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 V 
 
ELEMENTARY 
 
 G E O M E T RY 
 
 I 
 
 NRW EDITION. 
 
 INCLUDING PLANE, SOLID, AND SPHERICAL GEOMETRY, WITH 
 PRACTICAL EXERCISES. 
 
 BY 
 
 EDWARD OLNEY, 
 
 PROFESSOR OF MATHEMATICS IN TH^ UNIVERSITY OF MICHIGAN. 
 
 SHELDON AND COMPANY, 
 
 NEW YORK AND CHICAGO. 
 
 1883. 
 
OLNEY'S NEW SERIES 
 
 EMBRACES THE FOLLOWING BOOKS. 
 
 FIRST LESSONS IN ARITHMETIC, 
 PRACTICAL ARITHMETIC. 
 
 Two-Book Series. 
 
 This Series has more examples, and at less price, than any ever 
 published. _^«,,..,^ 
 
 (^ strictly High School Text-Book) 
 Send for full Circular of Olney's Abithmeticd. 
 
 FIRST PRINCIPLES OF ALGEBRA. 
 
 COMPLETE ALGEBRA. (Newly electrotyped in large type.) 
 NEW ELEMENTARY GEOMETRY. 
 
 Copyright, 1883, by Sheldon &= Co. 
 
 Smith & McDougal, Ei-kctbotvpekh. 
 82 Buekmiii) St., N. Y. 
 
^i^^^l^ 'l^ J ^1^ 'i^ '4^ 
 
 
 THE first edition of Olney's Special or Elementary 
 Geometry was issued nearly twelve years ago. It con- 
 tained many new features. The book has gone into use in every 
 State in the Union, and has been tested by practical teachers in 
 all grades of schools. This long and varied test has been watched 
 with care by the author, and it is with the greatest pleasure that 
 he has found that the general features of the book have been 
 well-nigh universally approved. 
 
 To make the book still more acceptable to the teachers and 
 schools of our country, and to keep it abreast with the real 
 advancement in science and methods of teaching, as well as to 
 make it a worthy exponent of the best style of the printer's art, 
 are some of the reasons which have led. to the preparation of this 
 edition. 
 
 1. The division into Chapters and Sections, instead of Books, 
 has been retained, as affording better means of classifying the 
 subject-matter, and also as conforming to the usage of modern 
 times in other literary and scientific treatises. , 
 
 2. Part First of the old edition has been omitted, and the 
 definitions and illustrations necessary to the integrity of the 
 subject ha.ve been incorporated with the body of the work. This 
 has been done solely in deference to the general sentiment of the 
 teachers of our country. The author can but feel that this senti- 
 ment is wrong. That the best way to present the subject of 
 Geometry is tb^present some of its leading notions and practical 
 facts with their uses in drawing and in common life, before 
 attempting to reason upon them, appears to him quite clear. It 
 is in accord with one of the settled maxims of teaching which 
 
 a c -i -^ 7 
 
IV PREFACE. 
 
 requires "facts before reasoning," and then it is in harmony 
 with the historic development of the science, and with the order 
 of mental development in the individual. Moreover, since this 
 method was presented to the American public in this treatise, the 
 author has received books on exactly the same plan, which are 
 in general use in Germany, and also "A Syllabus of Plane 
 Geometry, prepared by the Association for the improvement of 
 Geometrical teaching " in England, in which this principle 
 is recognized by recommending quite an extended course in 
 Geometrical constructions before entering upon the logical treat- 
 ment of the subject. The author hopes to revise his Part First, 
 and present it as a Httle treatise adapted to our Grammar or 
 lower schools ; as he can but think these subjects much more 
 interesting and usefuf to pupils of this grade than much of the 
 matter usually brought before them, especially the more advanced 
 portions of arithmetic, and as he is confident that they are the 
 proper preparation for the intelligent study of logical geometry. 
 
 3. The same general analysis of the subject is adhered to as in 
 the first edition. All must acknowledge it a reproach to the 
 oldest and most perfect of the sciences that, hitherto, no system- 
 atic classification of its subject-matter has been reached. That 
 the ordinary arrangement found in our Geometries is not 
 based upon a scientific analysis of the subject, and a systematic 
 classification of topics will be evident to any one who attempts 
 to give the subject-title of almost any so-called Book. A 
 glance at the table of contents of this volume will show that 
 the analysis of the subject-matter is simple and strictly philo- 
 sophical. There are two lines of inquiry in geometry, viz., 
 concerning position (from which form results) and magni- 
 tude. The concepts of Plane Geometry are the point, straight 
 line, angle, and circle. Kow, the measurement of magnitude is 
 either direct or indirect. The direct measurement and compar- 
 ison of magnitudes is a simple arithmetical operation, and is 
 presented, as regards straight lines, in Section 4. The direct 
 measurement of other magnitudes is effected in a similar manner, 
 but is unimportant from a scientific point of view. The indirect 
 measurement of magnitude, as when we find the third side of a 
 triangle from the other two and their included angle, the circum- 
 
PREFACE. V 
 
 ference or area of a circle from the radius, etc., is a somewhat 
 remote application of more elementary principles. There is then 
 left, as the natural first object of inquiry, the relative position of 
 two (and hence of all) straight lines. Here we have philosophi- 
 cally the first inquiry of logical geometry. This inquiry divides 
 into the three inquiries concerning perpendicular, oblique and 
 parallel lines. In a similar manner the topics of the succeeding 
 sections unfold themselves from the principles stated. 
 
 4. This analysis and classification of the subject-matter re- 
 quires that a somewhat larger number of propositions be demon- 
 strated from fundamental principles, that did the old method, of 
 proving first any proposition you could, and then any other, and 
 so on ; but who will consider this a defect ? On the other hand, 
 it gives almost absolute unity of method of demonstration in 
 the propositions of any one section. 
 
 5. The freedom with which revolution is used as a method of 
 demonstration, will be observed upon a cursory reading. Of 
 course it is assumed that the old repugnance to the introduction 
 of the notions of time and motion into geometry is outgrown. 
 Indeed, the old geometers could not get on without the super- 
 position of magnitudes, and this idea involves motion. Now, 
 revolution is but a systematic method of effecting superposition, 
 which is well-nigh the only geometrical method of proving the 
 equality of magnitudes. 
 
 6. The author has long desired to introduce the idea of same- 
 ness of direction in treating parallels ; but could not accept 
 what seemed to him the vague methods of writers who have 
 made the attempt. If we cannot define the notion of direction, 
 we certainly should have some method of estimating and measur- 
 ing it before it can be made a proper subject of geometrical 
 inquiry. This the author thinks he has secured, by giving the 
 necessary precision to certain very common and simple notions. 
 
 7. As to the introduction of the infinitesimal method into 
 mathematics (and if introduced at all, why not in the elements 
 where it will do most service ?), the author is confident that no 
 one thing would do more to simplify, and hence to advance, 
 elementary mathematical study, than the general and hearty 
 acceptance of this method. No writer has succeeded in getting 
 
VI PREFACE. 
 
 on far, even in pure mathematics, without openly or covertly 
 introducing the notion, and its directness, simplicity, if not 
 absolute necessity, in the applied mathematics make its intro- 
 duction into the elements exceedingly desirable. Nevertheless, 
 the author has given alternative demonstrations, either in the 
 body of the text or in the appendix, so that those who prefer 
 can omit the demonstrations involving the infinitesimal concep- 
 tion. 
 
 8. Thanks to the spirit of the times, no geometry can now 
 receive favor which does not give opportunity for the application 
 of principles and for independent investigation. As in the 
 former edition, so in this, large attention has been given to this 
 just demand of the times. As a help to independent thinking, 
 after the student has been fairly introduced to the methods, and 
 had time to imbibe somewhat of the spirit of geometrical reason- 
 ing, the references to the antecedent principles on which state- 
 ments in the demonstrations are based, are sometimes omitted, 
 and their place supplied by interrogation marks. 
 
 9. In the earlier part of the work, the demonstrations are 
 divided, according to the suggestion originally given by De 
 Morgan, into short paragraphs, each of which presents but a 
 single step. So, also, in this part, care has been taken to make 
 separate paragraphs of the statement of premises and the conclu- 
 sion, and to put the former in different type from the body of 
 the demonstration. But, in the latter part of the work, this 
 somewhat stiff and mechanical arrangement gives place to the 
 freer and more elegant forms with which the student will need to 
 be familiar in his subsequent reading. 
 
 10. In the preparation of the work the author has availed him- 
 self of the suggestions of a large number of the best practical 
 teachers in all parts of our country. His chief advisers have 
 been Professor Benjamin F. Clarke, of Brown University, K. I., 
 and Professor H. N. Chute, of the Ann Arbor High School, 
 Mich. To Professor Clarke he is indebted for valuable sugges- 
 tions on the whole of Chapter II., and especially on triedrals. 
 Indeed, whatever merit there may be in the general method of 
 treatment of triedrals, is due more to him than to the writer. 
 His ability as a mathematician, and his knowledge of what is 
 
PREFACE, yU 
 
 practical in methods of presentation, gained by long experience 
 in teaching the subject, appear on well-nigh every page of the 
 latter part of the work. Professor Chute, the able and accom- 
 plished teacher of geometry in the Ann Arbor High School, has 
 given me the free use of his careful and scholarly thought, and 
 long and successful experience as a teacher, by several readings 
 of the proofs, and by the use of the advance sheets of the entire 
 work in his classes. His logical acumen, practical skill, and 
 generous contribution of whatever he has found most valuable 
 in matter or method, have been of the highest service. The 
 same general acknowledgments are due to other authors as were 
 made in the earlier edition. To the taste and skill of the stereo- 
 typers, and the lavish expenditure of patience and money of the 
 PubHshers, the author is indebted for the elegant and beautiful 
 dress in which the book appears. 
 
 EDWAKD OLNEY. 
 University op Michigan, 
 
 Ann Arbor, September i, 1883. 
 
 N.B. — Part III. of the old edition will still be published for use in such 
 schools as wish to push the study of geometry still further than it is carried 
 in the ordinary treatises, and especially into the methods of what is called 
 the Modern Geometry. The topics embraced in that part are Exercises in 
 Geometrical Invention, including advanced theorems in Special or Elemen- 
 tary Geometry, Problems in the same, and Applications of Algebra to 
 Geometry ; and also an Introduction to Modern Geometry, including the 
 elements of the subjects of Loci, Symmetry, Maxima and Minima, Isoperi- 
 metry. Transversals, Harmonic Proportion, Pencils and Ratio, Poles and 
 Polars, Radical Axes and Centres of Similitude in respect to Circles. 
 
 The author's Trigonometry can also be had, bound separately or in con- 
 nection with the other parts of the Geometry, the same as formerly, 
 
 E. O. 
 
SUGGESTIONS TO TEACHERS. 
 
 1. Fix firmly in mind the fundamental definitions of the 
 science, in exact language, and illustrate them so fully that 
 the terms cannot be used in the hearing of the pupil, or by 
 him, without bringing before his mind, witJiout conscious 
 effort, the geometrical conception. 
 
 2. By numerous and varied applications of the fundamental 
 principles of plane geometry to the most famihar and homely 
 things in common life, divest the pupil's mind of the impression 
 that he is studying " higher mathematics " (as he is not), and 
 beget in him the habit of seeing the applications and illustrations 
 of these principles everywhere about him. 
 
 3. By means of much experience in the elements of geometri- 
 cal drawing, train the taste to enjoy, the eye to perceive, and the 
 hand to execute, geometrical forms, and by so doing fix indelibly 
 in the mind the " working facts " of geometry. 
 
 4. Have all definitions, theorems, corollaries, &c., memorized 
 with perfect exactitude, and repeated till they can be given with- 
 out effort. Demonstrations should not be memorized by the 
 pupil ; and considerable latitude may be allowed in the use of 
 language, provided the argument is brought out clearly. But 
 errors in grammar, and inelegancies in style, should be carefully 
 guarded against. One of the chief benefits to be derived from 
 class-room drill in mathematics is the ability to think clearly 
 and logically, and to express the thought in concise, perspicuous, 
 and elegant language. 
 
 5. The teacher should never give a theorem or corollary in 
 proper form, but by some such half -questions as the following, 
 suggest the topic : 
 
 The relation between the hypotenuse and the sides of a right- 
 angled triangle ? 
 
 The relative position of two circles when the distance between 
 the centres is less than the sum and greater than the difference 
 of the radii ? 
 
SUGGESTIONS TO TEACHERS. ix 
 
 The sum of the angles of a triangle ? 
 
 The relation between the angles and the sides of a triangle ? 
 etc. 
 
 In this manner the teacher should always designate the propo- 
 sition without stating it. The statement is one of the most 
 important things for the pupil to learn, and have at perfect com- 
 mand, and hence should not be given him by the teacher. 
 
 6. The construction of the figure is a necessary part of the 
 demonstration, and no assistance should be given the pupil, nor 
 aids allowed. 
 
 7. All figures in plane geometry should, upon first going over 
 the subject, be constructed by the pupils with strict accuracy, on 
 correct geometrical principles, using ruler and string ; and this 
 should be persisted in until it can be done with ease. In reviews, 
 free-hand drawing of figures may be allowed, and is even desir- 
 able. 
 
 8. The ordinary notation by letters should be used. 
 
 9. All the exercises in the book should be worked with care in 
 the study, and in the class, and be carefully explained by the 
 pupil ; and as many additional, impromptu exercises as may be 
 found necessary in order to render the pupil familiar with the 
 practical import of the propositions. 
 
 10. Little, if any, original demonstration of theorems not in 
 the book should be required of the pupil upon first going over 
 plane geometry. In review, more or less of such work may be 
 required. 
 
 11. Great pains should be taken that original demonstrations 
 be given in good, workmanlike form. For this purpose, they 
 should be written out with care by the pupil. Indeed, it is an 
 excellent occasional exercise, to have demonstrations written 
 out in full in class. 
 
 12. In review, much attention should be given to synopses of 
 demonstrations. They are the main reliancie for fixing in 
 memory the line of argument by which a proposition is demon- 
 strated. 
 
 I 
 

 
 INTRODUCTION. 
 SECTION I. 
 
 GENERAL DEFINITIONS 13-15 
 
 SECTION II. 
 THE GEOMETRICAL CONCEPTS 16-^0 
 
 SECTION III. 
 AXIOMS AND POSTULATES 30-33 
 
 SECTION IV. 
 MEASUREMENT OF RIGHT LINES 34-38 
 
 CHAPTER I. 
 
 PLANE GEOMETRY. 
 
 SECTION I. 
 
 PERPENDICULAR STRAIGHT LINES 38-46 
 
 Exercises 47-48 
 
 SECTION 11. 
 
 OBLIQUE STRAIGHT LINES 48-53 
 
 Exercises 63-54 
 
 SECTION III. 
 
 PARALLELS 55-66 
 
 Exercises 67 
 
 SECTION IV. 
 
 RELATIVE POSITIONS OF STRAIGHT LINES AND CIRCUMFERENCES 68-79 
 Exercises 79 
 
 SECTION V. 
 
 RELATIVE POSITIONS OF CIRCUMFERENCES 80-86 
 
 Exercises 87 
 
CONTENTS. XI 
 
 SECTION VI. 
 
 MEASUREMENT OF ANGLES 88-103 
 
 Exercises 103-104 
 
 SECTION VII. 
 
 ANGLES OF POLYGONS AND THE RELATION BETWEEN THE ANGLES 
 
 AND SIDES. 
 
 Op Triangles 104-109 
 
 Of QUADRn.ATERALS 110-118 
 
 Of Polygons of more than Four Sides 118-121 
 
 Of Regular Polygons 121-127 
 
 Exercises 127-128 
 
 Theorems for Original Investigation 129-130 
 
 SECTION VIII. 
 
 OF EQUALITY. 
 
 Op Angles 130-133 
 
 Op Triangles 133-139 
 
 Determination op Triangles 139-143 
 
 Determination op Quadrilaterals 143-147 
 
 Determination of Polygons 147-149 
 
 Propositions for Original Investigation 150-151 
 
 SECTION IX. 
 OF EQUIVALENCY AND AREAS 
 
 Equivalency 151-155 
 
 Area 155-163 
 
 Exercises 163-164 
 
 SECTION X. 
 
 OF SIMILARITY 164-181 
 
 Exercises 181-183 
 
 SECTION XI. 
 
 APPLICATIONS OF THE DOCTRINE OF SIMILARITY TO THE DEVELOP- 
 MENT OF GEOMETRICAL PROPERTIES OF FIGURES. 
 
 Op the Relations op the Segments on two Lines in- 
 tersecting EACH OTHER AND INTERSECTED BY A CIRCUM- 
 FERENCE 184-186 
 
 Op THE Bisector op an Angle of a Triangle. . . . . 186-188 
 
 Areas of Similar Figures 188-191 
 
 Perimeters and the Rectification op the Circumper- 
 
 ence 191-196 
 
 Area op the Circle 197-199 
 
 Divison in Extreme and Mean Ratio 199-201 
 
 Exercises 201-204 
 
XU CONTENTS. 
 
 CHAPTER II. 
 
 SOLID GEOMETRY. 
 
 SECTION I. 
 
 OF STRAIGHT LINES AND PLANES. p^gbs 
 
 Plane Determined 205-207 
 
 Perpendicular Lines to a Plane. 208-213 
 
 Oblique Lines to a Plane 213-216 
 
 Parallel Lines to a Plane 216-218 
 
 Parallel Planes 218-223 
 
 Exercises. 224 
 
 SECTION XL 
 OF SOLID ANGLES. 
 
 Of Diedrals 225-231 
 
 Of Triedrals 231-244 
 
 Of Polyedrals 244 
 
 Exercises 245-246 
 
 SECTION III. 
 
 OF PRISMS. AND CYLINDERS 247-263 
 
 Exercises 263-264 
 
 SECTION IV. 
 
 OF PYRAMIDS AND CONES 264-278 
 
 Of the Regular Polyedrons 279-282 
 
 Exercises 282-283 
 
 SECTION V. 
 
 OF THE SPHERE. 
 
 Circles of the Sphere 284-286 
 
 Distances on the Surface op a Sphere 287-293 
 
 Spherical Angles 293-296 
 
 Tangent Planes 297-298 
 
 Spherical Triangles 298-309 
 
 Polar or Supplemental Triangles 309-312 
 
 Quadrature of the Surface op the Sphere 313-316 
 
 Lunes 317-323 
 
 Volume of Sphere 324-327 
 
 Spherical Polygons and Pyramids 327-;J28 
 
 Exercises 329-330 
 
 APPENDIX 381-333 
 
/' 
 
 ^ E[eTnenCar 
 
 V ^:- OR, 
 
 -V 
 
 yiEOMETI^ 
 
 INTRODUCTION. 
 
 PRELIMINARY NOTIONS AND DEFINITIONS. 
 
 ^gCtlOH I. 
 
 GENERAL DEFINITIONS.* 
 
 1. A Proposition is a statement of something to be con- 
 sidered or done. 
 
 Illustration. — Thus, the comraon statement, "Life is short," is a 
 proposition; so, also, we make, or state a proposition, when we say, 
 "Let us seek earnestly after truth." — "The product of the divisor and 
 quotient, plus the remainder, equals the dividend," and the requirement, 
 " To reduce a fraction to its lowest terms," are examples of Arithmetical 
 propositions. 
 
 2. Propositions are distinguished as Axioms, Theorems, Lem- 
 mas, Corollaries, Postulates, and Problems. 
 
 * The terms here defined are such as are used in the science in conee- 
 quence of its logical character, hence tliey are sometimes called logico- 
 mathematical terms. The science of the Pure Mathematics may be con- 
 sidered as a department of practical logic. 
 
14 ELEMENTARY GEOMETRY. 
 
 3. An Axiom is a proposition which states a principle that 
 is so elementary, and so evidently true as to require no proof. 
 
 Illustration. — Thus, " A part of a thing is less than the whole of 
 it," " Equimultiples of equals are equal," are examples of axioms. If any 
 one does not admit the truth of axioms, when he understands the terms 
 used, we say that his mind is not sound, and that we cannot reason with 
 him. 
 
 4. A Theorem is a proposition which states a real or sup- 
 posed fact, whose truth or falsity we are to determine by 
 reasoning. 
 
 Illustration. — " If the same quantity be added to both numerator 
 and denominator of a proper fraction, the value of the fraction will be 
 increased," is a Theorem. It is a statement the truth or falsity of which 
 we are to determine by a course of reasoning. 
 
 6. A Demonstration is the course of reasoning by means 
 of which the truth or falsity of a theorem is made to appear. 
 The term is also applied to a logical statement of the reasons for 
 the processes of a rule. 
 
 A solution tells how a thing is done : a demonstration tells why it is 
 so done. A demonstration is often called proof. 
 
 6. A Lemma is a theorem demonstrated for the purpose 
 of using it in the demonstration of another theorem. 
 
 Illustration. — Thus, in order to demonstrate the rule for finding 
 the greatest common divisor of two or more numbers, it may be best first 
 to prove that " A divisor of two numbers is a divisor of their sum, and 
 also of their difference." This theorem, when proved for such a purpose, 
 is called a Lemma. 
 
 The term Lemma is not much used, and is not very important, since 
 most theorems, once proved, become in turn auxiliary to the proof of 
 others, and hence might be called lemmas. 
 
 7. A Corollary is a subordinate theorem which is sug- 
 gested, or the truth of which is made evident, in the course of 
 the demonstration of a more general theorem, or which is a 
 direct inference from a proposition, or a definition. 
 
 Illustration. — Thus, by the discussion of the ordinary process of 
 performing subtraction in Arithmetic, the following Gorolhry might be 
 
GENERAL DEFINITIONS. 15 
 
 suggested : " Subtraction may also be performed by addition, as we can 
 readily observe what number must be added to the subtrahend to pro- 
 duce the minuend." 
 
 8. A Postulate is a proposition which states that some- 
 thing can be done, and which is so evidently true as to require 
 no process of reasoning to show that it is possible to be done. 
 We may or may not know how to perform the operation. 
 
 Illustration. — Quantities of the same kind can be added together. 
 
 9. A Problem is a proposition to*do some specified thing, 
 and is stated with reference to developing the method of doing it 
 
 Illustration. — A problem is often stated as an incomplete sentence, 
 as, "To reduce fractions to forms having a common denominator." — This 
 incomplete statement means that " We propose to show how to reduce 
 fractions to forms having a common denominator." Again, the problem 
 " To construct a square," means that " We propose to draw a figure 
 which is called a square, and to tell how it is done." 
 
 10. A Rule is a formal statement of the method of solving 
 u general problem, and is designed for practical application in 
 solving special examples of the same class. 
 
 11. A Solution is the process of performing a problem or 
 an example. 
 
 A solution should usually be accompanied by a demonstration of the 
 process. 
 
 12. A Scholium is a remark made at the close of a dis- 
 cussion, and designed to call attention to some particular feature 
 or features of it. 
 
 Illustration. —Thus, after having discussed the subject of multipli- 
 cation and division in Arithmetic, the remark that " Division is the con- 
 verse of multiplication," is a scholium. 
 
 13. An Hypothesis is a supposition made in the state- 
 ment of a proposition, or in the course of a demonstration. 
 
 The Data are the things given or granted in a proposition. 
 The Conclusion is the thing to be proved. 
 
 The data of a proposition and the hypotheses are the same thing. 
 
16 
 
 ELEMENTARY GEOMETRY, 
 
 (i^^lrCtfiaM IL 
 
 THE GEOMETRICAL CONCEPTS.* 
 
 Points are designated 
 
 POINTS. 
 
 14. A Point is a place without size, 
 by letters. 
 
 Illustration. — If we wish to designate any particular point (place) 
 on the paper, we put a letter by it, and 
 sometimes a dot in it. Thus, in Fig. 1, the] 
 ends of the line, which are points, are desig- 
 nated as " point A," " point D ;" or, simply, 
 as A and D. The points marked in the I 
 line are designated as " point B," " point C," 
 or as B and C. F and E are two points 
 above the line. 
 
 Fig. I. 
 
 LINES. 
 
 16. A Line is the path of a point in motion. 
 
 Lines are represented upon paper by marks made with a pen or pen- 
 cil, the point of the pen or pencil representing the moving point. 
 
 A line is designated by naming the letters written at its ex- 
 tremities, or somewhere upon it. 
 
 Illustration. — In each case in Fig. 2, conceive a point to start from 
 A and move along the path indicated by the mark to B. The path thus 
 traced is a line. Since a point has no size, a line hcts no breadth, though 
 
 * A concept is a thing thought about ; — a thought-object. Thus, in Arith- 
 metic, number is the concept ; in Botany, plants ; in Geometry, as will 
 appear in this section, points, lines, surfaces, and solids. These may also be 
 said to constitute the subject-matter of the science. 
 
THE GEOMETRICAL CONCEPTS. 17 
 
 the marks by which we represent lines have some breadth. The first and 
 third lines in the figure are each designated as "the line AB." The sec- 
 ond line is considered as traced by a point starting from A and coming 
 
 around to A again, so that B and A coincide. This line may he desig- 
 nated as the line AmnA, or AmnE. In the fourth case, there are three 
 lines represented, which are designated, respectively, as AmB, AnB, and 
 AcB ; or, the last, as AB. 
 
 16. Lines are of Two Kinds, Straight and Curved. A 
 straight line is also called a Eight Line. A curved line is often 
 called simply a Curve. 
 
 17. A Straight Line is a line traced by a point which 
 moves constantly in the same direction. (See 46, a.) 
 
 The word " line " used alone generally signifies a straight line. 
 
 18. A Curved Line is a line traced by a point which con- 
 stantly changes its direction of motion. 
 
 Illustration. —Thus, in (1), Fig. 2, if the line AB is conceived as 
 traced by a point moving from A to B, it is evident that this point moves 
 in the same direction throughout its course ; hence AB is a straight line. 
 If a body, as a stone, is let fall, it moves constantly toward the centre of 
 the earth ; hence its path represents a straight line. If a weight is sus- 
 pended by a string, the string represents a straight line. 
 
 Considering the line represented by AtB, (3), Pig. 2, as the path of a 
 point moving from A to B, we see that the direction of motion is con- 
 stantly changing. 
 
18 
 
 ELEMENTARY GEOMETRY, 
 
 Sometimes a path like that rep- 
 resented in Fig. 3 is called, though 
 improperly, a Broken Line. It is not 
 a line at all ; that is, not one line : it 
 is a combination of straight lines. 
 
 SURFACES. 
 
 19. A Surface is the path of a line in motion. 
 
 20. Surfaces are of Two Kinds^ Plane and Curved. 
 
 21. A Plane Surface, or simply a Plane, is a surface 
 such that a straight line passing through any two of its points 
 lies wholly in the surface. Such a surface may always be con- 
 ceived as the path of a straight line in motion. 
 
 Illustration. — Let AB, Fig. 4, be supposed to move to the right, so 
 that its extremities A and B move at the 
 same rate and in the same direction, A 
 tracing the line AD, and B the line BC. The 
 path of the line, the figure ABCD, is a sur- 
 face. This page is a surface, and may be 
 conceived as the path of a line sliding like a 
 ruler from top to bottom of it, or from one 
 side to the other. Such a path will have '^' ' 
 
 length and breadth, being in the latter respect unlike a line, which has 
 only length. 
 
 22. A Curved Surface is a surface in which, if various 
 lin€8 are drawn through any point, some or all of them will be 
 curved. 
 
 Illustration. — Suppose a fine wire bent into the form of the curve 
 AmB, Fig. 5, and its ends A and B stuck into a rod XY. Now, taking 
 the rod XY in the fingers and rolling it, it is evident that the path of the 
 line represented by the wire AwB will be the surface of a ball (sphere). 
 
 Again, suppose the rod XY placed on the surface of this paper so 
 that the wire AmB shall stand straight up from the paper, just as it 
 
THE GEOMETRICAL CONCEPTS, 
 
 19 
 
 would if we could take hold of the curve at tn and raise it right up, 
 letting XY lie as it does in the figure. Now slide the rod straight up or 
 down the page, making both ends move at the same rate. The path ol 
 
 Fig. 6. Fig. 7. 
 
 A/?iB will be like the surface of a half-round rod (a semi-cylinder). Thus 
 we see how surfaces, plane and curved, may be conceiyed as the paths of 
 lines in motion. 
 
 Ex. 1. If the curve AnB, Fig. 6, be conceived as revolved 
 about the line XY, the surface of what object will its path be like? 
 
 Ex. 2. If the figure OMNP, Fig. 7, be conceived as revolved 
 about OP, what kind of a path will MN trace ? What kind of 
 paths will FN and OM trace ? 
 
 Ans. One path will be like the surface of a joint of stove- 
 pipe, L e., a cylindrical surface ; and one will be like a flat wheel, 
 i. e., a circle. 
 
 Ex. 3. If you fasten one end of a cord at a point in the ceil- 
 ing and hang a ball on the other end, and then make the ball 
 swing around in a circle, what kind of a surface will the string 
 describe ? 
 
 Ex. 4. If on the surface of a stove-pipe, you were to draw 
 various lines through the same point, might any of them be 
 straight ? Could all of them be straight ? What kind of a sur- 
 face is this, therefore? 
 
 Ex. 5. Can you draw a straight line on the surface of a ball ? 
 On the surface of an egg ? What kind of surfaces are these ? 
 
20 ELEMENTARY GEOMETRY. 
 
 Ex. 6. When the carpenter wishes to make the surface of a 
 board perfectly flat, he takes a ruler whose edge is a straight line, 
 and lays this straight edge on the surface in all directions, 
 watching closely to see if it touches at all points in all positions. 
 Which of our definitions is he illustrating by hig practice ^ 
 
 Ex. 7. How can you conceive a straight line to move so that 
 it shall not generate a surface ? 
 
 OF TH E CI RCLE. 
 
 23. A Circle is a plane surface bounded by a curved line 
 all points in which are equally distant from a point within. 
 
 24. The Circumference of a circle is the curved line 
 all points in which are equally distant from a point within. 
 
 25. The Centre of a circle is the point within, which is 
 equally distant from all points in the circumference. 
 
 26. An Arc is a part of a circumference. 
 
 27. A Radius is a straight line drawn from the centre to 
 any point in the circumference of a circle. 
 
 By reason of (24) all radii of the same circle are equal. 
 
 28. A Diameter of a circle is a straight line passing 
 through the centre and limited by the circumference. 
 
 A diameter is equal to the sum of two radii; hence, all diam- 
 eters of the same circle are equal. 
 
 Illustration. — A circle may be conceived as the path of a line, like 
 OB, Fig. 8, one end of which, 0, remains at the same point, while the 
 other end, B, moves around it in the plane of the paper. OB is the radius, 
 and the path described by the point B is the circumference. A Bis a diam- 
 eter. In Fig. 9, the curved line ABCDA is the circumference, is the 
 centref&n^ the surface within the circumference is the circle. Any part of 
 
THE GEOMETRICAL CONCEPTS, 
 
 31 
 
 Fig. 8. Fig. ». Fig. 10. 
 
 a circumference, as AB, or any one of tlie curved lines BB, Fig. 8, is an 
 arc. So also AM and EF, Fig. 10, are arcs. EF is an arc drawn from 0' 
 as a centre, with the radius O'B. 
 
 29. A Chord is a straight line joining any two points in a 
 circumference, as BC or AD, Fig. 9. The portion of the circle 
 included between the chord and its arc, as A/«D, is a Segfiiient. 
 
 30. A Tangent to a circle is a straight line which touches 
 the circumference, but does not intersect it, how far soever the 
 line be produced. 
 
 Two circles which touch each other in but one point are said 
 to be tangent to each other. A straight-line tangent is called a 
 Rectilinear Tangent. 
 
 31. A Secant is a straight line which intersects the circum- 
 ference. 
 
 ANGLES. 
 
 32. A Plane Angle, or simply an Angle, is the opening 
 between two lines which meet each other. 
 
 The point in which the lines meet is called the Vertex, and 
 the lines are called the Sides. 
 
 An angle is designated by pla. lui,' ;- letter at its vertex, and 
 one by each of its sides. In reaciiiig. we name the letter at the 
 vertex when there is but one vertex at the point, and the three 
 
5535 ELEMENTARY GEOMETRY. 
 
 letters when there are i^o or more vertices at the same point. 
 In the latter case, the letter at the vertex is put between the 
 other two. 
 
 Illustration. — Id com- 
 mon language an angle la 
 called a comer. The open- 
 ing between the two lines 
 AB and AC, in which the 
 ligure 1 stands, is called the 
 angle A ; or, if we choose, 
 we may call it the angle 
 BAC. At L there are two 
 vertices, so that were we to 
 say the angle L, one would 
 not know whether we meant 
 the angle (corner) in which 
 4 stands, or that in which 
 6 stands. To avoid this 
 ambiguity, we say the angle 
 HLR for the former, and 
 RLT for the latter. The 
 angle ZAY is the corner in 
 
 which 11 stands; that is, ^ 
 
 the opening between the '^' 
 
 two lines AY and AZ. In designating an angle by three letters, it is 
 immaterial which letter stands first, so that the one at the vertex is put 
 between the other two. Thus, PQS and SQP are both designations of 
 the angle in which 6 stands. An angle is also frequently designated by 
 putting a letter or figure in it and near the vertex. 
 
 33. The Size of an Angle depends upon the rapidity 
 with which its sides separate, and not upon their length. 
 
 Illustration. —The angles BAC and MON, Fig. 11, are equal, since 
 the sides separate at the same rate, although the sides of the latter are 
 more prolonged than those of the former. The sides DF and DE separate 
 faster than AB and AC, hence the angle EDF is greater than the angle 
 BAC. 
 
 34. AdUacent Angles are angles so situated as to have a 
 common vertex and one common side lying between them. 
 
THE GEOMETRICAL CONCEPTS. 
 
 Illustration. — In Fig. 12, angles 4 and 
 5 are adjdcenty since tliey have the common 
 vertex L, and the common side LR. Angles 
 9 and 10 are also adjacent. 
 
 35. Angles are distinguished as 
 
 Right Angles and Oblique Angles. 
 
 Oblique angles are either Acute or 
 Obtuse. 
 
 Fig. 12. 
 
 36. A Right Angle is an angle included between two 
 straight lines which meet each other in such a manner as to 
 make the adjacent angles equal. 
 
 37. An Acute Angle is an angle which is less than a 
 right angle, i. e., one whose sides sepamte less rapidly than those 
 of a right angle. 
 
 38. An Obtuse Angle is an angle which is greater than a 
 right angle, i. e.^ one whose sides separate more rapidly than those 
 of a right angle. 
 
 39. A Straight Angle is an angle whose sides extend in 
 opposite directions, and hence form one and the same straight line. 
 
 Illustrations. — In comnjon language, a right angle is called a 
 square corner, and an acute angle a sharp comer. 
 
 Fig. 13. 
 
 Angles BAD and BAC, Fig. 13, are right angles^ PST is an acme arigle^ 
 and HLR is an obtuse angle. 
 
 If HL were turned to the left until it fell in the dotted line, the angle 
 HLR would increase, and when HL fell in the dotted line, the angle would 
 become what is called a straight angle. 
 
24 ELEMENTARY GEOMETRY. 
 
 40. The Sum of Two Angles is the angle included 
 between their non-coincident sides, when the two angles are so 
 placed as to be adjacent angles, and their sides lie in the same 
 plane. 
 
 Fig. 14. 
 
 Illustration. — Let and M be any two angles. Make EPB = M, 
 and APE = 0, thus placing the two angles and M so that they become 
 adjacent angles (34). Then is APB the sum of and M, and we write, 
 
 + M = APB, or APE + EPB = APB. 
 
 That is, the sum of the angles and M, or APE and EPB, is APB. 
 
 41. The Difference between Two Angles is the 
 
 angle included by their non-coincident sides, when the angles 
 are so placed as to have a common vertex and side, the second 
 side of the less angle lying between the sides of the greater. 
 
 Fig. 15. 
 
 Illustration.— To find the difierence between the two angles and 
 S, we place the vertices and S at a common point, as at P, making 
 APB = RST, and APC = DOE. Then is CPB the difiference between 
 RST and DOE ; that is, 
 
 RST - DOE = CPB. 
 So also APB - APC = CPB, 
 and APB - CPB = APC. 
 
THE GEOMETRICAL CONCEPTS. 25 
 
 42. Corollary !.—(</) The sum of two right angles, 
 (6) Or, the sam of the two adjacent angles formed by 
 one straight line meeting another, 
 
 (c) Or, the sum of all the consecutive angles included 
 by several lines lying on the same side of a given line and 
 m^eeting it in a com^m^on point, is a straight angle. 
 
 Fig. 16. 
 
 Thus, ABP + PBC, or DEG + GEP, or HIL + LIM + MIN + NIK, 
 
 is a straight angle. 
 
 43. Corollary 2.— The sum of the four angles formed 
 by two intersecting lines, or the sum. of all the consecutive 
 angles formed by any numher of lines meeting in a com- 
 mon point is two straight angles, or four right angles. 
 
 Thus, the sum of the four angles ADC, CDB, BDE, and EDA is four 
 right angles, as also is the sum of AOB, BOC, COD, DOE, EOF, FOG, 
 and GOA. 
 
26 ELEMENTARY GEOMETRY, 
 
 44. A Solid is a limited portion of space. 
 
 Illustration. — Suppose you have a block of wood like that repre. 
 sented in Fig. 17. Hold it still in 
 your fingers a moment, and fix your 
 mind upon it. Now take the block 
 away and think of the space (place) 
 where it was. This space is an ex- 
 ample of what we call a Solid in 
 Geometry. In fact, the solids of 
 Geometry are not solids at all, in 
 the common sense of the word 
 solid ; they are only places of certain shapes. 
 
 Again, hold your ball still a moment in your fingers, then let it drop, 
 and think of the place it filled when you had it in your fingers. It is 
 this^Zace, shaped just like your ball, that we think about and talk about 
 as a solid in Geometry. 
 
 GENERATION OF LINES, SURFACES, 
 AND ANGLES. 
 
 45. When one geometrical concept is conceived to move so 
 that its path is some other concept, the former is said to generate 
 the latter, and the latter is called the locus of the former. 
 
 The Locus of a Point is the line (either straight or 
 curved) generated by the motion of the point according to some 
 given law. 
 
 In the same manner, a surface is conceived as the locus of a 
 line moving in some determinate manner. 
 
 46. A Line is generated by a moving point (15-18). Hence, 
 the locus of a point is a line. 
 
"TMB g:eoMetrical co^cjspts. %^ 
 
 (a) The came straight line may be conceived as generated by a point 
 moving in either of two opposite directions, or part of it may be con- 
 ceived as generated by a point moving in one direction, and part by a 
 point moving in the opposite direction. Thus, FA, Fig. 18, may be con- 
 ceived as generated by a point moving from F to A, or from A to F ; or 
 the part OA may be conceived as generated by a point moving from to 
 A, and the part OF by a point moving in the opposite direction, i. <?., 
 I from to F. 
 
 47. A Surface is generated by a moving line (19-22). 
 Hence, the locus of a line is a surface. 
 
 48. An Angle is generated by the revolution of a straight 
 line about one of its extremities, the line lying all the time in 
 the same plane. 
 
 Illustbation. — The angle BOA, 
 Fig. 18, may be considered as gen- 
 erated by the revolution of the line 
 BO from the position AO to its pres- 
 ent position. The angle COB may 
 be considered as generated by the 
 revolution of CO from the position 
 BO to its present position, etc. i^ 
 
 49. A Right Angle is generated by one-fourth of an en- 
 tire revolution, an Acute Angle by less than one-fourth of an 
 entire revolution, and an Obtuse Angle by more than one- 
 fourth. A Straight Angle is generated by one-half of a 
 revolution. 
 
 50. A Solid may be conceived as generated by the motion 
 of a plane, and hence may be defined as the path of a plane in 
 motion. 
 
 Illustration.— Thus the solid, Fig. 17, may be conceived as gener- 
 ated by the movement of the plane ABCD from its present position to the 
 position GHFE. 
 
 61. A Sphere may be conceived as generated by the revo- 
 lution of a semi-circle about its diameter. (See illustration at 
 the bottom of page 18.) 
 
38 ELEMENTARY GEOMETRY, 
 
 QUERIES. 
 
 ■ 1. If the surface OMNP, Fig. 19, is conceived as reyolyed 
 around OP, what is the path through which it moves ? 
 
 Caution. — The student should distinguish between the surface gener- 
 ated by the line MN, and the solid generated by the surface OMNP. 
 
 Fig. 19. Fig. 20. 
 
 S5. If the surface represented by CAB, Fig. 20, is conceived as 
 revolved about its side CA, what kind of a solid is its path ? 
 
 3. As you fill a vessel with water, what is the solid traced by 
 the surface of the water ? 
 
 Ans. The same as the space within the vessel. 
 
 4. If a circle is conceived as lying horizontally, and then 
 moved directly up, what will be the solid described, i. e., its path ? 
 Do not confound the surface described with the solid. What 
 describes the surface ? What the solid ? 
 
 EXTENSION AND FORM. 
 
 52. Extension means a stretching, or reaching out. 
 Hence, a Point has no extension. It has only position (place). 
 
 A Line stretches or reaches out, but only in length, as it has 
 no width. Hence, a line is said to have One Dimension, viz., 
 length. 
 
 A Surface extends not only in length, but also in breadth ; 
 and hence has Two Dimensions, viz., length and breadth. 
 
THE GEOMETRICAL CONCEPTS, 29 
 
 A Solid has Three Dimensions, viz., length, breadth, and 
 thickness. 
 
 Illustration. —Suppose we think of a point as capable of stretching 
 out (extending) in one direction. It would become a line. Now sup- 
 pose the line to stretch out (extend) in another direction — to widen. It 
 would become a surface. Finally, suppose the surface capable of thick- 
 ening, that is, extending in another direction. It would become a solid. 
 
 63. The Limits (extremities) of a line are points. 
 The Limits (boundaries) of a surface are lines. 
 The Limits (boundaries) of a solid are surfaces. 
 
 64. Magnitude (size) is the result of extension. lines, 
 surfaces, and solids are the geometrical magnitudes. A point is 
 not a magnitude, since it has no size. The magnitude of a line 
 is its length ; of a surface, its area ; of a solid, its volume. 
 
 55. Figure or Form (shape) is the result of position of 
 points. The form of a line (as straight or curved) depends upon 
 the relative position of the points in the line. The form of a 
 surface (as plane or curved) depends upon the relative position of 
 the points in it. The form of a solid depends upon the relative 
 position of the points in its surface. 
 
 QU ER I ES 
 
 1. Suppose a line to begin to contract in length, and continue 
 the operation till it can contract no longer, what does it become ? 
 That is, what is the minor limit of a line ? 
 
 2. If a surface contracts in one dimension, as width, till it 
 reaches its limit, what does it become ? 
 
 3. If a solid contracts to its limit in one dimension, what does 
 it pass into ? If in two dimensions ? If in three dimensions ? 
 
 4. What kind of a surface is that, every point in which is 
 equally distant from a given point? 
 
30 ELEMENTARY GEOMETRY. 
 
 56. Geometry is that science which treats of magnitude 
 and form as the result of extension and position. 
 
 The Geometrical Concepts are points, lines, surfaces 
 (including plane and spherical angles), and solids (including 
 solid angles).* 
 
 The Object of the science is the measurement and compari- 
 son of these concepts. 
 
 Plane Geometry treats of figures all of whose parts are confined to one 
 plane. Solid Geometry^ called also Geometry of Spaxx^ and Geometry of 
 Three IHmensions, treats of figures whose parts lie in different planes. 
 The division of this treatise into two chapters is founded upon this dis- 
 tinction. 
 
 0irct!0H m 
 
 AXIOMS AND POSTULATES, 
 
 67. There are very many axioms ; but, as they are truths 
 which the mind grants on the mere statement, it is not needful 
 to enumerate them all. We give a few of the more important, 
 with some illustrative remarks. 
 
 58. All demonstration is based upon definitions, axioms, or 
 previously demonstrated propositions. 
 
 * A plane angle may be conceived as a portion of a plane, and hence as 
 itself a surface, and thus capable of increase or diminution like the other 
 magnitudes. The angle thus considered becomes a sort of infinity deter- 
 mined relatively by the rate of separation of the lines. It is thus analogous 
 to an infinite series the law of which is determined by a few of its first terms, 
 
 ^e definitions 32, 33, and 48, with their illustrations. 
 
AXIOMS AND POSTULATES, 31 
 
 69. Axiom I. — A straight line is the slwrtest line be- 
 tween two points, 
 
 Illustbation. — If a cord is stretched across the table, it marks a 
 straight line. In this way the caipenter marks a straight line. Having 
 rubbed a cord, called a chalk-line, with chalk, he stretches it tightly from 
 one point to another on the surface upon which he wishes to mark the 
 line, and then raising the middle of the cord, lets it snap upon the sur- 
 face. So the gardener makes the edges of his paths straight by sti-etching 
 a cord along them. These operations depend upon the principle that 
 when the line between the points is the shortest possible, it is straight. 
 
 60. Axiom II. — Two points in a straight line deter- 
 mine its position. 
 
 Illustration. — If the farmer wants a straight fence built, he sets two 
 stakes to mark its ends. From these its entire course becomes known. 
 This is the principle upon which aligning (or sighting) depends. Two 
 points in the required line being given, by looking from one in the direc- 
 tion of the other, we look along a straight line, and are thus able to locate 
 other points in the 
 line. If the points A 
 and B are marked, by 
 putting the eye at A 
 and looking steadily 
 towards B, we can tell whether D and E are in the same straight line with 
 A and B, or not. So we can observe that C and C" are not in the line ; 
 but that C is. This process of discovering other points in a line with two 
 given points is called aligning, or sighting. In this way a row of trees is 
 made straight, or a line of stakes set. It is the principle upon which the 
 surveyor runs his lines, and the hunter aims his gun. In the latter case, 
 the two sights are the given points, and the mark, or game, is a third 
 point, which the marksman wishes to have in the same straight line as 
 the sights. 
 
 61. Axiom III. — Between the same two points there 
 is one straight line, and only one. 
 
 Illustration.— Let any two letters on this page represent the situa- 
 tion of two points ; we readily see that there is one, and only one, straight 
 path between them. A^ain, let a comer of the desk represent one point 
 
32 ELEMENTARY GEOMETRY. 
 
 and a comer of the ceiling of the room represent another point ; we per- 
 ceive at once that, if a point is conceived to pass in a straight line from 
 one to the other, it will always trace the same path. In short, as soon as 
 two points are mentioned, we think of the distance between them as a 
 single straight line, — for example, the centre of the earth and the centre 
 of the sun. 
 
 Once more, conceive A and B, Fig. 21, to be two points in the path of 
 a point moving from A in the direction of B. Now all the points in the 
 same direction from A that B is, are in this path ; and any point out of this 
 line, as C or C", is in a different direction from A. 
 
 In this manner we draw a straight line on paper by laying the straight 
 edge of a ruler on two points through which we wish the line to pass, 
 and passing a pen or pencil along this edge. 
 
 62. To Intersect is to cross ; and a crossing is called an 
 Intersection, 
 
 63. Corollary. — Two straight lines can intersect in hut 
 one point; for, if they had two points common, they would 
 coincide and not intersect. 
 
 Ex. 1. A railroad is to be run from the town A to town B. 
 Kit is made straight^ through what points will it pass ? Can it 
 pass through any points not in the same direction from A that 
 B is? 
 
 Ex. 2. If I live on the south side of a straight railroad, and 
 my friend on the north side, but five miles farther east and two 
 miles farther north, and the road from my house to his is 
 straight, how many times does it cross the railroad ? 
 
 Ex. 3. Can you always draw a straight line which shall cut a 
 curve (whatever curve it may be) in two points at least ? Try it. 
 
 64. Axiom IV. — The whole is greater than any of its 
 parts. 
 
 65. Axiom V. — The whole is equal to the sum of all 
 its parts. 
 
AXIOMS AND POSTULATES, 33 
 
 66. Axiom VI. — Things which are equal to the same 
 thing are e^ual to each other. 
 
 67. Axiom VII. — // equals he increased or dimin- 
 ished equally, the results will he equal. 
 
 68. Axiom VIII. — // unequals he increased or di" 
 minished equally, the greater will give the greater result, 
 i. e., the inequality wUl exist in the same sense. 
 
 POSTULATES 
 
 69. Postulatel! like axioms, are very numerous, and it would 
 be useless to attempt to enumerate them all. We give a few 
 simply as specimens. 
 
 70. Postulate I. — A line can he produced to any 
 length. 
 
 71. Postulate II. — From^ any point a straight line 
 can he drawn to any other point. 
 
 72. Postulate III. — Geometrical magnitudes can he 
 added, suhtracted, multiplied, or divided. 
 
 73. Postulate FV. — A geometrical figure can he con- 
 ceived as moved at pleasure, without changing its size or 
 the relation of its parts (shape). 
 
 74. Postulate V. — Any nurriber of lines can he drawn 
 making equal angles with a given line, 
 
 75. Postulate Yl. — With any point as a centre, a 
 circumference can he drawn with any radius. 
 
34 
 
 ELEMENTARY GEOMETRY, 
 
 jsiriCtlOM lY 
 
 MEASUREMENT OF RIGHT LINES. 
 
 76. The Measure of a line is another line which is con- 
 tained in it an exact number of times. 
 
 77. A Common Measure of two or more lines is a line 
 which measures each of them. 
 
 78. Commensurable Lines are lilies which have a 
 finite common measure. 
 
 79. The Sum of Two Lines is the line formed by uniting 
 them so that one shall be the prolongation of the other. 
 
 80. The Difference between Two Lines is the line 
 which remains after the length of the less has been taken from 
 the greater. 
 
 81. Problem. — To measure a straight line with the 
 dividers and scale. 
 
 Solution.— Let AB, Fig. 22, be the line to be measured. Take the 
 dividers, Fig. 2 (frontis- 
 piece), and placing the 
 sharp point A firmly 
 upon the end A of the 
 line AB, open the di- 
 viders till the other point 
 B (the pencil point) just 
 reaches the other end of 
 the line, B. Then letting Fig. 22. 
 
 the dividers remain open just this amount, place the point A on the 
 lower end of the left-hand scale, as at o, Fig. 1 (frontispiece), and notice 
 where the point B reaches. In this case, it reaches 3 spaces beyond the 
 
MEASUREMENT OF RIGHT LINES, 35 
 
 figure 1. Now, as this scale is inches and tenths of inches,* the line AB is 
 1.3 inches long. 
 
 Ex. 1. What is the length of CD ? Ans. .15 of a foot. 
 
 Ex. 2. What is the length of EF ? Ajis. .75 of an inch. 
 
 Ex. 3. What is the leng£h of GH ? Ans. 1\ inches. 
 
 Ex. 4. What is the length of IK ? Ans. .18 of a foot 
 
 Ex. 5. Draw a line 3 inches long. 
 
 Ex. 6. Draw a line 2.15 inches long. 
 
 Ex. 7. Draw a line 1.25 inches long. 
 
 Ex. 8. Draw a line .85 of an inch long. 
 
 [Note. — Suppose a fine elastic cord were attached by each of its ends 
 to the points A and B of the dividers; when they were opened so as to 
 reach from C to D, Fig. 22, the cord would represent the line CD. Now 
 applying the dividers to the scale is the same as laying this cord on the 
 scale. Without the cord, we can imagine the distance between the points 
 of the dividers to be a line of the same length as CD.] 
 
 Ex. 9. Find in the same way as above the length and width 
 of this page. Also the distance from one corner (angle) to the 
 opposite one (the diagonal). 
 
 82. Problem. — To find the suvi of two lines. 
 
 Solution.— To find the sum of AB and CD, If first draw the indefi- 
 nite line EX. With the di- 
 viders I obtain the length 
 of AB, by placing one point 
 on A and extending the 
 other to B. This length I 
 now lay off on the indefinite ^'S* 23. 
 
 line EX, by putting one point of the dividers at E and with the othei 
 marking the point F. EF is thus made equal to AB. In the same man- 
 
 * The next scale to the right is divided into lOths and lOOths of a foot. 
 Thus, from p to 10 is 1 tenth of a foot, and the smaller divisions are hun- 
 dredths. 
 
 f These elementary solutions are sometimes put in the singular, as the 
 more simple style. 
 
36 ELEMENTARY GEOMETRY. 
 
 ner, taking the length of CD 
 with the dividers, I lay it 
 off from F on the line FX. 
 Thus I obtain 
 
 EG = EF + FG ^. ,, 
 
 = AB + CD. 
 
 Hence, the sum of AB and CD is EG. 
 
 Ex. 1. Find the sum of AB and EF, Fig. 22. 
 Ex. 2. Find the sum of EF, CD, and GH, Fig. 22. 
 Ex. 3. Make a line twice as long as CD, Fig. 22. Three 
 times as long. 
 
 83. Problem.— To find the difference of two lines. 
 
 Solution.— To find the difference of AB and CD, I take the length of 
 the less line AB with the dividers; 
 and placing one point of the dividers 
 at one extremity of CD, as C, make 
 CE = AB. Then is ED the differ- 
 ence of AB and CD, since 
 
 ED = CD - CE = CD 
 
 Ex. 1. Find the difference of IK and EF, Fig. 22. 
 
 Ex. 2. Find the difference of GH and CD, Fig. 22. 
 
 Ex. 3. Find how much longer IK, Fig. 22, is than the sum 
 of EF, Fig. 22, and CD, Fig. 23. 
 
 Ex. 4. Find the difference of the sum of AB and GH, and 
 the sum of CD and EF, Fig. 22. 
 
 84. Problem. — To find the ratio of two commensurable 
 lines. 
 
 Solution. — Let AB and CD (Fig. 25) be the two lines whose ratio we 
 seek. 
 
 Apply the shorter (AB) to the longer (CD) as many times as the latter 
 will contain the former. If A B is contained an integral number of times 
 (say 3, or m) in CD, then AB is a common measure of AB and CD, and we 
 , AB 1 
 ^^'^^CD^^S'^"^- 
 
MEASUREMENT OF RIGHT LINES, 37 
 
 But if the shorter is not contained in the longer an integral number of 
 times, apply it as many times as it is contained, and note the remainder; 
 thus, AB is contained in CD once, with a remainder oD. 
 
 Now apply this remainder, «D, to AB as many times as AB will con- 
 tain it, which, in this case, is once with a remainder &B. 
 
 Fig. 25. 
 
 Again, apply this remainder, JB, to aD, the former remainder. In 
 this case, it is contained once with a remainder cD. 
 
 Again, apply cD to JB. It is contained twice, with a remainder dE. 
 
 ^nally, applying dE to cD, we find it contained 3 times, without any 
 remainder. 
 
 Hence, dE is the common measure of AB and CD. 
 
 Calling dE the unit of measure, 1, we have, 
 
 dE = \] 
 
 
 cD = ZdE = 3; 
 
 
 ^i = 2d) = 6; 
 
 
 M = })E = U + dE = 1', 
 
 
 aD = ac + cD = 10 ; 
 
 
 AB = A6 + ftB — aD ->r ac = 
 
 17; 
 
 CD = Co + aD = AB + aD : 
 
 = 27. 
 
 Hence the lines AB and CD are to each other as the numbers 17 and 
 27 ; AB is ^ of CD ; or, expressed in the form of a proportion, 
 
 AB _17* 
 CD ~ 27 • ' 
 
 [Note. — This process will be seen to be the same as that developed 
 in Arithmetic and Algebra for finding the greatest or highest Common 
 Measure of two numbers. See Practical Arithmetic, p. 362, and 
 Complete Algebra, (137).] 
 
 * This method will not always obtain the exact ratio, both because of 
 the imperfection of the measurement, and because some lines are incommen- 
 Burable by any finite unit, as will appear hereafter. 
 
38 ELEMENTARY GEOMETRY. 
 
 Fig. 26. 
 
 Ex. 1. Find, as above, the approximate ratio of AB to CD. 
 
 Ratio, — . 
 Ex. 2. Find, as above, the approximate ratio of CD and IK. 
 
 Ratio, -• 
 
 D 
 
 Ex. 3. Find, as above, the approximate ratio of EF to GH. 
 
 Ratio, jr« 
 Ex. 4. Find, as above, the approximate ratio of EF to CD. 
 
 Ratio, ~ 
 
 CONTINUOUS VARIATION. 
 
 85. A magnitude is said to vary continuously when in 
 passing from one value to another it passes through all interme- 
 diate values. 
 
 Illustbation. — Let the line EF, Fig. 26, be produced by placing a 
 pencil at F and tracing the line to the right, until it becomes equal to IK. 
 EF has thus been made to be successively of all intermediate lengths be- 
 tween its present length and the length of IK ; i. e., it has varied continu- 
 ously. 
 
 In like manner, an angle may be conceived to vary continuously from 
 one magnitude to another. Thus, in Fig. 27, the angle CPB may be made 
 greater or less by revolving CP about P. By such a revolution of CP the 
 angle CPB may be conceived to vary, or grow, continuoudy till it becomes 
 CPB. 
 
•*• CHAPTER i. 
 
 • ••••• • • •• • 
 
 PLANE GEOMETRY. 
 
 OF PERPENDICULAR STRAIGHT LINES. 
 
 86. A Perpendicular to a given line is a line which 
 makes a right angle (36) with the given line. 
 
 87. An Obliqvie line is a line which makes an oblique angle 
 with a given line. 
 
 PROPOSITION I. 
 
 88. Theorem. — At any point in a straight line, one 
 perpendicular can be erected to the line, and only one, 
 which shall lie on the same side of the line. 
 
 Demonstration. 
 
 Let AB represent any line, and P be any point therein. 
 
 We are to prove that, on the same side 
 of AB, there can be one, and only one, 
 perpendicular erected to AB at P. 
 
 From P draw any oblique line, as PC, 
 forming with AB the two angles CPB and 
 CPA. 
 
 Now, while the extremity P, of PC, 
 remains at P, conceive the line PC to re- pjg. 27. 
 
 volve so as to increase the less of the two 
 
 angles, as CPB, continuously. Since the sum of CPB and CPA remaillS 
 constant, CPA will diminish continuously. 
 
40 
 
 ELEMENTARY GEOMETRY, 
 
 Hence, for a certain position of CP, as 
 C'P, these angles will become equal. In 
 this position, the line is perpendicular to 
 AB (36, 86). Therefore, there can be one 
 perpendicular, C'P, erected to AB at P. 
 
 Again, if the line C'P revolve from the 
 position in which the angles are equal, one 
 angle will increase and the other diminish ; 
 hence there is only one position of the line 
 on this side of AB in which the adjacent angles are equal. 
 
 Therefore there can be only one perpendicular erected to AB at P, 
 which shall lie on the same side of AB. Q. e. d. 
 
 89. Corollary 1. — On the other side of the line a second 
 perpendicular, and only one, can he erected from the same 
 point in the line, 
 
 90. Corollary 2. — // one straight line meets another 
 so as to make the angle on one side of it a right angle, the 
 angle on the other side is also a right angle. 
 
 PROPOSITION II. 
 
 91. Theorem. — // two straight lines intersect so as to 
 mahe one of the four angles formed a right angle, the 
 other three are right angles, and the lines are mutually 
 perpendicular to each other. 
 
 Demonstration". 
 
 Let CD intersect AB, making CEB a 
 right angle. 
 
 We are to prove that CEA, A ED, and 
 DEB are also right angles, and that CD is 
 perpendicular to AB, and AB to CD. 
 
 By (90), since CEB is a right angle, 
 CEA is also a right angle. 
 
 In like manner, as BE meets CD, 
 making CEB aright angle, BED is a right 
 angle, by (90). 
 
 Again, since DEB is right, DE meets 
 AB, making one angle right; hence the 
 Other, AED, is right also (90). Q. e. d. 
 
 ii^^^^^^H 
 
 Fig. 28. 
 
OF PERPENDICULAR STRAIGHT LINES. 
 
 41 
 
 Finally, since CD meets AB, making AEC a right angle, CD is perpen- 
 dicular to AB (86) ; and, since AB meets CD, making AEC a right angle, 
 AB is perpendicular to CD. Q. e. d. 
 
 PROPOSITION III. 
 
 92. Theorem.— TFT^en. two straight lines intersect at 
 right angles, if the portion of the plane of the lines on one 
 side of either line he conceived to revolve on that line as 
 an axis until it coincides with the portion of the plane on 
 the other side, the parts of the second line will coincide,* 
 
 Demonstration. 
 
 Let the two lines AB and CD intersect at right angles at E ; and let 
 the portion of the plane of the lines on the side of CD on which B lies 
 be conceived to revolve on the line CD as an axis, until It falls In the 
 portion of the plane on the other side of CD.f 
 
 We are to prove that EB will fall in 
 and coincide with EA. 
 
 The point E being in CD, does not 
 change position in the revolution; and, 
 as EB remains perpendicular to CD, it 
 must coincide with EA after the revolu- 
 tion, or there would be two perpendicu- 
 lars to CD on the same side and from the 
 same point, E, which is impossible (88). 
 
 Hence, EB coincides with EA. Q. e. d. 
 
 Fig. 29. 
 
 •i^^^^^^H 
 
 PROPOSITION IV. 
 
 93. Theorem,— i^oTw. any point without a straight 
 line, one perpendicular can be let fall upon that line, and 
 only one. 
 
 * This has nothing to do with the lengths of EB and EA ; indeed, lines 
 are generally supposed indefinite in length, unless limited by the data. 
 
 f This revolution may be illustrated by conceiving the paper folded in 
 the line CD until EB is brought into EA. 
 
42 
 
 ELEMENTARY GEOGRAPHY, 
 
 
 ?' 
 
 P 
 
 \ 
 \ 
 \ 
 
 \ 
 — ) 
 
 P' 
 
 A D 
 
 /D"B A' D' B' 
 
 1 / 
 
 / 
 / 
 
 jP 
 
 Demoi^stration. 
 Let AB be any line and P any point without the line. 
 
 We are to prove that one per- 
 pendicular, and only one, can be 
 let fall from P upon AB. 
 
 Let A'B' be an auxiliary line; 
 and at any point in it, as D', let a 
 perpendicular P'D' be erected (88). 
 Now place A'B', bearing P'D' 
 with it, in AB, and move it to the 
 right or left till P'D' passes through 
 P, and when in this position let 
 D be the point in AB in which 
 ^''^- 3°- D' falls. 
 
 Connect P and D. 
 
 Then, since angle PDB coincides with the right angle P'D'B', PDB is a 
 right angle, and PD is a perpendicular from P to the line AB (86)- Q- e. d. 
 We are now to prove that PD is the only perpendicular from P to the 
 line AB. 
 
 Suppose that there can be another, and let it be PD". 
 Produce PD to P'", and take DP"=DP, and draw P"D". 
 Now let the portion of the plane above AB be revolved upon AB as 
 an axis until it falls in the plane on the opposite side of AB from its first 
 position. Then will DP' fall in DP'" (92), and since DP" is by construc- 
 tion equal to DP, P will fall in P". 
 
 Then, since PDB is a right angle BDP" is also a right angle, and PP" 
 is a straight line (42, a)- 
 
 For a like reason PD" P" is a straight line, and we have two straight 
 lines from P to P", which is impossible. 
 
 Hence there can be but one perpendicular, as PD, from P upon AB. 
 
 Q. E. D. 
 
 PROPOSITION V. 
 
 94. Theorem.^From a point laithout a straight line, 
 the -perpendicular is the shortest distance to the line. 
 
 Demonstration. 
 Let AB be any straight line, P any point without it, PD a perpen- 
 dicular, and PC any oblique line. 
 
 We are to prove th^t Pp is shorter than any oblique line, as PC. 
 
OF PERPENDICULAR STRAIGHT LINES. 
 
 43 
 
 Ist. Since the shortest distance from P to 
 any point in the line AB is a straight line 
 (69), we are to examine only straight lines. 
 
 2d. Produce PD, making DP' = PD, and 
 draw P'C. 
 
 Now let the portion ofthe plane of the lines 
 above AB be revohed upon AB as an axis until 
 it coincides with the portion below AB. 
 
 Since PP' and AB intersect at right 
 angles, PD will fall in DP' (92); and, since 
 PD = DP', P will fall in P', and PC = P'C, 
 since they coincide when applied. 
 
