$B 527 T^b LIBRARY OF THE University of California. GIFT OF ^. CL-...- V^(?r^^AAX^rinteb at TTbe Eclectic l^reee (Cincinnati. TIL S. a PREFACE. In this treatise, an endeavor is made to present the elements of geometry with a logical strictness approaching that of Euclid, while taking advantage of such improvements in arrangement and nota- tion as are suggested by modern experience. It has been carefully kept in mind that the purpose of such a work is only in a secondary degree the presentation of a system of useful knowledge. A much more important purpose is to afford those who study this subject the only course of strict reasoning with which the great majority of them will ever become closely acquainted. A mind that, by exer- cise in following and weighing examples of strict logical deduction, has learned to appreciate sound reasoning, and, by practice on suitable exercises, has been trained to reason out a sound logical deduction for itself, has gained what is of far greater importance than mere knowledge ; it has gained power. A treatise on rational geometry ought, accordingly, to have for guiding principles those laid down by Pascal as the chief laws of demonstration, substan- tially as follows : to leave no obscure terms undefined ; to assume nothing not perfectly evident ; to prove everything at all doubtful, by reference to admitted principles. In accordance with the first principle, great care has been taken in the wording of the definitions. In the case of some terms, such as straight line and angle, for which no definitions quite free from objection have as yet been proposed, those adopted have been chosen, not as theoretically perfect, but as best suited to the com- prehension of the beginner, and most available in deducing the properties of the things defined. The use of hypothetical constructions has been abandoned for several reasons. To assume them silently, as is now usually done, is unwarrantable in a treatise upon a science supposed, above all others, to consist of a series of rigorous deductions from admitted truths. Why state so carefully that we must assume the possibility 3 183969 4 PREFACE. of drawing or prolonging a straight line, and say nothing in regard to constructions so much less obvious? The author's experience in teaching geometry has convinced him that these stealthy assump- tions are decidedly adverse to the acquisition, on the part of the learner, of those habits of strict reasoning which it is the main object of geometrical study to impart. The teacher should have it in his power to inquire, at every step, not only why such a state- ment is true, but also why such a construction is allowable. On the other hand, the insertion of the few problems really needed as auxiliaries in the demonstration of theorems, imparts to their sequence a logical consistency that cannot obtain where the learner is continually required to perform operations, the possibility of which has neither been proved nor formally assumed. In regard to the use of circumferences as construction lines in the first book, it may be remarked that, of all lines, the circum- ference is the easiest to define, to construct, and to conceive, — far more so than is the case with the straight line. No property of the circle, not even its name, is introduced in the first book, but only that property of the circumference which is given in its definition, and some immediate consequences from that property. The employment, at the outset, of the simpler of the two lines treated of in elementary geometry would hardly require apology or explanation but for the force of custom. Yet, strangely enough, in treatises that scrupulously defer the definition of the terms cir- cumference and radius to a subsequent book, there seems to be no scruple against the employment of arcs in illustration of the nature of angles, and we see gravely laid down the postulate : A circum- ference may be described about any point as center, etc. The deviation in this work from the usual order of propositions is comparatively slight. In the early part of the first book, so important as the foundation of the science, the properties of tri- angles are introduced immediately after the discussion of the general properties of angles. This arrangement, especially as regards the different cases of equal triangles, presents several advantages : these propositions are immediately deducible from first principles, or from each other ; they are easily grasped by the beginner ; above all, they are of the highest utility as aids to further acquisition. It is in itself no slight advantage for the learner to become accustomed, from the first, to the use of these important auxiliaries in demon- PREFACE. 5 stration. From the second book, again, certain propositions that treat of proportional angles have been removed to the place where they belong, after the discussion of ratio and proportion. In the treatment of these subjects, while adhering to the now prevalent method, an endeavor has been made to obviate one frequent source of confusion by making a clear distinction between concrete quan- tities and their numerical measures. In regard to propositions and corollaries, the rule observed has been to admit only such as are important in themselves, or have a bearing on subsequent demonstrations and studies. In this era of over-crowded curricula, the aim of an elementary text-book should be to present the necessary rather than the novel or merely interesting. For this reason some subjects have been relegated to an appendix, where they may be studied or omitted according to circumstances.* The exercises have been carefully selected with a view to their bearing upon important principles, and are, with few exceptions, of such slight difficulty as not to discourage the learner of average ability. In the first sets of exercises, ample assistance is afforded the student by means of references and diagrams, — aids that are withheld from the point where the student should have learned to help himself. It is by no means expected that the average class will find time for all the exercises, but enough are given to afford the teacher full opportunity of choice. Suggestions. It is earnestly recommended that, before any book work is assigned to a beginning class, a lesson be devoted to the constructions given in Arts. 200-206. The teacher, having shown on the blackboard, for example, how to bisect a straight line, should set the class to doing the same, and require each pupil to bring in, next day, one or more neatly worked examples of the required constructions. The practical familiarity thus gained with the geometrical concepts involved will amply repay the time thus spent. The two great sources of difficulty to the beginner in geometry are the comparative novelty of the subject matter and the unaccustomed clearness of conception and exactness of expres- * The whole of the last part of Book IX., treating of spherical angles and polygons, may as well be omitted by pupils not to take up spherical trigonometry. 6 PREFACE. sion required in this new study ; it will be found that the second source of difficulty is most easily diminished by reducing the first to a minimum. The easy exercises at the foot of the page can be employed to the best advantage as material for impromptu work, the teacher giving as much aid, by suggestive questions, as will enable the class to solve them oifhand. Such as prove too difficult for this first attempt may then be assigned as work to be prepared. The questions found at the end of most books may either be taken up as the class progresses through the book, or be left for review. Such matter as that found on p. 105 is to be carefully read over in the class with a running comment by the teacher ; the gist of the ideas, not the words in which they are expressed, being what the pupil should try to retain. The pupil should be required to give the exact words of defini- tions, axioms, postulates, and enunciations. In other matters some latitude may be allowed, though occasion should be taken to point out in what respects the pupil's own wording may be objectionable. Outside of Euclid, it is of doubtful utility to exact from pupils a knowledge of all propositions by number. The axioms, postu- lates, and certain important propositions should be known by number, but in written or oral work, other references may be given by abbreviated quotations. CONTENTS. PAET I. PLANE GEOMETEY. INTRODUCTION. DEFINITIONS. PAGE Lines 13 Angles 15 Propositions 18 Axioms 19 Postulates .19 Method of Proof 20 BOOK I. LINES AND RECTILINEAR FIGURES. Angles 23 Triangles 28 Perpendiculars 46 Parallels 51 Quadrilaterals 61 Exercises 69 Geometrical Synthesis and Analysis 71 Exercises 74 BOOK II. THE CIRCLE. LOCI. PROBLEMS. Elementary Properties 78 Chords 81 Tangent and Secants 90 Exercises 93 Constructions .......... 95 Plane Loci 102 Exercises . 103 Analysis of Problems ........ 105 Exercises 107 7 8 CONTENTS. BOOK III. RATIO. PROPORTION. LIMITS. PAGE Measurement Ill Ratio 113 Proportion. Definitions ....... 114 Exercises 116 Limits 117 Theorems in Proportion . . 119 BOOK IV. PROPORTIONAL ANGLES AND LINES. SIMILAR POLYGONS. Proportional Angles 128 Exercises 136 Proportional Lines 138 Similar Polygons 144 Ratios of Certain Lines 151 Exercises 155 Constructions 157 Exercises 160 BOOK V. AREAS AND THEIR COMPARISON. Quadrilaterals 165 Exercises . . . . 186 Constructions . . . . 187 Numerical Applications 191 Exercises 194 BOOK VI. REGULAR PLANE FIGURES. Regular Polygons 198 Area op the Circle 208 Division of Circumference 212 Exercises 221 CONTENTS. 9 PART II. SOLID GEOMETRY. BOOK VII. PLANES AND POLYHEDKAL ANGLES. PAGE Planes and Perpendiculars . . . . . . . 225 Planes and Parallels 234 Dihedral Angles ......... 240 Polyhedral Angles . . 261 Exercises . 258 BOOK VIII. POLYHEDRONS. Prisms 262 Pyramids . . . . . ... . . . 275 The Regular Polyhedrons 286 Exercises 290 BOOK IX. THE THREE ROUND BODIES. Cylinders 293 Cones 297 Spheres . 300 Spherical Angles and Polygons ...... 309 Polar Triangles 313 Exercises . 330 BOOK X. MEASUREMENT OF THE THREE ROUND BODIES. Cylinders. 332 Cones 336 Exercises 342 Spheres 344 Exercises 352 APPENDIX. Symmetry 355 Symmetrical Polyhedrons 360 Maxima and Minima 362 INDEX 371 SYMBOLS. ••• because or since. .'. therefore. + plus. — minus. = is equal to. = is equivalent to. > is greater than. < is less than. ^ coincides with. II parallel.* ± perpendicular.* O circle. Z angle. St. Z straight angle. rt. Z right angle. A triangle. rt. A right triangle. AB, line or side ab. comp. complement. supp. supplement. resp. respectively. Req. required. Q.E.D. as teas ?o be proved. Q.E.F.' as was to be done. The symbols marked above with an asterisk are to be read as nouns when preceded by an article or similar word. Plurals are formed by adding 's to the singular. Thus, A's, ll's, etc. In reading such expressions as A A'B'C, O abed, it will be suffi- cient, in almost every case, to give the distinguishing epithet to the last letter only. Thus, the above expressions may be read respec- tively : triangle ABC prime, circle abed small or minor. It is, of course, seldom necessary to say the line or the side AB ; it is usually sufficient to say AB. 10 PART I. PLANE GEOMETRY. >i*;c INTRODUCTION. // We conceive space as extending without limit in every direction. Each physical body, a marble cube, for example, occupies a limited portion of space. Thus the cube AG is limited on all sides by bound- aries which are called faces; as AFj EG, etc. These faces, again, meet in edges, which are called lines; as ^5, i^G^, etc. ; and final- ly, these lines meet in extremi- ties, which are called points; as A, F, H, etc. If, now, leav- ing its material entirely out of consideration, we think only of the portion of space occupied by the cube, with its faces, edges, and points, we have before us a so-called geometrical solid. It is with such abstract solids as this that geometry is concerned. / / / F V DEFINITIONS. 1. A solid is a limited portion of space, and has three dimensions : length, breadth, and thickness. 2. Surfaces are the limiting boundaries of solids, and have two dimensions : length and breadth. n 12 PLANE GEOMETRY. 3. Lines are the boundaries or intersections of surfaces, and have but one dimension : length. 4. Points are the extremities or intersections of lines ; hence a point has no dimension, but only position. Just as we conceive a geometrical solid apart from mate- rial, so we may conceive of surfaces apart from solids, of lines apart from surfaces, and of points apart from lines. These abstract points, lines, surfaces, and solids are the so-called space-concepts, the elements of our notions of space. 5. Geometry is the science that treats of the proper- ties and relations of space-concepts, also called geometrical concepts. Each geometrical concept has position, determined by the location of its points in space ; and all except points have form or shape, determined by the relative position of their points, and extent or magnitude, determined by the nearness or remoteness of their bounding points. 6. A straight line is a line that does not change its direc- tion at any point ; as AB. 7. A Iroken line, as J 5 (7 or ^ J5 CDE, changes direction at one or more points. 8. A curved line, as MN, changes direction at every point ; i.e., no part of it is a straight line. The curve is said to be closed when it forms a continuous boundary. Thus P and Q are closed curves. CZ3 The term line, when used hereafter by itself, is to be understood as denoting a straight line; similarly, curve will signify curved line. INTRODUCTION. 13 9. A plane surface or plane is such that a straight line passing through any two points in that surface lies wholly in the surface. 10. A geometrical figure is any given combination of points, lines, or surfaces ; the representation of a figure is a diagram. Just as a diagram is the representation of a figure, so the lines of the diagram and the surface on which it is drawn are merely more or less imperfect representations of the ideally perfect lines and planes to be treated of. 11. K plane figure is such that all its points lie in the same plane. 12. Plane geometry is the geometry of plane figures. 13. A magnitude is a concept any part of which is of the same kind as the whole. Thus since any part of a line is a line ; any part of a surface, a surface ; any part of a solid, a solid, — lines, sur- faces, solids, and angles, presently to be defined, are geometrical magnitudes. Other magnitudes are time, weight, mass, etc. 14. Equal magnitudes are such as can be made to coin- cide exactly. Magnitudes coincide exactly when, one being placed upon the other, every point of the one lies upon a corresponding point of the other. LINES. From the definition of the straight line {fo),* it follows : — 15. Straight lines that coincide in part coincide throughout. 16. Two points determine a straight line. 17. Two straight lines cannot inclose a surface. * Numbers in parentheses throughout the book refer in general to paragraphs. 14 PLANE GEOMETRY. For if two lines ABC, ABD, coincide in a part AB only, or in two points A and B only, they cannot both be straight lines (6), since one, at least, must change direction. 18. Since the direction of one point from another may be regarded as the path of a point that passes along the ^ ^ straight line that connects these points, and as, if A and B are the points, the direction may evidently be either from A to B or from B to A, the same straight line may mark either of two opposite directions, or it may be regarded as extending in both directions from any point in it. 19. A line is said to be indefinitely produced when pro- longed as far as necessary, or without assigned limit. If two straight lines AB, CD, lie in the same plane, it is evident that, if indefinitely produced, they must either meet or not meet. Thus : — (a) If they can be produced so as to have more than one common point, they lie in the same line and have ^ the same directions. (6) If they can have hut one common point, they can intersect in that point, and they have dif- ferent directions. A B (c) If they can have no common point, — that is, cannot meet, — they are said to be parallel. INTRODUCTION. 15 20. If the straight lines drawn from A io B and from z> to E are equal, the distance from A to B is said to be equal to that from B to E. If the distances from ^ to ^ and C are equal, A is said to be equidistant from B and C, while 5 and C are said to be equally distant from ^. 21. A circumference is a closed curve described in a plane, and is such that all its points are equally distant from a point within the curve, called the center. 22. A radius is any straight line drawn from center to circumference. Thus OA, OB, OC, are radii of the circumference ACB, whose center is 0. 23. All radii to the same circumference are equal. It is also obvious that a point in its plane lies within or icithout a given circumference, according as its dis- tance from the center is less or greater than the radius. ANGLES. 24. A plane angle is the opening or difference of direction between two lines that meet or might meet. The point of meeting is called the vertex; and the lines, the sides of the angle. Thus the lines AB, AC, meeting in the vertex A, are said to form or include the angle called BAC, CAB, or sim- ply the angle A when no other angle has the same vertex. Even when sev- eral angles have a common vertex, it is sometimes convenient to designate each by a number or letter placed within, near the vertex. Thus the angles MAR, RAP, PAN, may be more briefly designated as the angles 1, 2, and A, respectively. 16 PLANE GEOMETRY. 25. Angles are equal if their sides can be made to coincide. Thus the angles BAC and EDF are equal if, DE being made to lie on AB, DF can also be made to lie on AC, which is possible only when the opening between DE and DF is the same as the opening between AB and AO. 26. Adjacent angles are such as have a common vertex and a common side that separates them. Thus the angles 1 and 2 in Art. 24, or 2 and A, are adjacent angles. Angles that have a common side, as D and C, but separate vertices, are sometimes called adjacent. It is better, however, to avoid pos- sible confusion by calling them including angles, as including the common side between their vertices. 27. Vertical angles are the opposite angles formed by the intersection of two straight lines. Thus the angles 1 and 2 are vertical angles, as are also the angles 3 and 4. 28. The sum of two angles is equal to the angle formed by placing them so as to be adjacent. The difference of two angles is the angle that added to the smaller angle makes an angle equal to the greater. Thus BAD is the sum of BAC and GAD ; BAC is the difference of BAD and CAD. 29. A straight angle is one whose sides lie in opposite directions from the vertex, so as to P be in a straight line. ....•'•"■' Thus 5^c is a straight angle if ..•■•"" . its sides AB, AC are in a straight ^ ^ ^ line. It may be regarded as formed by two lines drawn in opposite directions from A; or by opening out a smaller angle, BAD, until its sides lie in opposite directions; or as the INTRODUCTION. 17 sum of two adjacent angles, BAD, DAC, whose exterior sides lie in a straight line. 30. When the adjacent angles formed by one straight line meeting another are equal, each angle is called a right angle. Thus, if the angles BAC, CAD, formed by CA meeting BD in A, are equal, each of them is a right angle. d 31. A right angle is equal to one half of a straight angle. This follows from the preceding definitions ; for the two equal angles, CAB, CAD, make up the straight angle BAD. 32. An acute angle is less than a right angle ; an obtuse angle is greater than a right angle and less than a straight angle. Thus BAC is an acute, and cad an obtuse, angle. Acute angles and obtuse angles are called oblique angles; and lines are said to be j9er- dab pendicular or oblique to each other according as they meet at a right or an oblique angle. 33. If the sum of two angles is equal to a right angle, each is called the complement of the other, and the angles are said to be complementary. If the sum of two angles is equal to a straight angle, each is called the supplement of the other, and the angles are said to be supplementary. QUESTIONS. 1. In a line AB, take any point C ; then AB is the sum of what two lines ? '■ 1 ~ AC IS the difference of what two ? 2. If you fold a piece of note paper so as to form an edge, what sort of a line is formed ? 3. If you fold the paper again, so as to double that edge upon itself, what angle will the second edge thus formed make with the lirst? Geom.— 2 18 PLANE GEOMETRY. 4. If you suspend a weight by a string, in what sort of a line is the string stretched ? 5. If you whirl the weight round at the end of the string, in what sort of a line does the weight move ? 6. What sort of a surface is presented, roughly speaking, by the walls of a room ? By the surface of a floor ? By the surface of a slate ? Mention other like surfaces. 7. How would you apply a straightedge or ruler so as to ascertain whether a given surface is a plane ? What property of planes do you apply (Art. 9) ? 8. Straight lines can be drawn on the surface of a stovepipe, and yet it is not a plane : why not ? 9. Can a straight line be drawn on the surface of an eggshell? If not, what kind of a line can be drawn on such a surface ? 10. From a point draw lines OA^ OB, OC, OD, in one plane. Name each of the angles thus formed. Which are adjacent angles ? Name one that is the sum of two ; of three. 11. Can you draw two angles that have a common vertex and a common side, and yet are not adjacent ? 12. What sort of an angle is less than its supplement ? Is equal to its supplement ? Is greater than its supplement ? PROPOSITIONS. The truths of geometry are presented for consideration under the form of general statements called propositions. 34. A theorem is a proposition stating a geometrical truth. 35. A problem is a proposition stating a proposed con- struction. 36. A corollary is a theorem that follows so plainly as a consequence from a preceding proposition, or definition, that its formal proof is either omitted or is merely indicated. Thus in Arts. 15, 16, 17, are given certain important corollaries from the definition of the straight line; and in Art. 23 we have a very obvious corollary from the definitions of circumference and radius. INTRODUCTION. 19 37. An axiom is a theorem assumed as self-evident. 38. A postulate is a problem assumed as possible. 39. A scholium is a remark upon a preceding proposition. 40. The axioms and postulates, together with the defini- tions, constitute the logical basis of geometry. Axioms express certain simple truths in regard to magnitude in general, — truths so confirmed by all our experience that the mind cannot conceive their opposites as true. All the axioms except the last two, which are really definitions of the terms whole and part, might be deduced from the first. They are such obvious truths, however, that it is deemed sufficient to state them for convenience of reference. AXIOMS. 1. Magnitudes equal to the same or equal magnitudes are equal to each other. 2. If equals are added to equals, the sums are equal 3. If equals are taken from equals, the remainders are equal. 4. If equals are added to unequals, the sums are unequal. 5. If equals are taken from unequals, or unequals from equals, the remainders are unequal. 6. The doubles of equals are equal. 7. The halves of equals are equal. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. POSTULATES. Let it be assumed that, in a given plane, 1. A straight line can he drawn joining aity two given points. 2. A given straight line can be produced to any extent. 20 PLANE GEOMETRY. 3. On the greater of two straight lines a part can he laid off equal to the less. 4. A circumference can he described from any center, with any radius. 5. A figure can he moved unaltered to a new position. It is assumed in these postulates that we have at our disposal, (1) a plane extending indefinitely in all direc- tions, (2) some means of causing a marking point to move in a straight line in any given direction, (3) some means of causing a marking point to move in that plane so as always to remain at a given distance from a given point in it. The plane may be represented by a blackboard or flat piece of paper ; the second requirement is met by the use of a straightedge and marking point ; the third, by a pair of compasses, or other device. By means of Post. 5, con- taining the principle of superposition, we are enabled to apply the criterion of Art. 14 in order to prove the equality of two given magnitudes. Thus two straight lines would be proved equal by placing them so as to coincide end with end; two angles, by causing their sides to coincide; and so on. METHOD OF PR00F.=* In general, the statement and proof of a proposition con- sist of several distinct parts ; the enunciation, the construc- tion, and the demonstration. 1. The General Enunciation or statement consists of an hypothesis (or supposition) and a conclusion. Thus in Prop. I. we have, though in different words : — Hypothesis. If a line is perpendicular to a second line at a certain point, * To be read in connection with Prop. I. INTRODUCTION. ' 21 Conclusion. No other line in the same plane can be per- pendicular to the second line at that point. 2. The Particular Enunciation, again, refers us to a particular figure or figures fulfilling the given conditions. The hypothesis and conclusion of the particular enunciation will be distinguished by the headings Given, and To Prove, respectively. 3. In the Construction we apply the postulates, or problems that have been proved possible, to make such changes in the form or position of the given figures as may be needful for the demonstration. 4. In the Demonstration we deduce, by a train of reasoning, the proposition to be proved, from other propo- sitions already proved or granted. Thus, in Prop. I., we show, by means of Ax. 8 and Ax. 1 and the definition of right angles, that the angles formed by the line AE with BD must be unequal, and therefore cannot be right angles. In problems, the construction, not always a necessity in the proof of a theorem, is the essential part. Instead of the heading To Prove, however, we put the heading Required, as indicating what is required to be done. QUESTIONS. 1. What kind of a surface and what instruments are assumed as necessary in the constructions of plane geometry ? 2. How do you draw a straight line longer than your straightedge ? What property of straight lines do you apply ? 3. With what instrument do a B you lay off on a line AB a shorter line C2>? 4. Draw a line equal to the given line AB ; produce it to E, so that BE shall be equal to the line CD : AE is the sum, and BE the difference, of what lines ? 22 PLANE GEOMETRY, 5. The sum of a right angle and an acute angle is necessarily what sort of an angle ? 6. The difference of an obtuse angle and a right angle is neces- sarily what sort of an angle ? 7. In the annexed diagram, show that ^ ^ there are three different angles having the same vertex A. Name them. 8. If BA were produced through A to E, how many different angles would have the same vertex A? Name them. Kead the following expressions : 9. ZBAC+ZGAD = ABAD. 11. ABAD>/.CAD. 10. A BAD - /.B AC = ACAD. 12. A BAG < A BAD. Express in symbols each of the following statements : 13. The sum of the lines AB and BC is equal to the line ^ b C AC. »- ' ^ 14. The difference of the lines AC and BC is equal to the line AB. 15. The line AC i^ greater than the line BC. 16. The line AB is less than the line A C. 17. The sum of the angles BAD and BAG is a right angle. 18. The difference of the angles BAD and BAG is equal to the angle GAD. Book I. LINES AND RECTILINEAR FIGURES, ANGLES. Proposition I. Theorem. 41. At a given point in a straight line there can he in the same plane but one perpendicular to that line. c ,E A D F Given: CF perpendicular to 5D at ^; To Prove : No other line drawn through A can be X to BD. For any other line drawn from A must lie between FC and AB OY between FC and AD. Let it lie between FC and AD, as AE. Because ZcAB = ZcADj (Hyp.) but Z EAB > Z CAB, and Zead = angle BOC, then OA and OC are in the same straight line, as are also OB and OD; and the pairs of angles AOB, AOD, and BOC, DOC, are supplementary. 26 PLANE GEOMETRY. — BOOK I. Propositiox III. Theorem. 49. If two adjacent angles are supplementary, their exterior sides are in a straight line. Given: Two adjacent A, BAC, CAD, that are supplementary; To Prove : Their exterior sides, AB, AD, are in a straight line. Since the A BAC, CAD, are supplementary, (Hyp.) Z BAC + A CAD = a straight A, (33) .-. their sum, A BAD, is a straight A, .*. AB and AD, the sides of A BAD, are in a st. line. (29) Q.E.D. Proposition IV. Theorem. 50. If two straight lines intersect, the vertical angles are equal. -D Given: Two straight lines, AB, CD, intersecting in o; To Prove: Angle ^oc is equal to angle BOD, and angle AOD is equal to angle BOC. Since A aod is supp. to A AOC, and also to A BOD, (46) A AOC = A BOD. (44) Similarly, A AOD = A BOC, each being the supplement of A BOD. q.e.d. ANGLES. 27 Proposition V. Theorem. 51. From a point without a line, there can he hut one perpendicular to that line. Given: The point A, the line BC, AB A. to BC, and AC any other line from A%o BC; To Prove : ^C is not perpendicular to BC. Turn the figure ABC about BC till A takes the position A^ in the original plane. Mark the point A\ restore ABC to its first position, and join A^B, a'c. Then since AB and AC can be made to coincide with a'b and a'c, while BC retains its position, (Const.) Z ABC = Z a'bc, and Z ACB = Z a'cb. (25) But Z ABC is a right Z, (Hyp.) .-. Z ^'5C is a right Z, (30) .'. aba' is a straight line, (49) .*. ^C^' is not a straight line, (6) .-. Z AC A' is not a straight Z, (29) .*. Z ^C-B, the half of Z ^c^', is not a right Z. q.e.d. Exercise 9. If in the diagram for Prop. III., angle CAD is | of a right angle, what angle must BA make with ^C so that BA shall be in the same straight line with AD ? 10. If, in a line MN, a point P be taken, and two lines PQ, Plt^ be drawn so that angle QPM = angle BPN^ then QB is a straight line. 11. In the diagram for Prop. V., the angles A and A' are the sup- plements of what angles ? Show that these angles are equal. 12. In this diagram, show that the exterior angles at C are equal. 28 PLANE GEOMETRY. — BOOK I. TRIANGLES. 52. A triangle is a portion of a plane bounded by three straight lines. 53. The lines that bound a triangle are called its sides; the angles formed by the sides, its angles; and the vertices of the angles, the vertices or angular points of the triangle. If all the sides of a triangle be produced both ways (see triangle C below), nine new angles will be formed in addi- tion to those properly called the angles of the triangle, or by way of distinction, its interior angles. Of the nine outer angles, the six angles that are supplementary to the interior angles, are called exterior angles. Interior angles are always meant whenever we refer to the angles of a triangle without any distinguishing epithet. 54. The base of a triangle is the side on which it is sup- posed to stand ; the angle opposite the base is sometimes referred to as the vertical angle. 55. An equilateral triangle has three equal sides ; as A. 56. An isosceles triangle has two equal sides ; as ^. 57. A scalene triangle has no two sides equal ; as C. TRIANGLES. 29 58. A right triangle has a right angle ; as D. 59. An obtuse triangle has an obtuse angle ; as E. 60. An acute triangle has all its angles acute ; as F. While lines and angles can be equal only in one way, it is different in regard to triangles and other inclosed figures. For as will afterwards be seen, two triangles that are equal as regards surface may differ greatly as to their sides and angles. Hence it is necessary to define the term equal in regard to figures in general. 61. Equal figures are such as can be made to coincide exactly ; that is, are equal in every respect. 62. Theorem. Two triangles are equal if their angular points can he made to coincide. For if the angular points coincide, the sides terminated by those points must coincide (14) ; hence, also, the angles formed by the sides, and the surfaces bounded by them, must coincide. It is evident that the theorem may be extended so as to apply to figures bounded by any number of straight lines, the reasoning being exactly the same. Exercise 13. Show, by Prop. V., that no triangle can have two of its angles right angles. 14. If triangle A has all its angles equal, show that its six exterior angles are all e(iual. 15. If a triangle has two equal angles, those angles must be acute. What about the third angle ? 16. If a triangle has no two angles equal, it has three different pairs of equal exterior angles. 17. If, in triangle Z), the sides containing the right angle be pro- duced through its vertex, the sum of the exterior angles thus formed is equal to a straight angle. 18. If, in triangle E, the sides containing the obtuse angle be produced through its vertex, the sum of the exterior angles thus formed is less than a straight angle. 30 PLANE GEOMETRY. — BOOK I. Proposition YI. Theorem. 63. Two triangles are equal if a side and the including angles of the one are respectively equal to a side and the including angles of the other. Given: In triangles ABC, a'b'c', bc equal to B'c', angle B equal to angle B', and angle C equal to angle c' ; To Prove: Triangle ABC is equal to triangle a'b'c'. liAABC be placed upon A a'b'c', (Post. 5) so thsit BC=^ B'c', then since Z B = Z b', (Hyp.) BA will take the direction of b'A', (25) and A will lie on b'a' ov b'a' produced. Also since Zc = Zc', (Hyp.) CA will take the direction of c'a', (25) and A will lie on c'A' or c'a' produced. Then since A lies on both b'a' and C'A\ A must coincide with A', the only point common to b'a' and C'a', (6) .-. Aabc = Aa'b'c'. q.e.d. (62) 64. Scholium. If the including angles are all equal, i.e., if Z C = Z B = Z c' = Z b', as in the right hand pair of triangles, it is manifest that A ABC can be made to coincide with A A'b'c' in the reverse position also, so that AB will coincide with a'c', and AC with a'b'. OF THE UNIVERSITY ^^ ^ TRIANGLES. 81 Proposition VII. Theorem. 65. If two angles of a triangle are equal, the sides opposite those angles are also equal. Given: In triangle ABC, angle C equal to angle B ; To Prove: ^i? is equal to ^C. Turn A ABC about its base BC till A takes the position A' (Post. 5). Mark the point a\ restore ABC to its first position, and join A^B, A'C (Post. 1), so as to form the Aa'bc, Since their angular points can be made to coincide, (Const.) AABC = AA'BC, (62) and A'B = AB, A'C = AC,Z a'bc = Zb,Z A'CB = Zc. (14) Since these four angles are all equal, (Hyp. and Const.) A ABC can be made to coincide with Aa'bc in the reverse of the first position, (64) so that AB ^ A'C and AC =^ a'b. Then since AB = A'c, (14) and^C = ^'C, (Above) ABz=AC. Q.E.D. (Ax. 1) Exercise 19. In the diagram for Prop. VIL, if AA' be drawn, cut- ting BC in O, show that when triangle ABC is, folded over on triangle A'BC, AO must coincide with A'O. 32 PLANE GEOMETRY. — BOOK I. Proposition VIII. Theorem. 66. Two triangles are equal if two sides and the included angle of the one are respectively equal to two sides and the included angle of the other. Given: In triangles ABC, a'b'c', AB equal io A^B^, AC equal to A'c\ and angle A equal to angle A' ; To Prove: Triangle ABC is equal to triangle a'b'c'. HAabc be placed upon A a'b'c', (Post. 5) so that Z ^ r^ Z^', then since AB = A'b', (Hyp.) B=^B'. Also since yl (7 = ^'C', (Hyp.) c ^ c', .'. A ABC = A a'b'c'. q.e.d. (62) 67. Scholium. If the including sides are all equal, i.e., if AB — AC = A'b' = a'c', as in the right hand pair of trian- gles, it is manifest that A ABC can be made to coincide with A a'b'c' in the reverse position also, so that B will coincide with C', and C with B'. Exercise 20. Show (Exercise 19) that ^0 is perpendicular to EC, bisects angle A, and bisects base BC. 21. Prove, by Prop. VI., that all the triangles having their vertices at are equal. 22. Prove the same equalities by means of Prop. VIII. 23. In the triangles ABC, A'B'C, above, when AB is made to coincide with A'B', where would AC fall (1) if angle A were greater than angle A' ; (2) if less than angle A'f TRIANGLES. 33 Proposition IX. Theorem. 68. If two sides of a triangle are equal, the angles opposite them are equal. Given : In triangle ABC, AC equal to AB ; To Prove : Angle B is equal to angle C. Turn A ABC about its base BC till A takes the position A' (Post. 5). Mark the point A', restore ABC to its first position, and draw A^B, A'c (Post. 1). Since their angular points can be made to coincide, (Const.) AABC = AA'BC, (62) and A'b = AB, A'c = AC, Z a'bc = Zb,Z A'CB = Z C. (14) Since all these sides are equal, (Hyp. and Const.) A ABC can be made to coinciile with A A'BC in the reverse of the first position, (67) so that Zb =^ Z A'CB, and Z c =^ Z A'bc. Then since Zb = Z A'CB, (14) and Zc = Z A'CB, (Above) Zb=Zc. q.e.d. (Ax. 1) Exercise 24. If two sides of a triangle are unequal, the angles opposite to them are also unequal, and conversely. 25. A scalene triangle cannot have two equal angles. Geom. — 3 34 PLANE GEOMETRY. — BOOK I. Proposition X. Theorem. 69. Two triangles are equal if the three sides of the one are respectively equal to the three sides of the other. A Given: In triangles ABC, A'b'c', AB equal to a'b\ AC equal to A'&, and BC equal to B^& ; To Prove: Triangle ABC i& equal to triangle a'b'c'. Apply A^^C to A A'b'c', so that BC =^ b'c', and let A fall on the further side of B'c' from A'. Join AA', and first suppose AA' cuts B'c'. Since AB' = a'b', and AC' = A'c', (Hyp.) Z 1 = Z 2, and Z 3 = Z 4, (68) .-. Z1 + Z3 = Z2 + Z4, (Ax. 2) i.e., Z.BAC = Z.B'A 'c', .'. A ABC = A a'b'c'. q.e.d. (66) If A a' should pass through an extremity of B'c', or cut b'c' produced, the proof would be similar to that above. But, as will afterwards be seen, the triangles can always be applied to each other so that A A' shall cut B'c' between B' and C'. 70. Scholium. In equal triangles, sides that are equal to each other subtend equal angles, and equal angles are subtended by equal sides. TRIANGLES. 35 For when equal triangles are made to coincide, coinciding sides necessarily subtend coinciding angles. This important proposition is a scholium rather than a corollary, because it does not deduce a new truth, but merely calls attention to a truth, or set of truths, already virtually proved. See definition, p. 19. 71. Definition. In a right triangle, the side subtending the right angle is called the hypotenuse ; and the other sides are called arms. Proposition XI. Theorem. 72. Two right triangles are equal if the hypotenuse and an arm of the one are respectively equal to the hypotenuse and an arm of the other. Given: In right triangles ABC, A'b'c', hypotenuse AC equal to hypotenuse A'c', and AB equal to a'b' ; To Prove : Right triangle ABC is equal to right triangle A 'B 'c'. )ply A ABC to A A'b'c', so that AC ^ A'C', and B falls remote from b'. Join BB'. Since AB = a'b', (Hyp.) Za'b'b==Za'bb', (68) .-. Z c'b'b = Z c'bb'. (44) (being complements of equal A,) .: BC' OT BC = B'C', (65) .'. Aabc = aa'b'c'. Q.E.D • (69) 36 PLANE GEOMETRY. — BOOK I. Proposition XII. Theorem. 73. Two right triangles are equal if the hypote- nuse and an acute angle of the one are respectively equal to the hypotenuse and an acute angle of the other. Given: In right triangles ABC, A^b'c', hypotenuse BC equal to hypotenuse B'c', and angle B equal to angle B' ; To Prove : Eight triangle ABC is equal to right triangle A'b'c'. Place A ABC upon A A'b'c', so that BC =^ b'c'. Then since Z B = Z B', (Hyp.) BA will take the direction oi b'a' , (25) and A will fall upon A' ; since there can be but one ± from c' to b'A' ; (51) .-. AABC = AA'B'C'. Q.E.D. (62) The student is now in possession of all the theorems con- cerning equal triangles that are of practical importance, viz. — Two triangles are equal if they have, respectively : 1. One side and the including angles equal; Prop. VI. 2. Two sides arid the included avgle equal; Prop. VIII. 3. Three sides wMually equal. Prop. X. 4. If right, the hypotenuse and an arm, or the hypotenuse and an acute angle, mutually equal. Props. XI., XII. It may be easily proved that triangles are equal that have a side and any two angles equal ; the case is, however, of no importance, nor is the more diiRcult one concerning triangles having two sides and a not included angle mutually equal. TRIANGLES. 37 Proposition XIII. Theorem. 74. The line that joins the vertices of two isosceles triangles on the same base, bisects, 1°, the vertical angles, 2°, the common base at right angles. Given: A A' joining the vertices of isosceles triangles ABC, A'bc, and cutting BC in B ; To Prove : A A' bisects angles A and A', and is perpendicular to BC Sit its mid point. 1°. As AB = AC, A'b = A'C, and AA' is common,=^ (Hyp.) Abaa' = Acaa', (69) .-. Z BAD = Z CAD, and Zba'd = Z CA'd. q.e.d. (70) 2°. As AB = AC, AD is common, and Z BAD=Z CAD, (1°) A BAD = A CAD, (66) .'. BD = CD, and the A Sit D are equal, (70) .-. A A' is ± to BC Sit its mid point, q.e.d. (30) 75. Cor. 1. If two points are each equidistant from the extremities of a given line, the line joining these points is perpendicular to the given line at its midpoint. The points, it will be observed, may be on different sides of the given line, or both on the same side. * That is, is common to (or belongs to) both triangles. 38 PLANE GEOMETRY.— BOOK 1. 76. Definition. The line that divides an angle into two equal angles is called its bisector. 77. Cor. 2. The bisector of the vertical angle of an isosceles triangle bisects the base at right angles; and conversely. Thus far the constructions required as aids in the demon- stration of theorems have called for nothing beyond the postulates. We shall presently have occasion for other constructions, such as bisection of lines, angles, etc. The problem about to be given is of some importance as entering largely into subsequent problems. Exercise 26. In the diagram for Prop. IX., show that angle ABA' = angle AC A'. 27. Prove, by means of Prop. X., the equality proved in Exercise 22. 28. Draw the diagram for Prop. X. when the angle B is obtuse, and prove the proposition. 29. Prove that on the same side of the same base there can be but one isosceles triangle having its arms, or equal sides, eacli equal to a given line. 30. Prove that on the same side of the same base there can be but one isosceles triangle having a given vertical angle. 31. In the diagram for Prop. XIII., prove in regard to both figures that the angle ABA' is equal to the angle AC A'. 32. If the base of an isosceles triangle ^I?C be produced both ways to D and £', so that BD = CE, and A be joined with D and E, the triangle ADE will be isosceles. 33. If the exterior angles formed by producing both ways a side of a triangle are equal, the other sides are equal. 34. Right triangles are equal if their arms are respectively equal. 35. Right triangles are equal if they have equal acute angles at the extremities of equal arms. 36. In the diagram for Prop. XIV., if BA be produced to meet the circumference at Z>, show that DE is equal to the sum of the three sides of triangle CAB. 37. In the same diagram, BA being produced as above, and C joined with D and E, show that CAD, CBE, are equal isosceles triangles. 38. In the same diagram, if the circumferences intersect a second time at F, and CF cut AB in G, show that GE=3AG. TRIANGLES. 39 Proposition XIV. Problem. 78. To find a point equidistant froin the extremi- ties of a given line. A'L Given : A straight line AB ; Required : To find a point equidistant from A and Bx From A as center, with radius AB, describe the circumference BCD. From B as center, with radius BA, describe the circumference ACE. Produce ^^ to meet ACE in E. Since AE > AB, E lies without the circumf. BCD. But A lies within the circumf. BCD, .'. circumf. ACE must intersect circumf. BCD. Let them intersect in c. c is the point required. For since AC = AB^ and BC — BA or AB, .'. AC = BC, (Ax. 1) '. a point C has been found equidistant from A and B. q.e.f. J Post. (4) (Post. 2) (Const.) (23) (21) (23) 79. Scholium. Although, for convenience of demonstra- tion, the radii have been taken each equal to the given line, it is manifest that other points equidistant from A and B 40 PLANE GEOMETRY.— BOOK I. can be obtained by taking any equal radii greater than one half the given line. It is also obvious that, in practice, instead of entire circumferences, only such portions need be described as will give the points of intersection. Proposition XV. Problem. 80. To bisect a given straight line. i^ Given : A straight hne AB ; Required : To bisect AB ; that is, to find its mid point. Find two points C and D, each equidistant from A and B. (78) Join CD. CD will intersect ^S at its mid point. Since C and D are two points equidistant from A and B, CD is _L to AB at its mid point, say E ; (75) i.e., AB is bisected in E. q.e.f. Scholium. This construction gives not only the mid point of AB, but also the perpendicular to AB through its mid point. Exercise 39. Find a point X that shall be twice as far from each of two given points, A and B, as A is from B. 40. Find a second point Y that shall be three fourths the distance from each of those same points, A and JB, that A is from B. 41. Join AB, and prove that the perpendicular at the mid point of ^B will pass through X and Y. TRIANGLES. 41 Proposition XVI. Problem. 81. To bisect a given angle. c Given : An angle BAC; Required: To bisect angle ^^C. On AB, AC, lay off any equal parts AD, AE. (Post. 3) Join DE (Post. 1), and find a point F equidistant from D and E. (78) The line joining AF is the bisector of the given angle. Since A and F are each equidistant from D and E, (Const.) AF is ± to DE at its mid point, (75) .-. Z ^^(7 is bisected by ^-F. q.e.f. (77) 82. Scholium. By repeating the operations described in Props. XV. and XVI., a given line or angle can be divided into 4, 8, 16, •••, 2", equal parts. Exercise 42. In the diagram above, show that angle BDE = angle CED. 43. Divide a given angle ABC into four equal angles. 