second? 1178. The radius of a circle is 13 in. Through a point 5 in. from the center a chord is drawn. What is the product of the two THE CIRCLE 287 sects of the chord? What is the length of the shortest chord that can be drawn through that point? 1179. One sect of a chord through a point 3.5 units from the center of a circle is 2 units in length. If the diameter of the circle is 12 units, what is the length of the other sect of the chord? 1180. The radius of a circle is 2 units. If through a point P, 4 units from the center, secant PQR is drawn, and QR is one unit, what is the length of PQ? 1181. A ABC is inscribed in a circle of radius 5 in. Find the altitude to BC if AB is 4, and AC is 5 in. 1182. The sides of a triangle are 4, 13, and 15, respectively. Find the radius of the circumscribed circle. 1183. In Aa6c, a = 20, 6 = 15, and the projection of b upon c (p bc ) is 9. Find the radius of the circumscribed circle. 1184. In Aa&c, a = 9 and 6=12. Find c if the diameter of the circumscribed circle is 15. 1185. The sides of a triangle are 18, 9, and 21, respectively. Find the angle bisector corresponding to 21. 1186. The sides of a triangle are 21, 14, and 25, respectively. Find the angle bisector corresponding to 25. 1187. The sides of a triangle are 22, 11, and 21, respectively. Find the angle bisector corresponding to 21. 1188. The sides of a triangle are 6, 3, and 7, respectively. Find the angle bisector corresponding to 7. 1189. In a triangle the sides of which are 48, 36, and 50, where do the bisectors of the angles intersect the sides? What are the lengths of the angle bisectors? 1190. In each of the Exercises 1181 to 1189 what kind of triangle is involved? Construction 1191. Construct the mean proportional between two given sects, using in turn the methods suggested by the following propositions : (a) 39 Cor. 1, (6) 39 Cor. 4, (c) 55. 1192.* Construct a square equal in area to that of a given: (a) rectangle; (6) triangle; (c) trapezoid. * For discussion and illustrations of the type of analysis applicable see pp. 318 to 321, and Problems 17, 19, 21, 22, 26, 27, Chapter VII. 288 PLANE GEOMETRY 1193. Draw through a given external point P a secant PAR to a given circle so that AB 2 = PA-PB. VII. dl!94. Draw through one of the points of intersection of two given intersecting circles a common secant of given length. VII. dl!95. From a point outside a circle draw a secant whose exter- nal sect is equal to one-half the secant. Locus dl!96. Given the fixed base of a triangle and the sum of the squares of the other two sides, describe the locus of the vertex. dl!97. Repeat Ex. 1 196, given the difference of the squares of the other two sides. dl!98. Through P any PMN is drawn, cutting a circle K in M and N, and P moves so that the product of the sects PM PN has the constant value k 2 . Find the locus of P. Theoretic 1199. If chord AB bisects chord CD, either sect of chord CD is a mean proportional between the sects of AB. 1200. If two circles intersect, their common chord produced bisects the common tangents. 1201. In the diameter of a circle points A and B are taken equally distant from the center, and joined to a point P in the circumference. Prove that ^AP 2 +BP 2 is constant for all posi- tions of P. 1202. If a tangent is limited by two other parallel tangents to the same circle, the radius of the circle is the mean proportional between its sects. 1203. The tangents to two intersecting circles, drawn from any point in their common chord produced, are equal. d!204. The sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of the legs plus twice the product of the bases. 1205. In an inscribed quadrilateral the product of the diagonals is equal to the sum of the products of the opposite sides. (Ptolemy's Theorem.) d!206. If the opposite sides of an inscribed hexagon intersect, they determine three collinear points. (" Mystic Hexagram," discovered by Pascal when he was 16 years of age.) THE CIRCLE 289 d!207. If a circumference intersects tha sides a, b, c, of a AABC in the points Ai and A 2 , #1 and B 2 , Ci and C2, respectively, then Ad BAi CBi AC Z BAz CB 2 . , L (Camot Theorem 56. A circle may be circumscribed about, and in- scribed within, any regular polygon Cor. 1. An equilateral polygon inscribed in a circle is regular. Cor. 2. An equiangular polygon circum- scribed about a circle is regular. Cor. 3. The area of a regular polygon is A> equal to half the product of its apo- them and perimeter. Suggestion: What is the area of AAOBf ff Theorem 57. If a circle is divided into any number of equal arcs, the chords joining the successive points of division form a regular inscribed polygon; and the tangents drawn at the points of division form a regular circumscribed polygon. Cor. 1. Tangents to a circle at the vertices of a regular inscribed polygon form a regular circumscribed polygon of the same number of sides. Cor. 2. Lines drawn from each vertex of a regular inscribed polygon to the mid-points of the adjacent arcs sub- tended by its sides form a regular inscribed polygon of double the number of sides. Cor. 3. Tangents at the mid-points of the arcs between con- secutive points of contact of the sides of a regular circumscribed polygon, form a regular circumscribed polygon of double the number of sides. Cor. 4. The perimeter of a regular inscribed polygon is less than that of a regular inscribed polygon of double the number of sides; and the perimeter of a regular circumscribed polygon is greater than that of a regular circumscribed polygon of double the number of sides. 19 290 PLANE GEOMETRY Cor. 5. Tangents to a circle at the mid-points of the arcs sub- tended by the sides of a regular inscribed polygon, form a regular circumscribed polygon, of which the sides are parallel to those of the original polygon and the vertices lie on the prolongations of the radii of the one inscribed. Suggestions: Show that AB and AiBi are both perpendicular to OP and are, therefore, parallel. Show .