 Finally, PP' being a straight line, is shorter than PCP' which is a broken 
 line, since a straight line is the shortest distance between two points (59). 
 
 Now PD, the half of PP', is less than PC, the half of the broken line PCP . 
 
 Therefore, the perpendicular, PD, is the shortest distance from R to 
 the line AB. q. e. d. 
 
 95. The Distance between two points is the straight line 
 which joins them, and the Distance from a point to a line is 
 the perpendicular from the point to the line. 
 
 S^^IH^^E 
 
 Fig. 31. 
 
 PROPOSITION VI. 
 
 96. Theorem. — // a perpendicular is erected at the 
 middle point of a straight line, 
 
 1st. Any point in the perpendicular is equally distant 
 from the extremities of the line. 
 
 2d. Any point without the perpendicular is nearer the 
 extremity ofthe line on its own side of the perpendicular. 
 
 Demonstration^. 
 
 Let PD be a perpendicular to AB at 
 its middle point, D, any point in this 
 perpendicular, and 0' any point without 
 the perpendicular. 
 
 Draw OA, OB, O'A, and O'B. 
 
 We are to prove, 1st, that OA = OB; 
 and 2d, that O'B < O'A. 
 
 Ist. Revolve ODB on PD as an axis, 
 
 Fig. 32. 
 
44 ELEMENTARY GEOMETRY, 
 
 till B falls in the plane on the opposite 
 side of PD. 
 
 Then, since PD is perpendicular to AB, 
 DB will fall in DA (92)- And since DB 
 = DA by hypothesis, B will fall in A, and 
 OB will coincide with OA (61). 
 
 Hence OA = OB. q. e. d. 
 
 2d. 0' being on the opposite side of PD 
 from A, 0' A will cut PD at some point, as 0. 
 
 Draw CB. ^''^- ^^■ 
 
 Now, since C is a point in the perpendicular, CA = CB by the for- 
 mer part of the demonstration. 
 
 And, since O'B is a straight line and O'C 4- CB is a broken line, 
 
 OB < O'C + CB (59). 
 Whence, substituting CA for its equal CB, we have 
 OB < O'C + CA, 
 or O'B < O'A. Q. E. D. 
 
 97. Corollary. — Conversely, The locus of a point equi- 
 distant from the extremities of a given line is a perpen- 
 dicular to that line at its middle point, since any point in 
 such perpendicular is equidistant from the extremities of the line, 
 and any point not in the perpendicular is unequally distant from 
 the extremities. 
 
 PROPOSITION VII. 
 
 98. Theorem. — // each of two points in one line is 
 equally distant from the extremities of another line, the 
 former line is perpendicular to the latter at its middle 
 point. 
 
 Demonstration. 
 
 Every point equally distant from the extremities of a straight line lies 
 in a perpendicular to that line at its middle point, by (97)- But two 
 points determine the position of a straight line. Hence, two points, each 
 equally distant from the extremities of a straight line, determine the 
 position of the perpendicular at the middle point of the line. Q. e. d. 
 
OF PERPENDICULAR STRAIGHT LINES. 
 
 45 
 
 PROPOSITION VIII. 
 
 99. Problem. — To erect a perpendicular to a given 
 line at a given point in the line. 
 
 SOL[JTION. 
 Let XY be the given line, and A the given point. 
 
 We are to erect a perpendicu- 
 lar to XY, at A. 
 
 From A lay off on each side 
 equal distances, as AC = AB. 
 
 From C and B as centres, with 
 a radius sufficiently great to cause 
 the arcs to intersect at some point 
 without XY, describe arcs intersecting at 0. 
 
 Pass a line through and A, and it will be the perpendicular sought. 
 
 Fig. 33. 
 
 Demoxstkatiok of Solution. 
 
 Since OA has two points, and A, each equally distant from B and 
 C, OA is a perpendicular to BC at A, its middle point (98)* 
 
 But BC coincides with XY ; hence OA is perpendicular to XY at A. 
 
 100. Definition. — To Bisect anything is to divide it into 
 two equal parts. 
 
 PROPOSITION IX. 
 101. Problem. — To bisect a given line. 
 
 Solution. 
 Let AB be the given line. 
 
 We are to bisect it, that is, to divide it 
 into two equal parts. 
 
 From the extremities A and B as cen- 
 tres, with any radius sufficiently great to 
 cause the arcs to intersect without the line 
 AB, describe arcs intersecting in two points, 
 as m and n. 
 
 i^^^^^if^^^^: 
 
 Fig. 34. 
 
46 
 
 J^LEMBNTARt GEOMETRY. 
 
 Pass a line through m and n, intersect- 
 ing AB at 0. 
 
 Then is the middle point of AB, and 
 AO = OB. 
 
 Demonstration of Solution. 
 
 Since the line mn has two points, m and 
 n, eacli equally distant from A and B, it is 
 perpendicular to AB at its middle point (98) 
 
 Fig. 34. 
 
 PROPOSITION X. 
 
 102. Problem. — From a point without a given line, to 
 let fall a perpendicular upon the line. 
 
 Solution. 
 Let XY be the given line, and the point without the line. 
 
 We are to let fall a perpendicular 
 from to XY. 
 
 From as a centre, with a radius 
 sufficiently great to cause the arcs to in- 
 tersect, descrilje an arc cutting XY in two 
 points, as B and C. 
 
 From B and C as centres, with a ra- 
 dius sufficiently great to cause the arcs to 
 intersect without XY, describe arcs in- 
 tersecting at some point, as D. Fig. 35. 
 
 Pass a line through and D, meeting XY in A. Then is OA the 
 perpendicular sought. 
 
 Demonstration of Solution. 
 
 OA being produced through D has two points, and D, each equally 
 distant from B and C, and hence is perpendicular to BC, which coincides 
 with XY. Hence, OA passes through and is perpendicular to XY. 
 
OF PERPENDICULAR STRAIGHT LlJSES, 
 
 47 
 
 QU E R I ES. 
 
 103. 1. In the solution of Proposition IX, is it necessary 
 that the arcs which intersect at n should be struck with the same 
 radius as those which intersect at ?w ? Is it necessary that the 
 two intersections be on different sides of AB ? 
 
 2. In the solution of Proposition X, is it necessary that the 
 intersection D should fall on the opposite side of XY from ? 
 Why is it necessary to take the radius with which these arcs are 
 struck greater than half of BC ? 
 
 EXERCISES. 
 
 104. 1. A mason wishes to build a wall from (Fig. 36), in the 
 wall AB, '* straight across" (perpendicular) to the wall CD, which 
 is 8 feet from AB. He has only his 10-foot pole, which is subdi- 
 vided into feet and inches, with which to find the point in the 
 opposite wall at which the cross wall must join. How shall he 
 find it? What principle is involved? 
 
 
 
 Fig. 36. 
 
 Fig. 37. 
 
 2. Wishing to erect a line perpendicular to AB (Fig. 37) at 
 its centre, I take a cord or chain somewhat longer than AB, and 
 fastening its ends at A and B, take hold of the middle of the cord 
 or chain and carry it as far from AB as I can, first on one side 
 and then on the other, sticking pins at the most remote points, 
 as at P and P'. These points determine the perpendicular sought. 
 What is the principle involved ? 
 
48 
 
 ELEMENTARY GEOMETRY, 
 
 3. Bisect a line by making marks on only one side of it. 
 
 4. With a measuring-tape as an instrument, how would you 
 erect on the shore a perpendicular to the straight bank of a lake, 
 at a given point in the bank ? 
 
 -♦-•-♦- 
 
 0]&Ct!iaH 11 
 
 OF OBLIQUE STRAIGHT LINES. 
 
 105. The Supplement of an angle is the angle which 
 remains after it has been taken from a straight angle, or two 
 right angles. 
 
 106. Supplemental Angles are, therefore, two angles 
 whose sum is a straight angle, or two right angles (42, h). 
 
 107. Vertical, or Opposite Angles are the non-adja- 
 cent angles formed by the intersection of two straight lines. 
 
 PROPOSITION I. 
 
 108. Theorem.— Vertical, or opposite angles are equal. 
 
 Demokstration. 
 Let AB and CE intersect at D. 
 
 We are to prove that ADC = BDE, and 
 CDB =ADE. 
 
 ADC + CDB = a straight angle (42, ft); 
 and for the same reason CDB + BDE = a 
 straight angle. 
 
 Hence, ADC + CDB = CDB + BDE ; and 
 subtracting CDB from each member, we have 
 ADC = BDE. 
 
 In like manner, CDB + BDE — BDE + ADE ; whence, CDB = ADE. 
 q. E. D. 
 
 Fig. 38. 
 
OF OBLIQUE STRAIGHT LINES. 
 
 40 
 
 PROPOSITION II. 
 
 109. Theorem. — // two supplemental angles are so 
 placed as to he adjacent to each other, the two sides not 
 common fall in the same straight line. 
 
 Demonstration. 
 
 Let AOB and B'O'E' be two supplemental 
 angles, and let B'O'E' be placed so as to be 
 adjacent to AOB, e. e.^ as BOE. 
 
 We are to prove that AE is a straight line. 
 
 Before considering B'O'E' as placed adja- 
 cent to AOB, produce AO to E, forming AE. 
 
 By (42, ft), AOB + BOE = a straight 
 angle, i. e., two right angles, whence BOE is the 
 supplement of AOB. 
 
 Now, as by hypothesis B'O'E' is the supplement of AOB, B'O'E' 
 = BOE. 
 
 Place B'O'E' a^acent to AOB, 0' in 0, and O'B' in OB. Then will 
 O'E' fall m OE. 
 
 Therefore the two sides not common, i. e., AO and O'E', fall in the 
 same straight line AE- Q. e. d. 
 
 Fig. 39. 
 
 PROPOSITION III. 
 
 110. Theorem. — // from, a point without a line a 
 perpendicular is drawn to the line, and oblique lines are 
 drawn from, the same point, meeting the line at equal 
 distances from, the foot of the perpendicular, 
 
 1st. Tlie oblique lines are equal to each other. 
 
 2d. The angles which the oblique lines form ivith the 
 perpendicular are equal to each other. 
 
 3d. The angles formed by the oblique lines with the first 
 line are equal to each other. 
 
 Demonstration. 
 Let AB (Fig. 40) be any line, P any point without it, PD a perpendicular, 
 and PC and PE oblique lines meeting AB at and E, so that CD = DE. 
 
50 
 
 ELEMENTARY GEOMETRY. 
 
 We are to prove, 1st, that PC = PE ; 
 2d, that CPD = DPE ; and 3d, that 
 PCD = PED. 
 
 Revolve PDE on PD as an axis, until 
 E falls in the plane on the other side of 
 PD. 
 
 Now, since AB is perpendicular to 
 PD, DB will fall in DA (92). And since 
 DE = DC by hypothesis, E will fall in 
 C. Hence the two figures PDE and PDC 
 coincide, and we have, 1st, PC = PE ; 2d, CPD = DPE ; and 3d, PCD 
 = PED. Q. E. D. 
 
 Fig. 40 
 
 Query.- 
 
 (96) y 
 
 -How would the equality of PC and PE follo\y from 
 
 PROPOSITION IV. 
 
 111. Theorem. — // from a point i^vithout a line a 
 perpendicular is draian to the line, and from the same 
 point two oblique lines are drawn, making equal angles 
 with the perpendicular and, meeting the first line, 
 
 1st. The oblique lines are equal to each other. 
 
 2d. The oblique lines cut off equal distances from the 
 foot of the perpendicular. 
 
 3d. The oblique lines make equal angles with the first 
 
 line.* 
 
 Demonstratioit. 
 
 Let AB be a straight line, P any point without it, and PD a perpen- 
 dicular to AB ; and let PE and PC be drawn, making CPD = EPD. 
 
 We are to prove, 1st, that PC = PE ; 
 2d, that DE = DC; and 3d, that PED 
 = PCD. 
 
 Revolve PDE upon PD as an axis, 
 until E falls in the plane on the opposite 
 side of PD. 
 
 Then, since EPD = CPD by hypoth- 
 esis, PE will fall in PF, and the point E 
 will be found somewhere in PF. c. -, 
 
 Fig. 41. 
 
 * This proposition is the converse of tlie last. The significance of this 
 statement will be more fully developed farther on (128). 
 
OF OBLIQUE STRAIGHT LINES, 51 
 
 Again, DE will fall in DA (92), and E will fall somewhere in DA. 
 
 Now as E falls at the same time in DA and PF, it must fall at their 
 intersection C, and the figures PDE and PDC must coincide ; whence we 
 have, 
 
 1st, PC = PE ; 2d, DE = DC ; and 3d, PCD = PED. Q. e. d. 
 
 PROPOSITION V. 
 
 112. Theorem. — // from a point without a line a 
 perpendicular is drawn to the line, and from the sam^e 
 point two oblique lines are drawn making equal angles 
 with the first line, 
 
 1st. The oblique lines cut off equal distances from the 
 foot of the perpendicular, 
 
 2d. The oblique lines are equal to each other. 
 
 3d. The oblique lines make equal angles with the per- 
 pendicular. 
 
 Demonstration. 
 
 Let P be any point without the line AB, and PD a perpendicular from 
 P upon AB, and let PE and PC be drawn making the angle DEP = angle 
 DCP. 
 
 We are to prove, 1st, that DE = DC ; 
 2d, that PE = PC ; and 3d, that angle 
 DPE = angle DPC. 
 
 Conceive a perpendicular erected at 
 the middle point of CE, and let it intersect 
 CP, or CP produced, in some point as X. 
 Conceive X joined with E. 
 
 By (110, 3d.) XED = XCD, {i. c, PCD). Fig. 42. 
 
 But by hypothesis PED = PCD. Hence, XE falls in PE, and PD is the 
 perpendicular to CE at its middle point. 
 
 Therefore, DE = DC; and by (HO) PE = PC, and DPE = DPC. 
 Q. E. D. 
 
52 ELEMENTARY GEOMETRY. 
 
 PROPOSITION VI. 
 
 113. Theorem. — If from a point without a line a per- 
 pendicular is let fall on the line, and from the same point 
 two oblique lines are drawn, the oblique line which cuts off 
 the greater distance from the foot of the perpendicular is 
 the greater. 
 
 Demonstration. 
 
 Let AB be any straight line, P any point without it, and PC and PF 
 two oblique lines of which PF cuts off the greater distance from the foot 
 of the perpendicular PD; that is, DF > DC. 
 
 We are to prove that PF > PC. 
 
 If the two oblique lines do not lie on 
 the same side of the perpendicular, as in 
 the case of PF anfl PE, take DC = DE, 
 and on the side in which PF lies, 
 draw PC. Then PC will be equal to PE, 
 by (110, 1st). Hence, if we show the 
 proposition true when both oblique lines 
 lie on the same side of the perpendicular, 
 * it will be true in general. ^'9* *^' 
 
 Produce PD, making DP' = PD, and draw P'F and P'C, producing 
 the latter until it meets PF in H. 
 
 Revolve the figure FPD upon AB as an axis, until it falls in the plane 
 on the opposite side of AB. 
 
 Since PP' is perpendicular to AB, PD will fall in P'D; and, since 
 PD = P'D, P will fall at P'. Then P'C = PC and P'F = PF. 
 
 Now the broken line PCP' < than the broken line PHP', since the 
 straight line PC < the broken line PHC. 
 
 For a like reason, the broken line PHP' < PFP', since HP' < HFP'. 
 
 Hence PCP' < PFP', and PC (the half of PCP') < PF (the half of 
 PFP'). Q. E. D. 
 
 114. Corollary. — From a given point without a line, 
 there can he two, and only two, equal oblique lines draivn 
 to the line, and these will lie on opposite sides of the per- 
 pendicular drawn from the given point to the given line. 
 
 liH^^B! 
 
DF OBLiqUE STRAIGHT LINES. 
 
 PROPOSITION VII. 
 
 116. Theorem.: — // two equal oblique lines are drawn 
 from the same point in a perpendicular to a given line, 
 they cut off equal distances on that line from, the foot 
 of the perpendicular. 
 
 Let PD be perpendicular to AB, and 
 PE = PC. 
 
 We are to prove that DE = DC. 
 
 If DE were greater than DC, PE would 
 be greater than PC, and if DE were less than 
 DC, PE would be less than PC (113) ; but 
 both of these conclusions are contrary to 
 the hypothesis PE = PC. 
 
 Hence, as DE can neither be greater nor 
 equal to DC. Q. e. d. 
 
 Fig. 44. 
 than DC it must l^e 
 
 EXERCISES. 
 
 116. 1. Having an angle given, how can you construct its 
 supplement ? Draw on the blackboard any angle, and then con- 
 struct its supplement. What is the supplement of a right 
 angle ? 
 
 Fig. 45. 
 
 2. The several angles in Fig. 45 are such parts of a right 
 angle as are indicated by the fractions placed in them. If these 
 anghfs are added together by bringing the vertices together and 
 causing the adjacent sides of the angles to coincide, how will the 
 two sides not common lie ? Why ? 
 
54 ELEMENTARY GEOMETRY. 
 
 3. If two times A, B (Fig. 45), two times D, three times E, 
 three times C, three times G, and two times F, are added in order, 
 how will AM and GN lie with reference to each other ? Why ? 
 
 Ans, They will coincide. 
 
 4. If you place the vertices of any two equal angles together 
 so that two of the sides shall extend in opposite directions and 
 form one and the same straight line, the other two sides lying on 
 opposite sides thereof, how will the latter sides lie? By what 
 principle ? 
 
 5. If two lines intersect, show that tlie line which hisects one 
 of the angles will, if produced, bisect the opposite angle. 
 
 6. If one line meet another, show that the two lines bisecting 
 these supplemental angles are perpendicular to each other. 
 
 7. If two lines intersect, show that two lines bisecting the 
 two pairs of opposite angles are perpendicular to each other. 
 
PARALLELS. 
 
 I«» 
 
 0irct!0H m. 
 
 OF PARALLELS. 
 
 117. The Direction of a straight line is defined or deter- 
 mined by the plane in which it lies and the angle which it makes 
 with some fixed line, this angle being generated (48) from the 
 fixed line around in the same direction,* in the same argument. 
 
 118. The assumed fixed line is called the Direction Line, 
 
 and the angle which the line makes with the direction hne is 
 called the Direction Angle. 
 
 Illustration. — Thus the directions 
 of the several lines AB, CD, and EF may 
 be defined by referring them to some as- 
 sumed fixed line, as XY. 
 
 The direction of AB is defined by say- 
 ing that its direction angle is YOA, or its 
 equal XOB, this angle l)cing conceived as 
 generated /r^'7/i the direction line, as indi- 
 cated by the arrows. 
 
 So also the direction of CD is defined 
 by the angle YMC, or its equal XMD; and 
 the direction of EF is in like manner de- 
 fined by YNE, or XNF. 
 
 Fig. 46. 
 
 119. Witli reference to its generation, the same 
 line may be conceived as having either of two opposite directions, 
 or various parts of it may be conceived as having opposite direc- 
 tions. 
 
 Illustration.— Thus, the line AB (Fig. 47) may be considered as 
 generated by a point moving from A to B, whence its direction would 
 
 * Revolution around a fixed point is often designated as from left to 
 right, or from right to left. To comprehend these terms, one may conceive 
 himself in the centre of motion, and facing the moving point. Thus all the 
 motions represented by arrows in Fig. 46 will be seen to he from right to left. 
 
56 
 
 ELEMENTARY GEOMETRY. 
 
 be from A towards B ; or, it may be considered 
 as generated by a point moving from B to A, 
 whence its direction would be from B towards A. 
 In like manner, part of the line, as PB, may 
 be considered as having the direction from P 
 towards B, while the other part is conceived as 
 having the opposite direction, i. e., from P towards A. 
 
 Fig. 47. 
 
 120. Lines have the Same Direction when they lie in 
 the same plane and make equal direction angles with the same 
 line. 
 
 Any line may be assumed at pleasure as the direction line, 
 provided that in comparing the directions of different Hnes they 
 all be referred to the same direction line. 
 
 121. Parallel Lines are lines which have the same or 
 opposite directions. 
 
 .122. A Transversal is a line cutting a system of lines. 
 
 123. When two lines are cut by a transversal, the angles 
 formed are named as follows : 
 
 Exterior Angles are those without 
 the two lines, as 1, 2, 7, and 8. 
 
 Interior Angles are those within 
 the two lines, as 3, 4, 5, and 6. 
 
 Alternate Exterior Angles are 
 
 those without the two lines and on differ- 
 ent sides of the transversal, but not adja- 
 cent, as 2 and 7, 1 and 8. 
 
 Fig. 48. 
 
 Alternate Interior Angles are those within the two 
 lines and on different sides of the transversal, but not adjacent, 
 as 3 and 6, 4 and 5. 
 
 Corresponding Angles are one without and one within 
 the two lines, and on the same side of the transversal, but not 
 adjacent, as 2 and 6, 4 and 8, 1 and 5, 3 and 7. 
 
PARALLELS. 67 
 
 PROPOSITION I. 
 
 124. Theorem. — Through a given point one line can be 
 drawn parallel to a given line, and but one. 
 
 Demonstration. 
 Let AB be the given line, and P the given point. 
 
 We arc to prove that one line can 
 be drawn through P parallel to AB, 
 and but one. 
 
 Through P draw XY as the direc- 
 tion line, intersecting AB in E. 
 
 Also through P pass a line C'D', 
 making XPD' greater than XEB. 
 
 Then revolving CD' about P as a 
 centre, XPD' may be made to dimin- 
 ish continuously, and in some posi- Fig. 49. 
 tion, as CD, XPD will equal XEB. In this position, CD is parallel to AB 
 (120, 121). 
 
 Hence there can be one line drawn through P parallel to AB. q. e. d. 
 
 Again, there can be but one; since, if CD be revolved in either direc- 
 tion about P, the angle XPD will become unequal to XEB, and hence the 
 line CD will not be parallel to AB. Q. e. d. 
 
 PROPOSITION II. 
 
 125. Theorem. — // a transversal cuts two parallels, 
 
 1st. Any two corresponding angles are equal. 
 
 2d. Any two alternate interior, or any two alternate 
 exterior angles are equal. 
 
 3d. The sum of any two interior angles on the same 
 side of the transversal, or the sum of any two exterior 
 angles on the same side, is two right angles, or a straight 
 angle. 
 
 Demonstration. 
 
 Let AB and CD (Fig. 50) beany two parallels, and EF any transversal. 
 
58 
 
 ELEMENT A RT GEO METR F. 
 
 We are to prove, Ist. Of the cor- 
 responding angles, b = d^ a = e, 
 e = g^ and / = h. 
 
 2d. Of the alternate interior an- 
 gles, b =^f, and c = g; of the alter- 
 nate exterior angles, d = h, and 
 a = e. 
 
 3d. Of the interior angles on the 
 same side of the transversal, 
 
 & + c = 2 right angles, 
 and g -\- f = 2 right angles ; 
 
 of the exterior angles on the same side, 
 a + d = 2 right angles, and 
 
 Fig. 50. 
 
 e + h = 2 right angle 
 
 Let EF be taken as the direction line, the direction angles being esti- 
 mated from right to left (120, 121, and foot-note, p. 55). Then, 
 
 1st. Of the corresponding angles, b = d, these being the direction 
 angles, and AB and CD being parallel. 
 
 a = c, since they are supplements of the equal angles, b and d ; and 
 e = g, for the same reason. 
 
 Also, f = h^ since they are opposite angles to the equal angles, b 
 and d. 
 
 2d. Of the alternate interior angles, ft =/, since f — d (108) \ c — g^ 
 since they are supplements of b and d. 
 
 Of the alternate exterior angles, d = h^ since h = b (108); and 
 « = a, since they are supplements of b and d. 
 
 3d. Of the interior angles on the same side, « 
 
 b + c = 2 right angles (or a straight angle), 
 since d -\- c = 2 right angles (or a straight angle), (42, *), and b = d; 
 
 and 1 
 
 since 
 
 g + f z=z 2 right angles, 
 
 g -{. b = a straight angle, and b =f. 
 
 Of the exterior angles on the same side, 
 
 a + d = 2 right angles, 
 since a + b = b, straight angle, and b = d; 
 
 also 
 since 
 
 e -\- h = 2 right an^ 
 
 g -\- h = & straight angle, and e = g. Q. e. d. 
 
PARALLELS, 69 
 
 PROPOSITION III. 
 
 126. Theorem. — Conversely to Proposition II, When 
 two lines are cut by a transversal, the two lines are par- 
 allel, 
 
 |- 1st. // any two corresponding angles are equal, 
 
 2d. // a?nj two alternate interior, or any two alternate 
 exterior angles are equal. 
 
 3d. //' the sum of any two interior angles on the same 
 side, or the sum of any two exterior angles on the same 
 side is two right angles. 
 
 Demonstration. 
 
 Let AB and CD be two lines cut by the transversal EF, making any 
 pair of corresponding angles equal, as 6 = f/, a = r, y = e, h = /'; 
 or any two alternate interior angles, or any two alternate exterior 
 angles equal, as ft = /, f/ — c, (i = e, or h = d; or the sum of any 
 two interior angles on the same side, or of any two exterior angles on 
 the same side, equal to 2 right angles, 3S h + c, y + f, a ^ d, h + e, 
 equal to 2 right angles. 
 
 Wc are to prove that AB and CD 
 are parallel. 
 
 Let EF be the direction line, and b 
 and d the direction angles. If, then, 
 these are granted or proved equal, the 
 lines are parallel (121)- 
 
 Now, 1st. Of the corresponding an- 1 
 gles, if h = d, AB and CD are parallel 
 by definition; but, if a = c, h = d, 
 since & and d arc supplements of a and 
 
 c; or, if ^ = e, b = d, since b and d " " Fig. 51. 
 
 are supplements of g and <?; or, if ^ =/, h = d^ since h = h and d =/ 
 (108). Hence, in every case, h = d, and AB and CD are parallel. 
 
 2d. Of alternate interior angles, if ft = /, b = d, since / = d; or, if 
 g = c, b — d, since b and d are supplements of g and c. Hence, in either 
 case, b = d. and AB and CD are parallel. 
 
60 ELEMENTARY GEOMETRY. 
 
 Of alternate exterior angles, if 
 h = d,I} = dy since b = h (108) ; o^i 
 if a = e, J = df, since ft and d are 
 supplements of a and e. Heuc^, in 
 either case, h = d and, AB and CD 
 are parallel. 
 
 3d. Of interior angles on the 
 same side, if & + « = 3 right angles, 
 h = d, since d + c = 2 right angles 
 (42) ; or, if ^ + / = 2 right angles, 
 l> = d, since Pig, 5,^ 
 
 g + f +i + c = A right angles ; 
 
 hence, ft + c = 2 right angles, and as <f + c = 2 right angles, d = h. 
 Hence, in either case, ft = <f , and AB and CD are parallel. 
 
 Of exterior angles on the same side, if a + <Z = 2 right angles, h = d, 
 since a -{- b = 2 right angles ; or, if h -\- e = 2 right angles, b = d, since 
 h + e-\-a-{-d = 4: right angles; hence, a ^ d = 2 right angles, and, as 
 a + ft = 2 right angles, d = b. Hence, in either case, ft = (?, and AB and 
 CD are parallel, q. E. D. 
 
 127. Corollary. — Two lines which are perpendicular 
 to a third are parallel to each other. 
 
 For, in such a case, all the eight angles formed are equal ; hence, any 
 of the conditions of the proposition are met. 
 
 128. Scholium. — The last two propositions are the converse 
 of each other; i.e., the hypotheses and conclusions are exchanged. 
 Thus, in Prop. II, the hypothesis is that the two lines are parallel, 
 and the conclusion is certain relations between the angles ; while 
 in Prop. Ill the hypotheses are certain relations among the angles, 
 and the conclusion is that the lines are parallel. 
 
 The learner may think that, if a proposition is true, its converse is 
 necessarily true ; and hence, that when a proposition has been proved, its 
 converse may be assumed as also proved. Now this is by no means the 
 case. Although in a great number of mathematical propositions, it hap- 
 pens that the proposition and its converse are both true, we never assume 
 one from having proved the other ; and we shall occasionally find a prop- 
 osition whose converse is not true. 
 
PARALLELS. 
 
 ei 
 
 PROPOSITION IV, 
 
 129. Theorem. - When two straight lines are cut hy a 
 transversal, if the sum, of the two interior angles on either 
 side is less than two right angles, tiie two lines will meet 
 on this side of the transversal, if sufficiently extended. 
 
 Demoi^stration. 
 
 Let AB and CD be two lines cut by the transversal XY, making 
 BEP + EPD < 2 right angles. 
 
 We are to prove that AB and CD will meet on the side of XY on which 
 these angles lie. 
 
 Through P draw FG parallel to AB. 
 
 Take EH = EP and draw PH, and also ET perpendicular to PH By 
 (115), TH = TP. whence EHT = EPT (110) 
 
 But EHT = GPH 125) Hence GPH = ^GPE. 
 
 Again, take HI — PH and draw PI, and it may be shown in the same 
 manner that GPI -= ^GPH == ^GPE. 
 
 In this manner we may continue to draw oblique lines through P cut- 
 ing AB further and further from E, and may thus diminish at pleasure 
 the angle included by the oblique line and PG. Hence this angle may be 
 made less than GPD, the difference, between DPE and the supplement 
 of PEB, when the oblique line will fall between PD and PG. Call this 
 line PR. Now as PR and PE cut AB, and PD lies between them, it must 
 cat AB between E and R. q. e. d. 
 
6S5 ELEMENTARY GEOMETRY, 
 
 130. Corollary 1. — If a transversal cuts one of two 
 parallels, it cuts the other also. 
 
 131. Corollary ^^i. — JVon-jMirallel straight lines meety 
 if sufficiently produced. 
 
 132. Corollary 3. — Two straight lines in the same 
 plane which do not meet, however far they are extended, 
 are parallel. 
 
 For, let AB and CD be two such lines, and P any point in CD. Now 
 all lines through P which are not4)arallel to AB meet AB (131). Hence 
 as there can be one parallel to AB through P (124), it is the line which 
 does not meet AS. 
 
 PROPOSITION V. 
 
 133. Theorem. — A line which is perpendicular to one 
 of two parallels is perpendicular to the other also. 
 
 Demonstration. 
 Let AB and CD be two parallels, and let EF be perpendicular to AB. 
 
 We are to prove that EF is also 
 perpendicular to CD. 
 
 Since EF is a transversal cut- 
 ting AB and CD, angle EOB = angle 
 EMD (125, 1). 
 
 Now EOB is a right angle by 
 hypothesis (86), 
 
 Hence EMD is a right angle, 
 and EF is perpendicular to CD. 
 
 Q. E. D. 
 
 Fig. 53. 
 
 134. Corollary. — The shortest distance between two 
 parallels is the -perpendicular which joins them. 
 
 For, CM being a perpendicular from O to CD, is shorter than any 
 other line from to CD (94). 
 
PARALLELS. 63 
 
 135. The Distance between two parallels is the perpen- 
 dicular which joins them. 
 
 • 
 
 PROPOSITION VI. 
 
 136. Theorem. — Two parallels are everywhere equally 
 distant from each other, and hence never meet. 
 
 Demonstration. 
 
 Let E and F be any two points in the line CD, and EG and FH per- 
 pendiculars measuring the distances between the parallels CD and AB at 
 these points. 
 
 We are to prove EG = FH. 
 
 Let P t>e the middle point between 
 E and F, and PO a perpendicular at 
 this point. 
 
 Revolve the portion of the figure 
 on the right of PO, upon PO as an axis, f 'q- 54. 
 
 until it falls upon the plane of the paper at the left. 
 
 Then, since FPO and EPO are right angles, PD will fall in PC ; and, 
 as PF = PE, F will fall on E. As F and E are right angles, FH will 
 take the direction EG, and H will lie in EG or EG produced. Also, 
 as POH and POG are right angles, OB will fall in OA, and H, falling at the 
 same time in EG and OA, is at their intersection G. 
 
 Hence, FH coincides with and is equ.il to EG. Q. e. d. 
 
 Hence, also, CD cannot meet AB, since the distance from any point 
 in CD to AB is EG. Q. e. d. 
 
 PROPOSITION VII. 
 
 137. Theorem. —Conversely to Profwsition VI., // two 
 points in one straight line are equally distant from a 
 second straight line, and on the same side of it, the lines 
 are parallel to each other. 
 
 Demonstration. 
 
 Let AB and CD (Fig. 55) be two lines having the points P and S in CD 
 equally distant from AB, and on the same side of it. 
 
 We are to prove that CD and AB are parallel. 
 
64 
 
 ELEMENTARY GEOMETRY. 
 
 From P and S draw PE and SF perpendicular to AB. Then is PE = 
 SF, by hypothesis. 
 
 Through 0, the middle point of PE, 
 draw GH parallel to AB. 
 
 Since PE is parallel to SF, GH cuts 
 SF in some point as I (130)- 
 
 By (136), OE = IF ; and since SF = 
 PE and OE is |PE, IF is ^SF, that is, 
 IF = IS. *''S- 55. 
 
 Now, as PE and SF are perpendicu-ar to GH (133), if we revolye the 
 figure OAEFBI on GH, E will fall in P, and F in S (92), and AB will have 
 two points in common with CD, and hence will coincide with it. 
 
 Hence, DPO = BEO, and as the latter is a right angle by construc- 
 tion, AB and CD are perpendicular to PE, and hence parallel (127). 
 
 Q. E. D. 
 
 PROPOSITION VIII. 
 
 138. Theorem.— t/i pair of parallel transversals inter- 
 cept equal portions of two parallels. 
 
 Demonstratioi^. 
 
 Let ST and RL be two parallel transversals, cutting the two parallels 
 AB and CD. 
 
 We are to prove that GE = HF. 
 From E and F let fall the per- 
 pendiculars EM and FK. Then 
 
 EM = FK (136). 
 
 Now apply the figure GEM to 
 HFK, placing EM in its equal FK. 
 Since M and K are right angles, MG 
 will fall in KH. 
 
 With the figures in this position, 
 FH and EG are lines drawn from the 
 same point in the perpendicular to 
 ST and making equal angles with it 
 (125), and are hence equal (112). Q- E. D, 
 
 Fig. 56. 
 
PARALLELS. 
 
 «6 
 
 PROPOSITION IX. 
 
 139. Theorem. — Two straight lines which are parallel 
 to a third are parallel to each other. 
 
 Demonstration. 
 Let AB and CD be each parallel to EF 
 
 We are to prove that AB and CD are 
 parallel to each other. 
 
 Draw HI perpendicular to EF; then will 
 it be perpendicular to CD (133)- 
 
 For a like reason, HI is perpendicular 
 to AB. 
 
 Hence, CD and AB are both perpendic- 
 ular to HI, and consequently parallel (127)- Q. k. d 
 
 <» 
 
 Fig. 57. 
 
 PROPOSITION X. 
 
 140. Theorem. — // to each of two parallels perpendic- 
 ulars are drawn, then are the perpendiculars parallel. 
 
 Demonstration. 
 Let A and B be parallel lines, P be perpendicular to A, and Q to B. 
 
 We are to prove that P and Q are par- 
 allel to each other. 
 
 Q, which is perpendicular to B, one of 
 the two parallels, is perpendicular to A, 
 the other parallel, also (133). 
 
 Hence, P and Q are both perpendicu- 
 lar to A, and hence are parallel (127)- Q. e. d.| 
 
 Fig. 58. 
 
 
66 
 
 ELEMENT A R Y GEOMETR T. 
 
 PROPOSITION XI. 
 
 141. Theorem. — // to each of two non-parallel lines 
 a perpendiGwlar is drawn, the perpendiculars are non- 
 parallel. 
 
 Demonstration. 
 
 Let A and B be non-parallel, and P a perpendicular to A, and Q to B. 
 
 We are to prove that P and Q are 
 non-parallel. 
 
 If P and Q were parallel, then, by the 
 preceding proposition, A and B would be 
 parallel, which is contrary to the hy- 
 pothesis. Hence, P and Q are non-par- 
 allel. <i. B. D. 
 
 Fig. 59. 
 
 PROPOSITION XII. 
 
 142. Problem. — Through a given point to draw a 
 parallel to a giveri line. 
 
 Solution. 
 Let AB be the given line, and P the given point. 
 
 Fig. 60. 
 
 "We are to draw through P a parallel to AB. 
 
 Let fall PF, a perpendicular from P to AB (102). 
 
 At P erect CD, a perpendicular to PF (99). 
 
 Then is CD parallel to AB (127). [Pupil give proof.] 
 
PARALLELS. 
 
 67 
 
 EXE RCISES. 
 
 143. 1. How can a farmer tell whether the opposite sides 
 of his farm are parallel ? 
 
 2. If we wish to cross over from one of two parallel roads to 
 the other, is it of any use to travel farther in the hope that the 
 distance across will be less ? Why ? 
 
 3. If a straight line intersects two parallel lines, how many 
 angles are formed ? How many angles of the same size ? May 
 they all be of the same size ? When ? Wlien will they not be 
 all of the same size ? 
 
 4. Are the two opposite walls of a building which are carried 
 up by the plumb line exactly parallel ? Why ? 
 
 6. A hetad (Fig. 61) is an instni-l 
 ment much used by carpenters, and[ 
 consists of a main limb AB, in 
 which a tongue CD is placed, so as 1 
 to open and shut like the blade of 
 a knife. This tongue turns on the 
 pivot 0, which is a screw, and can 
 be tightened so as to hold the 
 tongue firmly at any angle with the 
 limb. The tongue can also be ad- 
 justed so as to allow a greater ori 
 less portion to extend on a given 
 side, as CB, of the limb. Now,' 
 suppose the tongue fixed in posi- 
 tion, as represented in the figure, and the side m of the limb to be placed 
 against the straight edge of a board, and slid up and down, while lines 
 are drawn along the side n of the tongue. What will be the relative 
 position of these lines ? Upon what proposition does their relative posi- 
 tion depend ? How can the carpenter adjust the bevel to a right angle 
 upon the principle in Prop. I, Sec. I ? At what angle is the bevel set, 
 when, drawing two lines from the same point in the edge of the board, 
 one with one edge m of the bevel against the edge of the board, and the 
 other with the other edge m', these lines are at right angles to each other? 
 
 Fig. 61. 
 
68 
 
 ELEMENTARY GEOMETRY, 
 
 
 ^^0]^cti0M ty 
 
 OF THE RELATIVE POSITIONS OF STRAIGHT LINES AND 
 CIRCUMFERENCES. 
 
 PROPOSITION I. 
 
 144. Theorem. — Any diameter divides a circle^ and 
 also its circumference, into two equal parts. 
 
 Demon^stkation. 
 Let AB be the diameter of the circle kin^n. 
 
 We are to prove that arc A»iB = arc 
 AwB, and that segment AwiB = segment AwB. 
 
 Revolve A^B upon AB as an axis, until 
 it falls in the plane on the opposite side of AB. 
 
 Then, since every point in AwB is at the 
 same distance from the centre C as every 
 point of A?nB (24), the arc AwB falls in AwB, 
 and both arcs and segments coincide; 
 whence, arc AwB = arc AmB, and segment 
 AnB = segment AwB. Q. e. d. 
 
 Fig. 62. 
 
 PROPOSITION II. 
 
 145. Theorem. — The diameter of a circle is greater 
 than any other chord of the same circle. 
 
 Demonstration. 
 
 Let AB (Fig. 63) be a chord meeting the circumrerence in A and B, 
 and not passing through the centre ; and let AC be the diameter. 
 
STRAIGHT LINES AND CIRCUMFERENCES, 
 
 69 
 
 We are to prove that AB is less than any 
 diameter, as AC (28). 
 
 Now as AB is not a diameter, it does not 
 pass through 0, or lie in AC Hence B is a 
 different point from C. 
 
 Draw OB. 
 
 Now AB being a straight line, is less than 
 A0 + OB, which is a broken line (69); hence, 
 as AO + OB = AC, AB < AC. Q. e. d. 
 
 146. An arc is said to be Siibteiicled by the chord which 
 joins its extremities, and the arc is said to subtend the angle in- 
 cluded by the radii drawn to its extremities. 
 
 PROPOSITION III. 
 
 147. Theorem. — ^ radius which is perpendicidar to 
 a chord bisects the chord, the subtended arc, and the sub- 
 tended angle* 
 
 Demonstration. 
 
 Let AB be a chord subtending the arc AB, which arc subtends the 
 angle AOB. Let the radius EO be perpendicular to AB, cutting it in D. 
 
 We are to prove that DA = DB, arc AE 
 = arc EB, and angle AOE = angle BOE. 
 
 Produce EO, forming the diameter EC. 
 
 Revolve the semicircle EBC on EC as an 
 axis, till it falls in the plane on the other side 
 of EC. 
 
 The semicircles will coincide (144), and 
 since AB is perpendicular to EO, DB will fall 
 
 in DA. 
 
 Moreover, as OA = OB, and there cannot Fig. 64. 
 
 * Such statements in Plane Geometry are generally limited to the con- 
 sideration of arcs less than a semi circumference, yet all the propositions 
 in this section, except Prop. VIII, are equally true whatever the arcs 
 may be. 
 
70 
 
 ELEMENTARY GEOMETRY. 
 
 be two equal oblique lines from a point to a line on the same side of a 
 perpendicular (114), OB falls in OA, and B falls in A. 
 
 Hence, DB coincides with DA, EB with EA. and angle BOE with angle 
 AOE, and we have DA = DB, arc AE = arc EB, and angle AOE = angle 
 
 BOE. Q. E. D. 
 
 PROPOSITION IV. 
 
 148. Theorem, — Conversely to Proposition Til, A radius 
 ivhich bisects an arc bisects the chord which subtends the 
 arCi, is perpendicular to the chord, and also bisects the 
 subtended angle. 
 
 Demonstration. 
 
 Let arc AB be bisected by the radius OE at E. Let the straight line 
 AB be the chord of this arc, and AOB the subtended angle. 
 
 We are to prove that OE bisects the chord 
 AB and is perpendicular to it, and also bisects 
 the angle AOB. 
 
 Produce EO, forming the diameter EC. 
 
 Revolve the semi-circumference EBC upon 
 EC as an axis, till it falls in the plane at the 
 left of EC. 
 
 Then will semi circumference EBC coincide 
 with EAC, and since arc BE = arc AE by hy- 
 pothesis, B will fall in A, and BD = AD. Fig. 64. 
 
 Hence, the line OE has two points, and D, each equally distant 
 from A and B, and is therefore perpendicular to AB (98)- 
 
 Furthermore, angle BOD coincides with AOD, and BOD = AOD- 
 
 Q. E. D. 
 
 ^ 
 
 PROPOSITION V. 
 
 149. Theorem.— Conversely to Propositions III and IV, 
 A radius which bisects the angle included by two other 
 radii bisects the arc subtending the angle, and the chord 
 of the arc, and is perpendicular to the chord. 
 
 [Let the student give the demonstration.] 
 
STRAIGHT LINES AND CIRCUMFERENCES. 
 
 PROPOSITION VI. 
 
 150. Theorem. — In the same circle, or in equal cU'- 
 ales, equal chords are equally distant from the centre. 
 
 Demonstration. 
 
 Let EF and GH be equal chords in the same circle or in equal circles ; 
 and OL and ON be the perpendiculars from the centre upon the 
 chords, and thus be the distances of the chords from the centre (95). 
 
 Fig. 65. 
 
 We are to prove OL = ON. 
 
 Since OL and ON are perpendiculars from the centre upon the equal 
 chords EF and GH, HN = FL (147). 
 
 Now apply the figure HNO to FLO, placing HN in its equal FL. Then 
 will NO coincide with LO (88)- 
 
 In this position, HO and FO are equal lines drawn from the same 
 point in the perpendicular FL to the line LO- Hence, LO = NO (116). 
 
 Q. E. D. 
 
 [Let the student state and prove other converses to Propositions III. 
 IV and VI.] 
 
 PROPOSITION VII. 
 
 151. Theorem. — In the same circle, or in equal cir- 
 cles, if two arcs are equal, the chords ivhich subtend them 
 are equal; and, conversely, if two chords are equal, the 
 subtended arcs are equal. 
 
72 
 
 ELEMENTARY GEOMETRY. 
 
 Demonstration. 
 
 Let AaitB and Cfi^D be equal arcs in the same circle or in equal 
 circles. 
 
 Fig. 66. 
 
 "We are to prove, first, that the chords AB and CD are equal. 
 
 Apply the figure C/iDO to AmBO, placing the radius CO in its equal 
 AO, and let the arc CD extend in the direction of arc AB. 
 
 Then, since each point in arc CD is at the same distance from the cen- 
 tre as each point in arc AB, arc CD falls in arc AB, and since arc CD = 
 arc AB by hypothesis, D falls in B. 
 
 Hence, chord AB = chord CD (61). Q. e. d. 
 
 Conversely, If chord AB = chord CD, arc AB = aro CD. 
 
 Draw the perpendiculars OL and ON. Then, since the chords are 
 equal, OL = ON (150). 
 
 Now apply the figure CnDO to At/iBO, placing ON in its equal OL. 
 Since CD is perpendicular to ON, and AB to OL, CD will fall in AB ; and, 
 since the chords are equal by hypothesis, and are bisected at N and L 
 (147), D falls in B and C in A. 
 
 Hence, arc CwD coincides with arc AmB, and arc CwD = arc AwB. 
 
 Q. E. D. 
 
 PROPOSITION VIII 
 
 152. Theorem.— /^i' the same circle, or in equal cir- 
 cles, if two arcs are unequal, the less arc has the less chord ; 
 and, conversely, if two chords are unequal, the less chord> 
 subtends the less arc. 
 
straight lines and circumferences, 73 
 
 Demonstration. 
 In the same circle, or in equal circles, let arc AmB < arc CtiD 
 
 Fig. 67. 
 
 We are to prove, first, that chord AB < chord CD. 
 Draw OA, OB, OD, and OC 
 
 Apply the figure CnDO to AmBO, placing OC in its equal OA, and the 
 arc n in the arc m. 
 
 Since arc CnD > arc AwB, D will fall beyond B, as at D'. Draw OD'. 
 AD' will evidently cut OB. Let N be the point of intersection. 
 
 Now AB < AN + NB (59), 
 
 and BO = D'O < ND' + ON (59). 
 
 Adding, AB + BO < AN + ND' + ON + NB, 
 
 or AB + BO < AD' + BO. 
 
 Subtracting BO from each member, we have 
 AB < AD'. Q. E. D. 
 
 Conversely, if chord AB < chord CD. 
 
 We are to prove that arc AmB is less than arc CnD. 
 
 For, if arc AwiB = arc C/iD, chord AB = chord CD (151). And, if 
 arc AwB > arc C/iD, chord AB > chord CD, by the former part of this 
 demonstration. But both of these conclusions are contrary to the hy- 
 pothesis. 
 
 Hence, as arc AmB can neither be equal to nor greater than arc CwD, 
 it must be less. q. e. d. 
 
74 ELEMENTARY GEOMETRY. 
 
 PROPOSITION IX. 
 
 163. Theorem. — In the same circle, or' in equal civ- 
 cles, of two unequal chords, the less is at the greater dis- 
 tance from the centre; and, conversely, of two chords 
 which are unequally distant from the centre, that which 
 is at the greater distance is the less. 
 
 Demonstration. 
 
 In the same circle, or In equal circles, let chord CE < chord AB, 
 and OD and OD' their respective distances from the centre. 
 
 Fig 68. 
 
 We are to prove, first, that OD > OD'. 
 
 From A, one extremity of the greater chord, lay oflF towards B, AE' = 
 CE. Since AE' < AB, arc AE' < arc AB (152), and E' falls somewhere 
 on the arc AB between A and B. 
 
 Draw OD" perpendicular to AE', and OD" = OD, since the equal 
 chords are equally distant from the centre (160). 
 
 Now OD" is a different line from OD', since OD" produced would 
 bisect arc AE', and OD' would bisect arc AB. Hence, as OD' is perpen- 
 dicular to AB, OD" must be oblique (93). 
 
 Again, OD" cuts the line AB in some point as H, since the chord AE' 
 lies on the opposite side of AB from the centre 0. 
 
 Hence, OH > OD' (94), and much more is OH + HD" (= OD") > OD'. 
 
 Q. E. D. 
 
 Conversely, let OD > OD'. 
 
 We are to prove that CE < AB. 
 
 If CE = AB, OD =r OD' (160), and if CE > AB, OD < OD', both 
 of which conclusions are contrary to the hypothesis OD > CD'. 
 
 Hence, as CE can neither be equal to nor greater than AB, it must be 
 less. q. E. D. 
 
STRAIGHT LINES AND CIRCUMFERENCES. 75 
 
 PROPOSITION X. 
 
 154. Theorem. — A straight line which intersects a 
 circumference in one point intersects it also in a second 
 point, and can intersect it in hut two points. 
 
 Demonstration. 
 
 Let LM (Fig. 69) intersect the circumference in A. 
 
 We are to prove that it intersects in finother point, as B, and in only 
 these two points. 
 
 Since LM intersects the circumference in A, I 
 it passes within it, and hence has points nearerj 
 the centre than A. OA is, therefore, anf 
 oblique line, and not the perpendicular from 
 upon LM (94). 
 
 Now two equal oblique lines can be drawn! 
 to a line from a point without (114). Let OBI 
 be the other oblique line equal to OA. But asl 
 OA is a radius, OB = OA must also be a radiu3,[ 
 and B is in the circumference. Q. e. d. '^" 
 
 Again, LM cannot have another point common with the circumfer- 
 ence, since if it had there could be more than two equal straight lines 
 drawn from to LM, which is impossible. Q. e. d. 
 
 155. Corollary. — Any line which is oblique to a radius 
 at its extremity is a secant line, since any such line has points 
 nearer the centre than the extremity of the radius, and hence 
 passes within the circumference. 
 
 PROPOSITION XI. 
 
 156. Theorem. — A straight line which is perpendicu- 
 lar to a radius at its outer extremity is tangent to the 
 circumference ; and, conversely, a tangent to a circumfer- 
 ence is perpendicular to a radius drawn to the point of 
 contact. 
 
 Demonstration. 
 
 A line perpendicular to a radius at its extremity touches the circum- 
 ference because the extremity of the radius is in the circumference. 
 
76 
 
 ELEMENTARY GEOMETRY. 
 
 Moreover, it does not intersect the circumference, since, if it did, it would 
 have points nearer the centre than the extremity of the radius ; but these 
 it cannot have, as the perpendicular is the shortest distance from a point 
 to a line. Hence, as a line which is perpendicular to a radius at its ex- 
 tremity touches the circumference but does not intersect it, it is a tan- 
 gent (30). Q. E. D. 
 
 Conversely, as a tangent to a circumference does not pass within, the 
 point of contact is the nearest point to the centre, and hence is the foot 
 of a perpendicular from the centre, q. e. d. 
 
 157. Corollary. — A perpendicular from the centre of 
 a circle to a tangent meets the tangent in the point of tan- 
 gencij (93). 
 
 PROPOSITION XII. 
 
 158. Theorem. — The arcs of a circumference inter- 
 cepted by two parallels are equal. 
 
 Demonstration. 
 
 There may be three cases, 1st. When one parallel Is a tangent and 
 the other a secant, as AB and CD ; 
 
 2d. When both parallels are secants, as CD and EF; and 
 3d. When both parallels are tangents, as AB and GH. 
 
 In the first case we are to prove 
 Ml = MK; in the second, IL = KR; 
 and in the third, MwiN = M;iN. 
 
 Through draw MN perpendicular 
 to one of the parallels, in any case, and 
 it will be perpendicular to the other 
 also (133); anfl as a perpendicular 
 from the centre upon a tangent meets 
 the tangent at the point of tangency 
 (157), M and N are points of tangency, 
 and MN is a diameter. 
 
 Now, since the parallels are perpendicular to MN, and the chords IK 
 and LR are bisected by it, if we fold the right-hand portion of the figure 
 on MN as an axis until it falls in the plane on the left of MN, K will fall 
 in I, and R in L. 
 
 Hence, Ml = MK, IL = KR, and MmN = MwN. Q. e. d. 
 
STRAIGHT LINES AND CIRCUMFERENCES. 
 
 77 
 
 PROPOSITION XIII. 
 159. Problem. — To bisect a given arc. 
 
 Solution. 
 
 Let ACB be the given arc. 
 
 We are to bisect it ; that is, find its mid- 
 dle point. 
 
 Draw the chord AB joining the extremi- 
 ties of the arc ; and bisect this chord by the 
 perpendicular 00' (101). Then will 00' bi- 
 sect the arc, as at C. 
 
 Demonstration of Solution. 
 
 Fig. 71. 
 
 00' being a perpendicular to the chord AB at its middle point, any 
 point in it is equally distant from the extremities. Hence chord BO - 
 chord AC, and arc BC = arc AC (151). Q- e. d. 
 
 PROPOSITION XIV. 
 
 160. Problem, — To find the centre of a circle whose 
 circumference is known, or of any arc of it. 
 
 Solution. 
 Let ACB be an arc of a circumference. 
 
 We are to find the centre of the 
 circle. 
 
 Draw any two chords of the arc, as 
 AC and BC, not parallel, and bisect each 
 by a perpendicular. Then will the in- 
 tersection of these perpendiculars, as 0, 
 be the centre of the circle. ^^^^^ 
 
 Fig. 72. 
 
 Demonstration op Solution. 
 
 OL being perpendicular to the chord AC at its centre, passes through 
 the centre of the circle, since if the centre were out of OL it would 
 be unequally distant from A and C (96). And for a similar reason, CM 
 
78 
 
 ELEMENTARY GEOMETRY. 
 
 being perpendicular to the chord BC at its centre, passes through the 
 centre of the circle. 
 
 Hence, as the centre of the circle lies at the same time in LO and MO, 
 it is their intersection 0. Q. e. d. 
 
 PROPOSITION XV. 
 
 161. Problem. — To pass a circumference through 
 three given points not in the same straight line. 
 
 SOLITTION. 
 
 Let A, B, and C be the three given points not in the same straight 
 line. 
 
 Join AB and BC. 
 
 Bisect AB by the perpendicular IVINj 
 (101), and BC by the perpendicular RS. 
 
 With 0, the intersection of MN and RS,| 
 as a centre, and any one of the distances 0A,| 
 OB, OC, say OA, as a radius, describe a cir- 
 cumference. 
 
 Then will this circumference pass] 
 through the three points A, B, and C. 
 
 Fig. 73. 
 
 Demonstration of Solution. 
 
 Since AB and BC are non-parallel by hypothesis, MN and RS are non- 
 parallel (141), and hence meet in some point, as (131). 
 
 Now as every point in MN is equally distant from the extremities of 
 AB (96), OA = OB. 
 
 In like manner, every point in RS is equally distant from B and C. 
 Hence, OB = OC. 
 
 Hence, OA = OB — OC, and a circumference struck from as a 
 centre, with a radius OA, will pass through A, B, and C. Q. e. d. 
 
 PROPOSITION XVI. 
 
 162. Problem. — To draw a tangent to a circle at a 
 given point in its circumference. 
 
STRAIGHT LINES AND CIRCUMFERENCES, 79 
 
 Solution. 
 
 Let it be required to draw a tangent to' 
 the circle whose centre is 0, at the point P 
 in its circumference. 
 
 Draw the radius OP, and produce it to 
 any convenient distance beyond the circle. 
 Through P draw MT perpendicular to OP. 
 Then is MT a tangent to the circle at P. 
 
 Demonstration of Solution. 
 
 Fig. 74. 
 
 MT being a perpendicular to the radius at its extremity, is a tangent 
 to the circle by (166). Q- b. d. 
 
 EXE RCISES. 
 
 163. 1. Draw a circle and divide it into two equal parts. 
 What proposition is involved ? 
 
 2. Given a point in a circumference, to find where a semi- 
 circumference reckoned from this point terminates. Wliat 
 proposition is involved ? 
 
 3. In a circle whose radius is 11 there are drawn two chords, 
 one at 6 from the centre, and one at 4. Which chord is the 
 greater ? By what proposition ? 
 
 4. In a certain circle there are two chords, each 15 inches in 
 length. What are their relative distances from the centre? 
 Quote the principle. 
 
 5. There is a circular plat of ground whose diameter is 20 
 rods. A straight path in passing runs within 7 rods of the cen- 
 tre. What is the position of the path with reference to the plat ? 
 What is the position of a straight path whose nearest point is 
 10 rods from the centre? One whose nearest point is 11 rods 
 from the centre ? 
 
 6. Pass a line through a given point, and parallel to a given 
 line, by the principles contained in (151), (147), (148), and (127). 
 
80 
 
 ELEMENTARY GEOMETRY, 
 
 OF THE RELATIVE POSITIONS OF CIRCUMFERENCES. 
 
 AXI OMS. 
 
 164. Two circles may occupy any one of five positions with 
 reference to each other : 
 
 1st. One circle may be wholly exterior to the other. 
 
 2d. One circle may be tangent to the other externally, the 
 circles being exterior to each other. 
 
 3d. One circumference may intersect the other. 
 
 4th. One circle may be tangent to the other internally. 
 
 5th. One circle may be wholly interior to the other. 
 
 PROPOSITION I. 
 
 165. Theorem. — WTien one circle is wholly exterior to 
 another, the distance between their centres is greater than 
 the sum of their radii. 
 
 Demonstration. 
 
 Let M and N be two circles 
 whose centres are and 0', 
 and whose radii are OA = It, 
 and O'B = r, respectively; and 
 let N be wholly exterior to M, 
 
 We are to prove that 00' 
 > R + r. 
 
 Draw 00', and let it inter- 
 sect circumference M in A, and 
 N in B. 
 
RELATIVE POSITIONS OF CIRCUMFERENCES, 
 
 81 
 
 Since N is wholly exterior to M, OB > OA. 
 
 Adding BO' to each memlxjr of this inequality, we have 
 OB + BO' > OA + BO', 
 or 00' > B + r, 
 
 since OB + BO' = 00', OA = i2, and OB = r. q. e. d. 
 
 PROPOSITION II. 
 
 166. Theorem. — When two circles arc tangent to each 
 other externally, 
 
 1st The distance between their centres is the sum of 
 their radii. 
 
 2d. They have a common rectilinear tangent at their 
 point of tangency. 
 
 3d. The point of tangency is in the straight line join- 
 ing their centres. 
 
 Demonstration. 
 
 Let M and N be two circles tangent to each other externally; let 
 and 0' be their respective centres, B and r their radii, D the point of 
 tangency, and TR a tangent to M at D. 
 
 prove, 1st. That 
 2d. That TR is 
 3d. That D is in 
 
 We are to 
 00' = B + r; 
 
 tangent to N ; 
 00'. 
 
 1st. Draw the radii OD = R, 
 and O'D = r. 
 
 If we show that OD + O'D = 
 R + r is the shortest path from 
 to 0', we show that it is a straight 
 line (59), and hence is the distance 
 from to 0' (95). 
 
 Consider any other path from to 0', crossing circumference N in 
 some other point than D, say in P. 
 
 Now the shortest path from to P is the straight line OP (59) ; and 
 the shortest path from P to 0' is the straight line PO'. Hence the 
 shortest path from to 0' passing through P is OP -f PO'. 
 
 Fig. 76. 
 
82 
 
 ELEMENTARY GEOMETRY. 
 
 But OP > i? (?)-*^, and PO' = r, whence OP + PO' > 5 + r. 
 
 Hence, as P is the point where 
 any other path from to 0' crosses 
 circumference N, OD + D'O = R 
 + r is the distance from to 0'. 
 
 Q. E. D. 
 
 2d. As TR is tangent to M at D, 
 by hypothesis, and as ODO' has 
 been shown to be a straight line, TR 
 is perpendicular to DO' (?) and 
 hence tangent to N (156). Q- e. d. 
 
 3d. As D is the point of tangency, and ODO is 00', D is in 00'. 
 
 Q. E. D. 
 
 Fig. 76. 
 
 PROPOSITION III. 
 
 167. Theorem.— Tw/o circumferences which intersect 
 in one point intersect also in a second point, and hence 
 have a cormnon chord. 
 
 DEMOi;rSTRATIOIT. 
 
 Let M and N be two circumferences intersecting in P. 
 
 Fig. 77. 
 
 We are to prove that they intersect in another point, as P', and hence 
 have a common chord PP'. 
 
 As M intersects N, it has points both without and within N. 
 
 Now consider the circumference M as generated by a point moving 
 from left to right, and let Y be a point within N. The generating point, 
 
 * Hereafter, minor references to principles on which a statement de- 
 pends will be omitted, and the interrogation mark substituted. This indi- 
 cates that the student is to give the principle. In this case. P is without M 
 since by hypothesis N is external to M. 
 
RELATIVE POSITIONS OF CIRCUMFERENCES. 
 
 83 
 
 in passing from Y, a point within N, to X, any point in the circumference 
 M without N, must cross circumference N at some point, as P', since this 
 is a closed curve. 
 
 Moreover, this second point, P', is a different point from P, since a 
 circumference of a circle does not cut itself, or become tangent to itself. 
 
 Hence, if circumference M cuts circumference N in P, it cuts it also in 
 a second point, as P'. (j. e. d. 
 
 Finally, since P and P' are common to both circumferences, the 
 circles M and N have a common chord PP'. q. e. d. 
 
 PROPOSITION IV 
 
 168. Theorem. — Wken two circumferences intersect, 
 1st. The line joining their centres is perpendicular to 
 
 their common chord at its middle point. 
 
 2d. The distance between their centres is less than the 
 
 sum, of their radii and greater than their difference. 
 
 Demonstration. 
 
 Let M and N be two circumferences intersecting at P and P' ; let 
 and 0' be their centres, and li and r their radii respectively, li being 
 equal to or greater than r. 
 
 We are to prove, 1st. That 
 00' is perpendicular to PP' at 
 its middle point ; and 2d. That 
 00' < R -k- r, and 00' > 
 R-r. 
 
 Draw OP and O'P. 
 
 Ist. Since is equally dis- 
 tant from P and P', and 0' is 
 also equally distant from P and 
 P' (?) , 00' is perpendicular to 
 PP' at its middle point (98). 
 
 Q. E. D. 
 
 2d. As P is not in 00', 00' < 
 Again, 00' + 
 
 OP + PO' (?), or 00' < 
 OP > OP, 
 
 R -^ r. 
 
 or 00' + r > R\ 
 
 whence, subtracting r from each member, 
 
 00' > iJ - r. 
 
 Q. E. D. 
 
u 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION V. 
 
 169. Theorem. — WTien the less of two circles is tangent 
 to the other internally, 
 
 1st. They have a common rectilinear tangent at the 
 point of tangency. 
 
 2d. Their centres and the point of tangency lie in the 
 same straight line. 
 
 3d. The distance between the cerCtres is equal to the dif- 
 ference of their radii. 
 
 Demonstration-. 
 
 Let M and N be two circles whose centres are and 0' respectively, 
 N being less than M and tangent to it internally; let H and r be their 
 radii, and D the point of tangency. 
 
 We are to prove, 1st. That they have 
 a coramon rectilinear tangent at D ; 3d. 
 That 0, 0', and D are in the same straight 
 line ; and 3d. That 00' = i^ — r. 
 
 1st. Draw TR tangent to IVI at D. 
 Draw also O'D, and any other line from 0' 
 to TR, as O'E. 
 
 Now, since E is without the circle IVI 
 (?), and IVI is without N (?), O'E > O'D, 
 and O'D is perpendicular to TR (94). 
 
 Hence, TR is tangent to N (156), and 
 is therefore a common tangent, q. e. d. 
 
 2d. Since both OD and O'D are perpendicular to TR at D (?), OD 
 and O'D coincide (88), and and 0' lie in the same straight line with D. 
 
 Q. E. D. 
 
 3d. Since 00' and D are in the same straight line, and 0' is between 
 and D, 00' = OD — O'D ; that is, 00' = -B — r, q. e. d. 
 
 Fig. 79. 
 
RELATIVE POSITIONS OF CIRCUMFERENCES. 85 
 
 PROPOSITION VI. 
 
 170. Theorem. — WTien the less of two circles is wholly 
 interior to the other, the distance between the centres is 
 less than the difference of their radii. 
 
 Demonstration. 
 
 Let M and N be two circles whose centres are and 0', and whose 
 radii are R and r respectively, and let N be wholly within M. 
 
 We are to prove that 00' < 22 — r. 
 