44. In the diagram above, if the point F were taken on the same side of DE as A, what three positions might F have in regard to ^ ^ In which of these positions would the point taken fail to assist in the required construction ? 45. In the diagram above, join F with D and E, and prove the proposition by Prop. X. 46. Show that in an equilateral triangle the three lines joining the vertices with the mid points of the opposite sides are (1) equal to each other ; (2) perpendicular to these sides ; (3) bisectors of the angles. 42 PLANE GEOMETRY.— BOOK L Proposition XVII. Theorem. 83. An exterior angle of any triangle is greater than either of the remote interior angles. Given: ACD, an exterior angle of triangle ABC; To Prove : Angle ACD is greater than angle A or angle B. Of the two, let Z A he not less than Z J?.* Bisect J.C in ^ (80) ; join BE, and produce BE to F, so that EF = BE (Posts. 1 and 3). Join FC. Then since EA = EC, and EF = zEB, (Const.) also Z AEB = Z CEF, (50) Aeab = Aecf, (66) .'. Zecf= Za. (70) But Zacd> Zecf, (Ax. 8) .'. Z ACD > Z A ; and since Zb is not greater than Z A, (Hyp.) Z ACD > Z i?, as well as > Z ^. Q.E.D. 84. Cor. 1. No two angles of a triangle can be supple- mentary. For each is less than the exterior angle which is the supplement of the other. 85. Cor. 2. If a triangle has an obtuse or a right angle, each of the others is acute. * That is, angle B is, at most, equal to angle A. TRIANGLES. 43 Proposition XVIII. Theorem. 86. A greater side of a triangle subtends a greater angle Given : In triangle ABC, BC greater than AC ; To Prove : Angle A is greater than angle B. Produce CA to n, so that en = cb ; and join BD. Since CB = CD, (Const.) Zd = Zdbc. (68) But Za>Zd, (83) (angle A being an exterior Z of A DAB,) .-. Z A > Zdbc, .-. Za>Zb. q.e.d. (a.f.)* 87. Cor. Conversely, a greater angle is subtended by a greater side. In A ABC suppose Z A> Z B. AC cannot be equal to J5 c ; for then Z B would be equal to Z A (65) : nor can ^C be greater than BC ; for then Z B would be greater than Z A (86). Hence, since BC can neither be equal to nor less than AC, it must be greater than AC. q.e.d. Exercise 47. Draw the diagram for Prop. XVII., when the exterior angle ACD is acute. 48. Prove that a perpendicular from a vertex to the opposite side of a triangle falls (1) within the triangle if the angles including that side are both acute ; (2) without the triangle if one of those angles is obtuse. * If m > n and n >p, then a fortiori (i.e., with greater reason), m >p. This is called the argument a fortiori, abbreviated a.f. 44 PLANE GEOMETRY. — BOOK I. Proposition XIX. Theorem. 88. Any side of a triangle is less than the sum of the other two. driven: Any side jBC of a triangle ABC; To Prove : BCis less than AB -{- AC. Produce BA to D, so that AD = AC ; and join DC. Since AC = AD, (Const.) /.acd = Zd. (68) But Zbcd> Zacd, (Ax. 8) .-. Z.BCD >Zd, .'. BD, or BA -i- AC,>BC. Q.E.D. (86) 89. Cor. Any side of a triangle is greater than the dif- ference of the other two. ¥oT since BC-\- AC>AB, (88) BC>AB —AC. (Ax. 5) Exercise 49. Prove Prop. XVIII. by cutting off on BC, CD equal to CA, joining AD, etc. 60. The perpendicular to the greatest side from the opposite vertex falls within the triangle. 51. Prove Prop. XIX. by supposing BO to be the greatest side, and drawing AD perpendicular to BC. 52. In a triangle ABC, having AB less than AC, prove that any point in the line joining A with the mid point ot BC is nearer to B than to C. TRIANGLES. 45 Proposition XX, Theorem. 90. // two triangles have two sides of the one respectively equal to two sides of the other, hut the included angles unequal, the greater angle is sub- tended hy a greater base. Given: Two triangles ABC, A'b'c', having AB ec[ual to A'b', AC equal to A'c', but angle A greater than A' ; To Prove : Base BC is greater than base b'c'. Place A a'b'c' upon A ABC, so that a'b' =^ AB. Then •.• Za>Za' (Hyp.), ^'C'' will fall between ^5 and^c, as^c'. Draw AD bisecting Z CAC' to meet BC in D, and join DC'. Since AC' = AC (Hyp.), and AD is common, also Zdac' = Zdac, (Const. ) .-. AADC' = AADC (66), 3ind DC' = DC. (70) But BD-\- DC' >BC', (88) and BD -^DC = BC, (Const.) .*. BC > BC' or b'c'. q.e.d. 91. Cor. Conversely, if triangles ABC, A'b'c', have AB, AC, equal to a'b', A'c', respectively, hut the base BC greater than the base B'c', then the angle A is greater than the angle A'. For Z A cannot be equal to Z.A', since then BC would be equal to B'c' (66); nor can Z A he less than Z A', since then BC would be less than B'c' (90). Hence Z A must be greater than Z A'. 46 PLANE GEOMETRY. — BOOK I. PERPENDICULARS. Proposition XXI. Problem. 92. To draw a perpendicular to a given line from a given point without it. F .■ ■■■-.. G D . H .-E "■■ -'C Given : A point P without a straight line AB of indefinite length ; * Required: To draw a perpendicular from P to AB. Take any point C on the side of AB remote from P. From P as center, with, radius PC, describe a circumfer- ence, CFG. (Post. 4) Since AB extends indefinitely between P and C, and the circumference may be described in either direction from C, it will intersect AB in two points, D, E. Bisect DE in H (80), and join PH. PH is ± to AB. Since D and E are points in the circumf., CFG, (Const.) P is equidistant from D and E ; also H is equidistant from D and E, .'. PH is ± to DE at its mid point H, i.e., PH is drawn A- to AB. 93. Cor. The perpendicular is the short- est line that can he drawn to a given line from a given point without it. For if PA is a perpendicular, and PB a line oblique to MN, then Z. B <, right Z A (85) ; hence PA<,PB (86). * That is, the given line may be produced to any extent, if necessary. PERPENDIC ULA RS. 47 94. Definition. The distance of a point from a line is the distance from the point to the foot of the perpendicu- lar to the line. Thus the distance of P from MN is the perpendicular distance PAy this being the least distance from P to any point in MN (93). Proposition XXII. Problem. 95. To draw a perpendicular to a given line at a given point in it. Given : A point P in a straight line AB ; Required : To draw a perpendicular to ^i? at P. On AB lay off, on each side of P, equal parts PC, PD. Find a point E that is equidistant from c and D. (78) t Join EP ; then EP is ± to ^IP at P. Since P is equidistant from C and D, ) . .... . . [ (Const.) and E IS equidistant from C and D, ) EP is ± to CD at its mid point P, (75) i.e., EP is drawn _L to AB at P. q.e.f. Scholium. In this problem we have what may be regarded as a special case of the bisection of an angle. What is re- quired is, in fact, to bisect a straight angle (compare 81). Exercise 53. At a given point in a line make an angle equal to half a right angle. 64. From a given point without a line draw a line making with the given line an angle equal to half a right angle. 48 PLANE GEOMETRY. — BOOK I. Proposition XXIIT. Theorem. 96. Any point in the perpendicular at the mid point of a line is equidistant from the extremities of the line. Given : Any point P in CD, the _L at the mid point of AB ; To Prove : P is equidistant from A and B. Join P with A and B. Then since DA = DB (Hyp.), and PD is common, also rt. Z PDA = rt. Z PDB, (42) A PDA = A PDB, (66) .-. PA = PB, (70) i.e., P is equidistant from A and B. q.e.d. 97. Cor. 1. Conversely, cuiy point that is equidistant from the extremities of a line will lie in the perpendicular through its mid point. For the line joining that point with the mid point of the line is perpendicular to the latter at its mid point (75). 98. Cor. 2. Aiiy point not in the perpendicular through the midpoint of a line is unequally distant from its extremities. For if it were equidistant from those extremities, it would be in that perpendicular (97). Exercise 55. In the diagram for Prop. XXIII., show that angle CPA = angle CPB. 56. In the same diagram, what condition must be fulfilled in order that triangle PAB may be equiangular ? PERPENDICULARS. 49 Proposition XXIV. Theorem. 99. If from any point in a perpendicular to a given line different oblique lines he drawn to the given line, then 1°. Oblique lines drawn to equal distances from the foot of the perpendicular are equal. 2°. Of oblique lines drawn to unequal distances from the foot of the perpendicular, the more remote is the greater. Given : PD ± to MN, and PA, PB, PC, drawn oblique to MN, so that AD is equal to BD, but CD is greater than AD ; To Prove : PA is equal to PB, but PC is greater than PA oi PB. 1°. Since P is in the ± at the mid point of AB, (Hyp.) PA = PB. Q.E.D. (96) 2°. Since exterior Zpac> right Z D, (83) but right Zd>Zc, (85) ZpaoZc, (a.f.) .-. PC > PA or its equal PB. q.e.d. (86) 100. Cor. Only two equal straight lines can be drawn from a point to a given straight line, one on each side of the perpendicular from that point. Exercise 57. From a point without a line, show that only two oblique lines can be drawn so as to make equal angles with the given line. 58. If oblique lines drawn from a given point without it make unequal interior angles with a line, that which makes the lesser angle is the greater. Geom. — 4 50 PLANE GEOMETRY. — BOOK I. Proposition XXV. Theorem. 101. Any point in the bisector of an angle is equi- distant from its sides; and conversely. 1°. Given : A point P in the bisector of angle BAC; To Prove : P is equidistant from AB and AC. Draw PD ± to AB and PE _L to AC. Then since AP is the hypotenuse of rt. A PDA, PEA, (Const.) and Z PAT) = Z PAE, (Hyp.) right A PAD = right A PAE, (73) .-. PD = PE. Q.E.D. (70) 2°. Given : In angle BAC,si point P equidistant from AB and AC ; To Prove: The line joining PA bisects angle BAC. Draw PD ± to AB, PE J- to AC, and join PA. Then since PD = PE (Hyp.), and AP is common, also right Z D = right Z E, (Const.) right A PAD = right A PAE, (72) .'. Zpad = Zpae, (70) .-. PJ[ bisects Z^^e. q.e.d. Exercise 59. Show that Prop. XXIII. is a particular case of XXV. 60. Prove that any point not in the bisector of an angle is unequally distant from its sides. 61. Enunciate and prove the converse of Exercise 60. 62. Prove that a perpendicular to the bisector of an angle makes equal angles with the sides. PARALLELS. 51 PARALLELS. 102. Parallel lines are such as lie in the same plane, but cannot meet, however far pro- duced in either direction. 103. A transversal is a straight line that is trans- verse to, that is, meets or in- tersects, a set of two or more straight lines. 104. When a transversal EF intersects two lines AB, CD, then {a) The four A 3, 4, 5, 6, within AB, CD, are interior A. (b) The four A 1, 2, 7, 8, without AB and CD are exterior A. (c) The pairs of interior A, 3-5, 4-6, situated on opposite sides of the transversal, are alternate-interior A. (d) The pairs of exterior A, 1-7, 2-8, situated on opposite sides of the transversal, are alternate- exterior A. (e) The pairs of A, 1-5, 4-8, 2-6, 3-7, situated on the same side of the transversal, but one exterior, the other interior, are corresponding A. Angles hereafter referred to simply as alternate are to be understood as being alternate-interior, those most frequently mentioned. E 105. Axiom 10. Tivo inter- a -^ — ■— ,__^^ — - secting lines cannot both be par- f allel to a third line. C D Thus, if AB is parallel to CD, EF cannot be so ; and vice versa. Exercise 63. Enunciate and prove the converse of Exercise 62. 64. Prove that the bisector of an angle of a triangle divides the opposite side equally or unequally according as the other sides are or are not equal. 62 PLANE GEOMETRY. — BOOK I. Proposition XXYI. Theorem. 106. If a transversal is perpendicular to two lines in the same plane, these lines are parallel. A E B C F D Given : EF perpendicular to AB and also to CD ; To Prove : AB is parallel to CD. Since AB and CD are each ± to EP, (Hyp.) AB cannot meet CD in any point on either side of EF, no matter how far produced, since from the same point there cannot be two Js to a line, (51) .-. AB is II to CD. Q.E.D. (102) Proposition XXVII. Theorem. 107. If a transversal is perpendicular to one of two parallels, it is perpendicular to the other also. E Given : Two parallels AB, CD, and ^C a transversal ± to AB ; To Prove: AC i^ perpendicular to CD. PARALLELS. At C suppose CE drawn 1. to AC. Then since CE is II to AB, 63 (95) (106) (Hyp.) and CD is II to AB, CD must coincide with CE, since otherwise there would be through C two intersecting parallels to the same line, which is impossible. (Ax. 10) .-. CD \^ A. to AC. Q.E.D. 108. CoR. If AB, CD, are each parallel to EF, then AB is parallel to CD. For a ± to EF must be ± to both AB and CD (107). A B C D E F Proposition XXVIII. Problem. 109. Through a given point without a line to draw a parallel to the line. Given : A point P without a given line AB ; Required: To draw through P a hne parallel to AB. From P draw PC ± to AB, or AB produced, (92) and from P draw PD A. to PC. (95) PD is the parallel required. For since PC is _L to AB, and also to PD, (Const.) .-. PD is II to AB. Q.E.F. (106) 54 PLANE GEOMETRY. — BOOK I. Proposition XXIX. Theorem. 110. Two lines are parallel if a transversal to those lines makes the alternate angles equal; and conversely. A B p R^ y^ ^ c y^F Q T> 1°. Given: EF transverse to AB, CD, making the alternate angles E, F, equal ; To Prove: AB i^ parallel to CD. Find O, the mid point of EF (80), draw OP ± to AB (92), and produce PO to meet CD in Q. Then since /.F = Z.E (Hyp.), and /.FOQ = /. EOF, (50) and OF = OE, (Const.) AOQF = AOPE, (63) .-. Zq = rt. Z P, (70) .-. AB is II to CD. Q.E.D. (106) 2°. Given: A transversal EF meeting the parallels AB, CD, in E, F, respectively ; To Prove : Angle E is equal to alternate angle F. Making the same constructions as in 1°, Since AB is II to CD (Hyp.), and PQ is _L to AB, PQ is also ± to CD (107), and Q is a rt. Z ; also Z EOF = Zfoq (50), and OE = OF, .: rt. A OPE = rt. AOQF, (Const.) (Const.) (73) .-. Ze = Z F. 111. Scholium. It is sufficient to prove the foregoing propositions, ~ as above, for one pair of alternate angles only ; since the equality of any one pair of the alternate angles Q.E.D. (70) jzi. 8^7 PARALLELS. 55 .formed by a transversal and two lines, entails the equality of every pair. For if Z 4 = Z 6, then Z 2 = Z 8 (50) ; also Z3 = Z5, andZl= Z7 (44). Proposition XXX. Theorem. 112. Two lines are parallel if a transversal to those lines makes the corresponding angles equal; and conversely. T 1°. Given : EF transverse to AB, CD, making the corresponding angles F, g, equal ; To Prove : AB h parallel to CD. Since Af = /.g (Hyp.), and Z ^ = Z G, (50) /.E = /.F, (Ax. 1) .-. AB is II to CD. Q.E.D. (110')* 2°. Given: A transversal EF meeting the parallels AB, CD, in E, F, respectively ; To Prove : Angle F is equal to the corresponding angle G. Since EF is transverse to the II 's AB, CD, (Hyp.) Ze = Zf. (110") But Z J5: = Z G, (50) .\Z.F = /.G. Q.E.D. (Ax. 1) * When, as here, reference is made to a proposition consisting of two parts, these parts will be distinguished by accents; thus, 110' and 110". 56 PLANE GEOMETRY. — BOOK I. Proposition XXXI. Theorem. 113. Two lines are parallel if a transversal to those lines mahes the interior angles on the same side supplementary; and conversely. A yo B /^' C yF D X 1°. Given: EF transverse to AB, CD, making the interior angles E, F, supplementary ; To Prove : AB is parallel to CD. Since Z F is the supplement of Z E, (Hyp.) and Z G is the supplement of Z E, (46) Z F = the corresponding Z G, (Ax. 1) .-. AB is II to CD. Q.E.D. (112') 2°. Given : A transversal EF meeting the parallels AB, CD, in E, F, respectively ; To Prove: The interior angles E and F are supplementary. Since EF is transverse to the ll's AB, CD, (Hyp.) Z F = corresponding Z G. (112") But Z jE; is the supplement of Z G, (46) .*. Z ^ is the supplement of Z F. q.e.d. (44) 114. Cor. If a transversal to two lines makes the sum of the interior angles on one side less than a straight angle, these lines will meet if produced on that side. For if they did not meet, they would be parallel, and those interior angles would be supplementary (113") ; that they must meet on that side of the transversal on which the interior A are less than a straight Z follows from Art. 84. PARALLELS. 57 115. Definition. Parallel lines are said to lie in the same or opposite directions, according as they are on the same or opposite sides of the transversal through the points from which they are supposed to be drawn. Proposition XXXII. Theorem. 116. Angles whose corresponding sides are parallel are equal or supplementary. Given: AB parallel to DE, and AC parallel to HF, forming angles at A and D ; To Prove : Angle A is equal to angle D, and angle A is supple- ment of angle EI)H. Let AC, DE, produced if necessary, meet in G. Since AB is II to DE, and ^C is transverse to both, Ag = Aa; (112") Since AG is II to DF, and DE is transverse to both, Z.G = Z.D, (112") .-. /.A — Ad; q.e.d. (Ax. 1) also Z D is supp. of Z EDH, (46) .-. Z ^ is supp. of Z ^DiT. q.e.d. (44) 117. Scholium. The angles are equal if both corre- sponding pairs of sides lie in the same or opi^osite direc- tions; they are supplementary if one pair has the same and the other opposite directions. 58 PLANE GEOMETRY. — BOOK I. Proposition XXXIII. Theorem. 118. Angles whose corresponding sides are perpen- dicular to each other are either equal or supple- mentary. F IK sB Given: AB perpendicular to DE and AC perpendicular to DF, forming angles at A and D ; To Prove : Angle A is equal to angle D, and angle A is the supplement of angle EDG. At A draw AH 1. to AB, and AK 1. to AC. (95) Since BAH and CAK are rt. A, (Const.) Z.A = /. HAK, (44) (each being the comp. of Z bak.) Since AB is ± DE and AH, and ^C is _L to I)F and AK, (Hyp. and Const.) DE is II to AH, and DF is || to AK ; (106) and they lie in the same directions, .-. Zd = Zhak, (117) .*. Za = Zd; q.e.d. (Ax. 1) also Zd is supp. of Z EDG, (46) .*. Z^ is supp. of Zedg. q.e.d. (44) 119. Scholium. The angles are equal if both are acute or both obtuse. Exercise 65. In the diagram for Prop. XXIX., show that PF being joined, a line drawn through Q parallel to PF will pass through E. PARALLELS. 59 Proposition XXXIV. Theorem. 120. The sum of the angles of any triangle is equal to a straight angle. Given: Any triangle ABC ; To Prove : Za-\-Zb-\-Zc[& equal to a straight angle. Through A draw DAE II to BC. (109) •.• DE is II to BC, and AB is transverse to both, (Const.) Zb = Z1; (110") ••• DE is II to BC, and ^C is transverse to both, ZC = Z2, (110") .'. Za + Zb + Zc = Za-{-Z1-{-Z2, (Ax.2) i.e., Za + Zb ■\-Zc = Z DAE, a st. Z. q.e.d. (47) 121. Cor. 1. Each angle of a triangle is the supplement of the sum of the other two. 122. Cor. 2. An exterior angle of a triangle is equal to the sum of the two remote interior angles. For its supplement is also the supplement of those two. 123. Cor. 3. The acute angles of a right triangle are com- plementary. 124. Definition. A plane polygon is a portion of a plane bounded by straight lines. Its sides, angles, and vertices are defined as for the triangle, which is a three-sided polygon. The polygons treated of are to be understood as being convex; i.e., such that no side, if produced, will fall within the polygon. 60 PLANE GEOMETRY. — BOOK I. Proposition XXXV. Theorem. 125. The Sinn of the interior angles of a polygon is equal to as many straight angles as the figure has sides, less two. E D Given : A polygon AB CDEF • • • of 7i sides ; To Prove :Za+Zb-{-Zc-\ = {n — 2) straight a; From any vertex, as A, draw lines to C, D, E, •••. Since, except the two that meet at A, each of the n sides of the polygon is the base of a triangle having its vertex at A, (Const.) the number of triangles = ?i — 2 ; .*. the sum of the int. A of the A= (n — 2) st. A, (120) •.• the sum of the int. A of the polygon = {n — 2) st. A. q.e.d. 126. Cor. 1. If the sides of a polygon be produced, going round in the same order, the sum of the exterior A thus formed = two straight A. For each interior Z A with its exterior Z a is equal to a straight Z (46) . Hence /A ^Ml the sum of all the angles, interior and exterior, = n straight A. But the interior A alone = (n — 2) straight A ; hence the exterior A = two straight A. 127. Cor. 2. Each interior angle of an equiangular poly- gon is equal to straight angles. QUADRILA TERALS, 61 QUADRILATERALS. 128. A quadrilateral is a polygon bounded by four sides. 129. A trapezium is a quadrilateral having no two of its sides parallel. 130. A trapezoid is a quadrilateral having two of its sides parallel. 131. A parallelogram is a quadrilateral having its oppo- site sides parallel. PARALLELOGRAM 132. A rectangle is a right-angled parallelogram. 133. A square is an equilateral rectangle. 134. A rhombus is a parallelogram which is equilateral but not equiangular. 135. A diagonal is a line joining any two nonadjacent vertices of any polygon. RECTANGLE SQUARE RHOMBUS Exercise 66. Prove that each angle of an equilateral triangle is one third of a straight angle. 67. Prove that if an angle at the base of an isosceles triangle is one half of the vertical angle, the latter is a right angle. 68. If an angle at the base of an isosceles triangle is n times the vertical angle, what fraction is the latter of a straight angle ? 62 PLANE GEOMETRY. — BOOK I. Proposition XXXVI. Theorem. 136. The opposite sides and angles of a parallelogram are equal. D c Given : A parallelogram AB CD, oi AC; To Prove : AB = CD, AD = BC; A A =Z C, and Z B =Z D. Since AB is li to CD, and BC\^ II to AD, (Hyp.) and the corresponding lines lie in opposite directions, (115) Z^ = Z c, and Z5 =Zi). q.e.d. (117) Draw the diagonal DB. Then since BD is transverse to the ll's AB, CD, Abdc = Aabd; (110") similarly, Z CBD = Z ADB, and BD is common, .: AABD = ACBD, (63) .-. AB = CD, and AD = BC. Q.E.D. (70) 137. Definition. An intercept is the straight line, or part of a line, intercepted between two other lines. Thus a diagonal is an intercept through the angular points. 138. Cor. 1. Parallel interceiyts between parallels are equal. 139. Cor. 2. Two parallels are everywhere equally distant. 140. Cor. 3. A diagonal divides a parallelogram into two equal triangles. This was proved in the demonstration of Prop. XXXVI. QUADRILA TERALS. 63 Proposition XXXVII. Theorem. 141. // the opposite sides or angles of a quadri- lateral are equal, the figure is a parallelogram. Given: In the quadrilateral ABCD, 1°, AB equal to CD, EC equal to AD ; 2°, angle A equal to angle c, angle B to angle D ; To Prove : In either case, AB CD is a parallelogram. Draw the diagonal BD. Then 1°. Since AB = CD, AD = BC, and BD is common, (Hyp.) A ABD = A CBD, (69) .-. Zabd = Zbdc, and Zadb=Zcbd, (70) .♦. AB is 11 to CD, and AD is II to BC, (HO') .-. ABCD is a parallelogram, q.e.d. (131) 2°. . Since Za = Zc, and Zb =Zd, (Hyp-) Za-\-Zb = Zc-^Zd. ■ (Ax. 2) But Z ^ + Z 5 -f Z (7 -f ^ i> = 4 rt. Zs, (125) .-. Z^+ZJ5 = 2 rt. Zs, (Ax. 7) .-. ^DislltoJSC; (113') similarly, AB is 11 to CD, .-. ABCD is a parallelogram, q.e.d. (131) Exercise 69. Show that an angle of a triangle is obtuse or acute, according as it is greater or less than the sum of the other angles. 70. Show that the bisectors of opposite angles of a parallelogram are parallel. 64 PLANE GEOMETRY. — BOOK I. Proposition XXXVIII. Theorem. 142. If two sides of a quadrilateral are parallel and equal, the figure is a parallelogram. D c Given: In the quadrilateral ABCD, AB equal and parallel to CD; To Prove: ABCD is a parallelogram. Draw the diagonal BD. Then since 5Z> is transverse to the parallels AB, CD, A ABB = alt. Z CDB ; (110") also AB = CD, and BD is common, (Hyp.) .'. AABD z=ACBD, (66) .♦. Z ADB = Z CBD, (70) .-. AD is II to BC, (110') .-. ABCD is a parallelogram, q.e.d. (131) Exercise 71. A quadrilateral whose diagonals bisect each other is a parallelogram. 72. In a quadrilateral ABCD, if angle A is supplementary to angle B, and angle B to angle C, then the figure is a parallelogram. 73. If two parallelograms have an angle of the one equal to an angle of the other, they are mutually equiangular. 74. A parallelogram whose diagonals are equal is a rectangle. 75. A parallelogram whose diagonals bisect the angles through whose vertices they pass is equilateral. 76. If one angle of a parallelogram is a right angle, all its angles are right angles. Q UA DRILA TERA LS. Q^ Proposition XXXIX. Theokem. 143. Tivo parallelogj^aiivs are equal if two sides and the included angle of the one are respectively equal to two sides and the included angle of the other. A' Given: In parallelograms AC and A^c\ AB equal to A^B\ AD equal ta A^D\ and angle A equal to angle A^ ; To Prove : Parallelogram ^ C is equal to parallelogram A'c'. Place AC upon A'c', so that Za^Za'. Since AB = a'b', and AD = a'd', (Hyp.) B will fall on B', and D on 7)'. (14) •.• DC, D'C' are II to the coinciding lines AB, a'b', (Hyp.) DC must take the direction of D'c' ; ) > (Ax. 10) similarly, BC must take the direction oi B'c^, ) .'. C, the common point of DC, BC, must fall on c', the common point of D'c', B'c', .-. parallelogram ^C = parallelogram ^'C'. q.e.d. (61) 144. Scholium. Hence two adjacent sides and the in- cluded angle are said to determine a parallelogram. Exercise 77. If both diagonals of a parallelogram are drawn, of the four triangles thus formed, those opposite are equal. 78. The bisectors of those angles of a parallelogram that have a side common, meet at right angles. 79. In the diagram for Prop. IV., if OA be made equal to OB, and OC to OD, and AC, AD, BC, BD, be drawn, ABCD'^flW be a paral- lelogram. What condition must be fulfilled so that the figure will be equilateral ? Geom. — 6 66 PLANE GEOMETRY.— BOOK I. Proposition XL. Theorem. 145. An intercept passing through the mid point of a diagonal of a parallelogram is bisected in that point, and cuts off equal parts on the intercepting sides, o A E Given : In parallelogram A c, an intercept EF tlirougli 0, the mid point of diagonal BD ; To Prove : OE is equal to OF, and EB to FD. Since OB = OD, (Hyp.) also Z BDF = Z DBF (110"), and ^ at are equal, (50) Aboe = Adof, (63) .-. OE— OF, and EB = FD. Q.E.D. (70) 146. Cor. The diagonals of a parallelogram bisect each other. Exercise 80. In the diagram for Prop. XV., if A and B be each joined with C and D, then ADBC will be a square or a rhombus according as EA is or is not equal to EC. 81. If two equal isosceles triangles are constructed on opposite sides of the same base, what sort of a figure is obtained ? 82. Each interior angle of an equiangular polygon of six sides is the double of each interior angle of an equilateral triangle. 83. How many sides has an equiangular polygon if an interior is equal to one half of an exterior angle ? 84. How many sides has an equiangular polygon if an interior is the double of an exterior angle ? 85. In the diagram for Prop. XL., show that DEAD = OFCB. ' QUADBJLATERALS. 67 Proposition XLI. Theorem. 147. An intercept parallel to the base of a triangle and bisecting one side bisects the other also. Given: In triangle ABC, an intercept OE parallel Xo AB and bisecting BC m ; To Prove : ^ C is bisected in E. Complete the parallelogram AD by drawing BD, CD, parallel to AC, AB, respectively, and produce EO to meet BD in F. Since EF is an intercept through the mid point oiBC, (Hyp.) BF = EC. (145) But BF = AE, (136) (since AF is a parallelogram by construction,) .'.AE=EC. Q.E.D. (Ax. 1) 148. Cor. Conversely, if OE bisects both AC and BC, then OE is parallel to AB, and OE is equal to i AB. For OE must coincide with the parallel to ^^ through 0, since that parallel must pass through E, the mid point of BC (147). Also OE is equal to ^fe (145), and FE is equal to AB (136). Exercise 86.. In the diagram for Prop. XLI., show that OEAB can be superposed on OFDC. 87. In the diagram for Prop. XLII., if BD, CG, be drawn, show that these lines will intersect in the mid point of FH. 88. In the same diagram, if AD = BC, then the angles A and B are equal. 68 PLANE GEOMETRY. — BOOK I. Proposition XLTI. Theorem. 149. An intercept parallel to the bases of a trape- zoid and bisecting one of the nonparallel sides bisects the other also. D c H Given: In trapezoid AC, EF parallel to AB, bisecting AD in E and meeting BC in F ; To Prove : ^C is bisected in F. Draw DG II to BC, meeting AB, EF, in G, H, resp. Since in A DAG, EH is II to AG and bisects AD, (Hyp.) DG i^ also bisected in //. (147) But BH, HC, are parallelograms, (Const.) .-. BF = GH, and FC = HD or GH, (136) .: BF = FC. Q.E.D. (Ax. 1) 150. Cor. Conversely, if EF bisects both AD and BC, then EF is parallel to AB and equal to ^ (AB -{-CD). For EF must coincide with the parallel to ^J5 through E, since that parallel must pass through F, the mid point of BC (149). Also EF = EH -\- HF = i AG -h i {BG + CD) = i{AB-\-CD). Exercise 89. In the diagram for Prop. XLII., if CK be drawn parallel to AD, then will BK be equal to AG, and GK=2DG - AB. 90. If through P, the mid point of AE, FQ be drawn parallel to AB to meet BF in Q, then will ^^ be one fourth of ^C, and PQ wiU be equal to ^ (3AB+ CD). EXERCISES. 69 Proposition XLIII. Theorem. 151. If a series of parallels wiahe equal intercepts on one transversal, they make equal intercepts on any transversal. Given : Transversals AE, ae, cut by a series of parallels Aa, Bh, Cc, Dd, Ee, so that, the intercepts AB, BC, CD, BE, on AE are equal ; To Prove : The intercepts ab, be, cd, de, on ae, are also equal. If AE is II to ae, the opposite intercepts are equal. (136) If AE is not II to ae, then in the trapezoid ACca, since Bb is II to the bases Aa, Cc, and bisects AC, (Hyp.) ab = be. (149) In the same way may be proved that be = cd, and cd = de. .-. ab = bc = cd = de. q.e.d. (Ax. 1) 152. Cor. Conversely, if Aa, Bb, etc., make equal inter- cepts on AE and also on ae, then Aa is II to Bb, Cc, etc. (loO) EXERCISES. QUESTIONS. 91 . Would a triangle constructed of rods hinged at their extremities be rigid ; that is, incapable of change of form ? Would a parallelo- gram similarly constructed be rigid ? 92. Triangles having their sides severally equal have their angles also severally equal. Is the converse true ? 93. An acute angle being given, by what construction can you find its complement ? Its supplement ? 70 PLANE GEOMETRY. — BOOK I. 94. By what angle is the supplement of an angle greater than its complement ? 95. A right angle being 90°, how many degrees are there in the complement of an angle of 36° ? Of 45°? Of 90°? ^, -,^ How many in their supplements ? 96. Bisect an obtuse angle (81) ; also a straight angle. Can you bisect a reflex angle, i.e., an angle greater than a straight angle, by the same process ? 97. In Proposition XXIII., can you prove Cor. 2 independently? 98. Are all straight lines that cannot meet parallel ? 99. If one angle of a triangle is 23°, what is the sum of the other angles ? 100. If one angle of a triangle is equal to the sum of the other two, how many degrees has that angle ? What is it called ? 101. If the vertical angle of an isosceles triangle is 50°, how many degrees in each of the base angles ? 102. If a base angle of an isosceles triangle is 45°, what is the vertical angle ? 103. In an isosceles triangle, if each base angle is (1) twice, (2) three times, (3) n times, the vertical angle, how many degrees in each ? 104. How many degrees in each angle of an equilateral triangle ? 105. An acute angle of a right triangle is f of the other acute angle. How many degrees in each ? Generalize by putting ^— for f . n 106. Arrange the following terms in order of generality : square, polygon, rectangle, quadrilateral^ parallelogram. 107. I wish to cut off half a rectangular field by a straight fence. Through what point must the fence pass ? 108. The base of a triangle is fifty feet. How long is the line join- ing the mid points of the two sides ? 109. How many degrees in the sum of the interior angles of a polygon of four sides ? Of five ? Of six ? Oi n sides ? 110. How many degrees in each interior angle of an equiangular polygon of four sides ? Of five ? Of six ? Of ten sides ? 111. One angle of a parallelogram is double the other. How many degrees in each ? How many, if one is — of the other ? n GEOMETRICAL SYNTHESIS AND ANALYSIS. 71 GEOMETRICAL SYNTHESIS AND ANALYSIS. In the demonstration of propositions we have usually proceeded by the method of synthesis, or direct proof; that is, taking as a basis certain admitted truths, we built upon these a demonstration of the truth we wished to establish. In other words, in synthesis we reason from admitted prin- ciples to consequences (see, for example, 72). This, how- ever,^ though usually the most convenient way of presenting the proof of a proposition, does not show how that proof was invented; we are given a result, not the process by which the result was reached. In the method of analysis we proceed in the opposite way ; that is, assuming as true the conclusion we wish to establish, we reason back to principles. If we are led back to principles already known as true, we can take these as the basis of a synthetic proof of the conclusion we wish to establish. If, on the other hand, we are led to a contra- diction of a known truth, we know that the assumed con- clusion was false. This indirect method is often made use of in the demon- stration of theorems for which it would be inconvenient or difficult to find direct proof. In Prop. XXVII., for exam- ple, we show that CD must coincide with CE because their noncoincidence would entail a consequence that had been shown to be impossible. Instead of proving directly the conclusion we wish to establish, we prove it indirectly by showing that any other conclusion would lead to a contra- diction of some known truth. Among the exercises about to be given, most are so easy that the principles on which to base a synthetic proof at once suggest themselves. The student should, however, in every case, base his synthetic proof upon a previous analy- sis, guided by the general directions given below. The diagrams and references given as aids should, as far as possible, be left as a last resort. 72 PLANE GEOMETRY. — BOOK I. ANALYSIS OF THEOREMS. 1. Assuming the theorem as true, construct a diagram, accordingly. 2. Deduce, with the aid of SKch constructions as may he necessary, any consequences that follow from that assumption, a}iplying such theorems already proved as are ajyplicahle to the diagram. 3. A consequence that contradicts some known truth shoivs the theorem to be false, and ive can proceed to prove it so by a direct ptroof. 4. On arriving at a. consequence known to be true, take this as the basis of a synthetic proof, retracing the steps by tvhich this consequence was reached. Oiveii, for example, the following theorem : 153. Theorem. TJie bisectors of the angles of a triangle meet in a point. A Given: In triangle ABC, AD, BE, CF, the bisectors of angles A, B, C, respectively ; To Prove : AD, BE, CF, have a common point. Analysis. Let O be the point common to the three bisectors AD, BE, CF ; then must be equidistant from AB, AC, and BC (101). But this condition is satisfied if there is a point common to BE and CF, since that point must be equidistant from AB and BC, from BC and AC (101). Now, in order that BE and CF may meet in a point, we must have /. EBC + Z. FCB < a straight angle (114). But GEOMETRICAL SYNTHESIS AND ANALYSIS. 73 these angles are less than a straight angle, since they are the halves of A ABC, ACB, respectively, which are less than a straight angle (84). Hence Synthesis. Since Z B -\- Z c < a st. Z, (84) Z EBC + Z FOB < a St. Z, .'. BE, CF, will meet in a point 0. (H^) Now, since is in the bisector BE of Z ABC, is equidistant from BA and BC ; similarly is equidistant from CA and BC, .'. o is in the bisector AD of Z BAC, q.e.d. (101") (since it is equidistant from AB and AC.) (101') 154. Scholium. The point of intersection of the bisec- tors of any two angles of a triangle is equidistant from all the sides. 155. Theorem. The perpendiculars at the mid points of the sides of a triayiyle meet in a poiiit. Given: In triangle .45(7, DE, FG, HK, perpendiculars to AB, AC, BC, at their respective mid points ; To Prove : DE, FG, HK, have a common point. The analysis and synthesis of this theorem are so similar to those of the preceding theorem that they may be fairly left for the student to supply, with the remark that the analysis depends upon Prop. XXIII. and its first Cor. much as that of the preceding theorem depends upon Prop. XXV. /' OF THE \ I l^N/VERsiTYf 74 PLANE GEOMETRY.— BOOK I. 156. Scholium. The point of intersection of any two of the perpendiculars at the mid points of the sides of a triangle is equidistant from all the vertices. 157. Definitions. In any triangle, the perpendicular from a vertex to the opposite side is called the altitude to that side; the line joining a vertex with the mid point of the opposite side is called the mediaii to that side; the sum of the sides is called the perimeter. In an isosceles triangle, the equal sides may be referred to as the arms, and the other side as the base. EXERCISES. THEOREMS. 112. If the sides of a right angle BAC are produced through A^ the new angles thus formed are all right angles. (30, 50) 113. The bisectors of adjacent supple- mentary angles are perpendicular to each other. (46) 114. Conversely, if the bisectors of two adjacent angles are perpendicular to each other, those angles are supplementary. (46) 115. The bisectors of vertical angles are in the same straight line. (49) 116. If a line bisect one of two vertical angles, it bisects the other also. (50) 117. The intercepts bisecting the base angles of an isosceles triangle are equal. (63) 118. Any intercept drawn perpendicular to the bisector of an angle cuts off equal parts from the sides. (63) 119. The altitudes to the arms of an isosce- les triangle are equal. (63) C A B EXERCISES. 76 (66) 120. The bisectors of the base angles of an isosceles triangle form, if they meet, an isosceles triangle. (65) 121. The medians to the arms of an isosceles trian- gle are equal. {66) 122. Enunciate and prove the converse of Exer- cise 118. (66) 123. Enunciate and prove the converse of Exer- cise 120. (66) 124. Lines drawn from the extremities of the base of an isosceles triangle to points equally distant from the vertex are equal, and divide the arms into parts that are mutually equal. {66) 125. Lines drawn from the vertex of an isosceles triangle to points in the base equally distant from its extremities are equal. (66) 126. If equal distances from the vertices of an equilateral triangle be laid off in the same order, the lines joining these points form an equilateral triangle. 127. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. (66) 128. If a quadrilateral has two adjacent sides equal and making equal angles with the other two sides, these are also equal. (68) 129. An equilateral triangle is equiangular ; and conversely. (68, 65) 130. Prove that in the diagram for Exercise 128, the diagonals divide the figure into pairs of equal triangles. (69) 131. The perpendiculars to a diagonal of a parallelogram from the opposite vertices are equal. (73) 132. The diagonals of a square or rhom- bus bisect each other at right angles, and bisect the angles whose vertices they join. (74) 133. If from any point P in A ABC, PB, PC, be drawn, then Z P> Z^l, but PB + PC, =, or < AC. (91) 138. The median to any side of a triangle is less than the semisum of the other two sides, but greater than the semidifference of their sum and the third side. (88) 139. A line drawn from any point in the bisec- tor of an angle, parallel to one side, meets the other in a point equidistant from the vertex and the point. (110) 140. If AB., an arm of the isosceles triangle ABC, be produced so that AD = AB, then will the line joining Z>C be perpendicular to BC (120) 141. If BD is the altitude upon AC, one of the arms of the isosceles triangle ABC, then the angle DBC is equal to one half the vertical angle. (120) 142. The sum of the medians to the sides of a triangle is less than the sum, but greater than the semisum, of the sides. (Exercise 138) 143. If AP is the altitude, and AM the bisec- tor, from A to BC, then ZPA3f=l(ZC~ZB). (120) B' MFC 144. The median to the hypotenuse of a right triangle is equal to one half the hypotenuse. (120) EXERCISES. 77 145. If the bisectors of two angles of an equilateral triangle meet, and, from the point of meeting, lines be drawn parallel to any two. sides, these lines will trisect the third side. (120) 146. The bisector of an exterior angle at the vertex of an isosceles triangle is parallel to the base. (122) 147. If, in a triangle ABC, the bisectors of an interior angle at B and of an exterior angle at C meet in D, then ZD = ^ZA. 148. The angle formed by the bisectors of any two consecutive angles of a quadri- lateral is equal to the sum of the other two angles. (125) 149. The sum of the distances of any point in the base of an isosceles triangle from the arms, is constant ; i.e., is always equal to the altitude upon an arm. (143) 150. The sum of the distances of any point within an equilateral triangle from the sides is constant; i.e., is equal to an altitude. (Exer- cise 148) 151. The lines joining the mid points of the sides of a triangle divide it into four equal triangles. (152) 152. The three altitudes of a triangle have a common point. (141, 155) 153. The lines joining the mid points of adjacent sides of any quadrilateral, form a parallelogram the sum of whose sides is equal to that of the diagonals of the quad- rilateral. (147) 154. The three medians of a triangle have a common point which cuts off one third of each median. (147) 155. If the exterior angles of a triangle are bisected, the three exterior triangles formed on the sides of the original triangle are equi- angular. 156. The vertices of all right triangles having a common base as hypotenuse, lie in the same circumference. (Exercise 144) Book II. the circle. loci. problems. »« Zaof. (Ax. 8) Now OA, OB — OA, OF, respectively, (162) .'. AB >AF or CB. Q.E.D. (90) 2°. Given: In circle ADB, chord AB greater than chord CD ; To Prove : Arc AFB is greater than arc CED. With the same construction as in 1°, since OA, OB = oA, OF, respectively, (162) but AB>AF, (Hyp.) Zaob>Zaof, (90) .-. F falls between A and B, ,\ arc AFB > arc AF or arc CD. q.e.d. (Ax. 8) CHORDS. 87 Scholium. If the arcs are in equal circles, the proof is similar to that above ; and, in general, the demonstration of a theorem concerning arcs, etc., in equal circles would be similar to that for arcs in the same circle, and vice versa. Propositiox IX. Theorem. 18^. In the same circle, or in equal circles, equal chords are equally distant froin the center; and conversely. 1°. Given : In circle ADB, chord AB equal to chord CD ; To Prove : AB and CD are equally distant from 0, the center. Draw OP J_ to ^5, OR 1. to CD, and join OA, OC. Since OP is ± to AB, and OR is ± to CD, (Const.) AP = 1- AB, and CR = ^ CD, (172) .-. AP = CR (Ax. 7), and OA = OC, (162) .-. rt. AOAP = rt. A OCR, (72) .-. OP = OR. Q.E.D. (70) 2°. Given : In circle ADB, chords AB, CD, equally distant from 9, the center ; To Prove : AB = CD. With the same construction as in 1°, since OP = OR (Hyp.), and OA = 00, (162) rt. A OAP = rt. A OCR, (72) .-. AP= CR, (70) ,-. 2ap or AB = 2CR OT CD. Q.E.D. (Ax. 6) 88 PLANE GEOMETRY. — BOOK II. Proposition X. Theorem. 183. In the same circle, or in equal circles, a greater chord is nearer the center; and conversely. 1°. Given : In circle ADB, op perpendicular to chord AB, OQ perpendicular to chord CD, and AB greater than CB ; To Prove : OP is less than OQ. Since AB > CD (Hyp.), arc AEB > arc CD, (181") On arc AEB Islj off arc ^^ = arc CD. (1'<'8) Join AE, and draw OR _L to AE ; (92) then AE= CD (174'), and 0R= OQ. (182') Since AB lies between and AE, (Const.) OR must cut AB in some point S, .'. OS < OR. (Ax. 8) But OP < OS, ' (93) .-. OP <0R or OQ. Q.E.D. (a.f.) 2°. Given: In circle ADB, OP perpendicular to chord AB, OQ perpendicular to chord CD, and OP less than OQ; To Prove: Chord AB is greater than chord CD. Chord AB cannot be equal to chord CD, for then OP would be equal to OQ (182') ; nor can AB be less than CD, for then OP would be greater than OQ (1°). Hence, since AB can be neither equal to nor less than CD, it must be greater than CD. 184. Cor. A diameter is greater than any chord not pass- ing through the center. CHORDS. 89 Proposition XI. Theorem. 185. Through any three given points not in the same straight line one circumference can be described, arid only one. Given : Three points A, B, C, not in the same straight line; To Prove: One circumference can be described through A, B, and C. , Join AB, BC, and at D, E, the mid points of AB, BC, respec- tively, draw DO ± to AB, and EO ± to BC* (80) ••• DO,EO, are ± to AB,BC, resp., at their mid points, (Const.) DO, EO, will meet at a point equidistant from A, B, and a; (156) .-. the circumference described from as center, with radius OA, will pass through A, B, and C. q.e.d. (162) Moreover, there can be but one such circumference. For the center of any circumference passing through A, B, and C, must lie both in DO and in EO (97), which lines can have but one common point. 186. CoR. Two circumferences can intersect in not more than two points. For if they could intersect in three points, there would be two different circumferences passing through the same three points, which is impossible. 187. Scholium. One circumference, and only one, can be described through the vertices of a given triangle. * The construction by wliich we find the mid point of a line, evidently also gives the perpendicular at that point. 90 PLANE GEOMETRY. — BOOK II. TANGENTS AND SECANTS. 188. A tangent is a straight line that touches a circum- ference in one point only; as ABC. The point B in which the tangent touches the ^i circle is called the point of contact. 189. A secant is a straight line that cuts a circumference in two points ; as DE. Proposition XII. Theorem. 190. A straight line perpendicular to a radius at its extremity is a tangent to the circle. Given: AB perpendicular to 0(7, a radius of circle CEF, at its extremity C ; To Prove : AB is a tangent to circle CEF. Join O with D, any point in AB except C. Then since oc is ± to AB, (Hyp.) OD is oblique to AB, (51) .-. 0D> oc, (93) .