*. that %.A = X t .A 1 , 4# = 4#i, .... What kind of n-gon is then circum- scribed? BP=BT and iPi=i7\ and, therefore, B and BI lie on bisector of X,POT. Radius OBi bisects 4. POT. Theorem 58. A regular polygon the number of whose sides is 3*2" may be inscribed in a circle. Theorem 59. If i n represent the side of a regular inscribed polygon of n sides and i 2n the side of one of 2n sides and r the radius of the circle, i 2n =\2r 2 -rV / 4r 2 -i n 2 . Theorem 60. // i n represent the side of a regular inscribed polygon of n sides, c n that of a regular circumscribed polygon of n 2ri, sides, and r the radius of the circle, c n = Theorem 61. The perimeters of regular polygons of the same number of sides compare as their radii and also as their apothems. Theorem 62. Circumferences have the same ratio as their radii. Cor. 1. The ratio of any circumference to its diameter is constant. Cor. 2. In any circle c=Zirr. Theorem 63. The value of TT is approximately 3.14159. Theorem 64. The area of a circle is equal to one-half the product of its radius and its circumference. Cor. 1. The area of a circle is equal to TT times the square of its radius. THE CIRCLE 291 Cor. 2. The areas of circles compare as the squares of their radii. Cor. 3. The area of a sector is equal to half the product of its radius and its arc. (Proof left to the student.) NOTE. Cor. 3 does not suggest the most convenient method of determining the area of a sector. Suggest a more convenient one. A segment of a circle is a portion of it bounded by an arc and its subtending chord. Similar sectors and similar segments are those of which the arcs contain the same number of degrees. Cor. 4. Similar sectors and similar segments compare as the squares of their radii. Suggestions: How do circles and Oi compare in area? How, then, would like parts of them compare? Why are similar sectors like parts? Why is AAOB AAtOiBJ How, therefore, do the triangles compare in area? Justify the folio wing alge- braic statements which form the basis of proof here: sec. AOB R 2 \ &AOB . sec.AOB . AOB _(R 2 \_ &AOB . sec.AOB ^iOiBi \fV~AArfS5 ' AAOB seg.AOB_seg.AiOiBi seg. AOB _ &AOB ^R* ' AAOB = AAiOitfi r seg. A&Bi " A4i0i#i ~~ r 2 ' EXERCISES. SET LXXXVIII. MENSURATION OF THE CIRCLE Numeric 1208. Find follows : (a) 9 in. (6) 12 in. 1209. Find follows : (a) 15 (6) 7T 2 the (c) (<*) the (c) (d) circumferences of circles 5.9 in. (e) 2| ft. 7.3 in. (/) 3|- in. diameters of circles with 2wr (e) 188.496 in. ?7ra 2 (/) 219.912 in. with diameters as (g) 29 centimeters (h) 47 millimeters circumferences as (g) 3361.512 in. (h) 3173.016 in. 292 PLANE GEOMETRY 1210. Find the diameter of a carriage wheel that makes 264 revolutions in going half a mile. 1211. The diameter of a bicycle wheel is 28 in. How many revolutions does the wheel make in going 10 mi.? 1212. Find the radii of circles with circumferences as follows: (a) 77r (c) 15.708 in. (e) 18.8496 in. (g) 345.576 ft. (6) 3^-TT (d) 21.9912 in. (/) 125.664 in. (h) 3487.176 in. 1213. Find the radius of a circle whose circumference is m units. 1214. An arc of a certain circle is 100 ft. long and subtends an angle of 25 at the center. Compute the radius of the circle correct to one decimal place. 1215. The circumference of a circle is 10. Find the circumfer- ence of one having twice the area of the original. 1216. Find the central angle of a sector whose perimeter is equal to the circumference. 1217. Find the areas of circles with diameters as follows: (a) 16a6 (c) 2.5ft. (e) 3f yd. (g) 3 ft. 2 in. (6) 24vr 2 (d) 7.3 in. (/) 4f yd. (h) 4 ft. 1 in. 1218. Find the area of circles with radii as follows: (a) 5x (c) 27 ft. (e) 3 in. (g) 2 ft. 6 in. (b) 27r (d) 4.8 ft. (/) 4| in. (h) 7 ft. 9 in. 1219. Find the radii of circles with areas as follows : (a) 7ra 2 6 4 (c)ir (e) 12.5664 (g) 78.54 (b) 47rmV 5 (d) 27T (/) 28.2744 (h) 113.0976 1220. Find the areas of circles with circumferences as follows : (a) 27r (c) ira (e) 18.8496 in. (g) 333.0096 in. (b) 47r (d) 147ra 2 (/) 329.868 in. (h) 364.4256 in. 1221. Find the area of a circle whose circumference is C. 1222. Find the area of a sector whose radius is 5 and whose central angle is 40. 1223. Find the area of a fan that opens out into a sector of 120, the radius of which is 9^ in. 1224. The arc of a sector of a circle 2j in. in diameter is If in, What is the area of the sector? 1225. Find the central angle of a sector whose area is equal to the square of the radius. THE CIRCLE 293 1226. Find the circumference of a circle whose area is S. 1227. A circle has an area of 60 sq. in. Find the length of an arc of 40. 1228. Find the radius of a circle equivalent to a square the side of which is 6. 1229. The circumferences of two concentric circles are 30 and 40, respectively. Find the area bounded by the two circumferences by the shortest method you know. 1230. In an iron washer here shown, the dia- meter of the hole is 1|- in., and the width of the washer is -f in. Find the area of one face of the washer. 1231. The area of a fan which opens out into a sector of 111 is 96.866 sq. in. What is the radius? (Use ir = 3.1416. Why?) 1232. The radius of a circle is 10 ft. Two parallel chords are drawn, each equal to the radius. Find that part of the area of the circle lying between the parallel chords. 1233. A square is inscribed in a circle of radius 10. Find the area of the segment cut off by a side of the square. 1234. Find a semicircle equivalent to an equilateral triangle whose side is 5. 1235. A kite is made as shown in the dia- gram, the semicircle having a radius of 9 in., and the triangle a height of 25 in. Find the area of the kite. 1236. Two circles are tangent internally, the ratio of their radii being 2 to 3. Compare their areas, and also the area left in the larger circle with that of each of the circles. 1237. A reservoir constructed for irrigation purposes sends out a stream of water through a pipe 3 ft. in diameter. The pipe is 1000 ft. long. How many times must it be filled if it is to discharge 10,000 acre-feet of water? (An acre-foot of water is the amount required to cover 1 acre to a depth of 1 ft.) 294 PLANE GEOMETRY 1238. Each side of a triangle is 2n centimeters, and about each vertex as center, with radius of n centimeters, a circle is described. Find the area bounded by the three arcs that lie outside the triangle, and the area bounded by the three arcs that lie inside the triangle. 1239. From a point outside a circle whose radius is 10, two tangents are drawn. Find the area bounded by the tangents and the circumference, if they include an angle of 120. Find both results. 