 Produce 00' till it meets both circumferences 
 on the same side of that 0' is ; and let the inter- 
 sections with N and M respectively be D and E. 
 
 Then, as 0, 0', D, and E lie in order in the 
 same straight line, 
 
 OD < OE; J-, 
 
 and subtracting O'D from each, and noticing that 00 — O'D = 00', 
 that OE = i?, and O'D = r, we have 
 
 00' < B—r. Q. E. D. 
 
 171. General Scholium. — The converse of each of Props. 1, 11, IV, 
 V, and VI is also true. Thus, if the distance between the centres is 
 greater than the sum of the radii, the circles are wholly exterior the one 
 to the other ; since if they occupied any one of the other four possible 
 positions, the distance between the centres would be equal to the sum of 
 the radii, less than their sum, equal to their diflference, or less than their 
 diflference ; any one of which conclusions vrould be contrary to the hy- 
 pothesis. 
 
 In like manner, the converse of any one of the five propositions may 
 be proved. 
 
 This method of proof is called The Reductio ad Absurdum, and 
 consists in showing that any conclusion other than the one stated would 
 lead to an absurdity. 
 
ELEME^TAR Y GEOMETR Y, 
 
 PROPOSITION VII. 
 
 172. Theorem. — All the circumferences which can be 
 passed through three points not in the same straight line 
 coincide, and are one and the same. 
 
 Demonstration. 
 Let A, B, and C be three points not in the same straight line. 
 
 We are to prove that all tb& circnmferences 
 which can be passed through them coincide, and 
 are one and the same circumference. 
 
 By (161) a circumftrence can be passed through 
 A, B, and C. 
 
 Now every point equally distant from A and B 
 lies in FD, a perpendicular to AB at its middle 
 point (?). And, in like manner, every point equally 
 distant from B and C is in HE, a perpendicular to 
 BC at its middle point. 
 
 But the two straight lines FD and HE can intersect in only one point. 
 
 Hence all circumferences which can pass through A, B, and C have 
 their centre in 0, and their radius OA, and therefore they constitute one 
 and the same circumference, q. e. d. 
 
 173. Cor. 1. — Through any three points not in the sam^e 
 straight line a circumference can he passed, and hut one. 
 
 174. Definition. — A circle is said to be determined when 
 the position of its centre and the length of its radius are known. 
 
 175. Cor. 2. — Three points not in the same straight line 
 determ^ine a circle. 
 
 176. Cor. 3. — Two circumferences can intersect in only 
 two points. 
 
 For, if they have three points common, they coincide, and form one 
 and the same circumference. 
 
RELATIVE POSITIONS OF CIRCUMFERENCES, 
 
 87 
 
 EXERCISES. 
 
 177. 1. The centres of two circles whose radii are 10 and 7, 
 are at 4 from each other. What is the relative position of the 
 circumferences ? What if the distance between the centres is 17 ? 
 What if 20 ? What if 2 ? What if ? What if 3 ? 
 
 2. Given two circles and 0' (Fig. 82), to draw two others, 
 one of which shall be tangent to these externally, and to the 
 other of which the two given circles shall be tangent internally. 
 Give all the principles involved in the construction. Give other 
 methods. 
 
 Fig. 82. 
 
 Fig. 83. 
 
 3. Given two circles whose radii are 6 and 10, and the dis- 
 tance between their centres 20. To draw a third circle whose 
 radius shall be 8, and which shall be tangent to the two given 
 circles. Can a third circle whose radius is 2 be drawn tangent to 
 the two given circles ? How will it be situated ? Can one be 
 drawn tangent to the given circles, whose radius shall be 1? 
 Why? 
 
 4. With a given radius, draw a circumference (Fig. 83) which 
 shall pass through a given point and be tangent to a given line. 
 
ELEMENTABr GEOMETRY, 
 
 ^^^tvnn^ yi 
 
 OF THE MEASUREMENT OF ANGLES. 
 
 178. Two angles are Commensurable when there is a 
 common finite angle which measures each. When they have no 
 such common measure, they are Incommensurable. 
 
 179. An Angle at the Centre is an angle included be- 
 tween two radii. 
 
 180. An Inscribed Angle is an angle whose vertex is in 
 a circumference, and whose sides are chords of that circumfer- 
 ence. 
 
 181. Angles are said to be measured by arcs, according to the 
 principles developed in the following f)ropositions. 
 
 PROPOSITION I. 
 
 182. Theorem. — In the same circle, or in eqnal circles, 
 two angles at the centre are in the same ratio as the arcs 
 intercepted between their sides. 
 
 Demonstration. 
 There are three cases : 
 
 CASE I. 
 When the angles are equal. 
 
 Let angle AOB = angle DOE (Fig. 84) in the same circle or in equal 
 circles. 
 
MEASUREMENT OF ANGLES, 
 
 89 
 
 We are to prove that 
 
 AOB _ arc AB * 
 DOE arc DE 
 
 Apply tbe angle DOE 
 to the angle AOB, placing 
 the radius OD in its equal 
 OA. By reason of the 
 equality of the angles 
 DOE and AOB, OE will 
 fall in OB, and E in B (?). 
 
 Hence DE coincides with AB, and 
 
 arc AB 
 arc DE 
 
 Fig. 84. 
 
 = 1. 
 
 But, by hypothesis, 
 Hence, 
 
 AOB 
 DOE 
 
 AOB 
 DOE 
 
 = 1. 
 
 arc AB 
 arc DE 
 
 (66). Q. E. D. 
 
 CASE II. 
 
 When tJie angles are commensurable. 
 
 Let AOB and DOE be two commensurable angles at the centre in the 
 same circle^ or in equal circles. . 
 
 Fig. 85. 
 
 * This method of writing a proportion is ado^ited in this book as the 
 more elegant, and as it appears to be coming into exclusive use. The above 
 is the same as 
 
 AOB : DOE :: arc AB : arc DE 
 and is to be read in the same manner. 
 
90 
 
 ELEMENTARY GEOMETRY, 
 
 Fig. 85. 
 
 We are to prove that 
 
 AOB _ arc AB 
 DOE ~ arc DE * 
 
 As the angles are commensurable by hyp()thesis, let m be their com- 
 mon measure, and let it be contained 5 times in AOB and 8 times in DOE. 
 
 8o that 
 
 AOB 5 
 
 DOE 8* 
 
 Conceive the angle AOB divided into 5 partial angles, each equal to 
 
 m, and the angle DOE divided into 8 such partial angles. 
 
 Now as these partial angles are equal, their intercepted arcs are equal 
 
 (?), and as AB contains 5 of them, and DE 8, 
 
 arc AB _ 5 
 
 Hence, 
 
 arc DE 
 AOB 
 DOE 
 
 arc AB 
 arc DE 
 
 (?). q. E. D. 
 
 CASE III. 
 
 When the angles are incoinmensurahle. 
 
 Let AOB and DOE (Fig. 86) be two incommensurable angles at the 
 centre, in the same circle, or in equal circles. 
 
 «r . 1- . AOB arc AB 
 
 We are to prove that ^^ = ^^^-^ ' 
 
 If the ratio ^ is not equal to the ratio ^^5^, let it be ^rea<^r ; 
 
 and let 
 
 in which DL is less than DE- 
 
 AOB _ arc AB 
 DOE ~ arcDL' 
 
MEASUREMENT OF ANGLES. 
 
 91 
 
 Fig. 86. 
 
 Draw OL, and divide AOB into equal parts, each less than LOE. 
 Apply this measure to DOE, beginning at DO. At least one line of di- 
 vision win fall between OL and OE. Let this be OK. 
 
 Now AOB and DOK are commensurable ; hence, by Case II, 
 
 but by hjrpothesis 
 
 AOB 
 DOK 
 
 AOB 
 DOE 
 
 arc AB 
 arc DK * 
 
 arc AB 
 arc DL* 
 
 -^. .,. AOB , AOB , arc AB , arcAB 
 Dmdmg DOK ^y DOE' ^"^ ai^DK ^^ ar^DL' ^" ^*^^ 
 
 DOE 
 DOK 
 
 arc DL 
 
 arc DK 
 But this conclusion is absurd, since 
 
 DOE 
 DOK 
 
 > 1, and 
 
 arc DL 
 arcDK 
 
 < 1. 
 
 AOB 
 
 Thus we show that the ratio ^-^ cannot be greater than the ratio 
 
 — =^p ; and in a similar manner we may show that -r^r^ cannot be less 
 arc DE ^ DOE 
 
 than 
 
 arc AB 
 arc^E 
 
 AOB . 
 DOE 
 
 Hence, as '^^ is neither greater nor less than ^^^^^ , it is equal to 
 
 arc AB 
 arc DE 
 
 arc AB . , AOB arc AB 
 
 =r^, and we have =r;^=. = ^^f. 
 
 arc DE ' DOE arc DE 
 
 Q. E. D. 
 
 [For other methods of demonstrating this important theorem, see 
 Appendix.] 
 
92 ELEMENTARY GEOMETRY. 
 
 188. Out of the truth developed in this proposition grows the 
 method of representing angles by degrees, minutes, and seconds, as given 
 in Trigonometry (Part IV, 3-6)- It will be observed, that in all cases, if 
 arcs be struck with the same radius^ from the vertices of angles as centres, 
 the angles bear the same ratio to each other as the arcs intercepted by 
 their sides. Hence the arc is mid to measure the angle. Though this lan- 
 guage is convenient, it is not quite natural ; for we naturally measure a 
 quantity by another of like kind. Thus, distance (length) we measure by 
 distance., as when we say a line is 10 inches long. The line is length ; and 
 its measure, an inch, is length also. So, likewise, we say the area of a 
 field is 4 acres: the quantity measured is a surface; and the measure, an 
 acre, is a surface also. Yet, notwithstanding the artificiality of the 
 method of measuring angles by arcs, instead of directly by angles, it is 
 not only convenient but universally used ; and the student should know 
 just what is meant by it. 
 
 189. A Degree is ^^ part of the circumference of a circle; a 
 Minute is ^^ of a degree, and a Second is ^V of a minute. This is the 
 primary signification of these terms. But as any angle at the centre sus- 
 tains the same ratio to any other angle at the centre as do their subtended 
 arcs, we speak of an angle as an angle of so many degrees, minutes, 
 and seconds. Thus, an angle of 45 degrees (written 45°) means an angle 
 at the centre 45 times as large as one which subtends ^^ of the circumfer- 
 ence, or half as large as one which subtends 90° of the circumference. 
 
 This idea, as well as the notation °, ', ", for degrees, minutes, and 
 seconds, has already been made familiar in Arithmetic. 
 
 190. As the vertex of any angle may be conceived as the centre of a 
 circle, the intercepted arc of whose circumference measures the angle, we 
 speak of all angles in the same manner as of angles at the centre. Thus, 
 a right angle is called an angle of 90°. one-half a right angle is an angle 
 of 45°, a straight angle is an angle of 180°, and the sum of four right 
 angles, being measured by the entire circumference, is an angle of 360°, etc. 
 
 PROPOSITION II. 
 
 191. Theorem. — An inscribed angle is measured by 
 half the arc intercepted between its sides. 
 
 Demonstration. 
 Let APB be an angle inscribed in a circle whose centre is 0. 
 
MEASUREMENT OF ANOLES. 
 
 J 
 
 We are to prove that tlie angle APB is measured by one-half the 
 arc AB. 
 
 There are three cases : 1st. When the centre is in one side ; 2d. When 
 the centre is witliin the angle; and 3d. When it is without. 
 
 CASE I. 
 When the centre, 0, is in one side, ds PB, 
 
 Draw the diameter DC parallel to AP. 
 By reason of the parallels AP and CD, 
 arc AC = arc PD (158) ; 
 and, since COB = POD (?), 
 
 arc CB = arc PD (?). 
 Hence, arc AC = arc CB, 
 and arc CB = i arc AB. ^, „ 
 
 * Fig. 87. 
 
 Again, since the parallels AP and DC are cut by the transversal PB, 
 the angles APB and COB are equal (126). 
 
 But COB is measured by arc CB (?). Hence, APB is measured by 
 arc CB = ^ arc AB. Q. e. d. 
 
 CASE II. 
 When the centre is within the angle. 
 
 Draw the diameter PC. 
 
 Now by Case I, APC is measured by ^ arc AC, 
 and CPB by ^ arc CB. Hence the sum of these 
 angles, or APB, is measured by ^ arc AC -I- ^ arc 
 CB, or ^ arc AB. q. e. d. 
 
 CASE III. 
 Wlien the centre is withont the angle. 
 
 Draw the diameter PC. 
 
 By Case I, APC is measured by ^ arc AC, and 
 BPC by ^ arc BC. Hence, APB, which is APC — 
 BPC, is measured by 
 
 ^ arc KQ — ^ arc BC 
 
 or \ arc AB. Q. e. d. 
 
 Fig. 88. 
 
 Fig. 89. 
 
94 ELEMENTARY GEOMETRY, 
 
 192. Corollary. — In the same circle or in equal circles, 
 all angles inscribed in the same segment or in equal seg- 
 ments intercept equal arcs, and are consequently equal. 
 If the segment is less than a semicircle, the angles are 
 obtuse; if a semicircle, right; if greater than a semi- 
 circle, acute. 
 
 Fig. 90. 
 
 Illustration. — In each separate figure the angles P are equal to 
 each other, for they are each measured by half the same arc. 
 
 In 0, each angle P is acute, being measured by ^m, which is less than 
 a quarter of a circumference. 
 
 In 0', each angle P is a right angle, being measured by ^m', which is 
 a quadrant (quarter of a circumference). 
 
 In 0", each angle P is obtuse, being measured by ^m", which is 
 greater than a quadrant. 
 
 PROPOSITION III. 
 
 193. Theorem.— Any angle formed by two chords in- 
 tersecting in a circle is measured by one-half the sum of 
 the arcs intercepted between its sides and the sides of its 
 vertical, or opposite, angle. 
 
 Demonstration. 
 
 Let the chords AB and CD (Fig. 91) intersect in P. 
 
 We are to prove that angle APD (= angle CPB ?) is measured by 
 
 ^ (arc AD + arc CB) ; 
 
 and that angle BPD (= angle CPA ?) is measured by 
 
 ^ (arc BD + arc CA). 
 
MEASUnSMENT OF ANGLES. 95 
 
 •I Draw CE parallel to AB. 
 
 Arc AE = arc CB (?) ; whence, arc ED = 
 arc AD + arc CB. 
 
 Now the inscribed angle ECD is measured 
 by I arc ED = 1^ (arc AD + arc CB). 
 
 But ECD = APD (0 ; hence, APD (= CPB) 
 is measured by |^ (arc AD 4- arc CB). q. e. d. 
 
 Finally, that ARC, or its equal BPD, is 
 measured by | (AC + BD), appears from the f^'Q- ^•• 
 
 fact that the sum of the four angles about P 
 
 being equal to four right angles, is measured by a whole circumference 
 (190). 
 
 But APD + CPB is measured by AD + CB ; whence APC + BPD, or 
 2APC, is measured by the whole circumference minus (AD + CB); that 
 is, by AC + BD. Hence APC is measured by \ (AC 4- BD). q. e. d. 
 
 194. Scholium. — The case of the angle included between two chords 
 passes into that of the inscribed angle in the preceding proposition, by 
 conceiving AB to move parallel to its present position until P arrives at 
 C and BA coincides with CE. The angle APD is all the time measured 
 by half the sum of the intercepted arcs; but, when P has reached C, CB 
 becomes 0, and APD becomes an inscribed angle measured by half its in- 
 tercepted arc. 
 
 In a similar manner we may pass to the case of an angle at the centre, 
 by supposing P to move toward the centre All the time APD is meas- 
 ured by ^(AD + CB); but, when P reaches the centre, AD = CB, and 
 |(AD + CB) = ^ (2 AD) = AD ; i. e., an angle at the centre is measured 
 by its intercepted arc. 
 
 PROPOSITION IV. 
 
 196. Theorem. — An angle included between two se- 
 cants meeting iidthout the circle is measured by one-half 
 the difference of the intercepted arcs. 
 
 Demonstration. 
 
 Let APB (Fig. 92) be an angle included between the secants PA 
 and PB ; and let the intersections with the circumference be C and D. 
 
96 
 
 ELEMENTARF GEOMETRY. 
 
 We are to prove that APB is measured by 
 ^ (arc AB — arc CD). 
 
 Draw CE parallel to PB. 
 
 Now arc CD = arc EB (?). Hence, arc AE 
 = arc AB — arc CD. 
 
 Again, ACE = APB (0- 
 
 But ACE is measured by | arc AE (?). 
 Hence APB is measured by 
 
 ^ arc AE = ^ (arc AB — arc CD), q. e. d. 
 
 196. Scholium. — This case passes into that 
 of an inscribed angle, by conceiving P to move Fig. 92. 
 
 toward C, thus diminishing the arc CD. When 
 
 P reaches C, the angle becomes inscribed ; and as CD is then 0, ^ (AB — 
 CD) = jr AB. Also, by conceiving P to continue to move along PA, CD 
 will reappear on the other side of PA, hence will change its sign,* and 
 |(AE — CD) will become ^ (AE + CD), as it should, since the anglejs 
 then formed by two chords intersecting within the circumference. 
 
 PROPOSITION V. 
 
 197. Theorem. — ^11 equal angles whose sides inter- 
 cept a given line, and wJiose vertices lie on the same side 
 of that line, are inscribed in the same segment of which 
 the intercepted line is the chord. 
 
 Demonstration. 
 
 Let APB, APB, AP"B, etc., be any number of 
 equal angles whose sides intercept the given line AB. 
 
 We are to prove that the vertices P, P', P", etc., 
 all lie in the same arc of which AB is the chord. 
 
 Through one of the vertices, as P, and A and B 
 describe a circumference. 
 
 Now the angle APB is measured by \ the arc 
 AwB, and as the other angles are equal to this, they 
 must have the same measure. 
 
 * In accordance with the law of positive and negative quantities as used 
 in mathematics, whenever a continuously varying quantity is conceived as 
 diminishing till it reaches 0, and then as reappearing by the same law of 
 change, it must change its sign. 
 
 Fig. 93. 
 
MEASUREMENT OF AXGLES. 
 
 97 
 
 But suppose any one of them, as P', had its vertex within the 
 segment. It would then be an angle included between two chords 
 drawn from A and B, and hence would be measured by ^AmB plus some 
 arc (193). 
 
 If, on the other hand, the vertex P' was without the segment, the 
 angle would be an angle included between two secants, and would be 
 measured by ^A^nB less some arc (195) 
 
 Hence, as P' can lie neither without nor within the arc APB, it lies in 
 it. Q. E. D. 
 
 198. Corollary. — All right angles whose sides inter- 
 cept a given line are inscribed in a semicircle whose 
 diameter is the given line. 
 
 PROPOSITION VI. 
 
 199. Theorem. — An angle included between a tan- 
 gent and a chord drawn from the point of tangency is 
 measured by one-half the intercepted, arc. 
 
 Demonstration. 
 
 Let TPA be an angle included be- 
 tween the tangent TM and the chord 
 PA. 
 
 We are to prove that TPA is mea- 
 sured by \ arc PnA. 
 
 Through A draw the chord AD 
 parallel to TM. 
 
 Then is PAD = TPA (?). 
 
 Now PAD is measured by \Pmli (?). 
 
 Whence TPA is measured by 
 ^PmD. But PwD equals PnA (?). 
 
 Hence TPA is measured by ^PnA. 
 
 Q. E. D. 
 
 Fig. 94. 
 
 Exercise. — Show that APM is measured by \ arc AwP. 
 
 Also, observe how the case of two secants (195), passes into this. 
 
98 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION VII. 
 
 200. Theorem, — J_n angle included between two tan- 
 gents is measured by one-half the difference of the inter- 
 cepted arcs. 
 
 DEMONSTRATrON. 
 
 Let APB be an angle included between 
 the two tangents PA and PB, tangent at 
 C and D. 
 
 We are to prove that APB is measured 
 
 by 
 
 ^ (arc CmD — arc CnD). 
 
 Draw the chord CE parallel to PB. 
 
 Now arc CnD = arc EwD (?). 
 
 Whence arc CE = arc CwD — arc OnD, 
 
 Again, ACE = APB (?). 
 
 But ACE is measured by \ arc CE = | 
 (arc CwiD — arc CnD). Hence APB is measured by ^ (arc CwD — arc 
 CwD). Q. E. D. 
 
 201. Scholium. — The case of two secants (195) becomes this by sup- 
 posing the secants to move parallel to their first position till they botk 
 become tangents. 
 
 Fig. 95. 
 
 PROPOSITION VIII 
 
 202. Tlieorem. — An angle 
 included between a secant and a 
 tangent is measured by one-half 
 the difference of the intercepted 
 arcs. 
 
 [Let the student write out the demon- 
 stration in form.] 
 
 Fig. 96. 
 
MEASUREMENT OF ANGLES. 
 
 99 
 
 PROPOSITION IX. 
 
 203. Problem, — From a given point in a given line to 
 draw a line luhich shall mahe with the given line a given 
 angle. 
 
 SOLUTIOiT. 
 Let A be the given point in the given line AB, and the given angle. 
 
 We are to draw from A a 
 line which shall make with 
 AB an angle equal to 0. 
 
 From as a centre, with 
 any convenient radius, de- 
 scribe an arc, as a6, measuring 
 the angle 0. 
 
 From A as a centre, with 
 the same radius, describe an 
 
 arc en cutting AB and extend p. ^^ 
 
 ing on that side of AB on 
 which the angle is to lie. Let this arc intersect AB in c. 
 
 From c as a centre, with a radius equal to the chord oJ, describe an 
 arc cutting cw, as at d. 
 
 From A draw a line through <Z, as AC. 
 
 Then will CAB be the angle required. 
 
 Demonstration of Solution. 
 
 Arc ab measures angle (?). 
 
 Arc cd = arc ah (?). 
 
 Hence, angle CAB = angle (?). 
 
 PROPOSITION X. 
 
 204. Problem. — Through a given point to draw a par- 
 allel to a given line, 
 
 SOLLTION. 
 Let P (Fig. 98) be the given point, and AB the given line. 
 
100 
 
 ELEMENTARY GEOMETRY. 
 
 We are to draw a line through 
 P which shall be parallel to AB. 
 
 From P as a centre, with any 
 i-adius sufficiently great, strike an 
 arc cutting AB, as at a, and ex- 
 tending on the same side of AB 
 that the parallel is to lie. Let the arc be ac. 
 
 From a as a centre, with the same radius, pass an arc through P, cut- 
 ting AB in some point, as &. 
 
 With the chord IP as a radius and a as a centre, strike an arc cutting 
 «c, as in 0. 
 
 Draw a line through and P, and it will be the parallel required. 
 
 Demonstratioi?^ of Solution. 
 
 The arcs Oa and P6 are arcs of circles with equal radii, and have 
 equal chords, and are hence equal arcs (?). 
 
 Tiie angles OPa and Pab are equal, since they are measured by the 
 equal arcs Oa and P6 (?). 
 
 Hence the transversal Pa cuts the two lines MN and AB, making the 
 alternate angles MP« and BaP equal. Wherefore MN is parallel to AB, 
 and as it passes through the given point P, it is the parallel required. 
 Q. E. D. 
 
 PROPOSITION XI. 
 
 205. Problem. — From a point without a circle to draw 
 a tangent to the circle. 
 
 Solution. 
 
 Let O be the centre and OT the radius 
 of the given circle, and P the given point. 
 
 We are to draw from P a tangent to 
 the circle. 
 
 Join P with the centre by a straight 
 line. 
 
 On the line OP describe a circle inter- 
 secting the given circle in T and T'. 
 
 Through the points P and T, P and T' 
 draw the straight lines PM and PM'. These will be the required tangents. 
 
 Fig. 99. 
 
MEASUREMENT OF ANGLES, 
 
 101 
 
 Demonstration of Solution. 
 
 Drawing OT, the angle OTP is a riglit angle, since it is inscribed in a 
 semicircle (192). 
 
 Hence PM is a tangent to the circle, as it is a perpendicular to a radius 
 at its extremity, and as it passes through P it fulfills the conditions of the 
 problem. 
 
 In like manner, PM' is seen to be a tangent passing through P, and 
 the problem has two solutions. Q. e. d. 
 
 206. Corollary. — Through any point mithout a circle 
 two tangents may he drawn to the circle. 
 
 PROPOSITION XII. 
 
 207. Problem. — On a given line to construct a segment 
 which shall contain a given inscribed angle. 
 
 Solution. 
 
 Let AB be the given line and the given angle. 
 
 We are to construct a segment 
 on AB which shall contain the as 
 an inscribed angle. 
 
 At one extremity of AB, as B, 
 construct an angle ABC equal to 0, 
 and on the side of AB opposite to 
 that on which the segment is to lie. 
 
 Erect a perpendicular to CB at 
 B, and one to AB at its middle point 
 E. Let F be the intersection of these Fig. lOO. 
 
 perpendiculars. 
 
 With FB (or FA) as a radius, describe a circle. Then will AWw"B be 
 the segment required ; and any angle inscribed in this segment, as AHB, 
 will be equal to 0. 
 
 Demonstration of Solution. 
 
 Since CB and AB are non-parallel lines, perpendiculars erected to 
 them will meet in some point as F (141) 131)- 
 
102 
 
 ELEMENTARY GEOMETRY, 
 
 F being a point in the perpendicu- 
 lar to AB at its middle point FA = 
 FB (96), and a circle struck with FB 
 as a radius and F as a centre will 
 pass through A. Moreover CB will 
 be a tangent to this circle, since it is 
 perpendicular to a radius at its ex- 
 tremity (156). 
 
 Now = ABC by construction, 
 and ABC being an angle included 
 between a tangent and a chord, is 
 measured by half the intercepted arc AwB (?). 
 
 But any angle inscribed in the segment Am'm"B is measured by ^ arc 
 AwiB (0, and hence ijquals ABC =,0. Q. e. d. \ ^ 
 
 
 cr.>L L 
 
 
 PROPOSITION XIII 
 208. Problem. — To bisect a given angle. 
 
 Solution. 
 
 Let BOA be the given angle. 
 
 We are to draw a line dividing BOA 
 into two equal angles. 
 
 With any convenient radius and as 
 a centre, describe an arc cutting the sides 
 OB and OA at h and a. 
 
 From a and & as centres, with equal 
 radii, strike arcs cutting in some point, 
 as P. 
 
 Through and P draw a straight line. 
 
 Then is the angle BOA bisected by OP, and BOP = POA. 
 
 
 Fig. lOf. 
 
 Demonstration of Solution. 
 
 OP being perpendicular to the chord of arc ab (?) bisects the arc 
 (147). Hence arc &D = arc aO. 
 
 But »°gl«BOP=''J^*° Therefore, BOP = POA. <i. e. d. 
 angle POA arc aD 
 
MEASUREMENT OF ANGLES, 
 
 103 
 
 EXE RCISES 
 
 209. 1. To find a point in a plane having given its distances 
 from two known points. 
 
 When are there two sohitions? 
 When but one solution ? 
 When no solution ? 
 
 2. In Fig. 102 there are 4 pairs of equal angles. Which are 
 they, and why ? 
 
 Show that COB = ABD + CDB. 
 Show that DOB = ABC + DAB. 
 
 Fig. 102. Fig. 103. 
 
 210. Concentric Circles are circles which have a com- 
 mon centre. 
 
 3. Draw two concentric circles (Fig. 103), such that the 
 chords of the outer circle which are tangent to the inner shall 
 be equal to the diameter of the inner. 
 
 4. From a point out of a given straight line to draw a line 
 making a given angle with the first line. 
 
 5. Prove that if two circles are concentric, any chord of the 
 outer which is tangent to the inner is bisected at the point of 
 contact. 
 
 6. Prove that if D and B (Fig. 104) are right angles, A and C 
 are supplementary. 
 
104 
 
 ELEMENTARY GEOMETRY. 
 
 7. Prove that if, in the adjoining 
 figure, the opposite sides AB and DC, 
 and AD and BC be produced till they 
 meet, the lines which bisect the in- 
 cluded angles will be perpendicular to 
 each other. 
 
 8. Draw a triangle, and then draw 
 a circle about it so that all its angles 
 shall be inscribed ; L e., circumscribe 
 a circle about a triangle. (See 161.) 
 
 Fig. 104. 
 
 ♦ <» 
 
 ^^^irCTfoJI VII. 
 
 OF THE ANGLES OF POLYGONS, AND THE RELATION 
 ' BETWEEN THE ANGLES AND SIDES. 
 
 OF TRIANGLES. 
 
 211. A Plane Triangle, or simply a Triangle, is a plane 
 figure bounded by three straight lines. 
 
 212. With respect to their sides, triangles are distin- 
 guished as Scalene, Isosceles, 
 and Equilateral. 
 
 A Scalene Triangle is 
 
 a triangle which has no two Fig. los. 
 
 sides equal, as (1) or (2). ^^^^^^ 
 
 An Isosceles Triangle is a triangle HHRB 
 which has two of its sides equal to each other, ^E^^^^k 
 
 as (3). Fig. 106. 
 
TRIANGLES. 
 
 105 
 
 An Equilateral Triangle is a triangle 
 which has all tliree of its sides equal each to each, 
 as (4). 
 
 213. With respect to their angles, triangles ^'9 io7. 
 are distinguished as acute angled, right angled, and obtuse 
 angled. 
 
 An Acute Angled Triangle is a triangle all of whose 
 angles are acute, as (4). 
 
 A Right Angled Triangle is a triangle one of whose 
 angles is right, as (2). 
 
 An Obtuse Angled Triangle is a triangle one of whose 
 angles is obtuse, as (1). 
 
 214. A circle Circumscribes a figure when all the angles 
 of the latter are inscribed. 
 
 PROPOSITION I. 
 
 215. Theorem. — The sum of the three angles of a tri- 
 angle is two right angles. 
 
 Demonstration. 
 
 Let ABC be any triangle. 
 We are to prove that 
 
 A + B + C = 2 right angles. 
 
 Circumscribe a circle about the triangle (161)- 
 
 Then the angle A is measured by ^ the arc 
 BaC (0, the angle B by | the arc C&A, and the 
 angle C by ^ the arc AcB. 
 
 Hence the sum of the three angles, or A + B + f^'9- '08. 
 
 C. is measured by ^ the sum of BaC + C&A 4 AcB, or ^ the circumference. 
 
 But a semi-circumference is the measure of two right angles (190)- 
 Hence A + B + C = 2 right angles, q. e. d. 
 
106 ELEMENTARY GEOMETRY. 
 
 216. CoROLLAKY 1. — A triangle can have only one right 
 angle, or one obtuse angle. Why ? 
 
 217. Corollary 2.— Two angles of a triangle, or their 
 sum, being given, the third may be found by subtracting 
 this sum from two right angles, i. e., any angle is the 
 supplement of the sum of the other two. 
 
 218. Corollary 3. — The sum> of the two acute angles of 
 a right-angled triangle is equal to one right angle ; i. e., 
 they are complements of each other. 
 
 219. Corollary 4. — If the angles of a, triangle are 
 equal each to each, any one is one-third of two right an- 
 gles, or two-thirds of one right angle. 
 
 PROPOSITION II. 
 
 220. Theorem.— IT^e sides of a triangle sustain the 
 same general relation to each other as their opposite an- 
 gles; that is, the greatest side is opposite the greatest 
 angle, the second greatest side opposite the second grea^test 
 angle, and the least side opposite the least angle. 
 
 Demonstration. 
 
 Let ABC be any triangle having the angle C greater than B. and B 
 greater than A. 
 
 We are to prove that AB opposite C is the 
 greatest side, AC opposite B the next greatest, 
 and BC opposite A the least. 
 
 Circumscribe a circle about the triangle (161). 
 
 If the triangle is acute-angled, the arc meas- 
 uring any angle is less than a quarter of a circum- 
 ference (191). 
 
 Now the angle C being greater than B, the 
 arc c is greater than arc & (?). Hence, the chord Fig. I09. 
 
 AB is greater than the chord AC. 
 
TRIANGLES. 
 
 107 
 
 In like manner, the angle B being greater than the angle A, the arc 
 I is greater thaa^ arc a (?). Hence the chord AC is greater than the 
 chord BC. 
 
 If the triangle has one right angle, as C, Fig. 110, this angle is 
 measured by ^ the semi-circumference AcB, and inscribed in the semi- 
 circumference ACB. Hence the order of magnitude of the arcs is still 
 c>l>a (?), and of the sides AB > AC > BC. 
 
 Fig. 110. 
 
 Fig. III. 
 
 If any angle of the triangle, as C, is obtuse. Fig. Ill, this angle is in- 
 scribed in a segment less than a semicircle (192), whence this arc ACB is 
 less than a semi-circumference, and greater than either a or J, as it is their 
 sum. 
 
 Hence the chord AB is greater than either AC or BC (?). 
 
 Thus we have shown that in all cases, the order of magnitude of the 
 angles being C > B > A, the order of magnitude of the sides is 
 AB > AC > BC. Q. E. D. 
 
 221. Corollary 1. — Conversely, TJie order of the Tnag- 
 nitudes of the sides being AB > AC > BC, the order of the 
 magnitudes of the angles is C> B > A. 
 
 [Let the student give the demonstration in 
 form.] 
 
 222. Corollary 2. — An equiangu- 
 lar triangle is also equilateral ; and, 
 conversely, an equilateral triangle is 
 equiangular. 
 
 Thus, if A = B = C, arc a = arc 6 = arc c, p. ,,2 
 
108 
 
 ELEMENTARY GEOMETRY. 
 
 and, consequently, chord BC = chord AC = chord AB. Conversely, if 
 the chords are equal, the arcs are, and hence the angles subtended by 
 these arcs. 
 
 223. Corollary 3. — In an isosceles triangle the angles 
 opposite the equal sides are ecfual ; and, 
 conversely, if two angles of a triangle 
 are equal, the sides opposite are equal, 
 and the triangle is isosceles. 
 
 Thus, if AB = BC, arc a = arc c ; and hence, 
 angle A, measured by ^ a, = angle C, measured 
 by|c. 
 
 Conversely, if A = C, arc a = arc c ; and 
 hence chord BC = chord AB. 
 
 224. Scholium. — It should be observed that the proposition gives 
 only the general relation between the angles and sides of a triangle. It is 
 not meant that the sides are in the same ratio 
 as their opposite angles: this is not true. 
 Thus, in Fig. 114, angle C is twice as great as 
 angle A ; but side c is not twice as great as side 
 «, although it is greater. Trigonometry dis- 
 covers the exact relation which exists between 
 the sides and angles. 
 
 PROPOSITION III. 
 
 225. Theorem. — // from any point ivithin a triangle 
 lines are drawn to the extremities of any side, the included 
 angle is greater than the angle of the 
 triangle opposite this side. 
 
 Demonstration. 
 
 Let ACB be any triangle, any point with- 
 in, and OB and DA lines drawn from this point 
 to the extremities of AB. 
 
 We are to prove that angle AOB > angle 
 
 ACB. Fig. 115. 
 
TRIANGLES, 
 
 109 
 
 Circumscribe a circle about the triangle (161), and produce AO and 
 BO till they meet the circumference. 
 
 Now ACB is measured by ^AnB (191); but AOB is measured by 
 |(AnB + EmD) (193). Hence^ AOB > ACB. Q. e. d. 
 
 226. An Exterior Angle of a triangle is an angle formed 
 by any side with its adjacent side produced, as CBD, Fig. 116. 
 
 PROPOSITION IV. 
 
 227. Theorem.— v^/i'i/ exterior angle of a triangle is 
 equal to the sum of the two interior non-adjacent angles. 
 
 Demonstration. 
 
 Let ABC be a triangle, and CBD be^n ex- 
 terior angle. 
 
 We are to prove that CBD = A + C. 
 ABC + CBD = a straight angle (?). 
 But ABC + A + C = a straight angle (?). 
 Hence, ABC + CBD = ABC + A + C (?). 
 Hence, subtracting ABC from each member, 
 
 CBD = A + C. Q. E. D. 
 
 Fig. 116. 
 
 228. Corollary. — Either angle of a triangle not adja- 
 cent to a specified exterior angle, is cqiuil to the differ- 
 ence between this exterior angle and the other non- 
 adjacent angle. 
 
 Thus, since CBD = A + 0, 
 
 by transposition, CBD — A = C, 
 
 and CBD — C = A. 
 
110 
 
 ELEMENTARY GEOMETRY. 
 
 OF QUADRILATERALS. 
 
 229. A Quadrilateral is a plane surface inclosed hj four 
 right lines. 
 
 230. There are three Classes of quadrilaterals, viz., Trape- 
 ziums, Trapezoids, and Parallelograms, 
 
 231. A Trapezium is a quadrilateral which has no two 
 of its sides parallel to each other. 
 
 232. A Trapezoid is a quadrilateral which has but two of 
 
 its sides parallel to each other. 
 
 233. A Parallelogram is a quadrilateral which has its 
 opposite sides parallel. 
 
 234. A Rectangle is a parallelogram whose angles are 
 right angles. 
 
 235. A Sqviare is an equilateral rectangle. 
 
 236. A Rhombus is an equilateral parallelogram whose 
 angles are oblique. 
 
 237. A Rhom- 
 boid is an oblique- 
 angled parallelogram 
 two of whose sides 
 are greater than the 
 other two. 
 
 III. — The figures in 
 the margin are all quad- 
 rilaterals. A is a trape- 
 zium. (Why ?) B is a 
 trapezoid. (Why?) C, 
 D, E, and F are paral- 
 lelograms. (Why?) D 
 and E are rectangles, 
 
 Fig. 117. 
 
Q UADRILA TERALS. 
 
 Ill 
 
 although D is the form usually referred to by the term rectangle. So C 
 is the form usually referred to when a parallelogram is spoken of, without 
 saying what kind of a parallelogram. C is also a rhomboid. (Why ?) 
 E is a square. (Why ?) F is a rhombus. (Why?) 
 
 238. A Diagonal is a line joining the vertices of two non- 
 consecutive angles of a figure. 
 
 239. The Altitude of a parallelogram is a perpendicular 
 between its opposite sides ; of a trapezoid, it is a perpendicular 
 between its parallel sides ; of a triangle, it is the perpendicular 
 from any vertex to the side opposite or to that side produced. 
 
 240. The Bases of a parallelogram, or of a trapezoid, are 
 the sides between which the altitude is conceived as taken ; of a 
 triangle, the base is the side to which the altitude is perpendiculai*. 
 
 PROPOSITION V. 
 
 241. Theorem. — ITie sum of the angles of a quadri- 
 lateral is four right angles.* 
 
 Demonstration. 
 
 Let ABCD be any quadrilateral. 
 
 We are to prove that 
 DAB + B + BCD + D = 4 right angles. 
 
 Draw either diagonal, as AC. 
 
 The diagonal divides the quadrilateral 
 into two triangles, and the sum of the an- 
 gles of the two triangles is the same as the 
 sum of the angles of the quadrilateral, since 
 
 BCA + ACD = BCD, 
 and BAC + CAD = DAB. 
 
 But the sum of the angles of the triangles is four right angles (?). 
 Hence the sum of the angles of the quadrilateral is four rii;ht angles. 
 
 <l. B.D. 
 
 * See (261). 
 
 Fig. 118. 
 
112 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION VI. 
 
 242. Theorem. — The opposite angles of any quadri- 
 lateral which can be inscribed in a circle are suppLe- 
 mentary. 
 
 DEMOKSTRATIOiq^. 
 Let ABCD be any inscribed quadrilateral. 
 
 We are to prove that 
 
 A + C = 2 right angles, 
 and also that D + B = 2 right angles. 
 
 A is measured by ^ the arc DCS, and C by ^ 
 the arc BAD. 
 
 Hence, A + C is measured by ^(DCB + BAD), 
 that is, by a semi-circumference, and is therefore 
 2 right angles (190). Q- e. d. 
 
 Fig. 119. 
 
 In like manner, B + D is measured by ^(ADC + CBA), and hence is 
 3 right angles. Q. e. d. 
 
 PROPOSITION VII. 
 
 243. Theorem. — The adjacent angles of a parallelo- 
 gram are supplemental, and the opposite angles are equal 
 to each other. 
 
 r Demokstration. 
 
 Let ABCD be any parallelogram. 
 
 We are to prove, 1st. That A + B, or 
 B + C, or C + D , or D + A is 2 right 
 angles ; and 2d. That A = C and D = B. 
 
 1st. Since, by definition (233), AD is ^'9- '^^• 
 
 parallel to BC, and the transversal AB cuts them, the sum of the two in- 
 terior angles on the same side, that is, A + B, is 2 right angles (126). 
 
 In like manner, B + C is two right angles, since they are the interior 
 angles on the same side of the transversal BC which cuts the parallels AB 
 and DC. 
 
QUADRILATERALS. 113 
 
 In the same way, C + D, or D + A maybe shown equal to 2 right 
 
 angles. 
 
 Hence the sum of any two adjacent angles of a parallelogram is 2 
 right angles, q. e. d. 
 
 2d. A + B = B + C, since each sum is 2 right angles, by the pre- 
 ceding part of this demonstration. 
 
 Hence, subtracting B from each member, we have A = C. 
 
 In a similar manner, we may show that B = D. 
 
 Hence, either two opposite angles are equal to each other, q. k d. 
 
 244. Corollary 1. — The two angles 
 of a trapezoid adjacent to either one 
 of the two sides not parallel are sup- 
 plemental. Fig. 121. 
 
 [Let the student show why.] 
 
 245. Corollary 2. — // one angle of a parallelogram is 
 right, the others are also, and, the figure is a rectangle. 
 
 PROPOSITION VIII. 
 
 246. Theorem. — Conversely to the last, If three consec- 
 utive angles of a quadrilateral are such that the first 
 and the second, and the second, and the third, are sup- 
 plemental, or if the opposite angles are equal, the figure is 
 a parallelogram. 
 
 Demonstratiok. 
 
 Let ABCD be a quadrilateral having D and A, and A and B supple- 
 mental, or having A = C and D = B. 
 
 We are to prove that, in either case, 
 the figure is a parallelogram. 
 
 1st. If we have D and A, and A and B 
 supplemental. ^.^ ,22. 
 
 Since the transversal AD cuts the lines 
 AB and DC, making A + D = 2 right angles, the lines AS and DC are 
 parallel (126). 
 
114 
 
 ELEMENTARY GEOMETRY, 
 
 Again, for a like reason, since A + B = 2 right angles, AD and BC are 
 parallel. 
 
 Hence the opposite sides of the quad- 
 rilateral are jDarallel, and the figure is a 
 parallelogram (23d)< Q. e. d. 
 
 Fig. 172. 
 
 2d. If A = C, and D = B, adding, 
 we have 
 
 A + D = C + B. 
 
 But A + D-l-C + B = 4 right angles. 
 
 Hence, substituting, we have 
 
 A + D + A + D = 4 right angles (?), 
 (Mr 2 (A + D) = 4 right angles, 
 
 or A + D = A + B (?) = 2 right angles, 
 
 and the figure is a parallelogram by the former part of the demonstration. 
 
 Q. B. D. 
 
 PROPOSITION IX. 
 
 247. Theorem. — // two opposite sides of a quadrilat- 
 eral are equal and parallel, the figure is a parallelogram, 
 
 Demokstratiok. 
 Let ABCD, in (a), be a quad- 
 rilateral having the sides AB and 
 DC equal and parallel. 
 
 We are to prove that AD and 
 BC are parallel, and hence that 
 the figure is a parallelogram. 
 
 Draw the diagonal AC. 
 
 Then, by reason of the paral- 
 lels AB and DC. the angles BAC 
 and DCA are equal (?) 
 
 Conceive the quadrilateral di- 
 vided in this diagonal into two 
 triangles, as in (b). 
 
 Reverse the triangle ACB and 
 place it as in (c). Since AC of 
 the triangle ADC = CA of the 
 triangle ABC, CA may be placed in AC, as in (c) 
 
 Fig. 123. 
 
Q UA DRILA TERAL8, 
 
 115 
 
 Now revolve the triangle CBA on CA as an axis. Since, as we have 
 shown, the angle BAC = angle DCA, BA will take the direction CD, and 
 being equal to it, by hypothesis, B will fall in D, and the angle BCA co- 
 incides with and is equal to DAC. 
 
 But in {a) the angles BCA and DAC are alternate interior angles made 
 by the transversal AC cutting AD and BC. Hence AD and BC are paral- 
 lel, and as AB and DC are parallel by hypothesis, the quadrilateral is a 
 parallelogram (233). Q- e. d. 
 
 PROPOSITION X. 
 
 248. Theorem. — // the opposite sides of a quadrilat- 
 eral are equal, the figure is a parallelogram. 
 
 Demonstration. 
 Let ABCD, (a), be a quadrilateral, having AD = BC and AB = DC. 
 
 We are to prove that ABCD 
 is a parallelogram, i. e., that 
 AB is parallel to DC, and AD to 
 BC. 
 
 Draw the diagonal AC, and 
 conceive the quadrilateral di- 
 vided in this diagonal into two 
 triangles, as in (&). 
 
 Reverse the triangle ABC, 
 and place it as in (c). Since 
 AC of the triangle ADC equals 
 CA of the triangle ABC, CA 
 may be placed in AC, as in (c). 
 
 Draw DB, intersecting CA 
 (or CA produced), in 0. 
 
 As CD = AB, and AD = 
 CB, by hypothesis, the line AC 
 has two points each equally distant from the extremities of DB, and AC 
 and DB are perpendicular to each other (98)- Moreover, since AB and 
 CD are equal oblique lines drawn from the same point in the perpendic- 
 ular to the line DB, angle BAC = angle DCA (98, 110, 2d). 
 
 Now in (a), as angles BAC and DCA are the alternate interior angles 
 made by the transversal with the lines AB and DC, the latter are parallel, 
 
 Fig. 124. 
 
116 
 
 ELEMENTARY GEOMETRY, 
 
 and as they are equal by hypothesis, the quadrilateral is a parallelogram 
 by the last proposition, q. e. d. 
 
 249. Corollary. — A diagonal of a parallelogranv di- 
 vides it into two equal triangles. 
 
 PROPOSITION XI. 
 
 250. Theorem. — Conversely to the last, The opposite 
 sides of a parallelogram are equal. 
 
 Demonstration. 
 
 Let ABCD be a parallelogram. 
 We are to prove that AD = BC, 
 and AS = DC. 
 
 AD and BC being parallel transver- 
 sals cutting the parallels AB and DC, their 
 intercepted portions, which are the oppo- 
 site sides of the parallelogram, are equal by (138) 
 
 For a like reason, AB = DC. 
 
 Hence, AD = BC and AB = DC. Q. e. d. 
 
 Fig. 125. 
 
 PROPOSITION XII. 
 
 251. Theorem. — The diagonals of a parallelogram 
 bisect each other. 
 
 Demonstration. 
 
 Let ABCD be a parallelogram whose 
 diagonals AC and DB intersect In Q. 
 
 We are to prove that AQ = QC, and 
 DQ = QB. 
 
 Angle QDC - angle QBA (?), angle 
 QCD = angle QAB (?), and DC = AB (?). ''■g- '26. 
 
 For distinctness, let Q' represent the vertex at Q of the triangle DQC. 
 
 Now apply the triangle AQB to DQ'C, placing the side BA in its equal 
 DC, with the extremity B in D, and A in C, and the vertex Q on the same 
 
Q UADRILA TERALS. 
 
 117 
 
 side of DC that the vertex Q' is, and the triangles will coincide. For, 
 since angle QAB = angle Q'CD, AQ will take the direction CQ', and the 
 vertex Q will fall somewhere in the line CQ'. In like manner, by reason 
 of the equality of angles QBA and Q'DC, the vertex Q will fall in DQ'. 
 Hence the vertex Q of the triangle AQB falling at the same time in CQ' 
 and DQ', falls at their intersection. 
 
 Hence, as these triangles coincide, AQ = Q'C, and DQ' = QB ; that is, 
 AQ = QC, and DQ = QB. q. e. d. 
 
 PROPOSITION XIII. 
 
 252. Theorem. — The diagonals of a rlLombus bisect 
 each other at Hght angles. 
 
 Demonstration. 
 
 Let ABCD be a rhombus, and AC and DB its diagonals intersecting 
 at Q. 
 
 We are to prove that DB and AC are per- 
 pendicular to each other. 
 
 Since AB = AD, and CD = CB (?), the 
 line AC has two points, A and C, each equally 
 distant from the extremities of DB. Hence AC 
 is a perpendicular to DB at its middle point Q 
 (98). Q. E. D. P5g- '27. 
 
 In like manner, DB may be shown to be perpendicular to AC at its 
 middle point. Q. e. d. 
 
 253. Corollary. — The diagonals of a rhombus bisect its 
 
 angles. 
 
 For, revolve ABC upon AC as an axis, and it will coincide with ADC. 
 Hence angles A and C are bisected. In like manner revolve DAB upon 
 DB, and it will coincide with DCB. Hence, D and B are bisected. 
 
 PROPOSITION XIV. 
 
 254. Theorem. — The diagonals of a rectangle are 
 equal. 
 
118 ELEMEyTARY GEOMETBT. 
 
 Demonstration. 
 
 Let AC and DB be the diagonals of the rectangle ABCD, 
 
 We are to prove that AC = DB; 
 
 Upon AC as a diameter describe a circle. 
 
 Since ADC and ABC are right angles whose 
 sides intercept AC, they are inscribed in the cir- 
 cumference of which AC is a diameter (198). 
 
 Again, since DCB is a right angle and is in- 
 scribed, DB is a diameter (?). 
 
 Hence AC and DB, being diameters of the same 
 circle, are equal, q. e. d. 
 
 Fig. 128. 
 
 255. Corollary. — Conversely, // the diagoTials of a par- 
 allelogram are equal, the figure is a rectangle. 
 
 By (251) the parallelogram is circuniscriptible ; whence, by (192) the 
 angles are right angles. 
 
 OF POLYGONS OF MORE THAN FOUR 
 
 SIDES. 
 
 256. A Polygon is a portion of a plane bounded by straight 
 lines. 
 
 The word polygon means many-angled ; so that with strict propriety 
 we might limit the definition to plane figures with five or more sides. 
 This limitation in the use of the word is frequently made. 
 
 267. A polygon of three sides is a triangle ; of four, a quad- 
 rilateral ; of five, a pentagon; of six, a hexagon; of seven, a 
 heptagon ; of eight, an octagon ; of nine, a nonagon ; of ten, a 
 decagon; of twelve, n dodecagon. 
 
 258. The Perimeter of a polygon is the distance around 
 it, or the sum of the bounding lines. 
 
 259. A Salient Angle of a polygon is one whose sides, 
 when produced, can only extend without the polygon. 
 
POLYGONS, 
 
 119 
 
 260. A Re-entrant Angle of a 
 
 polygon is one whose sides, when pro- 
 duced, can extend within the polygon. 
 
 Illustration.— In the polygon ABCDEFG, 
 all the angles are salient except D, which is 
 re-entrant. 
 
 261. A Convex Polygon is a 
 
 polygon which has only salient angles. F«fl- •«. 
 
 A polygon is always supposed to be convex, unless the contrary 
 
 is stated. 
 
 262. A Concave or Re-entrant Polygon is a polygon 
 
 with at least one re-entrant angle. 
 
 263. An Equilateral Polygon is a polygon whose sides 
 are equal, each to each ; and an Equiangular Polygon is 
 a polygon whose angles are equal, each to each. 
 
 PROPOSITION XV. 
 
 264. Theorem. — The sum of the interior angles of a 
 polygon is equal to twice as many right angles as the poly- 
 gon has sides, less four right angles. 
 
 Demonstration. 
 
 Let n be the number of sides of any polygon 
 
 We are to prove that the sum of its angles 
 is 71 times 2 right angles less 4 right angles. 
 
 From any point within, as 0, draw lines to 
 the vertices of the angles. As many triangles 
 will then be formed as the polygon has sides, 
 that is, n. 
 
 The sum of the angles of the triangles is n 
 times 2 right angles. 
 
 But this sum exceeds the sum of the angles 
 of the polygon by the sum of the angles around 
 the common vertex 0, that is, by 4 right angles. 
 
 f-iy. 130. 
 
120 
 
 ELEMENTARY GEOMETRY. 
 
 Hence the sum of the angles of the polygon is 
 
 7* times 2 right angles less 4 right angles, q. k. d. 
 
 265. Scholium 1. — The sum of the angles of a pentagon is 
 
 5 times 2 right angles - 4 right angles, or 6 right angles. 
 
 The sum of the angles of a hexagon is 8 right angles ; of a heptagon, 
 10 ; of an octagon, 12, etc. 
 
 266. Scholium 2. — This proposition is equally applicable to triangles 
 and to quadrilaterals. Thus, the sum of the angles of a triangle is 
 
 3 times 2 right angles — 4 right angles = 2 right angles. 
 So also the sum of the angles of a quadrilateral is 
 
 4 times 2 right angles — 4 right angles, or 4 riglit angles. 
 
 267. Scholium 3. — To find the value of an angle of an equiangular 
 polygon, divide the sum of all the angles by the number of angles. 
 
 PROPOSITION XVI. 
 
 268. Theorem. — // one of the sides of a polygon is 
 produced (and only one) at each vertex, the sum of the 
 exterior angles thus formed is four right angles. 
 
 Demonstration. 
 
 Let n be the number of the sides of any 
 polygon, and one side be produced at each 
 vertex. 
 
 We are to prove that the sum of the ex- 
 terior angles thus formed, &8 a + h •\- c -\- d, 
 etc., is 4 right angles. 
 
 At each of the n vertices there are two 
 angles, an interior and an exterior one, 
 whose sum, as A + a, is 2 right angles. 
 Hence the sum of all the exterior and inte- 
 rior angles is 
 
 n times 2 right angl 
 
 Fig. 131. 
 
REGULAR POLYGONS, 
 
 \%1 
 
 Now, from this sum subtracting the sum of the exterior angles, the 
 remainder is the sum of the interior angles. 
 
 But, by the preceding propositiim, 4 right angles subtracted from n 
 times 2 right angles leaves the sum of the interior angles. 
 
 Therefore the sum of the exterior angles is 4 right angles, q. e. d. 
 
 OF REGULAR POLYGONS. 
 
 269. A Regular Polygon is a polygon which is both 
 equilateral and equiangular (263). 
 
 270. An Inscribed Polygon is a polygon whose angles 
 are all inscribed in the same circumference. 
 
 271. A Circumscribed Polygon is a polygon whose 
 sides are all tangent to the same circle. The circumference is 
 said to be inscribed in the polygon. 
 
 PROPOSITION XVII. 
 
 272. Theorem. — The angles of an inscribed equilateral 
 polygon are equal ; and the polygon is regular. 
 
 Demonstration. 
 
 Let ABCDEF be an inscribed poly- 
 gon, having AB =r BC = CD, etc. 
 
 We are to prove that angle ABC = 
 angle BCD = angle CDE, etc. 
 
 The sides of the polygon being equal 
 chords, subtend equal arcs (151). 
 
 Now any angle of the polygon is 
 measured by ^ the diflference between 
 the circumference and the sum of two 
 of these equal arcs, as angle ABC meas- 
 ured by ^ (circumference — arc ABC) 
 = i arc AFEDC. 
 
 Hence all the angles are equal, and the polygon is regular (269). Q. E. D. 
 6 
 
 Fig. 132. 
 
122 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XVIII. 
 
 273. Theorem. — A circumference may he circum,- 
 scribed about any regular polygon. 
 
 Demonstration. 
 Let ABCDEF be a regular polygon. 
 
 We are to prove that a circumference can 
 be circumscribed about it. 
 
 Bisecting any two consecutive sides, as FA 
 and AB, by perpendiculars, as Oa and 06, pass 
 a circumference through the vertices F, A, and 
 B (161). 
 
 We will now show that this circumference 
 passes through all the other vertices. 
 
 Revolve the quadrilateral FO^A upon 06 as 
 an axis until it falls in the plane of C06B, 6A will fall in its equal IB (?) ; 
 and since angle A = angle B, and side AF = side BO, F will fall in C. 
 
 Thus it appears that the circumference described from 0, and pass- 
 ing through F, A, and B, also passes through 0. 
 
 In a similar manner it can be shown that the same circumference 
 passes through all the vertices, and hence is circumscribed. Q. e. d. 
 
 PROPOSITION XIX. 
 
 274. Theorem. — A circumference may be inscribed in 
 any regular polygon. 
 
 Demonstration. 
 Let ABCDEF be a regular polygon. 
 
 We are to show that a circumference may 
 be inscribed in it. 
 
 Let be the centre of the circumscribed 
 circumference (273); then the sides of the 
 polygon are equul chords of this circle, and 
 consequently equally distant from the centre 
 (160). Fig 134. 
 
REOULAR POLYGONS. 123 
 
 Now draw the perpendiculars 0«, 0&, Oc, Orf, etc. These perpendic- 
 ulars are all equal, and a circumference struck from as a centre, with 
 anyone of them, as Oa, as a radius, will pass through 5, c, d^ etc. 
 
 Moreover, the sides AB, BC, CD, etc., being perpendicular to the 
 radii Oa, OJ, etc., are tangents to this circumference, which is therefore 
 an inscribed circumference (271). Q. e. d. 
 
 275. Corollary. — The centres of the inscribed and cir- 
 ' curnscrihed circles coincide. 
 
 276. The Centre of a regular polygon is the common cen- 
 tre of its inscribed and circumscribed circles. 
 
 277. An Angle at the Centre of a regular polygon is 
 the angle included by two lines drawn from the centre to the 
 extremities of a side, as FOA, AOB (Fig. 133). 
 
 278. The Apotheni of a regular polygon is the distance 
 from the centre to any side, and is the radius of the inscribed 
 circle. 
 
 PROPOSITION XX. 
 
 279. Theorem. — The angles at the centre of a regular 
 polygon are equal each to each ; and any one is equal to 
 four right angles divided by the number of sides of the 
 polygon. 
 
 Demonstration. 
 
 Let P be a polygon ofn sides. 
 
 We are to prove, Ist. That the angles at the centre are equal ; and 
 
 2d. Tl.at any one of them is ^ ^ght angles _ 
 
 n 
 
 1st. Each angle at the centre intercepts one of the equal sides of the 
 polygon. But these sides are chords of equal arcs (?). Hence the sev- 
 eral angles at the centre have equal measures, and are therefore equal. 
 
 Q. E. D. 
 
124 
 
 ELEMENTARY GEOMETRY, 
 
 2d. The sum of all the angles at the centre is 4 right angles (?), and 
 as they are equal and n in number, any one is 
 
 4 right angles 
 
 2 5 Q. E. D. 
 
 PROPOSITION XXI. 
 
 280. Theorem. — Any side of a regular inscribed hex- 
 agon is equal to the radius. 
 
 Demonstration. 
 
 Let ABCDEF be a regular hexagon inscribed in a circle whose radius 
 is 11. 
 
 We are to prove that any one of the equal 
 sides, as AB, equals B. 
 
 Let be the centre of the polygon, and draw 
 OA, OB, etc. 
 
 Now in the triangle AOB, angle is ^ of 4 
 right angles, or ^ of 2 right angles (?). 
 
 Whence the sum of the angles OAB and OBA 
 is I of 2 right angles (?). 
 
 But the triangle AOB is isosceles, OA and 
 OB being radii of the same circle. Hence, each ^"'9- '35. 
 
 one of the angles at the base is ^ of f of 2 right angles, or | of 2 right 
 angles. Therefore the triangle AOB is equiangular and consequently 
 equilateral (222), and AB = OA = i2. Q. e. d. 
 
 281. A Broken Line is said to be Convex when a straight line 
 cannot be drawn which shall cut it in more than two points. 
 
 PROPOSITION XXII. 
 
 282. Theorem. — A convex broken line is less than any 
 broken line which envelops it and has the same extremities, 
 the former lying between the latter and a straight line 
 joining its extremities. 
 
REGULAR POLYGONS, 
 
 125 
 
 Demonstration. 
 
 Let ^hcd^ be a broken line enveloped by the broken line ACDEFB, 
 and having the same extremities A and B. 
 
 We are to prove that 
 
 Mcd^ < ACDEFB. 
 
 Produce the parts of A5aZB till they 
 
 meet the enveloping line, as Kb to e, he to 
 /, and cd to g. 
 
 Now, Kb-^he < ACe (?), 
 
 hc + cf <he + eDE/(0, 
 
 ed + dg <cf +ffg, 
 
 dB < dg -\- gB. 
 
 Fi{|. 136. 
 
 Hence, adding, and subtracting common terms, 
 
 Ab + be -^ cd -{- dB < ACe + cDEf + /Fg + gB, 
 or AbcdB < ACDEFB. Q. e. d. 
 
 angle. 
 
 PROPOSITION XXIII. 
 
 Problem. — To inscribe a circle in a given tri- 
 
 SOLUTION. 
 
 Let ABC be a triangle. 
 
 We are to inscribe a circle. 
 Bisect any two angles, as A and B (208). 
 From the intersection of the bisectors, as 0, 
 let fall a perpendicular, as OD* 
 
 Then is the centre of the inscribed circle, 
 and OD its radius. 
 
 Hence a circle described with as a centre 
 and OD as a radius will be inscribed. 
 
 Fig. 137. 
 
 Demonstration of Solution. 
 From let fall the perpendiculars OD, OE, and OG on the sides. 
 
126 
 
 ELEMENTARY GEOMETBY, 
 
 Now the triangle AOE = AOG (?), BEO = 
 BOD (?). 
 
 Hence OD = OE = OG, and the circum- 
 ference struck from as a centre with a radius 
 OD, passes through E and G. 
 
 Moreover, AC, AS, and BC are perpendicu- 
 lar to the radii OG, OE, and OD respectively, 
 and hence are tangents to the circle. 
 
 Therefore the circle is inscribed in the 
 triangle. Q. b. d. 
 
 Fig. 137. 
 
 PROPOSITION XXIV. 
 
 284. Problem. — In a given circle to inscribe a square, 
 and hence a regular octagon, and then a regular polygon 
 of 16 sides, etc. 
 
 [Let the pupil give the solution and demonstration.] 
 
 PROPOSITION XXV. 
 
 285. Problem. — In a given circle to inscribe a regular 
 hexagon, and hence an equilateral triangle and a dodec- 
 agon. 
 
 [Let the pupil give the solution.] 
 
 PROPOSITION XXVI. 
 
 286. Problem. — To circumscribe a square about a 
 given circle. 
 
 [Let the pupil give the solution.] 
 
EXERCISES, 
 
 127 
 
 PROPOSITION XXVII. 
 
 287. Problem. — To circumscribe an equilateral tri- 
 angle about a circle. 
 
 [Let the pupil give the solution.] 
 
 PROPOSITION XXVIII. 
 
 288. Problem. — To circumscribe a regular hexagon 
 about a given circle. 
 
 [Let the pupil give the solution.] 
 
 289. Query. — Given any regular inscribed polygon, how is 
 the regular cir(^m8cribed polygon of the same number of sides 
 constructed ? 
 
 EXERCISES 
 
 290. 1. Given two angles of a tri- 
 angle, to find the third. 
 
 Suggestions. — Tlie student should draw 
 two angles on the black lx)ard, as a and ft, and 
 then proceed to find the third. The figure 
 will suggest the method. The third angle 
 is c. 
 
 The solution is effected also by con- 
 structing the two given angles at the extrem- Fig. I38. 
 ities of any line, and producing the sides till they meet (?). 
 
 2. What part of a right angle is one of the angles of an equi- 
 lateral triangle ? From this fact, how can you obtain an angle 
 equal to | of a right angle ? 
 
 3. Two angles of a triangle are respectively | and J of a right 
 angle. What is the third angle ? 
 
128 ELEMENTARY GEOMETRY. 
 
 4. The angles of a triangle are respectively f , J, and | of a 
 right angle. Which is the greatest side? Which the least? 
 Can you tell the ratio of the sides ? 
 
 5. What is the value of one of the equal angles of an isosceles 
 triangle whose third angle is J of a right angle ? 
 
 6. Two consecutive angles of a quadrilateral are respectively 
 I and f of a right angle, and the other two angles are mutually 
 equal to each other. What is the form of the quadrilateral ? 
 What the value of each of the two latter angles? 
 