*. D lies without the circle CEF, (162) .-. AB is tangent to CEF at C, q.e.d. (188) (since C is the only point in AB not without CEF.) 191. Cor. 1. A tangent is perpendicular to the radius drawn to the point of contact. For OC must be less than any other line drawn from to AB (162). TANGENTS AND SECANTS. 91 192. Cor. 2. A perpendicular to a tangent at the point of contact passes through the center of the circle. For otherwise there could be two perpendiculars to the tangent at that point (191). 193. CoR. 3. A perpendicular from the center to a tangent meets it at the point of contact (192). 194. CoR. 4. At a given point of contact there can he hut one tangent. Proposition XII I. Theorem. 195. Parallels intercept equal arcs on a circum- ference. Given: Two parallels AB, CD, cutting or touching the cir- cumference whose center is ; To Prove: These parallels intercept equal arcs. 1°. Let AB, CD, be secants. Through 0, the center, draw the diameter MN _L to AB. Since AB is II to CD (Hyp.), and MN is _L to AB, (Const.) MN is ± to CD, (107) .-. arc MA = arc MB, and arc MC = arc MD, (175) .-. arc ^C = arc i5Z). q.e.d. (Ax. 3) 2°. Let AB be tangent at M, and CD a secant. Draw OM. Then OM is ± to AB (191) and also to its parallel CD ; (107) .-. arc i^/C = arc J/JD. q.e.d. (175) 92 PLANE GEOMETRY. — BOOK IL 3°. Let AB, CD, be tangents at M, N, respectively. ^e/^^^\^^^f Draw a secant EF II to AB ; it will also "/ V' be II to en. (108) ( ^ 1 Since arc me = ^yc MF, and arc NEz= \ / arc NF, (2°) c^^N^D arc 3fEN = arc MFN. q.e.d. (Ax. 2) 196. Definition. The line joining the centers of two circles is their Zme of centers; the distance between their centers is their central distance. 197. Two circles are tangent to each other when both are tangent to the same straight line at the same point. 198. A circle tangent to another is tangent to it internally or externally according as it lies within or without the other circle. Proposition XIV. Theorem. 199. If two circles are tangent to each other, their line of centers passes through the point of contact. Given: Circles with centers 0, O', tangent at P in ^5; To Prove : P lies in the line joining oo'. Through P draw a perpendicular to AB. (95) Since ^^ is tangent to both circles at P, (Hyp.) the J_ to AB at P passes through O and O', (192) i.e., P is in the line of centers 00'. q.e.d. EXERCISES, 93 EXERCISES. QUESTIONS. 180. Can you find the center of a circle without bisecting any straight line ? 181. If a chord of one circle is equal to a chord of another, are these circles necessarily equal ? If not, what other condition must be fulfilled that the circles may be equal ? 182. In the diagram for Prop. VII., if iHfiVwere equal to a diame- ter, through what point would AB necessarily pass ? 183. In equal circles, which is subtended by the greater chord, the greater or the less of two major arcs ? 184. Can a tangent be drawn to a circle from a point within it ? 185. How may a tangent be drawn at a given point in a circum- ference when the center is not known ? 186. How many points determine a circumference ? 187. Is there any limit to the number of circles that may be described through two given points ? 188. In what case is it impossible to pass a circle through three given points ? 189. Can a circumference always be described through the angular points of a rectangle ? 190. How many circles of equal radii may touch a given straight line at a given point ? 191. How many circles of any radii may touch a given straight line at a given point ? THEOREMS. 192. A circle is wholly without or within another circle, according as their central distance is greater than the sum, or less than the dif- ference, of their radii. (162) 193. A circle is tangent to another externally or internally, accord- ing as their central distance is equal to the sum or the difference of their radii. (199) 194. Two circles intersect if their central distance is less than the sum, and greater than the difference, of their radii. (88, 89)^ 94 PLANE GEOMETRY. — BOOK IL 195. If two circles intersect, their line of centers is perpendicular to their common chord at its mid point. (172) 196. The center is the only point from which more than two equal lines can be drawn to the circumference. (97, 171) 197. If two equal chords intersect, their segments are severally equal. (174, 69) 198. If through any point in a radius two chords be drawn making equal oblique angles with it, these chords are equal. (182") ,199. The chord drawn through any point of a radius perpendicular to it is the least chord that can be drawn through that point. (183) 200. The tangents drawn to a circle from any point without it are equal, and make equal angles with the line joining the point with the center. (192) 201. The sums of the opposite sides of a quadrilateral described about a circle are equal. (Exercise 200) 202. Tangents AD, BC, at the extremi- ties of a diameter AB, meet another tangent CD in C and D; join 00, OD. (1) COD is a right angle ; {2) CD = AD + BC. (Exercise 200) 203. If through the points of intersection of two circumferences, parallels be drawn to meet the circumferences, these parallels will be equal. (195, 174) 204. If two tangents drawn to a circle from the same point, inter- cept between them a third tangent touching any point of the inter- cepted arc, the perimeter of the triangle formed by the three tangents is constant. (Exercise 200) 205. If a circle is inscribed in a triangle, the distances of the vertex of any angle to the points of contact of its sides are equal to the semi- perimeter of the triangle less the side opposite the angle. (Exercise 200) 206. If a circle is inscribed in a trapezoid that has equal angles at its base, each nonparallel side is equal to half the sum of the parallel sides. (Exercise 200) 207. If two circles are each tangent to a pair of parallel lines and also to a transverse intercept between the parallels, the intercept is equal to the central distance of the circles. (Exercise 200) CONSTRUCTIONS. 95 CONSTRUCTIONS. Certain problems of construction have already been given as occasion arose for their use. The form in which they were presented was not, however, always the simplest pos- sible, but such as the means of demonstration at that stage admitted. Those already given are, accordingly, presented again in a simplified form, with the directions for their con- struction worded as briefly as possible. The demonstra- tions the student should complete for himself with the aid of the diagrams and references. 200. To bisect a given straight c .. ^ or arc, AB. From A and B, with any radius greater than half the distance AB, describe arcs intersecting in C and D. The line CD bisects the line AB va. X, and the arc AB in y (74, 175). i)- 201. To bisect a giveyi angle BAG. From A, with any radius, describe an arc cutting AB, AC, in D, E, re- spectively. From D and E, with the same radius, describe arcs inter- secting in X. The line AX is the bisector of Abac (101"). 202. To draw a perpendicular to a given line AB through a point C hi or without AB. ... X .. From C, with any suitable radius, describe an arc cutting AB inD and E. From D and E, with equal radii / of suitable length, describe arcs ^ — intersecting in X. The line CX is perpendicular to AB (74). 96 PLANE GEOMETRY. — BOOK IL It will be seen that the simplification in these cases arises from the use of arcs of any convenient radius instead of whole circumferences of prescribed radius, and in the omis- sion of construction lines needed only for demonstration. Proposition XV. Problem. 203. At a point C, in a given straight line AB, to construct an angle equal to a given angle 0. F JJ Given: A point C in a straight line AB, and an angle ; Required : To make at C an angle equal to angle 0. From as center, with any radius, describe an arc cutting the sides of Z in D and E. From C as center, with the same radius, describe an arc FG, and on FG set off an arc FX equal to arc DE. (179) Join ex. FCX is the required angle. Join FX. Then since FX and DE are equal arcs of equal circles, (Const.) chord FX = chord DE ; (174') also CF and CX = OD and OE, resp. (Const.) .-. AFCX = Adoe. .-. Za = Zo. Q.E.F. CONSTRUCTIONS. 97 Proposition XVI. Problem. 204. Through a given point to draw a parallel to a ^iven line. Given : A straight line AB and a point C ; Required: To draw tbrougli C a parallel to AB. Through C draw ECD, making any Z D with AB. At the point C in ED make Z ECF = /.B. (203) F^QF is the required parallel. For since Ac = Ab, (Const.) F^CF is II to AB. Q.E.F. (112') Exercise 208. The common chord of two intersecting circles may he a diameter of one, hut not of hoth. 209. The chords joining the extremities of equal parallel chords of the same circle form with them a rectangle. 210. In the diagram for Prop. IX., if BA, DC, be produced to meet in Q, the line joining OQ will bisect /.BQD. 211. Hence show that, in the diagram for Prop. VIII., if AC, DF, be joined, AC will be parallel to DF. 212. In the diagram for Prop. XI., show that Z is the supplement of Z J5. 213. In the same diagram, if OD, OE, be produced to meet the circumference in i^ and G, then will arc FBG = ^(arc AB + arc BC). 214. In the same diagram, if UK be tangent to the circle at B, then will Z HBD + Z KBE be equal to Z O. 215. Tangents at the extremities of a diameter are parallel, and conversely. If two radii are at right angles to each other, the tangents at their extremities are perpendicular to each other. Geom. — 7 98 PLANE GEOMETRY. — BOOK 11. Proposition XVII. Problem. 205. To construct a triangle having its sides respec- tively equal to three given straight lines, each less than the sum of the other two. Given : Three straight lines, A, B, c, each less than the sum of the other two ; Required : To construct a triangle having sides equal to A, B, C, respectively. As base, draw a line BE = A say. From B as center, with radius equal to B, draw an arc. From E as center, with radius equal to C, draw another arc. Since BF -{-FE> BE, the central distance, (Hyp.) the arcs will intersect in some point F. ' Join BF, EF. BEF is the required A ; its sides being equal to A, B, C, resp. q.e.f. (Const.) Scholium. If the given lines are all equal, the triangle will be equilateral; if two are equal, the triangle will be isosceles. Exercise 216. The angle formed by two tangents drawn to a circle from the same point, is supplementary to that formed by the radii to the points of contact. 217. Show that the angle formed by those radii is bisected by the line joining its vertex with the center of the circle. 218. If a tangent and a secant be drawn to a circle from the same point, the angle formed by them will be equal to, or supplementary to, that formed by radii perpendicular to them, according as they do or do not lie on the same side of the center. CONS TR UC Tl ONS. 99 Proposition XVIII. Problem. 206. To construct a parallelograin with adjacent sides respectively equal to two given lines, and includ- ing a given angle. Given : Two straight lines A and B, and an angle C ; Required : To construct a parallelogram with two sides equal to A, B, respectively, including an angle C. Draw a line DE = A, and at D make Z GDE = Zc. (203) On DG lay off DF = B. From F as center, with radius equal to DE, describe an arc HK. From E as center, with radius equal to DF, describe an arc HL. Let the arcs intersect in H, and jpin FH, EH; DII is the parallelogram required. Since FII = DE, EH = DF, and Zd = Zc, (Const.) DH is a parallelogram (141), with sides = A, B, resp., and with an angle = Z C. q.e.f. Scholium. When the given angle is a right angle, this construction gives a rectangle. When the given sides are equal, the construction gives a square or rhombus, according as the given angle is or is not a right angle. Exercise 219. Construct a rectangle having one side double the other. 220. Construct a square and circumscribe a circle about it, 100 PLANE GEOMETRY. — BOOK 11. Proposition XIX. Problem. 207. To divide a given straight line into any num- ber of equal parts. lu IS n h 10 .9 d... Given : A straight line AB ; Required : To divide AB into n equal parts. Let n = l. Draw AC, making any convenient Z with AB^ and draw BD II to AC. (204) On A C lay off 7 equal parts, A-1, • • • 6-7, of any convenient length. On BD lay off 7 equal parts, ^-8, ••= 13-14, of the same length as in AC. Join ^-14, 1-13, ••• 1-B. AB is divided into 7 equal parts. Since A-1 is equal and parallel to 13-14, (Const.) ^-14 is II to 1-13. (142) Similarly, all the lines 2-12, 3-11, ••. 1-B, are || to ^-14. Since these II 's cut off 7 equal intercepts on A-1, (Const.) these ll's cut off 7 equal intercepts on AB, (151) i.e., AB is divided into 7 equal parts. q.e.f. We proceed in the same way if n is any other integer. Scholium. In practice, it will be sufficient to lay off only one part on BD, as -B-8 ; then by joining 8 with the corre- sponding division on AC, as 6, we cut off one of the required equal parts on AB ; then lay off the others. CONSTRUCTIONS. 101 208. Definition. A circle is said to circumscribe sl polygon when its circumference passes through each of the angular points of the polygon, which is then said to be inscribed in the circle. 209. Definition. A circle is said to be inscribed in a polygon when it touches each of the sides of the polygon, which is then said to be circumscribed about the circle. Proposition XX. Problem. 210. To circuTYhscrihe a circle about, or to inscribe a circle in, a given triangle. Given : Two triangles AB C, A 'B 'c'. Required: 1°, To circumscribe a circle about ABC; 2°, To inscribe a circle m a'b^c\ 1°. When, by the construction given in Prop. XI., a cir- cumference has been passed through the three points A, B, C, we have a circle circumscribing A ABC (208). q.e.f. 2°. When, by the construction given in Art. 153, we have found a point o' that is equidistant from A'b', A'c', B'c', then the circle described from o' as center, with any of the equal perpendiculars from O' to the sides as radius, will touch each of the sides, and thus be inscribed in A a'b'c' (209). *Q.E.F. 102 PLANE GEOMETRY. — BOOK II. PLANE LOCI. 211. The nature of loci becomes clear when we conceive a line as being the path generated by a point moving ac- cording to some specified condition. If the point move so as to be always at a given distance from a given fixed point, the line thus traced will be the locus of all points that are at the given distance from the given point. Hence, A locus in a plane is the line, or lines, every point in which satisfies certain conditions fulfilled by no point not in that line or lines. The following theorems upon the subject are merely the enuncia- tion from a special point of view of certain theorems already proved. Though the proofs are here merely indicated by references, the student would find it useful to frame a demonstration of each of these theo- rems, keeping in mind that, in order that a line be a locus, (1) Evey-y point in the line must satisfy the given conditions; (2) No point not in the line should satisfy them. 212. Theorem. The locus of all the points situated at a given distance from a given point is the circumference described from that point as center with a radius equal to the given distance. (162) 213. Theorem. The locus of all the points that are equi- distant from two given points is the p)erpendicular at the mid point of the line joining those points. (20, 96) 214. Theorem. The locus of all the points situated at a given distance from a given straight line consists of the two parallels drawn at the given distance from the given line, one on each side. (139) 215. Theorem. The locus of all the points thai are equi- distant from two given parallels is the parallel through a point equidistant from both. (139) 216. Theorem. The locus of all the points that are equi- distant from two intersecting lines consists of the bisectors of the vertical angles formed by those lines. (101) EXERCISES. 103 217. Theorem. The locus of the mid points of parallel chords is the diameter perpendicidar to those 'chords. (172) 218. Theorem. The locus of the mid points of equal chords is the circumference conce7itric with the given circumference, and having a radius equal to the common distance of the chords. (182) Familiarity with these and similar theorems is of great imj)drtaiice, especially in the solution of problems, which very often depends upon the intersection of loci. The fol- lowing easy exercises will be found useful as an introduction to the application of loci to the solution of problems. EXERCISES. LOCI. Note. — The line AB referred to in these exercises is supposed to be a straight line of indefinite — that is, of any requisite — length. 221. In the line AB find a point that shall be at a given distance CB from a given point P. State the conditions under which no such point can be found ; one point ; more than one. 222. In the line AB find a point that shall be equidistant from two given points Pand Q. State the conditions, etc., as in Exercise 221. 223. In the line AB find a point that shall be at a given distance CD from a given line EF. State the conditions, etc. 224. In the line AB find a point that shall be equidistant from two given parallels CD, EF. State the conditions, etc, 225. In the line AB find a point that shall be equidistant from two intersecting lines CD, EF. State the conditions, etc. 226. In a given circumference ABM find a point that shall be at a given distance CD from a given point P. State the conditions, etc. 227. In a given circumference ABM find a point that shall be equi- distant from two given points P and Q. State the conditions, etc. 228. In a given circumference ABM find a point that shall be at a given distance CD from a given line EF. State the conditions, etc. 104 PLANE GEOMETRY. — BOOK II. 229. In a given circumference ABM find a point that shall be equi- distant from two given parallels CD, EF. State the conditions, etc. 230. In a given circumference AB3Ifm6. a point that shall be equi- distant from two intersecting lines CD, EF. State the conditions, etc. 231. Find a point equidistant from two given points Pand Q, and at a given distance CD from a given line AB, State the conditions, etc. 232. Find a point equidistant from two given points P and ^, and also equidistant from two parallels AB and CD. State the con- ditions, etc. ,233. Find a point equidistant from two given points P and ^, and also equidistant from two intersecting lines AB and CD. State the conditions, etc. 234. Within a given angle BAC find the point that is at a given distance DE from each of the sides. 235. Find a point that shall be at a given distance AB from a given point P, and at a given distance CD from a given point Q. State the con- ditions, etc. 236. Find the locus of a point P such that the sum of its distances from the sides of a given angle BAG is always equal to a given line DE. (101) 237. Find the locus of a point P such that the difference of its distances from the sides •• of a given angle BAC is always equal to a given line DE. (101) 238. Find the locus of the mid points of all the lines drawn from to a line AB. (147) 239. Find the locus of the mid point of a ladder whose lower end is being pulled away from a wall. (144) ANALYSIS OF PROBLEMS. 105 ANALYSIS OP PROBLEMS. Most of the foregoing simple exercises are problems depending so plainly for solution upon the intersection of certain loci that no previous analysis is necessary. In general, the solution of a problem will be found to depend upon several problems or theorems already solved or proved, and the solution will be greatly facilitated by a previous analysis. From the nature of the case, no precise rules can be given, but the general course of procedure is expressed by the following rules. 1. Construct a diagram in accordance tvith the statement of the problem, as if the required construction ivere effected. 2. Study the relations of the lines, angles, etc., in the dia- gram, so as to discover whether the assumed solution can be made to depend upon some known problem or theorem, espe- cially those concerning loci. 3. If such dependence cannot be found by means of the original diagram, make such additions to it as the case may suggest, by joining points, draiving parallels or perpendiculars, etc., and proceed as in 2. . 4. On discovering the dependence of the solution upon some known theorem or problem, make this the basis of a synthetic solution, proceeding in reverse order through the several steps of the previous analysis, till it is shown that the problem is solved. For example : 219. Problem. Through a given point within an angle draw a straight line intercepted between the sides and bisected in the given point. ^X Analysis. 1. Let P be the given yy point within Z BAC, and suppose xy 1)^^/ drawn through P so that Px = Py. y^ 7 2. The diagram as it stands is not y"^ / suggestive. Ay b 106 PLANE GEOMETRY.— BOOK 11. 3. Through P draw PD II to AB, to meet AC in D. 4. It is at once apparent that since PD is drawn through the mid point of xy and is II to AB, .'. Z) is the mid point of Ax. (1^7) Synthesis. Having thus found that the solution depends upon Prop. XLI. (147), we proceed as follows : Through P draw PD II to AB to meet AC in D. (109) In DC take Dx = DA, and through x draw xPy, meeting AB in y. xy is the required line. Since D is the mid point of Ax, ") /n i.\ and DP is II to AB, the base of AxAy, ) ^ "^ .-. xy is bisected in P. q.e.f. (147) The complete diagram required in the solution of a prob- lem contains : 1. The data, or given figure ; as point P in Z BAC, above. 2. The qiimsita, or things required ; as the line xPy. 3. Construction lines employed as auxiliaries in the solu- tion ; as PD. Of these, the data may conveniently be distinguished, as above, by the first letters of the -alphabet ; the qusesita, as far as requiring new letters, by the last letters; and the auxiliary constructions by dotted lines. We give another example in order to show the application of the principle of intersecting loci to the solution of problems. Pkoblem. Describe a circumference passing through two given points and having its center in a given straight line. Analysis. 1. Let AB be the given line, and C, D, the given points ; and suppose circle CDE, having its center at X in AB, passes through c and D. nXEkCtSES. 107 2. Since X is equidistant from c and i), it should be found in the locus of all points that are equidistant from C and D. Now that locus is (213) the' perpendicular at the mid point of the line joining C and D ; .: X is found at the intersection of that locus with AB. Synthesis. Join en. Draw the J-FH at the mid point of CD, and let FH intersect AB in X. From X as center, with a radius equal to the distance XC, describe the circum- ference CDE. CDE is the required circumference. Since X is a point in the ± at the mid point of CD, X is equidistant from C and D, (213) .-. a circumference passing through C will also pass through D, (164) .*. a circumference having its center in AB has been described through C and D. q.e.f. Scholium. The problem becomes impossible in a certain case. In what case ? EXERCISES PROBLEMS. 240. Construct an isosceles triangle having its sides each double the length of the base. 241. Upon a given base AB, construct a right isosceles triangle. 242. With a given line ^S as diagonal, con- struct a square. 24.3. Construct an equilateral triangle having a given altitude AB. 108 PLANE GEOMETRY. — BOOK IT. 244. In any side of a triangle, find the point which is equidistant from the other two sides. 245. In any side of a triangle, find the point from which the lines drawn parallel to the other side are equal. 246. Trisect a given right angle. 247. In a given line AB^ find a point X such that the angle formed by lines drawn from X to two given points C, 2), on opposite sides of AB^ shall be bisected by AB. 248. In a given line AB., find a point X such that the angles formed by lines drawn from Xto two given points C, 2), on the same side of AB, shall be equal. How may Exercises 247 and 248 be enunciated as one problem ? 249. From a point P, without a given line AB, draw a line PXsuch that BXA shall equal a given angle C 250. Find the bisector of the angle that would be formed by two given lines, without producing the lines. 251. Through a given point P, draw a line that cuts off equal parts from two intersecting lines. 252. Draw an intercept parallel to the base of a triangle such that it shall be equal to the sum of the intercepts between it and the base. 253. From a given isosceles triangle cut off a trapezoid having for base that of the triangle, and having its other three sides equal. 1 EXERCISES. 109 254. Three lines being given diverging from a point, draw a fourth line cutting them so that the intercepted segments shall be equal. 255. Construct an isosceles right triangle, the sum of the hypotenuse and a side being given. 256. Construct an isosceles right triangle, the difference of the hypot- enuse and a side being given. 257. > Two angles of a triangle being given, find the third angle. 258. Construct an isosceles triangle of given altitude, whose sides pass through two given points, and whose vertex is in a given straight line. Construct an isosceles triangle, having given : 259. The base and the vertical angle. 260. The base and a base angle. 261. An arm and the vertical angle. Construct a triangle, having given : 262. Two sides and the included angle. 263. The base and the base angles. 264. The three sides, AB, AC, BC, such that AC = ^AB, and BC = iAC. 265. Construct an isosceles right triangle, having given the sum and the difference of the hypotenuse and an arm. Hint. — It is useful to remember that ^ and B, being any two mag- nitudes, (A + B)-{-iA-B)=2A; (A-^ B) - {A- B) =2B. 266. Construct a right triangle, having given an arm and the alti- tude from the right angle upon the hypotenuse. 267. Construct a right triangle, having given the hypotenuse and the difference of the other sides. 110 PLANE GEOMETRY. — BOOK 11. Construct a parallelogram, having given : 268. Two adjacent sides and a diagonal. 269. A side and both diagonals. 270. Both diagonals and their included angle. Construct a trapezoid, having given : 271. The four sides. 272. The parallel sides and the diagonals. 273. The parallel sides, a diagonal, and the angle formed by the diagonals. 274. Through a point within a circle, draw a chord that is bisected in that point, and show it is the least chord through that point. 275. The position and magnitude of two chords of a circle being given, describe the circle. 276. In a given circle, draw a chord whose length is double its distance from the center. 277. Draw that diameter of a given circle, which, being produced, meets a given line at a given distance from the center. When is this impossible ? 278. Describe a circle with given radius, to touch a given line in a given point. How many such circles can be described ? 279. Describe a circle of given radius to touch two intersecting lines. How many such circles can be described ? 280. Describe a circle touching two intersecting lines at a given distance from their intersection. How many such circles can be described ? 281. Describe a circumference passing through a given point, and touching a given line in a given point. 282. Describe a circumference touching two given lines, and passing through two given points between those lines. 283. From a given center, describe a circumference that bisects a given circumference. 284. With a given radius, describe a circle touching two given circles. Book HI. RATIO. PROPORTION. LIMITS. MEASUREMENT. For many purposes, as in the propositions thus far con- sidered, it is sufficient to prove, in regard to two given magnitudes, that they are equal or unequal. Thus we proved, in Prop. XXIV. (99), that PA = PB, and PC> PA. We have now to consider how to proceed when we wish to estimate exactly the relative greatness of given magnitudes. 220. To measure a magnitude is to find out how many times it contains another magnitude of the same kind. Thus we measure a line by finding how many times ic contains another line called the U7iit of length, or linear unit. This unit may be either a standard unit, as an inch, a meter, etc., or a unit found by dividing a line into any desired num- ber of equal parts, as in Prop. XIX. (207). 221. A quantity is a magnitude conceived as consisting of some number of equal parts. Thus AB, regarded merely as a line, is a magnitude ; but when thought of, or re- ferred to, as 22 millimeters, or x linear units of any kind, it is a quantity meas- ured by millimeters, or some other unit. The angle BAO, again, if referred to as an angle of 31° 15' 47.2", is a quan- tity measured by tenths of seconds. Ill 112 PLANE GEOMETRY. — BOOK III. 222. The number that expresses how many times a quan- tity contains the unit is called the nmnerical measure of that quantity. Ths numerical meas- ure may be a number of any kind, -^— - — ■ r integral or fractional, rational or irrar ^ b tional, or any letter denoting number. Thus \i AB is divided into 11 equal parts, of which CD is found to contain 8|, and EF, x parts, then 11, 8|, and x are the numerical measures of AB, CD, and EF, respectively. If two quantities of the same kind, but expressed in dif- ferent units, are to be compared as quantities, it is evident that both must be expressed as consisting of units of the same kind before the comparison can be effected. Thus if AB were 8 rods and CD 15 yards, we cannot compare them until we have expressed them as 132 feet and 45 feet, for example. In every case, then, where numerical measures are compared, they are to be understood as refer- ring to the same unit. It may also be observed that though abstract numbers are properly only the numerical measures of quantities, yet they are conveniently regarded as quanti- ties whose unit is not expressed. 223. If a quantity is contained an exact number of times in each of two like quantities, it is called a common measure of these quantities, which are then said to be commensurable with each other. Thus a rod and a yard are commensurable, since they have 2, 3, 6, 9 inches as common measures. 1\ inches and 13J^ inches are commensurable, since they have f of an inch as a common measure ; and any two quantities are commen- surable when, referred to the same unit, their numerical measures are rational numbers. 224. If two quantities have no common measure, they are said to be incommensurable with each other. Thus every unit of our common system of measures is RATIO. 113 incommensurable with every like unit of the metric sys- tem ; and, as afterwards will be seen, the diagonal and side of a square are incommensurable with each other, as are also the diameter and circumference of a circle, etc. When such quantities are compared, their numerical measures are either approximate, as when we say 3 meters = 118.11237 -|- inches, or the numerical measures are such symbols as y'2, -^3, TT, etc. RATIO. 225. The ratio of two quantities is their relative greatness as expressed by the quotient of the one by the other. Thus the ratio of 15 inches to 7 inches is 15 in. -j- 7 in. = \^- ; the ratio of 8 rods to 15 yards is (reducing to a common measure) 132 ft. -j- 45 ft. = ^^-^- ; and, generally, if the nu- merical measures of two quantities are a and h, the ratio of these quantities is -, a and b being any numbers whatever. 226. The ratio of the quantity ^ to a like quantity B is denoted symbolically by either of the expressions A : B, or — , B each of which is read, the ratio of A to B. In each, A is called the first term or antecedent; B, the second term or consequent. 227. When A and B are commensurable quantities, the value of their ratio is expressed exactly by the fraction de- noting the quotient of the numerical measure of the ante- cedent by that of the consequent. But if the quantities are incommensurable, no rational fraction can express their ratio exactly, since then they would not be incommensurable. Yet, by taking the unit of measure sufficiently small, we can find a fraction that expresses the true value of the ratio to as near a degree of approximation as we please. Geom. — 8 114 PLANE GEOMETRY. — BOOK III. Thus suppose we have two lines A and B, whose nu- merical measures are -yj2 and 1, re- spectively. Now, carried to seven decimals, ^2 = 1.4142135 + ; that ^ is, V2 > 1.414213 and < 1.414214, so that the ratio of A to B or ^2 : 1 lies between TFoiroinj" ^^^ 1 o'O"? and must differ from either by less than one millionth. As we can find the value of -^2 to any number of decimals, we can find a fraction that differs from ^2 : 1 by less than any assignable quantity. To generalize, let A and B be any two incommensurable quantities. If we suppose B divided into any number of equal parts, so that B = nP, P denoting one of the parts, then A must contain some number m of such parts, with a remainder less than P ; or ^ > m P and < (m -}- 1)P. Thus B nP nP * A 'Yfh T/h I 1 that is, — lies between — and — -^--, and must differ from B n n 1 either of these fractions by less than -, a fraction that may be made less than any assigned quantity by taking n suffi- ciently great. Hence A. rational fraction can he found that expresses the ratio of any two given incommensurable quantities within any required degree of x>recision. PROPORTION. DEFINITIONS. 228. If two pairs of quantities have equal ratios, they are said to be proportionals or to be in proportion. Thus A G each of the equalities A: B — C:D, — = — , B D PROPORTION. 115 expresses a proportion that may be read, the ratio of A to B is equal to the ratio of C to D ; or more briefly, A is to B as C is to D. 229. The first and fourth terms of a proportion are called its extremes; the second and third, its means. The fourth term is also called a fourth proportional to the other three. 230. Although ratio, from its very nature, can exist between like quantities only, yet as we may have, for example, Z. A~ ^/.B, and also arcP = | arc Q, the pro- portion Z. A : Zb = arc P : arc Q, may be stated between these pairs of unlike quantities, since the proportion simply states that the angle A is just as great compared with the angle B as is the arc P compared with the arc Q. 231. Of the following theorems concerning proportions and their transformations, some apply to pairs of quantities whether like or unlike, and the given proportion will be stated under the form A:B =P:Q. Others apply only when the pairs of quantities are like, and the given proportions will be stated under the form A:B = C:B. Others, again, that apply properly to numbers only, will have the given proportion stated under the form a:b = c:d. 232. It follows at once, from the definitions of quantity, ratio, and propoj'tion, that (1) if four quantities are in pro- portion, their numerical measures are also in proportio7i ; i.e. if A : B = P : Q, then a:h = p:q, or - = -, ^ 'h q 116 PLANE GEOMETRY. — BOOK III. and conversely ; (2) if four numbers are in proportion, quan- tities of which these yiumhers are numerical measures are also inproporiion; i.e., .„ . a p if a:b = p:q, OY t = -? b q then A.B = P: Q. It also follows from the same definitions, and Ax. 1, that (3) ratios equal to the same ratio are equal to each other. EXERCISES. QUESTIONS. 285. Taking the inch as unit, what is the ratio of 1 ft. to 7 in. ? To 13in. ? Tol^ft. ? To 21 ft. ? To | yd. ? 286. A train goes at the rate of 112 miles in 3| hrs. ; a second train at the rate of 105 miles in 2\ hrs. What is the ratio of the speed of the first train to that of the second ? 287. What is the ratio of 1 lb. to 9 oz. ? To 33 oz. ? To 2| lbs. ? To 20| lbs. ? 288. That 216 grs. of silver may be worth 13| grs. of gold, what should be the ratio of the value of gold to that of silver ? 280. 180° F. = 100° C. = 80° R. What is the ratio of 1° F. to 1° C. and 1 R.°, respectively ? 290. The vertical angle of an isosceles triangle is 50°. What is the ratio of that angle, (1) to a right angle ? (2) to each of the base angles ? 291. A base angle of an isosceles triangle is 75°. What is the ratio of that angle, (1) to a right angle ? (2) to the vertical angle ? 292. The ratio of a base angle of an isosceles triangle to the vertical angle is f . What is that angle, and what is its ratio to a right angle ? 293. An acute angle of a right triangle is 35°. What is the ratio of that angle to the other acute angle ? 294. A certain angle has to an angle of an equilateral triangle the same ratio that the latter has to a right angle. How many degrees in the first angle ? LIMITS. 117 LIMITS. Thus far the magnitudes considered have been of fixed greatness only. In Art. 227 we had occasion, however, to discuss certain approximate ratios tending towards a value they can never exactly attain, though they can approach it indefinitely near. In these we have examples of what are termed a variable and its limit, now about to be defined. 233. A constant is a magnitude or quantity whose great- ness remains the same, neither iucreasing nor decreasing. 234. A variable is a magnitude or quantity whose great- ness goes on increasing or decreasing. Thus a chord .4J5, as long as it passes through the center O, remains constant in every position. But if it turn about an extremity A, we have a decreasing variable chord AC, Av', etc., while in the angles BAG, etc., and in the arcs BC, BC', etc., we have increasing variables. 235. If a variable increases or decreases so as to approach indefinitely near to an equality with a certain constant, this constant is called the limit of the variable. Thus in the figure above, the variable chord decreases towards zero as limit, the variable angle increases towards a right angle as limit, the variable arc increases towards ji semicircumference as limit, and the distance of the variable chord from the center increases towards the radius as limit, none of which limits, evidently, can be attained as long as there is any chord. Again, if a point P o p p' p" p'" ^ move along from towards N, the distances OP and PN are variables ; the one increasing towards ON, the other decreasing towards zero. If no condition were imposed upon the motion of P, it might 118 PLANE GEOMETRY. — BOOK 111. reach N, or even pass beyond it, and the distance ON would not be a limit of OP according to the definition. But if we impose the condition that at the end of the first second it reach p', half way to N, at the end of the o_ p p' p" p'" n next second reach P," half way between P' and JSf, and so on, it is evident that P could never reach N, though it might come indefinitely near to it. For the fraction of the distance passed over in n seconds would be the sum of the series J + -| + g- -j- yg- + a series that has 1 for its limit, a limit the series can never attain, no matter how many terms may be taken. Proposition I. Theorem. 236. If two variables tending towards limits are always equal, these limits are also equal. A M^ D P B N Q Given : Two equal variables A 31 and BN, tending towards limits AP and PQ; To Prove : PQ is equal to ^P. For if AP could be greater than BQ, some part of AP, as AD, would be equal to BQ. Now, however small the con- stant difference DP could be, since A3f can increase so as to approach AP nearer than any constant difference (Hyp.), A3I would become greater than AD or its equal B Q, while BN would always remain less than BQ (Hyp.). That is, BN would be both equal to and less than AM. In the same way it can be proved that B Q cannot be greater than AP. Hence B Q must be equal to AP. q.e.d. THEORY OF PROPORTION. 119 PROPORTION. THEOREMS. Proposition II. Theorem. 237. If four numbers are in proportion, the product of the extremes is equal to the product of the means. Given: a:b = c:d; To. Prove : ad = be. Since ^ = ^, (Hyp.) b d .'. multiplying both members by bd, ad = be. Q.E.D. 238. Cor. If the means are equal, that is, if a : b = b : c, then ac = b% (237) i.e., the product of the extremes is equal to the square of the mean. 239. Definition. Here b is said to be a 7nean propor- tional between a and c, and c is a third proportional to a and b. Exercise 295. If the bisector of an angle of a parallelogram passes through the opposite vertex, the figure must be equilateral. 296. Any parallelogram that can be circumscribed about a circle must be equilateral. 297. If each of three equal circles is tangent to the other two, their lines of centers form an equilateral triangle. 298. In triangle ABC, if D, E, the mid points of AB, AC, be joined, show that A ^5 C: trapezoid DBCE = 4:3. 299. If A is an angle of an equilateral triangle and J5 is a right angle, show that Z ^ : Z ^ = 2 : 3. 300. In the diagram for Art. 153, ii ABAC : AACB = 1 '.2>, and if Z BAC : Z ABC =7:3, show that /.AOC-.Z. ABC =7:2. 301. Hence deduce the ratio of angle BOG to angle BAC and of angle AOB to angle ACB. 120 PLANE GEOMETRY. — BOOK III. Proposition III. Theorem. 240. Conversely, if the product of two numbers is equal to the product of two others, either two may be made the extremes of a proportion, and the other two its means. Given: ad = bc; To Prove : a:b = c:d. Since ad = be, (Hyp.) .-. dividing both members by bd, 9: — 2. b~d! i.e., a:b = c: d. q.e.d. 241. Scholium. If, instead of by bd, we divide by ab, ac, or cd, we obtain from ad = be the proportions, each derivable from any of the others : d:b = e: a, or c: a — d:b ; d : c = b: a, or b: a = d: c; a : e = b : d, oy b: d = a: c. Exercise 302. In the first diagram for Prop. XIII. (195), prove that if AC and BD be joined, angle BAC will be equal to angle ABD. 303. In the same diagram, show that if AD and BC be joined, they will intersect in MN. 304. In the third diagram for the same proposition, if a perpen- dicular be drawn through the mid point of EF, it will pass through M and iV. 305. If two tangents to the same circle make equal angles with an intercept between them that passes through the center, they are equal. 306. If two secants to the same circle make equal angles with an intercept between them that passes through the center, they are equal. THEORY OF PROPORTION. 121 Proposition IV. Theorem. 242. The products of the corresponding terms of two or more numerical proportions are in proportion. Given : a:b = c: d, and e :f= g : h; To Prove : ae :bf=cg : dh. Since ^ = "^, and ^ = f, (Hyp.) h d f h .'. multiplying member by member, ae _ eg hf^dJi i.e., ae : hf= eg : dh. q.e.d. Tn the same way the theorem can be proved for any number of proportions. Proposition V. Theorem. 243. If four numbers are in proportion, like powers or Wee roots of those numbers are in proportion. Given : a:b = c:d; i_ 1 11 To Prove : a'* : 6" = c" : d'\ and a"" : 6" = c** : d"". Since ^ = |, (Hyp.) .'. taking like powers or like roots of both members, 1^ 1 a" c" T a" c** — = — and — f = — r, 1 i 1 i i.e., a" : b" = c" : d% and a" : 6" = c" : d" q.e.d. 122 PLANE GEOMETRY. — BOOK III. Proposition VI. Theorem. 244. If four like quantities are in proportion, they are in proportion taken alternately. Given : A:B = C:D; To Prove : A:C = B:D. Since - = -, (Hyp. and 232') h d ad=bc, (237) .-. a:c = b:d, (241) .-. A:C = B:D. Q.E.D. (232") Scholium. Alternation, as will be seen from the symboli- cal statement of the theorem, means taking the alternate terms of a given proportion so as to form a new one. It is also evident that this transformation can be applied only to a proportion in which the terms are all like terms, since, if A and P are unlike quantities, no ratio can exist between them. Proposition VII. Theorem. 245. If any four quantities are in proportion, they are in proportion taken inversely. Given: A:B = P : Q; To Prove : B:A = Q'.P. Since a:b = p:q, (Hyp. and 232') bjj = aq, (237) .-. b:a = q:p, (241) .-. B:A = Q:P. Q.E.D. (232") Scholium. Inversion, as may be seen from the symboli- cal statement, means taking the terms in inverse order. THEORY OF PROPORTION. 123 Proposition VIII. Theorem. 246. // any four quantities are in proportion, they are in proportion taJcen hy composition. Given : A:B = P:Q; To Prove : A-\-B:B = P -{- Q:Q. Since ^=^, b q (Hyp. and 232') hH^^' (Ax. 2) •• h - q ' .-. A-\-B'.B = P -\-Q:Q. Q.E.D. (232") Similarly, A -\-B:A = P + Q'.P. Proposition IX. Theorem. 247. If any four quantities are in proportion, they are in proportion taken hy division. Oiven : A:B = P:Q; To Prove: A — B:B = P—Q: Q. Since ^ = ^, b q (Hyp. and 232') b q • (Ax. 3) a—b_p—q '■ b g ' .-. A - B : B = P - Q . Q. Q.E.D. (232") Similarly, A— B :A = P ~ Q: p. 124 PLANE GEOMETRY. — BOOK III. Proposition X. Theorem. 248. // any four quantities are in proportion, they are in proportion taken hy composition and division. Given : A:B = P'.Q; To Prove : A-\-B:A — b = P-{-Q:P — q. Since a:h=i^:q, (Hyp. and 232') a-{-b_p + q and^-^^ = ^-^, (247) .*. dividing member by member, a-\-h _ p-{-q a — b p — q ,\ A -{- B : A — B = P -\. Q : p — Q. q.e.d. (232") Proposition XI. Theorem. 249. If two proportions have the same antecedents, the consequents are in proportion, and vice versa. Oiven : A:B=P:Q,SindLA c = P:.r; To Prove: B :C= Q:B. Since e=.^ I I \ (Hyp. and 232') and ^ ^l, b 4 ,'. dividing member by member, ^ = % c r .'. B :C= Q: R. Q.E.D. (232") 250. Definition. A continued proportion is a series of equal ratios. THEORY OF PROPORTION. 125 Proposition XII. Theorem. 251. If any number of like quantities are in con- tinued proportion, the sum of the antecedents is to the suin of the consequents as any antecedent is to its consequent. Given: A:B = C:D=E:F; To Prove : A -\- C -\- E : B -\- n -\- F = A : B. Since a:b = c:d = e:f, (Hyp. and 232') ab = ba, ad = be, of— be, (237) ... a(b-^d-^f)=:b(a-{-c-\-e), .'. a-]-c + e:bi-d-\-f=a:b, (240) .'. A-{- C -\-E:B -{-D -{-F = A:B. Q.E.D. (232") In the same way, the theorem may be proved for any number of ratios. Proposition XIII. Theorem. 252. If the antecedents of any four quantities in proportion be multiplied by any number, and the consequents by any, the results will be in propor- tion. Given : A:B = P:Q; To Prove : mA :nB = mP:nQ. Since a : b = p : q, (Hyp. and 232') d m:n = m: n, ma : nb = mp : nq, (242) ,-. mA : nB = mP : nQ. Q.E.D. (232") 126 PLANE GEOMETRY. —BOOK III. Proposition XIV. Theorem. 253. If both terms of a ratio he multiplied or divided by any numher, the ratio remains the same. Given : Any ratio A: B ; To Prove : mA : mB = A : B = -A :-B. n n Since a : h = a : b = a : b, and m:m = l:l = -:-, n n ma :mb = a:b = -a:-b, (242) .'