1240. Upon each side of a square as a diameter semicircles are described inside the square. If a side of the square is s, find the sum of the areas of the four leaves. A 1241. Find the area bounded by three arcs each of 60 and radius 5 if the convex sides of the arcs are turned toward the area. 1242. Find the area bounded by three arcs each of 60 and radius 5 if the concave sides of the arcs are turned toward the area. dt!243. The flywheel of an engine is connected by a belt with a smaller wheel driving the machinery of a mill. The radius of the flywheel is 7 ft., and of the driving wheel is 21 in. (a) How many revolutions does the smaller wheel make to one of the larger wheel? (6) The distance between the centers is 10 ft. 6 in. What is the length of the belt connecting the two wheels if it is not crossed? (c) If it is crossed? 1244. Given a circle whose radius is 16, find the perimeter and the area of the regular inscribed octagon. 1245. The following is Ceradini's approximate method of constructing a sect equal in length to a circle: Draw diameter AB and tan- gent BK at B. Draw OC making -K . Draw BE. Then BE equals the circle. Deter- mine the accuracy of this construction. 1247. In making a drawing for an arch it is required to mark off on a circle drawn with a radius of 5 in. an arc that shall be 8 in. long. This is best done by finding the angle at the center. How many degrees are there in this angle? 1248. The perimeter of the circumscribed equilateral triangle is double that of the similar inscribed triangle. 1249. Squares are inscribed in two circles of radii 2 in. and 6 in., respectively. Find the ratio of the areas of the squares, and also of the perimeters. 1250. Squares are inscribed in two circles of radii 2 in. and 8 in. respectively, and on their sides equilateral triangles are con- structed. What is the ratio of the areas of these triangles? 1251. A log a foot in diameter is sawed so as to have the cross- section the largest square possible. What is the area of this square? What would be the area of the cross-section of the square beam cut from a log of half this diameter? 1252. If r denotes the radius of a regular inscribed polygon, a its apothem, s a side, A an angle, and C the angle at the center, show that: (a) In a regular inscribed triangle s=rV3, a=|r, A =60, (b) In a regular inscribed quadrilateral s=rA/2, a== 4 ==90, C = 90. (c) In a regular inscribed hexagon s^=r, a=^-rv 3, A = 120, C=60. (d) In a regular inscribed decagon s=^r( V 5 1), a^ \r\10+2V5, A = 144, C = 36. 1253. If r is the radius of a circle, a the apothem of a regular inscribed n-gon, and i n one of its sides, i 2n a side of a regular inscribed PLANE GEOMETRY 2n-gon, c n a side of a regular circumscribed n-gon; A n the area of a regular inscribed n-gon, and A Zn that of a regular inscribed 2n-gon, fill out the accompanying table: Given Required Given Required 1. r, i n a 15. in, n = 3 An 2. r, i n fe. 16. r, n = 3 An 3. r, ijn in 17. in, n=6 An 4. r, in Cn 18. r, n = 6 An 5. r, n = 3 in, a 19. in, n = 12 An 6. r, n = 6 in, a 20. r, n = 20 An 7. r, n = 12 in, a 21. in, n = 12 An 8. r, n = 4 in, a 22. r, n = 12 An 9. r, n=8 in, a 23. in, n = 4 An 10. r, n = 10 in, a 24, r, n = 4 An 11. r, n = 5 in, a 25. in, n = 8 An 12. in, n, a An 26. r, n = 8 An 13. in, n, r An 27. r, n = 5 An 14. in, n, r Ain 28. r, n = 10 An Theoretic 1254. The area of a regular inscribed hexagon is a mean propor- tional between the areas of the inscribed and circumscribed equi- lateral triangles. d!255. An equilateral polygon circumscribed about a circle is regular if the number of its sides is odd. d!256, An equiangular polygon inscribed in a circle is regular if the number of its sides is odd. 1257. If C be a point in the straight line AB, the three semi- circles, drawn respectively upon sects AB, AC, and CB as dia- meters, bound an area equal to a circle of which the diameter is the perpendicular CD, D being in the largest semicircle. 1258. If upon three sides of a right tri- angle semicircles be drawn as indicated in the diagram, the area of the right triangle is equal to the sum of the two crescent- shaped areas, bounded by the semi- circles. (Hippocrates' Theorem.) A 1259. Give a simpler proof for Ex. 859. (b) Generalize this fact. CHAPTER VI METHODS OF PROOF There are two general methods of proving theorems, the direct or synthetic, and the indirect method. Each of these methods of proof may be in its nature geometric or algebraic. Further, indirect proofs, whether geometric or algebraic, may take different forms. Thus a theorem may be proved indirectly, either by means of exclusion or reduction to an absurdity, or by analysis. It is the object of this chapter to give an illustration of each of these methods of proof, together with several exercises that will be most naturally proved by that method. A. THE DIRECT OR SYNTHETIC METHOD In this method we employ either superposition or start with the data, and combining them with known truths proceed step by step until we arrive at the desired conclusion. I. GEOMETRIC PROOF.* B / Illustration: The bisector of the vertex angle of an isosceles triangle bisects the base. Given: AABC, AB^BC, X in AC so that Prove: AX=XC. In &ABX and ACBX. (1) AB=BC, 2S^ABX = (2) BX=BX. (3) .'. AABX^ACBX (A\ AY r'Y \*) ' <*" ^>- A PROOF (1) Data. (3) Two sides and the included angle determine a triangle. (4) Horn, sides of cong. A are equal. * The method of superposition should be very rarely used. For illustra- tions of it, recall the proofs of the fundamental theorems in congruence of triangles. 297 298 PLANE GEOMETRY EXERCISES. SET LXXXIX. SYNTHETIC METHODS OF PROOF Give a synthetic proof of each of the following: 1260. The bisector of the vertex angle of an isosceles triangle is perpendicular to the base. 1261. If the perpendicular bisector of the base of a triangle passes through the vertex, the triangle is isosceles. 1262. Any point in the bisector of the vertex angle of an isosceles triangle is equidistant from the ends of the base. II. ALGEBRAIC PROOF. Illustration: If one leg of an isosceles triangle is extended through the vertex by its own length, the sect joining its end to the end of the base is perpendicular to the base. Given: ABD a st. line, AB =BD =BC. Prove: DCJ.AC. PROOF (1) The sum of the 4 of a A is a st. 4. (2) Data. (3) Base 2^ of anisosceles Aareequal . (4) Quantities may be substituted for their equals in an equation. (5) The whole equals the sum of all its parts. (6) See (4). (7) Quotients of equals divided by equals are equal. (8) The numeric measure of a rt. 2$. is 90. (9) By def. of J_. (1) (2) (3) =BD. 4.DCB. (4) .'