 7. One of the angles of a parallelogram is f of a right angle. 
 What are the values of the other angles ? 
 
 8. The two opposite angles of a quadrilateral are respectively 
 I and i of a right angle. Can a circumference be circumscribed ? 
 If so, do it. 
 
 9. Two of the opposite sides of a quadrilateral are parallel, 
 and each is 15 in length. What is the figure ? Do these facts 
 determine the angles ? 
 
 10. Two of the opposite sides of a quadrilateral are 12 each, 
 and the other two 7 each. What do these facts determine with 
 reference to the form of the figure ? 
 
 11. What is the value of an angle of a regular dodecagon ? 
 
 12. What is the sum of the angles of a nonagon ? What is 
 the value of one angle of a regular nonagon ? Of one exterior 
 angle ? 
 
 13. What is the regular polygon, one of whose angles is l^nf 
 right angles ? 
 
 14. What is the regular polygon, one of whose exterior angles 
 is f of a right angle ? 
 
 15. Can you cover a plane surface with equilateral triangles 
 without overlapping them or leaving vacant spaces ? With 
 quadrilaterals? Of what form? With pentagons? Why? 
 With hexagons? Why? What insect puts the latter fact to 
 practical use ? Can you cover a plane surface thus with regular 
 polygons of more than 6 sides ? Why ? 
 
EXERCISES, 129 
 
 THEOREMS FOR ORIGINAL. INVESTI- 
 GATION. 
 
 [It is quite desirable that students have exercise, early in their course, 
 in the original demonstration of theorems. Those which are given in this 
 and the following lists are not such as are essential to the integrity of an 
 elementary course, and pupils may be encouraged to demonstrate more or 
 lesis of them, as their time and ability will allow. But all should do some 
 such work — it is the true test of mathematical ability and attainment.] 
 
 291. 1. Theorem. — The least chord that can he drawn 
 through a point within a circle is the chord which is per- 
 pendicular to a diameter passing through the same point. 
 
 2. Theorem. — Tlie shortest distance from a point 
 without a circle to the circumference is measured in a 
 line which passes through the centre. 
 
 3. Theorem. — The sum of the angles form,ed by pro- 
 ducing the alternate sides of any pentagon is two right 
 angles. 
 
 4. Theorem. — Prove that the sum^ 
 of the angles of a triangle is two right 
 angles, by producing two of the sides 
 about an angle, and through the vertex 
 of this angle drawing a line parallel to 
 the third side. 
 
 Prove the same by producing one 
 side of the triangle, and drawing a line 
 through the vertex of the exterior angle 
 parallel to the non-adjacent side. 
 
 Fig. 139. 
 
 5. Theorem.—// AB is any chord, AC a tangent at A, 
 and CDE a line parallel to AB and cutting the circumfer- 
 ence in D and E, the triangles ACD, CAE, and ADB are 
 mutually equiangular. 
 
 6. Theorem. — // from any point in the base of an 
 
130 
 
 ELEMENTAR Y GEOMETR Y. 
 
 isosceles triangle lines are drawn parallel to the equal 
 sides, a parallelogram is formed whose perimeter is equal 
 to the sum of the equal sides. 
 
 0UCTI0M H% 
 
 OF EQUALITY. 
 
 292. Equality signifies likeness in every respect. 
 
 293. The equality of magnitudes is usually shown by apply- 
 ing one to the other, and observing that the two coincide. 
 
 OF A NGLES. 
 
 PROPOSITION I. 
 
 294. Theorem.— rM;o angles whose corresponding 
 sides are parallel, and extend in the same or in opposite 
 directions from their vertices, are equal. 
 
 DEMONSTRATIOi?". 
 
 Mrst, In («) and {a'), let B and E be two 
 angles having BA parallel to ED and extending in 
 the same direction from the vertices, and also BC 
 parallel to EF and extending in the same direction 
 from the vertices. 
 
 We are to prove that angles B and E are equal. 
 
 Produce (if necessary) either two non-parallel 
 sides, as BC and ED, till they intersect, as in H. 
 
 ABC r= DHC (?), 
 and DHC = DEF (?). 
 
 Therefore, ABC = DEF. Q. e. d. 
 
 Fig. 140. 
 
JSQUALirr OF ANGLES, 
 
 131 
 
 Second. In (6) and (&'), let B' and E' have 
 B'A' parallel to ET', but extending in an oppo- 
 site direction from the vertices ; and in like man- 
 ner B'C parallel to^ but extending in an opposite 
 direction from E'D'. 
 
 We are to prove that B' and E' are equal. 
 
 Produce (it* necessary) either two non -parallel 
 sides, as A'B' and E'D', till they meet in some 
 point, as H'. 
 
 D'H B' = D E F' (?), 
 nnd D'H'B' = A'B'C (?). 
 
 Therefore D'ET' = A'B'C (?). H, e. d. 
 
 Fig. I4r. 
 
 PROPOSITION II. 
 
 295. Theorem. — Two angles having their corres- 
 ponding sides parallel, ivhile two extend in the same 
 direction, and the other two in opposite directions from 
 the vertices, are supplemental. 
 
 Demonstration. 
 
 Let ABC and DEF be two 
 
 angles whose corresponding 
 sides BA and EF are parallel 
 and extend in the same direc- 
 tion from B and E, while BC 
 and ED extend in opposite 
 directions from the vertices. 
 
 We are to prove that ABC 
 and DEF are supplemental. 
 
 Produce one of the two '^* * ' 
 
 sides having opposite directions as DE to H, in the same direction from 
 the vertex that BC extends. 
 
 Now DEF is supplemental to FEH (?), and FEH is equal to ABC (?) 
 
 Therefore, DEF and ABC are supplemental, q. e. d. 
 
132 ELEMENTARY GEOMETRY, 
 
 ^" PROPOSITION III. 
 
 296. Theorem. — If the sides of one angle are perpen- 
 dicular respectively to the sides of another, the angles are 
 either equal or supplemental. 
 
 Demonstration. 
 
 Let ABC be any angle and DE 
 and FH be two lines drawn through 
 any point 0, DE being perpendicular 
 to BC and FH to AB. 
 
 We are to prove that of the four 
 angles FOD, DOH, etc., two are 
 equal to ABC, and two are supple- 
 mental. 
 
 Draw BS bisecting ABC, and 
 from any point in this bisector, as L, 
 draw LM and LN, respectively paral- 
 lel to DE and FH. 
 
 Now, in the quadrilateral LNBM, the sum of the four angles is four 
 right angles (266) ; and, as LNB and LMB are right angles (?), NLM and 
 NBM (or ABC) are supplemental. 
 
 But NLM = FOD (?) = HOE C). 
 
 Therefore two of the four angles FOD, DOH, etc., namely, FOD and 
 HOE, are supplemental to ABC. Q. e. d. 
 
 Finally, FOE and DOH are supplements of FOD and HOE (?) and 
 hence equal to ABC. Q. k. d. 
 
 297. Scholium.— To determine whether the angles are equal, or 
 whether they are supplemental, we may consider one angle as moved 
 (if necessary) till its vertex falls in the bisector, its sides remaining 
 paraMel to their first position. Then, if both sides of one angle extend 
 towards, or both extend from the sides of the other, the angles are sup- 
 plemental, otherwise they are equal. 
 
 Fig. 143. 
 
 OF TRIANGLES. 
 
 PROPOSITION IV. 
 
 298. Theorem. — Two triangles which have two sides 
 and the included angle of one equal to two sides and the 
 included angle of the other, each to each, are equal. 
 
EQUALITY OF TRIANGLES, 
 
 133 
 
 Demonstration". 
 
 Let ABC and DEF be 
 two triangles, having AC 
 = DF, AB - DE, and angle 
 A = angle D. 
 
 We are to prove that 
 the triangles are equal. 
 
 Place the triangle ABC 
 in the position (6), the side 
 AB in its equal DE, and 
 the angle A adjacent to its 
 equal angle D. 
 
 Then revolving ABC upon DE, until it falls in the plane on the oppo- 
 site side of DE, since angle A = angle D, AC will take the direction DF; 
 and as AC = DF, C will fall at F. Hence BC will fall in EF, and the 
 triangles will coincide. Therefore the two triangles are equal, q. e. d. 
 
 Scholium 1. — We may also make the application of ABC to 
 DEF directly. The method here given is used for the purpose of uni- 
 formity in this and the following. We may observe that in this, as in 
 the other cases, DB is perpendicular to FC, and bisects it at 0. 
 
 300- Scholium 2. —This proposition signifies that the two triangles 
 are equal in all respects^ L «., that the two remaining sides are equal, as 
 CB = FE; that angle C = angle F, angle B = angle E, and that the 
 areas are equal. 
 
 PROPOSITION V. 
 
 301. Theorem. — Tivo triangles which have two angles 
 and the included side of the one equal to two angles and 
 the included side of the other, each to each, are equal. 
 
 Demonstration. 
 
 Let ABC and DEF be two triangles, having angle A = angle D, angle 
 B = angle E, and side AB = side DE. 
 
 We are to prove that the triangles are equal. 
 
134 
 
 ELEMENTARY GEOMETRY. 
 
 Place ABC in the po- 
 sition (5), the side AB in 
 its equal DE, the angle A 
 adjacent to its equal angle 
 D, and B adjacent to its 
 equal angle E. 
 
 Then revolving ABC 
 upon DE till it falls in the 
 plane on the same side as 
 DFE, since angle A = Fig. I45. 
 
 angle D, AC will take the 
 direction DF, and C will fall somewhere in DF, or DF produced. 
 
 Also, since angle B = angle E, BC will take the direction EF, and C 
 will fall somewhere in EF, or EF produced. 
 
 Hence, as C falls at the same time in DF and EF, it falls at their in- 
 tersection F. Therefore the two triangles coincide, and are consequently 
 equal. Q. e. d. 
 
 302. Corollary. — // one triangle has a side, its oppo- 
 site angle, and one adjacent angle, equal to the correspond- 
 ing parts in another triangle, the triangles are equal. 
 
 For the third angles are equal to each other, since each is the supple- 
 ment of the sum of the given angles. Whence the case is included in the 
 proposition. 
 
 303. Scholium. — A triangle may have a side and one adjacent angle 
 equal to a side and an adjacent angle in another, and the second adjacent 
 angle of the first equal to the angle opposite the equal side in the second, 
 and the triangles not be equal. Thus, in the figure, AB = C'A', A = A', 
 
 and B = B' ; but the triangles are evidently not equal. [Such triangles 
 are, however, simitar^ as will be shown hereafter.] 
 
EQUALITY OF TRIANGLES. 
 
 135 
 
 PROPOSITION VI. 
 
 304. Theorem. — Two triangles which have two sides 
 and an angle opposite one of these sides, in the one, equal 
 to the corresponding parts in the other, are equal, if of 
 these two sides the one opposite the given angle is equal to 
 or greater than the one adjacent. 
 
 Demonstration. 
 
 In the triangles ABC and DEF, let AC = DF, CB - FE, A = D, and 
 
 CB (= FE)> AC (= DF). 
 We are to prove that 
 the triangles are equal. 
 
 Apply the triangle 
 ABC to DEF, placing AC 
 in its equal DF, the point 
 A falling at D, and C at 
 F. 
 
 Since A = D, AB 
 will take the direction 
 DE. 
 
 Let fall the perpen- 
 dicular FH upon DE, or 
 DE produced. 
 
 Now, CB being ^ DF, cannot fall between it and the perpendicular, 
 but must fall in FD or beyond both (?). 
 
 But CB cannot fall in FD, since it is a different line from CA. 
 
 Again, as CB = FE, and both lie on the same side of FH, they must 
 coincide (114). 
 
 Hence, the two triangles coincide, and are consequently equal. 
 
 Q. E. D. 
 
 PROPOSITION VII. 
 
 305. Theorem. — Two triangles which have the three 
 sides of the one equal to the three sides of the other, each 
 to each, are equal. 
 
136 
 
 ELEMENTART GEOMETRY. 
 
 Demon^stration. 
 
 Fig. 148. 
 
 l-et ABC and DEF be 
 two triangles, in which AB 
 = DE, AC = DF, and BC 
 = EF. 
 
 We are to prove that 
 the triangles are equal. 
 
 Place the triangle ABC 
 in the position (5), with 
 the longest side, AB, in its 
 equal, DE, so that the 
 other equal sides shall be 
 adjacent, as AC adjacent to DF, and BC to EF. Draw FC cutting 
 DE in 0. 
 
 Now, since AC = DF, and BC = EF, DE is peipendicular to FC at 
 its middle point (?). 
 
 Hence, revolving ABC upon DE, it will coincide with DEF when 
 brought into the plane of the latter, since OC will fall in OF (?) and is 
 equal to it. 
 
 Therefore the two triangles coincide, and hence are equal, q. e. d. 
 
 306. Corollary. — In two equal triangles, the equal an- 
 gles lie opposite the equal sides. 
 
 PROPOSITION VIII. 
 
 307. Theorem.— //^m;o triangles have two sides of the 
 one respectively equal to two sides of the other, and the 
 included^ angles unequal, the third sides are unequal, 
 and the greater third side belongs to the triangle having 
 the greater included angle. 
 
 Demonstration. 
 
 Let ABC and DEF be two triangles having AC = DF, CB = FE, 
 and C > F. 
 
 We are to prove that AB > DE. 
 
EQUALITY OF TRIANGLES. 
 
 137 
 
 Make the angle ACE = DFE, 
 take CE = FE, and draw AE. 
 Then is the triangle ACE = DFE, 
 and AE = DE. 
 
 Bisect ECB with CH. 
 
 Now since angle DFE = ACE 
 < ACB by hypothesis, CE falls be- 
 tween CA and CB, and CH will 
 meet AB in some point, as H. 
 
 Draw HE. 
 
 The triangles HCB and HCE 
 have two sides and the included 
 angle of the one, equal to the cor- 
 responding parts of the other, 
 whence HE = HB (?). 
 
 Now AH + HE > AE 
 
 Fig. 149. 
 
 but 
 
 Therefore, 
 
 AH + HE = AH + HB = AB. 
 
 AB > AE, or AB > DE. q. e. d. 
 
 308. Corollary. —Conversely, // two sides of one tri- 
 angle are respectively equal to two sides of another, and 
 the third sides are unequal, the angle opposite this third 
 side is the greater in the triangle which has the greater 
 third side. 
 
 That is, if AC = DF, CB = FE, and AB > DE, angle C > angle F. 
 For, if C = F, the triangles would be equal, and AB = DE (298) ; and, 
 if C were less than F, AB would be less than DE, by the proposition. 
 But both these conclusions are contrary to the hypothesis. Hence, as C 
 cannot be equal to F, or less than F, it must be greater. 
 
 PROPOSITION IX. 
 
 309. Theorem. — Two right-angled triangles which 
 have the hypotenuse and one side of the one equal to the 
 hypotenuse and one side of the other, each to each, are 
 equal. 
 
138 
 
 ELEMENTARY GEOMETRY, 
 
 Demonstration. 
 
 In the two triangles ABC and DEF, right-angled at B and E, let AC 
 DF, and BC = EF. 
 
 We are to prove that the 
 triangles are equal. 
 
 Place FE in its equal CB, 
 with FD on the same side of 
 CB that AC is. 
 
 Then, since two equal 
 oblique lines cannot be drawn 
 from C to AB on the same side 
 of CB, FD will coincide with 
 CA, and DE with AB (?) 
 
 Hence the two triangles 
 are equal, as they coincide 
 throughout when applied (292, 293). Q. e. d. 
 
 Fig. 150. 
 
 PROPOSITION X. 
 
 310. Theorem. — Two right-angled triangles having 
 any side and one acute angle of the one equal to the 
 corresponding parts of the other are equal. 
 
 Demonstration. 
 
 One acute angle in one triangle being equal to one in the other, the 
 other acute angles are equal, since they are complements of the same 
 angles (218). The case then falls under (301). 
 
 EXERCISES. 
 
 Exercise 1. Given the sides of a triangle, as 15, 8, and 5, to 
 construct the triangle. 
 
 Ex. 2. Given two sides of a triangle, a = 20, ^> = 8, and the 
 angle B opposite the side h equal J of a right angle, to construct 
 the triangle. 
 
DETERMINATION OF TRIANGLES, 139 
 
 Ex. 3. Same as in the preceding example, except h = 12. 
 Same, except that b = 25. 
 
 Ex. 4. Construct a triangle with angle A = J of a right 
 angle, angle B = ^ of a right angle, and side a opposite angle 
 A, 15. 
 
 Ex. 5. Construct an isosceles triangle whose vertical angle 
 is 30^. 
 
 Ex. 6. Construct a right-angled triangle whose hypotenuse 
 is 12 and one of whose acute angles is 60^ 
 
 Ex. 7. Construct an equilateral triangle, and let fall a per- 
 pendicular from one vertex upon the opposite side. How is this 
 angle divided ? How many degrees measure the angle between 
 the perpendicular and one side ? 
 
 THE DETERMINATION OF POLYGONS. 
 
 311. A triangle, or any polygon, is said to be Determined 
 when a sufficient number of parts are known to enable us to 
 construct the figure, or to find the unknown parts. If two 
 different figures can be constructed, the case is said to be 
 Ambiguous. 
 
 312. Since, in such a case, if several polygons were to be con- 
 structed with the same given parts all would be equal, the condi- 
 tions which determine a polygon are, in general, the same as 
 those which insure equality (292). Hence, having shown that 
 certain given parts determine a polygon, we may assert that two 
 polygons having these parts respectively equal are equal, except 
 in the ambiguous cases. 
 
 PROPOSITION XI. 
 
 313. Theorem. — A triangle is determined in the fol- 
 lowing cases : 
 
 L W7ien two sides and the included angle are known. 
 II. WTien two angles and the included side are known. 
 
140 ELEMENTARY GEOMETRY, 
 
 III. When the three sides are known. 
 
 IV. When two sides and an angle opposite one of them 
 are known, 
 
 (a.) If the known angle is right or obtuse. 
 
 (&.) If the known angle is acute and the known side 
 opposite it is equal to the perpendicular upon the unknown 
 side ; or equal to or greater than the other known side. 
 
 {c.) But, if the known angle is acute and the known 
 side opposite it is intermediate in length between the other 
 known side and the perpendicular upon the unknown side, 
 the case is ambiguous, i. e., there are two triangles possible. 
 
 Demonstration. 
 
 The demonstration of this proposition is effected in the solution of the 
 following problems. 
 
 314. Problem. — Given two sides and the included 
 angle, to construct a triangle. 
 
 Solution. 
 
 Let A and B be the given (or known) sides, and the given angle. 
 
 We are to construct a triangle having 
 an angle equal to included between sides 
 equal to A and B. 
 
 Draw any line, as O'D, equal to either 
 of the given sides, as A. 
 
 Lay off at either extremity of O'D, as 
 at 0', an angle equal to (203), and make 
 O'E equal to B, and draw ED. 
 
 Then will EO'D be the triangle re- Fig. isi. 
 
 quired. 
 
 For, if two triangles (or any number) be constructed with the sftme 
 sides and included angle, they will^all be equal to each other (298). 
 
 315. Problem. — Given two angles and the included 
 side, to construct a triangle. 
 
DETERMINATTNN OF TRIANGLES. 
 
 141 
 
 Solution. 
 
 Let M and N be the two given 
 angles, and A the given side. 
 
 We are to construct a triangle 
 having a side equal to A and in- 
 cluded between the vertices of two 
 angles equal respectively to M 
 and N. 
 
 Draw DE equal to A. At one 
 extremity, as D, make angle FDE = 
 M, and at E make FED = N. 
 
 Then is DEF the triangle re- 
 quired (?). 
 
 QuKRY. — What is the limit of the sum of the given 
 
 316. Problem. 
 
 angle. 
 
 Given three sides, to construct a tri- 
 
 Solution. 
 
 Let A, B, and C be the three given sides. 
 
 We are to construct a triangle which 
 shall have its three sides respectively 
 equal to A, B, and C. 
 
 Draw DE = A. 
 
 With D as a centre and a radius 
 equal to B, strike an arc intersecting an 
 arc struck from E as a centre, with a 
 radius C. 
 
 The triangle DEF is the triangle 
 sought (?). 
 
 Fig. 153. 
 
 317, ScHOLiDM.— If any one of the three proposed sides is greater 
 than the sum or less than the difference of the other two, a triangle is 
 impossible (?). 
 
 318. Problem. — To construct a triangle, having given 
 two sides and the angle opposite one of them. 
 
142 
 
 ELEMENTARr GEOMETBr. 
 
 Solution. 
 
 There are three cases. 
 
 CASE (a). 
 When the given angle is right or obtuse. 
 
 Let be the angle, and A and B 
 the sides, the angle to be opposite 
 the side A. 
 
 Construct angle NDM =: (203), 
 and take FD = B. 
 
 From F as a centre, with A as a 
 radius, strike an arc cutting DM in 
 E, and draw FE. 
 
 Then is FDE the triangle sought. 
 
 For it has FD = B, FE = A 
 (since FE is a radius of a circle 
 struck with A as a radius), and angle 
 FDE, opposite FE, equal to 0. 
 
 If the given angle were right, the construction would be the same. 
 
 Fig. 154. 
 
 CASE (h). 
 
 When the given angle is acute, and, 1st, the side opposite 
 equal to the jterpendicular upon the unknoivn side, and, 2d, 
 when the side opposite is equal to or greater than the other 
 given side, 
 
 1st, Let A and B be the given sides 
 and the given angle opposite B. 
 
 Proceed exactly as in the preceding 
 case, but when the arc is struck from F 
 as a centre with a radius equal to B, 
 instead of intersecting DM it will be 
 tangent to it, since B = FE is the per- 
 pendicular, and a line which is perpen- 
 dicular to a radius at its extremity is 
 tangent to the arc (156). 
 
 2d, If the side opposite the given angle is equal to the other 
 given side, the arc struck from F tvith it as a radius will cut 
 
DETERMINATION OF QUADRILATERALS. 
 
 143 
 
 DM at an equal distance tuith FD from the foot of the perpen- 
 dicular (?), and tite trUmgle formed will he isosceles {?), 
 
 If the side opposite is greater than the other given side, it 
 tu^iU cut MD hut once (?) and there ivUl he hut one triangle. 
 
 CASE (r). 
 
 When the given angle is acute, and the given side opposite 
 it is intermediate in length hetween the other given side and 
 the perpendicular to the unknown siile. 
 
 Let A and B be the given sides 
 and the angle opposite B, B 
 being intermediate in length be- 
 tween A and the perpendicular 
 FH on the unknown side. 
 
 Proceed as in the two preced- 
 ing cases, but instead of tangency 
 we get two intersections of DM 
 by the arc struck from F with 
 radius B. as E and E', since two 
 equal oblique lines can be drawn 
 
 from F to DM (114), and B being p;g ,55^ 
 
 less than FD = A, FE will lie 
 between FD and FH, and FE' beyond FH (113). 
 
 Thus we have two triangles, DEF and DE'F, each of which fulfills the 
 required conditions. 
 
 319. Scholium. — In order that the triangle should be possible, the 
 side opposite the given angle must be equal to or greater than the per- 
 pendicular upon the unknown side. 
 
 OF QUADRILATERALS. 
 
 The subject of the conditions which determine a quadrilateral or 
 other polygon is quite an important and practical subject, especially in 
 surveying, and we treat the problem of the equality of polygons of more 
 than three sides in this way. (See 312.) 
 
144 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION XII. 
 
 320. Theorem. — J_ quadrilateral is determined when 
 there are given in their order : 
 
 I. The four sides and either diagonal, 
 
 II. The four sides and one angle. 
 
 III. 1st. Three sides and two included angles. 
 
 2d. When the two angles are not both included between 
 the known sides, the case may be ambiguous. 
 
 IV. Three angles and two sides, the unknown sides 
 being non-parallel. 
 
 Demonstration. 
 
 CASE I. 
 
 Let a, hf c, d (Fig. 157), be the sides in order, and e the diagonal 
 joining the vertex of the angle between a and d with the vertex between 
 b and c. 
 
 With LO = a, OM = J, MN = e, NL = d^ and LM = e, construct, by 
 (316), the triangles LOM and MNL, on LM as a common side. 
 Then is LOMN the quadrilateral sought. 
 
 Fig. 157. 
 
 Fig. 158. 
 
 CASE II. 
 
 Let <i, b, c, and d (Fig. 158), be the given sides in order, and the 
 angle included between a and b. 
 
 With the same notation as before, construct the triangle LOM by 
 (314), and then LMN by (316), and the quadrilateral is constructed, i.e., 
 all the parts are found. 
 
DETERMINATION OF QUADRILATERALS. 
 
 146 
 
 CASE III. 
 
 Let a, b, and c (Fig. 159) 
 be the given sides in order. 
 
 1st, Let both the given 
 angles and M be in- 
 eluded between the given 
 sides, being included by 
 by a and b, and !A by b 
 and c. 
 
 Construct an angle LOM 
 = 0, and take OL = o and 
 OM = 5. 
 
 Now lay off the angle OMN 
 = M, and taking MN = c, 
 draw LN. 
 
 Then is LOMN the quadri- 
 lateral Bought. 
 
 Fig. I5». 
 
 2d, Ambiguous Cases. — If three sides and two angles of 
 a quadrilateral are given, and both the given angles are not 
 included between given sides, the case may be Ambiguous. 
 
 There may be three cases : 1st. "When the two given angles are con- 
 secutive, and one only is included between given sides; 2d. When the 
 given angles are consecutive, and the includ- 
 ed side is unknown ; 3d. When the given 
 angles are opposite. 
 
 Fig. 160 shows how an ambiguous solu- 
 tion may arise under Case 1. The given 
 parts are a, &, c, and angles L and 0. 
 
 Fig. 161 shows how such solutions 
 may arise under cases 2 and 3. 
 
 Fig. 161. 
 
146 ELEMENTARY GEOMETRY, 
 
 CASE IV. 
 
 IsU Let the three given angles be 0, M, atid N, and,ftr8tf 
 let u and h he two consecutive given sides. 
 
 Since the sum of the angles of a quadrilateral is 4 right angles, and 
 0, M, and N are given, the fourth, L, can be found (241). 
 [Let the student make the construction.] 
 
 2d, Let 0, M, avid N he the given angles, and a and c the 
 given non-consecutive sides, d and h helug non-parallel, 
 i. e,, the angles L and iiot helng supplemental. 
 
 Find the fourth angle by subtract- 
 ing the sum of the three given angles 
 from 4 right angles. Whence all the 
 angles are known. 
 
 Lay off side a and at its extremities 
 make LOX = 0, and OLY = L. 
 
 Then draw any line, Arw, making 
 the angle m = M. 
 
 Take ink = c, and through A draw 
 AS parallel to OX. Let this intersect 
 LY in N. Through N draw NM parallel ^^^^^ ^. ^^^ 
 to Am. 
 
 Then is NM = the given side c (i), and OMN = the given angle M (?)^ 
 and LNM = the given angle N (?). 
 
 Hence LOMN is the required quadrilateral. 
 
 321. Scholium.— With a given set of parts, as above, the possibility 
 of constructing a quadrilateral can be determined on the same principle 
 as the possibility of a triangle. 
 
 1. In Case I, if the diagonal is less than the sum and greater than the 
 difference of the sides of either of the two triangles into which it divides 
 the quadrilateral, the quadrilateral is possible, but not otherwise. 
 
 2. In Case II, the two given sides and their included angle always 
 make a triangle possible; whence the possibility of the quadrilateral will 
 be determined by the relation of the other two sides to the third side of 
 this triangle, as (Fig. 158) when c -^ d > LM, and c-d < LM, the 
 quadrilateral is possible, but not otherwise. 
 
 3. In Case III, the Ist problem is always possible. The student will 
 be able to determine when the several cases in the 2d are possible by in- 
 specting Figs. 160 and 161. 
 
 4. In Case I\', the first problem is always possible when the sum of 
 
 ^?°f7^ 
 
 ^f] 
 
 
 K 
 
 -^0 
 
 
 U 
 
 -■-'"'^CA^ 
 
 
 
 
 
 
 Kv ^^ 
 
DETERMINATION OF POLYGONS. 147 
 
 the given angles is less than 4 right angles. In the second problem, if 
 the unknown sides are parallel, the problem is indeterminate, i. e., there 
 may be any number of solutions, if any. 
 
 Note. — In problems of this class, it is usually understood that the 
 given parts are such as to allow the construction ; i. c., that they are parts 
 of a possible polygon. 
 
 322. Corollary 1. — A parallelogrartv is determined 
 when two sides and their included angle are given. 
 
 Since the opposite sides of a parallelogram are equal (250), all the 
 sides are known when two are given, and the case falls under Case II of 
 the proposition, 
 
 323. Corollary 2. — Ihvo rectangles having equal bases 
 and equal altitudes are equal. 
 
 Exercise 1. Construct a quadrilateral three of whose con- 
 secutive sides are 20, 12, and 15, and the angle included between 
 20 and the unknown side f of a right angle, and that between 
 15 and the unknown side J a right angle. 
 
 Ex. 2. Construct a quadrilateral three of whose sides shall be 
 5, 4.2, and 4, and in which the angle between the unknown side 
 and the side 5 shall be J^ of a right angle, and that between the 
 unknown side and side 4, 1 J right angles. How many solutions 
 are there? How many solutions if the second side is made 1.2, 
 and the third 2 ? How many if the second side is made 1, and 
 the third 1.5 ? 
 
 OF PO L YGO NS 
 
 PROPOSITION XIII. 
 
 324. Theorem. — A polygon is determined when two 
 consecutive sides, the diagonals from the vertex of their 
 included angle, and the consecutive angles included he- 
 tween these lines are given. 
 
 [Let the student show how the construction is made, and thus demon- 
 strate the proposition.] 
 
148 ELEMENTARY GEOMETRY, 
 
 PROPOSITION XIV. 
 
 325. Theorem.—./^ polygon is determined by vieans 
 
 of its sides and angles, when there are given in order : 
 I. All the parts except two angles and their included side. 
 TI. All the parts except three angles. 
 III. All the parts except two non-parallel sides. 
 
 C o If s T R u c T I N s . 
 
 CASE I. 
 
 Beginning at one extremity of the unknown side, and constructing 
 the given sides and angles in order till all are constructed, and joining 
 the extremities of the broken line thus drawn, the polygon will be con- 
 structed. 
 
 CASE II. 
 Ist, When the three angles lire consecutive. 
 
 Suppose the polygon to be ABCDEFG, and the unknown angles A, G, 
 and F. Commencing with side AB, lay oflf the given sides and angles in 
 order till the unknown angle F is reached. Then from F as a centre, 
 with a radius equal to the known side FG, strike an arc intersecting an 
 arc struck from A as a centre with the side AG as a radius. This inter- 
 section determines the remaining vertex of the polygon. 
 
 Query — When does this case become impossible ? 
 
 2d, When two of the mi known angles are consecutive and 
 the third is separated from both the others. 
 
 Let A, B, and F be the unknown angles. 
 The two partial polygons AiHGF and 
 BCDEF can be constructed, and thus the 
 sides AF and BF will become known, as 
 also the angles AFG, lAF, BFE, and 
 FBC. 
 
 Then constructing the triangle ABF, 
 whose three sides are now known, the 
 angles AFB, ABF, and FAB become known. 
 Hence all the parts of the polygon are 
 found, for ^'9- '"• 
 
 the angle GFE = AFG + AFB -f- BFE, etc. 
 
DETERMINATION OF POLYGONS. 
 
 149 
 
 3d, When no two of the three unknown angles are con- 
 secutive. 
 
 Let A, C, and F be the unknown 
 
 Constructing the broken lines ABC, 
 CDEF, and FGHIA separately, and 
 apart from the position where the 
 polygon is to be constructed, the diago- 
 nals which form the sides of the triangle 
 ACF can be determined by joining the 
 extremities A and C, C and F, and F 
 and A. 
 
 This triangle can then be con- 
 structed in the position desired, and 
 
 Fig. 164. 
 
 the broken lines constructed on its sides, as in the figure 
 
 CASE III. 
 
 Under this case we have two problems: 
 
 1st, When the tivo unknotvn sides are consecutive, 
 
 2d, When the two unknown sides are separated. 
 
 [The student will be able to effect the construction. The first is 
 similar to that of Case II, 1st problem. The second is effected by 
 obtaining a quadrilateral similarly to the 
 construction in Case II, 3d problem. 
 
 326- In case the unknown parts are 
 two parallel sides, as a and 6, it is evident 
 that these may be varied in length at plea- 
 sure without changing the value of the other 
 parts. ' 
 
 327- It will be a profitable exercise for 
 the student to reduce the determination of polygons to that of quadri- 
 laterals, and both to that of triangles. 
 
150 ELEMENTARY GEOMETRY, 
 
 PROPOSITIONS FOR ORIGINAL SOLU- 
 TION AND DEMONSTRATION. 
 
 328. 1. Theorem,— IT^e sum of the exterior angles of 
 a polygon is four right angles. 
 
 Prove by drawing lines from a point and parallel to the sides of the 
 polygon. 
 
 2. Theorem. — The sum of the angles of a polygon is 
 twice as many right angles as the polygon has sides, 
 less four right angles. 
 
 Having proved the jweceding, base the proof of this upon that. 
 
 3. Theorem. — // the sum of two opposite sides of a 
 quadrilateral is equal to the sum of the other two op- 
 posite sides, show that a circle can be inscribed in the 
 quadrilateral. 
 
 4. Theorem. — // from, a 
 point without a circle two 
 tangents are drawn, and also 
 a chord joining the points of 
 tangency, the angle included 
 between a radius drawn to 
 either point of tangency and 
 the chord is half the angle 
 included between the tan- 
 gents. ^''^' "'• 
 
 5. Theorem. — In an isosceles triangle the line drawn 
 from the vertex to the middle of the base bisects the 
 triangle and also the angle at the vertex. 
 
 6. Problem. — With a given radius draw a circle tan- 
 gent to the sides of a given angle. 
 
 7. Problem. — Through a given point within a given 
 angle draw a line which shall make equal angles with 
 the sides. 
 
EQUIVALENCY AND AREA. 
 
 151 
 
 8. Problem. — To draw a circumference through two 
 given points and having its centre in a given line; or, 
 to find in a given line a point equally distant from two 
 points out of that line. 
 
 •9. Theorem.—// from the ^ 
 extremities of a diameter per- 
 pendiculars are let fall ot^ any 
 secant, the parts intercepted 
 between the feet of these per- 
 pendiculars and the circum- 
 ference are equal. 
 
 Fig. 167. 
 
 10. Problem. — To trisect a right angle. 
 
 Suggestion. — What is the value of an angle of an equilateral tri- 
 angle ? 
 
 OF EQUIVALENCY AND AREA. 
 
 329. Equivalent Figures are such as are equal in mag- 
 nitude. 
 
 330. The Area of a surface is the number of times it con- 
 tains some other surface taken as a unit of measure ; or it is the 
 ratio of one surface to another assumed as a standard of measure. 
 
 PROPOSITION I. 
 
 331. Theorem. — Parallelograms having equal bases 
 and equal altitudes are equivalent. 
 
152 elementary geometry. 
 
 Demonstration. 
 
 Let ABCD and EFGH be two parallelograms having equal bases, BC 
 and FG, and equal altitudes. 
 
 We are to prove 
 that the parallelo- 
 grams are equivalent. 
 
 Apply EFGH to 
 ABCD, placing FG in 
 its equal BC; and, 
 
 since the altitudes are ^'9' 
 
 equal, the upper base EH will fall in AD or AD produced, as E'H'. 
 
 Now, the two triangles AE'B and DH'C are equal, since they have two 
 sides and the included angle of the one equal to two sides and the in- 
 cluded angle of the other; viz., AB = DC, being opposite sides of a 
 parallelogram; and for a like reason BE' = CH'. Also, angle ABE' = 
 angle DCH', by reason of the parallelism of their sides (294)- 
 
 These triangles being equal, 
 
 the quadrilateral ABCH' — the triangle AE'B = ABCH' — DH'C. 
 But ABCH' - AE'B = E'BCH' = EFGH; 
 
 and ABCH' - DH'C = ABCD. 
 
 Hence, ABCD = EFGH. Q. e. d. 
 
 332. Corollary. — Any parallelogram is equivalent to a 
 rectangle having the same base and altitude. 
 
 PROPOSITION II. 
 
 333. Theorem. — A triangle is equivalent to one-half 
 of any parallelogram having an equal base and an 
 equal altitude with the triangle. 
 
 Demonstration. 
 
 Let ABC (Fig. 169) be a triangle. 
 
 We are to prove that ABC is equivalent to one-half a parallelogram 
 having an equal base and an equal altitude with the triangle. 
 
EQUIVALENCY AXD AREA 
 
 153 
 
 Consider AB as the base of the triangle, 
 and complete the parallelogram ABCD by 
 drawing AD parallel to EC, and DC to AB. 
 
 Now ABCD has the same base, AB, as 
 the triangle, and the same altitude, since 
 the altitude of each is the perpendicular 
 distance between the parallels DC and AB. 
 
 But ABC is half of ABCD (249), and as ^'9- '^^• 
 
 any other parallelogram having an equal base and altitude with ABCD is 
 equivalent to ABCD (331), ABC is equivalent to one-half of any parallel- 
 ogram having an equal base and altitude with ABC. Q. e. d. 
 
 334. Corollary 1. — J. tHangle is equivalent to one-half 
 of a rectangle having an equal base and an equal al- 
 titude with the triangle. 
 
 335. Corollary 2.~ Triangles of equal bases and equal 
 altitudes are equivalent, for they are halves of equivalent 
 parallelograms. 
 
 PROPOSITION III . 
 
 336. Theorem. — The square described on n times a 
 line is n^ times the square described on the line, n being 
 any integer. 
 
 Demonstration. 
 
 Let u be any line and AB a line n times as long, n being any integer. 
 
 We are to prove that the square de- 
 scribed on AB is V? times the square 
 on Aa. 
 
 Construct on AB the square ABCD. 
 
 Since m is a measure (76) of AB, by 
 hypothesis, divide AB into n equal parts 
 by applying m, and at the points of di- 
 vision a, ft, c, etc., draw parallels to AD. 
 
 In like manner divide AD, and draw 
 through the points of division a', 5', c', 
 etc., parallels to AB. 
 
 Then are the surfaces 1, 2, 3, 4, 5, 6, Pi^. 170 
 
154 
 
 ELEMENTAR Y GEOMETB T, 
 
 etc., squares, since their opposite sides 
 are parallel (139) and equal (138), and 
 their angles are right angles (125)^ 
 
 Now of these squares there are n in 
 each of the rectangles a'B, 5'E, etc. (?), 
 and as there are n divisions in AD, there 
 are n rectangles. 
 
 Hence there are n times n, or n^ 
 squares in ABCD. q. e. d. 
 
 Fig. 170. 
 
 337. Corollary. — The square described on twice a line 
 is four times the square described on the line ; that on 3 
 times a line is 9 times the square on the line, etc. 
 
 PROPOSITION IV. 
 
 338. Theorem. — A trapezoid is equivalent to two tri- 
 angles having for their bases the upper and' lower bases of 
 the trapezoid, and for their common altitude the altitude 
 of the trapezoid. 
 
 By constructing any trapezoid, and drawing either diagonal, the stu- 
 dent can show the truth of this theorem. 
 
 PROPOSITION V. 
 
 339. Problem.- 
 
 lent triangle. 
 
 To reduce any polygon to an equiva- 
 
 SOLUTION. 
 
 Let ABCDEF (Fig. 171) be a polygon. 
 
 We are to reduce it to an equivalent triangle. 
 
 Draw any diagonal, as EC, between two alternate vertices, and through 
 the intermediate vertex, D, draw DH parallel to EC and meeting BC pro- 
 duced in H. Then draw EH. 
 
EQUIVALENCY AND AREA, 
 
 155 
 
 In like manner, draw 
 FH, and through E draw El 
 parallel thereto, meeting 
 BH produced in I. Then 
 draw Fl. 
 
 Again, draw the diag- 
 onal FB, and through A 
 draw AG parallel thereto, 
 meeting BC produced in G. 
 Then draw FG. 
 
 Now FGI is equivalent to ABCDEF 
 
 N 
 
 
 K ^M\ 
 
 Fig. 171. 
 
 Demonstration of Solution. 
 
 Consider the polygon ABCDEF as diminished by ECD and then in- 
 creased by ECH. Since these triangles have the same base EC, and the 
 same altitude (as their vertices lie in DH parallel to EC, and parallels are 
 everywhere equidistant), the triangles are equivalent (335)- Hence, 
 ABHEF is equivalent to ABCDEF (?). 
 
 In like manner ABIF is equivalent to ABHEF, and FGI to ABIF. 
 
 Hence FGI is equivalent to ABCDEF. q. e. d. 
 
 A R E A. 
 
 340. An Infinitesimal is a quantity conceived under such 
 a law as to be less than any assignable quantity. 
 
 Illustration.— Consider a line of any finite length, as one foot. 
 Conceive this line bisected, and one-half taken. Again conceive this half 
 bisected, and one-half of it taken. By this process it is evident that the 
 line may be reduced to a line less than any assignable line. Moreover, if 
 the process be considered as repeated infinitely, the result is an infini- 
 tesimal. 
 
 Tins is the familiar conception of the last term of a decreasing infinite 
 progression, the last term of which is called zero. 
 
 341. Principle I. — In comparison with finite quanti- 
 ties, an infinitesimal is zero. 
 
156 ELEMENTARY GEOMETRY, 
 
 Thus, suppose — = a, 
 
 w, n, and a being finite quantities. Let i represent an infinitesimal ; then 
 
 m ± i m m ± i 
 
 , or ., or :, 
 
 n ' n ±1 n±i 
 
 is to be considered as still equal to a, for to consider it to difi*er from a 
 
 by any amount we might name, would be to assign some value to *'. 
 
 342. Principle II. — Any two geometrical magnitudes 
 of the sam^e kind are to he conceived as commensurable hy 
 an infinitesimal unit. 
 
 By the process for obtaimng the common measure of two lines (84), 
 the remainder may be made (in conception) less than any assignable quan- 
 tity, and hence in comparison with the lines should be considered zero. 
 
 The same conception may be applied to any geometrical magnitudes. 
 
 PROPOSITION VI. 
 
 343. Theorem.- Rectangles are to each other* as the 
 products of their respective bases and altitudes. 
 
 First Demonstration. 
 
 Lemma. — Two rectangles of equal altitudes are to each 
 other as their bases. 
 
 Let ABCD and abed be two rectangles having their altitudes AD and 
 ad equal. 
 
 Suppose rectangle ABCD gen- 
 erated by the movement of AD from 
 AD to BC,it remaining all the time 
 parallel to its first position, and 
 suppose ahcd generated in like man- 
 ner by the movement of ad. 
 
 Let these equal generatrices AD ^'3' '^^• 
 
 and ad move with uniform and equal velocities ; then it is evident that 
 the surfaces generated will be as the distances AB and ab. 
 
 That is, ABCp^AB. 
 
 abed ab 
 
 * This is a common elliptical form, meaning that surfaces, or areas, are 
 to each other. 
 
EQUIVALENCY AND AREA, 
 
 167 
 
 Now let M and N be any two rectan- 
 gles, the base of M being AB and the 
 altitude BC, and the base of N BE and 
 its altitude BG. 
 
 We are to prove that 
 M A B X BC 
 N "" BExBG* 
 
 Place the rectaDgles so that the 
 angles ABC and GBE shall be opposite, 
 i. e., so that AG and CE shall be straight Fig. I73. 
 
 lines (109) 
 
 Complete the rectangle CBGH, and call it 0. 
 
 Since M and have equal altitudes, 
 
 - = — . (1) 
 
 BG ^ ' 
 
 In like manner, since N and have equal altitudes, 
 
 N_ BE 
 0~ BC 
 
 (3) 
 
 Dividing the members of (1) by the corresponding members of (2), 
 we have 
 
 M AB X BC 
 
 n = beTbg- ^•^^• 
 
 Secon"d Demonstration. 
 Let ABCD and EFGH be any two rectangles. 
 
 We are to prove that 
 
 Fig. 174. 
 
 ABCD AB ) 
 
 AD 
 
 EFGH EF X EH 
 
 The bases and altitudes of the two rectangles are at least to be con- 
 sidered as commensurable by an infinitesimal unit (342)- 
 
158 
 
 ELEMENTARY GEOMETRY. 
 
 Fig. 174. 
 
 Let i be the common measure of AB, AD, EF, and EH, and suppose it 
 contained in AB m times, in AD ti times, in EF p times, and in EH q times. 
 
 AB AD 
 
 Whence, m = ^ , 7i = -;- 
 
 EF ^ EH 
 
 Now conceive the rectangles divided into squares by drawing through 
 the points of division of the bases and altitudes parallels to the altitudes 
 and bases, as in (336), whence the rectangles will be divided into equal 
 squares. 
 
 Of these equal squares, ABCD contains m x n, and EFGH p^-q. 
 
 Therefore 
 
 ABCD rnxn 
 
 AB AD 
 
 — r- X — r- 
 l I 
 
 ABxAD 
 
 EFGH ~ pxg - EF EH EFxEH 
 
 — :- X — r- 
 % % 
 
 Q. E. D. 
 
 PROPOSITION VII. 
 
 344. Theorem.— The area of a rectangle is equal to 
 the product of its base and 
 altitude. 
 
 Demonstration. 
 Let ABCD be a rectangle. 
 
 We are to prove that its area 
 is AB X AD. 
 
 Let the square u be the pro- 
 posed unit of measure, whose side 
 isl. 
 
 
 Fig. 175. 
 
 \^^V^l 
 
EQUIVALENCY AND AREA. 159 
 
 T, /«^A^ ABCD AB X AD .„ ._ 
 
 By (343), — ^- = , = AB X AD. 
 
 Hence, by (330), area ABCD = AB x AB. Q. E. D. 
 
 346. Corollary 1. — The area of a square is equal to the 
 second power of one of its sides, as in this case the base and 
 altitude are equal. 
 
 346. Corollary 2. — The area of any parallelogram is 
 equal to the product of its hose and altitude ; for any paral- 
 lelogram is equivalent to a rectangle of the same base and 
 altitude (332). 
 
 347. Corollary 3. — The area of a triangle is equal to 
 one-half the product of its base and altitude ; for a tnangle 
 is one-half of a parallelogram of the same base and altitude (333). 
 
 348. Corollary 4. — Parallelograms or triangles of 
 equal bases are to each other as their altitudes; of 
 equal altitudes, as their bases ; and in general they are 
 to each other as the product of their basses by their al^ 
 titudes, 
 
 349. Scholium. — The arithmetical signification of the theorem, The 
 area of a rectangle is equal to the product of its hose and altitude^ is this : 
 
 Let the base be 6 and the altitude a ; then we have, by the prop- 
 osition, 
 
 area = ah. 
 
 Now, in order that ah may represent a surface, one of the factors 
 must be conceived as a surface and the other as a number. Thus, we 
 may conceive h to represent h superficial units, i. e., the rectangle having 
 the base of the rectangle for its base and being 1 linear unit in.altitude. 
 
 The entire rectangle is, then, a times the rectangle which contains b 
 superficial units, or ah superficial units. 
 
 In the expression 
 
 area ABCD = AB x AD, 
 
 AB and AD may be given a similar interpretation. 
 
160 ELEMENTARY GEOMETRY, 
 
 PROPOSITION VIII. 
 
 360. Theorem. — The area of a trapezoid is equal to 
 the product of its altitude into one-half the sum of its 
 parallel sides, or, what is the same thing, the product 
 of its altitude into a line joining the middle points of 
 its inclined sides. 
 
 Demonstration^. 
 
 Let ABCD be a trapezoid, whose parallel sides are AB and DC, and 
 whose altitude is IK. 
 
 We are to prove, Ist, that 
 
 An^n. .., AB + CD 
 area ABCD = IK x — ^ — ' 
 
 and, 2d, that area ABCD - \K x db, ^ .^r 
 
 . ' ' r ig. I/O. 
 
 ab being a line joining the middle points of AD and BC. 
 
 Draw either diagonal, as AC. The trapezoid is thus divided into two 
 triangles, whose areas are together equal to one-half the product of their 
 common altitude (the altitude of the trapezoid) into their bases DC and 
 AB, or this altitude into ^ (AB + DC). Q. B. D. 
 
 At a and h draw the perpendiculars om and pn, meeting DC, pro- 
 duced, if necessary. 
 
 Now the triangles aoD and A«w are equal, since 
 
 Aa = aD, 
 
 angle o = angle m, 
 
 both being right, and angle oaD = Aa>n, being opposite. Whence 
 
 f^m = oD. 
 
 In like manner, we may show that 
 
 Cp = nB. 
 
 Hence, ab = ^(op + mn) (?) = ^ (AB + DC) ; and area ABCD, which 
 equals | (AB + DC) x IK, = a5 x IK. Q. E. D. 
 
EQUIVALENCY AND AREA, 
 
 161 
 
 PROPOSITION IX. 
 
 351. Theorem. — The area of a regular polygon is 
 equal to one-half the product of its apothem into its 
 perimeter. 
 
 Demonstration. 
 
 Let ABCDEFG be a regular polygon, whose perimeter is AB + BC + 
 CD + DE + EF + FG + GA, and whose apothem is Oa, 
 
 We are to prove that 
 
 area ABCDEFG = iOa(AB + BC + CD + DE + EF + FG + GA). 
 
 Draw the inscribed circle, the radii Oa, 05, 
 etc., to the points of tangency, and the radii of 
 the circumscribed circle OA, OB, etc. (273, 
 274). 
 
 The polygon is thus divided into as many 
 equal triangles as it has sides. 
 
 Now, the apothem (or radius of the in- 
 scribed circle) is the common altitude of these 
 triangles, and their bases make up the perimeter 
 of the polygon. *''9- "77. 
 
 Hence, the area = ^Oa (AB + BC + CD + DE + EF + FG -I- GA). q. e. d. 
 
 852. CoROLLART. — The area of any polygon in which 
 a circle can he inscribed is equal to one-half the 
 product of the radius of the inscribed circle into the 
 perimeter. 
 
 The student should draw a figure and observe the fact. It is espe- 
 cially worthy of note in the case of a triangle. See Fig. 187. 
 
 PROPOSITION X. 
 
 353. Lemma. — // any polygon is circum^scribed 
 about a circle and a second polygon is formed by draw- 
 ing tangents to the arcs intercepted between the con- 
 secutive points of tangency, thus forming a polygon of 
 double the number of sidles, the perimeter of the second 
 polygon is less than that of the first. 
 
162 
 
 ELEMENTARY GEOMETRY. 
 
 Demonstration. 
 
 Let ABODE be any circumscribed polygon, whose consecutive sides 
 are tangent at K, F, G, etc., and let a second polygon be formed by 
 drawing tangents at f, g, etc. 
 
 We are to prove that the 
 perimeter ab -^-hc -\- cd^ etc., is less 
 than the perimeter EA + AB + etc. 
 
 Observing the portions of the 
 perimeters from K to F, for the 
 first polygon we have 
 
 KA + AF = Ka + (aA + Aft)-f5F, 
 and for the second 
 
 But db < ak-\-fib (?). 
 
 Hence, 
 Ka + ab + W < KA + AF. 
 
 Now, as a similar reduction 
 will take place at each vertex, the entire perimeter of the second polygon 
 will be less than that of the first, q. e. d. 
 
 364. The Limit of a varying quantity is a fixed quantity 
 which it approaches by such a law as to be capable of b.eing 
 made to differ from it by less than any assignable quantity. 
 
 Such a varying quantity is often spoken of as reaching its 
 limit after an infinite number of steps of approach. 
 
 365. Corollary. — As the number of the sides of a cir- 
 cumscribed regular polygon is increased the perimeter 
 is diminished, andj approaches the circumference of the 
 circle as its limit, since tlie circle is the limit of such a poly- 
 gon. 
 
 Fig. 178. 
 
 PROPOSITION XI. 
 
 366. Tlieorem. — The area of a circle is equal to one- 
 half the product of its radius into its circumference. 
 
 Demonstration. 
 Let Oa (Fig. 179) be the radius ofthe circle. 
 
EXERCISES. 
 
 163 
 
 "We are to prove that the area of the circle 
 is ^Oa X the circumference. 
 
 Circumscribe any regular polygon. 
 
 Now the area of this j^olygon is one-half 
 the product of its apothem and perimeter. 
 
 Conceive the number of sides of the poly- 
 gon indefinitely increased, the polygon still 
 continuing to be circumscribed and regular. 
 
 The apothem continues to be the radius of 
 the circle, and the perimeter approaches the 
 circumference. 
 
 When, therefore, the number of sides of the polygon becomes infinite, 
 it is to be considered as coinciding with the circle, and its perimeter with 
 the circumference (366). 
 
 Hence the area of the circle is equal to one-half the product of its 
 radius into its circumference, q. e. d. 
 
 Fig. 179. 
 
 367. A Sector is a part of a circle included between two 
 radii and their intercepted arc. 
 
 368. Corollary 1. — The area of a sector is equal to 
 one-half the product of the radius into the arc of the 
 sector. 
 
 369. Corollary 2. — The area of a sector is to the area 
 of the circle as the arc of the sector is to the circumfer- 
 ence, or as the angle of the sector is to four right angles. 
 
 EXERCISES. 
 
 360. 1. What is the area in acres of a triangle whose base is 
 75 rods and altitude 110 rods ? 
 
 2. What is the area of a right-angled triangle whose sides 
 about the right angle are 126 feet and 72 feet ? 
 
 3. If two lines are drawn from the vertex of a triangle to the 
 base, dividing the base into parts which are to each other as 2, 3, 
 and 5, how is the triangle divided ? How does a line drawn 
 
164 ELEMENTARY GEOMETRY, 
 
 from an angle to the middle of the opposite side divide a tri- 
 
 V 
 
 le 
 
 4. What is the area of the largest triangle which can be in- 
 scribed in a circle whose radius is 12, the diameter being one 
 side? 
 
 5. What is the area of a cross section of a ditch which is 
 6 feet wide at the bottom, 9 feet at the top, and 3 feet deep ? 
 
 6. If one of the angles at the base of an isosceles triangle is 
 double the angle at the vertex, how many degrees in each ? 
 
 OF SI M I LARITY. 
 
 361. The primary notion of similarity is likeness of form. 
 Two figures are said to be similar which have the same shape, 
 although they may differ in magnitude. A more scientific defi- 
 nition is as follows : 
 
 362. Similar Figures are such as have their angles re- 
 spectively equal, and their homologous sides proportional. 
 
 363. Homologous Sides of similar figures are those 
 which are included between equal angles in the respective figures. 
 
 364. In similar triangles, the homologous sides are 
 those opposite the equal angles. 
 
 The student should be careful, at the outset, to mark the fact that 
 similarity involves two things^ equality of angles and proportionality 
 OP SIDES. It will appear that, in the case of triangles, if one of these 
 facts exists, the other exists also ; but this is not so in other polygons. 
 
 365. Two figures are said to be Mutually equiangular 
 when each angle in one has an equal angle in the other, and 
 Mutually equilateral when each side in the one has an equal 
 side in the other. 
 
SIMILARITY. 
 
 165 
 
 PROPOSITION I. 
 
 366. Theorem. — Triangles which are mutually equi- 
 angular are similar. 
 
 Demonstration. 
 
 Let ABC and DEF be two mutually equiangular triangles, in which 
 A = D, B = E, C = F. 
 
 We are to prove that the sides 
 opposite these equal angles are pro- 
 portional, and thus that the triangles 
 possess both the requisites of similar- 
 ity, viz., equality of angles and pro- 
 portionality of sides. 
 
 Lay off on CA CD' = FD, and on 
 CB CE' = FE, and draw D'E'. 
 
 Triangle CD'E' equals triangle 
 FDE (?). 
 
 Draw AE' and BD'. 
 
 Pig. 180. 
 
 Since angle CE'D' = CBA, D'E' is " 
 
 parallel to AB (?), and as the triangles D'E'B and D'E' A have a common 
 base D'E' and the same altitudes, their vertices being in a line parallel 
 to their base, they are equivalent (335). 
 
 Now the triangles CD'E' and D'E'A, having a common altitude, are to 
 each other as their buses (348). 
 
 or 
 
 or 
 
 Hence, 
 
 CD'E' CD' 
 D'E'A - DA* 
 
 For like reason, 
 
 CD'E' CE' 
 D'E'B - E B* 
 
 Whence, as 
 
 DEB = D E'A, 
 
 
 CD' CE' 
 DA ~ E'B 
 
 By composition, 
 
 CD' CE' 
 
 CD' ' [>'* CE' + E'B 
 
 
 CD CE' 
 
 
 CA 'CB ' 
 
 
 . - FE 
 
 
 CA CB 
 
166 ELEMENTA R Y GEOMETR Y. 
 
 In a similar manner, by laying off ED and EF in BA and BC respec- 
 tively, wenjan show that 
 
 FE ED 
 CB ~ BA * 
 
 „ FD FE ED 
 
 ^^^^^' CA = CB^BA-^-^-^- 
 
 367. Corollary 1. — // two triangles have two angles of 
 one respectively equal to two angles of the other, the tri- 
 angles are similar (?). 
 
 368. Corollary 2. — A transversal parallel to any side 
 of a triangle divides the other sides proportionally, and 
 the sides are in the ratio of either two corresponding 
 segments. 
 
 For in the demonstration we have D'E' parallel to AB, and 
 
 or 
 
 And also 
 or, by alternation, 
 
 CD' CE' 
 DA ~ E'B' 
 
 
 CD' DA 
 CE' ~ E'B ' 
 
 
 CD' CE' 
 CA ~ CB ' 
 
 
 CA CD' 
 CB ~ CE' ~ 
 
 DA 
 E'B 
 
 PROPOSITION II. 
 
 369. Theorem. — // any iwo transversals cut a series 
 of parallels, their intercepted segments are proportional. 
 
 Demonstration. 
 
 l8t. Let OA and O'B' (Fig. 181) be any two parallel transversals 
 cutting the series of parallels ah, cd, r/, yh, etc. 
 
 We are to prove that =-= = v? = -^ , etc. 
 oa af . fh 
 
SIMILARITY. 
 
 167 
 
 Now 
 Hence. 
 
 hd ' df ' 
 
 = 1. etc. (?) 
 
 ac_ce_eg^ 
 hd~ df~ fV 
 
 Q. E. D. 
 
 2ih Let OA and OB 
 beanynon-paralleltrans- 
 versals cutting ah, cd, 
 «/> O^i'9 etc. 
 
 "We are to prove 
 
 Fig. 181. 
 
 ae _ce_ _eg 
 
 and 
 
 Since OA and OB are non-parallel, they meet in some point, as 0. 
 Then, by (368), we have q^ ^ ^ 
 
 Oc _ C6 
 
 Od~df' 
 
 Whence, by equality of ratios, we have 
 
 ae _ M 
 bd~df' 
 
 ee 
 
 eg 
 
 Similarly, we may show that ^= tt^ 6^c. 
 
 Hence, also, by alternation, and by equality of ratios, 
 
 - = ^.. ^^ = '^^ a«d ^ = ^, etc. Q.E.D. 
 ee df hd fh' eg fh' 
 
 PROPOSITION III. 
 
 370. Theorem. — Conversely to Prop. I, If two triangles 
 have their corresponding sides proportional, they are sim- 
 ilar. 
 
 Demonstration. 
 
 AC OB BA 
 
 Let ABC and DEF have 
 
 OF FE ~ ED 
 
168 
 
 ELEMENTARY GEOMETRY, 
 
 We are to prove that ABC is 
 similar to DEF. 
 
 As one of the characteristics of 
 similarity, viz., proportionality of 
 sides, exists by hypothesis, we have 
 only to prove the other, i. e., that 
 
 A = D, C = F, and B = E. 
 
 Make CD' = 
 parallel to AB. 
 
 FD, and draw 
 
 D'E' 
 
 Then, by (368), 
 
 CA CB 
 CD' ~ CE' ' 
 
 
 Fig. 182. 
 
 and since by construction 
 and by hypothesis 
 
 CD' = FD, 
 
 CA 
 FD 
 
 CB 
 FE' 
 
 CE' = FE. 
 
 Again the triangles D'E'C and ABC are mutually equiangular, since C 
 is common, angle CD'E'= CAB (?), and angle CE'D' = CBA (?). 
 
 Whence 
 
 CA^ 
 CD' 
 
 AB 
 D'E' 
 
 But by hypothesis and construction 
 
 CA 
 CD' 
 
 CA 
 DF 
 
 AB 
 DE 
 
 Hence D'E' = DE, and the triangles CD'E' and DEF are equal (?). 
 
 Therefore ABC and D'E'C are similar; and as D'E'C = DEF, ABC and 
 DEF are similar, q. e. d. 
 
 371. Scholium. — As we now know that if two triangles are mutually 
 equiangular, they are similar ; or, if they have their corresponding sides 
 proportional, they are similar, it will be sufficient hereafter, in any given 
 case, to prove either one of these facts, in order to establish the similarity 
 of two triangles. For, either fact being proved, the other follows as a 
 consequence. 
 
SIMILARITY, 
 
 169 
 
 PROPOSITION IV. 
 
 372. Theorem. — Two triangles which have the sides 
 of the one respectively parallel or perpendicular to the 
 sides of the other, are similar. 
 
 Demonstration. 
 
 Let ABC and A'B'C be two triangles whose sides are respectively 
 parallel or perpendicular to each other. 
 
 We are to prove that tlie tri- 
 angles are similar. 
 
 Any angle in one triangle is 
 either equal or supplemental to the 
 angle in the other which is included 
 between the sides which are parallel 
 or perpendicular to its own sides. 
 Thus, A either equals A', or A + A' 
 = 2 right angles (294, 295, 296). 
 
 Now, if the corresponding angles 
 
 are all supplemental, that is, if 
 
 A + A' = 2 right angles, 
 
 B + B' = 2 right angles, 
 
 and C + C' = 2 right angles, 
 
 the sum of the angles of the two 
 triangles is 6 right angles, which is 
 impossible. 
 
 Again, if one angle in one triangle equals the corresponding angle in 
 the other, as A = A', and the other angles are supplemental, the sum is 
 4 right angles plus twice the equal angle, wiiich is impossible. Hence, 
 two of the angles of one triangle must be equal respectively to two 
 angles of the other. Therefore the triangles are similar (367)- Q. e. d. 
 
 PROPOSITION V. 
 
 373. Theorem. — Two triangles having an angle in 
 one equal to an angle in the other, and the sides about the 
 equal angles proportional, are similar, 
 
 8 
 
170 
 
 ELEMENT4.HY GEOMETRY. 
 
 AC 
 DF 
 
 CB 
 FE 
 
 Demonstration. 
 
 Let ABC and OEF have the angles C and F equal, and 
 
 We are to prove that ABC and DEF are similar. 
 Make CD' equal to FD, and draw 
 D'E' parallel to AB. Then is 
 
 angle CD'E' = angle CAB, 
 
 whence the triangles are similar (367), 
 and by (368), 
 
 AC CB^ 
 
 DC (= DF) ~ CE' ' 
 
 But, by hypothesis, 
 
 AC CB 
 DF ~ FE' 
 
 Whence CE' ■= FE. 
 
 Hence the triangle CD'E' is equal to the triangle FDE. Now, CD'E' 
 and ABC are mutually equiangular. Hence DFE and ABC are mutually 
 equiangular and consequently similar, q. e. d. 
 
 Fig. 184. 
 
 PROPOSITION VI. 
 
 374. Theorem. — In any right-angled triangle, if a 
 line is drawn from the vertex of the right angle perpen- 
 dicular to the hypotenuse : 
 
 1st. The perpendicular divides the triangle into two 
 triangles, which are similar to the given triangle, and 
 consequently similar to each other, 
 
 2d. Either side about the right angle is a mean propor- 
 tional between the jvhole hypotenuse and the adjacent 
 segment. 
 
 3d. The perpendicular is a jnean proportional between 
 the segments of the hypotenuse. 
 
 Demonstration. 
 
 Let ACB be a triangle right-angled at C, and CD a perpendicular 
 upon the hypotenuse AB ; then 
 
STMtLARlTY. 
 