. mA : mB = A:B = -A:~B. Q.E.D. (232") 254. Cor. If A . B = p : Q, then mA : mB = nP : 7iQ, \ 11 1 1 i (253) and —A:—B = -P:-Q.\ mm n n J Exercise 307. Any parallelogram that can be inscribed in a circle will have the intersection of its diagonals at the center of the circle. 308. Hence, show that rectangles are the only parallelograms that can be inscribed in a circle. 309. Any parallelogram that can be inscribed in a circle must be equiangular. 310. Describe a circumference passing through two given points and having its center in a given straight line. When is this impossible ? 311. Prove that all circumferences that pass through a given point and have their centers in a given straight line must also pass through a second given point. 312. From a given point as center describe a circle to touch a given circle. How many solutions are there ? THEORY OF PROPORTION. 127 Proposition XV. Theorem. 255. Two incommensurahle ratios are equal if their corresponding approximate values are always equal. A P Given : Two ratios — and — such that, when an approximate value of — is — , the corresponding value of — also is — ; To Prove : A : B = P : Q. The supposition is that when B and Q are each divided into n equal parts, if A contain m parts of B, with some remainder, then also P contains m parts of Q, with some remainder. ^. A m X ^ P m x' /XT \ Since - = - + - and -=- + -, (Hyp.) B n 71 Q n n m and n being integers, but x and x' each < 1, (235) — has for limit —, n B also — has for limit — , 71 Q X X as -J — each tend toward zero when ti is indefinitely great, .-. - = -, (236) B Q being the limits of variables always equal, i.e., A : B = P : Q. Q.E.D. Scholium 1. It is to be carefully noted that A : B and P : Q are here proved absolutely, not app7'oximately, equal. Scholium 2. In Props. II.-XIV., we found that, a pro- portion being given, certain transformations can be performed on it. In this proposition we have the criterion, or test, of proportionality as regards both incommensurable quantities and those of whose commensurability we know nothing. Book IV. PROPORTIONAL ANGLES AND LINES. SIMILAR POLYGONS. Xi^f^C PROPORTIONAL ANGLES. 256. Definition. An angle formed by two radii of a circle is called an angle at the center, and is said to intercept the arc that lies between its sides. Proposition I. Theorem. 257. In the same circle, or in equal circles, equal angles at the center intercept equal arcs; and con- versely. 1°. Given: At the centers of equal circles ADB, A'd^b', angle O equal to angle O' ; To Prove: Arc ACB is equal to arc A'c'b'. Join AB, A'b'. Then since OA = o'A', OB = o'b', and Z = Zo', (Hyp.) A GAB = A 0' A'B', (66) .-. AB=A'B', (70) .-. arc ACB = arc A'c'b'. q.e.d. (174") 128 PROPORTIONAL ANGLES. 129 2°. Given: In equal circles ADB, a'd'b', equal arcs ACB, a'c'b', intercepted by angles 0, o', respectively; To Prove: Angle is equal to angle o'. Join AB and A'b'. Then since arc ACB = arc a'c'b', (Hyp.) chord AB = chord A'b' ; (174") also OA =:0'A', and OB = o'b', (Hyp.) .-. Zo = Zo'. Q.E.D. (69,70) 258. Cor. 1. A radius bisecting an angle at the center bisects its arc; and conversely. 259. Cor. 2. If angle A is equal to m times ayigle B, then the arc of angle A is equal to m times the arc of angle B ; and conversely. Scholium. The conclusion proved in Prop. I. might be stated under the form Z 0:Z0' = 7)1 :m = avG ACB :A'c'b', such a ratio as m : m being called a ratio of equality. Exercise 313. In the same circle or in equal circles, the greater of two unequal angles at the center intercepts the greater arc. 314. An angle at the center is obtuse, right, or acute, according as its arc is greater than, equal to, or less than a quarter of a circumference. 315. Intersecting diameters intercept equal arcs between their extremities. 316. If the extremities of intersecting diameters be joined, the figure formed will be a rectangle. 317. In the diagram for Prop. I., if angle is f of a right angle, what is the ratio of arc ACB to arc ADB ? 318. In the same diagram, if vlO be produced to meet the circum- ference in E, what part will arc BE be of the circumference, suppos- ing, as before, that angle is f of a right angle ? 319. In the left-hand circle of the diagram for Prop. II., show how to bisect arc A CB without joining AB. Geom. — 9 130 PLANE GEOMETRY. — BOOK IV. Proposition II. Theorem. 260. In the same circle, or in equal circles, angles at the center are proportional to their intercepted arcs. Given: In equal circles with centers and 0\ respectively, angles AOB, A'o'b', with their respective arcs AB, A^B' ; To Prove : Angle AOB : angle ^'o'^' = arc ^5 : arc ^'5'. 1°. When the arcs AB, A'b', are commensurable. Let B'c' be a common measure of A'b' and AB, so that arc b'c' is contained 5 times in A'b' and 8 times in AB. Join O'C'. Since arc AB = S arc B'c', and arc A'b' =5 arc b'c', (Hyp.) Za'ob = SZ b'O'c', and Z a'o'b' = 5Z b'o'c', (259) .-. arc AB : arc A'b' = 8:5,] ^ndZA0B:ZA'0'B' = S'.5,\ ^^^^^ .'. ZAOBiZ A'O'B' = 2iVGAB:^TG A'B'. Q.E.D. (232'") 2°. When the arcs AB, A'b', are incommensurable. Suppose A'b' divided into any number of equal parts n, and that AB contains this nth. part of A'b' m times, with a remainder BC. Draw OC. Since AC and A'b' are commensurable arcs, (Const.) Zaoc _'wi_ arc^C Z A'O'B' ~'n~ SiVG A'b'' (1°) Z AOB m X , arc^B m x' —. — r-7— i = — h -) and - — . . = 1 — , Z A'o'b' n w diVcA'B' n n since AOB and AB are slightly > AOC and AC, respectively. PROPORTIONAL ANGLES. 131 Now when n is taken indefinitely great, - and — become indefinitely small, X and ic' being each < 1 ; /.AOB arc ^5 /or--s Za'o'b' sltga'b" ^ / being the limits of variables always equal. 261. Definition. One quantity is said to be measured by another of a different kind, if they are so related that the numerical measure of the one always expresses that of the other also. This may also be worded as follows : A quantity A is measured by another quantity ^', if A and its unit U are always in proportion to A' and its unit U' ; i.e., ii A:U — A':U'. Thus temperature is measured by the number of equal lengths in a column of mercury; time, by the number of equal arcs in the circumference of a dial ; etc. 262. Cor. An angle at the center is measured by its inter- cepted arc. For if A denote any angle, and u its unit-angle. A', the intercepted arc, and u' its unit-arc, then (260), ZA:Zu= arc A' : arc u' = m: 1, m being the numerical measure oi A' in terms of its unit u'. Scholium. Though any convenient arc may be taken as unit, that usually employed is the 90th part of a quad- rant or i of a circumference, and is called a degree. As a right Z is measured by a quadrant (262), the 90th part of a right Z is also called a degree. Each degree, again, is divided into 60 minutes, and each minute into 60 seconds. 263. Definition. The angle formed by two chords that meet in the circumference is called an inscribed angle. 132 PLANE GEOMETRY. — BOOK IV. Proposition III. Theorem. 264. An inscribed angle is measured hy half its intercepted arc. B, . B B ^ Given: An inscribed angle ^BC intercepting the arc AC ; To Prove : Angle ABC i^ measured by \ arc AC. Find o, the center of the circle. Then 1°. If lies in a side BC oi /.ABC, join AG. Since AO = B0, Za = Zb. But Zaoc=Za-{-Zb = 2Zb, .'. Z B =^ZA0C. Now Z ^oc is meas. by arc AC, Z B is meas. by ^ arc AC. 2°. If lies between BA and BC, draw BOD, a diam. Since Z ABD is meas. by |- arc AD, ^ and Z DBC is meas. by |-arc DC, ) Z ABD + Zdbc is meas. by i (arc AD + arc DC) ; i.e., Z ABC is meas. by i arc AC. q.e.d 3°. If lies without BA and 5C, draw BOD, a diam. Since Z ABD is meas. by ^ arc ^D, and Z 2)5(7 is meas. by i arc Z ABD — Z DBC is meas. by i (arc ^i) — arc DC) ; i.e., Z ABC is meas. by ^ arc ^c. q.e.d (162) (68) (122) (Ax. 7) (262) Q.E.D. (1°) AD, I DC, I (1°) PROPORTIONAL ANGLES. 133 265. Definition. A segment of a circle is the figure con- tained by a chord and its arc. 266. Cor. 1. All angles C,D,E, inscribed e in the segment AEDCB of a circle are equal. For each is measured by half the arc 2 AFB. 267. CoR. 2. An angle inscribed in a semicircle is a right angle. For it is measured by |- a semicircumference ; i.e., by \ a circumference, or a quadrant. 268. CoR. 3. The arc intercepted by an iyiscribed angle is double the arc intercepted by an equal angle at the center. Proposition IV. Theorem. 269. The angle formed hy a tangent and a chord meeting at the point of contact, is measured hy half the intercepted arc. Given : An angle BAC formed by a tangent AB and a chord AC ; To Prove : Angle BAC is measured by ^ arc AC. Through C draw CD II to AB (109). Then since CD is II to AB, (Const.) 3iTG AD=SiYG AC, (195") and Zc = Z J. (110") But Z C is meas. by i arc AD, (264) .-. Z ^ is meas. by i arc AD ov ^ arc AC. q.e.d. 134 PLANE GEOMETRY.— BOOK IV. Proposition V. Theorem. 270. The vertical angles formed hy intersecting chords are each measured hy half the sum of the intercepted arcs. a Given: Two chords AB, CD, intersecting in E ; To Prove : Angles AEC, bed, are each measured by \ (arc AC + arc BD). Draw the chord BC. Then since Z ^ is meas. by i arc AC, and Z (7 is meas. by i arc BD, (264) also Z AEC = Zb -\- Zc, (122) Z AEC or Z BED is meas. by ^(arc AC -{- arc BD). q.e.d. Exercise 320. In the diagram for Prop. III., if B is an angle of 32°, how many degrees are there in arc AC 9 321. In the diagram for Prop. IV., if CD is an arc of 102°, then BAC is an angle of how many degrees ? 322. In the diagram for Prop. V,, if AEG is an angle of 25° and J.C an arc of 30°, how many degrees in arc BD 9 323. Any three points of a circumference being given, how can we find other points of it without knowing the center ? 324. The opposite sides of an inscribed parallelogram divide in the same ratio the radii drawn perpendicular to them. 325. If two circles whose centers are and 0' have a common tangent AB, and OA, O'B, be joined, these lines will be parallel. 326. If two tangents, PA, PB, be drawn to a circle whose center is 0, and AB, AO, be drawn, then will angle BAG = ^ angle P. PROPORTIONAL ANGLES. 135 Proposition YI. Theorem. 271. The angle formed hy two secants meeting with- out the circle is measured hy half the difference of the intercepted arcs. B^ E )} ^^^^^ Given: Secants AB, AC, meeting in A, and intercepting arcs BC, DE ; To Prove : Angle A is measured by ^(arc 5C — arc DE). Draw the chord BD. Then since Z 52) a = Z ^ -hZ^, (122) /.A = /.BDC — Z.B. (Ax. 3) But Z 5I>C7 is meas. by l^arc J5a, and Z 5 is meas. by \ arc DE, .-. Z^ is meas. by |-(arc-BC — arcDjE:). q.e.d. Exercise 327. In the diagram for Prop. VI., if A is an angle of 17° and DE an arc of 36°, how many degrees in sltc BC ? 328. If a polygon of an even number of sides be inscribed in a circle, the sums of its alternate angles are equal. 329. Find a point equidistant from three given points. When is the problem impossible ? 330. Under what conditions is it possible to find a point equidis- tant from four given points ? 331. Prove the theorem given in Art. 155, by means of Prop. III.» (171). 332. If an inscribed triangle has unequal angles, the greater angle intercepts the greater arc. 333. If two circles intersect, their line of centers produced will bisect each of the four arcs. 136 PLANE GEOMETRY. — BOOK IV. Proposition VII. Theorem. 272. An angle formed by a tangent and a secant is measured by half the difference of the intercepted arcs. Given: Tangent AB and secant AC meeting in A and inter- cepting arcs BC, BD ; To Prove : Angle A is measured by i(arc BC — d^xoBD). Draw the chord BC. Then since /. B ^ A A -\- Z. c, (122) Z .4 = Z 5 - Z c. (Ax. 3) But Z j5 is meas. by ^ arc^C, (269) and Z c is meas. by \ arc BD, (264) .'. Z ^ is meas. by J(arc BC — arc BD). q.e.d. 273. Cor. The angle formed by tivo tangents is measured by half the difference of the intercepted arcs. EXERCISES. QUESTIONS. 334. If an angle A is measured by two tliirds of a quadrant, and an angle B = 50°, what is the ratio of Z ^ to Z ^ ? 335. By what fraction of a quadrant is the vertical angle of an isos- celes triangle measured, (1) if it is twice as great as a base angle ? (2) if it is I as great ? (3) if it is -th as great ? n EXERCISES. 137 336. Two angles of a triangle are measured by ^ and |^ of a circum- ference, respectively. How many degrees in each of the three angles ? 337. One angle of a right triangle is measured by /^ of a circum- ference. How many degrees in the other acute angle, and what is its ratio to the first ? 338. If, in the diagram for Cor. I. of Prop. III., ADB is an arc of 230°, how many degrees are there in each of the angles C, D, and^? 339. If, further, in the same diagram, AE is an arc of 30°, how many degrees in angles ABE and BAE, respectively ? 340. In the diagram for Prop. IV., if BAG is an angle of 65°, how many degrees are there in arc CD ? 341. In the same diagram, if arc CA = ^ arc CD, how many degrees in Z BAG? 342. In the diagram for Prop. V., if AEG is an angle of 40°, how many degrees are there in the sum of the arcs AD and BC? 343. In the diagram for Prop. VI., if A is an angle of 25°, and arc DE = ^ arc BC, how many degrees are there in the sum of arcs BE and CD ? 344. In the same diagram, if BD = AD, how many degrees are there in each of the arcs BC and DE, if ZA = n degrees ? 345. In the diagram for Prop. VII., if arc 5C= 3 arc 52), and Z.A — 30°, how many degrees are there in arc CD and in Z JS ? 346. In the same diagram, if BA — BC, what is the ratio of arc BG to arc BD ? 347. If the angle formed by two tangents is 30°, how many degrees are there in each of the intercepted arcs ? 348. If a tangent be drawn at a vertex of an inscribed equilateral triangle, what angle will it form with either adjacent side ? 349. If from the same point in a circumference, a side of a square and a side of an equilateral triangle be inscribed, the difference of the arcs subtended by them will be what part of the circumference ? 350. If the vertical angle of an inscribed isosceles triangle is 54°, what is the ratio of the arc opposite that angle to either of the other arcs? 351. If the vertical angle were 37° 15' 32", what would those ratios be? 138 PLANE GEOMETRY. — BOOK IV. PROPORTIONAL LINES. Proposition VIII. Theorem. 274. A line draivn parallel to one side of a triangle and meeting the other two sides, divides them pro- portionally. Given : DE parallel to 5C, a side of triangle ABC, and meeting AB, AC,m D, E, respectively; To Prove : DB : AD = EC: AE. 1°. When AD and DB are commensurable. Let BF \)Q 2i common measure oi DB and AD, so that BF can be laid oif 3 times on DB and 7 times on AD. Through the points of division draw lines FF^, etc., II to BC. Since the ll's FF\ etc., cut off 3 and 7 equal parts on DB, AD, respectively, ' (Const.) the ll's FF', etc., cut off 3 and 7 equal parts on EC, AE, respectively ; (1^1) .-. DB :AD = 3:7, and EC: AE = 3:7, (225) .'. DB : AD =EC :AE. Q.e.d. (232'") 2°. When AD and DB are incommensurable. Suppose AD divided into any number of equal parts n, and that DB contains this nth part of AD m times, with a remainder LB. Through L draw LL^ II to BC. PROPORTIONAL LINES. 139 Since AD and DL are commensurables, (Const.) DJL_m _ElJ_ .-|^ox AD~ n~ Ae' DB m , X T EC m , x' .: — = - + -, and —- = - + -, AD n n AE n n since DB and J?c are slightly >DL, EL', resp. X X Now when n is taken indefinitely great, -, — become in- definitely small, .-. ££ = E^ Q.E.D. (255) AD AE being the limits of variables always equal. 275. CoR. If two sides of a triangle are cut by a parallel to the base, 07ie side is to either of its parts as the other is to its corresponding part. E' T-D For since DB : AB z= EC : AC, (274) .-. DB -\- AB: DB, or AB, = EC-{-AC: EC, or A C, (246) ^ i.e., AD : AB, or DB, = AE: AC, or EC. D Scholium. It is obvious that the theorems (274, 275) hold good when DE meets AB, AC, produced in either direction. Exercise 352. Circumferences described on the arms of an isosceles triangle as diameters, will intersect in the mid point of the base. 353. Circumferences described on any two sides of a triangle as diameters, will intersect in the third side or the third side produced. 354. If an intersecting circumference pass through the center of another, the angle in the exterior segment of the latter is acute. 355. J^JS is a common exterior tangent to two circles touching at P; draw PA, PB ; APB is a right angle. 356. A tangent at the mid point of an arc is parallel to its chord. 140 PLANE GEOMETRY. — BOOK IV. Proposition IX. Theorem. 276. Conversely, if a straight line divide two sides of a triangle proportionally, that line is parallel to the third side. F G Given: In triangle ABC, DE meeting AB, AC, so that AD : DB = AE: EC ; To Prove : DE is parallel to BC. Produce AB to F so that BF = db, and through F draw FG II to DE to meet AC produced, in G. Since FG is II to DE, (Const.) AD : DF = AE :EG. (274) But AD : DB = AE : EC, (Hyp.) .'. DB '.DF = EC:EG. (249) Now DB = ^ DF, (Const.) .-. EC = ^EG, .'. BCis II to DE. Q.E.D. (150) 277. Definition. According as a point is taken in a given line, or in that line produced, the distances of the point from the extremities of the given line are called internal or external segments of the line. Hence the given line is the sum of any two internal segments, and the difference of any two ex- « ' ■ ternal segments. Thus \i AB is divided in C and produced to D, then ^5 is the sum of the internal segments AC, CB, and the differ- ence of the external segments AD, BD. PROPORTIONAL LINES. 141 Proposition X. Theorem. 278. The bisector of an interior angle of a triangle divides the opposite side into internal segments having the same ratio as the other two sides. E A..--' Given: In triangle ABC, AD bisecting angle BAG, and meeting BC'mD; To Prove : BDiDC = AB :AC. Through C draw CE II to AD, to meet ^^ produced, in E. Since AD is II to CE, (Const.) Z BAD =Ze, (112") and ZCAD =Za CE. (110") But Z BAD = Z CAD, (Hyp.) .'. Zace = Ze, (Ax. 1) .'. AE = AC. (65) ^ain, since AD is II to CE, BD :DC = BA : AE. (274) i.e.,BD :DC — AB : AC. Q.E.D. 279. CoR. Conversely, in A ABC, if D he a point in BC such that BD :DC= AB:AC, then AD bisects Z BAC. For the bisector of that angle must pass through D (278), and must therefore coincide with AD. 142 PLANE GEOMETRY. — BOOK IV. Proposition XI. Theorem. 280. The bisector of an exterior an£le of a triangle divides the opposite side into external segments having the same ratio as the other two sides. Given: In triangle ABC, AD bisecting exterior angle CAF and meeting BC m D ; To Prove : BD : DC = AB : AC. Through C draw CE II to AD, to meet BA in E. Since AD is 11 to CE, (Const.) Z FAD = Z AEC, (112") 2ind Z CAD = Z ACE. (110") But Z FAD = Z CAD, (Hyp.) .'. ZACEz=ZAEC, (Ax. 1) .'. AE = AC. {^5) Again, since AD is II to CE, BD :DC = BA : AE, (274) i.e., BD :DC — AB : AC. Q.E.D. 281. Cor. Conversely, in A ABC, if D he a point of BC produced, such that BD : DC= AB '.AC, then AD bisects the exterior Z CAF. For the bisector of that angle must pass through D, and therefore must coincide with AD. PROPORTIONAL LINES. 143 Proposition XII. Theorem. \. If three or more transversals passing through the same point mahe intercepts upon two parallels, these intercepts are proportional. Given: Transversals OA, OB, OC, cutting the parallels AC, A'C', in A, B, c, and A', B\ c', respectively; To Prove : AB : A'b' = BC: B'c'. Through A', c', draw A'd, &e, each II to OB. Then since A'& is II to AC, (Hyp.) AO: A'0=zBO :B^0 = CO.C'O. (274) Since A'n, c'e, are each II to OB, (Const.) DB = A^B', and BE = B'c'; (138) also AB : DB or A'b' = AO : A'O, \ [ (274) Sind BC: BE OT B'c' = C0: C'O, ) .'. AB:A'b' = BC:B'c'. q.e.d. (232'") As this holds true of the intercepts made by any three transversals through 0, it holds true of the intercepts made by any number of such transversals. 283. Cor. The parallel intercepts between any two con- verging transversals are proportional to the intercepts betweeti the parallels and the common point. YoT AB : A'b' = A0 :A'0 = B0 : B'o. (Above) 144 PLANE GEOMETRY. — BOOK IV. SIMILAR POLYGONS. 284. Similar polygons are such as are mutually equian- gular, and have the sides about the equal angles, taken in the same order, proportional. 285. In similar polygons, similarly situated points, lines, or angles are said to be homologous. Proposition XIII. Theorem. 286. Triangles that are inutually equiangular are similar. Given: In triangles ABC, A'b'c', angle yl equal to angle A', angle B to angle B', and angle C to angle C' ; To Prove : BC: B'& = AB : A'b' = AC : A^C'. Place A A'b'& upon A ABC, so that Z A' =^ Z A, and A a'b'c' takes the position AB'c'. Since Zb' = Zb, (HyP-) B'c' is II to BC, (112') .-. BC : B'c' = AB : AB' = AC : AC' ; (283) i.e., BC : B'c' = AB : a'b' = AC : A'c'. Q.E.D. 287. Cor. 1. Two triangles are similar, if tivo angles of the one are respectively equal to two angles of the other. (121) 288. Cor. 2. Ttco right triangles are similar, if an acute angle of the one is equal to an acute angle of the other. (123) 289. Scholium. In similar triangles, homologous sides lie opposite equal angles. SIMILAR POLYGONS. 145 Proposition XIV. Theorem. 290. Two triangles are similar, if an angle of the one is equal to an angle of the other, and the sides about these angles are proportional. B C B' c' Given: In triangles ABC, A'b'c', angle A equal to angle A^ ; also AB :A'b' = AC:A'c'; To Prove: Triangle ABC i^ similar to triangle A'b'c'. Place A a'b'c' upon A ABC, so that Z A' =^ Z A, and A a'b'c' takes the position AB'c'. Then since AB : A' b' = AC : AC', (Hyp.) B'C' is II to BC, (276) .-. Zb = Z b', and Zc = Z c'. (112") .-. A ^i? C is similar to A ^'iJ'c'. q.e.d. (286) Exercise 357. In the diagram for Prop. XI., if the bisector of angle BAC be drawn so as to meet the base in D', show that BC is divided into internal and external segments having the same ratio. Definition. — A straight line is said to be divided harmonically if it is divided into internal and external segments having the same ratio ; or if it is divided into three segments such that the whole line is to either of its outer segments as the other outer segment is to the inner segment. 358. Show that in the diagram for Exercise 357, BC is divided harmonically according to the first definition above, and BD^ accord- ing to the second. 359. Show that if D and D' divide any line MN harmonically according to the first definition, then also M and N divide DD' har- monically. 360. Also show that MD is divided harmonically according to the second definition. Geom. — 10 ' 146 PLANE GEOMETRY. — BOOK IV. Proposition XV. Theorem. 291. Triangles that have their sides mutually pro- portional are similar. Given: In triangles ABC, A'b'c', AB : A'b' = AC : A'c' = BC: B'c'; To Prove: Triangle A' B'c' is similar to triangle ABC. On AB, AC, take AD, AE = A'b', A'c', resp., and join DE. Since AB : AB = AC : AE, (Hyp. and Const.) DE is II to BC, (276) '. Zd = Zb, and Ze = Zc, (112") and AB : AD z=BC: DE. (283) But AB : A'b' = BC : B'c'. (Hyp.) .'. AD : a'b' = DE : B'C'. (249) Now AD = a'b'. (Const.) .'. DE = B'C'; and AE = A'c', (Const.) .: AADE = AA'B'C', (69) .-. Zb' = Zd = Zb, and Z c' = Z E = Z C, (Ax. 1) .-. A^'^'C' is similar to A^J?C. q.e.d. (286) Scholium. From this and Prop. XT. it is seen tliat mutually equiangular triangles have their homologous sides proportional ; and conversely. SIMILAR POLYGONS. 147 Proposition XVI. Theorem. 292. Two triangles are similar, if they have their sides respectively parallel or perpendicular to each other. Given: Triangles ABC, a'b'c', having their sides respectively parallel or respectively perpendicular to each other ; To Prove: Triangle ABC is similar to triangle a'b'c'. Since their sides are 1| or _L to each other, (Hyp.) any two homologous angles of these triangles must be either equal or supplementary. (116, 118) Hence there are three conceivable cases to consider. 1°. Suppose all the angles of the one triangle respectively supplemental to the homologous angles of the other. i.e., A-\-A'=2i st. Z, B-{-B'=Si St. Z, and also C+C' = a st. Z. 2°. Suppose two supplemental and one equal. i.e., A = A',B-{- b'= a St. Z, and C -\- C'= a st. Z. 3°. Suppose all the angles are mutually equal.* i.e., A== A', B =b', and .: C = c'. Each of the first two suppositions must be rejected, since * If two are right angles, they are equal and supplemental. 148 PLANE GEOMETRY. — BOOK IV. the sum of the angles of two triangles cannot exceed two straight angles. Hence the third alone is admissible ; .-. A^^C is similar to Aa'b'c'. q.e.d. (287) 293. Scholium. The homologous sides are those mutually parallel or perpendicular. Proposition XVII. Theorem. 294. Two polygons are similar if composed of the same number of triangles similar each to each and similarly placed. Given: In polygons P and P\ triangles ^JBJi), ADC, ACB, similar to triangles a'e'd', a'jd'c', A'c'b', respectively, and sim- ilarly placed; To Prove : P is similar to P'. 1°. P and P' are mutually equiangular. For their homologous A are either homologous A of similar A, or are like sums of homologous A of simi- lar A; (Hyp.) .-. Ae — A e\ a EDA 4- A ADC = A e'd'a' -f A a'd'c', etc. 2°. The homologous sides of P and P' are proportional. For since the A are similar, (Hyp.) AB : A'b' = BC : B'c' = AC : A'c' = CD : C'd', etc. ; .-. P is similar to P'. q.e.d. (284) SIMILAR POLYGONS, 149 Proposition XVIII. Theorem. 295. Conversely, two similar polygons may he di- vided into the same number of triangles, similar to each other and similarly placed. Given: ABODE, or P, and A'b'c'd'e', or P', two similar polygons; To Prove: P and P' may be divided into the same number of similar triangles similarly placed. From A and A', homologous vertices of P and P', draw diagonals AC, AD, and A'c', A'd', respectively. 1°. The number of triangles thus formed in P and P', respectively, is the same. For it is equal to the num- ber of sides in each, less two. 2°. These triangles are similar and similarly placed. For since P is similar to P' = A'P' : E'd', I (Hyp.) (284) Ze = Z p', and AE :ED /. Aaed is similar to Aa'e'd', (290) and Z EDA = Z e'd'a'. (289) But Z EDO = Ze'd'c', (Hyp.) .-. Zedc — Zeda = Ze'd'c' — Ze'd'a', (Ax. 3) i.e., Zadc = Za'd'c'. Again, since Aaed is similar to Aa'e'd', (Above) ED: DA =2 E'D' : D'a'. (285) But ED : DC = E'D': D'C', (Hyp.) 150 PLANE GEOMETRY. — BOOK IV. .'. DA:DC = D'A' : D'C', (249) .-. AADCis similar to A A'D^c'. (290) In the same way, the remaining triangles of P may be proved similar to the similarly placed triangles in P'. .-. P and P' may be divided as stated. q.e.d. Proposition XIX. Theorem. 296. The perijneters of two similar polygons have the same ratio as any two homologous sides. Given: AB, A'b', any two homologous sides of two similar polygons, P, P', of which j? andp' are the perimeters ; To Prove : p :p' = AB : A'b'. Since P and P' are similar polygons, (Hyp.) AB : A'b' = BC:B'C' = CD = C'd', etc., (284) .-. AB -\-BC-\-CD-\ : A'B' + B'C' + C'd' H = AB : A'b'. (251) But p = AB +BC -i- CD -\ , Sindp' = A'B'-{-B'C'+C'D'-\ , .-. p:p' = AB : a'b'. q.e.d. Exercise 361. Draw a diagram to show that two figures may be mutually equiangular though not similar ; and another to show that two figures may have their sides mutually proportional and yet not be similar. 362. In the diagram for Prop, XVII., if the numerical measures of BC and B'C are, respectively, 16 and 10, what is the ratio of j? top' ? RATIOS OF CERTAIN LINES. 151 RATIOS OF CERTAIN LINES. Proposition XX. Theorem. 297. In a right triangle, if a perpendicular he drawn from the vertex of the right angle to the hy- potenuse, i°. The perpendicular is a mean proportional between the segments of the hypotenuse. 2°. Each arm is a mean proportional between the hypotenuse and the adjacent segment. Given : In a right triangle ABC, AD perpendicular io BC from the right angle BAC ; BD :DA = DA : DC. BC:AB = AB:BD, and^C: AC=AC:DC. (1° To Prove : j Since BDA and BAC are rt. A, (Hyp.) and acute Z B is common to both, rt. A BDA is similar to rt. A BAC. (288) Since CD A and BAC are rt. A, (Hyp.) and acute Z C is common to both, rt. A CD A is similar to rt. A BAC, (288) .-. rt. A BDA is similar to rt. A CD A, (286) each being similar to rt. A BAC. Since A BDA is similar to A CD A, BD :DA=DA: DC. Q.E.D. (284) Since A^^C is similar to A BDA and CD A, BC:AB=AB:BD, ] mdBC:AC = AC:DC.] ^•^•^- ^^^^^ (297) 152 PLANE GEOMETRY. — BOOK IV. 298. Cor. If from any point A in a circumference, a perpendicular he drawn to a diameter BC, and chords AB, AC J be drawn, since BAC is a right angle, (267) DC: DA =z DA: DB, BC : BA = BA: BD, and BC: CA=CA: CD. Hence 1°. The perpendicidar from any point of a circumfer- ence to a diameter is a mean proportional between the segments. 2°. The chord drawn from the point to either extremity of the diameter is a mean proportional between the diameter and the adjacent segment. 299. Definition. Two ratios are said to be mutually inverse or reciprocal when the antecedent and the conse- quent of the one are, respectively, the consequent and the antecedent of the other. Thus B : A \^ the inverse oi A: B, and 11 : 7 is the inverse of 7 : 11. 300. Definition. If four quantities. A, B, C, D, are so related that A D A: C = D : B, or - = -, ' C B' i.e., if the first has to the third the inverse ratio of the second to the fourth, the quantities are said to be inversely proportional, while, as we know, if the first is to the third as the second to the fourth, the quantities are directly pro- portional. As will be seen in the next proposition, two lines have their segments inversely proportional if a seg- ment of the first is to a segment of the second as the re- maining segment of the second is to the remaining segment of the first ; while the segments would be directly propor- tional if a segment of the first were to a segment of the second as the remaining segment of the first to the remain- ing segment of the second. RATIOS OF CERTAIN LINES. 153 Proposition XXI. Theorem. 301. If two chords intersect, their segments are in- versely proportional. Given : In circle ABB, chords AB, CB, intersecting in 0; To Prove : OA: OB = OC : OB. Join AC and BB. Then since Za=Zb, and Zb = Zc, (264) A ^ C is similar to Ab OB, (286) .-. OA : OB = OC : OB. Q.E.D. (289) Scholium. In deducing such proportions between pairs of sides in similar triangles, the student may find it useful to remember that, as the homologous sides are opposite equal angles, the greater side of the one is to the greater side of the other as the lesser side of the one is to the lesser side of the other. Exercise 363. In the diagram for Prop. XI., if a parallel to EC cut AB, AC, and AD, in G, H, and L, respectively, show that AD is divided by GL so that AL:AD=GH:BC. 364. In the diagram for Prop. XII., show that the triangles A' AD and OA'B' are similar to each other and to triangle OAB. 365. In the same diagram, show that the perimeter of triangle OAB is to that of triangle OA'B' as OB is to OB'. 366. In the diagram for Prop. XX. , if Z J5 : Z (7 = 3 : 5, how many degrees in each of the acute angles at ^ ? 367. In the same diagram, if the numerical measures of BC and DC are 10 and 2, respectively, what is the numerical measure of AD ? 368. Find al§o the numerical measures oi AB and A C. 154 PLANE GEOMETRY— BOOK IV. Proposition XXIT. Theorem. 302. If two secants be drawn from a point with- out a circle, these secants and their external seg- ments are inversely proportional. Given : Two secants OA, OD, cutting circle A1)B in B, A, and C, D, respectively ; To Prove : OA: OD = OC : OB. Join AC and BD. Then since /.A = /. D, and Z is common to both, (264) A ^OC is similar to A DOB, (287) .-. OA:OD = OC: OB. Q.E.D. (289) Exercise 369. Prove Prop. XXI. by joining AD and BC^ and showing that OA: OD - OC: OB. 370. In the diagram for Prop. XXII., show that the triangles whose vertices are at the intersection of AC and BD are similar. 371. If two equal chords intersect, their segments are severally- equal. 372. If equal chords be produced to meet, the secants thus formed and their external segments will be severally equal. 373. In the diagram for Prop. XXI., show that if the chords are equally distant from the center, they are directly as well as inversely proportional. 374. In the diagram for Prop. XXII., show that if the secants are equally distant from the center, they are directly as well as inversely proportional. 375. If two circles intersect, tangents drawn to them from any point in their common chord produced, will be equal. EXERCISES. 155 Proposition XXIII. Theorem. 303. If a secant and a tangent he drawn from a point without a circle, the tangent is a mean proportional between the secant and its external segment. Given: A tangent OC toucliing circle ABC m C, and a secant OA cutting ABC in B, A; To Prove : OA : OC = OC : OB. Join AC and BC. Then :• Za and Z.OCB are each meas. by | arc BC, (264, 269) Z.A=/. OCB, and Z is common to A AC and OBC, .-. AOJLC is similar to AO^C, (287) .-. OA:OC=OC:OB. Q.E.D. (289) EXERCISES. QUESTIONS. 376. In the diagram for Prop. VIIL, if DB: AD = S:7, and AC = 65, what is the value of AE?* 377. If in the preceding question we substitute Vll for 3, what is the value of ^^? * Value, here and elsewhere, stands for numerical measure, the unit being left undetermined. If decimals occur in a result, it will be sufficient to have two places correct. 156 PLANE GEOMETRY. — BOOK IV. 378. In the diagram for Prop. X., if AB, AC, BC, = 9, 7, 12, respectively, what is the value of BD and of DC? 879. In the diagram for Prop. XI., if AB, AC, BD, = 9, 7, 20, respectively, what is the value ot BC? 380. If the angles at the base of A ABC in Prop XI. are equal, how is the proposition modified ? 381. If two triangles have an angle of the one equal to an angle of the other, and the sides about another angle proportional, are they necessarily similar ? 382. In the diagram for Prop. XII., if AB, AC, A'B,'= a, b, a', respectively, what is the value of A'C ? 383. In the diagram for Prop. XVII., if AD, DE, A C, A'D', = 4, 3, 5, 3.2, respectively, what are the values of D'E', A'C ? 384. In the diagram for Prop. XX., if AB, AC, BC, =4, 3, 5, respectively, what are the values of AD, BD, DC? 385. In the diagram for Prop. XXI., if ^0 = f OD, and OB = 8, what is the value of OC? 386. In the diagram for Prop. IX., if BD is — of AD, what part is Ca of AG ? '"^ 387. In the diagram for Prop. X., if AB : AC, = \0: 7, what is the ratio of ^Z> to JS'C ? 388. In the same diagram, if AB = A C, how many degrees are there in angle ^C^? 389. In the diagram for Prop. XI., if EC : AD, =2:3, what is the ratioof ^Bto^C? 390. In the same diagram, if AB is equal to AC, where will the point E fall ? 391. In the diagram for Prop. XII., if ^C = 12 and OB : OB' = 8:5, what is the value oi AD-\- EC? 392. In the same diagram, if OB bisects angle AOC, AB = 10, and BC=S, what is the ratio of OA' to OC ? 393. In the diagram for Prop. XIII., if BC : B'C> = m : n, what is the ratio ot AB - AC to A'B' -A'C? 394. In the diagram for Prop. XVII., if AB = 18, and A'B' = 12, what is the ratio of yl C to A' C ? 395. In the diagram for Prop. XX., if BD : DA = m : n, what is the ratio of the perimeter of triangle ADB to the perimeter of triangle ADC? CONSTRUCTIONS. 157 F G CONSTRUCTIONS. 304. To divide a gicen line AB proportionally to another given line CD divided in E, F, G. Upon ad', making any angle with AB, lay off AE', e'f',f'g', g'd',=ce, ef, fg, GD, respectively. Join BD', and through e', f\ g', draw lines parallel to BD', meeting AB in X, Y, Z, respectively. Then (274), AX: XY: YZ : ZB = AE' : E'F'iF'g': G'd'=CE : EF : FG : GD. If instead of a divided line we have numbers given, say 3, 7, 9, etc., we lay off on AD', ae'= 3, e'f'z= 7, etc., and AB will be divided proportionally to the given numbers. 305. To find a fourth proportional to three given lines A, B, C. Draw DE, DF, making any angle with each other. Upon DE lay off DG, GE, equal to A and B, respectively, and on DF lay off DH = C. Join GH, and through E draw EX parallel to GH, and meeting DF in X. Then DG: GE = DH: IIX, or A : B = C : HX. (274), It is obvious that if B — C, we take GE = DH, and we obtain by this construction a third proportional to A and B. 158 PLANE GEOMETRY. — BOOK IV. / X \ i C ED A B 306. To find a mean proportional between two lines, A, B. Upon an indefinite line lay off ED, CE, respectively equal to A and B. Upon CD as diameter describe a semicircle DXC; at E draw EX perpendicular to CD to meet the circumference in X. Then (312"), ED OV A:EX = EX: CE or B. 307. Definition. A straight line is said to be divided in extreme and mean ratio, when it is divided into two seg- ments such that the greater segment is a mean proportional between the whole line and the lesser segment. Thus if A 9 b AB is divided in C so that AB : AC =AC:BC, then AB ...•••■■ ~~~" •• .. is divided in extreme and / \d mean ratio. / .- ■ ' \ 308. To divide a line AB in extreme and mean ratio. At B draw 5 c JL to AB, and ..-•• make 5(7= 1^5. ^ From C as center, with radius CB, describe a circumf. BDE. Through A and C draw a secant meeting the circumference in D and E. From A as center, with radius AE, describe an arc EX meeting AB in X. Then AB : AX = AX : BX. For ••• AB is a tangent (191), and ^z> a secant to O £i)^, (Const.) AE : AB = AB : AD, (303) .-. AE : AB — AE = AB : AD — AB. (247) But AB = 2BC = ED, whence AD — AB =i AE = AX, .-. AX:BX = AB : AX, or AB : AX = AX : BX. (307) X CONSTR UCTIONS. 169 309. To divide a given line AB harmonically in the ratio of two lines C and D. Upon the line AM, making any angle with AB, lay off AE= C ; f and on each side of E lay off EF, EG, each ^ equal to D. Join FB, GB, and draw EX, EY, parallel to GB and FB, respectively. Then, AX : BX = AE :EG = C : D, and AY:BY= AE :FE (275) 310. Upon a straight line A'b' to construct a polygon simi- lar to a given polygon. Divide the given polygon ABODE into triangles by diagonals drawn from vertex A. At the extremities of A^B\ make Zb' = Zb, and Zb'A'c' = Z BAC. Then will AA^B'c'he similar to A^^c(288). In like manner con- struct A A'c'd' similar to Aacd, and A'd'e' similar to ADE. Polygon A'b'c'd'e' is similar to polygon ABODE. (294) 311. Upon a given line AB, to describe a segment of a cir- cle such that any angle inscribed in it shall equal a given Z O. At A, make Z BAD = Z (203); draw AG perpendicular to AD at A, and draw EG perpendicular to AB Sit its mid point E. From the intersection of AG, EG, de- scribe arc ABX. ABX is the required segment. (269) Proof. Z BAD = Z C, is measured by ^ arc ADB, as is also any angle inscribed in BXA. ■D 160 PLANE GEOMETRY. — BOOK IV. 312. From a given point A, in or without a given circum- ference BCD, to draw a tangent to BCD. Find 0, the center of BCD, (173) and join OA. Then 1°. If A is in the circumference, draw XY perpendicular to OA at A. XY is tangent to BCD at A. (191) 2°. If A is without the circumference, bisect OA in E. From E as center, with radius EA, describe a circumference AXY, cutting BCD in X and Y. Join AX, A y; then AX, A Y are tangents to BCD. (267, 191) Proof. Join OX, Y, and show that OX, Y are perpen- dicular to AX, A Y, respectiv^ely. EXERCISES. THEOREMS. 396. The chords that join the near extremities of equal chords are parallel. 397. The opposite angles of a quadrilateral inscribed in a circle are supplementary. Definition. Three or more points are said to be concyclic if a circum- ference can be described through them. Thus the preceding theorem could be enunciated thus : A quadrilateral whose vertices are concyclic has its opposite angles supplementary. 398. If two opposite angles of a quadrilateral are supplementary, its vertices are concyclic. 399. If AB, CD, intersect in 0, so that AG . OC - OD : OB, then A, B, C, D, are concyclic. 400. If OA, OD, are divided in B and C, respectively, so that OA: 0D= OC: OB, then A, B, C, D, are concyclic. EXERCISES. 161 401. If an arc be divided into three equal parts by chords drawn from one extremity of the arc, the middle chord bisects the angle formed by the other two. 402. If, from any point of a circumference, a tangent and a chord be drawn, the perpendiculars upon these lines from the mid point of the intercepted arc are equal. 403. The diagonals of a trapezoid cut each other in the same ratio. 404. If through one of the points of intersection of two equal cir- cles, any line be drawn to meet the circumferences, the extremities of this line are equally distant from the other point of intersection. 405. If, in a right triangle, the altitude upon the hypotenuse divides it in extreme and mean ratio, the lesser arm is equal to the farther segment. 406. In any right triangle, one arm is to the other as the difference of the hypotenuse and the second arm is to the intercept on the first arm between the right vertex and the bisector of the opposite acute angle. 407. The altitudes of a triangle are inversely proportional to the sides upon which they are drawn. 408. If from an angle of a parallelogram ABCD, a line be drawn cutting a diagonal in E and the sides in P and Q^ respectively, then will AE be a mean proportional between PE and QE. 409. If from the extremities of a diameter, perpendiculars be drawn to any chord of the circle, the feet of these perpendiculars will be equally distant from the center. 410. Show that there may be two, but not more than two, similar triangles in the same segment of a circle. 411. If two circles are tangent externally, a common exterior tan- gent is a mean proportional between their diameters. 412. The chord drawn from the vertex of an inscribed equilateral triangle to any point in the opposite arc is equal to the sum of the chords drawn to that point from the other vertices. 413. If two circles are tangent externally, lines drawn through the point of contact to the circumferences are divided proportionally at the point of contact. 414. If a circle is tangent to another internally, all chords of the outer circle drawn from the point of contact are divided proportionally by the circumference of the inner circle. Greom. — 11 162 PLANE GEOMETRY. — BOOK IV. 415. Chords drawn from the point of contact of a tangent have their segments made by any chord parallel to the tangent, inversely proportional. 416. If a tangent is intercepted between two parallel tangents to the same circle, its segments made by the point of contact have the radius as mean proportional. 417. If three circles intersect each other, their three common chords pass through the same point. 418. If through the mid point of a side of a triangle a line be drawn intersecting a second side, the third side produced, and a line parallel to the first through the opposite vertex, the line will be divided har- monically. PROBLEMS. 419. To a given circle draw a tangent that shall be perpendicular to a given line. 420. To a given circle draw a tangent that shall be parallel to a given line. 421. To a given circle draw two tangents including a given angle. 422. In a given straight line, find a point such that the tangents drawn from it to a given circle shall include the greatest angle. '■•••.. 423. In a chord produced, find a point such /^ "~~~\ .. ■•' that the tangent from that point shall be equal / ..-A to a given line. ••'■* \--'y^\ 424. From a given center describe a circle \^^'^^''''^ J^ tangent to a given circle. ^ -^ 425. Describe a circumference passing through a given point and touching a given circle in a given point. 426. Describe two circles with given radii, so intersecting that their common chord shall have a given length not greater than the lesser diameter. 427. With a given radius, describe a circle tangent to two given circles. 428. Draw a common exterior tangent to two given circles. 429. Draw a common interior tangent to two given circles. EXERCISES. 163 430. Find a point such that the tangents drawn from it to the outer sides of two tangent circles shall include a given angle. 431. Divide any side of a triangle into two parts proportional to the other sides. 432. Divide any side of a triangle into three parts proportional to the three sides. 433. From a given line cut off a part that shall be a mean propor- tional between the remainder and another given line. 434. Through a given point within a circle, draw a chord there divided in the same ratio as a given chord through that point. 435. From a given point without a circle draw a secant divided by the circumference in a given ratio. 436. From a given point in a given arc draw a chord bisected by the chord of the given arc. 437. In a given circle place a chord that shall be trisected by two given radii at right angles to each other. 438. In a given circle place a chord parallel to a given chord, and having to it a given ratio not greater than that of the diameter to the given chord. 439. Through one of the points of intersection of two given circles draw a secant forming chords that are in a given ratio. 440. Inscribe a square in a given triangle. 441. Inscribe a square in a given segment of a circle. 442. In a given semicircle inscribe a rectangle similar to a given rectangle. 443. In a given circle inscribe a triangle similar to a given triangle. 444. About a given circle circumscribe a triangle similar to a given triangle. 445. In a given triangle construct a parallelogram similar to a given parallelogram. . 446. Construct a triangle having given the base, the vertical angle, and the length of the bisector of that angle. 164 PLANE GEOMETRY. — BOOK IV. LOCI. 447. Find the locus of the center of each circumference that passes through two given points. 448. Find the locus of the center of each circle that is tangent to a given circle at a given point. 449. Find the locus of the center of each circle of given radius that is tangent to a given circle. 450. Find the locus of the center of each circle that is tangent to a given line at a given point. 451. Find the locus of the center of each circle that is tangent to t^ro given intersecting lines. 452. Find the locus of the points from which pairs of tangents of a given length may be drawn to a given circle. 453. Find the locus of the mid point of any chord that passes through a given point in a given circle. 454. Find the locus of the mid point of any secant that can be drawn from a given point to a given circumference. 