. substituting (3) in (1) (5) But (6) (7) . (8) .*. %-DCA is a rt. (9) .*. DCA.AC. This method is especially adapted to the proof of numerical relations between angles or sects. The following procedure is generally used in applying the alge- braic synthetic type of proof. 1. Observation of the numeric relations that immediately follow from the data. 2. Statement of these relations in algebraic form the equality or the inequality. 3. Reduction of these algebraic relations by the help of axioms until the desired conclusion is reached. METHODS OF PROOF 299 EXERCISES. SET LXXXIX (concluded) Give an algebraic synthetic proof of each of the following : 1263. The bisectors of two supplementary adjacent angles are perpendicular to each other. 1264. If the bisectors of two adjacent angles are perpendicular to each other, those angles are supplementary. 1265. If two sides of a triangle are unequal the angles opposite them are unequal in the same order. 1266. The sum of the altitudes of a triangle is less than its perimeter. 1267. The angle whose sides are the altitude from and the bisector of an angle of a triangle is equal to one-half the difference between the remaining angles of the triangle. B. THE INDIRECT METHODS I. GEOMETRIC; or II. ALGEBRAIC. a. By the Method of Exclusion. Two magnitudes of the same kind may bear one of three rela- tions to each other. The first may be less than, equal to, or greater than the second. If it can be shown that two of these relations are false, the third is of necessity true. Similarly the position of a point may be fixed by the method of exclusion. Illustration 1, Theorem 21c: If two angles of a triangle are unequal, the sides opposite them are B unequal in the same order. Given: &ABC in which %.A > %.C. Prove: a>c. PROOF o>c, a=c, or ac. Authorities left to the student. 300 PLANE GEOMETRY Illustration 2, Theorem 54a, Cor. 1: A quadrilateral whose opposite angles are supplementary is inscriptible. - Given: Quadrilateral ABCD in which 2LA Prove: A, B, C, D coney clic. PROOF (1) A circle may be passed through A, B, C. (2) D lies outside, inside, or on this circle. (3) Suppose D lies inside QABC in the position Di. Then ^.B^AXZ and (4) Authorities left to the student. (5) .*. D cannot lie inside the circle. (6) Suppose D lies outside QABC in the position I)j. Then %.B -L^/Pfr and ^D^H (CfiT-ZP). (7) /. 2$.B + 2H.D>*-M(APC+tfTA - ZP) = H(360-ZP)<180. (8) .*. D cannot lie outside the circle. (9) .*. D lies on the circle and ABCD is in- scriptible. EXERCISES. SET XC. PROOF BY THE METHOD OF EXCLUSION Prove the following facts by means of the method of exclusion : 1268. Knowing (1) that equal chords subtend equal arcs, and (2) that unequal chords subtend arcs unequal in the same order, prove the converse of each of these facts by the method of exclusion. 1269. In a fashion similar to that used to prove Ex. 1268, show (1) when a = b, c=d; that if (2) when a>b, c>d; then the converse of each of these (3) when a BC and take AX=BC. 1271. If upon a common base an isosceles u and a scalene triangle are constructed, the line joining their vertices does not bisect the vertex angle of the isosceles triangle. Hint: Assume that it does bisect it. 1272. If perpendiculars are drawn to the sides of an acute angle from a point inside the angle, they enclose an oblique angle. 1273. Prove that if two triangles resting on a common base have a second pair of sides equal, and the third vertex of each outside the other triangle, their third sides are unequal. c. By the Method of Analysis. The method of analysis, which is attributed to Plato, was undoubtedly used by Euclid, but was probably emphasized by the former. The analysis of a theorem is a course of reasoning, whether con- scious or unconscious, by means of which a proof is discovered. 302 PLANE GEOMETRY It consists of discovering the immediate condition under which the conclusion would be true, and continuing to do this with each new condition until one known to be true is reached. Analysis always takes the following form: Suppose we wish to prove that A=B if C=D. We start by saying: A=B if x=y. But x= y if ra=n. and m= n if C=D. But C=D by data. The proofs of theorems are usually put in the synthetic form, but they are derived analytically and then rearranged by retracing the steps taken. The analytic method might also be called the method of reduc- tionj or of successive substitutions. If we wish to discover how to prove a theorem, we should always use the analytic method. It is much more likely to suggest those helpful auxiliary lines which are frequently needed before any relation between the theorem to be proved and a known theorem is apparent. On the contrary, nothing in the synthetic method suggests the use of suitable auxiliary lines, and though we may continue to make deductions as they occur to us, we often waste time and energy without getting any nearer the conclusion. Illustration 1, Theorem 216: If two sides of a triangle are unequal, the angles opposite them are unequal in the same order. Given: a>c in AABC. Prove: Analysis : 2$. A > %.C if part of 4. A > 4<7. Part of 4. A > 4C if that part can be made an exterior 4 of a A in which 4C is a non-adjacent interior 4. This cannot be done as the 4 s are now placed. How, then, can we find an 4 which is equal to part of %-A and placed as desired? Since c< a, an isos. A can be formed by laying off BX=c on a. Hence the auxiliary line AX is suggested, and %-A > 4(7 if ^.BAX> 4.C. But X r BXA>X r C. :. We may reverse the steps and give a brief synthetic proof if desired. METHODS OF PROOF 303 The analytic method does not always lead at once to the shortest proof of a theorem, though it is far more likely to do so than the synthetic method. At times an analysis suggests several methods of proof, and our selection will depend upon which seems the shortest. Skill in selection can be acquired only by practice. Illustration 2: If one median in a triangle is intersected by a second, the sect between the vertex and the point of intersection is double the other sect. Given: Medians CD and AE intersecting at in A ABC. Prove: AO=20E, and CO=20D. Analysis: One sect may be proved double another by proving (1) half the longer =the shorter, or (2) double the shorter . _ =the longer. A The first method suggests that we take F