 171 
 
 Ist. The triangles ACD and ACB have the 
 angle A common, and a riglit angle in each ; 
 hence they are similar (367). For a like rea- 
 son, CDB and ACB are similar. Finally, as ACD 
 and CDB are both similar to ACB, they are 
 similar to each other, q. e. d. ^'9- '^^• 
 
 2d. By reason of the similarity of ACD and ACB, we have 
 
 AD AC 
 AC ~ AB' 
 
 DB CB 
 
 and from CDB and ACB, we have 
 
 CB AB 
 
 Q. E. D. 
 
 3d. By reason of the similarity of ACD and CDB, we have 
 
 AD 
 CD 
 
 CD 
 
 DB" ^' 
 
 E. D. 
 
 Queries. — To which triangle does the first CD belong ? To which 
 the second? Why is CD made the consequent of AD? Why, in the 
 second ratio, are CD and DB to be compared? 
 
 375. Corollary. — // a perpendicular is Let fall from 
 any point in a circumference upon a diameter, this per- 
 pendicular is a mean proportional between the segments 
 of the diameter. 
 
 Let CD be such perpendicular, and draw 
 AC and CB. Then, since ACB is a right angle 
 (192), we have, by Case 3d, the proportion 
 
 ^^§i. or CD-AOxDB. 
 
 ^" "° Fig. 186. 
 
 PROPOSITION VII. 
 
 376. Theorem. — The square described on the hypote- 
 nuse of a right-angled triangle is equivalent to the sum; 
 of the squares described on the other two sides. 
 
 First Demonstration. 
 Let ACB (Fig. 187) be any right-angled triangle. 
 
m 
 
 ELEMENTARY GEOMETRY. 
 
 We are to prove that AB^ =: AC^ + CbI 
 For, let fall the perpendicular CD, and by 
 (374, 3d) we have 
 
 AD 
 AC 
 
 AC 
 AB 
 
 v^ , and 
 
 DB 
 CB 
 
 CB 
 AB 
 
 and 
 
 ADxAB = AC'; 
 DB X AB = CB . 
 
 or 
 
 Fig. 187. 
 
 Adding, we have AB (AD + DB) = AC^ 4- CB^ 
 
 AB X AB = AB' = AC^+ CB'. q. e. d. 
 
 Second Demonstration. 
 Let ABC be any right-angled triangle, right-angled at B. 
 
 Describe the squares AE, AG, and CL on 
 the hypotenuse and the other sides respect- 
 ively. From the right angle let fall upon 
 DE the perpendicular BK intersecting AC in 
 I, and draw the diagonals BE, DB, HC, and 
 AF. 
 
 Now the triangles BAD and HAC are 
 equal, having two sides and the included 
 angle of one equal to two sides and the in- 
 cluded angle of the other; viz., BA = HA, 
 being sides of the same square, and for a 
 like reason AD = AC ; and the angle HAC 
 = BAD, since each is made up of a right 
 angle and the angle BAC. 
 
 Since ABG and ABC are right angles, BG is the prolongation of BC, 
 and the triangle HAC has the same base, HA, and the same altitude, AB, 
 as the square AG. Hence the triangle HAC is half the square AG. 
 
 Moreover, the triangle BAD has the same base, AD, as the rectangle 
 AK, and the same altitude as Al. Hence, 
 
 triangle BAD = ^ADKI. 
 
 Therefore, as the rectangle ADKI and the square AG are twice the 
 equal triangles BAD and HAC respectively, they are equivalent. 
 
 In like manner, the square CL may be shown to be equivalent to the 
 rectangle CK. 
 
 Whence we have ADKI = ABGH, 
 
 and IKEC ^ BCFL; 
 
 and adding, ADEC = ABGH f BCFL. Q. e. d. 
 
 Fig. 188. 
 
SIMILARITY. 173 
 
 377. Corollary 1. — The hypotenuse of a right-angled 
 triangle equals the square root of the sum of the squares 
 of the other two sides. 
 
 Also, either side about the right angle equals the square 
 root of the square of the hypotenuse minus tJie square of 
 the other side. 
 
 378. Corollary 2. — The diagonal of a square is a/2 
 times the side. 
 
 For, let S be the side. Drawing the diagonal, we have a right-angled 
 triangle of which the diagonal is the hypotenuse, and the sides about the 
 right angle are each S. Hence, by the proposition, 
 
 (diag.)« = S' + 8' = 25«, 
 or diag. = 8^'i. 
 
 379. Scholium. — Proposition VI with its corollary, and Prop. VII, 
 which is a direct result of Prop. VI, are perhaps the most Iruitful in 
 direct practical results of any in Geometry. Prop. VII is called the 
 Pythagorean Proposition, its original demonstration being attributed to 
 Pythagoras. 
 
 PROPOSITION VIII. 
 
 380. Theorem. — Regular polygons of the sam,e num- 
 ber of sides are similar figures. 
 
 Demonstration. 
 
 Let P and P' be two regular polygons of the same number of sides, 
 a, b, c, d, etc., being the sides of the former, and a', h' , c' , d' , etc., 
 the sides of the latter. 
 
 Now, by the definition of regular polygons, the sides a, ft, c, d, etc., 
 arc equal each to each, and also a\ b\ c', d\ etc. Hence, we have 
 a^ _ b _ c _ d 
 a'" V ~ d ~ W ' 
 Again, the angles are equal, since n being the number of angles of 
 each polygon, each angle is equal to 
 
174 ELEMENTARY GEOMETRY. 
 
 n X 2 right angles — 4 right angles (norj\ 
 n 
 Hence the polygons are mutually equiangular, and have their corres- 
 ponding sides proportional ; that is, they are similar, q. e. d. 
 
 PROPOSITION IX. 
 
 381. Theorem. — The corresponding diagonals of reg- 
 ular polygons of the same number of sides are in the same 
 ratio as the sides of the polygons. 
 
 [Let the student give the demonstration.] 
 
 PROPOSITION X. 
 
 382. Theorem. — TJie radii of the circumscribed, and 
 also of the inscribed circles, of regular polygons of the same 
 numjber of sides, are in the saine ratio as the sides of the 
 polygons, 
 
 Demonstkatioij. 
 
 Fig 189. 
 
 Let ABCDEF and abcdefbe two regular polygons of the same num- 
 ber of sides, and It and r be the radii of their circumscribed circles, 
 and II' and r' of their inscribed. 
 
 We are to prove that — ( = — , etc. J z= — = — - 
 
SIMILARITY. 176 
 
 Let and 0' be the centres of the polygons, and draw OA, OF, O'a, 
 and 0/, and also the apothenis 01 and O'i. 
 
 OA = R, and O'a = r (?); 
 also 01 = /?, and O'i = r' (?). 
 
 Now the triangles AFO and nfO' are equiangular (?), and hence 
 
 similar. 
 
 rru r AF/ FE ^ \ OA i? 
 
 Therefore, —-( = —-, etc. ) = -r- = - . q. e. d. 
 
 a/ V /« J O'a r ^ 
 
 Again, the triangles AlO and aiO' are mutually equiangular (?), and 
 hence similar. 
 
 Therefore, _ . — ,rvr»> 
 
 Ai_qi 
 
 ai ~ or 
 
 whence, doubling the terms of the first ratio, we have 
 ^F/ F£ ^ \ 0\ B' 
 
 383. Homologrous Altitudes in similar triangles are 
 perpendiculars let fall from the vertices of equal angles upon the 
 sides opposite. 
 
 384. Homologfous Diagonals in similar polygons are 
 diagonals joining the vertices of corresponding equal angles. 
 
 PROPOSITION XI. 
 
 385. Theorem.^-Hojnologous nZtitivdes in simUar trU 
 angles have the same ratio as the homologous sides. 
 [Let the student give the demonstration.] 
 
 PROPOSITION XII. 
 
 386. Theorem. — The bisectors of equal angles of simi^ 
 lar triangles are to each other as the homologous sides of 
 the triangles, hence as the homologous perpendiculars, 
 
 [Let the student ^ve the demonstration,] 
 
176 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XIII. 
 
 387. Theorem. — Homologous diagonals in similar 
 polygons have the same ratio as the homologous sides. 
 
 Demonstration. 
 
 Let ABCDEFG and ahcdefg be two similar polygons, having angle 
 A = angle a, B = 6, C = c, etc. 
 
 Fig. 190. 
 
 We are to prove that 
 
 AC AD ^ 
 
 — , or — 3 , etc. 
 oc ' ad 
 
 AB 
 
 AB 
 
 the ratio — r^ being the ratio of any two homologous sides of the polygons. 
 The triangles ABC and dbc are similar (?), and hence 
 
 AC^AB 
 
 ac ~ ab 
 
 Also, since triangle ABC is similar to a6c, 
 
 angle BCA = angle Jm, 
 
 and subtracting these respectively from the equal angles (?) BCD and Icd^ 
 we have 
 
 angle ACD = angle acd. 
 
 Hence the two triangles ACD and acd have an angle in each equal 
 and the including sides proportional (?), and are consequently similar. 
 
 AD ^ AC ^ AB 
 
 ad ac ~ ab 
 
 Therefore 
 
SIMILARITY. 
 
 177 
 
 In like manner, any homologous diagonals may be shown to have the 
 AB 
 
 ratio '-^^ , which is the ratio of any two homologous sides. 
 ab ' 
 
 Q. E. D. 
 
 388. Corollary 1. — Any two similar polygons are di- 
 vided by their homologous diagonals into an equal number 
 of similar triangles similarly placed, 
 
 389. Corollary 2.— Conversely, Tivo polygons which can 
 be divided by diagonals into the same number of mutually 
 similar triangles, similarly placed, are similar. 
 
 PROPOSITION XIV. 
 390. Theorem. — Circles are similar figures. 
 
 Fig. 191. 
 
 Demonstration. 
 Let 0« and OA be the radii of any two circles. 
 
 Place the circles so that they shall be con- 
 centric, as in the figure. Inscribe the regular 
 hexagons, as ahcdef, ABCDEF. 
 
 Conceive the arcs AB, BC, etc., of the outer 
 circumference bisected, and the regular do- 
 decagon inscribed, and also the corresponding 
 regular dodecagon in the inner circumference. 
 
 These are similar figures by (380). 
 
 Now, as the process of bisecting the arcs 
 of the exterior circumference can be conceived 
 as indefinitely repeated, and the corresponding regular polygons as in- 
 scribed in each circle, the circles may be considered as regular polygons 
 of the same number of sides, and hence similar, q. e. d. 
 
 391. Corollary. — Sectors which correspond to equal 
 angles at the centre are similar figures. 
 
 Since a radius is perpendicular to the circumference of its circle, such 
 sectors are mutually equiangular ; and by the proposition it is evident 
 that the arcs are to each other as the radii. 
 
 i e ^^^^ = ^ 
 ' ' arc FE OF' 
 
 Scholium. — The circle is said to be the limit of the inscribed polygon, 
 and the circumference the limit of the perimeter. By this is meant that 
 
178 ELEMENTARY GEOMETRY, 
 
 as the number of the sides of the inscribed polygon is increased it ap- 
 proaches nearer and nearer to equality with the circle. The apothem 
 approaches equality with the radius, and hence has the radius for its 
 limit. 
 
 PROPOSITION XV. 
 
 392. Problem.— To divide a given line into parts 
 which shall be proportional to several given lines. 
 
 Solution.* 
 
 Let it be required to divide OP into parts pro- 
 portional to the lines A, B, C, and D. 
 
 Draw ON making any convenient angle with 
 OP, and on it lay off A, B, 0, and D, in succession, 
 terminating at M. 
 
 Join M with the extremity P, and draw par- 
 allels to MP through the other points of division. 
 
 Then by reason of the parallels we shall have 
 A:B:C:D :: a : h : c : d (369). 
 
 Fig. 192. 
 
 393. The notation A : B : C : D : : a : 5 : c : c^ is of such frequent 
 occurrence in mathematical writing that we feel constrained to retain it. 
 It means that the successive ratios 
 
 ABC 
 BC D' 
 
 are equal to the successive ratios 
 
 a b c 
 b' c/ d' 
 
 We may read the expression thus : " The successive ratios A to B, 
 B to 0, to D = the successive ratios a to b.b to c, c to <Z." It does rwt 
 mean that the ratio A to B = B to 0, etc. 
 
 * Hereafter we shall change somewhat the style of our demonstrations, 
 from the elementary form hitherto used to the more common and free form 
 used by writers generally. In the " Solution " of a problem we shall here- 
 after usually include the " Demonstration of the Solution." 
 
siMiLAEirr, 
 
 179 
 
 PROPOSITION XVI. 
 
 394. Problem,— To find a fouHh proportional to three 
 given lines. 
 
 Solution. 
 
 Let it be required to find D, a fourth proportional to the lines A, B, 
 and C, so that we shall have g = q ' 
 
 From 8(1111 e point 0, draw two 
 indefinite lines OX, OY. Lay off on 
 OX, Oa = A, and Oc = B. Also, on 
 OY lay off 06 = C, and draw (th. 
 Through c draw cd parallel to oh. 
 Then is Od the fourth proportional, 
 D, which was sought. 
 
 For, since ah and cd are parallel, 
 we have, by (368), 
 
 Oa (or A ) _ Oh or ) ^ 
 Oc (or B) ~ Od (or D) ' 
 
 Hence D is the fourth proportional sought. 
 
 Fig. 193. 
 
 395. Scholium. — In speaking of the fourth proportional to three 
 given lines, it is necessary that the order in which the three are to occur 
 be specified. This order is usually understood to be that in which the 
 lines are named. Thus, a fourth proportional to A, B, and 0, is D, as 
 found above. But a fourth proportional to B, A, and C is quite a differ- 
 ent line from D. 
 
 PROPOSITION XVII. 
 
 396. Problem. — To find a third proportional to two 
 given lines. 
 
 Solution. 
 
 Let A and B be the two given lines. 
 
180 
 
 ELEMENTARY GEOMETRY, 
 
 We are to find a third proportional, 
 «, such that 
 
 A B 
 B~ x' 
 
 The usual solution is the same as the 
 last, C being equal to B. [Let the stu- 
 dent execute it.] Fig. I94. 
 
 Another Solution. 
 Let A and B be the two lines. 
 
 Draw an indefinite line AM, and take 
 AD = A 
 
 At D erect a perpendicular BD and 
 make it equal to B. 
 
 Join A and B, and bisect it by the per- 
 pendicular ON. 
 
 NO will intefsect AM ; since, as A is less Fig- '95. 
 
 than a right angle (?), the sum of the two angles ONA and OAN is less 
 than two right angles (129). 
 
 From as a centre, with OA as a radius, describe a semi-circumfer- 
 ence. It will pass through B (?). 
 
 Now 
 
 AD (or A) BD (or B) 
 
 (?). 
 
 BD (or B) DC (or x) 
 Hence, CD = a;, the required third proportional. 
 
 PROPOSITION XVIII. 
 
 397. Problem. — To find a mean proportional between 
 two given lines. 
 
 Solution. 
 
 Let it be required to find a mean 
 proportional, .r, between M and N, so 
 that 
 
 M _ « 
 
 a; ~ N 
 
 x= ^/Mx "N. 
 
 Fifl. 196. 
 
EXERCISES. 181 
 
 Draw an indefinite line, and on it lay off AD = M, and DB = N. On 
 AB as a diameter draw a semi-circumference, and erect DC perpendicular 
 to AB. Then CD = x, the mean proportional required. 
 
 [Let the student give the proof.] 
 
 PROPOSITION XIX. 
 
 I. Problem. — To construct a square equivalent to a 
 given tri angle. 
 
 Find a mean proportional between the altitude and half the base. On 
 this construct a square. 
 
 [Let the student execute the problem and demonstrate it.] 
 
 EXERCISES. 
 
 399. 1. Draw any line, and divide it into 3, 5, 8, or 10 equal 
 parts. 
 
 2. Draw any line and divide it into parts which shall be to 
 each other as 2, 3, and 5. 
 
 3. Construct the square root of 7, 11, 2. 
 Fig. 197 will suggest the construction 
 
 of y/1. 
 
 4. The diameter of a circle is 20. 
 What is the perpendicular distance to 
 the circumference from a point in 
 the diameter 15 from one extremity ? ^'fl- '^^• 
 
 What are the distances from the point where this perpendicular 
 meets the circumference to the extremities of the diameter ? 
 
 5. The sides of one triangle are 7, 9, and 11. The side of a 
 second similar triangle, homologous with side 9, is A^. What 
 are the other sides of the latter ? 
 
182 
 
 ELEMENTARY GEOMETRY. 
 
 6. DE being parallel to BC, prove that the tri- 
 angles DOE and BOC are similar, and hence that 
 
 OD OE 
 OC ~ OB 
 
 Are the following proportions true ? 
 OD OE OD 00 
 
 00 ~ OB DE ~ BC' 
 
 OD _ OC OB OE 
 
 OE ■ BC BC " DE Fig. 198. 
 
 7. Draw any triangle or polygon, and then construct a similar 
 one whose homologous sides shall be f as long. 
 
 8. Show that if ABCDEF is a regular 
 polygon, khcdef is also regular, he, cd, etc., 
 being parallel to BC, CD, etc. Show that 
 any two similar polygons may be placed 
 in similar relative positions, and hence 
 show that the corresponding diagonals are 
 in the same ratio as the homologous sides. 
 
 Fig. 199. 
 
 PROPOSITIONS FOR ORIGINAL INVES- 
 TIGATION. 
 
 400. 1. If two straight lines join 
 the alternate ends of two parallels, 
 the line joining their centres' is half 
 the difference of the parallels. 
 
 We are to prove that 
 
 EF = i(CD- AB). 
 ^CH = EF = HCD-AB). 
 
 Fig. 200. 
 
 2. To construct a square equivalent to a given polygon. 
 
 First reduce the polygon to a triangle (339). Then construct an 
 equivalent square (398)- 
 
EXERCISES. 183 
 
 3. The area of a regular inscribed dodecagon is three 
 times the square on, the radius. 
 
 4. // the sides of a quadrilateral he divided into m 
 equal parts, and the ii^^ points of division, reckoning from 
 two opposite vertices, he joined so as to form a quadri- 
 lateral, the quadrilateral will he a parallelogram,. 
 
 Fig 201. 
 
 5. The line drawn from the vertex of the right angle 
 of a right-angled triangle to the middle of the hypotenuse 
 is half the hypotenuse. 
 
 Prove from either figure. 
 
 6. In any triangle the rectangle of two sides is equiva- 
 lent to the rectangle of the perpendicular 
 let fall from their included angle upon the 
 third side, into the diameter of the circum- 
 scribed circle. 
 
 This proposition is an immediate consequence ot 
 the similarity of two triangles in the figure. pjg^ 202! 
 
184 ELEMENTAR Y GEOMETB Y, 
 
 APPLICATIONS OF THE DOCTRINE OF SIMILARITY TO 
 THE DEVELOPMENT OF GEOMETRICAL PROPERTIES 
 OF FIGURES. 
 
 401. The doctrine of similarity, as presented in the preceding 
 section, is the chief reliance for the development of the geomet- 
 rical properties of figures. This section will be devoted to the 
 investigation of a few of the more elementary properties of plane 
 figures, which we are able to discover by means of this doctrine. 
 
 OF THE RELATIONS 
 
 OFTHE SEGMENTS OF TWO LINES INTERSECT- 
 ING EACH OTHER, AND INTERSECTED BY A 
 CIRCUMFERENCE. 
 
 PROPOSITION I. 
 
 402. Theorem. — // two chords intersect each other in 
 
 a circle, • their segments are reciprocalhj proportioned ; 
 whence the product of the segments of one chord equals 
 the product of the segments of the other. 
 
 Demonstration. 
 Let the chords AC and BD (Fig. 203) intersect at 0. 
 
 We are to prove that OA ~ OD * 
 
 whence OB x OD = OA x OC. 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY, 
 
 185 
 
 Draw AD and BC. 
 
 The two triangles AOD and BOC are simi- 
 lar (?). 
 
 OB qc 
 
 OA ~ OD' 
 
 OB X OD = OA X 00. Q. E. D. 
 
 Hance, 
 whence 
 
 Queries. — Why is OB compared with OA? 
 Why 00 with OD ? Would AO : CO : : BO : DO Fig. 203. 
 
 be true ? Would AO : DO : : BO : CO ? What is the force of the word 
 " reciprocally," as used in the proposition ? 
 
 PROPOSITION II. 
 
 403. Theorem. — // from> a point ivithout a circle, two 
 secants are draivn terminating in the concave arc, the whole 
 secants are reciprocally proportional to their external seg- 
 ments ; whence the product of one secant into its external 
 segm^ent equals the product of the other into its external 
 segment. 
 
 Demonstration. 
 
 Let OA and OB be two secants intersecting the circumference in D 
 and C respectively. 
 
 We are to prove 
 
 OB _ OD 
 OA ~ OC ' 
 
 whence, OB x OC = OA x OD. 
 
 Draw AC and BD. 
 
 The two triangles AOC and BOD are simi- 
 lar (?). 
 
 OB OD 
 
 OA ~ OC ' 
 
 Hence, 
 
 whence, OB x OC = OA x OD. q. e. d. 
 
 Fig. 204. 
 
 Queries. —Same as under preceding demonstration. 
 
186 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION III. 
 
 404. Theorem. — // from a point without a circle a 
 tangent is drawn, and a secant terminating in the con- 
 cave arc, the tangent is a mean proportional between the 
 whole secant and its external segment ; whence the square 
 of the tangent equals the product of the secant into its 
 external segment. 
 
 Demonsteation. 
 
 Let OA be a tangent and OB a secant intersecting the circumference 
 
 in C. 
 
 We are to prove that 
 
 OB A 
 
 OA be * 
 
 whence, OB x OC = OA . 
 
 Draw AC and AB. 
 
 The two triangles AOB and AOC are simi' 
 lar, since angle is common, and angle OAC — 
 angle B (?). 
 
 „ OB OA 
 
 Hence, 
 
 Fig. 205. 
 
 whence, 
 
 OA 00* 
 
 OB X 00 = OAI q. e. d. 
 
 OF THE BISECTOR OF AN ANGLE OF 
 A TRIANGLE. 
 
 PROPOSITION IV. 
 
 405. Theorem. — A line which bisects any angle of a 
 triangle divides the opposite side into segments propor- 
 tional to the adjacent sides. 
 
 Demonstration. 
 In the triangle ABC (Fig. 206) let CD bisect the angle ACB. 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY, 
 Then is 
 
 187 
 
 AD AC* 
 DB ~ CB 
 
 Draw BE parallel to CD, and produce it 
 till it meets AC produced in E. 
 
 By reason of the parallels CD and EB, 
 angle ACD = AEB, 
 and DCB = CBE. 
 
 But, by hypothesis, ACD = DCB. 
 
 Therefore, AEB (or CEB) = CBE, 
 
 and CE = CB (?). 
 
 AD _ AC 
 
 DB ~ CE(= CB) 
 
 Hence, finally, 
 
 Fig. 206. 
 
 (368). Q. E. D. 
 
 PROPOSITION V. 
 
 406. Theorem. — // a line is drawn from any vertex 
 of a triangle bisecting the exterior angle and intersecting 
 the opposite side produced, the distances from the other 
 vertices to this intersection are proportional to the adjacent 
 sides. 
 
 Demonstration. 
 
 Let CD bisect the exterior angle BCF of the triangle ACB. 
 
 AD _ AC 
 BD ~ CB' 
 
 Then is 
 
 Fig. 207. 
 
 For, draw BE parallel to AC. 
 By reason of these parallels, 
 
 angle FCE = CEB, 
 and BCE = FCE, by hypothesis. 
 
 Hence, CEB = BCE, 
 
 and CB = BE. 
 
 Also, by reason of the similar triangles ACD and BED, 
 
 AD AC 
 
 BD ~ BE(orCB) 
 
 Q. E. D. 
 
 * See note at the bottom of p. 178, 
 
188 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION VI. 
 
 407. Theorem. — // a line is drawn bisecting any 
 angle of a triangle and intersecting the opposite side, the 
 product of the sides about the bisected angle equals the 
 product of the segments of the third side, plus the square 
 of the bisector. 
 
 Demokstration. 
 
 In the triangle ACB, let CD bisect the angle 
 ACB. 
 
 Then AC x CB = AD x DB + CD'. 
 
 For, circumscribe the circle about the trian- 
 gle, produce the bisector till it meets the circum- 
 ference at E, and draw EB. The triangles ADC 
 and CBE are similar, since angle ACD = ECB, by 
 hypothesis, and A = E, because each is measured 
 by ^ arc CB. 
 
 AC CD 
 
 CE ~ CB ' 
 
 Fig. 208. 
 
 Therefore, 
 
 whence, ACxCB = CE xCD = (DE + CD) CD 
 
 = DE X CD + CdI 
 For DE X CD, substituting its equivalent AD x DB (402), we have 
 AC X CB = AD X DB + CD^. Q. e. d. 
 
 AREAS OF SIMILAR FIGURES. 
 
 PROPOSITION VII. 
 
 408. Theorem. — The areas of similar triangles are to 
 
 each other as the squares described on their homologous 
 
 sides. 
 
 Demonstration. 
 
 Let ABC and EFG be two similar triangles, the homologous sides 
 being AB and EF, BC and FG, and AC and EG. 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY. 
 
 189 
 
 Then is 
 
 area ABC AC^ 
 
 AT BC=^ 
 
 area EFG e^^ e F^ FG^' 
 
 From the greatest * angle in each tri- 
 angle let fall a perpendicular upon the 
 opposite side. Let these perpendiculars 
 be BD and FH. 
 
 Now 
 
 and 
 
 BD _ AC (.. 
 FH-EG^'^' 
 
 jAC _ AC ,as 
 iEG"EG^^* 
 
 Fig. 209. 
 
 Multiplying the corresponding ratios 
 
 together, we have 
 
 iAC^BD AC^ 
 iEGxFH " eg' 
 
 and 
 
 But 
 
 Hence, 
 
 And, finally, as 
 
 ^AC X BD = area ABC, 
 
 iEG X FH ^ area EFG (?). 
 
 area ABC _ AC^ 
 area EFG ~ eq^ 
 
 AC 
 
 AB' BC" 
 
 we have 
 
 EG EF' 
 
 area ABC AC 
 
 area EFG 
 
 EG" 
 
 FG' 
 
 AT 
 EF^ 
 
 (?), 
 
 Bc; 
 
 FG^ 
 
 409. 
 
 PROPOSITION VIII. 
 Theorem. — The areas of similar polygons are to 
 
 each other as the squares of any two Jwmologous sides of 
 the polygons. 
 
 * The only object in taking the largest angles is to make the perpendic- 
 ular fall within the triangle. The demonstration is essentially the same 
 when the perpendiculars fall upon the opposite sides produced. 
 
190 
 
 ELEMENTARY GEOMETRY. 
 
 Demonstration. 
 
 Let ABCDEF and abcdef be two similar polygons, the homologous 
 sides being AB and a&, BC and 6c, CD and cd, DE and de, EF and ef, 
 FA and /«. 
 
 Let area ABCDEF = P, 
 and area abcdef = p. 
 
 Then is 
 
 BC; 
 
 67 
 
 Fig. 210. 
 
 or as the squares of any two 
 homologous sides. 
 
 Draw the homologous di- 
 agimals AC, AD, AE, and ac, ad^ and ae, dividing the polygons into the 
 similar triangles M and m^ N and w, and o. and S and s (388)* 
 
 r2 
 
 Now 
 
 - - ^^ (?) 
 
 ED" 
 
 -de' 
 
 S^CB' 
 
 But 
 
 whence, 
 
 cb; 
 
 a;' 
 
 Dc; 
 
 dc 
 = - 
 
 = %-%i^y, 
 
 ed 
 
 
 
 
 fe 
 
 Taking this by composition, we have 
 
 M+N+O+S 
 
 m+n+o+8 
 
 
 CB 
 
 And as the ratio ^ is the same as that of the squares of any two 
 
 cb' 
 homologous sides, P and p are to each other as the squares of any two 
 homologous sides. 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY. 191 
 
 Finally, as this argument can be extended to the case of any two 
 similar polygons, the areas of any two similar polygons are to each other 
 as the squares of any two homologous sides of the polygons, q. e. d. 
 
 410. Corollary 1. — Similar polijgons^ are to each other 
 as the squares of their corresponding diagonals. 
 
 In the demonstration we have - = — = — -. 
 
 P m cb^ 
 
 By (388, 408) we have ^=^1-^-^ 
 ^ ae^ adr a<r 
 
 P AE2 
 
 Hence - = ^=i , etc. 
 
 P ai 
 
 411. Corollary ^.—Regular polygons* of the same 
 number of sides are to each other as the squares of their 
 homologous sides. [They are similar figures (?)]. 
 
 412. Corollary 3. — Regular polygons of the same num- 
 ber of sides are to each other as the squares of their 
 apothems. 
 
 For their apothems are to each other as their sides. Hence the 
 squares of their apothems are to eacli other as the squares of their sides. 
 
 413. Corollary 4. — Circles are to each other as the 
 squares of their radii (390), and as the squares of their 
 diameters. 
 
 OF PERIMETERS AND THE RECTIFI- 
 CATION OF THE CIRCUMFERENCE. 
 
 414. The Rectification of a curve is the process of find- 
 ing its length. 
 
 The term rectification signifies making straight, and is applied as 
 above, under the conception that the process consists in finding a straight 
 line equal in length to the curve. 
 
 * This is a common elliptical form for " The a/reas of, etc" 
 
192 ELEMENTARY GEOMETRY. 
 
 PROPOSITION IX. 
 
 415. Theorem. — Tlie perimeters of similar polygons 
 are to each otlier as tJieir homologous sides, and as their 
 corresponding diagonals. 
 
 Demo:n^stration. 
 
 Let (If b, c, dy etc., and A, B, C, D, etc., be the homologous sides of 
 two similar polygons whose perimeters are^> and P. 
 
 and r and R being corresponding diagonals, 
 
 p _ r 
 P ~ R 
 
 Since the polygons are similar, 
 By composition, 
 
 a h c d ^ 
 
 a + h-\-c + d4- etc. (or p) _ a 
 A + B + C + D + etc. (orP) ~ A . 
 
 or as any other homologous sides. Also, as the homologons sides are to 
 each other as the corresponding diagonals (387), 
 
 p r 
 
 P R ^ 
 
 416. Corollary 1. — The perimeters of regular polygons 
 of the same number of sides are to each other as the apo- 
 thems of the polygons (382). 
 
 417. Corollary 2. — The circmnferences of circles are 
 to eojch other as their radii, and as their diam^eters (390). 
 
 PROPOSITION X. 
 
 418. Problem. — To find the relation between the 
 chord of an arc and the chord of half the arc in a circle 
 whose radius is v. 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY. 
 
 193 
 
 Solution. 
 
 Let be the centre of the circle, AB any chord, and CB the chord 
 of half the arc AB. 
 
 Let AB = C, and CB = c. 
 
 We are to find the relation between C and c. 
 Draw the radii CO and BO, and call each r. 
 CO is perpendicular to AB (?). 
 In the right-angled triangle BDO, 
 
 DO = VBO'-iC-^ (?), 
 or DO = V*-' - iC". 
 
 Fig. 211. 
 
 Hence, 
 
 CD = r - ^/r^' 
 
 (P, 
 
 Again, in the right-angled triangle CDB, 
 
 CB = Vcb' + bd' 
 
 = y 2r» - 2rv^r» — ;^(7» 
 = |/2r 
 
 iC« 
 
 .y'4r» - G\ 
 
 Therefore, e = y 2r' — r>v/4r'» — C is the relation desired. 
 419. ScHOLroM. — The formula 
 
 = V 2r'-rV'4r*- 
 
 (7« 
 
 is the value of the chord of half the arc in terms of the chord of the 
 whole arc and the radius. From this we readily obtain 
 
 C = - V4r' - c\ 
 
 which is the value of the chord in terras of the chord of half the arc and 
 the radius. 
 
T8I4 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XI. 
 
 420. Theorem. — The circumference of a circle whose 
 radius is 1, is 2-rT, the numerical value of n being approj^i- 
 matelyS.Uie. 
 
 Demonstration. 
 
 We will approximate the circumference of a 
 circle whose radius is 1, by obtaining, 1st, the 
 perimeter of the regular inscribed hexagon ; 2d, 
 the perimeter of the regular inscribed dodeca- 
 gon ; 3d, the perimeter of the regular inscribed 
 polygon of 24 sides ; then of 48, etc. 
 
 By varying the polygon in this manner, it is 
 evident that the perimeter approaches the cir- 
 cumference as its limit (282, 354), since at each '^' '^' 
 bisection the sum of two sides of a triangle is substituted for the third 
 side. Moreover, the perimeter can never pass the circumference, since a 
 chord is always less than its arc. 
 
 Now let AB = r (?) = 1 be the side of the inscribed hexagon. Then 
 by the formula (418), we have 
 
 = 1/2- 
 
 CB = c = y2-V4-l = .51763809, 
 
 which is therefore the side of a regular dodecagon. Hence the perimeter 
 of the dodecagtm is 
 
 .51763809 X 12 ^ 6.21165708. 
 
 Again, let the side of the inscribed regular polygon of 24 sides be c\ 
 and we have 
 
 5'-|/s 
 
 y^4 _ c« = y 2 - ^4 - (.51763809)« = .26105238; 
 
 and the perimeter, .26105238 x 24 = 6.26525722. 
 
 Carrying the computation forward in this manner, we have the fol- 
 lowing : 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY. 
 
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196 ELEMENTARY GEOMETRY, 
 
 It now appears that the first four decimal figures do not change as 
 the number of sides is increased, but will remain the same 7ww far soever 
 we 2>Toceed. When the foregoing process is continued till 5 decimals be- 
 come constant, we have 6. 28318 + . We may therefore consider 6.28318 
 as app'oximatdy the circumference of a circle whose radius is 1, 
 
 Hence, letting 27r stand for the circumference, we have 
 27r = 6.28318 + , 
 and TT — 3.1416, nearly, q. e. d. 
 
 421. Scholium.— The symbol ;r is much used in mathematics, and 
 signifies, primarily, the aemi-drcumference of a circle whose radius is 1. 
 ^TT is therefore a symbol for a quadrant, 90°, or a right angle. \n is 
 equivalent to 45", and 2n to a circumference, the radius being always 
 supposed 1, unless statement is made to the contrary. The numerical 
 value of TT has been sought in a great variety of ways, all of which agree 
 in the conclusion that it cannot be exactly expressed in decimal numbers, 
 but is approximately as given in the proposition. From the time of 
 Archimedes (287 B.C.) to the present, much ingenious labor has been 
 bestowed upon this problem. The most expeditious and elegant methods 
 of approximation are furnished by the Calculus. The following is the 
 value of n extended to fifteen places of decimals : 3.141592653589793. 
 
 PROPOSITION XII. 
 
 422. Theorem. — The circumference of any circle is 
 2rrr, r being the radius. 
 
 ' Demonstration. 
 
 The circumferences of circles being to each other as their radii (417), 
 and 27r being the circumference of a circle whose radius is 1, we have 
 
 27r _ 1^ 
 circf. r ' 
 
 whence, circf. = 27rr. Q. b. d. 
 
 423. Corollary. — The circumference of any circle is 
 ttD, D being the diameter. 
 
AFPLICATIONS OF DOCTRINE OF SIMILARITY. 197 
 
 AREA OF THE CIRCLE. 
 
 PROPOSITION XIII. 
 
 424. Theorem. — The area of a circle whose radius 
 is 1, is TT. 
 
 Demonstration. 
 
 The area of a circle is ^r x circf. (356). When r = 1, 
 circf. = 2t(420); 
 hence, area of circle whose radius is 1 = ^ x Stt = n^. q. e. d. 
 
 PROPOSITION XIV. 
 
 425. Theorem. — The area of any circle is nr^, r being 
 the radias. 
 
 Demonstration. 
 
 The areas of circles being to each other as the squares of their radii 
 (413), and tt being the area of a circle whose radius is 1, we have 
 
 n V 
 
 area of any circle r' 
 whence, area of any circle = tt? ■^ q. e. d. 
 
 426. Scholium 1. — Since the area of a sector is to the area of the 
 circle of the same radius as its angle is to 4 right angles (359), if we 
 
 d TT T^ 
 
 represent the angle of the sector by a°, we have for its area • 
 
 427. Scholium 2. — As the value of tt cannot be exactly expressed in 
 numbers, it follows that the area cannot. Finding the area of a circle has 
 long been known as the problem of " Squaring the Circle ;" i. e., find- 
 ing a square equal in area to a circle of given radius. Doubtless many 
 hare-brained visionaries or ignoramuses will still continue the chase after 
 the phantom, although it has long ago been demonstrated that the diam- 
 
198 ELEMENTARY GEOMETRY, 
 
 eter of a circle and its circumference are incommensurable by any finite 
 unit. It is, however, an easy matter to conceive a square of the same 
 area as any given circle. Thus, let there be a rectangle whose base is 
 equal to the circumference of the circle, and whose altitude is half the 
 radius ; its area is exactly equal to the area of the circle. Now, let there 
 be a square whose side is a mean proportional between the altitude and 
 base of this rectangle ; the area of the square is exactly equal to the area 
 of the circle. 
 
 PROPOSITION XV. 
 
 428. Theorem.—// a -perpendicular is let fall from 
 any angle of a triangle upon the opposite side (or on the 
 side produced), the difference of the squares of the segments 
 is equivalent to the difference of the squares of the other 
 two sides. 
 
 Demonstration. 
 
 Let ABC be any triangle, and CD be the perpendicular let fall from 
 C upon AB (or AB produced). Call the sides opposite the angles 
 A, B, and C, a, h, and c, respectively; and let the segment BD = m, 
 AD = ti, and CD = p. 
 
 Then is m'— n^ = «'—&». 
 For, from the right-angled tri- 
 angle BCD, 
 
 Also, from CDA, 
 
 Whence, o" — m^ = &^ — w 
 
 or m^ — n^ = a^ — 5' 
 
 429. Corollary, — Since 
 
 m' — n* = (m + n)(m — n), 
 
 and a'-h' = {a-\-l){a- I), 
 
 m + n (or c) a — b 
 
 we have — r — = - — - 
 
 a + m — n 
 
APPLICATIONS OF DOCTRINE OF SIMILARITY. 199 
 
 430. Scholium. — In case tlie perpendicular falls without, the dis- 
 tances BD and AD are still, for simplicity of expression, spoken of as 
 segments. 
 
 431. A line is said to be divided in Extreme and 
 Mean Ratio when it is so divided that the whole line is to the 
 greater segment as the greater segment is to tlie less, ?'. e.y when 
 the greater segment is a mean proportional between the whole 
 line and the less segment. 
 
 PROPOSITION XVI. 
 
 432. Problem. — To divide a line in extreme and 
 mean ratio. 
 
 Solution. 
 
 Let it be proposed to divide the line AB in extreme and mean ratio, 
 ue,, C being the point of division, so that 
 
 AB AC 
 AC ~ CB" 
 
 At one extremity of AB, as B, erect a 
 perpendicular BO, and make it equal to 
 iAB. 
 
 From as a centre, with OB as a ra- 
 dius, describe a circle. 
 
 Draw AO, cutting the circumference 
 in D. 
 
 Then is AD the greater segment, and taking AC = AD, AB is divided 
 in extreme and mean ratio at C. 
 
 Demonstbation.of Solution. 
 
 Produce AO to E. 
 
 Now ^ = ^('n 
 
 AB AO ^ ^' 
 
 or, by inversion, ^ = ^. 
 
 At AB 
 
 Fig. 214. 
 
200 ELEMENTARY GEOME'tRT, 
 
 By division, we have 
 
 AB AD 
 
 AE - AB ~ AB - AD 
 
 But, as DE = AB (?), 
 
 AE - AB = AE - DE = AD = AC; 
 
 and AB - AD = AB - AC = CB. 
 
 A Pi Kf\ 
 
 Hence, substituting, AC ~ CB ' ^ ^* ^• 
 
 PROPOSITION XVII. 
 
 433. Problem. — To inscribe a regular decagon in a 
 circle, and hence a regular pentagon, and regular polygons 
 of 20, 40, 80, etc., sides. 
 
 Solution. 
 
 Let it be required to inscribe a regular decagon in the circle whose 
 centre is and radius OA. 
 
 Divide the radius OA in 
 extreme and mean ratio, as 
 at («). 
 
 Then is ac, the greater 
 segment, the side of the in- 
 scribed decagon, ABODE, 
 etc. 
 
 To prove this, draw OA 
 and OB, and taking CM = 
 
 ° Fig. 215. 
 
 ac = AB, draw BM. 
 
 ^^^ OM ~ MA ' ^y construction ; and, as CM = AB, we have 
 
 0A_ AB 
 AB ~ MA * 
 
 Hence, considering the antecedents as belonging to the triangle OAB, 
 and the consequents to the triangle BAM, we observe that the two sides 
 about the angle A, which is common to both triangles, are proportional ; 
 hence the triangles are similar (373). 
 
EXERCISES. 201 
 
 Therefore, ABM is isosceles, since OAB is, and 
 angle BMA = A = OBA, 
 and MB = BA = OM. 
 
 This makes 0MB also isosceles, and 
 
 the angle = OBM. 
 Again, the exterior angle BMA = + OBM = 20; 
 hence, A (which equals BMA) = 20. 
 
 Hence, also, OBA (which equals A) = 20. 
 
 Wherefore, is | the sum of the angles of the triangle OAB, or | of 
 2 right angles, = y^ of 4 right angles. 
 
 The arc AB is therefore the measure of ^ of 4 right angles, and is 
 consequently ^V "^ t^e circumference. Hence AB is the chord of ^ of 
 the circumference, and if applied, as AB, BC, CD, DE, etc., will give an 
 equilateral inscribed decagon. 
 
 Moreover this inscribed polygon is equiangular, and hence regular 
 by (272). 
 
 To construct the pentagon, join the alternate angles of the decagon. 
 To construct the regular polygon of 20 sides, bisect the arcs subtended 
 by the sides of the decagon, etc. 
 
 MISCELLANEOUS EXERCISES. 
 
 434. f. Show that if a chord of a circle is conceived to re- 
 volve, varying in length as it revolves, so as to keep its extremities 
 in the circumference while it constantly passes through a fixed 
 point, the rectangle of its segments remains constant. 
 
 2. The two segments of a chord intersected by another chord 
 are 6 and 4, and one segment of the other chord is 3. What is 
 the other segment of the latter chord ? 
 
 3. Show how Propositions I, II, and III may be considered as 
 different cases of one and the same proposition. 
 
 Suggestions. — By stating Propositions I and IT thus, The distances 
 from the irUer section of the linen to their intersections with tlie circumference^ 
 what follows? In Fig. 204, if the secant AO becomes a tangent, what 
 does OD become ? 
 
202 ELEMENTARY GEOMETRY. 
 
 4. In a triangle whose sides are 48, 36, and 50, where do the 
 bisectors of the angles intersect the sides ? 
 
 5. In the last example, find the lengths of the bisectors. 
 
 6. A and B have farms of similar shape, with their homolo- 
 gous sides on the same road. A's is 150 rods on the road, and 
 B's 200 rods. How does A's farm compare with B's in size ? 
 
 7. Draw two similar triangles with their homologous sides in 
 the ratio of 3 to 5, and divide them into equal partial triangles, 
 showing that their areas are as 3^ to 5^, that is, as 9 to 25. 
 
 8. What are the relative capacities of a 5-inch and a 7-inch 
 stove-pipe ? 
 
 9. If a circle whose radius is 34 is divided into 5 equal parts 
 by concentric circumferences, what are the diameters of the sev- 
 eral circles ? 
 
 Solve geometrically as well as numerically. 
 
 10. The projection of one line upon another in the same plane 
 is the distance between the feet of two perpendiculars let fall 
 from the extremities of the former upon the latter. Show that 
 this projection is equal to the square root of the difference be- 
 tween the square of the line and the square of the difference of 
 the perpendiculars. 
 
 11. The three sides of a triangle being 4, 5, and 6, find the 
 segments of the last side made by a perpendicular from the op- 
 posite angle. Ans. 3.75 and 2.25. 
 
 12. Same as above, when the sides are 10, 4, and 7, and the 
 perpendicular is let fall from the angle included by the sides 10 
 and 4. Draw the figure. Why is one of the segments negative ? 
 
 13. What is the area of a regular octagon inscribed in a circle 
 whose radius is 1 ? What is its perimeter? What if the radius 
 is 10? 
 
 14. What is the side of an equilateral triangle inscribed in a 
 circle whose radius is 1 ? 
 
EXERCISES. 203 
 
 15. What is the side of a regular inscribed decagon in a circle 
 whose radius is 4 ? What the side of the inscribed pentagon? 
 What is the area of each ? 
 
 16. Draw two squares, and construct two others, one equal to 
 their sum, and the other to their difference. 
 
 17. Draw any two polygons, and construct two squares, one 
 equivalent to their sum, and the other equivalent to their differ- 
 ence. 
 
 18. Show that the length of a degree in any circle is -^y 
 
 and hence that the lengths of degrees in different circles are to 
 each other as the radii of the circles. 
 
 19. What is the length of a minute on a circle whose radius 
 is 10 miles ? 
 
 20. Calling the equatorial radius of the earth 3962.8 miles, 
 what is the length of a degree on the equator ? 
 
 21. How many degrees in the arc of a circle which is equal in 
 length to the radius ? 
 
 22. Compute the area of the triangle whose sides are 20, 30, 
 and 40. 
 
 Find the segments of the base (40) by (428)- Hence the perpendic- 
 ular. 
 
 23. Given the side of a regular inscribed pentagon, as 16, to 
 find the side of the similar circumscribed polygon. 
 
 24. Prove that if a triangle is circumscribed about a given 
 triangle by drawing lines through the vertices of the given tri- 
 angle and parallel to the opposite sides, the area of the circum- 
 scribed triangle is four times that of the given triangle. 
 
 25. Prove that the bisectors of the angles of a triangle pass 
 through a common point. 
 
 26. Prove that the perpendiculars to the three sides of a tri- 
 angle at their middle points pass through a common point. 
 
304: 
 
 ELEMENTA R Y GEOMETR Y. 
 
 27. The three perpendiculars drawn from the angles of a tri- 
 angle upon the opposite sides intersect in a common point. 
 
 Draw through the vertices of the 
 triangle lines parallel to the opposite 
 sides. The proposition may then be 
 brought under the preceding. 
 
 28. The following triangles are 
 similar — viz., BOE, BDC, AOD, and 
 AEC, each to each ; also EOF, BDA, 
 DOC, and CFA. Prove it. 
 
 Fig. 216. 
 
 435. The Medial Lines in a triangle are the lines drawn 
 from the vertices to the middle points of the opposite sides. 
 
 29. The three medial lines of a triangle mutually trisect each 
 other, and hence intersect in a common point. 
 
 To prove that OE = ^BE (Fig. 317), draw FC parallel to AD until it 
 
 meets BE produced. Then the triangles AEO and FEC are equal (?); 
 
 whence 
 
 EF = OE. 
 
 Also, BO = OF (?). 
 
 Having shown that 
 
 OE = iBE, 
 by a similar construction we can show that 
 OD = iAD 
 Finally, we may show that the medial line 
 from C to AB cuts off ^ of BE, and hence cuts BE 
 at the same point as does AD. 
 
 I- — ^c 
 
 Fig. 217. 
 
 Another Demonstration. — Lines through parallel to the sides 
 trisect the sides, etc. 
 
 Still Another.— Without EF and FC, draw ED, and prove by simi- 
 lar triangles. 
 
4- CHAPtfeR n; 
 
 SOLID GEOMETRY.* 
 
 OF STRAIGHT LINES AND PLANES. 
 
 436. Solid Geometry is that department of Geometry 
 in which the magnitudes treated are not limited to a single 
 plane. " ^ 
 
 437. A Plane (or a Plane Surface) is a surface such 
 that a straight line joining any two points in it lies wholly in the 
 surface. 
 
 PLANE, HOW DETERMINED. 
 
 438. A plane is said to be Determined by given conditions 
 which fix its position. 
 
 All planes are considered as indefinite in extent, unless the 
 contrary is stated. 
 
 * In some respects, perhaps, " Geometry of Space" is preferable to this 
 term ; but, as neither is free from objections, and as this has the advantage 
 of simplicity and long use, the author prefers to retain it. 
 
206 ELEMENTARY GEOMETRY, 
 
 PROPOSITION I. 
 
 439. Theorem. — Three points not in the same straight 
 line determine a plane. 
 
 Demonstration. 
 
 Let A, B, and C be three points not in the same straight line. 
 
 Then one plane can be passed through 
 them, and only one; i. c, they determine the 
 position of a plane. 
 
 For, pass a straight line through any two of 
 these points, as A and B. Now, conceive any 
 plane containing these two points ; then will the 
 line passing through them lie wholly in the 
 plane (437). Conceive this plane to revolve on ^'9- 2i8. 
 
 the line as an axis until the point C falls in the plane. Thus we have one 
 plane passed through the three points. 
 
 That there can be only one is evident, since when C falls in the plane, 
 if the plane be revolved either way, C will not be in it. The same may 
 be shown by first passing a plane through B and C, or A and C. There 
 is, therefore, only one position ot the plane in which it will contain the 
 third point, q. e. d. 
 
 440. Corollary 1. — A line and a point witJwut it de^ 
 termine a plane. ^ 
 
 441. Corollary 2. — Through one line, or two points, an 
 infinite number of planes can he passed. , 
 
 442. Corollary 3. — The intersection of two planes is a 
 straight line. 
 
 For two planes cannot have even three points, not in the same straight 
 line, common, much less an indefinite number, which would be required 
 if we conceived the intersection (that is, the common points) to be in any 
 other than a straight line. 
 
 443. The Trace of one plane in another is their intersection. 
 
STRAIGHT LINES AND PLANES. 207 
 
 PROPOSITION II. 
 
 444. Theorem. — Two intersecting lines determine the 
 position of a plane. 
 
 Demonstration. 
 
 For, the point of intersection may be taken as one of the three points 
 requisite to determine the position of a plane, and any two other points, 
 one in each of the lines, as the other two requisite points. Now, the 
 plane passing through these, points contains lx)th the lines, for it contains 
 two points in each. q. e. d. 
 
 PROPOSITION III. 
 
 445. Theorem. — Two parallel lines determine the po- 
 sition of a plane. 
 
 Demonstration. 
 
 For, pass a plane through one of the parallels, and conceive it revolved 
 until it contains some point of the second parallel. Now, if the plane be 
 revolved either way from this position, the point will be left without it. 
 Hence, it is the only plane containing the first parallel and this point in 
 the second. 
 
 But parallels lie in the same plane (120, 121), whence the plane of the 
 parallels must contain the first line and the specified point in the second. 
 
 Therefore, the plane containing the first line and a point in the second 
 is the plane of the parallels, and is fixed in position, q. e. d. 
 
 446. Scholium. — When a plane is determined by two lines, accord- 
 ing to either of the last two propositions, it is spoken of as the Plane of 
 the Lines. In like manner, we may speak of the Plane of Three Points. 
 
 RELATIVE POSITION OF A LINE AND 
 A PLANE. 
 
 447. A line may have one of three positions in relation to a 
 plane : {a) It may be perpendicular, (b) oblique, or (c) parallel 
 
208 
 
 ELEMENTARY GEOMETRY. 
 
 OF LINES PERPENDICULAR TO A PLANE. 
 
 448. A line is said to Pierce a plane at the point where it 
 passes through it. 
 
 449. The point where a perpendicular meets, or pierces, a 
 plane is called its Foot. 
 
 450. A Perpendicular to a Plane is a line which is 
 perpendicular to all lines of the plane passing through its foot, 
 and hence to every line of the plane. Conversely, the plane is 
 perpendicular to the line. 
 
 451. The Distance of a point from a plane is the length 
 of the perpendicular let fall from the point upon the plane. 
 
 PROPOSITION IV. 
 
 452. Theorem. — A line which is perpendicular to two 
 lines of a plane, at their intersection, is perpendicular to 
 the plane. 
 
 Demonstration^. 
 
 Let PD be perpendicular to AB and CF at D. 
 
 Then is it perpendicular to MN, the 
 plane of the lines AB and CF. 
 
 Let OQ be any other line of the 
 plane MN, passing through O. Draw 
 FB iutersecting the three lines AB, CF, 
 and OQ in B, E, and F. Produce PD 
 to P', making P'D = PD, and draw PF, 
 PE, PB, P'F, PE, P'B. 
 
 Then is PF ^ P'F, 
 and PB = P'B, 
 
 since FD and BD are perpendicular to 
 PP', and 
 
 PD = P'D(96) 
 
 Fig. 219. 
 
STRAIGHT LINES AND PLANES, 209 
 
 Hence, the triangles PFB and PTB are equal (305) ; and if PFB be 
 revolved upon FB till P falls at P', PE will fall in P'E. 
 
 Therefore, OQ has E equally distant from P and P', and as D is also 
 equidistant from the same points, OQ is perpendicular to PD at D (98). 
 
 Now, as OQ is any line, PD is perpendicular to any line of the plane 
 passing through its foot, and consequently perpendicular to the plane 
 (460). Q. E. D. 
 
 463. Corollary. — If one of two perpendiculars revolves 
 about the other as an axis, its path is avlane perpendicu- 
 lar to the axis, and this plane contains all the perpendicu- 
 lars to the axis at the common point. 
 
 ThUvS, if AB revolves about PP' as an axis, it describes the plane MN, 
 and MN contains all the perpendiculars to PP' at D. For, if there could 
 be a perpendicular to PP at D which did not lie in the plane M N, there 
 would be two perpendiculars to PP' at D, both'^' lying in the same plane, 
 which is impossible (88). ' ^ 
 
 PROPOSITION V. 
 
 454. Theorem. — M any point in a plane one perpen- 
 dicular can he erected to the plane, and only one. 
 
 Demonstration. 
 
 Let it be required to show that one perpendicular, and only one, can 
 be erected to the plane MN at D. 
 
 Through D draw two lines. of the plane, as 
 AB and CE, at right angles to each other. CE 
 being perpendicular to AB, let a line be con- 
 ceived as starting from the position ED to re- 
 volve about A B as an axis. It will remain per- 
 pendicular to AB (453). Conceive it to have 
 passed to P'D. Now, as it continues to revolve, 
 P'DC diminishes continuously, and at the same 
 
 , • . . rig. 220. 
 
 rate as P'DE increases; hence, in one position 
 
 of the revolving line, and in only one, as PD, PDE = PDC, and PD is 
 
 perpendicular to CE (86). 
 
 Again, any line which is perpendicular to MN at D is perpendicular 
 
210 
 
 ELEMENTA R Y GEOMETR Y. 
 
 to AB and CE (450). But the plane of the lines PD and DE contains all 
 lines perpendicular to AB at D. Hence, PD is perpendicular to the plane 
 (452), and is the only perpendicular, q. e. d. 
 
 PROPOSITION VI. 
 
 455. Theorem — From a point without a plane one 
 perpendicular can he drawn to the plane, and only one. 
 
 4 
 
 Demonstration. 
 
 Let It be required to show that one perpendicular can be drawn from 
 P to the plane MN, and only one. 
 
 Take RS as an aux- 
 iliary plane, and at any 
 point as C erect DC per- 
 pendicular to RS. 
 
 Now place the plane 
 RS in coincidence with 
 MN, and move it in MN 
 till the perpendicular DC 
 passes through P. 
 
 Then DC, which passes Fig. 221 
 
 through P and is pei-pen- 
 dicular to RS, is perpendicular to MN, with which RS is coincident. 
 
 Q. E. D. 
 
 To prove that there can be but one perpendicular from P to MN, sup- 
 pose that there could be two, as PA and PF. 
 
 Draw FA. 
 
 Then since FA is a line of the plane, and PF and PA are perpendic- 
 ulars to the plane, PFA and PAF are both right angles (?), and the tri- 
 angle PFA has two right angles, which is absurd. Hence there can be 
 but one perpendicular from P to MN. Q. e. d. 
 
 456. Corollary. — The perpendicular is the shortest line 
 that can he drawn to a plane from a point without. 
 
 Thus, let PA be a perpendicular and PF any oblique line, 
 PA < PF (?). 
 
STRAIGHT LINES AND PLANES, 
 
 211 
 
 PROPOSITION VII. 
 
 457. Theorem. — Conversely to the last. Through a given 
 point in a line, one plane can be passed perpendicular to 
 the line, and only one. 
 
 Demonstration. 
 
 Let D be the point in the line PG. 
 
 Pass two lines through D, as 
 EF and AB, each perpendicular to 
 PD ; the plane of these lines is per- 
 pendicular to PD. Q. E. D. 
 
 To show that but one plane can 
 be passed through D perpendicular 
 to PG, assupie that M'N' is another 
 plane passing through D, and per- 
 pendicular to PG, but not contain- 
 ing BD. Through PD and BD pass 
 a plane, and let BD be its intersec- 
 tion with M'N'. Then, on the 
 hypothesis that M'N' is perpendic- 
 ular to PG, B'DP is a right angle, and we have two lines in the same 
 plane with PG, and perpendicular to it at the same point, which is 
 absurd. Hence there can be but one plane perpendicular to PG and pass- 
 ing through D. Q. e. d. 
 
 PROPOSITION VIII. 
 
 458. Theorem. — // from the foot of a perpendicular 
 to a plane a line is drawn at right angles to any line of 
 the plane, and their intersection is joined with any point 
 in the perpendicular, the last line is perpendicular to the 
 line of the plane. 
 
 Demonstration. 
 
 From the foot of the perpendicular PD (Fig. 223) let DE be drawn 
 perpendicular to AB, any line of the plane MN, and E joined with 0, any 
 point of the perpendicular. 
 
212 
 
 ELEMENT A RY G E03IETR Y, 
 
 Then is OE perpendicular to AB. 
 
 Take EF = EC, and draw CD, FD, CO, and 
 FO. Now, 
 
 CD = DF (?), 
 
 whence CO = FO (?), 
 
 and OE has equally distant from F and C, and 
 also E. Therefore, OE is perpendicular to AB (?). 
 
 Q. E. D. 
 
 Fig. 223. 
 
 459. Corollary. — The line DE measures the shortest 
 distance between PD and AB. 
 
 For a line drawn from E to any other point in PD than D, as Ea, is 
 longer than DE (?). 
 
 Again, if from any other point in AB, as C, a line be drawn to D, it is 
 longer thari DE (?) ; and if drawn from C to «, any other point in PD 
 than D, Oa is longer than CD (?), and consequently longer than DE (?). 
 
 PROPOSITION IX. 
 
 460. Theorem. — // one of two parallels is perpendic- 
 ular to a plane, the other is perpendicular also. 
 
 Demonstration. 
 Let AB be parallel to CD and perpendicular to the plane MN. 
 
 Then is CD perpendicular to MN. 
 
 For, drawing BD in the plane MN, it is pcr- 
 pendiculaf to AB (?), and consequently to CD (?). 
 Through D draw EF in the plane and perpendic- 
 ular to BD, and join D with any point in AB, as 
 A; then is EF perpendicular to AD (?). 
 
 Now, EF being perpendicular to two lines, AD 
 and BD, of the plane ABDC, is perpendicular to F'g- 224. 
 
 the plane, and hence to any line of the plane passing through D, as CD. 
 
 Therefore, CD is perpendicular to BD and EF, and consequently to the 
 plane MN (?). Q. e. d. 
 
STRAIGHT LINES /iND PLANES. 213 
 
 461. Corollary.— T/fo lines which are perpendicular 
 to the same plane are parallel. 
 
 Thus, AB and CD being perpendicular to the plane MN are parallel. 
 For, if AB is not parallel to CD, draw a line through B which shall be. 
 By the Proposition, this line is perpendicular to MN, and hence must 
 coincide with AB (454)- 
 
 PROPOSITION X. 
 
 462. Theorem. —Ta^o lines parallel to a third not in 
 their own plane are parallel to each other. 
 
 D EMO N STR ATION-V 
 
 Let AB and CD be parallel to EF. 
 
 Then are they parallel to each other. 
 
 For, through F', any point in EF, pass a plane 
 MN perpendicular to EF. 
 
 Now AB and CD are respectively perpendicu- 
 lar to MN (?), and hence are parallel to each other 
 (?). Q. E. D. 
 
 Fig. 225. 
 
 OF LINES OBLIQUE TO A PLANE. 
 
 463. An Oblique Line is a line which pierces the plane 
 (if sufficiently produced), but is not perpendicular to the plane. 
 
 464. The Projection of a Point on a plane is the foot 
 of the perpendicular from the point to the plane. 
 
 465. The Projection of a Line upon a plane is the 
 locus of the projection of the point which generates the line. 
 
214 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION XI. 
 
 466. Theorem. — The projection of a straight line upon 
 a plane is a straight line. 
 
 Demonstration. 
 Let AB be any line and MN the plane upon which it is projected. 
 
 Then is the projection of AB in MN 
 a straight line. 
 
 Let P be a point in AB, and D its 
 projection in IMN. 
 
 Pass a plane, S, through AB and PD 
 (444), and let CE be its trace in MN. 
 
 Now let P' be any point in AB 
 other than P, and let D' be its projec- 
 tion in MN. 
 
 As PD and P'D' are perpendicular 
 to MN, they are parallel to each other 
 (461), and a plane may be passed throug'i them (445). 
 
 But the plane of PD and P'D' is S, since it contains PD and P' (440). 
 
 Therefore D' lies in S, and as it lies in MN, it is in the trace of S in 
 MN, which trace is a straight line (442). 
 
 Hence, as P' is any point in AB, the projection of every point of AB is 
 in a straight line. q. e. d. 
 
 •■■■^■^BfS^H 
 
 Fig. 226. 
 
 467. Corollary. — The projection of a line upon a plane 
 is the trace of a plane containing the line and the projec- 
 tion of any point of the line. 
 
 468. The Projecting Plane is the plane of a line and its 
 projection upon another plane. 
 
 469. The Plane of Projection is the plane upon which 
 a point or a line is projected. 
 
 470. The Inclination of a Line to a plane is the angle 
 included between the line and its projection. 
 
STRAIGHT LINES AND PLANES, 
 
 215 
 
 PROPOSITION XII. 
 
 471. Theorem. — If from any point in a perpendicular 
 to a plane, oblique lines are drawn to the plane, those which 
 pierce the plane at equal distances from the foot of the 
 perpendicular are equal ; and of those which pierce the 
 plane at unequal distances from the foot of the perpendic- 
 ular, those which pierce at the greater distances are the 
 greater. 
 
 Demonstration. 
 
 Let PD be a perpendiculdp to the plane MN, and PE, PE', PE",and 
 PE' " be oblique lines piercing the plane at equal distances ED, ED, E"D, 
 and E "D from the foot of the perpendicular. 
 
 Then PE = PE' = PE" = PE". 
 
 For each of the triangles PDE, PDE', etc., 
 has two sides and the included angle equal 
 to the corresponding parts in the other. 
 
 Again, let FD be longer than E'D. 
 
 Then is PF > PE'. 
 
 For, take ED = E'D ; then PE = PE', by 
 the preceding part of the demonstration. 
 But PF > PE, by (113). Hence, PF > PE'. Q. e. d. 
 
 472. Corollary 1. — The angles which oblique lines 
 drawn from a common point in a perpendicular to a 
 plane, and piercing the plane at equal distances from the 
 foot of the perpendicular, make with the perpendicular, 
 are equal ; and the inclinations of such lines to the plane 
 are equal. 
 
 Thus, the equality of the triangles, as shown in the demonstration, 
 shows that 
 
 EPD = E'PD = E'PD = E "PD, 
 
 and 
 
 PED = PE'D = PE"D = PE "D. 
 
 473. Corollary 2. — Conversely, // the angles which 
 oblique lines drawn from a point in a perpendicular to a 
 
216 ELEMENTARY GEOMETRY. 
 
 plane, make with the perpendicular, are equal, the lines 
 are equal, and pierce the plane at equal distances from 
 the foot of the perpendicular. 
 
 Thus, let E'PD = E"PD ; 
 
 then the right-angled triangles PDE' and PDE" are equal (?). Hence, 
 PE' = PE", and DE' = DE". 
 
 474. Corollary 3. — Lines drawn from the same point 
 in a perpendicular, and equally inclined to the plane, are 
 equal, and pierce the plane at equal distances from the 
 foot of the perpendicular. 
 