455. Find the locus of the vertex of any triangle constructed on a given base, with a given vertical angle. 456. Find the locus of a point whose distances from two given points are in a given ratio. 457. Find the locus of a point whose distances from two given straight lines are in a given ratio. 458. Find the locus of a point the sum of whose distances from two given straight lines is equal to a given line. 459. Find the locus of a point the difference of whose distances from two given straight lines is equal to a given line. 460. Find the locus of the points that divide the chords of a given circle so that the rectangle of their segments is equal to a given square. Book Y. areas and their comparison. 5i*i< QUADRILATERALS. 313. The area of a plane figure is the quantity of its surface as measured by the unit of surface, or is the numer- ical measure of that quantity. 314. Figures that are not similar but have equal areas are said to be equivalent. 315. The base of a polygon is any side on which we choose to regard it as constructed. 316. The altitude of a polygon is the perpendicular dis- tance to the base from the remotest vertex or from a side parallel to the base or the base produced. B D A P B A P B JP Thus in each of the figures above, the perpendicular CP is the altitude of the figure when ^^ is taken as base. It is obvious that two triangles having their bases in the same line and the opposite vertex common, as ACB and ACD, have the same altitude. 165 166 PLANE GEOMETRY. — BOOK V. Proposition I. Theorem. 317. Rectangles with equal altitudes are to each other as their bases. B A' B E D' B' C Given : Two rectangles AC, A'c', with equal altitudes AB, a'b'j and bases BC, b'c' ; To Prove : Kectangle AC : rectangle A'c'=bc : B'c'. 1°. When BC and B'c' are commensurable. Let BE be a common measure oi BC and B'c', so that BE can be laid off 5 times on B'c' and 8 times on BC. Through each point of division draw perpendiculars to the opposite side of the rectangle. The figures thus formed are equal rectangles. (144) Since BC and B'c' contain 8 and 5 parts, respectively, each equal to BE, (Const.) AC and A'c', resp., contain 8 and 5 parts, each equal to AE. .-. BC : B'C'= 8 : 5, and AC : A'c'= 8:5; (225) .-. rect. ^C: rect. ^'6''=i?C': j5'C''. q.e.d. (232'") 2°. When BC and B'c' are incommensurable. Suppose b'c' divided into any number of equal parts n, and that BC contains this nth part of B'c' m times with a remainder EC. Draw EF perpendicular to BC. Since BE and B'c' are commensurable, (Const.) BE Wc' m n rect. AE rect. A'C'^ (1°) QUADRILATERALS. 167 BC m X , rect. AC ma?' .*• —7—; = — h-) and — 7—, — — | — , B^G' n n rect. A'C' n n since BC and ^C are slightly > BE and AE, respectively. Now, when n is taken indefinitely great, - and — become dl; rect. AC BC indefinitely small ; Q.E.D. (236) rect. ^'C' ^'C' being the limits of variables always equal. 318. CoR. 1. Rectangles ivith equal bases are to each other as their altitudes. For (317) the equal bases may be taken as altitudes, and the altitudes as bases. 319. Scholium. Since a rectangle is determined by its base and altitude (144), that is, by any two adjacent sides, as AB, BC, we employ the expression, the rectangle contained by AB and BC, or more briefly, the rectangle AB, BC, to de- note the rectangle determined by AB and BC ; and use the symbol rect. AB • AC, or simply AB • BC, for either of these expressions. Since a square, again, is determined by its base, i.e., by a side, Ave employ the expression, the square of AB, or the symbol AB , an abbreviation for AB • AB, to denote the square whose base is AB. 320. Definition. The symbol =c=, to be read, is equiva- lent to, is the symbol of equivalence. 321. Cor. 2. If four lines, A, B, C, D, are in proportion, the rectangle contained by the extremes is equivalent to that contained by the means. B.n = A:B, I B'D = C :D = A:B,) ^ ^ For since rect. A-D : rect. B -D = A:B, and rect. B • C : rect. rect. A'B : rect. B • C ^veat. B-D : rect. B • D, (232'") .-. rect. A -D =0= rect. B • (7. 168 PLANE GEOMETRY. — BOOK V. 322. Cor. 3. If A, B, c, are lines such that A: B = B :C, then B^ = rect. A - C. (321) That is, the square of a mean proportional between two lines is equivalent to the rectangle contained by those lines. 323. The U7iit of area is the square having as base the linear unit. Thus if the base AB of the r. square ^C is equal to the linear unit, then the square AC is the unit with which all areas are compared. 324. As already defined, the numerical measure of a quantity is the number that ^ ^ shows how many times the quantity contains its unit ; in other words, it is the ratio of the quantity to its unit. As regards triangles, it is customary to denote the numerical measure of a side by means of the small letter corresponding to the capital designat- ing the opposite angle. Thus in A ABC, we employ a, b, c, to denote the numerical measures of BC, AC, AB, respectively. As regards polygons of more sides than three, there is no such convention, but we specify AB = a, BC = b, CD = c, and so on. Wherever it may occur, henceforth, such an expression as the product of A and B is to be understood as a con- venient abbreviation for the product of the numerical value of A by that of B. Great care should be taken, however, not to forget the real meaning of such abbre- viations. Beginners are often confused by the careless use of such expressions as, leiigth multiplied by breadth gives area, forgetting that what is meant is: the numerical measure of the length multijoled by that of the breadth gives as result the numerical measure of the area, as we find explained in Art. 326. QUADRILA TERALS. 169 Proposition II. Theorem. 325. Rectangles are to each other as the products of the numerical measures of their altitudes and bases. II c' Given: Two rectangles AC, A'c', with altitudes AB, A'b', and bases BC, B^c', respectively, whose numerical measures are, respectively, a, a', and b, b' ; To Prove : Eectangle AC : rectangle A'c' = a xb:a' x b'. Construct a rectangle EG with altitude EF = A'b', and base FG = BC, and let the numerical measures of AC, A'c', EG, be X, y, z, respectively. Since AC and EG are rectangles with equal bases, (Const.) AC:EG = AB : EF, OT X : z = a : a' ; (318, 232') '.'EG and a'c' are rectangles with equal altitudes, (Const.) EG : A'C' = FG : b'c', ovz:y=b:b'. (317, 232') From these numerical proportions we obtain (242) x:y= a xb:a' X b', .'. rect. AC : rect. A'c' = axb ; ft' x b' q.e.d. (232") 326. Cor. The area of a rectangle is measured by the product of its base and altitude. For if ^ Cbe any rectangle, and S the unit-square, then •.• area^C:S = a x 6:1 xl, area ^C= S xab; i.e., the area of AG is ab times the unit-square S. If, for example, the numerical measures of AB and BC are 5 and 7, respectively, then the 170 PLANE GEOMETRY. — BOOK V. numerical measure of the area of ^C is 35 ; that is, the area of AC is equal to 35 unit-squares. In the enunciation of this corollary, as elsewhere, the term area is for brevity used for numerical measure of the area; that is, the number of unit-squares to which the sur- face in question is equivalent. Proposition III. Theorem. 327. Any parallelogram is equivalent to the rect- angle having the same base and altitude. A E D F B C Given : A parallelogram AC and a rectangle EC, with the same base and altitude BC, EB ; To Prove: Parallelogram AC \^ equivalent to rectangle EC. Since AC and EC are parallelograms, (Hyp.) AB z= DC, and BE = CF, (1^6) .-. Tt./\ABE = itADCF. (72) But ABCF — A ^i?^^rect. EC, and ABCF — AZ)C'i"<>par'm AC, .'. par'm JC=o=rect. -EC. q.e.d. (Ax. 3) 328. Cor. 1. Parallelograms with equal bases and equal altitudes are equivalent. For each is equivalent to the same rectangle. (327) 329. Cor. 2. Parallelograms icith equal altitudes are to each other as their bases; and those ivith equal bases to each other as their altitudes. For they are as the rectangles hav- ing those bases or altitudes. 330. Cor. 3. The area of any parallelogram is equal to the product of its base and altitude. QUADRILATERALS. Proposition IV. Theorem. 171 331. Any triangle is equivalent to orie half the rectangle contained by its base and altitude. Given : A triangle ABC, having a base BC and altitude AD ; To Prove : Triangle ABC is equivalent to i rect. AD - BC. Complete the parallelogram ABCE. Then since ^C is a diagonal of par'm BE, Aabc=Aace, (140) .-. A ^^C =0=1 par'm ^^, Q.E.D. (327) .-. A ABC ^ i rect. AD - BC. 332. Cor. 1. Triangles with equal bases and equal alti- tudes are equivalent. 333. Cor. 2. Triangles with equal altitudes are to each other as their bases; and those with equal bases, as their altitudes. 334. Cor. 3. Triangles are to each other as the products of their bases and altitudes. Exercise 461. Prove Prop. III., when the upper base of the rectangle lies with- out that of the given parallelogram, as in the accompanying diagram. 462. Prove the same proposition when the upper and lower bases lie without each other, though in the same lines. 172 PLANE GEOMETRY.— BOOK V. Proposition V. Theorem. 335. The rectangle contained hy a line and the sum or difference of other two lines, is equivalent to the sum or difference of the several rectangles contained hy that line and the other two lines. Given : A line AB and another line BD equal %o BC ± CD ; To Prove : rect. AB - BD =c= rect. AB' BC ± rect. AB • CD. Suppose the rectangles AB • BDJ etc., duly constructed. (206) 1°. Let BD = BC-\- CD. Then, in left-hand Fig., rect. AD = rect. AC -\- rect. ED, (Ax. 9) i.e., rect. AB • BD = rect. AB - BC + rect. EC • CD, (319) .-. rect. AB' BDo= rect. AB - BC -\- rect. AB - CD , q.e.d. (since EC = AB). (136) 2°. Let BD = BC — CD. Then, in right-hand Fig., since rect. AB .. BD -\- rect. AB - CD<^ rect. AB - BC, (1°) rect. AB • BD ^rect AB • BC — rect. AB • CD. q.e.d. (Ax. 3) Scholium. Let a, b, c, denote the numerical measure of AB, BC, CD, respectively ; then by substitution we obtain the well-known algebraic formula a{b ±c) = ab ± ac. 336. Cor. 1. The square of the sum or difference of two lines is equivalerit to the sum of the squares of those lines plus or minus twice their rectwigle. QUADRILATERALS. 173 (319) (335) thus Let A and B be the lines. Then {A±By^ rect. {A±B)'{A± B), =0= rect. A(A ± B)± rect. B(A ±B) =(^A^± rect. ^ . ^ ± rect. B - A + B'^ ^A^-{-B^±2TeGt.A'B, Scholium. This result may be expressed algebraically {a±by=a'±2ab-\-b\ 337. Cor. 2. The rectangle contained by the sum and dif- ference of two lines is equivalent to the difference of the squares of the lines. Let A and B be the lines. Then {A-{-B)'{A— B)^ A{A — B)-\-B{A — B), ^ A^ — A-B + B'A— B% =^A'-B\ } (335) N Scholium. This result may be expressed algebraically thus : (a + 6) (a - 6) = a^- b\ The above corollaries, pure deductions from Prop. V., be- come obvious to inspection in the accompanying diagrams. In the first, we see that the square of MP, the sum of MN and NP, is made up of the squares Q and R and the two equal rectangles s, s. In the second, we see that the square of MP, the difference of MN and NP, is less than NMN't, the sum of the squares of MN and NP, by the figure NSN', which = 2 P r = 2 rect. MN • NP. In the third, we see that if from the square of MP we take the square of MN, we obtain the figure QTP, which is equiva- lent to QS or {MP + PN) • (MP — PN), where PN = MP — MN. N' M P' S M N T M N 174 PLANE GEOMETRY. — BOOK V. Proposition VI. Theorem. 338. A trapezoid is equivalent to the rectangle contained by its altitude and half the sum of its parallel sides. Given : A trapezoid ABCD, with bases AB, CD, and altitude CE ; To Prove : ABCD is equivalent to rect. CE • ^{AB + CD). Join AC. Since ABCD o^ A ABC -\- A ADC, hut A ABC <>=i rect. AB • CE, ^ and AADC^^ rect. CD - CE, > ABCD^lQdt. CE-i{AB + CD). (Ax. 9) (331) Q.E.D. (335) Scholium 1. The above proposition may be expressed under the form : The area of a trapezoid is measured by the product of its altitude by half the sum of its bases. 339. Scholium 2. The area of any polygon may be found by dividing it into triangles, and find- ing the areas of tliese (331). But the method generally employed is to draw the most convenient diagonal of the figure, and draw to it perpendiculars from the other angular points. The figure is thus divided into right triangles, rectangles, or trapezoids, the areas of which are easily found. Q UA DRILA TERALS. 175 Proposition YII. Theorem. 340. Any circuinscribed polygon is equivalent to the rectangle contained hy half its perimeter and the radius of the inscribed circle. Given: A polygon ABODE circumscribed about a circle with radius OP ; To Prove: ABODE ^iect.OP'i(AB+BC-{-CD-{-DE-{-EA). Join OA, OB, etc., and draw OP^ 1. to BO, OP^ ± to OD, etc. Since AAOB=o^ rect. OP • AB, >> and Aboo =0=^ rect. OP2 -DO, i (331) and similarly of the other triangles, J ABODES rect. OP ' ^(AB + BO+OD-\-DE-{-EA), Q.E.D. since ABODE = the sum of all the A, (Ax. 9) and OP = OP2 = OP3, etc. (162) Scholium. In the case of the triangle, the most frequent application of this theorem may be stated as follows : The area of a triangle is equal to the product of half its per- imeter and the radius of the inscribed circle. Exercise 463. lu diagram for Prop. VI. , show that rect. AB - (AE - EB) ^ AE^ - EB^. 464. In the same diagram, show that rect.^^-(^B+ EBy>SB^ —lElB^. 465. In the same diagram, if Z ^ = 62° 54' 23", what must Z Z> be in order that A^ B, C, and D may be concyclic points ? 176 PLANE GEOMETRY. — BOOK V. Proposition VIII. Theorem. 341. Triangles that have an angle of the one equal to an angle of the other, are to each other as the rectangles contained by the sides including those angles. Given: Two triangles, ABC, A'b'&, having angle A equal to angle A' ; To Prove: Triangle ^£C : triangle A'b'c' =: lect. AB - ACiiect. A'B' • A'c'. t. BD • AC = AB : BD, » [ (317) ^'D' 'A'& = A'B':B'D',) Draw BD, B'd', ± to AC, A'c', resp. Since Z A = Z A' (Hyp.), and A D, D', are rt. A, A BDA is similar to Ab'd'a' ; (286) .-. AB : BD = A'B' : B'd'. Since rect. AB -AC: rect. BD ■ AC = AB : BD, and rect. A'b' - A'c' : rect. B' Tect.AB 'AC:A'B''A'c'=vectBD'AC:TeGt.B'D'-A'c'. (232'") BntAABC :AA'B'c'=Tect. BD - AC : rect. B'd' - A'c'; (334) .-. A ABC: A A'b'C' = rect. AB - AC : rect. A'b' - A'c'. Q.E.D. (232"') Scholium. The theorem may be expressed also under the form : The areas of triangles that have an angle of the one equal to an angle of the other, are as the products of the sides about those angles. QUADRILATERALS. 177 Proposition IX. Theorem. 342. Similar triangles are to each other as the squares of their homologous sides. Given: Similar triangles ABC, A'b'c', having AB : AC = A^B'-.A'C'; To Prove : Triangle ABC : triangle A'b'c' = AB^iHF^ etc. Since Ib^ : rect. AB - AC= AB : AC, and A^' : rect Ct. AB ' AV= AB : AC, ") . A'B' 'A'C' = A'B':A'C') ^ AB : TW = rect. AB - AC : rect. A^B^ - A'c', (249, 244) {since AB : AC = A' B' : A' C'.) (Hyp.) But A ABC : A A'b'c' = Yect AB ' AC iTect. A'B' ' A'c', (341) (since Z ^ = Z ^';) .-. AABCiAA'B'c' = AB^:A^\ Q.E.D. (232'") 343. Cor. Similar triangles are to each other as the squares of any homologous lines. Exercise 466. Show that the square of the sum of two lines diminished by the square of their difference is equivalent to four times the rectangle contained by the two lines. 467. In the diagram for Prop. IX., if CD, CD', be drawn, making the same angle with AB, A'B', resp., /\ ABC: A A'B'C = VD'^ : C'D''^' 468. In the same diagram, what should be the ratio of AB to A'B' so that A ABC may have twice the area of A A'B'C F Geom. — 12 178 PLANE GEOMETRY.— BOOK V. Proposition X. Theorem. 344. Any two similar polygons are to each other as the squares of their homologous sides. D D' A B A' B' Given: Similar polygons ABODE or P, and A'b'c'd'e' or P', and AB homologous to a'b' ; To Prove : P : P' = AB^ : A^\ Draw diagonals from A and A', dividing the polygons into the same number of homologous triangles. (295) Since any homologous pair of these triangles, as ABC, A'b'c', or ACD, A'c'd', are as the squares of any homologous sides, (342) . AABC:Aa'b'& = AACD :Aa'&D' = AB^: A^^, etc.; .-. AABC-\-AACD-^AADE:AA'B'C'-\- AA'C'd' + AA'd'e' = Ab':A^'; (251) .-. P : P' ~ AB~ : J^l Q.E.D. 345. Cor. 1. Similar polygons are to each other as the squares of any homologous lines. 346. Cor. 2. The homologous sides of any two similar 2wly- gons are as the square roots of the areas of those polygons. Exercise 469. In the diagram for Prop. X., if AB : A'B' = m: 7i, what will be the ratio of the square described on AD to that described on A'D' ? 470. In the same diagram, if P: P'=m: n, what is the ratio of any two homologous lines in the figures, as ^O and A'C'f QUADRILATERALS. 179 Proposition XI. Theorem. 347. In a right triangle, the square of the hypot- enuse is equivalent to the sum of the squares of the arms. _ Given: A right triangle ABC, having the right angle at A ; To Prove : BG^ is equivalent to Zb -\- AC . Draw^D±to^C. since BC:AB — AB: BD, rect. BC 'BD=^AB^ ; ^mCQ BC : AC = AC '. DC, rect. BC ' DC=o=AC^,' rect. BC'BD -\- rect. BC » DC<^~AB^ + 7c' i.e., TeotBC'(BD+DC) AB^ -\-Ac\' .e., BG AB' -\-Ac\ (297) (321) (297) (321) (Ax. 2) (335) Q.E.D. 348. Cor. If any similar polygons are constructed on the sides of a right triangle, that on the hypotenuse is equivalent to the sum of those on the arms. 349. Definitions. The projection of a point upon an indefinite straight line is the foot of the perpendicular drawn from the point to the line. Thus P is the projection of the point A upon MN. 180 PLANE GEOMETRY. — BOOK V. TlnQ projection of a finite straight line upon an indefinite straight line is the intercept between the projections of the extremities of the line. Thus PQ is the projection of AB upon MN. If B, one of the extremities of AB, is in MN-, then B coincides with its own projection, and PB is the projection of AB. Propositiox XII. Theorem. 350. In any triangle, the square of a side subtend- ing an acute angle is less than the sum of the squares of the other sides by twice the rectangle con- tained hy either of these sides and the projection upon it of the other side. Given: c, an acute angle of triangle ABC, and PC, the pro- jection on BC oi AC; To Prove : AB^ is equivalent to bO^ + Ic^ — 2 rect. BC - PC. According as P is in J5C or in ^C produced, PB = BC — PC, or —PC — BC. In either case, PB^^BC^-\-P(f-2 rect. BC'PC; (336) .-. PB^ 4- AP^ =o PC^ + PC^ + AP^ — 2 rect. BC - PC. (Ax. 2) But PB^ + AP^^ ab\ and PG^ + Jp^ =o AC'; (347) .-. Zb^ =o=PC/^ + Ic^ — 2 rect. PC -PC. q.e.d. QUADRILATERALS. 181 Proposition XIII. Theorem. 351. In an obtuse triangle, the square of the side subtending the obtuse angle is greater than the sum of the squares of the other sides by twice the rect- angle of either of these sides and the projection upon it of the other side. Given : c, the obtuse angle of triangle ABC, and PC, the projec- tion on BC of AC; To Prove : AB^ is equivalent to 'bc^ + AC^ + 2 rect. BC • PC. Since PB = BC -{- PC, PB^^'BC^-\-PC^-h2Tect BC'PC; (336) .-. PB^ + AP^^BC^-\-PC^ + AP^-^2TeGt. BC-PC. (Ax. 2) But PB^ + ^^ =0= 'ab\ and Fc^ + AP^ = Ac\' (347) .-. AB^ =0 BC" + AC^ + 2 rect. BC • PC. Q.E.D. 352. Cor. If the square of one side of a triangle is equiva- lent to the sum or the difference of the squares of the other two sides, in either case the two lesser sides are at right angles to each other (347, 350, 351). Exercise 471. Show that the difference of the square of a line and the square of its projection on another line, is equivalent to the square of the difference of the perpendiculars that intercept the projection. 472. In the diagram for Prop. VI., show that AB^ -\- W]^ - Z€^ + 2 rect. AB'EB, and UD^ + AD^ = Ajf - 2 rect. CD . {AE - CD). « 473. Show, by means of the foregoing exercise, that 3^+ BV"-iUD''+ XZ)^)<>2(rect..4 B(AB- AE) + rect. CD{AE- CD)). 182 PLANE GEOMETRY. — BOOK V. Proposition XIV. Theorem. 353. The sum of the squares of any two sides of a triangle is equivalent to twice the square of the median to the third side plus twice the square of half this side. Given : In triangle ABC, AD the median drawn to bO; To Prove : AB^ + AC^ is equivalent to 2 AD^ + 2 'bd^. Draw AP ± to BC. 1°. li AB = AC, then AP coincides with AD, (74) AB^ ^ Bif 4- Aif, and Ajf =o= dc^ + ad^ ; (347) .-. ab'-\- Ac"^ =0=2 BD^-{- 2 Ad'\ q.e.d. (Ax. 2) (since DC = BD.) 2°. If AB> AC, Z ADB is obtuse and Z ADC, acute ; .: AB'^AD^ -\- BD^ -\-2BD • DP, (351) and AC^^AD^ + DC^ — 2 DC -dp; (350) .-. AB^ + ag^^2ad~ + 2bd\ q.e.d. (Ax. 2) Exercise 474. In the diagram for Prop. XII., left-hand figure, if BP' be drawn perpendicular to ^O, sliow that rect. ^ C • P'C^^rect.BG- PC. 475. In the same diagram, if AB^ ^ 370'^+ 3 PO^ in what way does P divide BG? 476. From the diagraip for Exercise 474 deduce a proof of the theorem : The altitudes of a triangle are inversely proportional to the sides to which they are drawn. QUADRILA TERALS. 183 Proposition XV. Theorem. 354. The rectangle of any two sides of a triangle is equivalent to the rectangle contained by the altitude upon the third side and the diameter of the cir- cumscribed circle. Given : Triangle ABC inscribed in circle ABBC, AB a diamfeter, and AP perpendicular to BC ; To Prove : Eect. AB - AC i^ equivalent to rect. AP - AB. Join BC. Since ACB is a semicircle, (Hyp.) rt. Z ACB = rt. Z APB; (267) 2i[so Zb=Zb; (266) .-. A APB is similar to A ACB ; (288) .'. AB'.AP = AB:AC ; (284) .-. rect. AB • AC ^ rect. AP • AB. q.e.d. (321) Exercise 477. If ^O is a diagonal of a parallelogram ABCD, having /LA equal to an angle of an equilateral triangle, show that ZU^^ZB'^ + W:!^ + rect. AB-BG. 478. If AC is a diagonal of a parallelogram ABCD, having ZA equal to twice an angle of an equilateral triangle, show that AjO^ <^AB'^ + BC'^ - rect. ABBC. 479. The square of the base of an isosceles triangle is equivalent to twice the rectangle contained by either of the arms and the projection of the base upon that side. 184 PLANE GEOMETRY. — BOOK V. Proposition XVI. Theorem. 355. The rectangle of any two sides of a triangle is equivalent to the rectangle of the segments ijvade on the third side hy the bisector of the opposite angle, plus the square of the bisector. Given: In triangle ABC, AD tlie bisector of angled, cutting BG'm D ; To Prove ': rect. AB - ACi^ equivalent to rect. BD - DC + AD^. Describe O ABEC about A ABC; ' (185) produce AD to meet circumference in £, and join EC. Since Z BAD —Z. CAE, (Hyp.) SindZB=:ZE, (266) A ABD is similar to A AEC; (287) .: AB: AD =AE:AC; (284) .-. rect. AB . ^C'=o= rect. AD • AE ; (321) i.e., rect. AB - AC=^ rect. AD • {AD + DE) ; i.e., rect. AB • AC^ AD^ + rect. AD • DE ; (335) i.e., rect. AB • .1C=<> rect. BD - DC -\- AD^. q.e.d. (301) Exercise 480. The sum of the squares of the sides of any paral- lelogram is equivalent to the sum of the squares of the diagonals. 481. A median divides a triangle into two equivalent triangles. 482. Three times the sum of the squares of the sides of a triangle is equivalent to four times the sum of the squares of the medians. QUADRILA TERALS. 185 Proposition XVII. Theorem. 356. If two chords intersect, the rectangle of the segments of the one is equivalent to the rectangle of the segments of the other. (See diagram for Prop. XXL, Book IV.) Since OA : OD = OC: OB, (301) rect. OA'OB :o rect. OC • OD. q.e.d. (321) Proposition XVIII. Theorem. 357. If from the same point a tangent and a secant he drawn to a circle, the square of the tan- gent is equivalent to the rectangle of the secant and its external segment. (See diagram for Prop. XXIII., Book IV.) Since OA:OC=OC: OB, (303) OC^^rect. OA ' OB. Q.E.D. (322) Scholium. The two foregoing propositions enunciate the last three of Book IV. from a different point of view. In these and similar theorems, we may substitute product for rectangle when regard is had to the numerical measures of the lines concerned. Exercise 483. In any triangle ABG^ if the altitudes BD^ CE, be drawn to AC, AB, respectively, show that BU^ <^ rect. AB - BE -^ rect.^C- CD. 484. If ABC is a scalene triangle, 'AB^ + AJ)'^-\-BV'^>AB-AC-\-AB-BC-^AC'BC. 485. The square of the median to the hypotenuse of a right tri- angle is equivalent to one fourth of the square of the hypotenuse. 186 PLANE GEOMETRY. — BOOK V. EXERCISES. THEOREMS. 486. A parallelogram is divided by its diagonals into four equiva- lent triangles. 487. If two triangles have two sides of the one severally equal to two sides of the other, and the included angles supplementary, the triangles are equivalent. 488. If any point within a parallelogram be joined with the ver- tices, the sums of the opposite pairs of triangles are equivalent. 489. If through any point in a diagonal of a parallelogram parallels to the sides be drawn, of the four parallelograms thus formed, the two through which the diagonal does not pass are equivalent. 490. A line joining the mid points of its bases bisects a trapezoid. 491. If the mid points of the sides of a quadrilateral be joined in order, a parallelogram is formed equivalent to one half the quadri- lateral. 492. The lines joining the mid point of a diagonal of a quadrilateral with the opposite vertices, cut off one half the quadrilateral. 493. The sum of the squares of the sides of any quadrilateral is equivalent to the sum of the squares of the diagonals, and four times the square of the line that joins their mid points. 494. If from any point P in the production of ^O, a diagonal of a parallelogram ABCD^ lines PB, PD, be drawn, the triangles PBC, PD C, will be equivalent. 495. If from a point P without a parallelogram ABCD, lines PA, PB, PC, PD, be drawn, then A PAB - A PCD is equivalent to one half the parallelogram. 496. Of the four triangles formed by drawing the diagonals of a trapezoid, (1) those having as bases the nonparallel sides are equiva- lent ; (2) those having as bases the parallel sides are as the squares of those sides. 497. If two triangles have a common angle and equal areas, the sides containing the common angle are inversely proportional. 498. If through its vertices lines be drawn parallel to the diagonals of any quadrilateral, the figure formed will be a parallelogram of twice the area of the quadrilateral. CONS TR UC TIONS. 1 87 499. In the diameter of a circle two points are taken equally dis- tant from the center ; if through one of these any chord be drawn, and its extremities joined with the other point, the sum of the squares of the triangle formed is constant. 500. The sum of the squares of the four segments of any two chords that intersect at right angles is constant. 501. The rectangle of the segments of chords intersecting at a given distance from the center is constant. 502. The square inscribed in a semicircle is to that inscribed in the circle as 2 is to 5. 503. If one side of a triangle is lengthened, and another shortened by the same length, the line joining the points of section is divided by the base in the inverse proportion of the sides. 504. If from the same point a secant and two tangents be drawn, the secant will be divided harmonically by the circumference and the chord joining the points of contact. CONSTRUCTIONS. 358. To construct a triangle equivalent to a given polygon AB-'-F. E Join the extremities of any two yyj ^ adjacent sides, AF, FE, by AE ; and "^ / / \ through i^ draw i^x parallel to ^^, to / y / ) meet BA produced in X. Join EX. / \ / / We have now a polygon BCDEX, hav- A^ / ing one side fewer than the given polygon, and equivalent to it, since A XEA ^ A FEA (332). By now joining BE, and proceeding as before, we obtain an equivalent polygon having one side fewer than BCDEX, and two fewer than the given polygon. Continuing the process, we evidently must at last obtain a triangle equiva- lent to the given polygon. 188 PLANE GEOMETRY. — BOOK V, 359. To construct a parallelogram equivalent to a given triangle ABC, and having an angle z y a equal to a given ayigle D. V \ 7\ Through A draw AZ parallel to \ \/ \ BC, and through X, the mid point of \ \ A \ BC, dmw XY,m2ikmg Zbxy=Zd, A \ / \ \ and meeting AZ in Y. Through B ^ ji q draw BZ parallel to xr to meet AZ in z. Then xz, which is equivalent to A^^C (331), is the parallelogram required. Scholium. If the given angle z> is a right angle, then XZ will be a rectangle. 360. To construct a square equivalent to a given rectangle A C. Produce AB to E, so that BE = BC, and upon AE as diameter describe a semicircumference AXE. Produce BC to meet the circumference in X. Then -BX is a side of the required square. For since AB :BX = BX: BE, (298) BX^^AB ' BE=o=AB ' BC. (322) Scholium. By means of the three foregoing construc- tions, we can construct a square equivalent to any given polygon by first transforming the given polygon into a triangle, then the triangle into a rectangle, and finally the rectangle into a square. 361. To construct a square equivalent to the sum of two given squares. q Draw BC ± to AB, and make AB, BC, respectively equal to the sides of the given squares. Join^C. Then 2 2 2 AG =c=AB -{- BC . CONSTRUCTIONS. 189 Scholium. It is obvious that, by a continuation of the process, we can obtain the side of a square equivalent to the sum of any number of given squares. 362. To construct a square equivalent to the difference of two given squares. Draw B3I ± to BN, and on BN lay off BA = the side of the lesser square. From A as cen- ter, with radius AX = the side of ^^^-.^ the greater square, describe an arc '\lr cutting BM in X. Then BX is a ...•••■■' \ side of the required square. For ....•••"" ^ BX^^^^AX^'-AB^ (347) ^ ^ i^ 363. To construct a square that shall be any given part of a given square. That is, L being the side of the given square, we have to find a line X, such that X:L = ^m:Vn, or x' = -L^. n Employing the construction of Art. 362, we take BA= ^^~^ x and AX = — '^^L ; that is, we divide L into 2n equal parts 2n (207), then take m — n and m-\-n such parts. \ 2n J \ 2n J n If, for example, we wish to find the side of a square that shall be equivalent to -y- of a given square whose side is a line L, we divide L into 12 equal parts (207), take BA = (11 — 6), or 5, of these parts, and AX = (11 + 6), or 17, of these parts. Then BX^ = ({iyL^-(^\yL'=l^l.L^, or X:i=Vll: V6. If, again, we wish to find a line whose numerical meas- ure shall be Vl3, supposing L to be the linear unit, we 190 PLANE GEOMETRY. — BOOK V. 1S — 1 1S4-1 take BA =^ — -L, or 6 L, and AX = "^ L, or 7 L. Then, 2 A since B^ = 7^L^ — 6^L^ = 13l^, we have BX = Vl3z ; ^.e., the numerical measure of BX is Vl3. 364. To construct a polygon P similar to a given polygon Q, so that we shall have P : Q = m : n. Let L be any side of Q ; then by Art. 363 find a line X such that X'.L = Vm : Vw. Upon X construct (310) a polygon P similar to the given polygon Q. Then P is the polygon required; since (344) P: Qz={^my: {^ny = m:n. 365. To construct a polygon similar to each of two given polygons, and equivalent to their sum or difference. We obtain an homologous side of the required polygon by proceeding with two homologous sides of the given poly- gons as we did with the sides of the given squares in Art. 361, when we wish to find a sura ; and as jve did with the sides of the given squares in Art. 362, when we wish to find a difference. Upon the base thus obtained we then construct (310) a polygon similar to the given polygons. 366. To construct a rectangle equivalent to a given square, and having the sum of two adjacent sides equal to a given line AB. Upon ^^ as diameter, describe a semicircumference ADB. At A draw AC A. to AB, and equal to the side of the given square. Through draw CD II to AB to meet ADB in D. From D draw DX ± to AB to meet AB in X. Then AX and BX are respec- tively the base and altitude of the required rectangle. For rect. AX'BX^ DX^ ^ ag\ (298) X B NUMERICAL APPLICATIONS. 191 367. To construct a rectangle equivalent to a given square, and having the difference of two adjacent sides equal to a given line AB. Upon AB as diameter, describe a circum- ference AXBY. At A draw AC A- to AB, and equal to the side of the given square. From c draw the secant CXY through the center to meet the circumference in x and Y. Then CX, CY are, respectively, the base and altitude of the required rectangle. For rect. CY -CX (357) 368. To construct a polygoii similar to a given polygon P, and equivaleyit to another given polygon Q. Find (360, Scholium) X and Y, the sides of squares equiv- alent to P and Q respectively. Then (305) find a fourth proportional Z, to X, F, and L, the base of P. The polygon constructed on z as base and similar to P (310) will be the required polygon. Proof by the student. NUMERICAL APPLICATIONS. 1°. To compute the altitude of any triangle in terms of its >:^ In the A ABC, let a, b, c, be the numerical measure of the sides opposite A, B, c, respec- tively, and h that of the altitude from A. Of the A B and c, one at least, say C, must be acute ; A h^=b^-Dc\' 3 d c^ = a ^ + 52 - 2 o^ op ; a^^b^-cK CD = 2a 192 PLANE GEOMETRY. — BOOK V. k' = b' ( ^' + ^' ~ "^' Y^ ^ ^'^' ~ ^''' ^ ^' ~ ^'^' ^ (2ab-\-a^-\-b'- c^) (2 ab - g^ - 6^ + c') 4a2 ^ ](a-^by-cmc'-ia-by\ 4.0? _ {a-\-b + c){a + b — c){c-\-a — b){c—a-{-b) ~ 4.0? Let a4-6 + c = 2s; le., let s denote half the perimeter; then a-\-b — c = 2(s — c) ; c + a-6 = 2(s-6); c — a + 6 = 2(s — a). Hence, simplifying and extracting the square root, 2 h = -Vs(s — a) (s — 6) (s — c). 2°. To compute the medians of a triangle in terms of the In A ABC, denoting the values of the sides as in 1°, and the median from Ahy m, ^ since ft^ + c" = 2 m^ + 2 (i a) \ (353) 3°. To compute the bisectors of a triangle in terms of the sides. In A ABC, denoting the values of the sides as in 1°, and the bisector from A, by d (see diagram for Prop. XVI.) ; since be = BD X DC -\-d^, (355) d^=bc-BD XDC. (1°) NUMERICAL APPLICATIONS. 193 ■r. , BD DC Bn-{-DC a .r^rjns But — = - = — — ^ = — -- ; (278) c b b.-\- c + c 6 + c b + c Substituting these values in 1°, and simplifying, 2 ^ = 7~, — ^bcs(s — a), b -he ^ ^ 4°. To compute the radius of the circumscribed circle in terms of the sides of a triangle ABC. In A ABC, denoting the values of the sides as in 1°, and the radius by R (see diagram for Prop. XV.) ; since bc = 2Rx AP, (354) 2 __^ and AP = ^-^s{s- a) {s -b){s- c), (1°) K a ahc 4 Vs(s — a) (s — b) {s — c) 5°. To compute the area of a triangle in terms of its sides. Denoting the values of the sides as in 1°, that of the alti- tude by h, and of the area by S, 2 since 8=1- ah (331), and 7i = - ^s{s-a) (s-b) (s-c), (1°) a S= Vs(s— a)(s — b){s — c). Scholium. If the triangle is equilateral, i.e., if a=b=c, a^ the formula reduces to *s = — ^^3, 6°. To compute the area of a triangle in terms of the sides and the radius of the circumscribing circle. Geom. — 13 194 PLANE GEOMETRY.— BOOK V. Denoting the values of the sides, altitude, and area as in 5°, since bG = 2 R -h, (354) abc = 2R'a-h = 4:R'S; . o _abc ~4^ s EXERCISES. QUESTIONS. 505. How many different altitudes can each of the following figures have : An equilateral triangle ? An isosceles triangle ? A scalene triangle ? A square ? A rectangle ? A trapezoid ? A trapezium ? 606. A side of an equilateral triangle is 6.* What is its»altitude ? 507. An arm and the base of an isosceles triangle are 18 and 16 respectively. What are its altitudes ? 508. The area of a triangle is 180 ; its sides are 30, 60, and 40, respectively. What are its altitudes ? 509. The area of a triangle is 252 ; its altitudes are 8, 12, and 14, respectively. What are its sides ? 510. The sides of a rectangle are 65 and 32 respectively. What are its area, perimeter, and diagonal ? 511. The altitude and base of a triangle being 23 and 10 respec- tively, what is its area ? 512. The area of a triangle is 221 sq. ft. ; its base is 5| yds. What is its altitude in inches ? 513. The bases of two parallelograms are 15 and 16 respectively ; their altitudes are 8 and 10 respectively. What is the ratio of their areas? 514. Two triangles of equal areas have their bases 26 in. and 3 ft. respectively. What is the ratio of their altitudes ? * Remember that the given abstract numbers are numerical measures. EXERCISES. 195 515. The bases of a trapezoid are 23 in. and 17 in. respectively, its altitude being 2 \ ft. What is its area ? 616. In A ABC, AB = i2, AG = 34. If DE cut off AD = 30 and AE = 15, what is the ratio of A^^Oto A^Z>^? 517. What should be the length of a ladder such that, having its foot 15 ft. from the wall, it may reach a window 20 ft. from the ground ? 518. Two chords intersect so that the segments of one are 12 and 7 respectively. If a segment of the other is 10, what is its second segment ? 519. What are the altitudes of a triangle whose sides are 12, 15, 9, respectively ? 520. What are the medians of the same triangle ? 521. What are the bisectors of the angles of the same triangle ? 522. What is the radius of the circle circumscribing the same triangle ? 528. What is the area of the same triangle ? 524. What is the area of a triangle whose sides are 6, 5, 5, respec- tively ? 525. What is the area of an equilateral triangle whose side is 3 ? 526. What is the area of an equilateral triangle whose altitude is 11 ? 527. The sides of a right triangle are 25, 24, 7. What are its medians and its altitude upon the hypotenuse ? 528. From the same point a tangent and a secant being drawn, if the secant and its external segment are as 27 to 3, what is the length of the tangent ? 529. If from the point just referred to a second secant be drawn, so that its external segment is 8, what will be the length of the secant ? 530. Two secants drawn from the same point have external seg- ments of 5 and 3 respectively. If the first secant is 27, what are the internal segments ? 196 PLANE GEOMETRY. — BOOK V. PROBLEMS. 531. Construct an isosceles triangle on the same base as a given triangle, and equivalent to it. • 532. Construct a right isosceles triangle equivalent to a given square. 533. Construct a parallelogram having a given angle upon the same base as a given square, and equivalent to it. 634. Divide a given line into two segments such that their squares shall be as 7 is to 5. 535. Bisect a given parallelogram (1) by a line passing through a given point ; (2) by a perpendicular to a side ; (3) by a line parallel to a given line. 536. Bisect a given triangle by a line drawn through a given point P in one of the sides. 537. Cut off one nth of a triangle by a line drawn through a given point in one of the sides. 538. Bisect a quadrilateral by a line drawn through one of the vertices. 539. Cut off from a quadrilateral one nth part by a line drawn through one of the vertices. 540. Bisect a triangle by a line parallel to the base. 541. Bisect a triangle by a line perpendicular to the base. 542. Find a point within a triangle such that lines joining the point with the vertices shall divide the triangle into three equivalent triangles. 543. Bisect a trapezoid by a line drawn parallel to the bases. 544. Bisect a trapezoid by a line drawn through a given point in one of the bases. 545. Construct a triangle equivalent to a given triangle, and having one side equal to a given line. 546. Construct a right triangle equivalent to a given triangle, and having one arm of a given length. EXERCISES. 197 547. Construct a right triangle equivalent to a given triangle, and having its hypotenuse of a given length. 548. Construct an isosceles triangle equivalent to a given triangle, and having its arms of a given length. 549. Construct an isosceles triangle equivalent to a given triangle, and having its base of a given length. 550. Construct an equilateral triangle equivalent to a given triangle. 651. Construct an equilateral triangle equivalent to a given square. 552. Construct a triangle similar to each of two given similar tri- angles, and equivalent to their sum. 553. Construct a triangle similar to each of two given similar tri- angles, and equivalent to their difference. 554. Construct a square that shall be to a given triangle as 5 is to 3. 555. Construct an equilateral triangle that shall be to a given square as 7 is to 5. Book VL regular plane figures. regular polygons. 369. A polygon of five sides is called a pentagon ; one of six sides is called a hexagon; of seven sides, a heptagon; of eight sides, an octagon; of fifteen sides, a pentadecagon ; and so on. 370. A regular polygon is both equilateral and equiangular. PiioposiTiox I. Theorem. 371. A regular polygon may he divided into as many equal isosceles triangles as the polygon has sides. E D Given : A regular polygon AB -" F, or P, having n sides ; To Prove : P may be divided into w equal isosceles triangles. Bisect A A and B by ^0, BO. (81) A A and B are each < a st. Z ; .-. ^Z A -\-}Zb < Si st Z; .*. AO, BO, must meet in some point 0. (114) 198 REGULAR POLYGONS. 199 Join Avitli C, D, E, F ; then since Z OAB = Z OB A, (Ax. 7) A OAB is isosceles. (65) Since AB = BC (Hyp.), OB is common, and Z OB A = ZOBC, (Const.) AOBa = AOAB. (66) In the same way it may be shown that AOCD = isos. A OBC, A OAF = isos. OAB, etc. q.e.d. 372. CoR. 1. The bisectors of any two angles of a regular polygon determine by their intersection a point equidistant from the vertices of the polygon. For the lines drawn from the vertices to that point are the sides of equal isosceles triangles (371). 373. Cor. 2. The point that is equidistant from the vertices, is also equidistant from the sides, of the pjolygoyi (101). For it is in the bisectors of all the angles of the polygon. 374. Cor. 3. A circle may be circumscribed about, or in- scribed in, any regular polygon, and both circles have the same center. For taking the intersection of the bisectors of any two angles as center, the circumference described through one vertex will pass through all (372), and that described tan- gent to one side will be tangent to all (373). 375. Definitiox. The center of a regular polygon is the common center of the inscribed and circumscribed circles. 376. Definition. The radius of a regular polygon is that of the circumscribed circle. 377. Definition. The opothem of a regular polygon is the radius of the inscribed circle; i.e., the distance from the center to any side. 200 • PLANE GEOMETRY. — BOOK VL 378. Definition. An angle at the center of a regular polygon is the angle formed by the radii drawn to the extremities of any side. 379. CoR. 4. // the number of sides is n, each angle at the center = 2 st. A-i-n. Proposition II. Theorem. 380. An equilateral polygon Uiscribed in a circle is regular. E^ — -^ D Given: An equilateral polygon P inscribed in (DACE; To Prove : P is a regular polygon. Since chd. AB = chd. PC = chd. CD, etc., (Hyp.) arc AB = arc PC = arc CD, etc. ; (174) .-. arc ABC = arc BCD = arc CDE, etc. ; (Ax. 2) .-. ZBz=Zc = Ad, etc., (266) (being inscribed in equal segments ;) .*. P, being equilateral and equiangular, is regular. Q.E.D. (370) 381. Cor. 1. If a circumference he divided into n equal arcs (n being > 2), the chords subtending these arcs ivillform a regular polygon of n sides. 382. Cor. 2. If the arcs subtended by the sides of a regu- lar polygon o/n sides be bisected, the chords drawn to subtend these arcs willforyn a regular polygon of 2n sides. REGULAR POLYGONS. 201 Proposition III. Theorem. 383. Regular polygons of the same number of sides are similar. Given: Two regular polygons, ABC or P, and A'b'c' or P', each of n sides ; To Prove: P is similar to P'. Since P and p' have each n angles, (Hyp.) each Z of P and p' = '-^^^ st. A resp. ; (127) n .'. P and 7^' are mutually equiangular. Since P and P' are each equilateral, (HyP-) AB :BC= A^B^:B'& ; similarly, all the sides about the equal A are proportional ; .*. P is similar to P'. q.e.d. (284) 384 Cor. The perimeters of regular polygons of the same number of sides are as their radii, or apothems; and their areas are as the squares of these lines. For the radii and apothems are homologous lines ; hence the perimeters are proportional to them (296), and the areas are proportional to the squares of those lines (344). ExERGiSE 556. The ratio of an interior angle of a regular polygon of n sides to an interior angle of a regular polygon of double the num- ber of sides, is expressed by n — 2 : ?i — 1. 202 PLANE GEOMETRY. — BOOK VI. Propositiox IV. Theorem. 385. A regular polygon being inscribed in a circle, a similar polygon may be circumscribed about the circle, L K M ^.6^ N Given : A regular polygon AB •-• F, or P, inscribed in O ACE ; To Prove: A similar polygon can be circumscribed about ACE. Through each vertex A, B, C, etc., draw tangents NG, GH, HK, etc. Since AB = BC = CD, etc., (Hyp.) Zgab = Z. gba = Aubc = A hcb, etc., (269) (being formed by tangents and chords of equal arcs.) Since each chord is less than a diameter, each of these equal A is less than a rt. Z ; (267) .*. the tangents through adjacent vertices will meet. (114) Let them meet in G, //, K, L, etc. Since isos. Agab = isos. A IIBC = isos. A KCD, etc., (63) Zg = Zr = Zk= etc. ; (70) then GHK '•• JV is equiangular. Also, GA = GB = HB =z lie = etc., (70) GH = HK = KL = etc. ; (Ax. 6) .-. GHK"'N is equilateral, with the same no. of sides as P ; .'. GHK ••• iV^ is a regular polygon similar to P. q.e.d. (370) 386. CoR. If a circumference he divided into n equal arcs (n being > 2), the tangents drawn at the points of division ivill form a regular circumscribed polygon of n sides. REGULAR POLYGONS. 203 Proposition V. Theorem. 387. Any regular polygon is equivalent to the rec- tangle of its apothem and half its perimeter. A R B Given : OB, the apothem of a regular polygon AB -•' f,ot p, of n sides ; To Prove : P is equivalent to rectangle ^n AB - OR. Draw AO, BO. Since A OAB ^i rect. AB • OR, (331) and P ^ ?i A OAB, (371) P =0= i n rect. AB - OR^ rect. ^n AB - OR. q.e.d. Scholium. The theorem may also be stated thus : The area of a regular polygon is measured by one half the product of its apothem and perimeter. Exercise 557. In the diagram for Prop. IV., if GR be the perpen- dicular from Q- to AB, the number of sides being n, show that GR^ is 4 n times the difference of the sums of the squares of the sides of the circumscribed and inscribed polygons. 558. In the same diagram, n being the number of sides, each angle formed like GAB by an interior and an exterior side, is - of a straight angle. ^ 659. In the diagram for 'Prop. V., if a parallel to AB be drawn through the mid point of 2?C, what lines will it bisect ? 