and G in AO and CO so that _ Now AO =20E and CO =20D if FO =OE and GO =OD. FO =OE and GO =OD if GD and FE are diagonals of a CJ. FE and GD are diagonals of^n^DE^FG and DE \\ FG. DE = FG and DE \\ FG if DE bisects AB and BC in AABC, and if FG bisects AO and CO in A AOC, for then DE= \(AC) = FG and DE \\ AC\\FG. .' . a synthetic proof can now readily be given. Success in this type of work very often depends on the selection of suitable auxiliary lines. Those which are most often of use are discovered by (a) joining two points, (6) drawing a line parallel to a given line, (c) drawing a line perpendicular to a given line, (d) bisecting given sects as in the last illustration, (e) producing a sect by its own length, as might have been done in the last illustration. Since we have not proved, even if it be true, that the algebraic processes employed in the proof of theorems in proportion are reversible, only a synthetic proof is valid. Of course, to suggest the synthetic line of argument it is desirable to give an analysis first if needed. 304 PLANE GEOMETRY Illustration: c d' Prove: ?- = Given: -, b ANALYSIS 2 _0 2 +C 2 (D Operation perfor med. fo2 )2_J_^2 if a 2 6 2 +a 2 d 2 ; 3a 2 6 2 +6 2 c 2 (2) (1) -6 2 (6 2 +d 2 ) (2) is true if a 2 d 2 = 6 2 c 2 (3) (2) -a 2 6 2 (3) is true if ad=bc (4) + V(3) (4) is true . f a c *i=d (5) (4) +bd But |=| by data. .' . the following synthetic proof a_c (D (1) Given. (D -M, :.ad=bc (2) (2) Products of equals multiplied by equals are equal (2) ', /. a 2 d 2 =6 2 c 2 (3) (3) Like powers of equals are equal. (3) +a 2 6, /. a 2 d 2 +a 2 52 = (4) Sums of equals added to equals 6 2 c 2 +a 2 6 2 (4) are equal. fA\ K2/75 +^1),... g2 = Q 2 2 + c2 (5) (5) Quotients of equals divided by (4; -rO f (& equals are equal. EXERCISES. SET XCII. ANALYTIC METHOD OF PROOF Give an analysis of each of the following exercises, and follow it by a concise synthetic proof. 1274. Prove the second illustra- tion under the analytic method by (a) proving ADOE^AFOG or ADOF^&GOE (p. 303). (6) Doubling DO and OE. Use each of the three methods sug- gested by the following figures. Fia. 2 Fio. 3 METHODS OF PROOF 305 1275. Prove Theorem 31, Cor. 4, analytically. 1276. Prove Theorem, 31 Cor. 5, analytically. 1277. If one acute angle of a right triangle is double the other, the shorter leg is one-half the hypotenuse. Prove first by drawing auxiliary lines outside the triangle, and then by drawing them inside the triangle. 1278. A median of a triangle is less than half the sum of the adjacent sides. 1279. Prove the theorem given under A II analytically (p. 298) . 1280. Prove the theorem given under B I a analytically (p. 299). 1281. Prove Ex. 1270 analytically. 1282. If t i * prove that r= -j. o a 1284. If J 4 prove that 19RR Tf a c fh a 2 _ 1285. If r = j, then - 1 r^ = > -J 6 d a 3 -6 3 c 3 -d a c While practice alone can give skill in the proof of theorems, the following suggestions may be of help to the student. First. Make as general a figure as possible. If a fact is to be proved concerning triangles in general the figure should be that of a scalene triangle, since many facts are true of isosceles or equilateral that are not true of scalene triangles. Again, if a fact is to be proved concerning quadrilaterals in general, it might be misleading to draw a parallelogram or even a trapezoid. Second. Always have clearly in mind what is given and what is to be proved. Third. If the proof is not readily seen, resort to analysis. Fourth. Give a proof by the method of reduction to an absurdity only as a last resort. 20 CHAPTER VII CONSTRUCTIONS. METHODS OF ATTACKING PROBLEMS The solutions of the following fourteen problems and the corol- laries to them are typical of one class of solution of construction problems, namely, those solutions which are at once found to rest directly upon some known theorem, and in addition, at times, upon some known construction. Problem 1.* Draw a perpendicular to a given line from a given point (a) outside the line, (b) on the line.] The constructions rest directly upon Theorem 40, Cor. 1, the fact that two points equidistant from the ends of a sect determine the perpendicular bisector of that sect. Problem 2. Bisect a given (a) sect, (b) angle, (c) arc. The construction of (a) rests immediately upon Problem 1. The construction of (6) rests immediately upon Theorem 5, the fact that three sides determine a triangle. The construction of (c) rests immediately upon Theorem 43, the fact that equal central angles intercept equal arcs, and Problem 2 (6). Problem 3. Reproduce a given angle. The construction rests directly upon Theorem 5. Problem 4. Draw a line through a given point, and parallel to a given line. The construction may rest directly upon Theorem 11, the fact that if when lines are cut by a transversal the alternate-interior angles are equal, the lines thus cut are parallel, and Problem 3. * The first thirteen problems of the syllabus were taken up as exercises in the First Study, but are repeated here (with only suggestions for their con- struction) as an integral part of a syllabus of constructions. t According to Proclus, this problem was first investigated by (Enopides, a Greek philosopher and mathematician of the 5th century B.C. Proclus speaks of such a line as a "gnomon" the common name for the vertical piece on a sundial. 306 METHODS OF ATTACKING PROBLEMS 307 Problem 5. Construct a triangle, given any three independent parts; (a) two angles and the included side, (b) two sides and the included angle, (c) three sides, (d) the hypotenuse and a leg of a right triangle. (What can you say in case (c) if one side is equal to or greater than the sum of the other two sides?) Problem 6. Divide a sect into n equal parts. The construction may rest upon Theorem 31, Cor. 3, the fact that parallels which intercept equal sects on one transversal do so on all transversals, and Problem 3. For further help see Ex. 429, p. 116. Problem 7. Find a common measure of two commensurable sects. Given: ~KB and CD commensur- 4r -^ able sects. ~ Required : A common measure ra. C - - - D Solution: Lay off A 7= CD on AB. Lay off CZ=2YB on CD. Layoff YX='ZD on YB. Lay off XB on ZD. It is found to be contained exactly in ZD. Then XB is the greatest common measure of AB and CD, and any integral part of XB is a common measure of them. Prove it. Problem 8. Pass a circle through three non-collinear points. The construction rests directly upon Theorem 416. Cor. 1. Circumscribe a circle about a triangle.