 475. Corollary 4. — Equal oblique lines from the same 
 point in the perpendicular, pierce the plane at equal dis- 
 tances fror)v the foot of the perpendicular, a>re equally 
 inclined to the plane, and also to the perpendicular. 
 
 Since the right-angled triangles PDE' and PDE" have their altitudes 
 and hypotenuses equal, the triangles are equal (309), and 
 
 DE' = DE", PE'D = PE "D, and E'PD == E "PD. 
 
 OF LINES PARALLEL TO A PLANE. 
 
 476. A Line is Parallel to a Plane when it is paral- 
 lel to its projection in that plane. 
 
 PROPOSITION XIII. 
 
 477. Theorem. — A line parallel to a plane is every- 
 where equidistant from the plane, and hence can never 
 meet the plane ; and, conversely, a straight line which can- 
 not meet a plane is parallel to it. 
 
 Demonstration. 
 
 The distance between a point in the line and the plane being the per- 
 pendicular (451), is also the distance between the point and the projec- 
 
STRAIGHT LINES AND PLANES. 217 
 
 tiori of the line (464). But this is everywhere the same (476, 136)- 
 Hence a line parallel to a plane is everywhere equidistant from it, and 
 therefore can never meet it. Q. e. d. 
 
 Conversely ; A line which meets a plane meets it in the projection of 
 the line in the plane, since the projecting plane contains all the per- 
 pendiculars, or shortest lines, from the line to the plane. Hence a line 
 which never meets a plane is parallel to its projection in that plane, that 
 is, to the plane itself (476). Q- e. d. 
 
 PROPOSITION XIV. 
 
 478. Theorem. — Either of two parallel lines is paral- 
 lel to every plane containing the other. 
 
 Demonstration. 
 
 Let AB and CD be two parallel lines, and MN a plane containing CD. 
 
 Then is AB parallel to the plane MN. 
 
 Since AB and CD are in the same plane 
 (?), and as the intersection of their plane 
 with MN is CD (?), if AB meets the plane 
 MN, it must meet it in CD, or CD produced. 
 But this is impossible (?). 
 
 Whence AB is parallel to MN (477). 
 
 Q. E. D. Fig- 228. 
 
 479. Corollary 1. A line which is parallel to a line of 
 a plane is parallel to the plane. 
 
 480. Corollary 2. — Through any given line, a plane 
 may he passed parallel to any other given line not in the 
 plane of the first. 
 
 For, through any point of the line through which the plane is to pass, 
 conceive a line parallel to the second given line. The plane of the two 
 intersecting lines is parallel to the second given line (?). 
 
 481. Corollary 3. — Through any point in space a plane 
 may he passed parallel to any two lines in space. 
 
 For, through the given point conceive two lines respectively parallel 
 to the given lines ; then is the plane of these intersecting lines parallel to 
 the two given lines (?). 
 10 
 
218 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XV. 
 
 482. Theorem. — Of two lines perpendicular to each 
 other, if one is perpendicular to a plane the other is par- 
 allel to the plane. 
 
 Demonstration. 
 
 Let AB and PD be perpendicular to each other, and PD perpendicu- 
 lar to the plane MN. 
 
 Then is AB parallel to MN. 
 
 If AB does not intersect PD, 
 through any point in PD, as G, 
 draw A'B' parallel to AB ; then is 
 it perpendicular to PD (32, foot- 
 note). 
 
 Let CE be the projection of 
 A'B' in the plane MN. Then is H 
 the point where PD pierces the 
 plane in CE (?). 
 
 Hence A'B' is parallel to its 
 projection CE (?), and consequently 
 parallel to the plane MN. 
 
 Therefore AB is parallel to CE (?), and consequently to the plane MN 
 (479). Q. E. D. 
 
 483. Corollary. — A line and a plane which are both 
 perpendicular to the same line are parallel. 
 
 
 Fia. 229. 
 
 RELATIVE POSITION OF TWO PLANES. 
 
 OF PARALLEL PLANES. 
 
 484. Parallel Planes are such that either is parallel to 
 any line of the other. 
 
 485. The Distance between Two Parallel Planes 
 
 at any point is measured by the perpendicular. 
 
STRAIGHT LINES AND PLANES. 
 
 219 
 
 PROPOSITION XVI. 
 
 486. Theorem. — Parallel planes are everywhere equi- 
 distant and hence can never meet. 
 
 Demonstration. 
 
 Let P and Q be two parallel 
 planes. 
 
 Then are they everywhere equi- 
 distant, and hence can never meet. 
 
 Let A and B be any two points 
 in P, and pass a line through them. 
 
 Since Q is parallel to P, it is 
 parallel to the line AB (484). And 
 since it is parallel to A B it is every- 
 where equidistant from AB. 
 
 Hence A and B, any two points 
 in P, are equidistant from Q, and 
 consequently P and Q can never 
 meet. q. e. d. 
 
 Fig. 230. 
 
 PROPOSITION XVII. 
 
 487. Theorem. — Two planes perpendicular to the 
 same line are parallel to eoA^h other. 
 
 Demonstration. 
 
 Let P and Q be two planes perpen- 
 dicular to the line AB. 
 
 Then are P and Q parallel. 
 
 For any line in one plane is parallel 
 to its projection in the other, since any 
 line in either plane is perpendicular to 
 AB (?). 
 
 Hence either plane is parallel to any 
 line of the other (476), and therefore the 
 planes are parallel to each other, q. e. d. Fig. 231. 
 
 ^ 
 
220 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XVIII. 
 
 488. Theorem.— 7/ a plane intersects two parallel 
 planes, the lines of intersection are parallel. 
 
 Demonstration. 
 
 Let RS intersect the parallel planes MN and PQ in AB and CD. 
 
 Then is AB parallel to CD. 
 
 For, if AB and CD could meet, the planes 
 MN and PQ would meet, as every point in A B is 
 in MN, and every point in CD in PQ. Hence, 
 AB and CD lie in the same plane, and do not 
 meet how far soever they be produced (132) ; 
 they are therefore parallel, q. e. d. 
 
 489. Corollary. — Parallel lines in- 
 tercepted between parallel planes are equal. Fig. 232. 
 
 Thus, AC = BD, if they are parallel. For, the intersections AB and 
 CD, of the plane of these parallels, are parallel (?), and the figure ABDC 
 is a parallelogram ; whence, AC = BD (?). 
 
 
 PROPOSITION XIX. 
 
 490. Theorem.— t^ line which is perpendicular to 
 one of two parallel planes, is perpendicular to the other 
 
 also. 
 
 Demonstration. 
 
 Let MN and PQ be two parallel planes; and 
 let AB be perpendicular to PQ. 
 
 Then is AB perpendicular to MN. 
 
 For, pass any plane through AB, and let AC 
 and BD be its intersections with MN and PQ re- 
 spectively Then are AC and BD parallel (?). 
 Now, AB is perpendicular to BD (?), and hence 
 to AC (?). Thus, AB is shown to be perpendic- 
 ular to any line of MN passing through its foot, 
 and hence perpendicular to MN (0- Q- e. d. 
 
 Fig. 233. 
 
STRAIGHT LINES AND PLANES, 
 
 221 
 
 PROPOSITION XX. 
 
 491. Theorem. — TJvrough any point without a plane, 
 one plane can be passed parallel to the given plane, and 
 only one. 
 
 Demonstration. 
 
 Let MN be a plane, and B any point without MN. 
 
 Let BA be a perpendicular from B 
 upon MN. 
 
 Through B draw DE and FG per- 
 pendicular to AB. Then is the plane of 
 DE and and FG parallel to MN (452, 
 487). Q. E. D. 
 
 Again, as any plane parallel to MN 
 is perpendicular to AB, and as only one 
 plane can be passed through B perpen- 
 dicular to AB (457), only one plane can 
 be passed through B parallel to MN. Q. e. d. 
 
 Fig. 234. 
 
 PROPOSITION XXI. 
 
 492. Theorem. — Two angles lying in different planes, 
 but having their sides parallel and extending in the 
 same direction, or in opposite directions, are equal, and 
 their planes are parallel. 
 
 Demonstration. 
 
 Let A and A' lie In the different planes 
 MN and PQ, and have AB parallel to AB , 
 and AC to AC. 
 
 Then A = A', and MN and PQ are 
 parallel. 
 
 For, take AD = A'D', and AE = A'E', 
 and draw AA', DD', EE', ED, and E'D'. 
 Now, AD being equal and parallel to 
 AD', 
 
 AA' = DD' (?) 
 
 Fig. 235. 
 
222 
 
 ELEMENTARY GEOMETRY. 
 
 EE'; 
 
 Again, since EE' 
 
 For like reason, 
 
 AA' = 
 therefore EE' = DD'. 
 and DD' are respectively parallel to AA', 
 they are parallel to each other (?) ; whence 
 EDO'E' is a parallelogram (?), and ED = 
 E'D'. Hence the triangles ADE and A'D'E' 
 are mutually equilateral, and A, opposite 
 ED, is equal to A', opposite E'D', equal to 
 ED. Q. E. D. 
 
 Again, the plane of the angle BAC, MN, 
 is parallel to PQ, the plane of B'A'C. 
 For, let a plane be passed through A and revolved until it is parallel 
 to PQ. It must cut DD' which is parallel to AA', and EE' which also is 
 parallel to AA', so that DD' and EE' shall equal AA' (?); hence it must 
 pass through D. Hence the planes of the angles are parallel, q. e. d. 
 
 493. Corollary l.—If two inter sectiiig planes are cut 
 by parallel planes, the angles formed hy the intersections 
 are equal. 
 
 Thus, AS' and AC being cut by the parallel planes MN and PQ. AD is 
 parallel to A'D' (0, and extends in the same direction from vertex A that 
 A'D' does from A'; and the same may be said of AC and A'C. Hence, 
 BAC = B'A'C (?). 
 
 494. Corollary 2. — // the corresponding extremities 
 of three equal parallel lines not in the same plane are 
 joined, the triangles formed are equal, and their planes 
 parallel. 
 
 Thus, if AA' = DD' = EE', the sides of the triangle AED are equal 
 to the sides of A'E'D', since the figures AD', DE', and EA' are parallelo- 
 grams (?), and the corollary comes under the proposition {i). 
 
STRAIGHT LINES AND PLANES, 
 
 PROPOSITION XXII. 
 
 495. Theorem. — // two lines are cut by three parallel 
 planes, the corresponding intercepted segments are propor- 
 tional. 
 
 Demonstration. 
 
 Let AB and CD be cut by the three parallel planes M, N, and P, AB 
 piercing the planes in A, E, and B, and CD in C, F, and D. 
 
 ^ . AE CF 
 ^^^^^EB = FD- 
 
 Join the points A and D by the 
 straight line AD, and eoneeive planes 
 passing through AD and DC, and 
 through AB and AD. 
 
 Let EH and BD be the intersec- 
 tions of the planes N and P with the 
 plane BAD, and AC and HF the in- 
 tersections of M and N with ADC- 
 
 Now, since EH is parallel to BD (?), 
 
 AE _ AH 
 
 EB ~ HD ^'' " 
 
 Fig. 236. 
 
 In like manner, by reason of the parallelism of HF and AC, 
 
 CF_ AH 
 FD ~ HD' 
 
 Hence, by equality of ratios, 
 
 AE CF 
 
 Ei = FD- ^"•^- 
 
 [Note. — Planes perpendicular or oblique to each other give rise to one 
 species of solid angles; hence their consideration is reaeired for the next 
 Section.] 
 
234 ELEMENTARY GEOMETRY, 
 
 EXERCISES. 
 
 496. 1. Designate any three points in the room, as one cor- 
 ner of the desk, a point on the stove, and some point in the 
 ceiling, and show how you can conceive the plane of these points. 
 
 2. Show the position of two lines which will not meet, and 
 yet are not parallel. 
 
 3. Conceive two lines, one line in the ceiling and one in the 
 floor, which shall not be parallel to each other. 
 
 4. The ceiling of my room is 10 feet above the floor. I have 
 a 12-foot pole, by the aid of which I wish to determine a point 
 in the floor directly under a certain point in the ceiling. How 
 can I do it ? 
 
 Suggestion.— Consult Proposition XII. 
 
 5. Upon what principle in this section is it that a stool with 
 three legs always stands firm on a level floor, when one with four 
 may not ? 
 
 6. By the use of two carpenter's squares you can determine a 
 perpendicular to a plane. How is it done ? 
 
 7. If you wish to test the perpendicularity of a stud to a level 
 floor, on how many sides of it is it necessary to measure the 
 angle which it makes with the floor? By applying the right 
 angle of the carpenter's square on any two sides of the stud, to 
 test the angle which it makes with the floor, can you determine 
 whether it is perpendicular or not ? 
 
 8. If a line is drawn at an inclination of 23° to a plane, what 
 is the greatest angle which any line of the plane, drawn through 
 the point where the inclined line pierces the plane, makes with 
 the line ? Can you conceive a line of the plane which makes an 
 angle of 50° with the inclined line ? Of 80° ? Of 15° ? Of 170° ? 
 
SOLID ANGLES, 225 
 
 ^^^gcfToM \l 
 
 OF SOLID ANGLES. 
 
 497. • A Solid Angle is the opening between two or more 
 planes, each of which intersects all the others. The lines of in- 
 tersection are called Edges, and the planes, or the portion of 
 the planes between the edges where there are more than two, 
 are called Faces. 
 
 498. Solid Angles are of Three Species, viz., Diedral, 
 Triedral, and Polyedral, according as they have two, three, 
 or more than three faces. 
 
 OF DIEDRALS. 
 
 499. A Diedral Angle, or simply a Diedral, is the 
 
 opening between two intersecting planes. 
 
 600. A Diedral (Angle) is Measured by the plane 
 angle included by lines drawn in its faces from any point in the 
 edge, and perpendicular thereto. 
 
 A diedral angle is called Right, Acute, or Obtuse, 
 according as its measure is right, acute, or obtuse. 
 
 Two diedrals are said to be Supi)leinentary, when their 
 measures are supplementary. 
 
 Of course the magnitude of a solid angle is independent of the dis- 
 tances to which the edges may chance to be produced. 
 
 Illustrations. — The opening ])etween the two planes CABF and 
 DABE (Fig. 237) is a Diedral (angle), AB is the Edge, and CABF and 
 DABE are the Faces. Let MO lie in the plane AF, perpendicular to the 
 edge ; and NO in AE, and also perpendicular to the edge ; then the plane 
 angle MON is the measure of the diedral. 
 
ELEMENTARY GEOMETRY. 
 
 Fig. 237. 
 
 Fig. 238. 
 
 Fig 239. 
 
 501. A diedral may be read by the letters on the edge, when 
 there would be no ambiguity, or otherwise by these letters and 
 one in each face. 
 
 Thus, the diedral in Fig. 237 may be designated as AB, or as C-AB-D. 
 
 502. A diedral may be considered as generated by the revolu- 
 tion of a plane about a line of the plane, and hence we may see 
 the propriety of measuring it by the angle included by two lines 
 in its faces perpendicular to its edge, as stated in the preceding 
 article. 
 
 Illustration. — Let AB (Fig. 338) be a line of the plane GB. Con- 
 ceive ^B perpendicular to A'B. Now, let the plane revolve upon AB as 
 an axis, whence grB describes a circle (?) ; and at any position of the re- 
 volving plane, as /BAF, since f^g measures the amount of revolution, it 
 may be taken as the measure of the diedral f-E^-g. When gB has made 
 ^ of a revolution, the plane will have made ^ of a revolution, and the 
 diedral will be right. 
 
 503. When two planes intersect, four diedrals are formed, any 
 two of which are either At\jaceiit to each other, or Opposite. 
 
 504. Adjacent Diedrals are on the same side of one 
 plane, but on the opposite sides of the other. 
 
 As D-AB-C and D-AB-c, or c-AB-D and c-AB-d (Fig. 239). 
 
 Opposite Diedrals are on opposite sides of both planes. 
 As D-AB-C and ^AB-c, or D-AB-c and <^AB-C (Fig. 239). 
 
SOLID ANGLES. 
 
 227 
 
 PROPOSITION I. 
 
 506. Theorem. — W%en two planes intersect, the op- 
 posite diedrals are equal, and the adjacent ones are 
 supplementary. 
 
 Demonstration. 
 Let the planes DE and CF intersect in AB. 
 Then D-AB-C = d-M-c, 
 
 and D-AB-r = <?-AB-C; 
 
 and D-AB-C + D-AB-C* = 180°, 
 
 c-AB-D + c-AB-</ = 180°, etc. 
 
 Through 0, any point in AB, let Mm J>e drawn 
 in the plane CF, and Nw in the plane DE, each 
 perpenrlicular to AB. Tlien is MON, the mcHsure 
 of D-AB C (?), = mOn, the measure of d-hB-c (?), 
 etc. Q. E. D. 
 
 Also, MON + NOm = 180° (?), 
 
 H0m-\-7n0n = 180°, etc. 
 
 Fig. 240. 
 
 q. E. D. 
 
 PROPOSITION II. 
 
 606. Theorem.—^/z,^ Une in one face of a right die- 
 dral, perpendicular to its edge, is perpendicular to the 
 other face. 
 
 Demonstration. 
 
 in the face CB of the right diedral C-AB-D^ let 
 MO be perpendicular to the edge AB. 
 
 Then is MO perpendicular to the face DB. 
 
 For, draw ON in the face DB, and perpendicu- 
 lar to AB. Now, since the diedral is right, and 
 MON measures its angle, MON is a right angle; 
 whence MO is perpendicular to two lines of the 
 plane DB, and consequently perpendicular to the 
 plane. Q. e. d. 
 
 Fig. 241. 
 
 *"■ By this is meant the measure of the diedral. 
 
228 
 
 ELEMENTARY GEOMETRY. 
 
 607. OoKOLLARY 1. — Conversely, If one 
 plane contains a line which is perpen- 
 dicular to another plane, the dtiedral is 
 right. 
 
 Thus, if MO is perpendicular to the plane 
 DB, C-AB-D is a right diedral. For MO is perpen- 
 dicular to every line of DB passing through its 
 foot (?) ; and hence is perpendicular to ON, drawn 
 at right angles to AB. When C-AB-D is a right 
 diedral, for it is measured by a right plane angle. ^'9- ^*'- 
 
 608. Two planes are Perpendicular to each other when 
 they intersect so as to make the adjacent diedrals equal. In this 
 case, all four of the diedrals are right. 
 
 609. Corollary 2.— The plane which projects a line 
 upon a plane (468) is perpendicular to the plane of projec- 
 tion. 
 
 PROPOSITION III. 
 
 610. Theorem. — // each of two intersecting planes is 
 perpendicular to a third, their intersection is perpendicu- 
 lar to the third plane. 
 
 Demonstration. 
 
 Let EF 3"'* CD bo twu planes perpendicular to the third plane MN, 
 and J^yt AB be the intersection of EF and CD. 
 
 Then is AB perpendicular to MN. 
 
 For, EF being perpendicular to MN, 
 D-FG-E is a right diedral, and a line in EF 
 perpendicular to FG at B is perpendicular to 
 MN ; also a line in the plane CD, and perpen- 
 dicular to DH at B, is perpendicular to MN (?). 
 
 Hence, as there can be one and only one 
 perpendicular to MN at B, and as this perpen- 
 dicular is in both planes, CD and EF, it is 
 their intersection, q. e. d. 
 
 Fig. 242. 
 
SOLID ANGLES. 
 
 229 
 
 PROPOSITION IV. 
 
 511. Theorem.— me angle Uicluded by perpendicu- 
 lars drawn from any point within a diedral to its faces, 
 is the supplement of the diedral. 
 
 Demonstration. 
 
 Fig 243. 
 
 Let P' be any point within the diedral F-AB-C, and let the perpendic- 
 ulars P'D' and P'E' be drawn to the faces. 
 
 Then is D'P'E' the sup- 
 plement of F-AB-C. 
 
 From P, any point in 
 the plane which Insects the 
 diedral F-AB-C, draw PD 
 and PE perpendicular to 
 the same faces respectively 
 as P'D' and P'E'. Then is 
 DPE = D'P'E'. 
 
 Now pass a plane 
 through PE and PD, and 
 let EG and DO be its inter- 
 sections with FB and CB 
 respectively. Then, by 
 
 (507), FB and CB are perpendicular to the plane PEOD. Hence, A5 is 
 perpendicular to PEOD (?), and EOD is the measure of F-AB-C (?). But 
 in the quadrilateral PEOD, P is the supplement of EOD (?), and hence of 
 F-AB-C. 
 
 Hence, D'P'E' is the supplement of F-AB-C. Q. e. d. 
 
 512. Corollary 1. — If from, a point in the edge of a 
 diedral perpendiculars are erected to the faces on the same 
 sides of the planes respectively as the perpendiculars let 
 fall from a point within, the included angle is the sup- 
 plem^ent of the angle of the diedral. 
 
 513. Corollary 2. — The angle DPE is the supplement 
 of the opposite diedral H-AB-I, and equal to each of the ad- 
 jacent diedrals C-AB-I and F-AB-H- 
 
230 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION V. 
 
 514. Theorem. — Between any two lines not in the 
 same plane one line, and only one, can he drawn which 
 shall he perpendicular to both, and this line is the shortest 
 distance between them. 
 
 Demonstration. 
 
 Let AB and CD be two lines not in the same plane. 
 
 Then one line, as HG, and only 
 one, can be drawn which is perpen- 
 dicular to both AB and CD, and HG 
 measures the shortest distance be- 
 tween AB and CD. 
 
 Through either line, as CD, pass 
 a plane MN parallel to AB (480). 
 From any point in AB, as E, let fall 
 EF perpendicular to MN. 
 
 Let EK be the plane of the lines 
 EF and EB, and let FK be its trace 
 
 Fig. 
 
 m MN. 
 
 Now, as AB and CD are not in the same plane, EK, and hence its 
 trace FK, cuts CD in some point, as G. 
 
 From G draw GH perpendicular to AB. 
 
 laf nu i;-- :-> the plane EK (?) widch is perpendicular to MN (?), and 
 being perpendicular to AB is perpemUcuIar to FK , and hence to the 
 
 GH, which is perpendicular to AB, is perpendicular to 
 
 244. 
 
 ■ia. an IS the only line which is perpendicular U) oth AB and CD. 
 
 For ftny line which is prrpendicular to A5 aua C^ is perpendicular 
 to FK (?), and hence to MN (?). 
 
 Now every perpendicular from AB to the plane MN meets this plane 
 in FK (?). 
 
 But FK and CD have only one point common, viz., G. Hence, GH is 
 the only perpendicular from AB to CD. 
 
 3d. GH is the shortest distance between AB and CD. For a line from 
 any point in AB to any other point in CD, as LS, would be oblique to 
 MN (?), and hence longer than the perpendicular LR, = HG. 
 
BOLID ANGLES, 
 
 231 
 
 PROPOSITION VI. 
 
 515. Theorem. — // one of two parallel planes is per- 
 pendicular to a third plane, the other is also. 
 
 Demonstration. 
 
 Let PO and QE be two paral 
 lei planes; and let PD be perpen- 
 dicular to the third plane MN. 
 
 Then is QE perpendicular to 
 MN. 
 
 Through PD and QE pass the 
 plane RS perpendicular to MN, 
 and let FK be its trace in QE, and 
 HI in PD. 
 
 Then is FK perpendicular to 
 MN (?). 
 
 And, as HI is parallel to FK 
 (?), it is perpendicular to MN 
 (460). 
 
 Hence, QE is perpendicular 
 to MN (507). Q. K. D. 
 
 Fig. 245. 
 
 OF TRIEDRALS. 
 
 516. As diedrals result from the intersection of two planes, 
 so triedrals result from the intersection of three planes. 
 
 Fifl. 24d. 
 
232 ♦ ELEMENTARY GEOMETRY. 
 
 517. Three planes may intersect in three principal ways: 
 
 1st. Their intersections may all coincide, as in (a). 
 
 2d. They may have three parallel intersections, as in (h), 
 
 3d. They may have three non-parallel intersections, as in (c). 
 In this case the three intersections meet in a common point, 
 as at S. 
 
 In the first case the three planes have an infinite number of 
 common points. In the second case they have no common point. 
 In the third case they have but one common point. 
 
 The third case gives rise to Triedrals. 
 
 618. A Triedral is the opening between three planes which 
 meet in a common point. 
 
 519. When three planes meet so as to form one triedral, they 
 form also eight, as planes are to be considered indefinitely ex- 
 tended, unless otherwise stated. 
 
 520. The planes enclosing a particular triedral are called its 
 Faces, and their intersections its Edges. The common point 
 is called the Vertex. 
 
 521. A triedral may be designated by 
 naming the letter at the vertex and then 
 three other letters, one in each edge. 
 
 Thus, in the figure, the opening between the 
 three planes ASC, CSB, and BSA is the triedral 
 S-ABC. ThQ faces are ASC, CSB, and BSA. 
 
 Fig. 247. 
 
 522. The plane angles enclosing a solid angle are called 
 Facial Angles. 
 
 523. In every particular triedral there are six parts, Three 
 Facial Angles and Three Diedrals. 
 
SOLID ANGLES, 
 
 233 
 
 524. Our study of triedrals will be confined to the relations 
 of the facial angles and the diedrals, and the comparison of dif- 
 ferent triedrals. 
 
 525. Triedrals are Rectangular, Bi-rectang^ular, or 
 
 Tri-rectangular, according as they have one, two, or three 
 right diedral angles. 
 
 Illubtbation. — The comer of a cube is a 
 Tinrectcmgular triedral, as S-ADC. Conceive the 
 upper portion of the cube removed by the plane 
 ASEF ; then the angle at S, i. e., S-AEC, is a Bi- 
 rectangular triedral, A-SC-E and A-SE-C being 
 right diedrals. Pig. 248. 
 
 626. An Isosceles Triedral is one that has two of ito 
 facial angles equal. An Equilateral Triedral is one that 
 has all three of its facial angles equal. 
 
 527. Opposite Triedrals are such as lie on opposite sides 
 of each of the intersecting planes, as S-ABC and %-ahc. 
 
 Opposite triedrals have mutually equal facial and equal 
 diedral angles, but these being differently disposed, such 
 triedrals are not in general capable of superposition. 
 
 Illustration. — Let the edges of the triedral S-ABC be 
 produced beyond the vertex, forming the opposite triedral 
 S-aZw. Now, the faces are equal plane angles, but disposed 
 in a different order. Thus, ASB = aS&, ASC = aSc, and 
 BSC = 6Sc, and the diedrals are also equal ; but the 
 triedrals cannot be superimposed, or made to coincide. To 
 show this fact, conceive the upper triedral detached, and 
 the face aSc placed in its equal face ASC, Sa in SA, and Sc ^'9- ^*^- 
 in SC. Now the edge S6, instead of falling in SB, in front of ASC, will 
 fall behind the plane ASC. 
 
 Or, otherwise, if %-ahc be revolved on S by bringing it forward and 
 turning it down on S-ABC, since the diedrals A-SB-C and o-Sh-a are 
 equal, they will coincide ; but, as facial angle aSh is not necessarily equal 
 to CSB, S/i will not necessarily fall in SC. For a like reason, %c will not 
 necessarily fall in SA. 
 
234 
 
 ELEMENTARY GEOMETRY, 
 
 528. Symmetrical Triedrals are triedrals in which each 
 part in one has an equal part in the other ; but the equal parts 
 not being similarly disposed, the triedrals may not be capable of 
 superposition. 
 
 Symmetrical solids are of frequent occurrence : the two hands form 
 an illustration ; for, though the parts may be exactly alike, the hands 
 cannot be placed so that their like parts "will be similarly situated ; in 
 short, the left glove will not fit the right hand. 
 
 529. Atljaceiit Triedrals are such as lie on different 
 sides of one of the intersecting planes, and on the mme side of 
 two of them. 
 
 Thus, S-ADE is adjacent to 
 8.DRE. 
 
 In adjacent triedrals, two of 
 the facial angles of one are the 
 supplements of two of the other, 
 each to each, and one is equal in 
 each 
 
 Thus, in the adjacent triedrals 
 S-DRE and S-ADE, ASE and ASD 
 are supplements respectively of ESR 
 and DSR, while DSE is common to 
 both. 
 
 Fig. 250. 
 
 530. Of the eight triedrals formed by the intersection of 
 three planes, each has its Oi)i)Osite or Symmetrical 
 triedral, and each has three Adjacent triedrals. 
 
 631. Two triedrals are Supplementary when the facial 
 angles of the one are the supplements of tiie measures of the 
 corresponding diedrals of the other. 
 
 532. Equality, as has been before defined, means, in Geom- 
 etry, equality in all respects ; and two figures that are said to be 
 equal are capable of being so applied the one to the other that 
 they will coincide throughout. This absolute equality is hence 
 
SOLID ANGLES, 
 
 235 
 
 often called Equality by Superposition^ in distinction 
 from Equality by Symmetry. 
 
 533. Two figures are said to be Equal by Symmetry, or 
 Symmetrically Equals or simply Symmetrical, when 
 each part in one has an equal part in the other ; but these equal 
 parts being differently arranged in the two figures, the one may 
 not be capable of being superimposed upon the other. (See 527.) 
 
 PROPOSITION VII. 
 
 534. Theorem. — Opposite triedrals are equal and may 
 be syninietrlcal. 
 
 Demonstration. 
 
 Let S-ABC and S-ahc be two opposite triedrals. 
 
 Then are the triedrals equal or syiiiiiietricul. 
 
 For the facial angle ASC = the facial angle aSc (?); 
 also, BSC = bSc, and ASB = aSh. 
 
 Again, the diedra' A-SB-C = a-Sft-c, since they are op- 
 posite diedrals. 
 
 For like reason, B SA-C = b-Sa-c, and A-SC-B = a-Sc-b. 
 
 Hence all the parts in one triedral have equal parts in 
 the other. 
 
 But, in general, these triedrals cannot be superimposed. 
 (See illustration, 527.) '** 
 
 If, however, ASB = CSB, then aSb = cSb, and the triedrals can be 
 superimposed. 
 
 Thus, conceive the triedral S-ahc revolved on S, being brought over 
 towards the observer until Sb falls in SB. 
 
 Tiien, since CSB = ASB = aSb, aSb may be made to coincide with 
 BSC, and as the diedrals A-SB-C and ct-Sb-c are equal, cSb will fall in 
 ASB, and the triedrals will coincide, and will be equal. 
 
 Hence, opposite diedrals are equal and may be symmetrical, q. e. d. 
 
 535. 
 equal. 
 
 Corollary 1. — Opposite isosceles triedrals are 
 
236 
 
 ELEMENTABY GEOMETRY. 
 
 PROPOSITION VIII. 
 
 536. Theorem. — Two syrmnetrical triedrals may al- 
 ways be conceived to be placed as opposite triedrals. 
 
 Demonstration. 
 
 Let S-ABC and S'-A'B'C be two symmetrical triedrals, B and B' 
 being in front of the planes ASC and A'S'C, ASB = A'S'B', ASC = 
 A'S'C, BSC = B'S'C', A-SB-C = A'-S'B-C', A-SC-B =- A'-SX'-B', and 
 B-SA-C = B'-S'A'-C. 
 
 Then may S-ABC and S'-A'B'C 
 
 be placed as opposite triedrals. 
 
 Produce the edges of either 
 triedral, as S'-A'B'C, beyond the 
 vertex, forming the opposite tri- 
 edral S'-aftc. 
 
 Then can S-ABC be super- 
 imposed upon ^'-dbc, and the latter 
 fulfills the requirements of the 
 proposition. 
 
 The application is made as 
 follows : 
 
 Since B' is in front of the plane Fig. 252. 
 
 A'S'C, I is behind the plane aS'c. 
 
 Now conceive S-ABC inverted and reversed so that B shall fall 
 behind the plane ASC. 
 
 Then apply ASC to its equal aS'c, SA falling in S'a, and SC 
 in S'c. 
 
 By reason of the equality of A-SC-B and a-S'c-& (= A'-S'C'-B'), the 
 plane BSC will fall in 5S'c, and for a like reason ASB will fall in 
 aS'5 ; and since the planes coincide, their intersections SB and S'ft must 
 coincide. 
 
 Hence, S-ABC = S'-a5c, the opposite to S'-A'B'C Q. E. d. 
 
SOLID ANGLES, 
 
 237 
 
 PROPOSITION IX. 
 
 537. Theorem. — Two triedrals which have two facial 
 angles and the included diedral equal, each to each, are 
 either equal or symmetrical, 
 
 Demonsteation. 
 
 Let I, 2, 3, be triedrals having the facial angle ASC = A'S'C = aS"c, 
 CSB = C'S'B' = cS "6, and A-SC-B = A -S'C'-B' = a-^'c-h. 
 
 Fig. 253. 
 
 Then are the triedrals either equal or symmetrical. 
 
 Ist. When the equal facial angles are on the same sides of the respec- 
 tive equal diedrals, as in Figs. 2 and 3, the triedrals may be appliecf the 
 one to the other. 
 
 Thus, let the facial angle A'S'C be placed in its equal oS"c, A'S' in 
 aS, and S'C in S"c; whence, by reason of the equality of the diedrals 
 A'-S'C'-B' and a-S"c-6, and since the facial angles B'S'C and hS"c lie on 
 the same sides respectively of their diedrals A'-S'C -B' and a-S"c-h, the 
 plane of B'S'C falls in the plane of 6S"c, and since angle B'S'C = angle 
 6S"c, B'S' falls in 6S", and A'S'B' coincides with aS"&. 
 
 Hence the triedrals coincide and are equal, q. e. d. 
 
 2d. But if the equal facial angles lie on different sides of the equal 
 diedrals, as in Figs. 1 and 3, let the opposite of S-ABC be drawn (627), 
 and call it S-a'Vcf. Then may 1 be applied to S-a'h'd. 
 
 [Let the student draw the figure and make the application.] 
 
338 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION X. 
 
 538. Theorem. — Two triedrals which have two die- 
 drals and the included facial angles equal each to each, 
 are either equal or symmetrical, 
 
 Demoi^stbation. 
 [Same as preceding. Let the student draw figures like those for the 
 preceding, and go through with the details of the application.] 
 
 539. Corollary. — In equal or in symmetrical triedrals, 
 the equal facial angles are opposite the equal diedrals. 
 
 PROPOSITION XI. 
 
 540. Theorem. — The sum of any two facial angles of 
 a triedral is greater than the third- 
 
 Demonstration. 
 This proposition needs demonstration only in case of the sum of the 
 two smaller facial angles as compared with the greatest {(). 
 Let ASB and BSC each be less than ASC ; then is 
 
 ASB + BSC > ASC. 
 
 For, in the face ASC, make the angle AS J' = ASB, 
 and S5' = S&, and pass a plane through 5 and 6', cut- 
 ting SA and SC in « and c. 
 
 The two triangles aSh and a%V are equal (?), whence 
 
 Now, ab ■\-'bc> a^ (?), Fig, 254. 
 
 and subtracting ah from the first member, and its equal cikl from the sec- 
 ond, we have he > ft'c. 
 
 Whence the two triangles 5Sc and V%c have two sides in the one 
 equal to two sides in the other, each to each, but the third side he > than 
 the third side ft'c, and consequently angle BSC > h SC. Adding ASB to 
 the former, and its equal AS&' to the latter, we have 
 ASB + BSC > ASC. Q. E. D. 
 
 541. Corollary. — The difference hetween any two facial 
 angles of a triedral is less than the third facial angle (?). 
 
SOLID ANGLES. 239 
 
 PROPOSITION XII. 
 
 542. Theorem. — Two triedrals which have two facial 
 angles of the one equal to two facial angles of the other, 
 each to each, and the included diedrals unequal, have the 
 third facial angles unequal, and the greater facial angle 
 belongs to the trledral having the greater included diedi'al. 
 
 Demonstr'ation. 
 
 Let ASC = asc, and ASB = asb, while the 
 diedral C-SA B > c-sa-b. 
 
 Then CSB > eab. 
 
 For, divide the diedral C-SA-B by a plane 
 ASO, making the diedral C-SA-0 = o-sa-b; 
 end taking ASO = aid, bisect the diedral 
 0-SA-B with the plane ISA. Conceive the 
 planes OSI and OSC. F«g- 255. 
 
 Now, the triedrals S-AOC and s-ahc are equal or symmetrical, having 
 two facial angles and the included diedral equul eael^^to each (637). 
 
 For a like reason, S-AIO and S-AIB are symmetrical, and the facial 
 angle OSI = ISB. 
 
 Again, in the triedral S-IOC> 
 
 OSI + ISC > OSC (640), 
 and substituting ISB for OSI, we have 
 
 ISB -I- ISC (or CSB) > OSC, or its equal csb. q. e. d. 
 
 643. Corollary.— Conversely, // tJte two facial angles 
 are equal, each to each, in two. triedrals, and the third 
 facial angles unequal, the diedral opposite the greater 
 facial angle is the greater. 
 
 That is, if ASB = asb, and ASC = anCy 
 
 while BSC > bsc, 
 
 the diedral B-AS-C > b-as-r. 
 
 For, if B-AS-C ^ b-as-c, BSC = bse (637, 639) ; 
 
 and if B-AS C < b-aa-c, BSC < bse, by the proposition. 
 
 Therefore, as B-AS-C cannot be equal to nor less than b-as-e, it must 
 be greater, q. e. d. 
 
240 
 
 ELEMENTARY OEOMETRT, 
 
 PROPOSITION XIII. 
 
 M 544. Theorem. — Two triedrals which have the three 
 facial angles of the one equal to the three facial angles of 
 the other, each to each, are either equal or symmetrical, 
 
 Demokstration. 
 
 Let A, B, and C represent the facial angles of one, and a, 5, and c the 
 corresponding facial angles of the other. If A = a, B = &, and C :^ c, 
 the triedrals are equal or symmetrical. 
 
 For A being equal to a, and B to &, if, of their included diedrals, SM 
 were greater than «m, C would be greater than c (?) ; and if diedral SM 
 were less than diedral sm^ C would be less than c (?). Hence, as diedral 
 SM can neither be greater nor less than diedral sm, it must be equal to it. 
 
 Therefore the triedrals have two facial angles and the included diedral 
 equal, each to each, and are consequently equal or symmetrical. Q. e. d. 
 
 PROPOSITION XIV. 
 
 545. Theorem,— ///roT^Z/ any point within a triedral 
 perpendiculars are drawn to the faces, they will be the edges 
 of a supplementary triedral. 
 
 Demonstratioit. 
 
 From S' within the triedral S-ABC, let S'A' be drawn perpendicular 
 to ASB, SB' to ASC, and S'C to BSC. 
 
 Then is S'-A'B'C supple- 
 mentary to S-ABC. 
 
 For the facial angle A'S'B' 
 is the supplement of the di- 
 edral B-AS-C (511); and for 
 like reason B'S'C is the sup- 
 plement of A-SC-B, andA'S'C 
 of A-SB-C. 
 
 Again, since S'A' is per- 
 pendicular to the face ASB, 
 and S'B' is perpendicular to 
 
 ASC, the plane of S'A' and ^. „^^ 
 
 rig. Z5o. 
 
SOLID ANGLES. 241 
 
 S'B' is perpendicular to ASB and ASC, and therefore to SA. Hence SA 
 is perpendicular to the face A'S'B'. 
 
 For a similar reason, SC is perpendicular to B'S'C Hence ASC is 
 the supplement of A'-S'B'-C 
 
 In like manner, it may be shown that BSC is the supplement of 
 A'-S'C'-B', and ASB of B'-S'A'-C. Q. e. d. 
 
 546. Scholium 1. — If perpendiculars were drawn from the point S, 
 or any other point, parallel to those from S', and in the same directions 
 respectively from S that S'A', etc., are from S', they would also l)e per- 
 pendicular to the faces of the diedral, and would form a supplementary 
 triedral. 
 
 547. Scholium 2. — The triedral S'-A'B'C is also supplementary to 
 the triedral opposite to S-ABC. 
 
 548. Scholium 3. — The triedral S'-A'B'C will not be supplementary 
 to the triedral adjacent to S-ABC, but one facial angle will be supple- 
 mentary to the corresponding diedral in the other, and the other facial 
 angles will be equal to their corresponding diedrals. 
 
 549. Scholium 4. — One triedral adjacent to S'-A'B'C will be sup- 
 plementary to one of those adjacent to S-ABC. 
 
 PROPOSITION XV. 
 
 550. Theorem. — In an isosceles triedral the diedrals 
 opposite the equal facial angles are equal ; and, 
 
 Conversely, If two diedrals of a triedral are equal, the 
 triedral is isosceles. 
 
 Demon^stration. 
 
 In the triedral S-ABC, let ASC = CSB. 
 
 Then is C-SA-B = C-SB-A. 
 
 For, pass the plane CSD through the edge 
 SC, bisecting the diedral A-SC-B. Then the 
 two triedrals S-ACD and S-CBD have two facial 
 angles of one equal to two facial angles of the 
 other, each to each ; that is, ASC = CSB, by 
 hypothesis, and CSD common; and the in- P, 257 
 
242 ELEMENTARY GEOMETRY. 
 
 eluded diedral8 equal by construction. Hence the triedrals are sym- 
 metrical, and 
 
 C-SA-B = C-SB-A (537, 639). Q. k. d. 
 
 Convereely, if C-SA-B = C-SB-A, 
 
 ASC = CSB. 
 
 For the supplementary triedral is isosceles ; whence the diedrals op- 
 posite those equal facial angles are equal. But ASC and CSB are the 
 supplements of these equal diedrals, and hence equal, q. e. d. 
 
 651. Corollary 1. — The plane which bisects the angle 
 included by the equal facial afigles of an isosceles triedral 
 is perpendicular to the opposite face, and bisects the oppo- 
 site facial angle. 
 
 662. Corollary 2.—// the^ three facial angles of a tri- 
 edral are equal, each to each, the diedrals are also equal, 
 each to each, and conversely. 
 
 PROPOSITION XVI. 
 
 663. Theorem. — Two triedrals which have the three 
 diedrals of the one equal to the three diedrals of the other, 
 each to each, are equal or symmetrical. 
 
 Demonstration. 
 
 In the two supplementary triedrals, the facial angles of the one are 
 equal to the facial angles of the other, each to each, since they are sup- 
 plements of equal diedrals (546). Hence, the supplementary triedrals 
 are equal or symmetrical (644). 
 
 Now, the facial angles of the first triedrals are supplements of the 
 diedrals of the supplementary ; whence the corresponding facial angles, 
 being the supplements of equal diedrals, are equal. Therefore, the pro- 
 posed triedrals have their facial angles equal, each to each, and are con- 
 sequently equal j or symmetrical. Q. E. D. 
 
 664. Corollary. — All tri-rectangular triedrals are 
 equal. 
 
SOLID ANGLES, 343 
 
 ^< V 
 
 PROPOSITION XVII. 
 
 565. Theorem. — The sum of the facial angles of a 
 -^ triedral may he anything between zero and four right 
 angles. 
 
 Demonstration. 
 
 Let ASB, BSC, and ASC be the facial angles 
 enclosing a triedral. 
 
 Then, as each must have some value, the sum 
 is greater than zero, and we have only to show 
 that ASB + ASC + BSC is less than 4 right angles. 
 
 Produce either edge, as AS, to D. Now, in the 
 triedral S-BCD, BSC is less than BSD + CSD (?). '"'s- '^^^^ 
 
 To each member of this inequality add ASB + ASC, and we have 
 
 ASB + ASC + BSC less than ASB + ASC + BSD + CSD (?). 
 
 But ASB + BSD = 2 right angles (?), 
 
 and ASC + CSD = 2 right angles; 
 
 whence ASB 4- ASC + BSD + CSD = 4 right angles, 
 
 and consequently ASB + ASC + BSC is less than 4 right angles, q. k. d. 
 
 PROPOSITION XVIII. 
 
 656. Tlxeorem.—The sum of the diedrals of a triedral 
 may he anything hetween two and six right angles. 
 
 Demonstration. 
 
 Each diedral being the supplement of a facial angle of the supple- 
 mentary triedral (531), the sum of the three diedrals is 3 times 2 right 
 angles, or 6 right angles, minus the sum of the facial angles of the sup- 
 plementary triedral. 
 
 But this latter sum may be anything between and 4 right angles (?). 
 Hence the sum of the diedrals may be anything between 2 and 6 right 
 angles, q. e. d. 
 
 \ 
 
244 
 
 ELEMENTARY GEOMETRY. 
 
 OF POLYEDRALS. 
 
 557. A Convex Polyedral is a polyedral none of the 
 faces of which, when produced, enter tlie solid angle. A sec- 
 tion of such a polyedral made by a plane cutting all its edges is 
 a convex polygon. (See Fig. 259. ) 
 
 PROPOSITION XIX. 
 
 558. Theorem. — Tke sum of the facial angles of any 
 convex polyedral is less than four right angles. 
 
 Demonstration. 
 Let S be the vertex of any convex polyedral. 
 
 Then is the sum of the angles ASB, BSC, CSD, 
 DSE, and ESA less than 4 right angles. 
 
 Let the edges of this polyedral be cut by any 
 plane, as ABODE, which section will be a convex 
 polygon, since the polyedral is convex. 
 
 From any point within this polygon, as 0, 
 draw lines to its vertices, as OA, OB, 00, etc. 
 There will thus be formed two sets of triangles, 
 one with their vertices at S, and the other with 
 their vertices at 0; and there will be an equal 
 number in each set, for the sides of the polygon 
 form the bases of both sets. 
 
 Now, the sum of the angles of each of these two sets of triangles is 
 the same. But the sum of the angles at the bases of the trian«;les having 
 their vertices at S is greater than the sum of the angles at the bases of 
 the triangles having their vertices at 0, since SBA + SBO is greater 
 than ABO, SOB + SOD is greater than BOD, etc. (540). 
 
 Therefore the sum of the angles at S is less than the sum of the angles 
 at 0> ». 6-, less than 4 right angles, q. e. d. 
 
 Fig. 259. 
 
i 
 
 EXEKCISES, 246 
 
 EXERCISES 
 
 659. 1. I have an iron block whose corners are all square 
 (edges right diedrals, and the vertices tri-rectangular, or right, 
 triedrals). If I bend a wire square 
 around one of its edges, as c S't?, at 
 what angle do I bend the wire ? If 
 I bend a wire obliquely around the 
 edge, as aSh, at what angle can I 
 bend it ? If I bend it obliquely, as 
 e S"/, at what angle can I bend it ? 
 
 2. Fig. 260 represents the ap- 
 pearance of a rectangular parallelo- Fi7"26o7 
 piped, as seen from a certain position. 
 
 Now, all the angles of such a solid are right angles : why is it 
 that they nearly all appear oblique ? Can you see a right paral- 
 lelopiped from such a position that all the angles seen shall 
 appear as right angles ? 
 
 3. The diedral angles of crystals are measured with great 
 care, in order to determine the substances of which the crystals 
 consist. How must the measure be taken? If we measure 
 obliquely around the edge, shall we get the true value of the 
 angle ? 
 
 4. Prove that if three planes intersect so as to make two 
 traces parallel, the third is parallel to each of these. 
 
 5. From a piece of pasteboard cut two figures of the same 
 size, like ABCDS and ahcds (Fig. 261). Then drawing SB and 
 SC so as to make 1 the largest angle and 3 the smallest, cut the 
 pasteboard almost through in these lines, so that it will readily 
 bend in them. Now fold the edges AS and DS together, and a 
 triedral will be formed. From the piece ahcds form a triedral in 
 like manner, only let the lines sc and sh be drawn so as to make 
 
24:6 ELEMENTARY GEOMETRY. 
 
 the angles 1, 2, and 3 of the same size as be- 
 fore, while they occur in the order given in 
 (ibcds. Now, see if you can slip one triedral 
 into the other, so that they will fit. What 
 is the difficulty? 
 
 6. In the last case, if 1 equals | of a right 
 angle, 2 = ^ of a right angle, and 3 = | of 
 a right angle, can you form the triedral? 
 Why ? If you keep increasing the size of 1, 
 2, and 3, until the sum becomes equal to 4 
 right angles, will it always be possible to Fig. 26i. 
 form a triedral ? How is it when the sum equals 4 right angles? 
 
 7. What is the locus of a point in space equidistant from 
 three given points ? 
 
 To demonstrate that such a locus is a straight line, pass a plane 
 through the three points, and also a circumference. Now, 1st, a perpen- 
 dicular to this circle at its centre has every point equidistant from the 
 three points; and, 2d, any point out of the perpendicular is unequally 
 distant from the points. Hence this perpendicular is the locus sought. 
 
 Notice that in demonstrating such a proposition the two points should 
 both be proved. 
 
 8. The locus of a point equidistant from two planes is the 
 plane which bisects the diedral included between them. [Give 
 proof.] 
 
 9. What is the locus of a point in space equidistant from the 
 faces of a triedral ? [Give proof.] 
 
 10. If each of the projections of a line upon three intersecting 
 planes is a straight line, the line is a straight line. 
 
 11. To find the point in a plane such that the sum of its dis- 
 tances from two given points without the plane, and on the 
 same side of it, shall be a minimum. 
 
 Solution. — Let the two points be P and P'. Let fall a perpendicular 
 from either point, as P, upon the plane, and call it PD. Produce PD on 
 the opposite side of the plane to P", making P"D = PD. Join P" and P'. 
 The point where P"P' pierces the plane is the point sought. [Give proof.] 
 
PRISMS, 
 
 247 
 
 0irCtlOM m. 
 
 OF PRISMS AND CYLINDERS. 
 
 660. A Prism is a solid, two of whose faces are equal, par- 
 allel polygons, while the other faces are parallelograms. The 
 equal parallel polygons are the Bases, and the parallelograms 
 make up the Lateral or Convex Surface. Prisms are triangular, 
 quadrangular, pentagonal, etc., according to the number of sides 
 of the polygon forming a base. 
 
 561. A Right Prism is a prism whose lateral edges are 
 perpendicular to its bases. An Oblique Prism is a prism 
 whose lateral edges are oblique to its bases. 
 
 562. A Regular Prism is a right prism whose bases are 
 regular polygons ; whence its faces are equal rectangles. 
 
 563. The Altitude of a prism is the perpendicular distance 
 between its bases : the altitude of a right prism is equal to any 
 one of its lateral edges. 
 
 564. A Truncated Prism is a portion of a prism cut off 
 by a plane cutting the 
 lateral edges, but not 
 parallel to its base. A 
 section of a prism made 
 by a plane perpendicu- 
 lar to its lateral edges is 
 called a Right Section. 
 
 Illustrations. — In the 
 figure, (a) and (&) are both 
 prisms : (a) is oblique and 
 (b) right. PO represents 
 the altitude of (a) ; and '"'S- 262. 
 
 any edge of (J), as JB, is its altitude. ABCDEF and abcd^ are lower and 
 
248 
 
 ELEMENT A It Y GEOMETR Y. 
 
 upper bases, respectively. Either portion of (b) cut off by an oblique 
 plane, as a'l'c'd'e', is a truncated prism. 
 
 565. A Parallelepiped is a prism whose bases are paral- 
 lelograms ; its faces, inclusive of the bases, are consequently all 
 parallelograms. If its faces are all rectangular, it is a rectangu- 
 lar parallelopiped. 
 
 566. A Cube is a rectangular parallelopiped whose faces are 
 all equal squares. 
 
 567. The Volume or Contents of a solid is the number 
 of times it contains some other solid taken as the unit of meas- 
 ure ; or it is the ratio of one solid to another taken as the stand- 
 ard of measure. 
 
 In applied geometry the unit of volume is usually a cube de- 
 scribed on some linear unit, as an inch, a foot, a yard, etc. To 
 this the perch and the cord are exceptions. 
 
 PROPOSITION I. 
 
 568. Theorem. — Parallel plane sections of any prism 
 are equal polygons. 
 
 Demoksteation^. 
 Let ABODE and ahcde be parallel sections of the prism MN* 
 
 Then are they equal polygons. 
 
 For, the intersections with the lateral faces, as 
 ab and AB, etc., are parallel, since they are inter- 
 sections of parallel planes by a third plane (488)- 
 
 Moreover, these intersections are equal, that 
 is, db = AB, Ic = BC, cd = CD, etc., since they 
 are parallels included between parallels (138)- 
 
 Again, the corresponding angles of these 
 polygons are equal, that is, a = A, J = B, c = C, 
 etc., since their sides are parallel and lie in the 
 same direction (492)- 
 
 Therefore the polygons ABODE and abcde are 
 mutually equilateral and equiangular ; that is, they are 
 
pJiiS3£S. 249 
 
 569. Corollary. — Any plane section of a prism, paral- 
 lel to its base, is equal to the base ; and all right sections 
 are equal. 
 
 PROPOSITION II. 
 
 570. Theorem. — // two prisms have equivalent bases, 
 any plane sections parallel to the bases are equivalent. 
 
 Demonstration. 
 
 Let M and N be any two prisms having equivalent bases B and B'; 
 and let P and Q be sections parallel thereto. 
 
 Then, by the preceding proposition, 
 
 P = B, 
 and Q = B' = B , 
 
 whence, P = Q. Q. b.d. 
 
 PROPOSITION III. 
 
 571. Theorem. — // three faeces including a triedral 
 of one prism — complete or truncated — are equal respective- 
 ly to three faces including a triedral of the other, and 
 similarly placed, the prisms are equal. 
 
 Demonstration. 
 
 In the prisms ^d and ^'d' (Fig. 264), let ABCDE equal A'B'C'D'E, 
 ^Bba = ^'B'b'a', and BCcb = B'CV6'. 
 
 Then are the prisms equal. 
 
 For, since the facial angles of the triedrals B and B' are equal, the 
 triedrals are equal (544), and being applied they will coincide. 
 
 Now, conceiving ^'d' as applied to A^, with B' in B, since the bases 
 are equal polygons, they will coincide throughout ; and for like reason 
 aB will coincide with a'B', and cB witli c'B'. 
 
250 
 
 ELEMENTARY GEOMETRY, 
 
 Furthermore, since the 
 coincide, CD' falls in CD, and as 
 C'c' falls in Cc, and D'd' is parallel 
 to CV, and Ud to Cc (?), [y'd' falls 
 in Dd. 
 
 In like manner, £'e' can be 
 shown to fall in Ee. 
 
 Finally, since the upper bases 
 have the angles a'l'c' and ahc co- 
 incident, they coincide (444). 
 
 Hence the prisms can be super- 
 imposed, and are therefore equal. 
 
 Fig. 264. 
 
 Q. E. D. 
 
 572. Corollary. — Two right prisms having equal bases 
 and equal altitudes are equal. 
 
 If the faces are not similarly arranged, as the edges are perpendicular 
 to the bases, one prism can be inverted and then superimposed on the 
 other. 
 
 PROPOSITION IV. 
 
 573. Theorem. — Any oblique prism is equivalent to a 
 right prism, whose bases are right sections of the oblique 
 prism, and whose edge is equal to the edge of the oblique 
 prism. 
 
 Demonstration. 
 
 Let LB be an oblique prism, of which abode 
 an6ff//iil are right sections, and gb = GB. 
 
 Then is lb equivalent to LB. 
 
 For the truncated prisms ^G and eB have the 
 faces including any two corresponding triedrals, as 
 G and B, respectively, equal and similarly placed 
 (?), whence these prisms are equal (571)- 
 
 Now, from the whole figure take away prism 
 ZG, and there remains the oblique prism LB ; also, 
 from the whole take away the prism cB, and there 
 remains the right prism lb. 
 
 Therefore, the right prism lb is equivalent to 
 the oblique prism LB. q. e. d. 
 
 Fig. 265. 
 
PRISMS, 
 
 251 
 
 PROPOSITION V. 
 
 574. Theorem. — The opposite faces of a parcdlelopiped 
 are equal and parallel. 
 
 Demonstration. 
 Let Ac be a parallelopiped, AC and ac being its equal bases (560) 
 
 Then are its opposite faces equal and parallel. 
 
 Since the bases are parallelograms, AB is equal 
 and parallel to DC ; and, since the faces are paral- 
 lelograms, ah, is equal and parallel to d[). Hence, 
 
 angle aAB = d[>0, 
 
 and their planes are parallel, since their sides are 
 parallel and extend in the same directions. 
 
 Therefore, aS and dC are equal (322) and parallel parallelograms. 
 In like manner it may be shown that aO is equal and parallel to M). 
 
 Q. E. D. 
 
 Fig. 266. 
 
 PROPOSITION VI 
 
 575. Theorem.— me 
 
 bisect each other. 
 
 diagonals of a parallelopiped 
 
 «. ^ 
 
 AVw--^f 
 
 Demonstration. 
 
 Let ABCD-A be a parallelopiped whose diagonals are frD, f/B, cA, 
 and aO. 
 
 Then do ftD, <iB, cA, and aC bisect each other. 
 
 Pass a plane through two opposite edges, as 
 &B and ^D. 
 
 Since the bases are parallel (?), hd and BD will 
 be parallel (488), and &BD<Z will be a parallelo- 
 gram. Hence, ftD and d^ are bisected at o (?). 
 
 For a like reason, pnssing a plane through dc 
 and AB, we may show that <iB and cA bisect each 
 other, and hence that cA passes through the com- 
 mon centre of c?B and &D. 
 
 So also aO is bisected by 5D, as appears from 
 passing a plane through ab and DC. 
 
 Hence, all the diagonals are bisected at o. q. b. d. 
 
 Fig. 267. 
 
252 ELE3IENTARY GEOMETRY. 
 
 576. Corollary. — The diagonals of a rectangular par- 
 allelopiped are equal. 
 
 PROPOSITION VII. 
 
 577. Theorem.— ITz/e diagonal of a right parallelo- 
 piped is equal to the square root of the sum of the squares 
 of the three adjacent edges of the parallelepiped. 
 
 Demonstration. 
 
 Let a, b, c be the three adjacent edges of a right parallelepiped, d 
 the diagonal of the face whose edges are h and c, and D the diagonal of 
 the parallelopiped. 
 
 Then ^» = &« + C (?), 
 
 and D' = 0? -\-d' = a' + ¥ -vc^ (?), 
 
 or D = ^a? + &' + c^ q. e. d. 
 
 578. Corollary. — The diagonal of a cube is Vs times 
 its edge. 
 
 PROPOSITION VIII. 
 
 579. Theorem.— ^e area of the lateral surface of a 
 right prism is equal to the product of its altitude into the 
 perimeter of its base. 
 
 Demonstration. 
 
 The lateral faces are all rectangles, having for their common altitude 
 the altitude of the prism (563). Whence the area of any face is the 
 product of the altitude into the side of the base which forms its base ; 
 and the sum of the areas of the faces is the common altitude into the sum 
 of the bases of the faces, that is, into the perimeter of the base of the 
 prism. Ci. '^. d. 
 
CYLINDERS. 
 
 253 
 
 580. A Cyliiitlrical Siu'face is a surface traced by a 
 straight line moving so as to remain constantly parallel to its 
 first position, while any point in it traces 
 some curve. The moving line is called the 
 Generate ixy and the curve traced by a point 
 of the line the Directrix. 
 
 Illustration.— Suppose a line to start from 
 the position AB, and move towards N in such a 
 manner as to remain all the time parallel to its 
 first position AB, while A traces the curve 
 
 A123456....M. 
 
 The surface thus traced is a Cylindrical Surface ; 
 AB is the Geneiatrix^ and the curve ANM the 
 Directrix. 
 
 Fig. 268. 
 
 581. A Circular Cylinder, called also a Cylinder of 
 
 Revolution, is a solid generated by the revolution of a rectan- 
 gle around one of its sides as an axis. 
 
 Illustration. — Let COAB be a rectangle, 
 and conceive it revolved about CO as an axis, 
 taking successively the positions COAB', 
 COA'B', etc.; the solid generated is a Circular 
 Oylinder, or a cylinder of revolution. The re- 
 volving side AB is the generatrix of the surface, 
 and the circumference A A' A" (or BB'B") is the 
 directrix. This is the only cylinder treated in 
 Elementary Geometry, and is usually meant 
 when the word Cijlinder is used without specify- 
 ing the kind of cylinder. '"•a- 269. 
 
 682. The Axis of the cylinder is the fixed side of the rectan- 
 gle. The side of the rectangle opposite the axis generates the 
 Convex Surface ; while the other sides of the rectangle, as 
 OA and CB, generate the Bases, which in the cylinder of revo- 
 lution are circles. Any line of the surface corresponding to some 
 position of the generatrix is called an Element of the surface. 
 
 583. Any section of a cylinder of revolution made by a plane 
 parallel to its base is equal to its base, since such a section would 
 be a circle with a radius equal to OA. 
 
254 
 
 ELEMENTARY GEOMETRY, 
 
 584. A Right Cylinder is one whose elements are perpen- 
 dicular to its base. In such a cylinder any element is equal to 
 the axis. A Cylinder of Revolution (581) is right. 
 
 685. A prism is said to be inscribed in a cylinder, when the 
 bases of the prism are inscribed in the bases of the cylinder, and 
 the edges of the prism coincide with elements of the cylinder. 
 
 PROPOSITION IX. 
 
 586. Theorem. — The area of the convex surface of a 
 cylinder of revolution is equal to the product of its axis 
 into the circumference of its base, i. e., 277 RH, H being the 
 axis and R the radius of the base. 
 
 Demonstration. 
 
 Let AD be a cylinder of revolution, whose axis HO = H^ and the 
 radius of whose base is OB = It, 
 
 Then is the area of its convex surface ZnliH. 
 
 Let a right prism, with any regular polygon for 
 its base, be inscribed in the cylinder, as h-ahcdef. 
 
 The area of the lateral surface of the prism is 
 HO (= Kb) into the perimeter of its base, i. e., 
 
 Y\Oy.(ah^'l)c-\-cd + de + ef-\-fa). 
 
 Now, bisect the arcs a5, &c, etc., and inscribe a 
 regular polygon of twice the number of sides of the 
 preceding, and on this polygon as a base construct 
 the right inscribed prism with double the number of 
 faces that the first had. The area of the lateral sur- 
 face of this prism is 
 
 HO X the perimeter of its base. 
 
 In like manner, conceive the operation of inscribing right prisms with 
 regular polygonal bases continually repeated ; it will always be true that 
 the area of the lateral surface is equal to 
 
 HO X the perimeter of the base. 
 
 Fig. 270. 
 
CYLINDERS. 256 
 
 By continually increasing the number of the sides of the inscribed 
 polygon in this manner, the perimeter of the polygon may be made to 
 differ from the circumference by less than any assignable quantity, i. e., 
 by an infinitesimal, which is therefore in comparison with the perimeter 
 (341), and the prism of an infinite number of faces is to be considered as 
 the cylinder. 
 
 Therefore, the area of the convex surface of the cylinder is HO into 
 the circumference of the base. 
 
 Finally, if E is the radius of the base, 27ri2 is its circumference. This 
 multiplied by H, the altitude, i. e., Hx %-R, or 'inRH, is the area of the 
 convex surface of the cylinder, q. e. d. 
 
 PROPOSITION X. 
 
 587. Theorem. — Rectangular parallelopipeds are to 
 eoAih other* as the products of any three adjacent edges. 
 
 Demon^stration. 
 
 Let the adjacent edges of one rectangular parallelopiped, P, be three 
 lines, which we will call A, B, and C^ and of another, Q, the three lines 
 a, bf c. 
 
 Then ^ = ^^^4^- 
 
 Q axoxc 
 
 For Aj jP, C, a, 5, and c are at least commensurable by an infinitesimal 
 unit. Let the common measure of the edges be i ; and let it be contained 
 in A m times, in ^ n times, in C p times, in a ^ times, in 5 r times, and in 
 e 8 times, so that 
 
 ABC 
 
 a ^ J c 
 
 g = T , - r = T , and « = t • 
 
 Now let A and 5 be the sides of the rectangular base of P, and G its 
 altitude, and a, 5, and c corresponding edges of Q. The base of P may be 
 conceived as divided into mn units of surface. If upon each of these we 
 conceive a cube described, there will be mn such cubes. Now, of these 
 layers of cubes there will be p in the entire parallelopiped P. Hence P 
 will be composed of mnp equal cubes. In like manner, Q may be shown 
 
 * This means that their volumes are to each other. 
 