560. In the same diagram, if the mid points of the radii be joined, what figure will be formed, and what ratio will its area have to that otPf 204 PLANE GEOMETRY. — BOOK VI. Proposition VI. Theorem. 388. As the miwiber of sides of a regular in- scribed polygon indefinitely increases, the apothem increases towards the radius as its limit. Given : AB, a side, and OP, the apothem, of a regular polygon of n sides inscribed in a circle whose radius is OA ; To Prove : As n increases, OP increases towards OA as limit. Since OP is ± to AB, (377) OpVoI'-Zp'. (347) Now as n increases, AB, and therefore AP, decreases (181), and when n becomes indefinitely great, AP becomes in- definitely small, while OA, the radius, is constant ; .♦. 'op'' has for limit Oa\' (235) .-. OP has for limit OA. q.e.d. Exercise 561. In the diagram for Prop. VI., if AO be produced, show that it will pass through a vertex of the polygon. 562. In any polygon of an even number of sides, the lines joining opposite vertices are diameters of the circumscribed circle. 663, In the diagram for Prop. VI., if the inscribed polygon is a regular hexagon, what is the ratio of OP to AP ? 564. In the same diagram, if OB be joined, and PC, PD, be drawn to the mid points of OA, OB, resp., OCPD will be a rhombus, unless Z AOB is a right angle. In what case will Z AOB be a right angle ? 565. The area of the regular inscribed hexagon is half the area of the circumscribed equilateral triangle. REGULAR POLYGONS. 205 Proposition VII. Theorem. 389. A straight line is the least of all the lines that terminate in two given points. ^,^-<<- J2/ ^ p^^ ^>^ Given: A straight line AB, and another line AHG -•' Bf termi- nating in A, B ; To Prove : AB is less than AHG • • • j5. Join alternate points AG, GE, EC. Since AG < AH-\-HG, GE < GF -{- FE, EC s, hut P'= 27ri?; />=-, R = ~- IT 2-^ Proposition X. Theorem. 397. A circle is equivalent to one half the rectan- gle contained hy its radius and circumference. Given : C, the circumference, and R, the radius, of a circle ABE ; To Prove : Circle ABE \^ equivalent to \ rectangle R • C. Let r denote the apothem, and p the perimeter, of a regu- lar polygon P inscribed in ABE. Then P =c= i rect. r • p. (387) Conceive the number of sides by continual duplication to become indefinitely great. Then P has for limit O ABEf and rect. r • j) has for limit rect. R - C, C being the limit of p and R that of r; (392) .-. O^j^J^^i rect. R ' c, Q.E.D. (236) (they being the limits of variables always equal.) Geoin. — 14 210 PLANE GEOMETRY. — BOOK VL « Scholium, This theorem may be stated under the form : The area of a circle is measured by one half the product of its radius and perimeter. 398. Cor. 1. The area of a circle is equal to ir times the square of its radius. For denoting the area of the circle by s, since S^^R- c, (397), and C = 27r • i? (396), Sz=^R X 27r- R = 7rR'\ 399. CoR. 2. The areas of circles are to each other as the squares of their radii. For S = TT • R", and S'^ir- R'; (398) .-. S:S' = 7rR':7rR"=R':R". 400. Definition. A segment of a circle is the figure bounded by an arc and its chord. 401. Definition. A sector of a circle is the figure bounded by two radii and their intercepted arc. 402. Definition. Similar segments and sectors in differ- ent circles are such as have arcs measuring equal angles at the center. 403. Cor. 3. The area of a sector is measured by one half the product of its radius and arc. For the sector is to the circle as the arc of the sector is to the circumference. 404. Cor. 4. Similar sectors are as the squares of their radii. For they are like parts of their respective circles. Exercise 575. What figure is both a segment and a sector ? 576. The area of one circle is twice that of another. What is the ratio of their radii ? 577. The radii of two similar segments is as 3 to 5. What is the ratio of their areas ? REGULAR POLYGONS. 211 Proposition XI. Theorem. 405. Similar segments are to each other as the squares of their radii. o o' Given : OA = R, O'A^ = R', radii of similar To Prove : B : b' = R- : R'^ 8 B, b' ; Since Zo = Zo' (Hyp.) (402) sector OABC is similar to sector O'A'b'c'. Since OA = OC, O'A' = O'c', and Zo = Zo', A AC is similar to A O'A'c'. Since sect. OABC -.sect. o'A'b'c' = r"^: R'^, and AOACiA OWC' = OA^ : O'A'^ = R^ : R'^, sect. OABC : A OAC = sect. oWb'c' : A O'A'c' ; (232'") .-. OABC— OAC: OAC = O'A'b'c' — 0W&: O'A'c'; (247) .-. seg. B : seg. B' =: A OAC : A O'a'c' = R^ : R'^. q.e.d. (244) (290) (404) (342) Exercise 578. In the diagram for Prop. XI., what must be the ratio of OA to O'A' if segment J5 is f of segment B'9 579. In the same diagram, if the segment is f of its circle, how many degrees are there in arc AC? 580. In this same diagram, if A were joined with the mid point D of arc AC, and Z CAD were found to be 38°, how many degrees in ZAOC? 581. In the same diagram, if a line ^F were drawn perpendicular to OA, and Z CAF were found to be 41°, how many degrees in arc AC? How many in the angle formed by lines drawn from A and C to any point in arc AC? 212 PLANE GEOMETRY. — BOOK VL DIVISION OF CIRCUMFERENCE. Proposition XII. Theorem. 406. A circumference can be divided into 2, 4, 8, 16, " equal arcs. Given : A circumference ACBD ; To Prove : ACBD can be divided into 2, 4, 8, 16, •••equal arcs. Find the center of ACBD. (173) 1°. Through draw a diameter AB. AB bisects the circumf. into the equal arcs ACB, ADB. (169) 2°. Through draw a diam. CD A. to AB. (95) Since the angles at are equal, being rt. A, (Const.) arc ^C = arc CB — arc BD — arc DA = \ACDB. (257) 3°. Bisect each of the zi at by diameters EF, GH. (81) The arcs ^^,J5:C',CG^, etc., each = 1 a quad. —\ACBD. (Const.) By successive bisections we can obtain, in the same way, the 16th, 32d, •••2"th part of ACBD, and hence can con- struct regular polygons of 4, 8, 16, ••• 2** sides. q.e.d. Exercise 582. In the diagram for Prop. XII., if A and E be joined with any point F in arc AE, how many degrees in the angle thus formed ? 583. How many degrees in the a,ngle formed by joining A and G with any point in arc ADBG, and what is its ratio to the angle found in the preceding exercise ? DIVISION OF CIRCUMFERENCE. 213 Proposition XIII. Theorem. 407. J. circumference can he divided into S, 6, 12, "• equal arcs. Given: A circumference ACBD ; To Prove: ACBD can be divided into 3, 6, 12, ••• equal arcs, Find the center (173), and draw any diameter AOB. 1°. From B draw a chord BC = BO (178), and join CO. Since CO = BO = BC, (Const.) Zb = Zboc. (68) But Zb is inscribed, and Z ^OC is at the center ; .-. arc ^C = 2 arc ^C; (268) i.e., SiTcAC = f arc BCA, a semicircumf. ; i.e., arc ^C = ^ACBD, the circumf. Since arc ^c = ^ arc ^C, arc 5C = I ACBD '} <'°^ 3°. By bisecting arc BC again, we obtain -^ACBD, thence 2V ACBD, and so on to any arc denoted by - — —ACBD. o X. Z Q.E.D. 408. Cor. The side of a regular inscribed hexagon is equal to the radius of the circle. For BC, which subtends ^ACBD, is equal to BO (Const.). We are thus enabled to construct any regular polygon of 3x2" sides. 214 PLANE GEOMETRY. — BOOK VI. Proposition XIV. Theorem. 409. A circumference can he divided into 5, 10 y 20, ■■• equal arcs. Given : A circumference A CED ; To Prove: AC ED can be divided into 5, 10, 20, ••• equal arcs. Find the center (173), and draw any radius OA. 1°. Divide OA in S, so that OA : OB = OB : BA. (308) From A draw a chord AC = OB (178), and draw CBD, to meet the circumference in D. Draw the diameter COE. Since OA: OB = OB: BA, > \ (Const.) and AC= OB, and OC = OA, ) we have (1) OA: AC = AC: BA, and (2) OC : AC = OB : BA. From (1), since Z ^ is com. to both A, A BAC is similar to A CA ; (290) .-. Zacbot ACD = Zaoc. (289) From (2), since the sides of A CAO are as the segments of the base OA, Z ACB = Z OCB ; (279) .-. arc ED = arc AD. (257) But arc AD — 2 arc AC, (268) (since Z ACD is inscribed and Z AOC is at the center;) .-. arc AD 4- arc DE -f- arc AC =^^ arc AC ; .'. arc AD — \ arc CDE = ^ ACED. 2°. Since arc AC=^\ arc AD, (1°) arc AC = A^ arc ACED, DIVISION OF CIRCUMFERENCE. 215 3°. By bisecting AC we obtain -^^ ACED, and so on. Thus we can divide the circumference ACED into 5, 10, 20, ••• 5 X 2" equal arcs. q.e.d. 410. CoR. Bij taking the difference of two arcs respectively equal to ^ and j\ of the circumfereyice, since i — yL = J^, we can find an arc that is J^ of the circumference ; and thence, by repeated bisections, arcs that are -f^, -^^, -" of the circum- ference. Till the beginning of this century, the division of the circumfer- ence by means of the straight line and circle could be effected only in the cases covered by the preceding propositions ; that is, the circumfer- ence could be thus divided into 1 x 2», 3x2", 5 x 2", 15 x 2", equal arcs. The celebrated geometer Gauss proved, however, that the cir- cumference can be divided into any number of equal arcs expressed by the formula 2" -}- 1 v^^hen this is a prime number. Thus, as we have seen, we can effect the division into 2° -f- 1 = 2, 2^ -|- 1 = 3, 22 4- 1 = 5, equal arcs, while 2^ 4- 1 = 9, not being a prime number, does not come under the rule. The division into 2* -f 1 = 17 equal arcs, not to speak of greater numbers, involves a process too intricate to come within the scope of an elementary work. 411. Stellate Polygons. If a circumference be divided into n equal parts, and, beginning at one of the points of division, we go round the circumference drawing chords subtending these arcs, m and m, m being prime to n and < ^ n, we obtain a polygon of starlike form called a stellate polygon. We say m is to be < ^ n, so as to confine our discussion to the minor of the arcs subtended by each chord, and con- sisting respectively of m and n — m of the equal parts. If m were a factor of n, say am = n, then we should return to the starting point after placing a chords, forming a regu- lar polygon of a sides. But, if m is prime to n, then as mn is the least number that divided by m gives n as quotient, we shall go round the circumference m times before return- ing to the starting point, and in doing so shall place n chords. 216 PLANE GEOMETRY. — BOOK VI. There is, accordingly, no stellate polygon of 3, 4, or 6 vertices, since each of these numbers has no prime to it less than its half. Since 2 is the only number prime to 5 and less than its half, there is but one stellate pentagon ; as ^. Since' 3 is the only number prime to 8 and less than its half, there is but one stellate octagon ; as B. For the like reason there is but one stellate decagon, and one stellate dodecagon. But since there are three numbers, 2, 4, and 7, that are prime to 15 and less than its half, there are three stellate pentadecagons, of which two, C and B, are given. In like manner, as there are seven numbers, 8, 7, 6, 5, 4, 3, 2, that are prime to 17 and less than its half, there are seven stellate polygons having 17 vertices. Among the many methods that have been devised for obtaining an approximate value of the important constant denoted by tt, that based upon the following proposition is perhaps the simplest, enabling us to find, in succession, the value of a side of a polygon of 2 n, 4 n, • • • sides. DIVISION OF CIRCUMFERENCE. 217 Proposition XV. Theorem. 412. The radius and the value of the chord of an arc being given, we can find the value of the chord of half that arc. Given : In circle ACBE, a radius OC perpendicular at D to AB, the chord of arc'^C5; To Prove: The value of the chord of one half arc ACB can be found. Since oc is ± to AB at D, (Hyp.) AD = l AB, and arc ^(7 = ^ arc ACB. (172) Join OA, AC. Then since AC < Si quadrant,* Zaoc< a rt. Z; . AC^ oOA^-\-Oc'^ — 2 0C'OD; (350) i.e., ac'^2r^-2R'OD. Since Z ADO is a rt. Z, (Hyp.) Olf ^ OA^ — Alf =o 7^2 _ 1 AJi^ ; (.347) .-. OD=^R''-\AB- = ^W^R^-AB'' (OD, R, AB, here denoting numerical measures;) .-. Jc''= 2 i?2 _ 2 7? .^^Ir^-Ab\' .'. ac=\2r-- ie V4 R^ - ab\ If, as is usually done, we take 7i' = 1, then AC =^2 -^4. ab': * At most AC = k of the circumference, since ACB =, at most, I of the circumference. 218 PLANE GEOMETRY. — BOOK VI. Pkoposition XVI. Problem. 413. To find an approximate value of it, the ratio of circumference to diameter. Formula : Sgu = ^2 — V4 — s„, where s„ denotes the vakie of a side of a regular inscribed polygon of n sides, R being taken = 1. If we take n = 6, then s„=l (408). Computation. Val. of sibe. Value of PERIM. = n.S V2 - V4 - 1 = .51763809. 6.21165708. s,, =V2-\/4-(.5176809)--2 =.26105238. 6.26525722. V2 - V4 - (.26105238/^ = .13080626. 6.27870041. s^ ^ V2 - \/4 - (. 13080626)2 = .06543817. 6.28206396. 8^^^ = V2 - V4 - (.06543817)=^ = .03272346. 6.28290510. V2 - V4 - (.03272346)2 = .01636228. 6.28311544. s,^ = V2 - V4 - (.01636228)2 = .00818126. 6.28316954. Taking this last value of the perimeter as an approxima- tion towards the value of the circumference whose radius is 1, since tt = — (396), we obtain tt = 1(6.28317) = 3.14159 Z R nearly, a result correct to the last decimal. By the useless expenditure of much time and labor, the value of TT has been calculated as far as 700 places of deci- mals. The first fifteen, more than sufficient for any useful purpose, are tt = 3.141592653589793. Ten decimals are suffi- ,cient to give the circumference of the earth to the fraction of an inch, and thirty decimals would give the circumference of the whole visible universe to a quantity imperceptible with the most powerful microscope. DIVISION OF CIRCUMFERENCE. Proposition XVIT. Problem. 219 414. The diameter being given, to find a line ap- proximately equal to the circumference. N C..-' M Given : D, the diameter of a circle ; Required: To find a straight line approximately equal to the circumference of that circle. Draw indefinite lines AM, AN, making any angle. Upon AM lay off ^^ = 113, and ^(7=355, units of length, =* and upon AN lay off AD =B. Join BD and draw CX II to BD. Since AB : AC = AD : AX = 113 : 355, (274) AX = ff-| AD = D X 3.141592 . • •. Q.E.F. The line thus obtained is greater than the circumference by about one four millionth of the diameter, as the student may easily prove for himself. 415. In the inscribed equilateral tri- angle ABC, draw AD Jl to BC, and join CD. Then AD bisects arc BDC (172), and DC =: R, the radius (408) ; .-. AC^ = da'-dc^ = 4:R'-r'- = 3 r'; .: AC = R^3. * Lengths respectively equal to 1.13 and 3.55 units can be taken from a diagonal scale ; or see Exercise 836. 220 PLANE GEOMETRY. — BOOK VI. 416. In the inscribed square ADBC, joining AB, CD, AC^z=Ad^+0C^ = 2R^ (347) ; .-. ac=bV2. 417. In the inscribed pentagon, the relation of the side to the radius is most readily obtained from the expres- ^ sion for the side of the inscribed deca- gon (see 420) . . From this we obtain AB = ^R VlO - 2 V5. 418. In the inscribed hexagon, the side DC—R (408), a result that can be obtained also by means of the formula referred to in the next article. 419. In the inscribed octagon (see diagram for Art. 416) the value of- a side is most readily found by means of the formula obtained in Prop. XV. ; that is AC = yJ2R^-E-^4:B'' "ab\ In this, putting AE for AC, and R^2, the value of a side of the inscribed square, for AB, we obtain AE = V2 i?2 _ 22 V4 R' - f¥; i.e., AE = V2 R' - R''^2 = R-^2- V2. 420. In the inscribed decagon AF •" E (see diagram for Art. 417), since R: BF=BF:R — BF, (409) BF — 7?2 R • BF; (322) .-. BF R{v5 — 1), solving as a quadratic. EXERCISES. 221 In the formula AC = yl2 r"" — R^^ R^ — Ab"^, putting -|-ie(V5 — 1), the value of a side of the inscribed deca- gon, for AC, and squaring, we obtain ^ R\-y'o -iy = 2R'- R^4. R"- - Ab\' whence 2^4: r'- AB^ = R{V5 + 1); whence AB = ^R(10 — 2 VS) . EXERCISES. QUESTIONS. 584. Into how many equal parts, up to 15, are we able to divide a circumference by means of the straight line and the circle ? 585. What is the ratio of the radius of the circle circumscribed about an equilateral triangle to that of the inscribed circle ? 586. By which of the constructions of this book can a right angle be divided into five equal parts ? 587. How many different stellate polygons can be formed if we have a circumference divided into 7 equal parts ? 9 equal parts ? 11 equal parts ? 13 equal parts ? 588. If two nonadjacent sides of a regular pentagon be produced to meet, what angle will they contain ? 589. What is the value of the interior angles of a regular octagon ? Of a regular decagon ? Of a regular pentadecagon ? What is the limit towards which each of the interior angles of a regular polygon tends as the number of sides increases indefinitely? What is the limit for each exterior angle ? 590. What is the circumference and area of a circle whose diameter is 10 inches, supposing tt = 3.1410? 591. What is the circumference and area of a circle whose radius is 2 ft. 6 in. ? 692. What is the circumference of a circle whose area is 100 sq. in. ? 222 PLANE GEOMETRY. — BOOK VL 593. What is the radius of a circle whose area is 6 sq. ft. ? 594. What is the radius of a circle whose circumference is 12 in, ? 595. What is the radius of a circle whose area is ii times that of a circle with radius B ? 596. What is the area of the ring between two concentric circles whose radii are 5 ft. and 7 ft. respectively ? 597. Within a circle whose radius is i?, a circle is drawn so as to cover I of the surface of the first circle. What is the radius of the second circle ? 598. The radii of two similar segments are as 4 to 7 ; if the first segment contains 25 sq. in. , what does the other contain ? 599. In a white circle of 3 in, radius, an inscribed square is painted black. How much white surface will remain ? 600. In a white square whose side is 4 in., an inscribed circle is painted red. How much white surface will remain ? 601. If the radius of a circle is 10 in., what is the side of the inscribed equilateral triangle ? 602. If the side of an inscribed equilateral triangle is 10 in,, what is the radius of the circle ? 603. If the radius of a circle is 8 in., what is the side of a regular inscribed pentagon ? 604. If the side of a regular inscribed pentagon is 9 in., what is the radius of the circle ? 605. If the radius of a circle is 6 in. , what is the side of a regular inscribed octagon ? 606. What must be the radius of a circle so that a side of a regular inscribed octagon shall be 10 in. ? 607. If the radius of a circle is 10 in., what is the side of a regular inscribed decagon ? 608. What must be the radius of a circle so that a side of a regular inscribed decagon shall be 3 in. ? THEOREMS. 609. An angle of a regular polygon of n sides is to an angle of a regular polygon of w 4- 2 sides, as n^ — 4 is to n^. 610. If the bisectors of all the angles of a polygon meet in a point, a circle can be inscribed in that polygon. EXERCISES. 223 611. An inscribed equilateral triangle is equivalent to half the regular hexagon of the same radius. 612. The altitude of an equilateral triangle is equal to the side of an equilateral triangle inscribed in a circle whose diameter is the base of the first. 613. The altitude of an equilateral triangle is to the radius of the circumscribing circle as 3 is to 2. 614. The area of the regular hexagon is a mean proportional be- tween the areas of the inscribed and circumscribed equilateral triangles. 615. The square of a side of an inscribed equilateral triangle is equivalent to three times the square of a regular hexagon inscribed in the same circle. 616. If the arcs subtended by two sides of an equilateral triangle be bisected, the chord joining those points will be trisected by those sides. 617. The diagonals drawn from a vertex of a regular pentagon to the opposite vertices, trisect that angle. 618. The diagonals drawn from a vertex of any regular polygon of n sides to the opposite vertices, divide the angle into n — 2 equal parts. 619. The diagonals joining alternate vertices of a regular pentagon form by their intercepts a regular pentagon. 620. If the alternate sides of a regular pentagon be produced to meet, the points of meeting will be the vertices of another regular pentagon. 621. The intersecting diagonals of a regular pentagon divide each other in extreme and mean ratio. 622. In a regular pentagon ABODE, diagonals AC, BE, are drawn, intersecting in F; show that FD is a parallelogram. 623. A ribbon is folded into a flat knot of five edges ; show that these edges form a regular pentagon. 624. If P, H^ and D denote respectively a side of a regular in- scribed pentagon, hexagon, and decagon ; then P'^=o=H'^ 4- D^. 625. If from any point within a regular polygon of n sides, perpen- diculars be drawn to the sides, the sum of these perpendiculars will be equal to n times the apothem. 224 PLANE GEOMETRY. — BOOK VI. 626. The radius of an inscribed regular polygon is a mean propor- tional between its apothem and the radius of the similar circumscribed polygon. 627. The area of a circular ring, i.e., the space between two con- centric circumferences, is equal to that of a circle having for diame- ter a chord of the outer circle tangent to the inner circle. 628. If, on the hypotenuse and the arms of a right triangle as diame- ters, semicircles be described, the curvilinear figures bounded by the greater and the two lesser semicircumferences will be equivalent to the triangle. PROBLEMS. Inscribe in a given circle : 629. An equilateral triangle. 631. A square. 630. A regular pentagon. 632. A regular octagon. Circumscribe about a given circle : 633. An equilateral triangle. 635. A regular pentagon. 634. A regular hexagon. 636. ,A regular decagon. 637. Describe a square about a given rectangle. 638. Inscribe an equilateral triangle in a given square, so as to have a vertex of the triangle at a vertex of the square. 639. Construct an equilateral triangle that shall be double the area of a given equilateral triangle. • 640. Construct a square that shall be | of a given square. 641. Construct a regular pentagon that shall be | of a given regular pentagon. 642. Construct a regular hexagon that shall be | of a given regular hexagon. 643. Describe a circle equivalent to | of a given circle. PART 11. SOLID GEOMETEY. Book Vll. PLANES AND POLYHEDRAL ANGLES. AB^ ; .: AP is ± to PB ; .'. AP is ± to itfiV, (Ax. 2, Ax. 7) (352) Q.E.D. (427) (being ± to any line in MN through P) PLANES AND PERPENDICULARS. 229 Proposition IV. Theorem. 429. A perpendicular to a plane can he drawn from any given point without or in that plane. 1°. Given: A point A without a plane MN ; To Prove : A perpendicular to MN can be drawn from A. In the plane passed through A and any line BC in MN, draw AD ± to BC. From D draw, in the plane MN, DE ± to BC ; and, in the plane of AD, DE, draw AP _L to DE. Then ^P is ± to MN. For since ADC, PDC, and APD are rt. A, (Const.) AC AC" o^AD" + DC^,\ and PC" =0= PD^ 4- -OC^' ) (347) - PC" =0= ^Z)' — P/)" =o ^P"; (Ax. 3, 347) .-. ^C''=o= JP" + PC'^• (Ax. 2) .^ AP is _L to PC; (352) .-. AP is ± to MN, Q.E.D. (428) (being ± to both PD and PC.) 2°. Given: A point P in plane ifiv; To Prove : A perpendicular to MN can be drawn at P. From P draw PD _L to PC, any line in MN ; and in any plane intersecting MN in BC, draw Z>.4 _L to BC. Then in 230 SOLID GEOMETRY. — BOOK VIL the plane of PD, DA, draw PA _L to PD. PA is ± to MN, as may be proved in the same way as in 1°. 430. CoR. 1. There can be drawn hut ¥- — one perpendicular to a given playie from / a point without the plane. /_ Z iV For if there could be two perpen- diculars, AP, AQ, from A to MN, draw PQ. Then, in the plane APQ, there would be two perpendiculars from A to the same line PQ, which is impossi- ble (51). M 431. CoR. 2. At a given point in a / plane there can he draivn hut one per- / ^ pendicidar to that plane. N For if there could be two perpendiculars, PA, PB, from P in MN, suppose a plane to pass through PA, PB, inter- secting MN in PQ. Then from the same point in PQ, and in the same plane, there would be two perpendiculars to PQ, which is impossible (41). 432. Cor. 3. The perpendicidar is the shortest line that can he draimi to a plane from a given p)oint ivithout the plane. For AP is shorter than any oblique line, A Q, drawn from A to PQ, in the same plane APQ (93). 433. Definition. The distance of a point from a plane is the length of the perpendicular from the point to the plane. Exercise 648. In the diagram for Prop. III., the planes determined by AB and P, AC and P, AD and P, intersect in what line ? 649. In those planes, if circumferences be described on AB, AC, and AD, as diameters, they will pass through what point and have what common chord ? 650. In the diagram for Prop. V., if AB and AD be joined, and PD = AB, show that {AD + AP) (AD- AP) = 35^ PLANES AND PERPENDICULARS. 231 Proposition V. Theorem. 434. All perpendiculars drawn to a given straight line at a given point, lie in a plane perpendicular to the line. Given: PB, PC, PD, any perpendiculars to ^P at P; To Prove : PB, PC, PD, all lie in a plane that is ± to ^P at P. Through PB, PC, pass the plane 3IN. Since AP is ± to PB and PC, (Hyp.) AP is ± to 3IN, and 3fN is ± to AP. (428, 427) Through ^P, PD, suppose aj^lane passed cutting 3/iV^ in PD'. Since ^P is ± to 3IM, AP is ± to PIJ', a line in 3fN. (427) But AP is ± to PB ; (Hyp.) .♦. PD must coincide with PD' and lie in 3IN, q.e.d. (since, in plane AP, PD, there can be but one _L to AP at P.) (41) 435. Cor. 1. If an indefinite line PB he made to revolve so as to remain always perpendicular to AP at P, it ivill generate the plane MN at right angles to AP. 436. Cor. 2. Through a given point in or without a straight line can be passed but one plane perpendicular to the line. 232 SOLID GEOMETRY. — BOOK VII. Proposition VI. Theorem. 437. Obliques drawn from a point so as to meet a plane at equal distances from the perpendicular, are equal ; of obliques meeti?i£ the plane at unequal dis- tances from the perpendicular, the more remote is the greater. Given: AB, AC, AD, obliques drawn from A to MN, so that B and C are equally distant from the perpendicular AP, but PB is greater than PB ; To Prove : AB is equal to AC, but AB is less than AD. On PD lay off PE = PB, and join AE. Since PB = PC = PE, AP is common, ) , , / X c (Hyp. and Const.) and the z§ at P are rt. A, ) AB = AC = AE. (66,70) But PE < PD ; (Const. ) .■.AE ./ . /■ Given: Planes MN and PQ, each perpendicular to AB ; To Prove: Plane MN is parallel to plane PQ. For if MN could meet P Q, then through any point com- mon to both would be passed two planes perpendicular to the same straight line AB. But this is impossible (436). Hence MN cannot meet PQ. q.e.d. Proposition XI. Theorem. 451. The intersections of two parallel planes with a third plane are parallel. A/ Given: Two parallel planes, MN, PQ, intersecting plane BB in AB, DC; To Prove : AB h parallel to DC. Since plane MN cannot meet plane PQ, (Hyp.) AB cannot meet DC, though in the same plane BD ; .'. AB is II to DC. Q.E.D. (102) PLANES AND PARALLELS. 237 452. Cor. Parallel lines that are intercepted between paral- lel planes are equal. For since the plane of the parallels AD, BC, intersects the parallel planes 3IN, FQ, in parallel lines AB, DC, the figure ^C is a parallelogram (131); whence AD = BC (136), Proposition XII. Theorem. 453. A straight line perpendicular to one of two parallel planes is perpendicular to the other also. M' 7f B '■■•■ 7 Given: Two parallel planes, MN, PQ, and a straight line AB perpendicular to MN ; To Prove: AB is perpendicular to PQ. Through AB pass any plane ad, intersecting MN in AC, and PQ in BD. Since plane MN is II to plane PQ, (Hyp.) ACis II to bd; (451) .'. AB is ± to AC Sind BD; (Hyp., 107) .-. AB is ± to plane PQ, q.e.d. (427) (being ± to any line through B in PQ.) ' 454. CoR. Tliroxigh a giuen point, A, one plane can be passed parallel to a given plane, PQ, and only one. For from A, a perpendicular AB can be drawn to PQ (429) ; and through A, a plane can be passed ± to AB, and hence II to plane PQ (450). Moreover, since from^ but one perpendicular can be drawn to PQ (430), there can be but one plane passed through A II to PQ. 238 SOLID GEOMETRY.— BOOK VI I. Proposition XIII. Theorem. 455. If two angles not in the same plane have their sides respectively parallel and drawn in the same direction, they are equal. Given : Two angles, BAC, b'a'c', lying in planes MN and PQ, respectively, so that BA and b'A', ca and &A', are respectively parallel and drawn in the same direction ; To Prove : Angle BAC is equal to angle b'a'c'. In AB, AC, take any points B and C, and lay oE A'b'=AB, A'C'=AC; join AA', BB', CC'. Since AB, AC, are resp. II and = to A'b', a'c', (Hyp. and Const.) AB' and AC' are parallelograms ; (142) BB', CC', are each II and = to AA' ; (^36) .-. BB' is II and = to CC' ; (445, Ax. 1) .-. BC is II and = to b'c' ; .: ABAC = A B'A'C'; (69) .-. Z A = Za'. q.e.d. (70) 456. Cor. 1. If two angles lying in different planes have their sides respectively parallel, their planes are parallel. For the intersecting lines that determine the one plane, being parallel to the intersecting lines that determine the other, the planes are parallel. 457. Cor. 2. Iftivo parallel planes, MN and PQ, are inter- sected by two other planes, AB', AC', the angles A, A', formed by their intersections, are equal. PLANES AND PARALLELS. 239 458. Cor. 3. If three lines, A A', BB', CC', not in one plane, are equal and parallel, the triangles ABC, A'b'c', formed by joining their extremities, are equal, and their planes are parallel. Proposition XIV. Theorem. 459. If two straight lines are cut hy three parallel planes, the intercepts are proportional. M^ eI )::-3-^^ tN eI± Given: A line AB meeting parallel planes MN, PQ, RS, in A, E, B, respectively ; and a line CD meeting the same planes in C, G, D, respectively ; To Prove : AE : EB z= CG : GD. Draw AD, meeting PQ in 7^; join AC, EF, FG, and BD. Since planes PQ, RS, are II, and plane ABD cuts them, EF is 11 to BD ; (451) .'. AEiEB = AF'.FD. (274) Since planes PQ, MN, are II, and plane DAC cuts them, FG is II to ^C; (451) .-. CG : GD = AF : FD ; (274) .-. AE'.EB = CG : GD. (232'") 460. Cor. If n straight lines are cut by m parallel planes, the intercepts are jjroportional. 240 SOLID GEOMETRY. — BOOK VIL DIHEDRAL ANGLES. 461. A dihedral angle is the opening between two planes that meet. The line in which the planes meet is called the edge of the angle, and the two planes are called its faces. Thus the faces A AC, BD, meeting in the edge AB, con- X \ tain the dihedral angle DABC. / A To designate a dihedral angle, four ^ V y \ \ letters are generally necessary, two at VX /_ \ the edge and one on each face, the two \ / \ at the edge being placed between the b\/— ^ other two. If the edge belongs to only one angle, the letters at the edge will suffice to designate the angle. Thus the dihedral angle DABC may be referred to as dihedral angle AB, or simply as the dihedral AB. 462. The plane angle of a dihedral angle is the angle con- tained by the two perpendiculars drawn, one in each face, to any point in the edge. Thus hac is the plane angle of the dihedral DABC. It is evident that the plane angle is the same at whatever point of the edge it is constructed (455). A dihedral angle may be conceived as gerierated by a plane BD turning from coincidence with plane AC about the edge AB as axis, till it reaches the position where its plane angle is Z bac; which, again, may be conceived as generated by the revolution of the line ab from an initial position, ac. 463. Two dihedral angles are equal when they can be placed so that their faces coincide. 464. A right dihedral angle has its plane angle a right angle, and its faces are said to be perpendicular to each other. In the same way, dihedral angles are acute or obtuse, and pairs of dihedral angles are adjacent, complementary, supplemen- tary, alternate, corresponding, vertical, etc., according as their plane angles are acute, etc. .DIHEDRAL ANGLES. 241 Proposition XV. Theorem. 465. Two dihedral angles are equal if their plane angles are equal. Given: Two dihedral angles, CABD, C'a^b^d\ having equal plane angles, CAD, cWd' ; To Prove: Dihedral angle c^ 5 Z) = dihedral angle c'a'b'd\ Apply c'a'b'ij' to CABD so that Z c'A'd' =^Z CAD. Then the planes of these angles will coincide (423), and A'b', AB, will coincide, both being perpendicular to the same plane at the same point (431) ; hence the planes B'c' and BC, b'd' and BD, will coincide (423) ; .-. dihed. Z CABD = dihed. Z c'a'b'd'. q.e.d. (463) Proposition XVI. Theorem. 466. Dihedral angles are to each other as their plane angles. D A D Given : Two dihedral angles, CABD, c'a'b'd', and their respec- tive plane angles, CAD, c'a'd' ; To Prove: Dihedral angle CABD : dihedral angle cU'i?'Z)'= angle CAD : angle C'a'd'. Geom. — 16 242 SOLID GEOMETRY. — BOOK VIL 1°. When A CAD and C'A'd' are commensurable. Let Z CAE be a common measure of these angles, so that Z CAE is contained 5 times in CAD and 4 times in c'a'd'. Draw lines AE, A'e', etc., dividing A CAD, C'A'd', into 5 and 4 equal parts respectively, and through these lines and the edges AB, a'b', pass planes so as to divide the dihedral A CABD, c'a'b'd', into 5 and 4 equal dihedral A respectively. Since Z CAD : Z C'A'd' = 5: 4, (Hyp.) and dihed. Z CABD : dihed. Z c'a'b'd' = 5:4, (Const.) dihed. Z CABD : dihed. Z a'^'i?'// = Z aiT) : C'.l'i)'. (232'") 2°. When Z C^D and C'a'd' are incommensurable, we can prove by the method of limits, as in (260"), that always dihed. Z CABD : dihed. Z C'a'b'd' = Z CAD : Z cU'i)'. q.e.d. 467. Cor. Vertical dihedral angles are equal. For they have the ratio of their equal vertical plane angles. 468. Scholium. In like manner may be established the following properties of dihedral angles by means of the cor- responding properties of plane angles. (1) All right dihedral angles are equal. (2) Ti(jo adjacent dihedral angles, formed by one plane meet- ing another, are supplementary ; i.e., are equal to ttvo right dihedral angles. (3) Of the dihedral angles formed by a plane intersecting parallel planes, the alternate and corresponding angles are equal, and the interior angles on the same side of the trans- verse plane are supplementary. (4) Dihedral angles having their faces mutually parallel, or, if their edges are parallel, respectively perpendicular to each other, are either equal or supplementary. DIHEDRAL ANGLES, 243 Proposition XVII. Theorem. 469. If a straight line is perpendicular to a given plane, every plane passed through that line is per- pendicular to the given plane. Given : AP perpendicular to plane MN, and J5D, a plane pass- ing through AP ; To Prove : Plane BD i^ perpendicular to plane MN. Draw PE ± to 5(7, the intersection of BD and MN. Since AP is _L to plane MN, (Hyp.) ^P is ± to 5C and PE ; (427) .-. APE, a right angle, is the plane angle that measures the dihedral angle formed by the planes intersecting in BC ; (462) .*. plane BD is ± to plane MN. (464) 470. Cor. A plane perpendicular to the edge of a dihedral angle is perpendicular to each of its faces. Exercise C51. In the diagram for Prop. XIII., if from any point D in A ABC, DD' be drawn parallel to AA> to meet A A' B'C in D\ and D be joined with A and C, D' with A' and O', the triangles DAC and D'A'C are equal. 652. If planes are passed through the sides of a triangle perpen- dicular to its plane, of the inner dihedral angles formed, no two are equal, two are equal, or all are equal, according as the triangle is scalene, isosceles, or equilateral. 244 SOLID GEOMETRY. — BOOK VIL Proposition XVIII. Theorem. 471. If two planes are perpendicular to each other, any line in the one plane perpendicular to their intersection is perpendicular to the other. Given : In plane MN, perpendicular to plane BD, EP perpen- dicular to BC\ the intersection of MN and BD ; To Prove : EP is perpendicular to plane BD. From P draw, in plane BD, PA J_ to 5C. Since EP, PA, are each X to ^C7 at P, (Hyp. and Const.) APE is the plane angle that measures the dihedral angle contained by planes BD, MN, intersecting in BC. (462) But plane BD is ± to plane MN; (Hyp.) .-. APE is a rt. Z ; .-. EP is ± to plane BD, q.e.d. (428) (since it is J_ to BC and to PA in that plane.) 472. Cor. 1. If two planes, BD, MN, are perpendicular to each other, a straight line PE drawn at any point P of their intersection, so as to he perpendicular to one of the planes, as BD, ivill lie in the other plane MN. For in the plane of AP and PE, only one perpendicular can be drawn to AP at P (431). 473. Cor. 2. If two planes are perpendicular to each other, a perpendicular from any point of the one plane to the other must lie in the first plane. DIHEDRAL ANGLES. 245 Proposition XIX. Theorem. 474. Through any straight line not perperidicular to a plane^ one plane, and only one, can be passed perpendicular to that plane. In Given: AB, any straight line not perpendicular to plane MN ; To Prove : Only one plane perpendicular to MN can be passed through AB. Through A draw ^P ± to plane MN, (429) and pass through AB and AP a plane, BP. Since AP is ± to plane MN, (Const.) plane BP is _L to plane MN. (469) But any plane passing through AB so as to be perpendic- ular to MN, must contain the perpendicular AP ; (472) .-. there can be but one such plane, q.e.d. (423) Exercise 653. In the diagram for Prop. XIV., show that FG : FE = ACDF:BDAF. 654. Ill the diagram for Prop. XVII., if in plane DB, GG' be drawn parallel to C'jB, and through GO' planes be passed intersecting plane MNin HH' and KI^, show that HW is parallel to KK'. 655. Show that if HH' and KK' are on opposite sides of CB, the dihedral angles formed by the plane GH' and GK' with J/JV, are or are not equal according as CB is or is not equidistant from HH' and KK'. 656. Show that, if BC is equidistant from HH' and /i/i', plane DB bisects the dihedral angle HGG'K'. 657. If HH' and KK' are on the same side of CB, show that the dihedral angle formed by GH' with MN is the difference of that formed by GK' with MN and that formed by GH' with GK'. 246 SOLID GEOMETRY. — BOOK VII. Propositiox XX. Theorem. 475. If two intersecting planes are each perpendic- ular to a third plane, their intersection is also per- pendicular to the third plane. Given: AB, the intersection of two planes, PQ, RS, each per- pendicular to plane MN ; To Prove : AB [b perpendicular to plane MN. At B erect a perpendicular 7? .4 to plane MN. (429) Then BA lies in eacli of the planes PQ, BS ; (472) .*. BA must coincide with the intersection of PQ, RS ; .'. AB is ± to 3fN. Q.E.D. 476. CoR. 1. A plane perpendicular to each of ttvo inter- secting planes is perpendicular to their intersection. 477. Cor. 2. If a plane is perpendicular to two planes per- pendicular to each other, the intersection of any tivo of these planes is perpendicular to the third plane, and each intersection is perpendicular to the other two. Exercise 658. Prove Cor. 2 of Prop. XX. by means of the diagram for Prop. XX., assuming that plane PQ is perpendicular to plane RS. 659. In the same diagram, if Z. SBQ were 140°; what would be the ratio of dihedral angle SBAQ to dihedral angle SBAP ? 660. In the diagram for Prop. XXI., show that the points P, E, O, P, are coney clic. DIHEDRAL ANGLES. 247 Proposition XXI. Theorem. 478. Every point in the plane that bisects a dihedral angle is equidistant from the faces of that angle. Given : P, a point in plane MA, bisecting dihedral angle CABD ; To Prove: P is equidistant from ^C and -Bi>. From P draw PE A_to AC, and PF ± to BD ; (429) through PE and PF pass a plane intersecting AC in OE, BD in OF, and therefore AM in OP. Since PE is ± to ^C, and PF to BD, (Const.) plane P^T^' is ± to ^^; (476) .-. POE and POF are the plane angles that measure the equal dihedral Am ABC, MABD ; (462) .-. Z POE = Z. POF ; .'. rt. A PEO = rt. A PFO ; (73) .-. PE = PF. Q.E.D. 479. Definition. The projection of a x>oint on a plane is the foot of the perpendicular from the point to the plane. 480. Definition. The projection of a line on a plane is the locus of the pro- jections on that plane of all the points * in the line. 248 SOLID GEOMETRY. — BOOK VII . Proposition XXII. Theorem. 481. The projection of a straight line on a plane is a straight line. Given : A straight line AB and a plane MN ; To Prove : The projection oi AB upon MN is a straight line, Through AB pass a plane AD ± to plane MN, which it intersects in CD. Since AD is ± to MN, (Const.) AD contains all the perpendiculars from points in AB upon MN ; (473) .-. the feet of all these perpendiculars must meet MN in CD. But CD is a straight line ; (426) .*. the projection of AB on MN is a straight line, q.e.d. Exercise 661. In the diagram for Prop. XXL, how many degrees must there be in the plane angle of dihedral angle CABD, that FO may be equal to FP f 662. In the same diagram, how many degrees are there in that j)lane angle if FF is equal to the line joining FE ? 663. Prove that if a line is equal to its projection on a given plane, it is parallel to the plane. 664. Prove that parallel lines have their projections on the same plane in lines that are coincident or parallel. 6G5. In the diagram for Prop. XXII., show that if Bxi be produced to meet MN in F, E is a, point in DC produced. 666. In the diagram for the preceding exercise, if BD : AC = m:n, what is the ratio of CE to CD f DIHEDRAL ANGLES. 249 Proposition XXIII. Theorem. 482. The acute angle formed by a straight line with its own projection on a plane^ is the least angle it makes with any line in that plane. Given: Angle ABC, formed hj AB with its projection SC on plane MN ; To Prove : Angle ABC is less than angle ABD, formed by AB with any other line in MN than BC. Lay off BD = BC, and join AD. In A ABC, ABD, AB = AB, BC = BD, (Const.) hut AC < AD, (432) (since ^C is _L to plane i/^;) (Hyp.) .'. Z ABC , two straight lines not in the same plane ; To Prove : A common perpendicular can be drawn to AB and CD. Through CB pass a plane MN that is II to AB, (454) and let ah be the projection of AB upon MN. Since ab is II to AB, (449) ab is not II to CD, (since AB and CD cannot be parallel ;) (Hyp.) .-. ab will meet CD, say in b. At b draw bB _L to ab in the projecting plane of AB. Since AB is II to ab in the same plane, bB is ± to AB ; (107) and since bB is ± to MN, (471) bB is ± to CD also. q.e.d. (427) 485. Cor. 1. Only one perpendicular can be drawn com- mon to two straight lines not in the same plane. For if there could be two such perpendiculars, then, as can easily be shown, there could be two perpendiculars drawn in the same plane at the same point in a straight line, which is impossible (41). 486. Cor. 2. The common perpendicular is the shortest distance betiveen two straight lines not in the same plane. POLYHEDRAL ANGLES. 251 POLYHEDRAL ANGLES. 487. A polyhedral or solid angle is the angle formed by three or more planes meeting in a common point. The point in which the planes meet is called the vertex; the intersections of the planes, ^ the edges; and the portions of the planes //W bounded by the edges, the faces of the angle. / / \\ Thus, in the polyhedral angle S-ABCDE, S V^ ■//:\\ is the vertex; SA, SB, etc., are the edges; / \,/ \.A and ASB, BSC, etc., are faces or face angles. j \ It is to be noted that, in a polyhedral angle, every two adjacent edges form a face angle ; and every two adjacent faces, a dihedral angle. It is also to be noted that the faces and edges of a polyhedral angle may be supposed to extend indefinitely. As a convenience in demonstration, however, portions of the faces and edges may be repre- sented as cut off by a plane. The section formed ,by the intersection of the plane with the faces is a polygon, sometimes called the base of the polyhedral angle. 488. A polyhedral angle is convex if any section made by a plane cutting all its faces is a convex polygon; as ABODE. It is to be understood that the polyhedral angles about to be treated of are convex. 489. A polyhedral angle is trihedral, tetrahedral, etc., according as it has three, four, etc., faces. 490. A trihedral angle is rectangular, hirectangular, or trirectangular, according as it has one, two, or three right dihedral angles. The ceiling and walls of a room form trirectangular angles. Exercise 070. In the diagram for Prop. XX., how many trihedral angles are represented, with what common vertex ? 071. In the same diagram, if SBQ is a right angle, of what class, according to Art. 490, is each of the four trihedral angles ? 252 SOLID GEOMETRY.— BOOK VII. Proposition XXV. Theorem. 491. In a trilwdral angle^ the sum of any two of the face angles is greater than the third face angle. CHven: In trihedral angle s-Abc, face angle CSA greater than angle ASB or angle BSC; To Prove : Angle ASB + angle BSC is greater than angle CSA. In face CSA make Z ASB = Z ASB ; (203) through any point D of SB draw ABC in plane CSA; take SB = SB, and join AB, BC. Since SB = SB, SA = SA, and Z ASB = Z ASB, (Const.) A ASB — A ASB, and AB = AB. (66, 70) lliAABC,AB-{-BC>AC; (88) .'. BC>{AC — AB) ov BC. (Ax. 5) In A BSC, BSC, since SB = SB, sc = sc, but BC > BC, Zbsc> Z BSC; (91) .-. Z ASB + Z BSC > Z ^5fz> 4- Z D^ic; (Ax. 4) i.e., Z ASB + ZbsoZ CSA. Q.E.D. Exercise 672. A plane can be passed perpendicular to only one edge or to two faces of a polyhedral angle. 673. If three lines in space are parallel, or meet in a common point, how many planes may they define, taken two and two ? POLYHEDRAL ANGLES. 253 Proposition XXVI. Theorem. 492. The sum of the face angles of any convex poly- hedral angle is less than four right angles. s Given: ASB, BSC, CSD, etc., face angles of a polyhedral angle S-ABCDE ; To Prove : The sum of the angles ASB, Bsa, CSD, etc., is less than four right angles. Pass a plane so as to cut the edges in A, B, C, I), E, and the faces in AB, BC, CD, DE, EA ; then ABODE is a convex polygon. (Hyp.) Taking any point in ABODE, join OA, OB, 00, OD, OE, As the number of triangles having their vertices at S is equal to the number of triangles having their vertices at O, (they having the same bases AB, BO, etc.,) the sum of the interior A of the one set is equal to that of the other set. But in the trihedral A formed at A, B, O, etc., Z SB A 4- A SBO > A ABO, and A SOB + A sod > A bod, etc. ; } (491) .-. the sum of the A at the bases of the triangles whose vertices are at S, is greater than that of the triangles whose vertices are at ; '. the sum of the Ant s < the sum of the A at o. (Ax. 5.) But the sum of the AdA, O = four rt. A^ ; ,% the sum of the A2X s <, lour rt. A q.e.d. 254 SOLID GEOMETRY. — BOOK VII. Proposition XXVII. Theorem. 