* Problem 9. Divide a given sect into parts proportional to n given sects. The construction rests directly upon Theorem 31, Cor. 2. Problem 10. Divide a given sect harmonically in the ratio of two given sects. Use (1) the method suggested in Problem 9, or (2) the method suggested by Theorem 326, Cor. 1. * The center of a circle circumscribed about a polygon is called its circum- center. 308 PLANE GEOMETRY Problem 11. Find a fourth proportional to three given sects. The construction rests directly upon Theorem 31. The solution is given, but the proof is left to the pupil. a e Given: Sects a, b, c. Required: Sect x, so that ~== & 6 d' Solution: Construct any angle RST. x- - N On ST, lay off SV=a. ^ \ On SR, lay off SW m b. /" \ On FT, lay off VQ=c. ;\ \ DrawFF. ,' \ \ Draw QY \\ VW and cutting SR ' Tf-V x Vfe at Then WY is the required sect x. Cor. 1. Find a third proportional to two given sects. What is the only modification needed in the construction of Problem 11 in order to find x, so that 4- = ? b x Problem 12. Upon a given sect as homologous to a designated side of a given polygon construct another similar to the original polygon. The construction rests directly upon Theorem 366 and Problem 11. Problem 13. Construct a square equal to the sum of two or more given squares. Use the Pythagorean Theorem. Cor. 1. Construct a square equal to the difference of two given squares. In this case the larger of the given squares will be the square on which side of the right triangle? The construction therefore rests upon Problem 5 (d) Cor. 2. Construct a polygon similar to two given similar poly- gons and equal to (a) their sum, (b) their difference. How do the areas of similar polygons compare? Problem 14. Inscribe in a circle, regular polygons the number of whose sides is (a) 3 *2 n , (b) 4 -2". (a) The construction rests directly upon Theorem 58 and the construction of an equilateral triangle given its side (here the radius of the circle) in order to obtain a central angle of 120. (b) What is the central angle in the case of the square? METHODS OF ATTACKING PROBLEMS 309 THE SYNTHETIC METHOD OF ATTACKING A PROBLEM The preceding type of construction problem is the simplest, and the solution of one of that nature is usually so readily seen, that without further explanation (except for one or two suggestions appended to the first exercises) the pupil is asked to solve the fol- lowing set of exercises. EXERCISES. SET XCIII. SYNTHETIC SOLUTIONS 1286. Trisect a right angle. (We know that each angle of an equilateral triangle is two-thirds of a right angle. What construc- tion does this therefore suggest?) 1287. Divide an equilateral triangle into three congruent tri- angles. (We know that the bisectors of the angles of a triangle are concurrent, and that triangles are determined by two angles and the included side.) 1288. Construct an equilateral triangle with a given sect as altitude. (What fact about the altitude of an equilateral triangle suggests the construction?) 1289. Construct a square having given its diagonal. 1290. Through two given points draw straight lines which shall make an equilateral triangle with a given straight line. 1291. On a given sect construct a rhombus having each of one pair of opposite angles double each of the other pair. 1292. On a given base construct a rectangle equal to (a) a given square, (b) another given rectangle, (c) a given triangle, (d) a given trapezoid. 1293. The sides of a polygon are 5, 7, 9, 11, 13. Construct one similar to it having the ratio of similitude 3 to 5. 1294 Construct a polygon similar to the accompanying polygon, having the ratio of simili- tude equal to that of the two given sects a and b. 1295. Construct a poly- gon similar to two (or more) given similar polygons and equivalent to their sum (or difference). 310 PLANE GEOMETRY / \ 1296. Construct a circle equal to the sum of two given circles. 1297. Construct a circle equal to <^ the difference of two given circles. Problem 15. Construct a triangle given two sides and the angle opposite one of them. Given: Sides a and c] 2^. A. Required: AABC. Construction: Construct 2^.PAQ= %.A. OnAQ, layoff AB=c L With B as center and a as radius, strike an arc cutting AP in C and Ci, touching it in only C , or not at all, according asa>hb,a=hb, or afo, the perpendicular from B to A P. II. Conditions under which there is but one solution. (1) 2$_A acute, a=/i& or c. acute, right, or obtuse, and a>c. A / x ^ A III. Conditions under which there is no solution. , (1) 2$. A right or obtuse | \f and ah b but c. (3) a>c. (4) CIEEEC. (5) a>c. or, I. a c. (Fill in all the necessary subheads.) Further illustration of what is meant by the discussion of a problem in construction: To find a point X which shall be equidistant from points Pi and P 2 and at a given distance d from P 3 . Given: Points PI, P 2 , P 3 ; sect d. d Required: Point X such that PiX = Construction: (1) Draw PiP 2 and Q v *- bisectitatA. ^ ffl ^~j^X$l . (2) Erect QARl.PJ^. / \ (3) With P 3 as center and d \ p I \ as radius describe a circle cut- a / }jp> 1 ting QR at X, -X"i. Then X, Xi are the required points. PROOF (1) and V X, (2) v X and Xi are on Pi, (1) The locus of points equidistant from the ends of a sect is the perpen- dicular bisector of that sect. (2) Const., and the locus of points at a given distance from a given point is a circle of which the center is the given point, and the radius the given distance. Discussion: Draw PzN A.QR- I. Two solutions: If d>P 3 N, for then a portion of QR will be a chord of the circle P 3 . II. One solution: If d = P 3 N, for then QR will be tangent to OP 3 at X. III. No solution: If d