256 ELEMENTARY GEOMETRY, 
 
 to be composed of qr* equal cubes, each equal to one of the mnp cubes 
 which compose P. 
 
 P mnp 
 
 and substituting their values for w, /i, 2*1 ?, ^% and s, we have 
 
 - - X - 
 P t i» AxBxG „ 
 
 ^ = -^ i: = J. Q- E- J^- 
 
 O a b c axo X.C 
 
 V X T X T 
 I I t 
 
 PROPOSITION XI. 
 
 588. Theorem. — The volume of a rectangular paral- 
 lelopiped is equal to the product of its three adjacent 
 edges. 
 
 Demonstration. 
 
 Let P be any rectangular parallelopiped whose adjacent edges are 
 Af Bf and C, and let Q be the proposed unit of measure, whose edges 
 are each I. 
 
 Then, by the last proposition, 
 
 P _ Aj<B^G 
 Q "" 1x1x1 ' 
 
 or, P z= {AxByC)yQ 
 
 Thus, P contains the unit Q AxBx C times. Hence, AxBx G is 
 the volume of P. q. e. d. 
 
 589. Corollary 1. — The volume of a cube is the third 
 power of its edge, 
 
 590. Scholium. — This fact gives rise to the term cube, as used in 
 arithmetic and algebra, for " third power.'* 
 
 591. Corollary 2. — The volum^e of a rectangular par- 
 allelopiped is equal to the product of its altitude into the 
 area of its base, the linear unit being the same for the 
 measure of all its edges. 
 
 * For other demonstrations see Appendix. 
 
CYLINDERS, ^67 
 
 PROPOSITION XII. 
 
 692. Theorem. — Tlie volume of amj prism, or of any 
 solid whose plane sections parallel to the base are all equal 
 to the base, is equal to that of a rectangular pardUelopipecl 
 having an equivalent base and the same altitude, and 
 hence is equal to the product of its base into its altitude. 
 
 Demonstration. 
 
 Let Q be any prism or solid whose plane sections parallel to its base 
 are equal to its base, and P a rectangular parallelepiped of the same 
 altitude, and whose base li = B', the base of the first solid. 
 
 Then is volume Q = volume P. 
 
 If Q be a prism, any plane section parallel to its base is equal to its 
 base (?) ; hence the case is the same whether Q be a prism or any other 
 solid having its plane sections parallel to its base equal to its base. 
 
 Now conceive two planes to start from coincidence with B and B* at 
 the same time, and move upward at the same rate, generating the solids 
 P and Q. As these sections are always equivalent to each other, since 
 each is constantly equal to /? or 5', they generate equal volumes in equal 
 times, and by reason of the equal altitudes of the two solids, both 
 volumes are generated in the same time. Hence the two volumes are 
 equivalent, q. e. d. 
 
 593. Corollary 1. — The volume of a right prism is 
 equal to the product of its edge into its base. 
 
 594. Corollary 2. — Prisms of the same altitude are to 
 each other as their bases ; and prisms of the sam^e or equiv- 
 alent bases are to each other a§ their altitudes ; and, in 
 general, prisrns are to each other as the products of their 
 bases and altitudes. 
 
258 ELEMJEIiTAMY GEOMETRY. 
 
 PROPOSITION XIII. 
 
 696. Theorem. — The volume of a cylinder of revolu- 
 tion is equal to the product of its base and altitude, i. e., 
 ttR^H, H being the altitude and R the radius of the base. 
 
 Demonstration. 
 
 By (592) the volume of sucli a cylinder is equal to the product of its 
 base into its altitude, since all plane sections parallel to its base are equal 
 thereto (583). 
 
 But the base is a circle whose radius is i2, the area of which is ttR^ (?). 
 Hence the volume of the cylinder is ^x TTJS^ or TrijJ^fi; q. e. d. 
 
 596. Corollary. — The volume of any cylinder is equal 
 to the product of its base into its altitude. 
 
 This can be demonstrated in a manner altogether analogous to the 
 case given in the proposition. 
 
 697. Similar Solids are such as have their corresponding 
 solid angles equal and their homologous edges proportional. 
 
 698. Similar Cylinders of revolution are such as have 
 their altitudes in the same ratio as the radii of their bases. 
 
 699. Homologous Edges of similar solids are such as 
 are included between equal plane angles in corresponding faces. 
 
 Illustration. — The idea of similarity in the case of solids is the 
 same as in the case of plane figures, viz., that of likeness of form. Thus, 
 one would not think such a cylinder as one joint of stovepipe similar to 
 another composed of a hundred joints of the same pipe. One would be 
 long and very slim in proportion to its length, while the other would not 
 be thought of as slim. But, if we have two cylinders the radii of whose 
 bases are 2 and 4, and whose lengths are respectively 6 and 13, we readily 
 recognize them as of the same shape : they are similar. 
 
PRISMS AND CYLINDERS. 
 
 PROPOSITION XIV. 
 
 600. Theorem, — The altitudes of two similar prisms 
 are to each other as any two homologous edges, and the 
 areas of corresponding faces are to each other as the 
 squares of any two homologous edges, or as the squares of 
 the altitudes. 
 
 Demonstration. 
 
 Let P and /> be any two similar prisms, 11 and h their altitudes, Ka 
 and k'a' two homologous edges, and A6 and Mb' two corresponding 
 faces. 
 
 Then is 
 
 H Ha 
 
 T- = jTi > or as aTiy other two homologouB edges : 
 
 and 
 
 Aft 
 
 A'6' 
 
 Aa 
 
 H' 
 
 > / ,i= -^ ' *• *M as the squares of any 
 A' a'* h^ n .7 
 
 other two homologous edges, or as the squares of the altitudes. 
 
 From the homologous vertices a and a' let fall the perpendiculars a\ 
 and a' I', and draw Al and AT. 
 
 a\ = H,m(\a'\' = hO). 
 
 Now, since the prisms are 
 similar, they may be so placed 
 that their homologous edges 
 wHl be parallel ; hence, let AB 
 be parallel to A'B', AE to A'E', 
 and ^A to a'A'. Then is al 
 parallel to a'!', and Al to AT, 
 and the triangles aAl and a' AT 
 are similar. 
 
 Whence we have 
 
 H 
 h 
 
 Aa 
 
 k^a" 
 
 Fig. 271. 
 
 or as any other two homolo- 
 gous edges, since by definition 
 any two homologous edges bear the same ratio. Q. e. d. 
 
260 
 
 ELEMENTARY GEOMETRY, 
 
 Again, since the corre- 
 sponding faces Kb and A'^' 
 have their homologous sides 
 proportional (597), and their 
 homologous angles, as aAB and 
 a'A'B', equal, being the ho- 
 mologous facial angles of equal 
 triedrals, the faces are similar 
 plane figures, and 
 
 
 ka 
 
 IP 
 
 or as the squares of any two 
 homologous edges, q. b. d. 
 
 Fig. 271. 
 
 601. Corollary. — Tlie corresponding faces of any two 
 similar solids are to each other as the squares of any two 
 homologous edges of the solids. 
 
 PROPOSITION XV. 
 
 602. Theorem. — The lateral surfaces of similar prisms 
 are to each other as the squares of any two homologous 
 edges, or as the squares of the altitudes of the prisms. 
 
 Demonstration. 
 
 Let Af B, C, I>, etc., and a, b, c, d, etc., be the corresponding 
 faces of two similar prisms, and M and in any two homologous edges, 
 and H and h the altitudes. 
 
 By the last proposition, 
 
 A 
 
 a " 
 
 Jf2 
 
 B M' 
 
 CM' D 
 
 ce, 
 
 
 A_B_G_ 
 a ~ b c 
 
 — , etc. = -—^ , 
 d VI- 
 
 M^ 
 
 etc. 
 
PRISMS AND CYLINDERS. 261 
 
 and, by coiuposition, 
 
 a + 6 + c + rf, etc. ~ m^ ~ h^ ^^' ^•^^^■ 
 
 603. Corollary. — The entire surfaces of any two simi- 
 lar solids are to each other as the squares of any two 
 homologous edges. 
 
 PROPOSITION XVI. 
 
 604. Theorem. — The volumes of similar prisms are to 
 each other as the cubes of their homologous edges, and as 
 the cubes of their altitudes. 
 
 Demonstration. 
 
 Let V and v be the volumes of any two similar prisms, M and m 
 any two homologous edges, and ^ and h their altitudes. 
 
 Then is -=-: = ^'- 
 
 Let B and h be the bases of the prisms ; whence their volumes are 
 5xJy and hxh respectively (592). 
 
 By (600), 
 
 B ^M^^H^ 
 
 b ~ W ~ A^ 
 
 Bat f=-=f(')- 
 
 h m h ^ ^ 
 
 Multiplying, -r-T- ^ — - -^ ^ Ti' Q. E. D. 
 
262 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XVII. 
 
 605. Theorem. — The convex surfaces of similar cylin- 
 ders of revolution are to each other as the squares of their 
 altitudes, and as the squares of the radii of their bases. 
 
 Demonstration. 
 
 Let H^rxA h be the altitudes, and R and r the radii of the bases of 
 two similar cylinders. 
 
 The convex surfaces are 27ri2^and ^tttK (586). 
 
 T^ 2nRH RH R H 
 
 •Now, — — - = — - = - X ^ . 
 
 2irrA rh t h 
 
 By hypothesis, — = — • 
 
 Whence, by substitution, we have 
 
 27rrh ~ h? ' 
 
 , 2itRh m 
 
 and — — ~ - —• Q. E. D. 
 
 2nrh T^ 
 
 PROPOSITION XVIII. 
 
 606. Theorem. — The volumes of similar cylinders of 
 revolution are to each other as the cubes of their altitudes, 
 or as the cubes of the radii of their bases. 
 
 Demonstration. 
 
 Let H and fi be the altitudes of two similar cylinders of revolution, 
 M and r the radii of their bases, and V and v their volumes. 
 
 V H^ R^ 
 
PRISMS AND CYLINDERS 263 
 
 For, by (595), V = itHR', 
 
 and V = nhr^. 
 
 Hence, — = — ir-r — ^tt 
 
 and, since 
 
 - = -(?) 
 
 we have, by substitution, — = — = -—. q. e. D. 
 
 607. Scholium.— It is a general truth, that the surfaces of similar 
 solids, of any form^ are to each other as the squares of homologous lines; 
 and their volumes are as the cubes of such lines. These truths will be 
 further illustrated in the following section, but the methods of demon- 
 stration will be seen to be the same as used in this. 
 
 EXERCISES. 
 
 1. A farmer has two grain bins which are parallelo- 
 pipeds. The front of one bin is a rectangle 6 feet long by 4 high, 
 and the front of the other a rectangle 8 feet long by 4 high. 
 They are built between parallel walls 5 feet apart. The bottom 
 and ends of the first, he says, are " square " (he means,* it is a 
 rectangular parallelopiped), while the bottom and ends of the 
 other slope, i. e., are oblique to the front. What are the rela- 
 tive capacities of the bins ? 
 
 2. How many square feet of boards in the walls and bottom 
 of the first bin mentioned in Ex. 1 ? 
 
 3. An average sized honey bee's cell is a right hexagonal 
 prism, .8 of an inch long, with faces ^ of an inch wide. The 
 width of the face is always the same, but the length of the cell 
 varies according to the space the bee has to fill. Are honey bees* 
 cells similar ? Is a honey bee's cell, of the dimensions given 
 above, similar to a wasp's cell, which is 1.6 inches long, and 
 whose face is .3 of an incli wide? What are the relative capaci- 
 ties of the wasp's cell and the honey bee's? 
 
264 ELEMENTARY GEOMETRY. 
 
 4. How many square inches of sheet iron docs it take to make 
 a joint of 7-inch stovepipe 2 feet 4 inches long, allowing an inch 
 and a half for making the seam ? 
 
 5. A certain water-pipe is 3 inches in diameter. How much 
 water is discharged through it in 24 hours, if the current flows 
 3 feet per minute ? How much through a pipe of twice as great 
 diameter, at the same rate of flow ? 
 
 6. What is the ratio of the length of a hogshead holding 125 
 gallons, to the length of a keg of the same shape, holding 
 8 gallons ? 
 
 7. What are the relative amounts of cloth required to clothe 
 three men of the same form (similar solids), one being 5 feet 
 high, another 5 feet 9 inches, and tlic other 6 feet, provided they 
 dress in the same style? If the second of these men weighs 
 156 lb., what do the others weigh ? 
 
 8. If a man 5 J feet high weighs 160 lb., and a man 3 inches 
 taller weighs 180 lb., which is the stouter in proportion to his 
 height ? 
 
 9. I have a prismatic piece of timber, from which I cut two 
 blocks, both 5 feet long measured along one edge of the stick ; 
 but one block is made by cutting the stick square across (a right 
 section), and the other by cutting both ends of it obliquely, 
 making an angle of 45° with the same face of the timber. Which 
 block is the greater ? Which has the greater lateral surface ? 
 
 10. How many cubic feet in a log 12 feet long and 2 feet and 
 
 5 inches in diameter ? How many square feet of inch boards 
 can be cut from such a log, allowing one-quarter for waste in 
 slabs and sawing ? 
 
 11. How many square feet of sheet copper will it take to line 
 the sides and bottom of a cylindrical vat (cylinder of revolution) 
 
 6 feet deep, if the diameter of the bottom is 4 feet? How many 
 barrels does such a vat contain ? 
 
 12. What are the relative capacities < - of rovolntion 
 of the same diameter, but of different len.. \'5.. < of those 
 of the same length, but of different diamuttrs 
 
PYRA 
 
 W CONES, 
 
 265 
 
 ^irCTiOH lY, 
 
 OF PYRAMIDS AND CONES. 
 
 609. A Pyramid is a solid having a polygon for its base, 
 and triangles for its lateral faces. If the base is also a triangle, 
 it is called a triangular pyramid, or a tetraedron (i. e., a solid with 
 four faces). The vertex of tlie polyedral angle formed by the 
 lateral faces is the Vertex of the pyramid. 
 
 610. The Altitude of a pyramid is the perpendicular dis- 
 tance from its vertex to the plane of its base. 
 
 611. A Right Pyramid is a pyramid whose base is a regu- 
 lar polygon, and the perpendicular from whose vertex falls at 
 the centre of the base. This perpendicular is called the axis, 
 
 612. A Frustum of a pyramid is a portion of the pyramid 
 intercepted between the base and a plane parallel to the base. 
 If the cutting plane is not parallel to the base, the portion inter- 
 cepted is called a Truncated pyramid. 
 
 613. The Slaut Height of a right pyramid is the altitude 
 of one of the triangles which form its faces. The Slant Height 
 of a Frustum of a right pyramid is the portion of the slant height 
 of the pyramid intercepted between the bases of the frustum. 
 
 Fig. 272. 
 
 Illustrations.— The student will be able to find illustrations of the 
 definitions in the above figures. 
 
266 ELEMENTARY GEOMETRY. 
 
 614. A Conical Surface is a surface traced by a line 
 which passes througli a fixed point, while any other point traces 
 a curve. The line is the Generatrix, and the curve the Direc- 
 trix. The fixed point is the Vertex. Any line of the surface 
 corresponding to some position of the generatrix is called an 
 Element of the surface. 
 
 615. A Cone of Revolution is a solid generated by the 
 revolution of a right-angled triangle around one of its sides, 
 called the Axis. The hypotenuse describes the Convex Surface 
 of the cone, and corresponds to the generatrix in the preceding 
 definition. The other side of tlie triangle describes the Base. 
 This cone is right, since the perpendicular (the axis) falls at the 
 centre of the base. The Slayit Height is the distance from the 
 vertex to tlie circumference of the base, and is the same as the 
 hypotenuse of the generating triangle. 
 
 616. The terms Frustum and Truncated are applied to 
 the cone in the same manner as to the pyramid. 
 
 617. A pyramid is said to be Inscribed in a cone when the 
 base of the pyramid is inscribed in the base of the cone, and the 
 edges of the pyramid are elements of the surface of the cone. 
 The two solids have a common vertex and a common altitude. 
 
 618. If the generatrix be considered as an indefinite straight 
 line passing through a fixed point, the portions of the line on 
 opposite sides of the point will each describe a conical surface. 
 These two surfaces, which in general discussions are considered 
 but one, are called Nappes. The two nappes of the same cone 
 are evidently alike. 
 
 Illustration. —In Fig, 273, (a) represents a conical surface which 
 has the curve ABC for its directrix^ and SA for its generatrix. The nu- 
 merals indicate the successive positions of the point A, as it passes around 
 the curve, while the point S remains fixed, (b) represents a Gorie of Rev- 
 olution^ or a right cone with a circular base. It may be considered as 
 generated in the general way, or by the right-angled triangle SOA revolv- 
 ing about SO as an axis. SA describes the convex surface, and OA the 
 
PYRAMIDS AND CONES. 
 
 267 
 
 Fig. 273. 
 
 base. The figure (c) represents the Frustum of a cone, the portion above 
 the plane abc being supposed removed. Figure {d) represents the two 
 Nappes of an oblique cone. 
 
 PROPOSITION I. 
 
 619. Theorem. — Any section of a pyramid made by a 
 plane parallel to its base is a polygon similar to the base. 
 
 Demonstratiox. 
 
 Let ahcde be a section of the pyramid 
 S-ABCDE made by a plane parallel to ABODE. 
 
 Then is dbaJe similar to ABODE. 
 
 Since AB and ab are intersections of two 
 parallel planes by a third plane, they are paral- 
 lel (0. So also Ic is parallel to BO, cd to OD, 
 etc. Hence, angle 5 = B, c = 0, etc. (?), and the 
 polygons are mutually equiangular. Again, 
 
 ab Sb , be Sb 
 
 AB-SB- ""^ Be = is<^)- 
 
 Fig. 274. 
 
 Hence 
 
 (?)• 
 
 c^ _ bc^ a5 _ AB 
 
 AB ~ BO ' ^^ 6c ~ BC 
 In like manner, we can show that 
 
 bc_ 
 
 ed~ CD' 
 
 Therefore, dhcde and ABODE are mutually equiangular, and have 
 tlu'ir corresponding *tles proportional, and are consequently similar. 
 
 Q. E. D. 
 
 etc. 
 
268 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION II. 
 
 620. Theorem. — // two pyramids of equal altitudes 
 are cab by planes equally distant from and parallel to 
 their bases, the sections are to each other as the bases. 
 
 Demonstration^. ^ 
 
 Let S-ABC and S'-A'B'C'D'E' be two pyramids of the same altitude, 
 cut by the planes abc and a'b'dd'e' , parallel to and at equal distances 
 from their bases. 
 
 Then is 
 
 abc 
 
 ABC 
 
 < 
 
 a'h'c'd'i 
 
 ABODE' 
 
 For, conceive the bases in the 
 same plane. Let SP and S'P' be the 
 equal altitudes, and Sp = S'p' the 
 distances of the cutting planes from 
 the vertices. 
 
 Conceive a plane passing through 
 
 the vertices parallel to the plane of 
 
 the bases. This plane, together with 
 
 the plane in which the sections lie, 
 
 and that in which the bases lie, make 
 
 three parallel planes which cut the 
 
 lines SA, SB, S'A', S'B', SP, and S'P', 
 
 whence 
 
 SB SP _ S^B 
 
 S6' " Sp~ S'V 
 
 g. 275. 
 
 ST' 
 
 S'p' 
 
 Also, since the planes ASB and A'S'B' are cut by parallel planes in 
 AB, (il, A'B', and a'V, ab is parallel to AB, and a'h' to A'B' ; whence. 
 
 Now 
 
 and 
 
 AB SB 
 
 ab ~ S6 ' 
 
 A'B' SB' 
 
 ABC 
 
 abc ~ 
 
 ^' ^'^ - S^'' ^ ^' 
 
 A'B'C'D'E 
 
 a'h'c'd'e' 
 
 :' AM3'' s^' 
 
PYRAMIDS AND CONES. 
 Hence, by equality of ratios, 
 
 269 
 
 ABC A'B'C'D'E' ahc 
 
 — r- = — ,., ,,, -r, or 
 
 aJbc ah'cde 
 
 ABC 
 
 a'h'c'd'ef A'B'C'D'E' 
 
 (?). Q. E. D. 
 
 621. Corollary. — // two pTjramids having equivalent 
 bases and equal altitudes are cut by planes parallel to and 
 equidistant from their bases, the sections are equivalent. 
 
 PROPOSITION III. 
 
 622. Theorem. — The area of the lateral surface of a 
 right pyramid is equal to the perimeter of the base multi- 
 plied bij one-half the slant height. 
 
 Demonstration. 
 
 The faces of such a pyramid are equal isosceles tri- 
 angles (?), whose common altitude is the slant height of 
 the pyramid (?). 
 
 Hence, the area of these triangles is the product of 
 one-half the slant height into the sum of their bases. But 
 this sum is the perimeter of the base. 
 
 Hence the area is equal to the perimeter of the base 
 multiplied by one-half the slant height, q. e. d. 
 
 Fig. 276. 
 
 623. Corollary. — The area of the lateral 
 surface of the frustum of a right pyrajnid is 
 equal to the product of its slant height into 
 half the sum, of the perimeters of its bases. 
 
 The proof is based upon (350) and definitions. 
 
 /<w^ 
 
 Fig. 277. 
 
270 ELEMEi^TARY GEOMETRY, 
 
 PROPOSITION IV. 
 
 624. Theorem. — The area of the convex surface of a 
 cone of revolution (a j^ight cone with a circular base) is 
 equal to the product of the circumference of its base and 
 one-half its slant height, i. e., -nRH', R being the radius of 
 the base, and H' the slant height. 
 
 Demonstration^. 
 
 In the circle which forms the base of the cone, 
 conceive a regular polygon inscribed, as abcdef. Join- 
 ing the vertices of the angles of this polygon with the 
 vertex of the cone, there will be constructed a right 
 pyramid inscribed in the cone. Now, if the arcs sub- 
 tended by the sides of this polygon be bisected, and 
 these are again bisected, etc., and at every step a right 
 pyramid is conceived as inscribed, it will always remain 
 true that the lateral surface of the pyramid is the pe- 
 rimeter of its base into half its slant height. 
 
 But, as the number of faces of the pyramid is in- '^' 
 
 creased, the perimeter of the base approaches the circumference of the 
 base of the cone as its limit, and hence the slant height of the pyramid 
 approaches the slant height of the cone, and the lateral surface of the 
 pyramid approaches the convex surface of the cone as their limits, and all 
 reach their limits simultaneously. 
 
 Therefore, at the limit we still have the same expression for the area 
 of the convex surface, that is, the circumference of the base multiplied 
 by half the slant height. 
 
 Finally, if E is the radius of the base, its circumference is 2;r^, and 
 if being the slant height, we have for the area of the convex surface 
 2jR X ^H\ or ttBH'. q. e. d. 
 
 625. Corollary 1. — The area of the convex surface of a 
 cone is also equal to the product of the slant height into 
 the circumference of the circle parallel to tJie base, and 
 midway between the base and vertex. 
 
 This follows directly from the fact that the radius of the circle mid- 
 way between the base and vertex is one-half the radius of the base, i. e., 
 ^B (?), whence its circumference is tt/?. Now, tvRxII' is the area of the 
 convex surface, by the proposition. 
 
^IDS AND CONES. 
 
 271 
 
 LLAEY '^. — rhe area of the convej&surfaceof 
 the fru of a cone is equal to the product of its slant 
 
 licight r%to half the sum of the circumferences of its 
 bases; i e., T{R-\-r)jr. R and r being the radii of its 
 bases, and M' its slant height. 
 
 From the corresponding property of the frustum of a pyramid, the 
 student will be able to deduce the fact that ^ {2nB-\-%Tn') H' or tt {H-\-r) IT 
 is the area of this surface by the same line of argument used in the 
 demonstration of the main theorem. 
 
 627. Corollary 3. — The area of the convex surface of 
 the frustum of a cone is equal to the product of its slant 
 lieight into the circumference of the circle midway between 
 the bases. 
 
 The radius of the circle midway between the bases is ^ (/•+/?), whence 
 its circumference is 7r(r + i2). Now, Tr{r-\-R)xH' is the area of the con- 
 vex surface of the frustum, by the preceding corollary. 
 
 PROPOSITION V. 
 
 628. Theorem. — Two pyram^ids having equivalent 
 bases and the same altitudes are equivalent, i. e., equal in 
 volume. 
 
 First Demonstration. 
 
 Let S-ABCD and S'-A'B'C'D'E' be two pyramids having the same 
 altitudes, and base ABCD equivalent to base ABODE', /. e,, equal in 
 area. 
 
 Then is pyramid S-ABCD 
 equivalent to S'-A'B'C'D'E', L ^., 
 equal in volume. 
 
 For, conceive the bases to be 
 in the same plane, and a plane to 
 start from coincidence with the 
 plane of the bases, and move to- 
 ward the vertices, remaining all 
 the time parallel to the bases. 
 
 Fig. 279. 
 
2^2 ELEMENTARY GEOMETRY. 
 
 Now each of the sections of the pyramids made by this plane may be 
 conceived as a varying polygon which generates its respective pyramid. 
 And as these polygons are always equivalent, and move at tbe same rate, 
 they generate equal volumes in equal times. Moreover, as the bases of 
 the pyramids are in the same plane, and their altitudes are equal, the 
 polygons generate their respective pyramids in the same time. Hence 
 these volumes are equal, q. e. d. 
 
 Second Demonstkation. 
 
 Consider the pyramids divided into an infinite number of laminae of 
 equal but infinitesimal thickness, as mc, m'c\ parallel to the bases. Now 
 each lamina in one will have a corresponding lamina in the other at the 
 same distance from the base since the laminae are of equal thickness, and 
 hence equivalent to it. 
 
 Hence the pyramids are composed of an equal number of equivalent 
 laminae, and are consequently equivalent, q. e. d. 
 
 PROPOSITION VI. 
 
 629. Theorem. — The volume of a triangular pyramid 
 is equal to one-third the product of its base and altitude, " 
 
 >i 
 
 Demonstration^. 
 
 Let S-ABC be a triangular pyramid, whose altitude is H, 
 
 Then is the volume equal to 
 ^H X area ABC. 
 
 For, through A and B draw Aa and B& 
 parallel to SC; and through S draw Sa 
 and S& parallel to CA and CB, and join a 
 and J); then Sa6-ABC is a prism with its 
 base equal to the base of the pyramid. 
 
 Now, the solid added to the given pyr- 
 amid is a quadrangular pyramid with abB^ 
 as its base, and its vertex at S. 
 
 Divide this into two triangular pyra- '^'9- 280. 
 
 mids by drawing aB and passing a plane through SB and aE. These tri- 
 angular pyramids are equivalent, since they have equal bases, aAB and 
 abS, and a common altitude, the vertices of both being at S. 
 
PYRAMIDS AND CONES. 
 
 273 
 
 Again, S-«JB may be considered as having abS (equal to ABC) as its 
 base, and the altitude of the given pyramid (equal to the altitude of the 
 prism) for its altitude, and hence as equivalent to the given pyramid. 
 
 Thus the prism KBCdbS is divided into the three equivalent pyramidsi, 
 S-ABC, B-a&S, and S-«BA. 
 
 Hence, the pyramid S-ABC is one-third the prism Sa5-ABC, which 
 has the same base and altitude. 
 
 But the volume of the prism is 
 
 H X area ABC. 
 Therefore the volume of the pyramid S-ABC is 
 
 ^H X area ABC. Q. e. d. 
 
 630. Corollary 1. — The volume of any 
 pyramid is equal to one-third the product 
 of its base and altitude. 
 
 Since any pyramid can be divided into triangular 
 pyramids by passing planes through any one edge, as 
 SE, and each of the other edges not adjacent, as SB 
 and SC, the volume of the pyramid is equal to the sum 
 of the volumes of several triangular pyramids having 
 the same altitude as the given pyramid, and the sum 
 of whose bases is the base of the given pyramid. 
 
 Fig. 281. 
 
 631. CoROLLART 2. — Pyramids having equivalent bases 
 are to each other as their altitudes ; such as have equal 
 altitudes are to each other as their bases ; and^ in general, 
 pyramids are to each other as the products of their bases 
 and altitudes. 
 
 Exercise.— A Regular Tetraedron is a triangular pyr- 
 amid whose base is an equilateral triangle and each of whose 
 lateral faces are equal to the base. What is the volume of such 
 a tetraedron whose edge is 1 inch ? Ans. iVV ^ cu- in. 
 
 What is the entire area of the surface of this tetraedron ? 
 
274 
 
 ELEMENTARY GEOMETRY, 
 
 PROPOSITION VII. 
 
 632. Theorem, — The volume of the frustum of a tri- 
 angular pyramid is equal to the volume of three -pyramids 
 of the same altitude as the frustum, and whose bases are 
 the upper base, the lower base, and a mean proportional 
 between the two bases of the frustum. 
 
 Demonstration. 
 Let M&c-ABC (Fig. 282) be the frustum of a triangular pyramid. 
 
 Through db and C pass a plane cutting off the pyramid 0-abc. This 
 has for its base the upper base of the frustum, and for its altitude the 
 altitude of the frustum. 
 
 Again, draw Aft, and pass a plane through 
 Aft and ftC, cutting off the pyramid ft-ABC, 
 which has the same altitude as the frustum, 
 and for its base the lower base of the frustum. 
 
 There now remains a third pyramid, ft-AC«, 
 to be examined. 
 
 Through ft draw ftD parallel to «A, and 
 draw DC and «D. 
 
 The pyramid D-ACa is equivalent to 
 ft-ACrt, since it has the same base and the same 
 altitude (?). But the former may be considered 
 as having ADC for its base, and the altitude of the frustum for its altitude, 
 i. e., as pyramid «-ADC. 
 
 We are now to show that ADC is a mean proportional between ahc 
 and ABC. 
 
 Fig. 282, 
 
 ABC 
 
 abc 
 
 AEr 
 
 ^1 
 
 ad" 
 
 (?). 
 
 Also, J— — 
 
 ABC 
 ADC 
 
 AB 
 AD 
 
 (?); or 
 
 abc;; ^ AT 
 
 ADC' ad' 
 
 (?). 
 
 Tj ,., „ ,. ABC ABC 
 
 By equality of ratios, — j— = 5 ; 
 
 '-''- ADC' 
 
 abc 
 
 whence, ADC" = aftcxABC 
 
 i.e.^ ADC is a mean proportional between the upper and lower bases of 
 the frustum. 
 
 Hence the volume of the frustum is equal to the volume of three 
 pyramids, etc. q. e. d. 
 
PYRAMIDS AND CONES. 275 
 
 633. Corollary. — The volume of the frustum of any 
 pyramid is equal to the volume of three pyramids hav- 
 ing the same altitude as the frustum, and for bases, the 
 upper base, the lower base, and a mean proportional 
 between the two bases of the frustum. 
 
 For, the frustum of any pyramid is equivalent to the corresponding 
 frustum of a triangular pyramid of the same altitude and an equivalent 
 base (?) ; and the bases of the frustum of the triangular pyramid being 
 both equivalent to the corresponding bases of the given frustum, a mean 
 proportional between the triangular bases is a mean proportional between 
 their equivalents. 
 
 PROPOSITION VIII. 
 
 634. Theorem. — The volum^e of a cone of revolution is 
 equal to one-third the product of its base and altitude ; 
 i. e. , ^-n R^H, R being the radius of the base and H the alti- 
 tude. 
 
 Demonstration. 
 
 The volume of a pyramid is equal to one-third the product of the 
 base and altitude, and the cone being the limit of the pyramid, the vol- 
 ume of the cone is one-third the product of its base and altitude. 
 
 Now, R being the radius of the base of a cone of revolution, the 
 base (area of) is tt^, whence ^Tr^-fiTis the volume, -S" being the altitude. 
 
 Q. E. D. 
 
 636. Corollary 1. — The volum^e of any cone is equal 
 to one-third the product of its base and altitude. 
 
 636. Corollary 2. — The volume of the frustum of a 
 cone is equal to the volum^e of three cones having the 
 same altitude as the frustum,, and for bases, the upper 
 base, the lower base, and a mean proportional between 
 the two bases of the frustum. 
 
 The truth of this appears from the fact that the frustum of a cone is 
 the limit of the frustum of a regular inscribed pyramid. 
 
276 ELEMEJSTARY GEOMETRY. 
 
 PROPOSITION IX. 
 
 637. Theorem. — The lateral surfaces of similar pyra- 
 mids are to each other as the squares of their homologous 
 edges, or of their altitudes. 
 
 Demokstration. 
 
 Let A, A', A", etc., and a, a\ a", etc., be homologous sides of the 
 bases of two similar pyramids, E, E', E", etc., and e, e', e", etc., 
 homologous lateral edges, H and h the altitudes of the pyramids, and 
 let S and &* be the lateral surfaces. 
 
 C A2 A/2 Kll-l 
 
 _ ^ _ ^' _ ^' f 
 
 ~ e" ~ e" ~ e'"" ' ' 
 
 Since the pyramids are similar, the corresponding facial angles are 
 equal, and the homologous edges proportional (597, 532), hence the 
 bases are similar polygons, and the corresponding lateral faces are simi- 
 lar triangles. 
 
 Now let F, F', F", etc., and/,/',/", etc., be the corresponding lateral 
 faces, of which triangles. A, A', A", etc., and «, a', a'\ etc., are the bases 
 respectively, and E, E', E", etc., and e, e', e", etc., other homologous sides. 
 
 A A' A" F F' F" U 
 
 ™™ * = * = *" • ^'<=- = J = 7 = ?-' '^*<=-' = " (')■ 
 
 A* A'"^ A"^ E^ E" E'" H^ 
 
 ^'^'^"'^^ a'=^ = ^' "^^ = ¥ = ^ = 7^' «'<=•' = F (^> 
 
 F A^ F' A'2 F" A"' 
 Moreover, -^=^, ^ = - _ = _, , etc. 
 
 Whence ^ + ^' + ^"^ ^'^- _ « _ A^ _ A'« _ A'"^ 
 
 "^ i^ "^ ^ "^ e^ ' ®*^-' "" hF' ^^^'^' 
 
 638. Corollary. — The lateral surfaces of similar right 
 pyramids are to each other as the squares of any homolo- 
 gous lines, as slant heights, altitudes, or of correspond- 
 ing diagonals of the bases.. 
 
PYRAMIDS AND CONES. 277 
 
 PROPOSITION X. 
 
 639. Theorem. — The convex surfaces of similar 
 cones of revolution are to each other as the squares of 
 their slant heights, the radii of their bases, or their 
 altitudes; i. e., as the squares of any two homologous 
 dimensions. 
 
 Demonstration. 
 
 Let H' and h' be the slant heights of two similar cones of revolution, 
 R and r the radii of their bases, and H and h their altitudes. 
 
 Their convex surfaces are ttRH' and TrrA'. 
 
 Now, since the cones are similar, 
 
 r h> ^' '' 
 
 Multiplying the terms of this proportion by the corresponding 
 terms of 
 
 vH' _ E' 
 
 ttA' ~ h' ' 
 
 . irRH' H'^ 
 
 we have — ^7- = -^7^ • 
 
 Hence the convex surfaces are as the squares of their slant heights. 
 
 Q. E. D. 
 
 But, as _=_(?) = _, 
 
 nRH^ _^ ^H* 
 
 Trrh! ~ r^ ~ h* 
 
 That is, the convex surfaces are to each other as the squares of the 
 radii of the bases, or as the squares of the altitudes, q. e. d. 
 
SJ78 ELEMENTARY GEOMETRY, 
 
 PROPOSITION XI. 
 
 640. Theorem, — The volumes of similar pyramids 
 are to each other as the cubes of their homologous dimen- 
 sions. 
 
 Synopsis of Demonstration. 
 
 Let A and a be homologous sides of the bases of two similar pyra- 
 mids, B and b their bases, and J^and h their altitudes. 
 
 We have 
 
 B 
 
 b 
 
 
 = f«' 
 
 
 
 
 W 
 
 P 
 
 _A 
 a 
 
 = > 
 
 
 
 
 \BxH 
 
 A^ 
 ~ a? 
 
 = f«- 
 
 Q. 
 
 B. 
 
 D. 
 
 PROPOSITION XII. 
 
 641. Theorem. — The volumes of similar cones are to 
 each other as the cubes of their altitudes, or as the cubes 
 of the radii of their bases. 
 
 Synopsis of Demonstration. 
 Let J? and r be the radii of their bases, and ^Tand h their altitudes. 
 
 We have ^ ^ 'M^'^' 
 
 •• ^Trr^xh ~ h^^'' 
 
 B? 
 or =^- QE.D. 
 
REGULAR POLYHDRONS. 279 
 
 OF THE REGULAR POLYEDRONS. 
 
 642. A Polyedron is a solid bounded by plane surfaces. 
 A Regular Convex Polyedron is a polyedron whose faces are all 
 equal regular polygons, and each of whose solid angles is convex 
 outward, and is enclosed by the same number effaces. 
 
 PROPOSITION XIII. 
 
 643. Theorem.— There are five and only five regular 
 
 convex polyedrons, viz. : 
 
 The TsTRAEDRONf whose faces are four equal equilat- 
 eral triangles ; 
 
 The Hexaedrox, or Cube, whose faces are six equal 
 squares ; 
 
 The OcTAEDRON, whose faces are eight equal equilateval 
 triangles ; 
 
 The DoDECAEDRON, whose faces are twelve equal regular 
 pentagons; and 
 
 The IcosAEDRON, whose faces are twenty equal equilat- 
 eral triangles. 
 
 Demonstration. 
 
 We demonstrate" this proposition by showing — Ist, that such solids 
 can be constructed ; and 2d, that no others are possible. 
 
 The Regular Tetraedron.— Taking three equilat- 
 eral triangles, as ASB, ASC, and BSC, it is possible to 
 enclose a solid angle, as S, with them, since the sum 
 of the three facial angles is (what ?) (555). 
 
 Then, since AC = AS = CB (?), considering ACB 
 the fourth face, we have a regular polyedron whose 
 four faces are equilateral triangles. ^'9- ^^^- 
 
 N 
 
1^80 
 
 ELEMENTARY GEOMETRY, 
 
 The Regular Hexaedron op Cube.— This is a familiar solid, but for 
 purposes of uniformity and completeness we may conceive it constructefi 
 as follows: Taking three equal squares, as ASCB, 
 CSED, and ASEF, we can enclose a solid angle, as S, 
 with them (?). 
 
 Now, conceive the planes of CB and CD, AB and 
 AF, EF and ED produced. The plane of CB and CD 
 being parallel to ASEF (?), will intersect the plane of 
 EF and ED in HD parallel to FE (?). In like manner, 
 FH can be shown parallel to ED, BH to CD, and HD to 
 BC. Hence the solid has for its faces six equal squares. ^^^' ^^ 
 
 The Regular Octaedron. — At the intersection, p, 
 of the diagonals of a square, ABCD, erect a perpendic- 
 ular SP to the plane of the square, and making SP = 
 AP (half of one of the diagonals) draw SA, SD, SC, 
 and SB. 
 
 Making a similar construction on the other side of 
 the plane ABCD, we have a solid having for faces eight 
 equal equilateral triangles (?). 
 
 Fig. 185. 
 
 The Regular Dodecaedron: — Taking twelve equal regular pentagons, 
 we may group them in two sets of six each, as in the figure. Thus, 
 around we may place five, forming five triedrals at the vertices of 0. 
 These triedrals are possible, 
 since the sum of the facial 
 angles enclosing each is 3f 
 righ! angles (?) — i.e. , between 
 and 4 right angles (555). 
 
 In like manner, the other 
 six may be grouped by 
 placing five of them about 
 0'. 
 
 XT ' ' ^x. '''9. 286. 
 
 Now, conceiving the con- 
 vexity of the group in front and the concavity o^ grovip 0', we may place 
 the two together so as to inclose a solid. Thus, placing A at &, the three 
 faces 5, 7, 1, will inclose a triedral, since the diedral included by 5 and 1 
 is the diedral of such a triedral. Then will vertex B fall at c, and a like 
 triedral will be formed at that point, and so of all the other vertices. 
 Hence we have a polyedron having for faces twelve equal regular penta- 
 gons. 
 
REG ULAR P OL YEDR ONS. 
 
 281 
 
 Fig. 287. 
 
 The Regular Icosaedron.— Taking twenty equal equilateral triangles, 
 they can be grouped in two sets, as in the figure, in a manner altogether 
 similar to the preceding case. 
 The solid angles in this case are 
 included by five facial angles 
 whose sum is 3^ right angles 
 (:•), which is a possible case 
 (555). As before, conceiving 
 the convexity of group in 
 front, and the concavity of 0', 
 we can place them together by 
 placing A at «, thus enclosing 
 a solid angle with five faces, whence B will fall at J, etc. Thus we obtain 
 a solid with twenty equal equilateral triangles for its faces. 
 
 That there can be no other regular polyedrons than these five is evi- 
 dent, since we can form no other convex solid angles by means of regular 
 polygons. Thus, with equilateral triangles (the simplest polygon) we 
 have formed solid angles with three faces (the least number possible), as 
 in the tetraedron ; with four, as in the octaedron ; and with five, as in the 
 icosaedron. Six such facial angles cannot enclose a solfd angle, since 
 their sum is four right angles (?), and much less can any greater number. 
 Again, with squares (the next most simple polygon) we have formed 
 solid angles with three faces, as in the hexaedron, and can form no other, 
 for the same reason as above. With regular i)entagons we can enclose 
 only a triedral, as in the dodecaedron, for a like reason. With regular 
 hexagons we cannot enclose a solid angle (?), and much less with any 
 regular polygon of more than six sides. 
 
 Fig. 288. 
 
282 ELEMENTARY GEOMETRY. 
 
 644. Scholium. — Models of the regular polyedrons are easily formed 
 by cutting the preceding figures from cardboard, cutting half-way through 
 the board in the dotted lines, and bringing the edges together as the 
 forms will readily suggest. 
 
 PROPOSITION XIV. 
 
 645. Theorem. — Any regular polyedron is inscriptihle 
 and circumscriptible by a sphere. 
 
 Outline of Demonstration. 
 
 From the centres of any two adjacent faces, as c and 
 c', let fall perpendiculars upon the common edge, and 
 they will meet it in the same point o (?). The plane of 
 these lines will be perpendicular to this edge (?), and 
 perpendiculars to these faces from their centres, as cS, 
 c'S, will lie in this plane (?), and hence will intersect at 
 a point equally distant from these faces (?). 
 
 In like manner c"S = c'S, and the point S can be ^'^" ^^^' 
 
 shown to be equally distant from all of the faces, and is therefore the cen- 
 tre of the inscribed sphere. 
 
 Joining S with the vertices, we can readily show that S is also the 
 centre of the circumscribed sphere. 
 
 EXERCISES. 
 
 646. 1. What is the area of the lateral surface of a right 
 hexagonal pyramid whose base is inscribed in a circle whose 
 diameter is 20 feet, the altitude of the pyramid being 8 feet ? 
 What is the volume of this pyramid ? 
 
 2. What is the area of the lateral surface of a right pentago- 
 nal pyramid whose base is inscribed in a circle whose radius is 
 G yards, the slant height of the pyramid being 10 yards? What 
 is the volume of this pyramid ? 
 
EXERCISES. 383 
 
 3. How many quarts will a can contain, whose entire height 
 is 10 inches, the body being a cylinder 6 inches in diameter and 
 Q^ inches high, and the top a cone? How much tin does it take 
 to make such a can, allowing nothing for waste and the seams ? 
 
 4. If very fine dry sand is piled upon a smooth horizontal 
 surface, without any lateral support, the angle of slope (i. e., the 
 angle of inclination of the sloping side of the pile with the plane) 
 is about 31°. Suppose two circles be drawn on the floor, one 
 4 feet in diameter and the other 3, and sand piles be made as 
 large as possible on these circles as bases, no other support being 
 giyen. What is the relative magnitude of the piles ? 
 
 5. In the case of sand piles, as given in the last example, the 
 ratio of the radius of the base to the altitude of the pile is |. 
 How many cubic feet in each of the above piles ? 
 
 6. The frustum of a right pyramid was 72 feet square at the 
 lower base and 48 at the upper ; and its altitude was 60 feet. 
 What was the lateral surface ? What the volume ? [Such a 
 solid is called a Prismoid.] 
 
 7. Find the area of the surface, and the contents of a regular 
 tetraedron, one of whose edges is 10 inches. What is the diam- 
 eter of the inscribed sphere ? Of the circumscribed ? 
 
 647. A Wedge is a sohd 
 bounded by three quadrilaterals 
 and two triangles. 
 
 Thus, ABCD is a rectangle, and is 
 called the Head of the wedge, the two 
 Irianj^les AED and FBC are the Ends, 
 and the two trapezoids ABFE and 
 DCFE are the Sides. The Altitude is 
 the perpendicular to the head from the Fiq. 290. 
 
 edge opposite. 
 
 8. The base of a wedge being 18 feet by 9 feet, the edge 20 
 feet, and the altitude 6 feet, what are the contents ? 
 
 Ans. 504 cu. ft. 
 
284 
 
 ELEMENTARY GEOMETRY. 
 
 OF THE SPHERE.* 
 
 648. A Sphere is a solid bounded by a surface every point 
 in which is equally distant from a point within called the Cejitre. 
 
 The distance from the centre to the surface is the Radius, 
 and a line passing through the centre and limited by the surface 
 is a Diameter. The diameter is equal to twice the radius. 
 
 CIRCLES OF THE SPHERE, 
 
 PROPOSITION I. 
 
 649. Theorem. — Every section of a sphere made hy a 
 plane is a circle. 
 
 Demonstration. 
 
 Let AFEBD be a section of a sphere, 
 whose centre is 0, made by a plane; then 
 is the section AFEBD a circle. 
 
 For, let fall from the centre a per- 
 pendicular upon the plane AFEBD, as OC, 
 and draw CA, CD, CE, CB, etc., lines of the 
 plane, from the foot of the perpendicular 
 to any points in which the plane cuts the Fig. 29i. 
 
 * A spherical blackboard is almost indispensable in teaching this section 
 as well as in teaching Spherical Trigonometry. A sphere about two feet in 
 diameter, mounted on a pedestal, and having its surface slated or painted as 
 a blackboard, is what is needed. It can be obtained of the manufacturers 
 of school apparatus, or made in any good turning-shop. 
 
OF THE SPHERE. 285 
 
 surface of the sphere. Join these points with the centre, 0, of the 
 sphere. 
 
 Now OA, OD, OB, OE, etc., being radii, are equal; whence, CA, CD, 
 CB, CE, etc., are equal; i. e., every point in the line of intersection of a 
 plane and surface of a sphere is equally distant from a point in this 
 plane. Hence, the intersection is a circle, q. e. d. 
 
 650. A circle made by a plane not passing through the centre 
 is a Small Circle ; one made by a plane passing through the 
 centre is a Great Circle. 
 
 651. Corollary \.—A perpendicidar from the centre 
 of a sphere upon any small circle pierces the cii'de at its 
 centre; and, conversely, a perpendicular to a small circle 
 at its centre passes through the centre of the sphere. 
 
 652. A diameter perpendicular to any circle of a sphere is 
 called the Axis of that circle. The extremities of the axis are 
 the Poles of the circle. 
 
 663. Corollary 2. — The pole of a circle is equally dis- 
 tant from every point in its circumference. 
 The student should give the reason. 
 
 654. Corollary 3. — Every circle of a sphere has two 
 poles, which, in case of a great circle, are equally distant 
 from every point in the circumference of the circle ; hut, 
 in case of a small circle, one pole is nearer any point in 
 the circumference than the other pole is. 
 
 655. Corollary 4. — A small circle is less a; it.< dis- 
 tance from the centre of the sphere is greater ; ) enc£ tlie 
 circle whose plane passes through the centre is th greatest 
 circle of the sphere. 
 
 For, its diameter, being a chord of a great circle, is less as it i 
 ther from the centre of the great circle, which is also the cenfrr i < . 
 sphere. 
 
 656. Corollary 5. — All great circles of the same sphere 
 are equal (?). 
 
286 ELEMENTARY GEOMETRY. 
 
 PROPOSITION II. 
 
 667. Theorem. — Any great circle divides the sphere 
 into two equal parts. 
 
 DEMOlfSTRATIOlir. 
 
 Conceive a sphere as divided by a great circle, i. e., by a plane pass- 
 ing through its centre, and let the great circle be considered as the base 
 of each portion. These bases being equal, reverse one of the portions and 
 conceive its base placed in the base of the other, the convex surfaces 
 being on the same side of the common base. Since the bases are equal 
 circles, they will coincide, and since all points in the convex surface of 
 each portion are equally distant from the centre of the common base, the 
 convex surfaces will coincide. Therefore, the portions coincide through- 
 out, and are consequently equal. Q. e. d. 
 
 657, a.— A Hemisphere is one of the two equal parts into 
 which a great circle divides a sphere. 
 
 PROPOSITION III. 
 
 658. Theorem.— The intersection of any two great cir- 
 cles of a sphere is a diameter of a sphere. 
 
 Demonstration. 
 
 The intersection of two planes is a straight line ; and in the case of 
 the two great circles, as they both pass through the centre of the sphere, 
 this is one point of their intersection. Hence, the intersection of two 
 great circles of a sphere is a straight line which passes through the cen- 
 tre. Q. E. D. 
 
 659. Corollary. — The intersections on the surface of a 
 sphere of two circumferences of great circles are a semi 
 circumference, or 180°, apart, since they are '^* /..r,v.c// 
 extremities of a diameter. 
 
OF THE SPHERE. 
 
 287 
 
 DISTANCES ON THE SURFACE OF A 
 SPHERE. 
 
 660. Distances on the surface of a sphere are always to be 
 understood as measured on the arc of a great circle, unless it is 
 otherwise stated. 
 
 PROPOSITION IV. 
 
 661. Theorem. — The distance, measured on the sur- 
 face of a sphere, from the pole of a circle to any point in 
 the circumference of that circle, is the same. 
 
 Demonstration. 
 
 Let P be a pole of the small circle AEB. 
 
 Then are the arcs PA, PE, PB, etc., 
 which measure the distances on the surface 
 of tlie sphere, from P to any points in the 
 circumference of circle AEB, equal. 
 
 For, by (653), the straight lines AP, PE, 
 PP, etc., are equal, and these equal chords 
 subtend equal arcs, as arc PA,^rc PE, arc 
 BB, etc., the great circles of which these 
 lines are chords and arcs being equal (656). 
 
 Thus, for like reasons, 
 
 Fig. 292. 
 
 arc P'QA = arc P'LE = arc P'RB, etc. q. e. d. 
 
 622. Corollary. — The distance from the pole of a great 
 circle to any point in the circumference of the circle is a 
 quadrant (a quarter of a circumference). 
 
 Since the poles are 180° apart (being the extremities of a diameter), 
 PAQP' = PELP' = a serai-circumference. But, in case of a great circle, 
 chord PL = chord P'L (= chord PQ = chord P'Q), whence arc PEL = 
 arc P'L = arc PAQ = arc P'Q. Hence, each of these arcs is a quadrant. 
 
288 
 
 • ELEMENTARY GEOMETRY. 
 
 663. Scholium. — By means of the facts 
 demonstrated in this proposition and corollary, 
 we are enabled to draw arcs of small and great 
 circles, in the surface of a sphere, with nearly the 
 same facility that we draw arcs and lines in a 
 plane. Thus, to draw the small circle AEB (Fig. 
 292), we take an arc equal to PE, and placing one 
 end of it at P, cause a pencil held at the other 
 end to trace the arc AEB, etc. To describe the 
 circumference of a great circle, a quadrant must 
 be used for the arc. By bending a wire into an arc of the circle, and 
 making a loop in each end, a wooden pin can be put through one loop 
 and a crayon through the other, and an arc drawn as represented in 
 Fig. 293. 
 
 Fig. 293. 
 
 PROPOSITION V. 
 
 664. Problem. — To pass a circumference of a great 
 circle through any two points on the surface of a sphere. 
 
 Solution. 
 
 Let A and B |3e two points on the surface of a sphere, through which 
 it is proposed to pass a circumference o/a great circle. 
 
 From B as a pole, with an arc equal to a quad- 
 rant, strike an arc on, as nearly where the pole of 
 the circle passing through A and B lies, as may be 
 determined by inspection. Then, from A, with 
 the same arc, strike an arc st intersecting on at P. 
 Now, P is the pole of the great circle passing 
 through A and B (?). Hence, from P as a pole, with 
 a quadrant arc drawing a circle, it will pass 
 through A and B ; and it will be a great circle, 
 since its pole is a quadrant's distance from its circumference. 
 
 [The student should make this construction on the spherical black- 
 board.] 
 
 Fig. 294. 
 
OF THE SPHERE. 289 
 
 PROPOSITION VI. 
 
 665. Theorem, — Through any two points on the sur- 
 face of a sphere, one great circle^ can always he made to 
 pass, and only one, except when the two points are at the 
 extremities of the sam^e diameter, in which case an infi- 
 nite number of great circles can he passed through the two 
 points. 
 
 Demonstration. 
 
 This proposition may be considered a corollary 
 to the preceding. Thus, in general, the two great 
 circles struck from A and B as poles, with a quad- 
 rant arc, can intersect in only two points (?), which 
 are the poles of the same great circle (?). 
 
 But, if the two given points were at the extrem- 
 ities of the same diameter, as at D and C, the arcs 
 st and m would coincide, and any point in this ^'8- 295. 
 
 circumference being takdn as a pole, great circles can be drawn through 
 D and C. 
 
 [The student should trace the work on the spherical blackboard.] 
 
 666. Scholium. — The truth of the proposition is also evident from 
 the fact that three points not in the same straight line determine the 
 position of a plane. Thus. A, B, and the centre of the sphere, fix the 
 position of one, and only one, great circle passing through A and B. 
 Moreover, if the two given points are at the extremities of the same diam- 
 eter, they are in the same straight line with the centre of the sphere, 
 whence an infinite number of planes can be passed through them and the 
 centre. The meridians on the earth's surface afford an example, the poles 
 (of the equator) being the given points. 
 
 667. Corollary. — // two points in the circumference of 
 a great circle of a sphere, not at the extremities of the 
 sam^e diameter, are at a quadrant's distance from a point 
 on the surface, this point is the pole of the circle, 
 
 "■ The word circle may be understood to refer either to the circle proper, 
 or to its circumference. The word is in constant use in the higher mathe- 
 matics in the latter sense. 
 13 
 
290 ELEMENTARY GEOMETRY. 
 
 PROPOSITION VII. 
 
 668. Theorem. — The shortest distance on the surface 
 of a sphere, between any two points in that surface, is 
 measured on the arc less than a semi-circumference of the 
 great circle which joins them. 
 
 Demonstration. 
 
 Let A and B be two points in the surface of a sphere, AB the arc of a 
 great circle joining them, and kiuCn^ any other path in the surface be- 
 tween A and B. 
 
 Then is arc AB less than AmCwB. 
 
 Let C be any point in AmCwB, and pass the 
 arcs of great circles through A and C, and B and 
 C. Join A, B, and C with the centre of the sphere. 
 The angles AOB, AOC, and COB form the facial 
 angles of a triedral, of which angles the arcs AB, 
 AC, and CB are the measures. 
 
 Now, angle AOB < AOC + COB (540) ; Fig. 296 
 
 whence arc AB < arc AC + arc CB (?), 
 
 and the path from A to B is less on arc AB than on arcs AC, CB. 
 
 In like manner, joining any point in AmC with A and C by arcs of 
 great circles, their sum will be greater than AC. So, also, joining any 
 point in CwB with C and B, the sum of the arcs will be greater than CB. 
 
 As this process is indefinitely repeated, the path from A to B on the 
 arcs of the great circles will continually increase, and also continually 
 approximate the path AmCwB. Hence, arc AB is less than the path 
 AmCwB. Q. E. D. 
 
 669. Corollary. — The least arc of a circle of a sphere 
 joining any two points in the surface, is the arc less than 
 a semi-circumference of the great circle passing through 
 the points; and the greatest arc is the circumference 
 minus this least arc. 
 
OF TEE SPHERE. 
 
 291 
 
 Thus, let Km^n be any small circle passing 
 through A and B, and ABD^^C the great circle; 
 then, as just shown, Ajt>B < A77iB. 
 
 Now, circf.*ABDoC > circf. AmB/i (655). 
 
 Subtracting the former inequality from the 
 latter, we have BDoCA > B/iA. Q. e. d. 
 
 670. Two arcs of great circles are said 
 to be perpendicular to each other when 
 their circles are. 
 
 Fig. 297. 
 
 • PROPOSITION VIII. 
 
 671. Theorem. — If at the middle point of an aro of a 
 great circle a perpendicular is drawn on the surface of a 
 sphere, the distances being measin^ed on great circles, 
 
 1st. Any p9^^ ^^ ^^^^ perpendicular is equally distant 
 from the extremities of the arc, 
 
 2d. Anij point out of the perpendicular is unequally 
 distant from the extremities of the arc. 
 
 Demonstration". 
 
 Let AB be any arc of a great circle, D its middle point, and PD a 
 perpendicular. 
 
 Then is PB = PA, the ares being all arcs of 
 great circles. 
 
 From 0, the centre of the sphere, draw OP, 
 OD, OB, and OA. The rectangular triedrals 
 0-PDB and 0-PDA are symmetrically equal (?) ; 
 whence PB = PA. Q. e. d. 
 
 Fig. 298. 
 
 Again, let P' be a point out of PD. Pass 
 arcs of great circles through P' and A, and P' and 
 B, as P'A, P'B. From P, where one of these 
 cuts PD, draw the arc of a great circle PB. Then is 
 
 P'B < P'P + PB (668), 
 whence, P'B < P'P + PA (?), and P'B < P'A (?). Q. e. d. 
 
292 
 
 ELEMENTARY GEOMETRY. 
 
 672. Corollary 1. — The perpendicular at the middle 
 point of an arc contains all the points in the surface of 
 the sphere which are equally distant from tJJn>e extremities 
 of the arc, 
 
 673. Corollary 2. — An arc which has each of two points, 
 not at the ejotremity of the same diameter, equally distant 
 from the extremities of another arc of a great circle, is 
 perpendicular to the latter at its middle point. 
 
 This is apparent, since by Corollary 1 such points are in the perpen- 
 dicular, and two such points with the centre determine a great circle. 
 
 PROPOSITION IX. 
 
 674. Theorem. — The shortest path on the surface of a 
 hemisphere, from any point therein to the circumference 
 of the great circle forming its base, is the arc not greater 
 than a quadrant of a great circle perpendicular to the 
 base, and the longest path, on any arc of a great circle, is 
 the supplement of this shortest path* 
 
 Dbmokstration. 
 
 Let P be a point in the surface of the hemisphere whose base is 
 ADCBC, and DPmD' an arc of a great circle passing through P and per- 
 pendicular to ADCBC. 
 
 Then is PD the shortest path on the sur- 
 face from P to circumference ADCBC, and 
 P?7iD' is the longest path from P to the cir- 
 cumference, measured on the arc of a great 
 circle. 
 
 For, the shortest path from P to any point 
 in circumference ADBC is measured on the 
 arc of a great circle (?). Now, let PC be any 
 oblique arc of a great circle. We will show 
 that Fig. 29! 
 
 arc PD < arc PC. 
 
OF THE SPHERE. 
 
 293 
 
 Produce PD until DP' = PD; and pass a great circle through P' 
 and C. 
 
 Then is the arc PC = arc P'C. 
 
 And, since PC + PC > PP', 
 
 PC, the half of PC 4- P'C, is greater than PD, the half of PP'. q. e. d. 
 
 Secondly, PmD' is the supplement of PD, and we are to show that it 
 Ml greater than any other arc of a great circle from P to the circumference 
 ADBC. Let FnC be any arc of a great circle oblique to ADCBC. Pro- 
 duce C'wP to C. Now CPtiC is a semi-circumference and consequently 
 equal to DPmD'. But we have before shown that 
 
 PD < PC, 
 
 and subtracting these from the equals CPnC and DPwD', we have 
 
 PwD' > PnC. Q. E. D. 
 
 675. Corollary. — From any point in the surface of a 
 hemisphere there are two perpendiculars to the circumfer- 
 ence of the great circle which forms the base of the hemi- 
 sphere; one of which perpendiculars measures the least 
 distance to that circumference, and the other the greatest, 
 on the arc of any great circle of the sphere. 
 
 SPHERICAL ANGLES. 
 
 676. The angle formed by two arcs 
 of circles of a sphere is conceived as 
 the same as the angle included by the 
 tangents to the arcs at the common 
 points. 
 
 Illubtbation. — Let AB and AC be two 
 arcs of circles of the sphere, meeting at A ; 
 then the angle BAC is conceived as the same 
 as the angle B'AC, B'A being tangent to the 
 circle BADm, and C'A to the circle CAEn. 
 
 Fio. 300. 
 
294 
 
 ELEMENTARY GEOMETRY. 
 
 677. A Spherical Angle is the angle included by two 
 arcs of grmt circles. 
 
 Illustration.— B AC is a spherical an- 
 gle, and is conceived as the same as the 
 angle B'AC, B'A and C'A being tangents to 
 the great circles BADF and CAEF. [The stu- 
 dent should not confound such an angle as 
 BAC Fig. 800) with a spherical angle.] 
 
 Fig. 301. 
 
 PROPOSITION X. 
 
 678. Theorem. — td spherical angle is equal to the 
 measicre of the diedral included by the great circles whose 
 arcs form the sides of the angle. 
 
 Demon^stration. 
 
 Let BAC be any spherical angle, and BADF and CAEF the great cir- 
 cles whose arcs BA and CA Include the angle. 
 
 Then is BAC equal to the measure of the 
 diedral C-AF-B. 
 
 For, since two great circles interaect in a 
 diameter (?), AF is a diameter. 
 
 Now B'A is a tangent to the circle BADF, 
 that is, it lies in the same plane and is per- 
 pendicular to AO at A. 
 
 In the like manner, C'A lies in the plane 
 CAEF and is perpendicular to AO. Hence 
 B'AC' is the measure of the diedral C-AF-B (?). 
 
 Fig. 302. 
 
 Therefore the spherical angle BAC, which is the same as the plane 
 angle B'Ab', is equal to the measure of the diedral C-AF-B. q. e. d. 
 
OF THE UPHERE, 
 
 295 
 
 PROPOSITION XI 
 
 679. Theorem.—// one of two great circles passes 
 through the pole of the other, their circinriferences inter- 
 sect at right angles. 
 
 Demonstration. 
 
 Thus, P being the pole of the great circle 
 CABm, PO is its axis, and any plane passing 
 through PO is perpendicular to the plane 
 CABm. (?). 
 
 Hence, the diedral B-AO-P is right, and 
 the spherical angle PAB, which is equal to the 
 measure of the diedral, is also right, q. e. d. 
 
 Fig. 303. 
 
 680. Corollary 1. — A spherical angle is measured by 
 the arc of a great circle intercepted between its sides, and 
 at a quadrant's distance from its vertex. 
 
 Thus, the spherical angle CPA is measured by CA, PC and PA being 
 quadrants. For, since PC is a quadrant, CO is a perpendicular to PO, 
 the edge of the diedral C-PO-A, and for the like reason AO is perpendic- 
 ular to PO. Hence, COA is the measure of the diedral, and consequently 
 CA, its measure, is the measure of the spherical angle CPA. q. k. d. 
 
 681. Corollary 2. — The angle included by two arcs of 
 sm,all circles is the same as the angle included, bi/ two arcs 
 of great circles passing through the vertex and having the 
 same .tangents. 
 
 Thus, 
 
 BAC = B AC 
 
 For the angle BAC is, by definition, the 
 same as B'AC, B'A and CA being tangents 
 to BA and CA. Now, passing planes 
 through CA, B'A. and the centre of the 
 sphere, we have the arcs B"A, C'A, and B'A, 
 C'A tangents to them. Hence, B"AC" is 
 the same as B'AC, and consequently the 
 same as BAC. q. e. d. 
 
 Fig. 304. 
 
296 
 
 ELEMENTARY GEOMETRY, 
 
 682. Scholium. — To draw an arc of a great circle which 
 shall he perpendicular to another ; or, what is the same 
 thing, to construct a right spherical angle. 
 