493. // the edges of a trihedral angle he produced through the vertex, they will form the edges of a second trihedral angle, called the symmetrical trihedral of the first, with its face and dihedral angles respec- tively equal to those of the first, hut arranged in reverse order. Given: Trihedral angle s-A'b*&, formed by producing the edges of trihedral angle S-ABC ; To Prove: The face angles and dihedral angles of S-A'b'c' are respectively equal to those of S-ABC, but in reverse order. 1°. The face A A' SB', ASB, etc., are respectively equal, (50) (being vertical A in planes determined by AA', bb\ etc.) 2°. The dihedral A SA', SA, SB', SB, etc., are respectively equal, (467) (being vertical dihedral angles.) 3°. The angles of both kinds are in reverse order. For to meet the faces of the first trihedral in the order ASB, BSC, OS A, we must turn from right to left,* while to meet the faces of the second trihedral in the order A' SB', B'SC', C'SA', we must turn from left to right. * In order to see this, look at each trihedral having the vertex a.bove^ POLYHEDRAL ANGLES. 255 The matter, which is of some importance, may be made clearer by the following illustration. Let x\BC, A'B'C, abc, be three triangles having their sides and angles respectively equal ; ABC, A'B'C, hav- ing theirs in the same order, but abc in reverse order. It is obvious that ABC can be made to slide over and coincide with A'B'C without reversing, while abc must be taken up and reversed in order to apply it to A'B'C, either from above or below, so as to make them coincide. S S' s If, now, we conceive planes to be passed through the sides of the tri- angles and through points S, S', s, similarly situated with respect to their vertices, it is evident that the trihedral angles thus formed will have their face angles and dihedral angles respectively equal, but those of s-abc in reverse order from those of S'-A'B'C and 8-ABC. Hence, if A ABC be made to coincide with A A'B'C, 8-ABC will coincide with S'-A'B'C ; since S will then coincide with S'. But if abc be made to coincide with A'B'C, s will be on that side of A'B'C which is remote from S', as in the figure. 494. Two polyhedral angles are equal and can be made to coincide if their face angles and dihedral angles are respec- tively equal and arranged in the same order ; if these parts are equal but not arranged in the same order, the poly- hedral s are said to be symmetrical. Exercise 074. If four lines in space are parallel, or meet in a common point, how many planes may they define, taken two and two ? 675. A line parallel to two intersecting planes is parallel to their intersection. 676. If two unequal similar triangles not in the same plane have their sides respectively parallel, the lines joining their homologous vertices will, if produced, meet in one point. 256 SOLID GEOMEriiY. — BOOK VII. Pkoposition XXVIII. Theorem. , 495. Trihedral angles that have their face angles respectively equal are equal or syimnetrical. Given: Two trihedral triangles, S-ABC, S^-A^b'c', having ASB equal to A's'b', bsc equal to B's'c', CSA equal to C's'j'; To Prove : S-ABC and s'-a'b'c' are equal or symmetrical. On the six edges lay off SA = s'a', SB = s'b', SC = s'c\ and join AB, AC, BC, a'b', a'c', b'c'. Since A SAB, SBC, SCA = As'a'b', s'b'c', s'c'a', re- spectively, (66) AB, AC, BC = A'b', A'c', b'c', respectively ; .-. A ABC = A A'b'c' Siiid Z B AC =Zb'a'c'. (69) At any point D in SA draw 1)E, DF, each _L to SA, in the faces ASB, ASC respectively. These lines will meet AB and. AC respectively, (114) (since the A SAB, SAC, are acute, being base A of isos. A.) Let them meet AB, AC, in E and F, respectively, and join FF. On s'a' lay off a'd' = AI), and construct d'e'f' as DEF was constructed. Then since AD = A'b' (Const.) and Z DAE = Z d'a'e', rt. AADE = rt. Aa'd'e'; (63) .-. AE = A'E' and BE = D'E'. POLYHEDRAL ANGLES. 257 In like manner it may be shown that AF = .4'^' and DF = D^F\ Since AE = A'e\ AF=A'f'. and Z BAC=Z b'a'c', (Above) A AEF = AA 'e'f', and EF = E'f'. (6(j) Since DE = J)'e', T)F = d'f', and EF = e'f', A DEF = A d'e'f', and Z i^Z»^ = Z F'd'e' ; (69) .-. dihedral Z ^S = dihedral Z ^'s', (465) (being measured by equal A FDE, f'd'e'.) In the same way it may be proved that dihed. Zbs = dihed. Z B's', and dihed. Z CS = dihed. Z C's'. Hence the trihedrals S-ABC, s'-A'b'c', are equal or sym- metrical according as the equal parts are or are not arranged in the same order. q.e.d. (494) 496. CoR. 1. If two trihedral angles have the three face angles of the one respectively equal to the three face angles of the other, the dihedral angles of the one are respectively equal to the dihedral angles of the other. 497. Cor. 2. An isosceles trihedral angle, that is, one having two of its face angles equal, is equal to its symmetrical trihedral. For if in S-ABC, we have Z ASB = Z BSC, then in s'-a'b'c' we shall have Z a's'b' = Z B's'c' ; and the A CSA, C's'a' will have equal faces on each side of them ; then also the dihedral angles will be similarly arranged, ami the trihedrals will be equal (495). Exercise 677. If a line makes an acute angle with a plane, every plane with which it makes the same angle is parallel to the first. 678. Parallel lines that intersect the same plane make equal angles with it. 679. If the projections of any number of points upon a plane lie in one straight line, these points are in one plane. What plane ? Geom. — 17 258 SOLID GEOMETRY. — BOOK VII. EXERCISES. QUESTIONS. 680. What space concepts are determined in position by one point, by two, by three, respectively ? 681. To what theorem in Book I. does Prop. V. correspond ? 682. What is the locus of all the points in space that are equidis- tant from a given circumference ? 683. What is the locus of all the points in space that are equidis- tant from the vertices of a triangle ? 684. What is the locus of all the points in space that are equidis- tant from the mid points of the sides of a triangle ? 685. To what theorem in Book I. does Cor. 3 of Prop. VI. corre- spond ? 686. What is the locus of all the lines that are perpendicular to a given line at a given point ? 687. To what theorems in Book I. do Prop. VIII. and its first corollary correspond ? 688. What is the locus of all the lines that have a given line in a given plane as their projection ? 689. What is the locus of all the points in space that are at a given distance from a given plane ? . 690. To what theorems in Book I. do Prop. XII., Cor., and Prop. XIII. correspond ? 691. To what theorem in Book IV. does Prop. XIV. correspond ? 692. To what theorem in Book I. does Prop. XIX. correspond ? 693. What is the locus of all the points in space that are equidis- tant from two intersecting planes ? From two parallel planes ? THEOREMS. 694. If a line is perpendicular to one of two intersecting planes, its projection on the other plane is perpendicular to the intersection. 695. A plane parallel to two sides of a quadrilateral in space, — that is, a quadrilateral having its sides two and two in different planes, — divides the other two sides proportionally. EXERCISES. 259 696. The mid points of the sides of a quadrilateral in space are the angular points of a parallelogram. 697. If the intersections of any number of planes are parallel, the perpendiculars drawn to these planes from the same point in space are in the same plane. 698. If a line is equally inclined to both faces of a dihedral angle, the points in which it meets the faces are equally distant from the edge of the dihedral. 699. If a line makes equal angles with three lines in the same plane, it is perpendicular to that plane. 700. If a plane be passed through one diagonal of a parallelogram, the perpendiculars to that plane from the extremities of the other diagonal are equal. 701. If from a point within a dihedral angle, perpendiculars are drawn to its faces, the angle contained by these perpendiculars is equal to the plane angle of the adjacent dihedral angle formed by pro- ducing one of the planes. 702. If three planes have a common intersection, perpendiculars to these planes from any point in the intersection are in the same plane. 703. In any trihedral angle, the three planes bisecting its dihedral angles intersect in the same line. 704. In any trihedral angle, the three planes passed through its edges perpendicular to the opposite faces, intersect in the same line. 705. In any trihedral angle, the three planes passed through the edges and the bisectors of the opposite face angles, intersect in the same line. 706. In any trihedral angle, the three planes passed perpendicularly through the bisectors of the face angles, intersect in the same line. LOCI. Find the loci of the points in space that respectively satisfy the following conditions : 707. Are equidistant from two given points. 708. Are equidistant from two given intersecting lines. 709. Have their distances from two given planes in a given ratio. 260 SOLID GEOMETRY. — BOOK VI I. 710. Are equidistant from the vertices of a given triangle. 711. Are equidistant from the sides of a given triangle. 712. Are equidistant from the vertices of a quadrilateral whose opposite angles are supplementary. 713. Are equidistant from the circumference of a given circle. 714. Are equidistant from three given planes. 715. Are equidistant from the edges of a given trihedral angle. 716. Are equidistant from two given planes and two given points in space. PROBLEMS. In the construction of the following problems, it is assumed that, besides the constructions of Plane Geometry, we are able : (1) to pass a plane through any given line and any point or line that can be in the same plane with it (422, 423, 424) ; (2) through any given point in or without a given plane, to draw a perpendicular to that plane. 717. Through a given line in a plane pass a plane making a given angle with that plane. 718. Through a given line without a given plane pass a plane mak- ing a given angle with that plane. 719. Through the edge of a given dihedral angle pass a plane bisect- ing that angle. 720. Through a given point without a given plane pass a plane parallel to that plane. 721. At a given distance from a given plane pass a plane parallel to that plane. 722. Through a given point pass a plane perpendicular to a given straight line. 723. Through the vertex of a given trihedral angle draw a line mak- ing equal angles with the edges. 724. In a given plane find a point such that the lines drawn to it from two given points without the plane, shall make equal angles, with the plane. (Two cases.) 725. In a given plane find a point equidistant from three given points without the plane. 726. In a given straight line find a point equidistaat from two given points not in the same plane as the line. Book VIIL polyhedrons. 498. A polyhedron is a solid bounded by four or more poly- gons. The bounding polygons are called the faces; their intersections, the edges; and the intersections of the edges, the vertices of the polyhedron. The least number of planes that can form a solid angle — the trihedral — is three. As the faces and edges extend indefinitely, the space within the angle is of indefinite extent, so that in order to cut off a definite portion of that space, a fourth plane must be passed intersecting the faces. Hence four is the least number of planes that can inclose a space. 499. A polyhedron of four faces is called a tetrahedron; one of six faces, a hexahedron; of eight faces, an octahe- dron; of twelve faces, a dodecahedron; of twenty faces, an icosahedron. 500. A polyhedron is convex when every section of it by a plane is a convex polygon. As none but convex poly- hedrons are to be treated of in what follows, the term poly- hedron will always signify convex polyhedron. 501. The volume of a polyhedron is its quantity as meas- ured by the polyhedron taken as unit of volume, or is the numerical measure of that quantity. In every-day life, volume is expressed by stating how many stated solid measures a given solid contains; as 356 cu. in. In abstract discussions, however, by volume is 261 262 SOLID GEOMETRY. — BOOK VIII. meant the numerical measure of a solid, or the ratio of a given solid to the solid unit. 502. Similar polyhedrons \i2iYQ fhQ ^2imQ form ; equivalent polyhedrons have the same volume; equal polyhedrons have the same form and volume. PRISMS. 503. A prism is a polyhedron two of whose faces are equal polygons having their homologous sides parallel, and whose other faces are parallelograms. 504. The bases of a prism are its equal paral- lel faces ; the other faces are called lateral faces. The intersections of the bases and lat- eral faces are called basal edges; the intersec- tions of the lateral faces are called lateral edges. 505. Prisms are triangular, quadrangular , pentagonal, etc., according as their bases are triangles, quadrangles, penta- gons, etc. 506. A right prism has its lateral edges perpendicular to its bases; all other prisms are called oblique prisms. 507. A regular prism is a right prism whose bases are regular polygons. CoR. The lateral faces of a regular prism are equal rectangles. 508. The altitude of a prism is the perpendicular distance between its bases. Cor. Each lateral edge of a right prism is equal to the altitude ; each lateral edge of an oblique prism is greater than the altitude. 509. A right section of a prism is a section perpendicular to its lateral edges. PRISMS. Proposition I. Theorem. 510. The sections of a prism made by parallel planes are equal polygons. Given: A prism MN intersected by parallel planes ABD, dbd; To Prove: ^5Ci)^ is equal to a6cde. Since plane ABB is II to plane abd, (Hyp.) AB, BC, CD, etc., are II to ab, be, cd, etc., resp., (451) and these lines are similarly directed ; .-. ZA = Za, ZB=Zb, Zc = Zc, etc. ; (455) .*. ABCDE and abode are mutually equiangular. Since AB = ab, BC= be, CD = cd, etc., (136) ABCDE and abcde are mutually equilateral; .-. ABCDE = abcde. q.e.d. (61) 511. Cor. Any section of a prism parallel to the base is equal to the base ; also all right sections of a prism are equal. 512. Definition. The lateral or convex stirface of a prism is the sum of its lateral faces. Exercise 727. In the diagram above, if the prism MNia pentag- onal, what is the sum of the plane angles of the lateral dihedral angles ? 728. Supposing, as above, that MN is pentagonal, how many faces, face angles, dihedral and trihedral angles, has MN9 729. If the base have n sides, how many faces, etc., has MN ? 264 SOLID GEOMETRY. — BOOK VI IL Propositiox II. Theorem. 513. The lateral surface of a prism is equivalent to the rectangle contained hy a lateral edge and the perimeter of a right section of the prism. B c Given: A right section ahcde, and AA\ ?i, lateral edge of prism AD' ; To Prove: The lateral surface of AD' is equivalent to rect- angle A A' • (ah -\-hc -\- etc.). Since AD' is a prism, (Hyp.) AB', bc', CD', etc., are parallelograms. (503) Since abcde is a right section, (Hyp.) ab, he, cd, etc., are ± to AA', BB', cc', etc., resp. ; (509) .-. AB' ^ AA' ■ ah; BC' =o BB' ' bc, etc. ; .-. AB' + BC' 4- CD', etc., ^ AA' - ah + BB' . bc + cc' •_ cd, etc. ; (Ax. 2) .-. lat. surf, of ad' ^ rect. AA' • (ah + 6c + etc.), q.e.d. (since AA' = BB' = CC' = etc.) (136) 614. CoR. The lateral surface of a right prism is equivalent to the rectangle of its altitude and the perimeter of its base. Scholium. Prop. II. may also be expressed under the form : The lateral area of a prism is equal to the product of a lateral edge and the perimeter of a right section. In the same way, in the corollary to Prop. II. and similar theorems, area may be substituted for surface, and p)roduct for rectangle, as explained in Art. 324. PRISMS. 265 Proposition III. Theorem. 515. Two prisms are equal, if three faces including a trihedral angle of the one are respectively equal to three similarly arranged faces including a trihedral angle of the other. Given: In the prisms Ad, A'd', the faces AD, Ah, Ae, re- spectively equal to the faces A'd', A'h\ A'e', and similarly arranged; To Prove : Prism Ad is equal to prism A'd'. (Hyp.) (495) . Since AD = A'd', Ah = A'h', Ae = A'e', and they are similarly arranged, trihedral A A •= trihedral Z A'. Hence, applying trihed. Z A to trihed. Z A', they will coincide ; then since AD ^ A'd', Ah ^ A'h', and Ae ^ A'e', edge ah =^ edge a'h', and edge ae =^ a'e' ; .'. base ad ^ base a'd', (being equal polygons having two sides coinciding ;) .-. all the lateral edges will coincide, (since their extremities coincide;) .-. the prisms coincide and are equal, q.e.d. (61) 266 SOLID GEOMETRY.— BOOK VIII. 516. Definition. A truncated prism is a part of a prism cut off by a plane not parallel to the base. 517. CoR. 1. Two truncated prisms are equal, if the three faces including a trihedral angle of the one are respectively equal to the three faces ijicluding a trihedral angle of the other, and are similarly arranged. ^^^ 518. CoR. 2. Tioo right prisms are equal if they have equal bases and equal altitudes. Proposition IV. Theorem. 519. An oblique prism is equivalent to a ri£ht prism having as base a right section, and as altitude a lat- eral edge, of the oblique prisma. Given: In oblique prism AD', a right section, ahcde, and a lat- eral edge, AA^ ; To Prove: AD' is equivalent to a right prism having as base ahcde, and an altitude equal to AaK Produce AA' to a', so that aa' = AA' ; and through a' pass a plane X to aa', and intersecting all the faces of the prism produced, thus forming a second right section, a'b'c'd'e', par- allel and equal to ahcde. The prism ad' thus formed is a right prism (506), whose base is the given right section, ahcde, and whose altitude aa'=^AA' (Const.). PRISMS. 267 Now the figures ABCDE-d and A'B'c'D'E^-d' are equal truncated prisms (517) ; and, if to each of these we add the figure abcde-D', we obtain (Ax. 2) prism ad' <^ rt. prism abcde-d'. q.e.d. 520. Definition. A parallelopiped is a prism whose bases are parallelograms. Cor. Any two opposite faces of a paral- lelopiped are equal parallelograms. 521. Definition. A right parallelopiped has its lateral edges _L to its bases. 522. Definition. A rectangular paral- lelopiped is a right parallelopiped whose bases are rectangles. Cor. Hence all its faces are rectangles. Proposition V. Theorem. 523. The plane passed through two diagonally op- posite edges of a parallelopiped divides it into two equivalent triangular prisms. Given: A plane AC' through diagonally opposite edges of paral- lelopiped bd' ; To Prove: Triangular prism ABC-B' is equivalent to tri- angular prism ADC-n'. 268 SOLID GEOMETRY.— BOOK VIIL Pass a plane ± to AA^ so as to form a rt. section ahcd of the parallelopiped, and intersecting AG' in ac. Since ac is a diagonal of par'm ahcd, A ahc = A adc. Now ABC-B' ^ a rt. prism having for base ~ A abc, and alt. = BB' ; and ADC-d' ^ a rt. prism having for base A adc, and alt. = BB' ; .'. prism ABC-B' =0= prism ADC-D', q.e.d. (140) (519) (Ax. 1) (since they are equivalent to equal rt. prisms.) 524. Cor. Any triangidar prism is equivalent to one half the parallelopiped having the same altitude and a base twice as great. Proposition^ VI. Theorem. 525. Any parallelopiped is equivalent to a rect- angular parallelopiped having the same altitude and an equivalent base. Given: A parallelopiped ABCD-C\ having an altitude H; To Prove: ABCD-c' is equivalent to a rectangular parallelo- piped having a base equivalent to ABCDy and altitude equal to H. PRISMS. 269 Through A and B pass planes each ± to AB. By the intersections of these planes with the faces, or the faces produced, a new parallelopiped, ABcd-c', will be formed, of the same altitude as the given figure, and hav- ing an equivalent base (Const., 328). Now ABcd-c' ^ABCD-& (519); for taking AD' as the base of the latter, then ABcd-c' is a rt. prism whose base, Ad*, is a rt. section of the given prism, and whose altitude is AB, a lateral edge of the same. Again (Fig. 2), through the edges AB and dc pass planes Af, dg, 1. to Ac, the base. By the intersection of these planes with the faces, pro- duced if necessary, of ABcd-c', a rectangular parallelopiped, ABcd-g, will be formed, having the same base and altitude as ABcd-c', or ABh'a'-c'. Now, by regarding Ah' as base of ABh'a'-c', and Af as a right section, it is seen that ABcd-g <> ABcd-c' (519). Hence (Ax. 1) ABCB-c' ^ ABcd-g, a rect. p'ped, etc., q.e.d. (since each is equivalent to ABcd-c'.) To obviate a frequent cause of difficulty to the student, it may be well to remark that the necessity of the double construction originates in the fact that the given solid ABCD-C' being any parallelopiped, we have to assume the possibility that both pairs of faces, AB' and DC', AD' and BC', are oblique to the base AC. If, in the given solid, we suppose AB' perpendicular to the base, then ABcd-c', the parallelopiped first constructed, will be rectangular^ and no further construction be necessary. Exercise 730, In the diagram for Prop. VI., left-hand figure, show that the solid AA'D'D-d' is equivalent to the solid BB'C^C-c'. 731. In AA'D'D-d', what kinds of polyhedral angles are those having their vertices at A and Z), respectively ? 270 SOLID GEOMETRY. — BOOK VII I. Proposition VII. Theorem. 526. Rectangular parallelopipeds with equal bases are to each other as their altitudes. \L i^i\ \ \ N i •■■.. E :■■••. X C'-... D\ F' \ L' = \ "T--.. -\ i'--.. E :•••.. \ C'-:. Given: Two rectangular parallelopipeds, AF, a'f', with equal bases, and altitudes AL, a'l' ; To Prove : p'ped AF : p'ped a'f' = al : A'l'. 1°. When AL and A'lJ are commensurable. Let ^'J5; be a common measure of AL and A'l', so that A'f can be laid off 7 times on AL and 5 times on A'L'. Through the points of division pass planes J_ to the edges. Having equal bases (510) and equal altitudes, (Const.) the p'peds thus formed are equal. (518) Since AL:A'l'=7:5, (Hyp.) and p'ped AF : p'ped A'F' =7 : 5, (Const.) p'ped^2^:p'ped^'j^' = ^i:^7.'. q.e.d. (P. 232'") 2°. When AL and A^L' are incommensurable. Suppose a'l' divided into any number of equal parts, n, and that AL contains m such parts with a remainder NL. Through N pass a plane perpendicular to the edges and cut- ting off the parallelopiped AK, PRISMS. 271 Since AN and A^L^ are commensurable, (Const.) AN _ m _ p'ped AK . A^L^~ n "p'ped ^'J^'' X (1°) and n n AF m ic' — — I — A'F' n n since AL and ^iT are slightly greater than AN and AK resp. Kow when n is taken indefinitely great, - and - become indefinitely small ; p'ped AF _ AL p'ped A'F'~ A'L'' (being the limits of variables always equal.) Q.E.D. (255) 527. Cor. If two rectangular parallelopipeds have two dimensions in common, they are to each other as their third dimensions. Proposition VIII. Theorem. 528. Rectangular parallelopipeds having equal alti- tudes are to each other as their bases. Given : Two rectangular parallelopipeds, CA, CD, having equal altitudes, and bases B G, CE ; To Prove : P'ped CA : p'ped CD — base BG : base CE. 272 SOLID GEOMETRY.— BOOK VIIL Place the parallelepipeds so that the edge CF may be common, and the right dihedral angles at CF vertical. Produce the faces AG, BG, AF, Dl, so as to meet, and form a third rectangular parallelopiped, CH. Since CA, CH, have the same base, FG, and the altitudes BC, CI, CA : CH = BC : CI = BC ' CG : CI - CG. (526, 318) Since CD, CH, have the same base, FI, and altitudes CK, CG, CD : CH = CK : CG = CI ' CK : CI ' CG ; (526, 318) .-. CA: CD = rect. BC - CG: rect. CI • CK ; (249) i.e., p'ped CA : p'ped CD = base BG : base CE. q.e.d. 529. Scholium. The foregoing proposition may be ex- pressed as follows : If two rectangular parallelopipeds have one dimension in common, they are to each other as the products of their other two dimensions. Proposition IX. Theorem. 530. Rectangular parallelopipeds are to each other as the products of their three dimensions. A ^ I ^ A ' i \ ^ i A / i p' c / r / Given : Two rectangular parallelopipeds, P, P', with dimensions a, b, c, and a', b', c', respectively; To Prove : P : P' = a x b x c : a' x b' x c\ PRISMS. • 273 Let Q be a third rectangular parallelepiped whose dimen- sions are a', b, c. Since P, Q, have two dimensions b, c, in common, (Hyp.) P:Q=a:a', ' (527) Since Q, P', have the dimension a' in common, (Hyp.) Q:P' = bxc:b' xc'. (529) .-. P : P' = a X b X c : a' X b' X c'. q.e.d. (242) 531. Definition. A cube is a rectangular parallelopiped whose faces are all squares. CoR. The edges of a cube are all equal. 532. Definition. The U7iit of volume is the cube whose edge is the linear unit, and whose base is, consequently, the unit of area. 533. Cor. 1. The volume of a rectangtdar parallelopiped is measured by the product of its three dimeyisions. For if a, b, c, be the dimensions of a rectangular paral- lelopiped P, and C7be the unit cube, then (530), P: U=abc: 1x1x1; .-. P—Ux abc. This conld be expressed at greater length as follows : The number of unit cubes in any rectangular parallelopiped is equal to the number of units in the product of the numeri- cal measures of its length, breadth, and thickness. 534. Cor. 2. The volume of a cube is measured by the cube of its edge. Thus if the edge of a cube be 7 linear units, the cube contains 7^ = 343 unit cubes ; if the edge be a linear units, the cube contains a^ times the unit cube. Geom. — 18 274 SOLID GEOMETRY. — BOOK VIII. Proposition X. Theorem. 535. The volume of any prism is measured hy the product of its base and altitude. 1°. Any parallelopiped is equivalent to a rectangular parallelopiped having the same altitude and an equivalent base (525) ; and the volume of the latter is measured by the product of its three dimensions, — that is, of its base and altitude (533) ; hence the volume of any parallelopiped is measured by the product of its base and altitude. 2°. Any triangular prism is equivalent to one half the parallelopiped having the same altitude and a base of twice the area (524) ; now, the volume of the latter being meas- ured by the product of its base and altitude (1°), the volume of the triangular prism is also measured by the product of its base and altitude. S°. By passing planes through its lat- eral edges, any prism can be divided into triangular prisms whose altitudes are the same as that of the given prism, and whose triangular bases together form the base of the given prism. As the volume of each of these triangular prisms is meas- ured by the product of its base and altitude (2°), the volume of any prism is measured by the product of its base a7id altitude, q.e.d. 536. Cor. Prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes are to each other as their bases ; and prisms are to each other as the products of their bases and altitudes. Exercise 732. Two triangular prisms, A and _B, liave tlie same alti- tude. A has for base a right-isosceles triangle ; B, for base an equi- lateral triangle of side equal to the hypotenuse of the base of A. Find the ratio of the volume of A to that of B.- 733. Find the ratio of the lateral area of A to that of B. PYRAMIDS, 275 PYRAMIDS. 537. A pyramid is a polyhedron bounded by a polygon called the base, and by triangular planes meeting in a com- mon point called the vertex. A plane intersecting the faces of any polyhedral angle cuts off a pyramid. The terms lateral face, lateral surface, lateral edge, basal edge, are defined as for prisms (504). 538. The altitude of a pyramid is the perpendicular distance from its vertex to the base ; as SP. 539. A regular pyraynid has for base a regular polygon, and has its vertex in the perpendicular at the center of the base, which perpendicu- lar is called the axis of the pyramid. 540. The slant height of a regular pyramid is the alti- tude of any lateral face. 541. A pyramid is triangular, quadrangular, pentagonal, etc., according as its base is a triangle, quadrangle, p)enta- gon, etc. In the triangular pyramid, or tetrahedron (499), any one of the faces may be regarded as the base. 542. A truncated pyramid is the portion /C^^nA of a pyramid included between the base and / / JA \ a plane that intersects all the lateral faces. \ / \ y 543. A frustum of a pyramid is a trun- cated pyramid in which the intersecting plane is parallel to the base. The base of the pyramid is called the lower base of the frustum ; the parallel section, the upper base. 544. The altitude of a frustum is the perpendicular dis- tance between its bases ; the slant height of a frustum of a regular pyramid is the altitude of any lateral face. 276 SOLID GEOMETRY.- BOOK VIIL Proposition XI. Theorem. 545. If a pyramid be cut by a -plane parallel to the base : 1°. The edges and altitude will be divided pro- portionally. 2°. The section is a polygon similar to the base. Given : A pyramid S-ABD, whose altitude SP is cut in p by a plane abd parallel to the base ; To Prove : 1°, SA : Sa = SB : Sb = SP : SjJ, etc. ; 2°, abd is similar to ABD. 1°. Suppose a plane passed through S II to ABD. Since the edges and altitude are cut by II planes, (Hyp. and Const.) SA : Sa = SB : Sb = SC :Sc = SP: Sp, etc. q.e.d. (459) 2°. Since plane abd is II to plane ABD, (Hyp.) ab is II to AB, be is II to BC, cd is II to CD, etc., (451) and they are similarly directed, (H^) abd and ABD are mutually equiangular. (455) Since ab is II to AB, and be is II to BC, A Sab is sira. to A SAB, and A sbc to A SBC; (291) .-. ab:AB = Sb: SB, and be: BC = Sb: SB ; (284) .-. ab:AB = bc:BC. PYRAMIDS. 277 In the same way we show that be : BC = cd: CD = de: BE, etc. ; .-. abode is similar to ABCBE. q.e.d. (284) 546. Cor. 1. The area of any section of a pyramid paral- lel to the base, is proportional to the square of its distance from the vertex. For parallel sections being similar to the base (545), their areas are proportional to the squares of their homologous sides (344). Thus abd : ABB = «6^ : AB^ = Ja :SA^ =^Sp : JP. 547. Cor. 2. If two pyramids, S-ABB, s'-A^b'b', having equal altitudes, SF, s'p', are cut by planes parallel to their bases, and at equal distances, Sp, s'p', from their vertices, the sections, abd, a'b'd', will be to each other as the bases. For abd : ABB = Sp : SP", ) and a'b'd' : A'b'b'= s'p' : s'p' ; \ (546) but Sp = S'p' and SP = s'p' ; (Hyp. ) .-. abd : ABB = a'b'd' : A'b'b'. 548. CoR. 3. If tivo pyramids have equal altitudes and equivalent bases, sections made by planes parallel to their bases, and at equal distances from the vertices, are equivalent. Exercise 734. Show that a plane perpendicular to the axis of a regular pyramid forms equal dihedral angles with all the faces of the pyramid. 735. In order that a plane intersecting the faces of a polyhedral angle may cut off a regular pyramid, what conditions must be fulfilled in regard to the form of the polyhedral and the inclination of the plane ? 736. In the diagram for Prop. XI., if a plane be passed through the mid point t)f pP, parallel to the base, show that the perimeter of the section thus formed will be equal to half the sum of the perimeters of ABD and abd. 278 SOLID GEOMETRY. — BOOK VIIL Proposition XII. Theorem. 549. The lateral surface of a regular pyramid is equivalent to one half the j^ectangle contained hy its perimeter and slant height. A B Given : A regular pyramid S-ABD, and SF its slant height ; To Prove : The lateral surface of S-ABD is equivalent to J-rect. Si^- {AB-{-BC-\ EA). Since ABD is a regular polygon, (Hyp.) AB =BC^CD =DE = EA; (370) since A, B, C, D, E, are equally distant from P, (539) SA = SB = SC = SD = SE; (437) .-. isos. A5fyli?=isos.A,Si?(7=isos. A-SCZ), etc. (69) But A SAE o= I rect. SF • AE ; . (331) .-. ASAB ^ ASBC -\ A SAE ^^SF -(AB -^BC+-"AE); (335) i.e., lat. surf, of S-ABD ^ }j SF ' perimeter. q.e.d. 550. Cor. 1. The lateral surface of a frustum, of a regular pyramid is equivaleyit to one half the rectangle con- tained hy the slant height of the frustum and the sum of the perimeters of the bases. (338) For it is the sum of as many trapezoids as the base has sides, having for common altitude the slant height of the frustum. (544) 551. Cor. 2. The dihedral and trihedral angles at the base of a regular pyramid are all equal. PYRAMIDS. 279 Proposition XIII. Theorem. 552. Triangular pi/ramids having equivalent bases and equal altitudes are equivalent. 1 k g'L- :rj^ ■■■■■J^ 4.\^l ^^ / \ ^^^^^^^ Given: Two triangular pyramids, S-ABC and s'-a'b'c', with equivalent bases, ABC and a'b'c', and the same altitude AL; To Prove : S-ABC is equivalent to s'-a'b'c'. Place the pyramids so that they shall be in the same plane, and AL be their common altitude. Divide AL into n equal parts, Aa, etc., and through the points of division pass planes parallel to the plane of the bases. The corresponding sections thus formed are equivalent ; (548) that is, DEF ^ d'e'f', ghi ^ g'h'T, etc. On the triangles ABC, DEF, etc., as lower bases, construct prisms D-ABC, G-DEF, etc., whose lateral edges are parallel to SA, and whose altitudes are each equal to Aa. On the triangles d'e'f', g'h'i', etc., as upper bases, con- struct prisms a'-d'e'f', d'-g'h'i', etc., whose lateral edges are parallel to S'a', and whose altitudes are each equal to Aa. 280 SOLID GEOMETRY. — BOOK VIII. Now prism G^-D^i^ =c= prism A'-d'e'f^ (536) because they have equivalent bases and the same altitude. Similarly, K-GHI o d'-g'h'i', etc. That is, corresponding to each triangular prism con- structed upon a section of S-ABC as lower base, is an equiva- lent prism constructed upon the corresponding section of S'-A'b'c' as upper base. Hence, the sum of all the prisms circumscribing S-ABC differs from the sum of the prisms inscribed in s'-a'b'c' by the prism D-ABC. But the sum of all the prisms circumscribing S-ABC is greater than that pyramid ; and the sum of all the prisms inscribed in S'-A'b'c' is less than S'-A'b'c'. Hence, these pyramids differ in volume by a volume less than prism D-ABC. Now if n be taken indefinitely great, the altitude Aa becomes infinitesimal, and therefore the prism D-ABC becomes infinitesimal. Hence, as they cannot differ by even an infinitesimal volume, pyramid S-ABC ^ pyramid S'-A'b'c'. q.e.d. Exercise 737. In the diagram for Prop. XIL , if the axis is equal to the apothem of the base, what is the inclination of each face to the base ? 738. In the same diagram, if the base is a regular pentagon, and Z SAB = 70°, what is the sum of the face angles ? 739. In the same diagram, show that the mid point of SF is equi- distant from S and P. 740. In a regular pyramid, the sum of the squares of the lateral edges is equivalent to one fourth of the sum of the squares of the basal edges, and n times the square of the slant height. 741. The perpendicular from the foot of the axis of a regular pyra- mid to the slant height is a mean proportional between the segments into which it divides it. 742. The axis and the slant height of a regular pyramid being given, find the apothem of the base. PYRAMIDS. 281 Proposition XIV. Theorem. 553. A triangular pyramid is one third of a tri- angular prism having the same base and altitude. Given: A triangular pyramid B'-ABC, and a triangular prism ABC-C', on the same base. To Prove : B'-ABC is equivalent to ^ ABC-C'. Take away the pyramid B'-ABC; there remains the quadrangular pyramid whose vertex is B' and whose base is the parallelogram AC'. Through B', A', C, pass a plane. It will divide the pyra- mid b'-AA'c'C into two triangular pyramids, which are equivalent (552) since the bases are halves of the parallelo- gram A c', and they have the same altitude, — the perpen- dicular from b' upon the base AC'. But pyramid b'-A'c'C, or C-A'b'c', ^ pyramid B'-ABC (552); for they have equal bases, A'b'c', ABC (503), and the same altitude, namely, that of the prism. Hence ipjT. B'-ABC + pyr. B'-AA'C 4- pyr. B'-A'c'C o 3 pyr. B'-ABC ; i.e., prism ABC-C' o= 3 pyr. B'-ABC. q.e.d. 554. OoR. The volume of a triavgular pyramid is meas- ured by one third the product of its base and altitude. 282 SOLID GEOMETRY. — BOOK VIII. Proposition XV. Theorem. 555. Any pyramid is one third of a prism having the same hase and altitude. Given : Any pyramid S-ABD, with base JBD and altitude SP ; To Prove: S-ABD is equivalent to one third the prism on base ABD with altitude SP. Througli SA pass planes SAD, SAC, etc. These planes divide S-ABD into triangular pyramids, whose bases make up the base ABD, and whose common altitude is SP. Each of these triangular pyramids is one third of the triangular prism with altitude SP that could be constructed on the same base (553). Hence the sum of the triangular pyramids that make up the given pyramid, is one third of the prism with altitude SP that could be constructed on the ba&e ABD. q.e.d. Scholium 1. The proposition could be expressed under the form : The volume of any pyramid is measured by one third the product of its base and altitude. 556. Scholium 2. The volume of any polyhedron may be found by dividing it into prisms or pyramids, and comput- ing the sum of their volumes. 557. Cor. Pyramids of equivalent bases are as their alti- tudes; pyramids of equal altitudes are as their bases; any two pyramids are as the products of their bases and altitudes. PYRAMIDS. 283 Proposition XVI. Theorem. 558. A frustum of a triangular pyramid is equiv- alent to the sum of three pyramids of the same alti- tude as the frustuTVii and whose bases are those of the frustum and a mean proportional between them. G c Given: A frustum D-ABC, with bases ABC, DEF; To Prove: D-ABC is equivalent to three pyramids having the altitude of the frustum, and having as bases ABC, DEF, and a mean proportional between ABC and DEF. Pass planes through the points F, A, B, and through i^, D, B. We thus divide the frustum into three triangular pyra- mids, ^ F-ABC, B-DEF, and F-ADB. The first two of these evidently have the same altitude as the frustum, and have for bases ABC and DEF respectively. The pyramid F-ADB, we shall show, is equivalent to a trian- gular pyramid having the same altitude as the frustum, and a base that is a mean proportional between ABC and DEF. In the. plane DC, draw i^G^ || to DA, and pass a pjane through FOB. FGi^W to plane ABED, (448) whence F and G are equally distant from that plane ; . •. py r. G-A DB=o F-A D B. (552) If, again, we take D as the vertex, and A GB 2is the base of G-ADB, its altitude is the same as that of the frustum. 284 SOLID GEOMETRY. — BOOK VIII. Draw GH II to CB. Then Z A GH= Z dfe, and Zgah=Zfde. (455) Since ^ i^ is a par'm, AG = DF; .: AAHG=ADEF (63), Siiid AH=DE. Since A ABC and ABG, Aabg and AHG, have equal alts., AABC:AABG=:AC:AG = AC:DF,} 2indAABG:AAHG = AB:AH=AB:DE.\ ^ ^ But AC:DF=zAB:DE, {A ABC and DEF being similar.) (545") .-. ABC: ABG = ABG: AHG = ABG: DEF; (232'") I.e., A ^5 G is a mean prop, between A ABC and DEF. 569. Cor. 1. T/ie volume of a frustum of a triangular pyramid is measured by the product of one third its altitude into the sum of its bases and a mean proportional between them. 560. Cor. 2. The volume of a frustum of any pyramid is measured by the product of one third its altitude into the sum of its bases and a mean proportional between them. For planes that divide the complete pyramid into trian- gular pyramids will divide the frustum into triangular frus- tums having the altitude of the given frustum. If, now, a plane be passed so as to cut the given frustums in a section that is a mean proportional between the bases, it will cut each triangular frustum in a section that is a mean propor- tional between its upper and lower bases ; that is, the mean proportional between the bases of the given frustum is the sum of those between the bases of the triangular frustums. The volume of the given frustum is the sum of the volumes of the triangular frustum ; hence it is measured by the product of one third their common altitude into the sum of the upper and lower bases of the • triangular frustums and the mean proportionals between them. PYRAMIDS. 285 Proposition XVII. Theorem. 561. Tetrahedrons with a trihedral angle of the one equal to a trihedral angle of the other, are to each other as the products of the edges of these trihedral angles. Given : V and F', the volumes of two tetrahedrons having trihe- dral angle A of the one equal to trihedral angle A' of the other ; To Prove : V: V' = AB X AC X AS: A'b' X A'c' X A's'. Apply one tetrahedron to the other so that A' =^ A. From S and -S' draw SP and S'p', Js to ABC Sind A'b'c', " and let their plane intersect ABC in AP'f. V: V" = ABC X SP: A'b'C' X S'P'. But ABC: A'B'C' = AB X AC: A'b' X A'C', (557) (341) (289) (288) ^i\d SP: S'P' = AS: A's', (since A SAP is similar to A S'A'P'.) Therefore, making the proper substitutions, we have, V: V' = AB X AC X AS: A'b' X A'C' X A'S'. q.e.d. 562. Cor. Similar tetrahedrons are as the cubes of their homologous edges. For let S-ABC, s'- a'b'c' be the similar tetrahedrons. Since V: V'= SAXSBX SC:S'A' X S'b' X S'c', (561) and SA : S'A' = SB : S'b' = SC:S'C', (Hyp.) V:r' = SAX SAXSA: S'A' X S'A' X S'A' = 'SA^ : S^l 286 SOLID GEOMETRY. — BOOK VIII. THE REGULAR POLYHEDRONS. 563. Definition. A regular polyhedron has its faces all equal, regular polygons, and its polyhedral angles all equal. Proposition XVIII. Theorem. 564. There are only five possible regular polyhedrons. The faces of a regular polyhedron must be regular poly- gons (563) ; at least three are necessary to form each poly- hedral angle (487) ; and the sum of the face angles of each polyhedral angle must be less than four right angles (492). 1°. The simplest regular polygon is the equilateral trian- gle, each of whose angles is 60°. Now 60° X 3 = 180°, 60° X 4 = 240°, and 60° x 5 = 300° ; but 60° X 6 = 360° = 4 rt. A. Hence only three regular polyhedrons can have equilateral triangles as faces. 2°. The regular four-sided polygon, or square, has each of its angles 90°. Now 90° X 3 = 270°, but 90° X 4 = 360° = 4 rt. A. Hence only one regular polyhedron can be formed having squares as faces. 3°. The regular pentagon has each angle 108°. Now 108° X 3 = 324°, but 108° x 4 = 432° > 4 rt. A. Hence only one regular polyhedron can be formed having pentagons as faces. 4°. The regular hexagon has each of its angles 120°. Now 120° X 3 = 360° = 4 rt. A. Hence no regular polyhe- dron can be formed having as faces regular polygons of six or more sides. Hence only five regular polyhedrons can be formed: three having equilateral triangles as faces, — namely, the tetrahedron, the octahedron, and the icosahedron ; one having squares as faces, — the hexahedron or cube; and one having pentagons as faces, — the dodecahedron. THE REGULAR POLYHEDRONS. 287 Proposition XIX. Problem. 565. To construct the regular polyhedrons, an edge being given. Given : A straight line AB slb edge ; Required : To construct the regular polyhedrons. 1°. Upon AB construct an equilateral triangle ABC, and find its center 0. At draw OD ± to AB C, and in OD take a point D such that AB = AB. Join BA, BB, BC ; B-ABC is the required tetrahedron. For the four faces are equilateral triangles (Const.) ; and the trihedral angles A, B, C, B, are equal (495), since their faces are all equal. H c B 2°. Upon AB construct a square AC, and upon the sides of this square construct squares AF, BG, CH, BE, in planes perpendicular to the plane of AC ; AG is the required cube. 288 SOLID GEOMETRY. — ROOK VIIL For the six faces are squares (Const.), and the trihedral angles A, B, C, D, E, F, G, H, are equal (495), since their faces are all equal. 3°. Upon ^J5 construct a square AC, and at its center 0, draw EF JL to plane AC ; make OF = OE = OA, and join EA, EB, EC, ED, FA, FB, FC, FD. These edges are equal to each other and to OA, since AGE, AGF, etc., are equal right triangles; hence the faces of the figure are equal equilateral triangles. Also, since the triangles DEB, DFB, are equal, DEBF is a square; whence it follows that the pyramid A-DEBF is equal to the pyramid E-ABCD ; hence the poly- hedral angles A and E are equal ; hence all the polyhedral angles of the figure are equal, and the figure is a regular octahedron. 4°. Upon AB construct a regular pentagon ABODE, and to each side of ABODE apply an equal pentagon, so inclined to the plane of ABCDE as to form trihedral angles at A, B, THE REGULAR POLYHEDRONS. 289 C, D, E. We thus obtain a convex surface FHLNP, com- posed of six regular pentagons. Construct a second con- vex surface f'h'l'n'p', equal to the first, and apply it so as to form a single convex surface. This will contain the figure required. For the faces are all equal pentagons (Const.), and the trihedral angles are equal, being contained by equal faces, and there being twelve such faces, the figure is a dodeca- hedron. 5°. Upon AB construct a regular pentagon ABODE, and at its center 0, draw OS ± to abode. Take s so that SA==AB, and join SA, SB, etc., thus obtaining a regular pyramid S-ABCDE, having each of its faces an equilateral triangle. Now take the vertices A and B as the vertices of two other pyramids A-BSEFG and B-ASOHG, having in common with S-ABODE the faces ASB,ASE, and ^Q>), conical (576), and spheri- cal surfaces (583), have been defined as generated by the motion of certain lines moving in a certain way. From another point of view, the plane generated as above men- tioned may be regarded as the locus of all perpendiculars to the given line at the given point ; a circular cylindrical surface, as the locus of all parallels to a given line at a given distance ; a circular conical surface, as the locus of all lines inclined to a given line, the axis, at a given angle. SPHERES. 303 Proposition VI. Theorem. 596. All points in the circumference of a circle of a sphere are equally distant from each pole. Given: P, P', poles of a circle ABC of the sphere whose center is 0; To Prove: All points in the circumference ABC are equally distant from P or P'. Join P with A, B, C, any points in circumf. ABC. Since OP is ± to circle ABC, . (Hyp.) OP passes through the center of ABC ; (589) .-. PA = PB = PC, (437) (being obliques from P to points A, B, C, in plane ABC, that are equally distant from the ± from P;) i.e., all points in circumference ABC are equally distant from P. Q.E.D. For like reasons they are equally distant from P'. 597. Cor. 1. All arcs of great circles draicn from a pole of a circle to poiiits in its circumference are equal. For their chords are equal chords (596) of equal circles (591). 