n I \ Required: Sects a and b so that (1) - =-~ and (2) . -a 6 s+b These suggest f a tangent as the i# __ _. mean proper- ~j tional between a secant and its external sect or the leg of a rt. A as the mean proportional be- tween the hypotenuse and its projection upon the hypotenuse. As the propor- tions now stand there is but one known term. But by composition we get (1) lib?= - and (2) ^^= J. S Q S b This suggests s as the tangent to a circle, +a as the entire secant, and a as its external sect on the one hand, and b as the entire secant and bs as its external sect on the other hand. The most natural secant to select is that which passes through the center. Hence the following: Construction: (1) Const. O# of diameter s tangent to XY. (XY=s) at Y. (2) Draw secant XQOP. (3) On XY lay off XQ^XQ, and on YX produced lay off XP^XP. (4) Then Qi and PI are the required points. (Proof left to the student.) Problem 22. Inscribe in a circle a regular decagon. Given: 0. Required: AB, the side of the regular in- scribed decagon. Analysis: If AB=side of reg. decagon, I OfJ _ = 72 and if PB bisects 4. ABO, %.PBO =36 = 40 and 2piP5 = 180 -(36 +72)=72 . /. AAPB> AABO and AOPB is PB=OP. AP AB . /.to find OP=AB divide CM, the radius of QO, into C/P C/^4. extreme and mean ratio, and use the mean as AB. (Proof left to student.) * A painting is said to be most artistically arranged when its center of inter- est is so placed that it divides the width of the picture into extreme and mean ratio. If the student is especially interested in the topic, he will find references in Chapter X. METHODS OF ATTACKING PROBLEMS 317 Cor. 1. Inscribe in a circle regular polygons the number of whose sides is (1) 5.2", (2) 15.2". (1) Left wholly to the student. * (2) Hint: ^i- T V To transform a polygon means to change its shape but not its C* area. Problem 23. Transform a polygon into a triangle. Given: Polygon ABCDEF. Required: A A= ABCDEF. Hint: With CA as base, what is the locus of the vertices of triangles whose area = area AABCt To eliminate side A B, by what particular A shall we then replace AABCt Similarly, what A is to replace ADEFf Similarly, how dispose of side DH? Problem 24. Construct the square equal to a given (a) parallelogram, (b) triangle, (c) polygon. Problem 25. Construct a parallelogram equal to a given square having given (a) the sum of base and altitude equal to a given sect, (b) the difference of base and altitude equal to a given sect. Given: Square S; sect a. Required: Ad =S / ^i \ with (a) base + s / alt. =a, (b) base \ i j * *-v , alt. =a. j (a) From Fig. 1 show that any parallelo- gram with QR as base and PQ as altitude = square S, and that it also f ulfills the second condition. (6) From Fig. 2 show that any parallelo- gram with QR as base and PQ as altitude = square s, and that it also fulfills the second condition. This problem solves geometrically the algebraic problems, given (a) x+y = a and xy=s, (6) xy=a and xy = s, find x and y. FIG. / Q FIQ. 2 318 PLANE GEOMETRY Problem 26. Construct a polygon similar to one and equal to another of two given polygons. Given: Polygons P and Q. .---'\ >/^^v Required: Polygon R so p \\ R \ \ Q \ Analysis: IfjRcopandz \ N \ \ homologous to s, then V X \ T-\ n If P and Q equal ra 2 and n 2 respectively, and Q =R, then = . P m' , P s 2 n * *' V Q^ and ]T:F 2 " s , /. x is the fourth proportional to the side of a square =P, the side of a square =Q, and the side in P homologous to x. (Construction and proof left to the student.) Problem 27. Construct a square which shall have a given ratio to a given square. D ^ Given: Square S of side s, ratio - E Required: Square R of side r so that N v H- o q \ Analysis: If a rectangle were to be found \ ,. A rect. AE so that - square square S would simply have to be divided so that alt - of rect - = P. s q Then it remains to convert rect. AE into a square. (Construction and proof left to the student.) Cor. 1.* Construct a polygon equal to any part of a given polygon and similar to it. (Construction left to the student.) FORMAL ANALYSIS OF A PROBLEM When problems .(as is usually the case) are of such nature that the application of known theorems and problems to their solution is not at once apparent, the only way to attack them is by a method * This corollary is included in the syllabus because of its practical value in drafting, since it is by this method that the draftsman finds his scale, in enlarg- ing or reducing a diagram of any sort. METHODS OF ATTACKING PROBLEMS 319 resembling the analytic method of proof of a theorem. This method is called the analysis of the problem. In brief, the directions for following this method are : 1. Suppose the construction completed. 2. Draw a figure showing all the parts concerned. (Given parts in heavy lines, and required parts in dotted lines.) 3. Study the relations of the parts, and try to find some relation which will suggest a possible construction. 4. If a first attempt fails, introduce new relations by means of auxiliary lines, and continue the study of relations until a clue to the correct construction is derived. 5. Look for that clue in a rigid part of the figure usually a triangle. A few more illustrations to show just what is meant by these directions will be helpful. 1. Construct a triangle, having given the base, a base angle, and the sum of the remaining sides. Suppose ABC to be the completed triangle. In it we know AC=b, ^A, and c+a. .'.produce AB to X so that BX = a. Then ABCX is an isos. A. But AAXC is determined (6, But ^XCB= $X V BX=BC. .*. AABC is determined. 2. Construct a triangle, having given one side, the median to that side, and the altitude to a second side. Given: b, mb, h c . Analysis: Suppose ABC to be the required A. Then the known parts are AC . _ 5 _ B _ :. AE =EC =*!. CD =h. BE = h D A 320 PLANE GEOMETRY 8. Construct a triangle having given its medians. _ Given: ma, ffib, Me. J? Required: AABC. ,' / \ Analysis: Suppose .m ABC to be the required A. Evidently no 3. within the A is fixed by the medians. But it is known that OC = %m e , OEa and .. / that OE bisects AC. .'. if OE is produced to X, so that XE=EO, a parallelogram is deter- mined whose sides and one diagonal are known. Hence the following: Construction: (1) Trisect m a , mb, m e . (2) Construct ACXO with %m c , %m a , and 2 / 3 mb as ita sides. (3) Complete the EJOCXA. (4) Produce EO by %mi>, and vertex B is determined. 4. Construct a trapezoid given the bases and the base angles. Analysis: Suppose ABCD to be the com- pleted trapezoid. We know AB, DC, 2$.D, and %.C. Is, then, any part of the figure de- termined? If we draw BD l || AD, then ABD& will be a parallelogram and DDi =AB. %.BDiC = 2$.ADC and D^C^DC-AB and ABD,C is thus determined. (Construction, proof and discussion are left to the student.) The preceding exercises are illustrative of analysis geometric in form, but the analysis of a problem in construction may also be algebraic in form, as, for example, in Problems 21, 26, and 27. NOTE 1. These diagrams show the symbolism used in reference to tri- angles and trapezoids throughout this text. -4 A A METHODS OF ATTACKING PROBLEMS 321 NOTE 2. Construction of trapezoids. The relations given by the following constructions are often useful. Let ABCD be any trapezoid. Draw CF and BH \\ AD, CG and AH || DB. Join H (the in- tersection of AH and BH) with F, C, G. The figures ACGH, ADBH, FCBH, are parallelograms. BF=a W - B, 180- 1318. a, h at m a . 