 Let it be required to erect an arc of a great circle perpendicular to 
 CAB at A. Lay oflFfrom A, on the arc CAB, 
 a quadrant's distance, as AP', and from P' as 
 a pole, with a quadrant describe an arc pass- 
 ing through A. This will be the perpendic- 
 ular required. 
 
 In a similar manner we may let fall a per- 
 pendicular from any point in the surface, 
 upon any arc of a great circle. To let fall a 
 pei-pendicular from P" upon the arc CAB, 
 from P" as a pole, with a quadrant describe 
 an arc cutting CAB, as at P'. Then, from P' 
 
 as a pole, with a quadrant describe an arc passing through P" and cutting 
 CAB, and it will be perpendicular to CAB. 
 
 Fig. 305 
 
 PROPOSITION XII. 
 
 683. Problem. — To pass the circumference of a small 
 circle through any three points on the surface of a sphere. 
 
 Solution-. 
 
 Let A, B, and C be the three points in the surface of the sphere 
 through which we propose to pass the circumference of a circle. 
 
 Pass arcs of great circles through the points, 
 thus forming the spherical triangle ABC (664)- 
 
 Bisect two of these arcs, as BC and AC, by 
 arcs of great circles perpendicular to each (673, 
 682)- The intersection of these perpendiculars, 
 t>, will be the pole of the small circle required (?). 
 
 Then from <?, as a pole, with an arc oB draw 
 the circumference of a small circle: it will pass 
 through A, B, and C (?), and hence is the circum- 
 ference required. 
 
 QpERY. —If the three given points chance to be in the circumference 
 of a great circle, how will it appear in the construction? 
 
 Fig. 306. 
 
OF THE SPHERE, 
 
 297 
 
 OF TANGENT PLANES. 
 
 684. A Tangent Plane to a curved surface at a given 
 point is the plane of two lines respectively tangent to two plane 
 sections through the point. 
 
 Illustration. — Let P be any 
 point in the curved surface. Pass any 
 two planes through the surface and 
 the point P, and let OPQ and MPN 
 represent the intersections of these 
 planes with the curved surface. 
 Draw UV and ST in the planes of 
 
 the sections, and tangent respectively to OPQ and MPN at P. 
 the plane of UV aud ST the tangent plane at P. 
 
 Fig. 307. 
 
 Then is 
 
 PROPOSITION XIII. 
 
 686. Theorem. — «^ tangent plane to a sphere is per- 
 pendicular to the radiios at the point of tangency. 
 
 Demonstration. 
 
 Let P be any point in the surface of a sphere ; pass two great circles, 
 as Pa A, etc., and P///AR, through P, and draw ST tangent to the arc 
 mP, and UV tangent to the arc a?. 
 
 Then is the plane SVTU a tangent 
 plane at P, and perpendicular to the ra- 
 dius OP. 
 
 For, a tangent (as ST) to the arc mP 
 is perpendicular to the radius of the cir- 
 cle, i. e., to OP, and also a tangent (as VU) 
 to the arc AP is perpendicular to the ra- 
 dius of tliis circle, i. e., to OP. 
 
 Hence, OP is perpendicular to two 
 lines of the plane SVTU, and consequent- 
 ly to the plane of these lines (?). q. e. d. 
 
ELEMENTARY GEOMETRY. 
 
 686. OoBOLLARY 1. — Jijvery point in a tangent plane to 
 a sphere, except the point of tangency, is without the 
 spherB, 
 
 For, OP, the perpendicular, is shorter than any line which can be 
 drawn from to any other point in the plane (?) ; hence any other point 
 in the plane than P lies farther from the centre of the sphere than the 
 length of the radius, and is, therefore, without the sphere. 
 
 687. Corollary 2. — A tangent through P to any circle 
 of the sphere passing through this point lies in the tan- 
 gent plane* 
 
 Thus, MN, tangent to the small circle PwRft through P, lies in the tan- 
 gent plane. 
 
 For, conceive the plane of the small circle extended till it intersects 
 the tangent plane. This intersection is tangent to the small circle, since 
 it touches at one point, but cannot cut it ; otherwise the tangent plane 
 would have another point than P common with the surface of the sphere. 
 
 But there can be only one tangent to a circle at a given point. Hence 
 this intersection is MN, which is consequently in the tangent plane. 
 
 OF SPHERICAL TRIANGLES. 
 
 688. A Spherical Triangle is a portion of the surface of 
 a sphere bounded by three arcs of great circles. In the present 
 treatise these arcs will be considered as each less than a semi- 
 circumference; and the triangle considered will be the one which 
 is less than a hemisphere. 
 
 The terms scalene, isosceles, equilateral, right-angled, and 
 oblique-angled, are applied to spherical triangles in the same 
 manner as to plane triangles. 
 
OF THE SPHERE. 
 
 299 
 
 PROPOSITION XIV. 
 
 Theorem. — The sum of any two sides of a 
 spherical triangle is greater than the third side, and 
 their difference is less than the third side. 
 
 Demonstration. 
 Let ABC be any spherical triangle. 
 Then is BC < BA + AC, 
 
 and BC — AC < BA ; 
 
 and the same is true of the sides in any order. 
 
 For, join the vertices A, B, and C with the cen- 
 tre of the sphere, by drawing AO, BO, and CO. 
 There is thus formed a triedral 0-ABC, whose '"'S- ^°^- 
 
 facial angles are measured by the sides of the triangle (188). Now, 
 angle BOC is less than BOA + AOC (?), whence BC is less than BA + AC ; 
 and substracting AC from each member, we have BC— AC less than BA. 
 
 Q. E. D. 
 
 PROPOSITION XV. 
 
 690. Theorem. — The sum of the sides of a spherical 
 triangle may be anything between and a circumfer- 
 ence' 
 
 Demonstration. 
 
 The sides of a spherical triangle are measures of the facial angles of a 
 triedral whose vertex is at the centre of the sphere. Hence their sum 
 may be anything between and the measure of 4 right angles, as these 
 are the limits of the sum of the facial angles of a triedral (?). q. e. d. 
 
 691. Scholium. — As the sides of a spherical triangle are arcs, they 
 can be measured in degrees. Hence, we speak of the side of a spherical 
 triangle as 30°, 57°, 115°, 10', etc. In accordance with this, we say that 
 the limit of the sum of the sides of a spherical triangle is 360°. 
 
30u ELEMJ^NTARY GEOMETRY, 
 
 PBOrOSITXON XVI. 
 
 692. Theorem. -I?! i^ ci^ni ?f *h o angles of a spheri- 
 cal triangle may be anything between two and six right 
 angles. 
 
 Demonstration. 
 
 The sum of the angles of a spherical triangle is the same as the sum 
 of the measures of the diedrals of a triedral having' its vertex at the centre 
 of the sphere, as in (?). Now the limits of the sum of the measures of 
 these diedrals are 2 and 6 right angles (?). Hence the sura of the angles 
 of any spherical triangle may be anything between 2 and 6 right angles. 
 
 Q. E. D. 
 
 6^. Corollary. — A spherical triangle may have one, 
 two, or even three right angles; and, in fact, it may 
 have one, two, or three obtuse angles; since, in the 
 latter case, the sum^ of the angles will not necessarily be 
 greater than 540°. 
 
 694. A Trirectaiig^ular Spherical Triangle is a 
 
 spherical triangle which has three right angles. 
 
 695. Scholium.— It will be observed that the sum of the angles of a 
 spherical triangle is not constant, as is the sum of the angles of a plane 
 triangle. Thus, the sum of the angles of a spherical triangle may be 
 200°, 290°, 350°, 500°, anything between 180° and 540°. 
 
 696. Spherical Excess is the amount by which the sum 
 
 of the angles of a spherical triangle exceeds the sum of the 
 angles of a plane triangle ; i. «., it is the sum of the spherical 
 angles — 180°, or tt. 
 
 Exercise. — Prove that if from any point within a spherical 
 triangle arcs of great circles be drawn to the extremities of any 
 side, the sum of these two arcs is less than the sum of the other 
 two sides of the triangle. 
 
OF THE SPHERE, 301 
 
 PROPOSITION XVII. 
 
 697. Theorem. — The trirectangular triangle is one- 
 eighth of the surface of the sphere. 
 
 Demonstration. 
 
 Pass three planes through the centre of a sphere, respectively perpen- 
 dicular to each other. They will divide the surface into eight tri- 
 rectangular triangles, any one of which may be applied to any other. 
 
 Thus, let ABA'B', ACA'C, and CBC'B' be 
 the great circles formed by the three planes, 
 mutually perpendicular to each other. The 
 planes being perpendicular to each other, the 
 diedrals, as A CO-B, C-BO-A, C-AO-B, etc., are 
 right, and hence the angles of the eight tri- 
 angles formed are all right. 
 
 Also, as AOB is a right angle, AB is a quad- 
 rant ; as BOC is a right angle, CB is a quadrant, 
 etc. Hence, each side of every triangle is a 
 quadrant. 
 
 Whence any one triangle may be applied to any other. [Let the stu- 
 dent make the application.] 
 
 Hence the trirectangular triangle is one-eighth of the surface of the 
 sphere, q. e. d. 
 
 698. Corollary. — The trirectangular triangle is 
 equilateral and its sides are quadrants. 
 
 Exercise 1. What is the spherical excess in a spherical tri- 
 angle whose angles are 117°, 84°, and 96°, expressed in degrees ? 
 Expressed in right angles ? Expressed in tt ? 
 
 Ans. 117°, 1A» and ^n. 
 
 2. Can there be a spherical triangle whose sides are 78°, 113°, 
 and 31° ? Can there be one whose sides are 152°, 136°, 148° ? 
 
 3. Can there be a spherical triangle whose sides are 52^, 
 126°, and 140°? 
 
ELEMENTARY GEOMETRY, 
 
 PROPOSITION XVIII. 
 
 699. Theorem. — In an isosceles spherical triangle, 
 the angles opposite the equal sides are equal ; and, con- 
 versely, // two angles of a spherical triangle are equal, 
 the triangle is isosceles. 
 
 Demonstration. 
 Let ABC be an isosceles spherical triangle, in which AB = AC. 
 
 Then 
 
 angle ABC = ACB. 
 
 For, draw the radii AO, CO, and BO, form- 
 ing the edges of the triedral 0-ABC. 
 
 Now, since AB = AC, the facial angles AOB 
 and AOC are equal, and the triedral is isosceles. 
 Hence the diedrals A-OB-C and A-OC-B are equal 
 (550), and consequently the spherical angles 
 ABC and ACB are equal (678). Q. e. d. 
 
 Fig. 311. 
 
 Again, if angle ABC = angle ACB, side AC 
 = side AB. For in the triedral 0-ABC, the 
 
 diedrals A-OB-C and A-OC-B are equal, whence the facial angles AOB 
 and AOC are equal (550), and consequently the sides AB and AC, which 
 nieasure'these angles. Q. e. d. 
 
 700. Corollary.— ./^W/ equilateral spherical triangle 
 is also equiangular; and, conversely, ^n equiangular 
 spherical triangle is equilateral. 
 
 Queries. — 1. What is the greatest angle which an equilateral 
 spherical triangle can have ? 
 
 2. What is the greatest side which an equilateral spherical 
 triangle can have ? 
 
OF THE SPHERE. 
 
 303 
 
 PROPOSITION XIX. 
 
 701. Theorem. — On the same sphere, or on equal 
 spheres, two isosceles triangles having two sides and 
 the included angle of the one equal to two sides and 
 the included angle of the other, each to each, can be 
 superimposed, and are consequently equal. 
 
 Demonstration. 
 
 In the triangles ABC and AB C . let AB = AC, AB' = AC ; and let 
 AB = AB', BC = B'C, and angle ABC = ABC . 
 
 Then can the triangle A B'C be superimposed 
 upon ABC. 
 
 For, since the triangles are isosceles, we have 
 
 angle ABC = ACB, 
 
 AB'C = ACB' (699), 
 and, as by hypothesis 
 
 ABC = AB'C, 
 
 these four angles are equal, each to each. 
 
 For a like reason, AB = AC = AB' = AC. 
 
 Now, applying AC to its equal AB, the extremity A at A, and C' at 
 B, with the angle B' on the same side of AB as C, the convexities of the 
 arcs AC and AB being the same, and in the same direction, the arcs will 
 coincide. Then, as 
 
 angle ACB' = ABC, 
 
 C'B' will take the direction BC, nnd since these arcs are equal by hypoth- 
 esis, B' will fall at C. Hence B'A will fall in CA, as only one arc of a 
 great circle can pass between C and A, and the triangle AB'C is super- 
 imposed upon ABC ; wherefore they are equal. Q. e. d. 
 
 Fig. 312. 
 
 702. Symmetrical Spherical Triangles are such as 
 have the parts of one respectively equal to the parts of the other, 
 but arranged in a different order ; hence such triangles are not 
 capable of superposition. 
 
304 
 
 ELEMENTARY GEOMETRY, 
 
 Fig. 313. 
 
 Fig. 314. 
 
 Illustration. — In Fig. 313, ABC and A'B'C represent symmetrical 
 spherical triangles. In these triangles, 
 
 A = A', 8 = 8', C = C', 
 
 AC = AC, AB = A'8', and 8C = BC; 
 
 nevertheless we cannot conceive one triangle superimposed upon the 
 other. Thus, were we to make the attempt by placing A' 8' in its equal 
 AB, A' at A, and B' at 8, the angle C would fall on the opposite side of 
 AB from C. Now, we cannot revolve A'C'B' on AB (or its chord), and 
 thus make the two coincide, for this would bring their convexities to- 
 gether. Nor can we make them coincide by reversing A'B'C, and placing 
 8' at A, and A' at 8. For, although these two arcs will thus coincide, as 
 the angle 8' is not equal to A, B'C will not fall in AC ; and, again, if it 
 did, C would not fall at C, since B'C and AC are not equal. 
 
 But, considering the triangles ABC and A'B'C in Fig. 314, in which 
 
 = 0', 
 
 A 
 AC 
 
 A', 
 AC, 
 
 8 = 8', 
 AB = A'B', 
 
 and 80 = B'C, 
 
 we can readily conceive the latter as superimposed upon the former. 
 [The student should make the application.] Now, the two triangles are 
 equivalent in each case, as will subseljueiltly appear ; and the former are 
 equal. Such triangles as those in Fig. 313 are called symmetrically equals 
 while the latter are said to be equal by superposition. 
 
 Fig. 315 represents the same triangles as Fig. 314, and exhibits a com- 
 plete projection* of the semi-circumferences of which the sides of the 
 
 * To understand what is meant by the projection of these lines, conceive 
 a hemisphere with its base on the paper, and represented by the circle abc, 
 and all the arcs raised up from the paper as they would be on the surface of 
 such a hemisphere. Thus, considering the arc a^Bb (Fig; 315), the ends a 
 and 6 would be in the paper just where they are, but the rest of the arc 
 would be off the paper, as though you could take hold of B and raise it from 
 
OF THE SPHERE. 
 
 305 
 
 triangles are arcs. The student should become perfectly familiar with 
 i#, and be able to draw it readily. Thus, aAB6 is the projection of the 
 semi-circumference of which AB is an arc, aACc of the semi-circumfer- 
 ence of which AC is an arc, etc., etc. 
 
 PROPOSITION XX. 
 
 703. Theorem. — Two symmetrical spherical tri- 
 angles are equivdlent, i.e., equal in area. 
 
 DEMONSTRATIOlf. 
 
 Let ABC and MB'C be two symmetrical spherical triangles, with AB 
 =. AB', AC = AC, BC := B'C, A = A, B = B , and C = C. 
 
 Then are they equivalent. 
 
 Pass circumferences of small circles 
 through the vertices A, B, C, and A', B', C, 
 as abc and o'6V, of which o and o' are the 
 respective poles. 
 
 Now, by reason of the mutual equality 
 of the sides, 
 
 the chord AC = chord A'C, 
 chord AB = chord A'B', 
 and chord BC = chord B'C, Fig- 3i6. 
 
 and as the small circles are circumscribed about the equal plane triangles 
 ABC and A'B'C, these circles are equal Hence, 
 
 o\ = o'^' =oB = o'B' = oC = o'C\ 
 
 and th^ triangles AoB and A'o'B', BoC and B'o'C, AoC and A'o'C are 
 isosceles. 
 
 Now call the centre of the sphere, and draw the radii OA, OB, DC, 
 Go, OA', OB', OC , and Oo'. 
 
 the paper while a and b remain fixed. The lines in the figure are represen- 
 tations of lines on the surface of such a hemisphere, as they would appear 
 to an eye situated in the axis of the circle abc, and at an infinite distance 
 from it ; that is, just as if each point in the lines dropped perpendirvlarlp 
 down upon the paper. Arcs of great circles perpendicular to the base are 
 projected in straight lines passing through the centre, and oblique arcs are 
 projected in ellipses. See Spherical TrigonoTnetry (97-109). 
 
306 
 
 ELEMENT A RY G EOMETR Y. 
 
 Considering the triedrals 0-AoB and 
 0-AVB', their facial angles are equal, being 
 measured by equal arcs ; hence the diedral 
 A-oO-B ^ A'-o'O'-B' (0, and the spherical 
 angle AoB = AVB' (?). Therefore, the 
 isosceles triangle AoB = k'o'W (701). 
 
 In like manner, we may prove the tri- 
 angles <?BC and o'B'C equal, as also AoC 
 and k'o'C. 
 
 Hence, ABC is equivalent to A'B'C, as '^'9- 3i6 
 
 the two are composed of parts respectively equal. Q. e. d. 
 
 If the poles of the small circles fell without the given triangles, ABC 
 would be equivalent to the sum of two of the partial triangles minus the 
 third. What if the pole fell in a side? 
 
 PROPOSITION XXi. 
 
 704. Theorem.— 0«. the same sphere, or on equal 
 spheres, two spherical trian,gles having two sides and the 
 included angle of the one equal to two sides mid the in- 
 cluded angle of the other, each to each, are equal, or sym- 
 metrical and equivalent. 
 
 DEMONSTRATrOIsr. 
 
 Let ABC and A'B'C be two spherical triangles, having B = B', 
 BA = B'A', and BC = B'C. 
 
 Then are they either 
 equal, or symmetrical and 
 equivalent. 
 
 For, passing planes 
 through the sides of each 
 triangle and the centre of 
 the sphere, two absolutely 
 or symmetrically equal tri- 
 edrals will be formed (?). ''»9- 3I7. Fig. 3i8. 
 Whence the facial angles AOC and A'OC are equal, and consequently 
 AC ^ A'C (?). Also, the diedrals C-OA-B and C'OA'-B a - -' — ' 
 B-OC-A = B'-OC'-A' (?). Whence A = A' and C = C (?) 
 
 Hence the parts of ABC are respectively equal to the pan- .u A o , 
 and the triangles are equal, or symmetrical and equivalent, according as 
 the equal parts are arranged in the same or in a different oner. t^. jl d, 
 
OF THE SPHERE, 307 
 
 PROPOSITION XXII. 
 
 705. Theorem. — On the sctme sphere, or on equal 
 spheres, two spherical triangles having two angles and the 
 included side of the one equal to two angles and the in- 
 cluded side of the other, each to each, are equal, or sym- 
 metrical and equivalent. 
 
 This is a direct consequence of a proposition concerning triedrals. 
 Let the student give the deduction. 
 
 PROPOSITION XXIII. 
 
 706. Theorem. — On the sam^e sphere, or on equal 
 spheres, if two spherical triangles have two sides of the one 
 equal to two sides of the other, each to each, and the in- 
 cluded angles unequal, the third sides are unequal, and 
 the greater third side belongs to the triangle having the 
 greater included angle. 
 
 Conversely, // the two sides are equal, each to each, and 
 the third sides unequal, the angles included by the equal 
 sides are unequal, and the greater belongs to the triangle 
 having the greater third side. 
 
 Demonstration. 
 
 In the triangles ABC and ABC, let AB = A'B', AC = A'C, and 
 A>A. 
 
 Then is BC > B'C 
 
 For, join the vertices with the centre, form- 
 ing the two triedrals 0-ABC and 0-A'B'C'. 
 
 In these triedrals, AOB — A' OB', AOC = 
 A'OC, being measured by equal arcs; and 
 C-AO-B > C'-A'O-B', having the same measure 
 as A and A' (678). Hence COB > COB' (?). 
 
 Therefore CB, the measure of COB, > CB', 
 the measure of COB'. Fig. 3I9. 
 
 In like manner, the same sides of tVie triangles, and consequently the 
 same facial angles of the triedrals, being granted equal, and BC > B'C, 
 A > A'. For, BC being greater than B'C, COB > COB'; whence 
 B AO-C > B'-A'O-C (?), or A is greater than A'. 
 
^08 ELEMENTARY GEOMETBY, 
 
 PROPOSITION XXIV. 
 
 707. Theorem. — On the same sphere, or on equal 
 spheres, two spherical triangles having the sides of the one 
 respectively equal to the sides of the other, or the angles of 
 the one respectively equal to the angles of the other, are 
 equal, or symmetrical and equivalent. 
 
 Demonstration. 
 
 The sides of the triangles being equal, the facial angles of the triedrals 
 at the centre are equal, whence the triedrals are equal or symmetrical (?). 
 Consequently, the angles of the triangles are equal, and the triangles are 
 equal, or symmetrical and equivalent. 
 
 Again, the triangles being mutually equiangular, the triedrals have 
 their diedrals mutually equal ; whence the triedrals are equal or sym- 
 metrical (?). Therefore, the sides of the triangles are mutually equal, and 
 the triangles are equal, or symmetrical and equivalent. (See Figs. 313, 
 814.) 
 
 PROPOSITION XXV. 
 
 708. Theorem. — On spheres of different radii, mu- 
 tually equiangular triangles are similar (not equal). 
 
 Demonstration. 
 
 Let ABC and (ibc be two mutually equiangular spherical triangles on 
 spheres whose radii are respectively li and r, and let angle A = a^ 
 B = 6, C = c. 
 
 r^. . AB BC CA 
 
 Then is -—=-—= — . 
 
 cu) oc ca 
 
 For, joining the vertices of the triangles with the centres of the 
 spheres, and C, the triedrals 0-ABC and O'-ahc have their diedrals 
 mutually equal (?), whence their facial angles are mutually equal (?). 
 Therefore sector AOB is similar to sector aO% sector BOC to JO'c, and 
 sector COA to cO'a. 
 
OF THE SPHERE, 309 
 
 From the similarity of these sectors, we have 
 
 db ~ r ^■'^' be ~ r ' ca ~ r ' 
 
 AB BC CA 
 
 and hence, —i: = ir — Q. e. d. 
 
 ' ab be ca 
 
 709. Scholium. — In Spherical Trigonometry we are taught to find 
 the sides of a splierical triangle having the angles given. But in such a 
 case the sides are found in degrees, etc., which does not determine their 
 absolute lengths. The length of an arc of any number of degrees is not 
 known unless the radius of the sphere is known. 
 
 POLAR OR SUPPLEMENTAL TRIAN- 
 GLES. 
 
 710. One spherical triangle is Polar to another when the 
 vertices of one are the poles of the sides of the other, and the 
 corresponding vertices lie on the same side of the side opposite. 
 (For illustration, see 713.) 
 
 Such triangles are also called supplemental, since the angles 
 of one are the supplements of the sides opposite in the other, as 
 will appear hereafter. 
 
 PROPOSITION XXVI. 
 
 711. Problem. — Having a spherical triangle given, 
 to draw its polar. 
 
 Solution. 
 
 Let ABC (Fig. 320) be the given triangle.* From A as a pole, with 
 
 * This should be executed on a sphere. Few students get clear ideas of 
 polar triangles without it. Care should be taken to construct a variety of 
 triangles as the given triangle, since the polar triangle does not always lie 
 in the position indicated in the figure here given. Let the given triangle 
 have one side considerably greater than 90", another somewhat less, and the 
 third quite small. Also, let each of the sides of the given triangle be 
 greater than 90°. 
 
310 
 
 t:LEMENTA R Y GEOMETR Y. 
 
 Fig. 320. 
 
 Fig. 321. 
 
 a quadrant strike an arc, as C'B'. From B as a pole, with a quadrant 
 strike the arc C'A'; and from C, the arc A'B'. Then is A'B'C polar to 
 ABC. 
 
 712. Corollary. — If one triangle is polar to another, 
 conversely, the latter is polar to the former ; i. e., the rela- 
 tion is reciprocal. 
 
 Thus, A'B'C (Fig. 320) being polar to ABC, reciprocally, ABC is polar 
 to A'B'C ; that is, A' is the pole of CB, B' of AC, and C of AB. For 
 every point in A'B' is at^ quadrant's distance from C, and every point in 
 A'C is at a quadrant's distance from B. Hence, A' is ai a quadrant's dis- 
 tance from the two points C and B of CB, and is therefore its pole. 
 
 [In like manner, the student should show that B' is the pole of AC, 
 and C of AB.] 
 
 713. Scholium.— By producing (Fig. 331) each of the arcs struck 
 from the vertices of the given triangle sufficiently, four new triangles 
 will be formed, viz., A'B'C, QC'B', PCA', and RA'B'. Only the first of 
 these is called polar to the given triangle. Thus, in A'B'C, A', corre- 
 sponding to A, lies on the same side of CB or C'B' that A does, and so of 
 any other corresponding vertices. 
 
 It is easy to observe the relation of any of the parts of the other three 
 triangles to the parts of the polar. Thus, 
 
 QC = 180° - &', 
 
 QB' = 180° - c', 
 
 QC'B' = 180° - B'CA', 
 
 QB'C = 180° — CB' A', 
 
 and Q =-- A' = 180° - a, 
 
 as will appear hereafter. 
 
OF THE SPHERE, 
 
 311 
 
 PROPOSITION XXVII. 
 
 714. Theorem. — Any angle of a spherical triangle is 
 the supplement of the side opposite in its polar triangle ; 
 and any side is the sicpplement of the angle opposite in 
 the polar triangle. 
 
 First Demonstration. 
 
 Let ABC and A'B'C be two spherical triangles polar to each other; 
 and let the sides of each be designated as a, b, c, a', h', d, a being op- 
 posite A, <e' opposite A', h opposite B, etc. 
 
 Then 
 
 and 
 
 A = 180° - a'^ 
 P = 180° - l\ 
 C = 180° - </, 
 a = 180° - A', 
 b = 180° - B', 
 c = 180° - C. 
 
 Let be the centre of the sphere, and 
 draw OA, OB, OC, OA', OB', and 00'. 
 
 The angles B'OA and B'OC being right 
 (?), B'O is perpendicular to the face AOC (?). 
 
 For like reasons, CO is perpendicular to 
 the face AOB. 
 
 Hence B'OC is the supplement of the 
 diedral B-AO-C (512). 
 
 But a' is the measure of B'OC, and 
 B-AO-C has the same measure as A. 
 
 Hence, A = 180° — a'. 
 
 In like manner, we may show that 
 
 B = 180° - h\ and C = 180° - d. 
 
 Again, since the edges AO, BO, and CO are perpendicular to the faces 
 R'OC, A'OC, and A'OB', we can show in like manner that 
 
 a = 180° - A', 
 
 h = 180° - B', 
 «*" 1 e = 180° — C'. Q. E. D. 
 
 Fig. 322. 
 
 Q. E. D. 
 
312 
 
 ELEMENTARY GEOMETRY, 
 
 Second Demonsteatiok. 
 
 Let ABC and A'B'C be two polar triangles. Let BC, CA, and AB be 
 represented by a, 6, and c respectively, and B'C, C'A', and AB' by a', 
 
 b', and d. 
 
 To show A = 180° — a', produce h and c, 
 if necessary, till they meet the side a' of the 
 triangle polar to ABC in e and d. 
 
 Now A is measured by ed (?). But, since 
 ^'e -= 90°, and C'd = 90°, 
 
 B'e + C'd, or B'C + ed = 180° ; 
 
 whence, transposing, and putting a/ for B'C, 
 we have 
 
 ed = ^ = 180° - a' 
 
 Fig. 323. 
 
 In like manner, 
 whence 
 So, also, 
 
 C'g + A'f= CA'+/^ = 180°; 
 fg =B = 180° - C'A', or 180^ 
 = 180° - c'. 
 
 To show that A' = 180° — a, consider that A' being the pole of CB, 
 fi is measure of A'. 
 
 whi 
 
 Now 
 ence, 
 
 B/=90°(?), and Ci = 90°; 
 Bf+Ci= 180°. 
 
 But B/+ C^* =fi + a, wherefore fi-\-a = 180° ; 
 and transposing, and putting A' for/i, we have A' = 180° — a. 
 
 In like manner, we may show that 
 
 B' = 180" - h, and C = 180° — e. q. e. d. 
 [The student should give the details.] 
 
 714, a. Corollary. — The sum of the supplements of 
 any two angles of a spherical triangle is greater than ^' 
 supplem^ent of the third angle. (Consider 714, 689.) 
 
OF THE SPHERE. 
 
 313 
 
 QUADRATURE OF THE SURFACE OF 
 THE SPHERE. 
 
 715. The Quadrature* of a surface is the process of find- 
 ing its area. The term is applied under the conception that the 
 process consists in finding a square which is equivalent to the 
 given surface. 
 
 PROPOSITION XXVIII. 
 
 716. Ijeiiiina. — The surf ace generated hy the revolwtiofi 
 of a regular semi-polygon of an even number of sides, 
 ahout the^diameter of the circumscribed circle as an axis, 
 is equivalent to the circumference of the inscribed circle 
 multiplied by the axis. 
 
 Demonstration. 
 
 Let ABODE be one-half of a regular octagon, AE being the diameter 
 of the circumscribing circle. 
 
 If the semi-perimeter ABODE be revolved about AE 
 as an axis, the surface generated is 27Tr x AE, r being the 
 radius of the inscribed circle, as aO, or hO. 
 
 This surface is composed of the convex surfaces of 
 cones and fi-ustums of cones. Thus, AB generates the 
 surface of a cone, BO the frustum of a cone, etc. 
 
 Let a and 6 be the middle points of AB and BO re- 
 spectively, and draw ow, Be, Z>», and 00 perpendicular to 
 the axis, and B<Z parallel to it. Also draw the radii of the 
 inscribed circle, aO and JO. Indicate the surfaces gener- '9* ^ ** 
 
 ated by the sides as Surf. BO, etc. The areas of these surfaces are : ^ 
 
 Surf. AB = 27r X am x AB (?), (1) 
 
 :^v 
 
 Surf. BO = 2- X hn x BO, etc. (?). 
 
 * Latin quadrcUus, squared. 
 
 14 
 
314 
 
 ELEMENTA R Y QEOMETR Y. 
 
 Now, from the similar triangles 0am and BAc, we 
 have 
 
 aO _ am 
 AB " Ac^' 
 
 or 
 
 2n X «0 _ 27rx am 
 ~AB~ " ~Ac~" ' 
 
 Stt X am X AS = 2Kr x Ac, 
 
 whence, 
 putting r for aO. 
 
 Also, from the similar triangles Obn and CB^, 
 
 we have 
 
 whence, 
 putting r for 50. 
 
 60 
 
 bn 
 
 BC~ Bd{=cO) 
 
 2'iTxbO _2nxbn 
 
 27r X J» X BO = 27rr x cO, 
 
 Fig. 324. 
 
 Substituting these values in (1) and (2), we obtain 
 
 Surf. AB = 27rrxAc, 
 
 Surf. BC = 2TTr x cO. 
 
 And, in like manner, Surf. CD = 27rr x Op, 
 
 and Surf. DE = 2T:rxpE. 
 
 Adding, 
 
 Surf. ABODE = 27rr(Ac + eO + Op +pE) 
 = 27rr X AE. 
 
 Finally, since the same course of reasoning is applicable to the semi- 
 polygons of 16, 32, 64, etc., sides, the truth of the proposition is estab- 
 lished. 
 
 717- Scholium. — This proposition is only a particular case of sur 
 faces generated by any broken line roTolving about an axis; and fy*p 
 general proposition can be established in a manner altogether similar ■ 
 the method given above. But this case is all that we need for our pres- 
 ent purpose. 
 
OF THE SPHERE. 
 
 Z\{ 
 
 PROPOSITION XXIX. 
 
 718. Theorem, — The surface of a sphere is equivalent 
 to four great circles ; that is, to ^nR^, R being the radius 
 of the sphere. 
 
 Demonstration. 
 
 Let the semi-circumference ABODE revolve upon the diameter AE, 
 and thus generate the surface of a sphere. 
 
 Conceive the half of a regular octagon inscribed in 
 the semicircle ABODE; and let both the semi-polygon 
 and the semi-circumference be revolved about AE as an 
 axis. 
 
 Call the radius of the inscribed circle, as aO, r, and 
 let AO = R. 
 
 The surface generated by the broken line ABODE is, 
 by the last proposition, Stt?* x 2R = AnrB. 
 
 Now, conceive the arcs AB, BO, etc., bisected, and the 
 chords drawn, and let r' be the radius of the circle in- 
 
 Fig. 325. 
 
 scribed in the regular polygon thus formed. The surface generated by 
 the revolution of this semi-polygon is ^ttt'R 
 
 By repeating the bisections, the broken line approximates to the 
 semi-circumference, the radius of the inscribed circle to R, and the sur- 
 face generated to the surface of the sphere, the three quantities reaching 
 their limits at the same time. Hence, at the limit we have 
 
 Surf, of sphere = 2nRx2R = 4.7rR\ q. e. d. 
 
 719. Corollary 1. — The area of the surface of a sphere 
 is equivalent to the circumference of a great circle multi- 
 plied by the diameter, that is, to 2nR x 2R, as above. 
 
 720. Corollary 2. — The surfaces of spheres are to each 
 other as the squares of their radii. 
 
 Thus, if R and R' are the radii of two spheres, the surfaces are 47ri?* 
 and ^ttR". Now, 
 
316 
 
 ELEMENTARY QEOMETRT. 
 
 721. A Zone is the portion of the surface of a sphere in- 
 cluded between the circumferences of two paral- 
 lel circles of a sphere. The altitude of a zone 
 is the distance between the parallel circles 
 whose circumferences form its bases. 
 
 Illustration. — The surface generated by CB, or 
 any arc of the circle ABODE, etc.. as the semicircle 
 revolves about AE as an axis, conforms to the defini- 
 tion, and is a zone. Such a portion of the surface as is 
 generated by AB is called a zone with one hase, the 
 circle whose circumference would form the upper base 
 having become tangent to the sphere. The altitude of 
 the zone generated by CB is a5, and of that generated by AB the alti- 
 tude is Aa. 
 
 Fig. 32G. 
 
 PROPOSITION XXX. 
 
 722. Theorem. — The area of a zone is equal to 27TaR, 
 a being the altitude of the zone and R the radius of the 
 sphere. 
 
 Demonstration. 
 
 It is evident that in passing to the limit, the surface generated by 
 such a portion of the broken line as lies between C and B, Fig. 326, is 
 measured by the circumference of the inscribed circle multiplied by db. 
 Hence, at the limit, the zone generated by arc BC is measured by 
 
 2rrifxaJ, that is, 3n-ai?, 
 representing db by a. Q. e. d. 
 
 723. Corollary. — On the same sphere, or on equal 
 spheres, zones are to each other as their altitudes, and any 
 zone is to the surface of the sphere as the altitude of the 
 zone is to the diameter of the sphere. 
 
OF THE SPHERE. 
 
 ai7 
 
 OF LU N ES. 
 
 724. A Ijuiie is a portion of the surface of a sphere- included 
 by two semi-circumferences of great circles. 
 
 The surface km^n is a lune. 
 
 725. The Angle of the Lune is 
 
 the angle included by the arcs which form \ 
 its sides ; or, what is the same thing, the 
 measure of the diedral included between the 
 great circles. 
 
 Thus, the spherical angle wAn, or the measure of the diedral w-AB-n 
 is the augle of the lune A^nB/i. 
 
 726. An Ungula, or Spherical Wedge, is that portion 
 of a sphere included between two semi-great-circles, as AwB and 
 knB. It has a lune for its convex surface and a diameter for its 
 edge. 
 
 PROPOSITION XXXI. 
 
 727. Theorem. — On the same sphere, or on equal 
 spheres, lunes which have equal angles are equal. 
 
 Demonstration. 
 
 [This is readily effected by applying one to the other. Let the stu- 
 dent make the application.] 
 
 Exercise. — Can there be a spherical triangle whose angles are 152°,- 
 136°, and 148°? One whose angles are 152°, 136°, and 168'/ (See 
 714, a.) 
 
318 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XXXII. 
 
 728. Theorem. — The area of a lune is to the area of 
 the surface of the sphere on which it is situated as the 
 angle of the lune is to four right angles. 
 
 First Demon^stration. 
 
 Let S represent the area of the surface of the sphere generated by 
 the revolution of the semicircle IV1AN about MN as an axis, and L the 
 area of the lune whose angle is AMD, or AOD. 
 
 Then is — = 
 
 AOD 
 
 4 riglit angles 
 
 In the generation of 8 and L by the 
 semi-circumference MAN, the middle 
 point, A, of the semi-circumference gen- 
 erates the great circle ACDBF, on which 
 the angles of the lunes are measured (?). 
 
 Now A generates equal and coinci- 
 dent parts of arc AD and circumference 
 ACDBFA, in the same time that MAN 
 generates corresponding equal and co- 
 incident parts of L and 8. 
 
 arc AD 
 
 Fig. 328. 
 
 Hence, if 
 
 circf. ACDBF ""£' 
 
 and 
 
 arc AD 
 
 AOD 
 
 circf. ACDBF 4 right angles 
 
 (?). Q. E. D. 
 
 Second Demonstration. 
 
 Let S represent the area of the surface of the sphere generated by 
 the revolution of the semicircle MAN about MN as an axis, and L the 
 area of the lune whose angle is AMD^ or AOD. 
 
 Now the angles AOD and the sum of the four right angles AOD, DOB^ 
 BOF, FOA are at least commensurable by an infinitesimal unit. Let i be 
 
OF THE SPHERE. 
 
 319 
 
 their common measure, and let it be contained in AOD n times, and* in 
 the four right angles m times, so that 
 
 AOD ^ n 
 
 4 right angles ~" m 
 
 Now conceive the circumference divided into m equal parts, and 
 radii drawn to the points of division ; and through their extremities let 
 serai-circumferences be drawn. Then is L divided into n lunes, each 
 equal to one of the m equal lunes into which S is divided (727), so that 
 
 Hence, 
 
 
 L _n 
 
 8 m 
 
 L 
 
 AOD 
 
 8 ~ 
 
 4 right angles 
 
 Q. E. D. 
 
 Third Demonstration. 
 
 Let 5 be the surface of the sphere, and ACEB = X be a lune whose 
 angle is the spherical angle CAB, or what is the same thing, the plane 
 angle BOC measured by the arc CB, of which A is the pole. 
 
 Then is 7; = 
 
 CAB 
 
 4 right angles 
 
 For, first, suppose the arc CB commensurable 
 with the circumference BCwDra, and suppose that 
 they are to each other as 5 : 24. 
 
 Divide CB into five equal arcs, and the entire 
 circumference BCmDn into twenty-four arcs of the 
 same length, and pass arcs of great circles through 
 A and these points of division. Thus the lune is ^'9- ^29. 
 
 divided into five equal lunes, and the entire surface into twenty-four 
 equal lunes of the same size. These lunes are equal to each other (727). 
 
 Hence, 
 Now, 
 
 COB 
 
 4 right angles 
 
 24" 
 
 CB 
 
 BCmDw. 
 
 _5^ 
 24' 
 
 Therefore 
 
 L _ COB (or C AB) 
 
 8 ~ ^ right angles 
 
 % E. D, 
 
320 ELEMENTARY GEOMETRY. 
 
 • If the angle of the lune is incommensurable with four right angles, 
 or, what is the same thing, if the arc BC is not commensurable with the 
 circumference, let us assume 
 
 - = -?^ (1) 
 
 in which BL < BC. 
 
 Conceive the circumference BCmDTi divided 
 into equal parts, each of which is less than CL, 
 the assumed difference between BC and BL. 
 Then conceive one of these equal parts applied 
 to BC as a measure, beginning at B. Since the 
 measure is less than LC, one point of division, at ^'9' ^^°' 
 
 least, will fall between L and C. Let I be such a point, and pass the arc 
 of a great circle through A and I. 
 
 XT lune AIEB Bl 
 
 8 BCmDri ^ ' 
 
 since the arc Bl is commensurable with the circumference. In (1) and 
 (2), the consequents being equal, the antecedents should be proportional ; 
 
 L BL 
 
 hence we should have 
 
 lune AIEB Bl 
 
 But this is absurd, since lune ACEB > lune AIEB, whereas BL < Bl, 
 that is, an improper fraction equals a proper fraction. 
 
 In a similar manner, we may reduce the assumption to an absurdity, 
 if we assume BL > BC. 
 
 L BC 
 
 Hence, as the ratio of -^ can neither be greater nor less than , 
 
 it is equal thereto, and 
 
 L BC BOC 
 
 8 ' EQrnXin 4 right angles 
 
 Q. E. D. 
 
 ^ 
 
 729. Scholium. — To obtain the ourea of a lune whose angle is known, 
 find the area of the sphere, and multiply it by the ratio of the angle of 
 the lune (in degrees) to 360°. Thus, B being the radius of the sphere, 
 4i7rB^ is the surface of the sphere ; and the lune whose angle is 30° is ^jj 
 or Y^2^ the surface of the sphere, i. e., ^^ oi A^nB^ = ^tzB^. 
 
 730. Corollary. — The sum of several lunes on the same 
 sphere is equal to a lune whose angle is the sum of the 
 angles of the lunes ; and the difference of two lunes is a 
 lune whose angle is the difference of their angles. 
 
OF THE SPHERE. 
 
 321 
 
 731. Corollary. — Ungulas hear the same ratio to the 
 volume of the sphere that the corresponding lunes do to 
 the area of the surface. 
 
 PROPOSITION XXXIII. 
 
 732. Theorem. — If two semi-circumferences of great 
 circles intersect on the surface of a hemisphere, the sum 
 of the two opposite triangles thus formed is equivalent to 
 a lune whose angle is that included by the semi-circum- 
 ferences. 
 
 Demonstration. 
 
 Let the semi-circumferences CEB and DEA intersect at E on the sur- 
 face of the hemisphere whose base is CABD. 
 
 Then the sum of the triangles CED and 
 AEB is equivalent to a lune whose angle is 
 AEB. 
 
 For, let the semi-circumferences CEB and 
 DEA be produced around the sphere, inter- 
 secting on the opposite hemisphere, at the 
 extremity F of the diameter through E. 
 
 Now, FBEA is a lune whose angle is AEB. 
 
 Moreover, the triangle AFB is equivalent 
 to the triangle DEC ; since F'g- 33'- 
 
 angle AFB = AEB = DEC, 
 
 side AF = side ED, 
 
 each being the supplement of AE ; and 
 
 BF = CE, 
 
 each being the supplement of EB. 
 
 Hence, the sum of the triangles CED and AEB is equivalent to the 
 lune FBEA. Q. e. d. 
 
ELEMENTARY GEOMETRY. 
 
 PROPOSITION XXXIV. 
 
 733. Theorem. — The area of a spherical triangle is to 
 the area of the surface of the hemisphere on which it is 
 situated, as its spherical excess is to four right angles, or 
 360°. 
 
 Demonstration. 
 
 Let ABC be a spherical triangle whose 
 angles are represented by A, B, and C; let T 
 represent the area of the triangle, and H the 
 area of the surface of the hemisphere. 
 
 T _ A + B + C-180° 
 R ~ 360° 
 
 Then is 
 
 Let lune A represent the lune whose angle is 
 the angle A of the triangle, i. e., angle CAB, and 
 in like manner understand lune B and lune C. 
 
 Fig. 332. 
 
 Now, triangle AHG + AED = lune A (732), 
 
 BHI + BEF = lune B, 
 
 CGF + CDI = lune C. 
 
 Adding, 3ABC + hemisphere = lune (A + B + C)* 
 
 (1) 
 
 by (730), and since the six triangles AHG, AED, BHI, BEF, CGF, and CDI 
 
 make the whole hemisphere and 2ABC besides, ABC being reckoned three 
 times 
 
 From (1) we have, by transposing, and remembering that a hemi- 
 sphere is a lune whose angle is 180° (730), and dividing by 2, 
 
 ABC = i lune (A + B + C - 180°).t 
 
 But, by (728), 
 
 
 
 
 
 
 
 i lune (A 
 
 + B + 
 H 
 
 C- 
 
 - 180°) 
 
 A + B 
 
 + C- 
 
 360° 
 
 180° 
 
 Therefore, 
 
 T 
 
 H ~ 
 
 . A 
 
 + B + C 
 • 360^ 
 
 - 180° 
 
 Q. E, 
 
 D. 
 
 * This signifies the lun<^whose angle is A -f B + C, which is of course 
 the sum of the three lunes whose angles are A, B. and C. 
 
 f This signifies one-half the lune whose angle is A + B + C— 180°. 
 
OF THE SPHERE. 323 
 
 734. Scholium 1. — To find the area of a spherical triangle on a given 
 sphere, the angles of the triangle being given, we have simply to multiply 
 the area of the hemisphere, *. e. , 2n-i2^, by the ratio of the spherical excess 
 to 360". Thus, if the angles are 
 
 A = 110°, B = 80°, and C = 50°, j s^ 
 
 we have ^"^ 
 
 area ABC = a.if' x ^^L?4^=J^1 = 2riP x ^ = ^nS?. 
 
 735. Scholium 3. — This proposition is often stated thus: Tlie area 
 of a spherical triangle is equal to its spherical excess multiplied by the trirec- 
 tangular triangle. When so stated, the spherical excess is to be estimated 
 in terms of the right angle ; i. e., having subtracted 180^ from the sum of 
 its angles, we are to divide the remainder by 90°, thus getting the spheri- 
 cal excess in right angles. In the example in the preceding scholium, 
 the spherical excess estimated in this way would be 
 
 110° + 80^ + 50^ - 18 0° _ 2 
 90° ~3 
 
 and the area of the triangle would be | of the trirectangular triangle. 
 Now, the trirectangular triangle being |- of the surface of the sphere (?), 
 is ^ of 47ri?', or ^7ri^^ This multiplied by | gives ^Tri?*, the same as 
 above. 
 
 The proportion 
 
 ABC ^ A + B + C - 180° 
 
 surf, of hemisph. 360' ' 
 
 is readily put into a form which agrees with the enunciation as given in 
 this scholium. Thus, 
 
 surf, of hemisph. = 27ri?» ; 
 whence, 
 
 ARP -c)^m ^ A + B + C-180^ , _- . A+B + C-180° 
 ABC - 2^i? X ^— = ^.m X 
 
324 
 
 ELEMENTARY GEOMETRY. 
 
 VOLUME OF SPHERE. 
 
 PROPOSITION XXXV. 
 
 736. Theorem. — The volume of a sphere is equal to 
 the area of its surface multiplied by one-third of the ra- 
 dius, that is, f 7Ti23j R being the radius. 
 
 Demonstration. 
 
 Let OL = i^ be the radius of a sphere. 
 
 Conceive a circumscribed cube, that is, a 
 cube wliose faces are tangent planes to the 
 sphere. Draw lines from the vertices of each of 
 the polyedral angles of the cube to the centre 
 of the sphere, as AO, BO, DO, CO, etc. These 
 lines are the edges of six pyramids, having for 
 their bases the faces of the cube, and for a com- 
 mon altitude the radius of the sphere (?). Hence ^^' 
 the volume of the circumscribed cube is equal to its surface multiplied 
 by ^B. 
 
 Again, conceive each of the triedral angles of the cube truncated by 
 planes tangent to the sphere. A new circumscribed solid will thus be 
 formed, whose volume will be nearer that of the sphere than is that of the 
 circumscribed cube. Let a5c represent one of the tangent planes. Draw 
 from the polyedral angles of this new solid, lines to the centre of the 
 sphere, as^^O. 50, and cO, etc. ; these lines will form the edges of a set of 
 pyramids whose bases constitute the surface of the solid, and whose com- 
 mon altitude is the radius of the sphere (?). Hence the volume of this 
 solid is equal to the product of its surface (the sum of the bases of the 
 pyramids) into \R. 
 
 Now, this process of truncating the angles by tangent planes may be 
 conceived as continued indefinitely ; and, to whatever extent it is carried, 
 it will always be true that the volume of the solid is equal to its surface 
 multiplied by ^R. Therefore, as the sphere is the limit of this circum- 
 scribed solid, we have the volume of the sphere equal to the surface of 
 the sphere, which is 47ri2' multiplied by |i2, i. e., to ^nW. q,. e. d. 
 
OF THE SPHERE. 325 
 
 737. CoEOLLARY. — The surface of the sphere may he 
 conceived as consisting of an infinite nuinher of infinitely 
 small plane faces, and the volume as composed of an infi- 
 nite number of pyramids having these faces for their 
 bases, and their vertices at the centre of the sphere, the 
 common altitude of the pyramids being the radius of the 
 sphere. 
 
 738. A Spherical Sector is a portion of a sphere gener- 
 ated by the revolution of a circular sector about the diameter 
 around which the semicircle which generates the sphere is con- 
 ceived to revolve. It has a zone for its base ; and it may have 
 as its other surfaces one or two conical surfaces, or one conical 
 and one plane surface. 
 
 Illustration. — Thus, let ab be the diam- 
 eter around which the semicircle aEl revolves 
 to generate the sphere. The solid generated 
 by the circular sector AOB will be a spherical 
 sector ha^^ng the zone generated by AB for its 
 base; and for its other surface, the conical sur- 
 face generated by AO. The spherical sector 
 generated by COD has the zone generated by 
 CD for its base; and for its other surfaces, 
 the concave conical surface generated by DO, 
 and the convex conical surface generated by CO. The spherical sector 
 generated by EOF has the zone generated by EF for its base, the plane 
 generated by EO for one surface, and the concave conical surface gener- 
 ated by FO for the other. ^ ' ., 
 
 739. A Spherical Segrment is a portion of the sphere 
 
 included by two parallel planes, it being understood that one of 
 the planes may become a tangent plane. In the latter case, the 
 segment has but one base ; in other cases, it has two. A spheri- 
 cal segment is bounded by a zone and one, or two, plane surfaces. 
 
326 
 
 ELEMENTARY GEOMETRY. 
 
 PROPOSITION XXXVI. 
 
 740. Theorem. — The volume of a spherical sector is 
 equal to the product of the zone which forms its base into 
 one-third the radius of the sphere. 
 
 Demonstration. 
 
 A spherical sector, like the sphere itself, may be conceived as consist- 
 ing of an infinite number of pyramids whose bases make up the base of the 
 sector, and whose common altitude is the radius of the sphere. Hence, 
 the volume of the sector is equal to the sum of the bases of these pyra- 
 mids, that is, the surface of the sector, multiplied by one-third their com- 
 mon altitude, which is one-third the radius of the sphere, q. e. d. 
 
 741. Corollary. — The volumes of spherical sectors of 
 the same sphere, or of equal spheres, are to each other as 
 the zones which form their bases ; and, since these zones are 
 to each other as their altitudes (723), the sectors are to each 
 other as the altitudes of tJie zones luhich form> their bases. 
 
 PROPOSITION XXXVII. 
 
 742. Theorem.— 17^6 volume of a spherical segment 
 of one base is ttA^{R — \A), A being the altitude of the 
 segment, and R the radius of the sphere. 
 
 Demonstration. 
 Let AG = JB, and CD = A. 
 
 Then is the volume of the spherical segment 
 generated by the revolution of ACD about CO 
 equal to ttA^ {R — ^A). 
 
 For, the volume of the spherical sector gener- 
 ated by AOC i^ the zone generated by AC, multi- 
 plied by \B, or ^TTAIty.^R = ^t:AR\ From 
 this we must subtract the cone, the radius of 
 whose base is AD, and whose altitude is DO. 
 
 Fig. 335. 
 
OF THE SPHERE, 
 
 To obtain this, we have 
 
 DO = R-A\ 
 
 whence, from the right-angled triangle ADO, 
 
 327 
 
 AD = V^ -{E- Af = ^2AB - A\ 
 
 Now, the volume of this cone is ^OD x ttAD^ or 
 
 ^TT (B — A) i2AB - A"") = ^n {2AR' — 3A'B + A^). 
 
 Subtracting this from the volume of the spherical sector, we have 
 ^kAB^ - ^TT (2^if - dA^'B + A') = n (A'R - ^A^ 
 
 = nA^B-^A). Q. E. D. 
 
 743. Scholium. — The volume of a spherical 
 segment with two bases is readily obtained by 
 taking the difference between two segments of 
 one base each. Thus, to obtain the volumes of 
 the segment generated by the revolution of JCAc 
 about aO, take the difference of the segments 
 whose altitudes are ac and ab. 
 
 Fifl. 336. 
 
 SPHERICAL POLYGONS AND SPHERI- 
 CAL PYRAMIDS. 
 
 744. A Spherical Polygon is a portion of the surface of 
 a sphere bounded by several arcs of great circles. 
 
 745. The Diagonal of a spherical polygon is an arc of a 
 great circle joining any two non-adjacent vertices. 
 
 746. A Spherical Pyramid is a portion of a sphere hav- 
 ing for its base a spherical polygon, and for its lateral faces the 
 circular sectors formed by joining the vertices of the polygon 
 with the centre of the sphere. 
 
328 ELEMENTARY GEOMETRY. 
 
 747. The elementary properties of spherical polygons and 
 spherical pyramids are so readily deduced from the corresponding 
 properties of polyedral angles, spherical triangles, etc., that we 
 leave them for the pupil to demonstrate, merely stating a few 
 fundamental theorems. 
 
 748. Theorem.— ^6 angles of a spherical polygon 
 and its sides sustain the same general relations to each 
 other as the diedral and facial angles of a polyedral 
 angle having for its edges the radii of the sphere drawn 
 to the vertices of the polygon. 
 
 749. Theorem. — T^^e sum of the sides of a convex 
 spherical polygon may be anything between 0° and 360°. 
 
 760. Theorem. — The sum of the angles of a spherical 
 polygon may be anything between ^n — 4 and Qn — 12 
 right angles, n being the number of sides. 
 
 751. The Spherical Excess of a spherical polygon is the 
 excess of the sum of its angles over the sum of the angles of a 
 plane polygon of the same number of sides. 
 
 752. Theorem. — The spherical excess of a spherical 
 polygon of n sides, the sum of whose angles is S, is 
 
 ^+ 360° -71.180°. 
 
 753. Theorem. — The area of a spherical polygon is to 
 the area of the surface of the hemisphere on which it is 
 situated as its spherical excess is to four right angles. 
 
 754. Theorem.— IT^e volume of a spherical pyramid 
 is the area of its base multiplied by one-third the radius 
 of the sphere on which it is situated. 
 
EXERCISES, 329 
 
 EXERCISES. 
 
 755. 1. What is the circumference of a small circle of a sphere 
 whose diameter is 10, the circle being at 3 from the centre ? 
 
 Ans. 25.1328. 
 
 2. Construct on the spherical blackboard a spherical angle of 
 60°. Of 45°. Of 90°. Of 120°. Of 250°. 
 
 Suggestions.— Let P be the point where the vertex of the required 
 angle is to be situated. With a quadrant strike an arc passing through 
 P, which shall represent one side of the required angle. From P as a 
 pole, with a quadrant strike an arc from the side before drawn, which 
 shall measure the required angle. On this last arc lay off from the firet 
 side the measure of the required angle,* as 60°, 45% etc. Through the 
 extremity of this arc and P pass a great circle (?). 
 
 3. On the spherical blackboard construct a spherical triangle 
 ABC, having AB = 100°, AC = 80°, and A = 58°. 
 
 4. Construct as above a spherical triangle ABC, having AB = 
 75°, A =. 110°, and B = 87°. 
 
 5. Construct as above, having AB = 150°, BC = 80°, and 
 AC = 100°. Also having AB = 160°, AC = 50°, and CB = 85°. 
 
 6. Construct as above, having A = 52°, AC = 47°, and CB 
 = 40°. 
 
 Suggestions. — Construct the angle A as before taught, and lay off 
 AC from A equal to 47'', with the tape. This determines the vertex C. 
 From C, as a pole, with an arc of 40°, describe an arc of a small circle; 
 in this case this arc will cut the opposite side of the angle A in two 
 places. Call these points B and B'. Pass circumferences of great circles 
 through C, and B, and B'. There are two triangles, ACB and ACB'. 
 
 7. Construct on the spherical blackboard a spherical triangle 
 ABC, having A = 59°, AC = 120°, and AB = 88°. 
 
 ♦ For this purpose, a tape equal in length to a semi-circumference of a 
 jrreat circle of the sphere used, and marked off into 180 equal parts, will be 
 convenient. A strip of paper may be used. 
 
330 ELEMENTARY GEOMETRY. 
 
 8. Construct a triangle whose angles are 160°, 150°, and 140°. 
 
 9. Can there be a spherical triangle whose angles are 85°, 
 120°, and 150° ? Try to construct such a triangle by first con- 
 structing its polar. 
 
 10. What is the area of a spherical triangle on the surface of a 
 sphere whose radius is 10, the angles of the triangle being 85°, 
 120°, and 110° ? Ans. 235.6-j-. 
 
 11. What is the area of a spherical triangle on a sphere whose 
 diameter is 12, the angles of the triangle being 82"", 98°, and 
 100°? 
 
 12. A sphere is cut by five parallel planes at 7 from each other. 
 What are the relative areas of the zones ? What of the segments ? 
 
 13. Considering the earth as a sphere, its radius would be 
 3958 miles, and the altitudes of the zones. North torrid = 1578, 
 North temperate = 2052, and North frigid = 328 miles. What 
 are the relative areas of the several zones ? 
 
 Suggestion. — The student should be careful to discriminate between 
 the width of a zone and its altitude. The altitudes are found from their 
 widths, as usually given in degrees, by means of Trigonometry. 
 
 14. The earth being regarded as a sphere whose radius is 
 3958 miles, what is the area of a spherical triangle on its surface, 
 the angles being 120°, 130°, and 150° ? What is the area of a 
 trirectangular triangle on the earth's surface ? 
 
 15. In the spherical triangle ABC, given A = 58°, B = 67°, 
 and AC = 81° ; what can you afiirm of the polar triangle ? 
 
 16. What is the volume of a globe which is 2 feet in diameter ? 
 What of a segment of the same globe included by two parallel 
 planes, one at 3 and the other at 9 inches from the centre, the 
 centre of the sphere being without the segment? What if the 
 centre is within the segment? 
 
 17. Compare the convex surfaces of a sphere and its circum- 
 scribed cylinder. 
 
 18. Compare the volumes of a sphere and its circumscribed 
 cube, cylinder, and cone, the vertical angle of the cone being 60°. 
 
 19. If a and b represent the distances from the centre of a 
 sphere whose radius is r, to the bases of a spherical segment, show 
 that the volume of the segment is tt [r^ (b — a) — \{h^ — a^)]. 
 
THE INFINITESIMAL METHOD. 
 
 The author is a firm believer in both the logical soundness and the 
 practical advantages of the strict infinitesimal method. Hence he has intro- 
 duced it — though generally as an alternative method — in those cases in 
 which the incommensurability of geometrical magnitudes by a finite unit 
 makes the old demonstrations cumbrous. 
 
 As to the logical soundness of the method, he has not the shadow of a 
 doubt. The well-known logical principle, that, if we create a certain cate- 
 gory of concepts, under certain definite laws, use them in our argument in 
 accordance with these laws, and finally eliminate them, the argument being 
 conducted according to correct logical principles, the final results are 
 correct, covers the entire case. Now the two essential laws of infinitesimals 
 are, (1) Infinitesimals of the same order have the same relations among 
 themselves as finite quantities ; and (2) Infinitesimals in comparison with 
 finites, are zero. 
 
 But the simple exposition given in the text (340-342) is quite adequate 
 to show that the method can introduce no conceivable error. Thus, if 
 
 — — <r, all the quantities being finite, and if i is an infinitesimal, = a 
 
 must be true, and i must be in the relation. Otherwise solving the equa- 
 
 tion we have i = an—m, a finite quantity, unless a — —- 
 
 Of the immense practical utility of the method there can be no question. 
 All, from Lagrange down, have acknowledged it. I know of no extended 
 treatise which does not in some way imply it. Why, then, should not the 
 pupil become familiar with it early in his course ? 
 
 As to the method of limits it is not at all difficult to show that it is 
 identical with the infinitesimal method, in its fundamental principles. 
 Moreover, there is a sort of jugglery in the very first step in the method of 
 limits which quite transcends any difficulty that the method of infinitesimals 
 presents. Thus, we give the variable an increment, assume that the func- 
 tion takes a related increment, manipulate the function, and then make the 
 increment of the variable zero (whence the increment of the function 
 becomes zero), and, presto, we have a finite relation between two zeros! 
 And this is the " simple " fundamental conception which the tyro is sup- 
 posed to see at a glance ! 
 
332 APPENDIX. 
 
 NOTE ON (182), (343), (587), (628), AND (728). 
 
 These propositions are of a class in wliich the incommensurability by a 
 finite unit of certain lines introduces particular diflficulty, which difficulty 
 disappears at once if we admit, as in the infinitesimal theory, that these lines 
 are commensurable by an infinitesimal unit. Also, by the introduction of 
 the principle of the generation of one magnitude by the motion of another, 
 very simple demonstrations are afforded. 
 
 In the text the author has given illustrations of the three sorts of 
 demonstrations. In (182) we have the old method of avoiding the difficulty 
 which grows out of the incommensurability, by the reductio absurdum. 
 The objection to this is not any objection to the reductio absurdum as a 
 method of reasoning. But why use so cumbrous a method, when other 
 exceedingly simple methods are at hand, and methods involving principles 
 so necessary to subsequent use ? 
 
 In (728) tlie three methods are given. In (343) and (628) the methods 
 involving generation by motion, and the infinitesimal method, are given. 
 
 NOTE ON (182). 
 
 1. To prove this proposition by means of the conception of the genera- 
 tion of magnitudes by the motion of other magnitudes, we do not need the 
 
 AOB 
 
 Lemma. Thus, referring to Fig. 85, p. 89, we are to prove that — - = 
 
 arc AB 
 arc DE 
 
 Let the sector AOB be applied to DOE. OA being placed in OD. By 
 reason of the equality of the circles the arc AB will fall in DE. 
 
 Conceive the angles AOB and DOE as generated by a radius mov- 
 ing from the position -OA (which is now also OD) to OB and OE, 
 with uniform motion. Let the time of generating AOB be r, and that of 
 
 AOB r 
 
 generating DOE be s. Whence —^ = - (48, 49). 
 
 Again, the extremity of the radius, as A (or D), describes equal and (as 
 far as the less extends) coincident parts of AB and DE in equal times, whence 
 
 arcA B _ r „ , ,... _. __.: ,,„_ AOB _ arc AB 
 
 arc DE 
 
 Hence, by equality of ratios, we have pQ£ = a7c qE 
 
 2. To prove the same proposition by the infinitesimal method, we pro 
 ceed exactly as in Case II., pp. 89, 90. simply conceiving vi as infinitesimal 
 when the angles are incommensurable by a finite unit, and for 5 putting the 
 indefinite number r, and for 8 the indefinite number s. 
 
INFINITESIMAL METHOD. 333 
 
 NOTE ON (343). 
 
 By the old method the Lemma on which this demonstration is based is 
 proved in two cases. 1st. When the bases are commensurable ; 2nd. When 
 the bases are incommensurable. Dividing the bases into equal parts and 
 erecting perpendiculars at the points of division the argument in the first 
 case proceeds exactly like the argument in Case II. of (182). When the 
 bases are incommensurable, we apply abed to ABCD placin«ir ad in its 
 equal AD, whence ab falls in AB, as far as it extends, and dc in DC. Then 
 
 ^, ^ .- ABCD . ^ , AB .^ . , AB , . .^, 
 
 assume that, if — ;; — ~ is not equal to -^r , it is equal to — , ag being either 
 abed ^ ab ag ^ ° 
 
 greater or less than ab. Now divide the base AB into equal parts, each of 
 
 which is less than bg, and erect perpendiculars at each of the points of 
 
 division. We may then show, as in Case III. of (182), the absurdity of 
 
 supposing ag greater or less than ab. 
 
 FINIS. 
 
'i^-A^ 
 
 '^CX. 
 
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