1321. 6, c, h* 1324. ft., ho, EXERCISES. SET XCV. PROBLEMS CALLING FOR ANALYSIS Construct a triangle, having given : 1316. a, b, m b . 1317. a, b, h b . 1319. a, m a , <#. 1320. m a , h a , 1322. a, ft a , h c . 1323. ft., 1326. a, 6, %A + 3:B. 1326. The base, the altitude, and an angle at the base. 1327. The base, the altitude, and the angle at the vertex. Construct an isosceles triangle, having given: 1328. The base and the angle at the vertex. 1329. The base and the radius of the circumscribed circle. 1330. The base and the radius of the inscribed circle. 1331. The perimeter and the altitude. Let ABC be the required A, E ~A D B J EF the given perimeter. The alti- tude CD passes through the mid- point of EF, and the As EAC, BFC are isosceles. 1332. Construct a triangle, having given two angles and the sum of A two sides. Can the third < be found? Assume the problem solved. If AX = AB+BC, what kind of triangle is ABXCt What does equal? Is <&X known? How can C be fixed? B 322 PLANE GEOMETRY 1333. Construct a triangle, having given a side, an adjacent angle, and the difference of the other sides. If AB, <&A, and AC -BC are known, /'\ what points are determined? Then can /'' \ XB be drawn? What kind of triangle is I AXBCt How can C be located? I 1334. Construct an isosceles triangle, J having given the sum of the base and an arm, and a base angle. 1335. Construct a triangle, having given the base, the sum of the other two sides, and the angle included by them. 1336. Construct a triangle given the mid-points of its sides. 1337. Draw between two sides of a triangle a sect parallel to the third side, and equal to a given sect. If PQ = d, what does AR equal? How will you reverse the reasoning? A R B 1338. Draw through a given point P between the sides of an angle A OB a sect terminated by the sides of the angle and bisected at P. If PM = PN, and PR \\ AO, what can you say about OR and RNt Can you now reverse this? Similarly, if PQ \\BO, isOQ = QM? ^ 1339. Given two perpendiculars, AB and CD, intersecting in 0, and a line intersecting these perpendiculars in E and F; construct a square, one of whose angles shall coincide with one of the right angles at 0, and the vertex of the opposite angle of the square shall lie in EF. (Two solutions.) 1340. Draw from a given point P in the base AB of a triangle ABC B, line to AC produced, so that it may be bisected by BC. Construct a rectangle, having given: 1341. One side and the diagonal. 1342. One side and the angle formed by the diagonals. 1343. The perimeter and the diagonal. METHODS OF ATTACKING PROBLEMS 323 Construct a parallelogram, having given: 1344. Two independent sides and one altitude 1345. Two independent sides and an angle. 1346. One side and the two diagonals. 1347. One side, one angle, and one diagonal. 1348. The diagonals and the angle formed by the diagonals Construct a rhombus, having given: 1349. The two diagonals. 1350. The perimeter and one diagonal. 1351. One angle and a diagonal. 1352. The altitude and the base. 1353. The altitude and one angle. 1354. Construct a square, having given the diagonal. Construct a trapezoid, having given : 1355. The four sides. 1356. The bases, another side, and one base angle. 1357. The bases and the diagonals. 1358. One base, the diagonals, and the angle formed by the diagonals. 1359. Inscribe a square in a given triangle. 1360. In a given triangle inscribe a rectangle similar to a given rectangle. Hint: Let ABC be the given triangle and R the given rectangle. On the altitude CH construct a A F ,B L rectangle CL similar to the given rectangle. The line AK will determine a point E which will be one of the vertices of the required rectangle. Why? 1361 Inscribe a square in a semicircle. 1362. Divide a given triangle into two equal parts by a line parallel to one of the sides. 1363. Bisect a triangle by a line parallel to a given line. 1364. Bisect a triangle by a line drawn perpendicular to the base. Construct a circle which shall be tangent to a given : 1365. Line, and to a given circle at a given point. 1366. Circle, and to a given line at a given point. 1367. Transform a triangle ABC so that AD. 1410. If the bisector of the right angle A of a right isosceles triangle BAC cuts BC in D and is produced to K so that DK=AD, then KC 1412. If the side BC of the square ABCD is produced to P and P is joined with A, prove that $MSR', also that 1416. If in parallelogram ABCD, side BOAB, and the diago- nals intersect in P, prove that %BPA. TRIANGLES SIMILAR. Show that they have (a) a center of similitude, (6) two angles respectively equal, SUMMARIES AND APPLICATIONS 345 (c) two pairs of sides proportional and the included angles equal, or, that (d) their sides bear a constant ratio, (e) they are triangles formed by homologous diagonals of similar polygons, (/) they are triangles formed by the altitude upon the hypo- tenuse of a right triangle. EXERCISES. SET GUI. SIMILARITY OF TRIANGLES 1417. If through the point of contact of two tangent circles three secants are drawn, cutting the circumferences in A, B, C, and A i, BI, Ci, respectively, then triangles ABC and AiBiCi are similar. 1418. If from the point P outside a circle two secants are drawn to meet the circumference in B and C, and D and E, respectively, the triangles PBE and PCD are similar. 1419. If the bisector AD of 5 it gives a very close approximation to the exact value of the side required. Inscribe a hexagon, a heptagon, an octagon, a nonagon, and a decagon. E SUMMARIES AND APPLICATIONS 355 1533. Draw a line meeting the sides CA, CB, of AABC in D, E, respectively, so that: (a) DE || AB, BE=ED. N.B. Bisect (6) DE || AB, DE^AD+BE. (c) DE=CD^BE. (See figure.) Analysis: Suppose the problem solved, and draw BD. The A BDE and DCE are isosceles, whence %.DBE = ^BDE=^DEC = ^DCE, which is known. This deter- mines the point D, and E is easily found, since DE^DC. Examine this problem for the special cases when