598. Definitions. The distance between two given points on the surface of a sphere is the arc of a great circle join- ing those points ; the polar distance of a point A in the cir- 304 SOLID GEOMETRY. — BOOK IX. cumference of a circle ^^c is the arc of a great circle joining A and the nearer pole of ^^ C 599. CoR. 2. The polar distance of a great circle a'b'c' is a quadrant; that is, the fourth part of the circumference of a great circle. For the polar distance PmA^ measures the rt. A poa^ (262). 600. CoR. S. If a point P on the surface of a sphere is at the distance of a quadrant from any two points, A', £', in an arc of a great circle, then P is the pole of that circle. For the arcs PA\ PB', being quadrants, the angles at O are right angles ; therefore P is .L to OA' and to OB' ; hence it is ± to the plane of arc a'b' (428); whence P is the pole oisLTGA'B' (588). 601. Scholium. A pole of a circle, great or small, whose circumference is to pass through a given point, being known, the circumference may be described on the surface of the sphere. For by revolving Pa, an arc of a V^V^^^^^o great circle, about the pole P, the ex- /^^^--^.,y\ tremity a will describe the small circle 5^" [ "Jo ahc; while the extremity of PA, a quad- y^^-Ld ^j rant, will describe the great circle ABC. \^ ^/ Or, placing one foot of the spherical compasses* at P, with an opening between the feet equal to the chord of the polar distance of the point a, we turn the second foot round the sphere so as to describe through a the required circumference. Exercise 785. Denoting by a and r the numerical measures of the axis and radius of the cylinder rolling as in Exercise 783, what must be the ratio of r to a so that the surface generated in one revolution shall be a square ? * Compasses with the feet inclined inwards. SPHERES. 305 Proposition VII. Problem. 602. To find the radius of a given sphere, p ....;p /b ^.""•••- !o Given : A sphere APP' ; Required : To find a radius of APP\ Take any point P on the surface as pole, and with any opening of the compass describe a circumference AB c on the sphere. Take any three points, A, B, C, in this circumference, and with the compasses take off the chord distances AB, AC, BO. On any plane construct the triangle A'b'c', having its sides equal to AB, AC, BC, respectively (205), and circumscribe a circumference about this triangle (185). This circle will be equal to circle ABC, and its radius A'O will be equal to the radius of circle ABC. With A'o as an arm, and PA' = the chord of the arc join- ing PA, as hypotenuse, construct the rt. A PA'o. Draw A'p' ± to A'P, and produce it to meet PO in P'. It is evident that PP', thus determined, is equal to the diameter of the sphere, and its half, PQ, is the required radius. Exercise 786. If a cone of revolution roll upon a plane, its vertex remaining fixed, what kind of a surface is generated by the axis of the cone? 787. If a cone of revolution roll on the surface of a second cone, so that their vertices coincide, what kind of a surface is generated by the axis of the first cone ? Geom. — 20 306 SOLID GEOMETRY. — BOOK IX. 603. Scholium 1. The radius of the sphere being known, we can obtain the chord of a quadrant of that sphere by finding the hypotenuse h of a right triangle having its arms each equal to r, the radius. 604. Scholium 2. The chord of a quadrant of a sphere being found, we can describe a circumference of a great circle through any two points. A, B, on the sur- face of the sphere, by describing from A and B as centers, quadrants intersecting in P, which will be the pole of the required circumference (600). Proposition VIIT. Theorem. 605. The intersection of the surfaces of two spheres is the circumference of a circle perpendicular to the line of centers of the spheres. Given : 00', the line of centers of two intersecting spheres ; To Prove : The intersection of the surfaces is the circumference of a circle perpendicular to 00'. Through the centers, O, 0,' let a plane be passed, cutting the two spheres in great circles (587), which intersect in A and B. (Hyp.) SPHERES. 307 Draw the chord AB, and produce 00' to meet the circum- ferences. 00' is X to ^5 at its mid point C. (75) If we now revolve the upper part of the figure about 00', the two semicircles will generate the two spheres (583), while the point A will generate the line of intersection of the surfaces. Also, since AC during the revolution remains ± to 00', AC will generate a circle whose center is C ; i.e., the intersection of the surfaces of the spheres is the circumference of a circle _L to 00'. q.e.d. 606. Definition. Two spheres are tangent if their sur- faces have but one common point. 607. CoR. If tivo spheres are tangent to each other, the point of tangency is in their line of centers. For if we conceive the centers, 0, O', to remove farther from each other till the circumferences of the great circles become tangent, the points A and B will coincide with, and the circumference of intersection be reduced to, a point C. 608. Scholium. Two spheres being given in any relative position, a plane passed through their centers will cut them in great circles; and according as these circles are within or without, are tangent or intersecting, the spheres will have corresponding positions in regard to each other. 609. Definition. A plane is tangent to a sphere when it has but one point in common with the surface of the sphere. Exercise 788. Denoting by a and r the altitude and radius of a cone rolling as in Ex. 784, find the ratio of the base of the cone to the entire circle generated. 789. What fraction of that circle is described by one revolution of the cone ? 308 SOLID GEOMETRY. — BOOK IX. Proposition IX. Theorem. 610. A plane perpendicular to a radius of a sphere at its ejctremity is tangent to the sphere. Given : OP, a radius of a sphere, and a plane MN _L to OP at its extremity ; To Prove : MN is tangent to the sphere. Take any point except P in MN, as A, and join OA, PA. Since OP is ± to AP, (Hyp., 427) OA > op; (99") .•. A lies without the sphere ; .-. MN is tangent to the sphere, q.e.d. (609) (since every point in MN, except P, lies without the sphere.) 611. Cor. 1. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. 612. Cor. 2. Any straight line drawn in the tangent plane through the point of contact is tangent to the sphere. 613. Cor. 3. Any two straight lines tangent to the sphere at the point of contact determine the tangent plane at that point. SPHERICAL ANGLES AND POLYGONS. 309 SPHERICAL ANGLES AND POLYGONS. 614. Definition. The angle of two intersecting curves is the angle contained by the two tangents to the curves at the common point. This definition applies, whatever be the surface upon which the curves are described. 615. Definition. A spherical angle is the angle included between two arcs of great circles of a sphere. The arcs are its sides; their intersection is its vertex. Proposition X. Theorem. 616. td spherical angle is measured hy the arc of a great circle described from its vertejc as pole, and included hy its sides, produced if necessary. Given : ^5, an arc of a great circle described from the vertex of spherical angle APB as pole, and included between the sides of angle APB ; To Prove : Spherical angle APB is measured by mc AB. Draw PT, pt', tangents to PAP'^ pbp' respectively, and radii OA, OB. [] (106) 310 SOLID GEOMETRY. — BOOK IX. Since PT is ± to PP' in plane PAP', (Const., 191) and OA is ± to PP' in plane PAP', (PA being a quadrant,) (Hyp.) PTis II to 0.4; similarly PT' is II to OB . ^ TPT' = Z AOB. (455) But Z AOB i^ meas. by arc ^5; (262) le., spher. Z .IPi? is meas. by arc AB. q.e.d. (614) 617. CoR. 1. A spherical angle has the same measure as the dihedral angle formed by the planes of its sides. 618. Cor. 2. All arcs of great circles drawn through the pole of a given great circle are perpendicular' to its circumfer- ence. For their planes are each perpendicular to its plane (469). 619. Scholium. The foregoing corollary enables us, through any giv^en point P on the surface of a sphere, to describe an arc of a great circle perpendicular to a given arc ABC of a great circle. From P as pole describe (428) an arc of a great circle cutting ABC in C, and from C as pole describe an arc of a great circle passing through P and cutting ABC in B ; then arc PB is _L to arc ABC (618). 620. Definition. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the sides of the polygon; the angles formed by these sides are the angles, and the vertices of the angles are the vertices, of the polygon. SPHERICAL ANGLES AND POLYGONS. 311 621. Since the planes of all great circles pass through the center of the sphere, the planes of the sides of a spheri- cal polygon form at the center a polyhedral angle whose edges are radii drawn to the vertices of the polygon ; the face angles of the polyhedral angle are angles at the center measured by the arcs that form the sides of the polygon ; while the dihedral angles of the polyhedral angle have the same measure as the angles of the polygon (616). Thus the planes of the sides of the polygon An CD -{see diagram for Prop. XII.) form at 0, the center of the sphere, the polyhedral angle 0-ABCD. The face angles, AOB, BOC, etc., are measured by the sides AB, BC, etc., of the poly- gon; and the dihedral angle whose edge is the radius OA, has the same measure as the spherical angle BAD, etc. 622. From the relations thus established between poly- hedral angles and spherical polygons, it clearly follows that from any known property of polyhedral angles we may infer a corresponding property of spherical polygons; and con- versely. 623. Definition. A diagonal of a spherical polygon is an arc of a great circle passing through two nonadjacent vertices. 624. Definition. A spherical triangle is a spherical poly- gon having three sides. Like plane triangles, spherical tri- angles may be right or oblique, scalene, isosceles, or equilateral. Exercise 790. Out of a circle with radius B a sector of 60° is cut, and the edges of the remaining sector are joined so as to form the lateral surface of a cone of revolution. Find the radius r, and the altitude a, of the cone thus formed. 791. Find general formulas for the radius r and the altitude a of a cone of revolution whose lateral surface is formed from a sector that is ~i\\^ of a circle with radius B. 312 SOLID GEOMETRY. — BOOK IX. Proposition XI. Theorem. 625. Any side of a spherical triangle is less than the suin of the other two. c ■...,A O-.;::;- Given : A spherical triangle ^^c, on a sphere whose center is o ; To Prove : AB -\- AC is greater than BC. In the trihedral angle 0-ABC, Zaob-{-Zaoc>Zboc; (491) .-. arc^5 +arc^C>arc5(7. q.e.d. (616) 626. Cor. 1. Any side of a spherical triangle is greater than the difference of the other two. 627. Cor. 2. The shortest path 07i the surface of a sphere between two given points, A and B, is the arc A MB of a great circle passing through those points. For let ACB be any other curve join- ing A and B. Take in it any point C, and through A, G, and C, B, describe arcs of great circles (594). Then AMB , on a sphere with center ; To Prove : AB-\-BC-{-CD-{-DA H-arc i)^ 180° and < 540°. Construct A'b'c', the polar triangle of ABC. Then denot- ing (as in Art. 632) the number of degrees in B'c', A'C', A'b', resp. by a', b', c', since Za = 180° -a', Zb = 180° -6', Z c =180°- c', (632) Z A-]-Zb -{-Z c = 540° - (a' + 6' + c'). (Ax. 2) But a' + b'-hc' < 360° and > 0°; (492) .'.Za+Zb + Zc> (540° - 360°) or 180°, andZ A + Z B -\- Z c < (540° - 0°) or 540°. q.e.d. 634. Cor. A spherical triangle may have two, or even three, right angles; also two, or even three, obtuse angles. 635. Definition. A spherical triangle having two right angles is said to be birectangular ; a spherical triangle hav- ing three right angles is said to be trirectangular. 636. Definition. The excess of the sum of the angles of a spherical triangle over two right angles is called fhe spherical excess of the triangle. POLAR TRIANGLES. 317 Thus, denoting its spherical excess by E, we have for A ABC, e=Za-^Zb-\-Zc- 180°. 637. Definition. The spherical excess of any spherical polygon of n sides is equal to the excess of the sum of its angles over 2{n — 2) right angles; that is, is equal to the sum of the spherical excess of the {n — 2) spherical tri- angles into which the polygon can be divided by means of arcs drawn from any vertex to the opposite vertices. 638. Definition. Symmetrical spherical triangles are such as have their sides severally equal, but arranged in reverse order; as ABC, A'b'c'. A'b'c' may be regarded as formed by pro- ducing AO, BO, CO, to meet the surface of the sphere in A', B', c', and joining the points thus obtained by arcs of great circles. We can conceive A'b'c', when thus formed, as moved to any other position on the spheri- cal surface; and vice versa. Since the face angles of the trihedrals 0-ABC, 0-A'b'c', are equal, but arranged in reverse order, the trihedrals are symmetrical (493), and cannot be made to coincide unless isosceles (497). Hence the spherical triangles whose sides are the intersections of the planes of the faces of the trihedrals, cannot be made to coincide unless these trihedrals can be made to coincide ; that is, unless they are isosceles. 639. CoR. An isosceles spherical triangle can he made to coincide with its symmetrical triangle. For as the trihedral angles formed by the planes of their sides can be made to coincide (497), the sides formed by the intersections of those planes with the surface can be made to coincide. 318 SOLID GEOMETRY. — BOOK IX. Proposition XVI. Theorem. 640. Two triangles on the saine sphere are either equal or syminetrical, if a side and the including angles of the one are respectively equal to a side and the including angles of the other. Given: In spherical triangles ABC, a'b'c', ab equal to A^B\ angle A equal to angle A\ and angle B equal to angle B' ; To Prove: ABC and A'b'c' are either equal or symmetrical. For A ABC may be placed either upon A A'b'c' or upon A abc, symmetrical with A'b'c', so as to coincide, as may be shown by the same course of reasoning as that employed in Prop. VI., Book I. Hence A ^J5C is either equal to A a'b'c' or symmetrical with it. q.e.d. 641. Cor. If a spherical triangle has two equal angles, it is isosceles. For ii Z A = Z b = Z A' = Z b', then, also, since Z A—Z a and Zb ^ Zh, A ABC can be made to coincide with both A a'b'c' and with its symmetrical triangle abc. Hence a'b'c' and ahc must be isosceles (639); whence also ABC is isosceles. Exercise 792. If two sides of a spherical triangle are quadrants, the third side measures the opposite angle. 793. A spherical triangle ABC has Z^ = 83o, / ^ = 50°, and AC — 97° ; find how many degrees there are in the sides of the polar triangle that are respectively opposite to those angles. POLAR TRIANGLES. 319 Proposition XVII. Theorem. 642. Two triangles on the same sphere are either equal or symmetrical, if two sides and the included angle of the one are respectively equal to two sides and the included angle of the otJier. Given: In spherical triangles ABC, a'b'c', AB equal to A^b\ AC equal to A'c\ and angle A equal to angle A^ ; To Prove : ABC and a'b'c' are either equal or symmetrical. For A ABC can be made to coincide either with A A'b'c', or with its symmetrical triangle abc, as may be shown by the same course of reasoning as that employed in Prop. VIII., Book I. Hence A ABC is either equal to Aa'b'c' or symmetrical with it. q.e.d. 643. Cor. 1. In an isosceles spherical triangle, the angles opposite the equal sides are equal. For if ABC is isosceles, then both A'b'c' and its sym- metrical triangle abc must be isosceles. Hence, as Z B can be made to coincide with Z c, and Z c = Z c' = Zc, Zb = Zc. 644. Cor. 2. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the mid point of the base, bisects the vertical angle, is perpendicular to the base, and divides the triangle into two symmetrical triangles. Exercise 794. If a spherical triangle ABC is isosceles, its polar triangle A'B'C is also isosceles. 320 SOLID GEOMETRY. — BOOK IX. Proposition XVITI. Theorem. 645. Two triangles on the same sphere are either equal or symmetrical, if the three sides of the one are respectively equal to the three sides of the other. For as the respectively equal sides must be arranged either in the same or in the reverse order, the one triangle can be made to coincide either with the other or with its symmetrical triangle, as may be shown by a course of reason- ing similar to that employed in Prop. X., Book I. q.e.d. 646. Scholium. This and other theorems in regard to triangles, etc., on the same sphere, will evidently hold true in regard to triangles, etc., on equal spheres ; since, if the cen- ters of the equal spheres be made to coincide, their surfaces will also coincide. Proposition XIX. Theorem. 647. If two triangles on the same sphere are mu- tually equiangular, they are mutually equilateral and are either equal or syxnmetrical. Given : Two mutually equiangular spherical triangles A and A^ ; To Prove : A and A^ are mutually equilateral and either equal or symmetrical. POLAR TRIANGLES. 321 Construct P, P', the polar triangles of A, A' respectively. Since A and A' are mutually equiangular, (Hyp.) P and P', their polar A, are mutually equilateral, (631) (their corresponding angles being measures of equal angles ;) .-. P and P' are mutually equiangular; (645) .-. A and A', the polar A of P and P', are mutually equi- lateral; (631) .-. A and A' are equal or symmetrical, q.e.d. (645) Scholium. Mutually equiangular spherical triangles are equilateral only when on the same sphere or equal spheres. When the equiangular triangles are on unequal spheres, their homologous sides are proportional to the radii of their spheres ; the triangles are then said to be similar. 648. CoR. If three planes are passed through the center of a sphere, each perpendicular to the other two, they divide the surface into eight equal trirectangular triangles (621, 647). Exercise 795. Birectangular triangles on equal spheres are equal if their acute angles are equal. 796. Birectangular triangles on unequal spheres are similar if their acute angles are equal. 797. Trirectangular triangles on equal spheres are equal. 798. Trirectangular triangles on unequal spheres are similar. 799. The polar triangle of a birectangular triangle is birectangular. 800. In a birectangular triangle, the sides that are opposite to the right angles are quadrants. 801. Each side of a trirectangular triangle is a quadrant. 802. The polar triangle of a trirectangular triangle is a trirect- angular triangle coinciding with the triangle itself. Geom. — 21 322 SOLID GEOMETRY. — BOOK IX. Proposition XX. Theorem. 649. Symmetrical spherical triangles are equiv- alent. .■■:/tA y^ . p:. Given: In symmetrical triangles ABC, A'b'c', AB equal to A's', AC equal to A'c', BC equal to £'c'; To Prove: Triangle ABC is equivalent to triangle A'b'c'. Find P and P', the poles of small circles passing through A, B, c, and A', b'j c' respectively. Since arcs AB, AC, 5C = arcs a'b', A'c', b'c' resp., (Hyp.) the chords of arcs AB, AC,BC = the chords of arcs A'b', A'c', b'c' resp. ; .*. the plane triangles formed by these chords are equal; (69) .-. the small circles through A, B, C, and A', B', c', are equal. Through PA, PB, PC, P'A', p'b', p'c', pass arcs of great circles. Since arcs PA, PB, PC, P'A', p'b', p'c', are equal, (596) (being polar distances of equal circles on the same sphere,) A PAB, p'a'b', are symmetrical and isosceles; .-. A PAB = A P'A'B'. (639) Similarly, A PBC = A p'b'c', and A PAC = A P'A'c' ; .'. PAC -{-PBC— PAB ^ P'A'c' -\- p'b'c' — p'a'b' ; (Ax. 2, Ax. 3.) .-. A ABC =0^ A a'b'c'. Q.e.d. POLAR TRIANGLES. 323 If pole P should lie within A ABC, then pole P' would also lie within Aa'b'C', and we should have PAB+ PAC-}- PBC^P'A'b' -{- P'A'C'-\-P'B'C'; .'. A ABC^ AA'B'C'. Proposition XXI. Theorem. 650. In a spherical triangle, a greater side is oppo- site a greater angle. A Given: In spherical triangle ABC, angle A greater than angle B; To Prove : i? C is greater than A c. Through A describe AD, an arc of a great O, so that Zbad = Zb. Bince Zbad= Zb, BD=DA; (641) .'. BD -\-DC=DA-\- DC; ^ .: BC> AC. Q.E.D (625) 651. Cor. Conversely, if, in triangle ABC, BC is greater than AC, then angle A is greater than angle B. For these angles cannot be equal, since BC and AC are not equal (Hyp.). Nor can Z Ahe less than Z B, for then BC would be less than AC (650) j hence Z A must be greater than Z B. 324 SOLID GEOMETRY. — BOOK IX. 652. Definition. A lu7ie is the part of the surface of a sphere that is included between two semi- circumferences of great circles; as AMBN.' The angle of the lune is that between the semicircumferences that form its sides. 653. Cor. Lunes on the same sphere hav- ing equal angles are equal. For they evidently can be made to coincide. 654. Definition. A spherical ungula or wedge is the part of a sphere bounded by a lune as base, and by the planes of its sides ; as AON MB. The angle of the ungula is the same as the angle of its base ; the diameter AOB is called the edge of the ungula. Proposition XXIL Theorem. 655. If two arcs of great circles intersect on the surface of a hemisphere, the sum of the two opposite triangles thus formed is equivalent to a lujie whose angle is equal to that formed hy the arcs. Given: On the hemisphere A-BDCE, two arcs of great circles, BAC and DAE, intersecting at A, and forming the A ABD, ACE ; To Prove : A ABD -i- A ACE ^ el lune whose Z = Z CAE. POLAR TRIANGLES. 325 Produce arcs AC, AE, till they meet at A'. Since BAC = ^ semicircumf. = ACA', (Hyp.) BAG — AC= ACA' — AC; (Ax. 3) i.e., AB =A'C. In the same way it may be shown that AD = a'e and BD = EC; .-. Aabd, a'ce, are mutually equilateral; .-. AABD ^ A A'ce; (645, 649) .-. AABD -}- A ACE -0= A A'CE + A ACE; i.e., A ABD -[-AACE<^ lune ACA'e, whose Z is Z CAE. q.e.d. 656. Definition. Just as the angle which is the nine- tieth part of a right angle is called a degree, so the birectangular triangle which is the ninetieth part of a trirectangular triangle, and has for base the arc that meas- ures one degree, is called a spherical degree. Denoting by T the surface of a trirectangular triangle, the surface of the sphere = 8 T = 720 spherical degrees. It is also obvious that the lune Avhose angle is one degree will contain two spherical degrees. It is important to keep in mind the distinction between the three different senses in which the term degree is employed. A degree of angular measure is the 90th part of a right angle, a relative position of two straight lines ; an arc degree is a line, the 90th part of a quadrant or 360th part of a circumference ; a spherical degree is a surface, the 90th part of a trirectangular triangle or the 360th part of the surface of a hemisphere. Exercise 803. If a spherical triangle has one right angle, the sum of the acute angles is greater than a right angle. 804. Lunes with equal angles on unequal spheres are similar. 326 SOLID GEOMETRY. — BOOK IX. Proposition XXIII. Theorem. 657. A lune is to the surface of the sphere as the angle of the lune is to four right angles. Given: A lune L upon a sphere whose surface is S, and AOB, the angle of the lune, whose poles are P, P' ; To Prove : L: S = angle AOB :4: right angles. 1°. When arc AB and circumference ABC are commensu- rable. Let arc AE be a common measure of arc AB and the cir- cumference, so that AE can be laid off 9 times on AB and 73 times on ABC. Suppose arcs of great circles to be passed through the points of division and the poles P, P'. The lunes whose angles are measured by the equal arcs are equal (653), and each is contained 9 times in L and 73 times in S. Since Z ^ 05 : 4 rt. Zs = 9 : 73, (Hyp. ) and i:S = 9:73, (Const.) L:S = ZA0B:4:Vt. A. Q.E.D. (232'") 2°. When arc AB and the circumference are incommensu- rable, we can prove by the method of limits (as in Prop. II., Book IV.) that always L:S = Zaob : 4: rt A. q.e.d. POLAR TRIANGLES. 327 658. Cor. 1. On the same sphere, lunes are to each other as their ayigles. 659. Cor. 2. Denoting by A the number of degrees in the angle of a lune, since X : 8 r = ^ : 360, 45 660. Cor. 3. The spherical ungula AB — PP' : sphere = ZaoB:S60. For ungulas are equal if their lunes are equal, since they^ can be made to coincide. Hence in equal spheres, ungulas are to each other as their lunes; or, employing V to denote the volume of the sphere, and u that of the ungula, we have U: V = A:360; 45 661. Scholium. In the above formulas, the symbols L, T, U, V, denote only relative values. In order to obtain the absolute value of L or U, we must know that of T or V, which, again, depend upon the radius of the sphere; in what manner will be seen further on. Exercise 805. The angle of a lune is 36°. What fraction is the lune of the surface of its sphere ? 806. A lune comprises one hundredth part of the surface of a sphere. What is the angle of that lune ? 807. A trirectangular triangle is to a lune on the same sphere as 12 is to 5. What is the angle of the lune, and what fraction is it of the whole spherical surface ? 808. The dihedral angle formed by the plane faces of a spherical wedge is 25°. What fraction is that wedge of the whole sphere ? 809. In order that a spherical wedge shall be J^th of its sphere, what must be the angle of its base ? 328 SOLID GEOMETRY. — BOOK IX. Proposition XXIV. Theorem. 662. A spherical triangle is equivalent to as many spherical degrees as there are angular degrees in its spherical excess. Given : A spherical A AB c, whose spherical excess is E degrees ; To Prove : Triangle AB C is equivalent to E spherical degrees. Produce the sides to meet DKH, a great circle described about ABC. Since A ADF + A AEG <^ a lune whose Z= Za, (655) A ADF -{-A AEG ^2 A spherical degrees. (656) In the same way it may be proved that A BEK + A BDH <^2 B spher. deg., A CGH + A CFK =0= 2 (7 spher. deg. Now the sum of these six triangles exceeds the surface of the hemisphere, or 360 spher. deg., by twice A ABC ; .: 360 spher. deg. -{-2 A ABC ^2(A -\- B -{- c) spher. deg. ; .-. AABC^{A-\-B-{-C — 180) spher. deg. ; I.e., A ^7? (7 <>^ spher. deg. q.e.d. (636) 663. Scholium. E, it must be remembered, is here em- ployed as a numerical measure, an abstract number. If, for example, the angles of the spherical triangle ABCSive 73°, 87°, and 100°, respectively, then J^ = 73 + 87 + 100 - 180 = 80. Hence we know that A ABC^SO spherical degrees =o=|^To=-J^ of the surface of the sphere. POLAR TRIANGLES. 829 Proposition XXV. Theorem. 664. Any spherical polygon is equivalent to as 7)%any spherical degrees as there are angular degrees in its spherical excess. Given: A spherical ^olj^on ABODE, whose spherical excess is E degrees ; To Prove : Polygon ABODE is equivalent to E spherical degrees. Through any vertex A and the opposite vertices, describe arcs of great circles, AO, AD, dividing the polygon into spher- ical triangles. Now the area of each triangle is equal to as many spheri- cal degrees as there are angular degrees in its spherical excess (661). Hence the area of the polygon is equal to as many spherical degrees as there are degrees in the sum of the spherical excesses of the triangles ; that is, as many degrees as there are in the spherical excess of the polygon; (637) .'. polygon ^£C2)-£J :o=^ spherical degrees. q.e.d. 665. Cor. The area of any spherical polygo7i is to the sur- face of the sphere as E is to 720. For (636, 637) E being the spherical excess of any poly- gon P, whether of three or more sides, and S the surface of the sphere, P'.S=:E'. 720. Exercise 810. If the angles A, B, C, D, E, of the spherical pen- tagon ABODE are 140°, 90°, 93°, 120°, and 117°, respectively, what part of the spherical surface is ABODE ? 330 SOLID GEOMETRY. — BOOK IX. EXERCISES. QUESTIONS. 811. What is the locus of all the points in space that are at a given distance from a given straight line ? 812. The cylinder may be regarded as the limiting form of what solid ? 813. What is the locus of all the straight lines in space that make a given angle with a given straight line at a given point ? 814. The cone may be regarded as the limiting form of what solid ? 815. What is the locus of all the points in space that are at a given distance from a given point ? 816. Two plane triangles that are mutually equiangular are not necessarily equilateral ; but spherical triangles on equal spheres, if mutually equiangular, are also mutually equilateral. Why so ? 817. If straight lines be drawn from any point in a spherical surface to the extremities of a diameter, what angle will those lines contain ? 818. The plane that is tangent to a sphere at a given point is the locus of what lines ? 819. What is the locus of all the points in space that have their dis- tances from two given parallel lines in a given ratio ? 820. What is the locus of all the points in space such that the dis- tances of each from a given straight line and a given point in that line have a given ratio ? 821. What is the locus of all the points in space at a given distance from a given plane ? 822. What is the locus of all the points in space at a given distance from a given spherical surface ? Under what circumstances will the locus consist of one surface only ? 823. What is the locus of all the points in space at a given distance from a given circular cylindrical surface ? Under what circumstances will the locus consist of one surface only ? 824. What is the locus of all the points in space at a given distance from a given circular conical surface ? EXERCISES. 331 THEOREMS. 825. The locus of all the points such that lines drawn from it to the extremities of a given straight line form a right angle, is a spherical surface. 826. If any number of lines in space pass through a given point, the feet of the perpendiculars from any other point to these lines lie upon a spherical surface. 827. If any number of lines in a plane pass through a point, the feet of the perpendiculars to those lines from any point without the plane lie in a circle. 828. If from a point on the surface of a sphere as pole, with a polar distance equal to one third the chord of a quadrant, a circle be described, the radius of this circle will be one half the radius of the sphere. 829. Any lune is to a trirectangular triangle as its angle is to half a right angle. 830. Spherical polygons on equal spheres are as their spherical excesses. 831. In any right spherical triangle, if one side be greater than a quadrant, there must be a second side greater than a quadrant. 832. In any right spherical triangle, a side less than a quadrant subtends an acute angle ; a side greater than a quadrant subtends an obtuse angle. 833. Two right spherical triangles are equal or symmetrical if the hypotenuse and an adjacent angle of the one are severally equal to the hypotenuse and an adjacent angle of the other. 834. Two right spherical triangles are equal or symmetrical if the hypotenuse and an arm of the one are severally equal to the hypot- enuse and an arm of the other. 835. The bisector of the angle contained by arcs of great circles is the locus of all points within the angle and equidistant from its sides. Book X. MEASUREMENT OF THE THREE ROUND BODIES. ^^*^< CYLINDERS. 666. A prism is inscribed in a cylinder when its bases are inscribed in the bases of the cylinder, and its lateral edges are elements of the lateral surface of the cylinder. 667. A prism is circumscribed about a cylinder when its bases are circumscribed about the bases of the cylinder. 668. The lateral area of a cylinder is the area of its lateral surface. Proposition I. Theorem. 669. The lateral area of a cylinder of revolution is measured hy the product of the circumference of its base hy its altitude. ^.^= c A' \ B ^^' d\ .. c\ y^d \P Given : C, the circumference of the base of a cyhnder of revo- lution ABCD-c' ; II, its altitude ; and s, its lateral surface ; To Prove : S is equal to C x //. CYLINDERS. 333 Inscribe in the cylinder a regular prism ABCD-C', whose bases are regular polygons inscribed in the bases of the cylinder. If we denote by s the lateral surface of the prisixi, and by p the perimeter of each of the base polygons; then, H being also its altitude, s=pxH, (514) whatever be the number of lateral faces of the prism. Let the number of lateral faces be indefinitely increased by continually doubling the number of sides of the base polygons. Then as p has for limit C (392), and the lateral edges of the prism are, if indefinitely increased in number, the elements of the surface s, s has for limit S ; .: S = C X H. Q.E.D. (23(5) 670. Cor. 1. If the cylinder be generated by a rectangle whose sides are H and R, revolving about II, then H is the alti- tude of the cylinder and R the radius of the base. Hence the perimeter of the base is 2 7r'R (396), and for the lateral area S we obtain the expression, S — 27r • R - H. 671. CoR. 2. Since the area of each base is tt-R^ (398), we obtain for T, the total area of a cylinder of revolution, the expression, t=2tt - R{H -\- R). 672. Definition. Similar cylinders of revolution are generated by similar rectangles revolving about homolo- gous sides. 673. CoR. 3. The lateral areas, or the total areas, of similar cylinders of revolution, are as the squares of their altitudes or radii 334 SOLID GEOMETRY.— BOOK X. For let S and S' denote the lateral areas of two similar cylinders of revolution; R and R\ the radii of their bases; H and h\ their altitudes ; T and T\ their total areas ; then, since the generating rectangles are similar, (Hyp.) H:H' = R:R' = H-\-H':R + R'; (246) .-. S:S' = 27r-R-H:27r'R^'H^ = H^ : h" = R- : R'% and T:T'=27r'R{H+R):27r'R'{H'-{-R') = H':H''=R^:R'\ 674. Scholium. The lateral area of any cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of its surface. This may be proved by a method similar to that employed in the proof of Prop. I., assuming that the bases of the cylinders, when not circular, are still the limits of inscribed polygons, the number of whose sides is indefinitely great. Proposition II. Theorem. 675. The volume of a cylinder of revolution is meas- ured hy the product of its base by its altitude. Given : F, the volume of a cylinder of revolution AB, whose base is B and altitude H; To Prove : V is equal lo B x H. Let f' denote the volume of a regular inscribed prism of any number of faces ; B ', its base ; H will also be its altitude. CYLINDERS. 335 Now whatever be the number of sides of the prism, r = B' XH. (535) But when the number of faces is indefinitely increased, B^ has for limit B, and F' has for limit V ; .'. V = B X H. Q.E.D. (236) 676. CoR. 1. Let v be the volume of the cylinder, R the radius of its base, and H its altitude ; then, since B = ir • R^, V=7r' R^ ' H. 677. CoR. 2. The volumes of similar cylinders of revolution are to each other as the cubes of their altitudes or radii. For if V and F' be the volumes of two similar cylinders of revolution, R and R^ the radii of their bases, H and H^ their altitudes ; since the generating rectangles are similar, (Hyp.) H:II'=R:R'; .'. V:V' = 7r'R^'H:7r- R''-'H=H^:H'^=R^: R^\ 678. Scholium. The volume of any cylinder is meas- ured by the product of its base by its altitude. This may be proved by the same method as that employed in Prop. II., making the assumption before referred to (674). Exercise 836. Show that, by the following construction, two lines, Jf and iV, can be found such that Jf : iV= 355 : 113. Take AB= 10 units of any convenient length, and on it lay off ^ C = 5, and AD = 3. Draw BE ± to AB and = 1. Join AE, and draw CF, DG, each ± to AB, and ^meeting AE in F, G resp. Take M = 3 AB + AC + CF, and N=AB-hBE+ DG. Then M:N=Sdd'. 113. 336 SOLID GEOMETRY. — BOOK X. CONES. 679. A pyramid is inscribed in a cone when its base is inscribed in tbe base of the cone, and its vertex is that of the cone, the lateral edges of the pyramid thus being ele-, ments of the surface of the cone. 680. A pyramid is circicmscribed about a cone when its base is circumscribed about the base of the cone, and its vertex is that of the cone. 681. The altitude of a cone is the perpendicular distance from its vertex to its base. 682. The slant height of a cone of revolution is equal to the hypotenuse of the generating triangle. 683. The lateral area of a cone is the area of its lateral surface. Proposition III. Theorem. 684. The lateral area of a cone of revolution is meas- ured hy the -product of the circumference of its base hy one half its slant height. Given : S, the lateral area ; C, the circumference of the base ; and L, the slant height of a cone of revolution S-ADF ; To Prove : S is equal to ^ C x L. Inscribe in the base any regular polygon ADF; and upon this polygon as base construct the regular inscribed pyra- CONES. 337 mid S-ADF. If we denote by s the lateral area of the pyramid, by p the perimeter of its base, and by I its slant height, s^\pxl, (549) whatever be the number of lateral faces of the pyramid. Conceive the number of lateral faces to be indefinitely increased by continually doubling the number of sides of the base polygon ; then, since s, p, and I have for limits S, C, L, respectively, s = \c X L. Q.E.D. (536) 685. CoR. 1. If R denote the radius of the base, then C = 2 TT . /? (396), and S = ^(2 7r • R - L) = 7r - R - L. Also, since the area of the base =ir - R^ (398), the total area, T, of the surface of a cone, is expressed by T=7r' R-L + TT' R^=7r- R(L + R). 686. Definition. Similar cones of revolution are gen- erated by similar right triangles revolving about homologous arms as axes. 687. CoR. 2. The lateral areas, or the total areas, of sim- ilar cones of revolution, are to each other as the squares of their altitudes, or as the squares of their radii. This may be proved as was Cor. 3 of Prop. I. 688. Definition. A truncated cone is the portion of a cone intercepted between the base of the cone and a plane cutting its lateral surface. 689. Definition. A frustum of a cone is a truncated cone that has the cutting plane parallel to the base. The section made by the cutting plane is the upper base of the frustum ; the perpendicular distance between its bases is the altitude of the frustum; and the portion of the slant height of the cone that is intercepted between the bases is the slant height of the frustum. Geom.— 22 338 SOLID GEOMETRY.— BOOK X. Proposition TV. Theorem. 690. The lateral area of a frustum of a cone of revolution is measured hy the product of its slant height by half the sum of the circumferences of its hases. Given : S, the lateral surface, C and c, the circumferences of its bases, and L, the slant height of ABC-c, a frustum of a cone of revolution; To Prove : s is equal io ^L{c -\-c). Inscribe in the frustum ABC-c a frustum of a regular pyramid. If we denote its lateral surface by s, the perime- ters of its upper and lower bases by p and P respectively, and its slant height by I, s = i,l{P+p), (550) whatever be the number of lateral faces of the pyramid. Conceive the number of lateral faces of the pyramidal frustum to be indefinitely increased by continually doubling the number of sides of its base polygons ; then. since s, p, F, I, have for limits S, c, C, L, respectively, S = ^L{C-{-C). Q.E.D. (236) 691. Cor. The lateral area of a frustum of a cone of revolution is measured by the product of its slant height by the circumference of a section equidistant from the bases. CONES. 339 For if oa be the radius of a sec- y, -^k .0 tion equidistant from the bases, whose radii are OA and O'A' re- spectively, "^ since oa = ^{OA+ O^A'), (150) .-. circumf. oa = |(circumf. OA -f- circumf. O'a') ; .-. circumf. oax^^'=^ (circumf. 0^ + circumf. 0'a')xAA\ -♦ Proposition V. Theorem. 692. The volume of a cone of revolution is jneasured hy one third the product of its base by its altitude. 8 Given: V, the volume, n, the base, //, the altitude, of a cone of revolution S-ABC; To Prove : V is equal to ^ i? x ^. Inscribe in the cone a regular pyramid, S-ABC; then denoting its volume by v', its base by B', H being also its altitude, F' = i i?' X //, (555) whatever be the number of lateral faces of the pyramid. 340 SOLID GEOMETRY. — BOOK X. Conceive the number of lateral faces of the pyramid to be indefinitely increased ; since F' and B^ have for limits V and B resp., V = \ B X H. Q.E. D. (236> 693. CoR. 1. If R denote the radius of the base ; then, since B^iT' R\ (398) 694. CoR. 2. Similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of their radii. For if R and R' are the radii of two similar cones of revo- lution, H and h' their altitudes, V and r' their volumes, since the generating triangles are similar, (Hyp.) H:H'=R:R'; .-. F: F' = i TT . i2- . iT: i TT • R" • H' = H^ : H'^ = R^ : R'\ 695. Scholium. The volume of any cone is measured by one third the product of its base by its altitude. This may be proved by the same method as that em- ployed in the proof of Prop. V., making the assumption before referred to (674). Exercise 837. If, by the method of Exercise 836, a straight line be found approximately equal to the circumference of a circle one yard in diameter, by what fraction of an inch will the line be too great, taking only two significant figures ? 838. The diameters of the bases of a frustum of a cone are 10 in. and 8 in. respectively, and its slant height is 12 in. Find its lateral area. 839. Find the area of a section of that same cone equidistant from its bases. 840. Find the volume of a cone of revolution the radius of its base being 10 in. and its altitude 20 in. 841. What is the altitude of a similar cone of twice the volume ? CONES. 341 Pkoposition VI. Theorem. 696. The volume of a frustum of a cone of revolu- Hon is measured hy one third the product of its alti- tude hy the sum of the hases of the frustum and a mean proportional between those bases. Given : r, the volume, B and B', the bases, and H, the altitude, of a frustum ABC-c; To Prove : V is equal to \ H {B ^ b'- -{- Vb - B') . Inscribe in the frustum a frustum, ABC-c, of a regular pyramid ; then, denoting its volume by v, its bases by b and b', H being its altitude, 1? = 1 ^ (6 + 6' + V5T6'), (559) whatever be the number of the lateral faces of the frustum. Conceive the number of lateral faces of the inscribed frustum to be indefinitely increased ; since v, b, and b', have for limits F, B, and B', resp., V=ill {B + B' -\-^B- B') . Q.E.D. (236) 697. CoR. If R and R' denote the radii of the bases of the frustum, as B = TT ' R^, B' = TT ' R'% and V^ • B' ==7r - R- R', V=^7r'H {R^ + R"'-^- R-R'). 698. Scholium. The volume of a frustum of any cone is measured by one third the product of its altitude, etc. The same remark applies here as in Arts. 674 and 695. 342 SOLID GEOMETRY. — BOOK X. EXERCISES. NUMERICAL. 842. Find the lateral area and the total area of a cylinder of revolu- .tion whose altitude is 18 in. and the diameter of its bases 12 in. 843. What is the volume of the same cylinder ? 844. The total area of a cylinder of revolution is 700 sq. in., its altitude is 14 in. What is the diameter of a base ? 845. What should be the altitude of a cylinder of revolution, the diameter of which is 5 in., so that the lateral area shall be a square foot ? 846. What should be the diameter of a cylinder of revolution whose altitude is 10 in., so that its total area shall be 500 sq. in.? 847. The altitude of a cylinder of revolution is three times its diame- ter ; the total area is 1200 sq. in. Find the altitude and diameter. 848. The altitudes of two similar cylinders of revolution are as 7 to 5. What is the ratio of their total areas ? Of their volumes ? 849. What formula expresses the total area of a cylinder of revolu- tion whose altitude and radius are equal ? 850. What formula expresses the volume of the same cylinder ? 851. What is the ratio of the volume of the same cylinder to the volume of a cube having the same altitude ? 852. Find the lateral area and the total area of a cone of revolu- tion whose altitude is 15 in., and the diameter of whose base is 12 in. 853. Find the volume of the same cone. 854. The total area of a cone of revolution is 400 sq. in. ; its alti- tude is 10 in. What is the diameter of its base ? 855. What should be the altitude of a cone of revolution whose base has a diameter of 10 in., so that the lateral area may be a square foot? 856. What should .be the radius of the base of a cone of revolution whose altitude is 10 in., so that its total area shall be 100 sq. in. ? 857. The altitude of a cone of revolution is four times the radius of its base ; the lateral area is 500 sq. in. Find the radius and alti- tude. EXERCISES. 343 858. The altitudes of two similar cones of revolution are as 11 to 8. Wliat is the ratio of tlieir total areas ? Of their volumes ? 859. What formula expresses the total area of a cone of revolution whose altitude is equal to tlie radius of its base ? 860. What formula expresses the volume of the same cone ? 861. What is the ratio of the volume of the same cone to the vol- ume of a regular tetrahedron liaving the same altitude ? 862. What should be tlie altitude of such a cone, that its lateral area may be 100 sq. in. ? 863. What should be the altitude of such a cone, that its volume may be 1000 cu. in. ? 864. What is the lateral area and the total area of a frustum of a cone of revolution whose altitude is 20 in., and the diameters of whose bases are 6 in. and 14 in, respectively ? 865. What is the volume of the same frustum ? 866. The diameters of the bases of a frustum of a cone of revolution are 10 in. and 16 in. respectively ; its volume is 575 cu. in. What is its altitude ? 867. How far from the base must a cone, whose altitude is 16 in., be cut by a plane so that the frustum shall be equivalent to one half the cone ? 868. What is the ratio of the lateral surfaces of a right circular cyl- inder and a right circular cone of the same base and altitude, if the altitude is three times the radius of the base ? 869. The diameter of a right circular cylinder is 10 ft., and its alti- tude 7 ft. What is the side of an equivalent cube ? 870. The altitude of a cone of revolution is 15 in., and the radius of its base 5 in. What should be the diameter of a cylinder of revo- lution liaving the same altitude and lateral area ? 871. What should be tlie ratio of the exterior to the interior diame- ter of a hollow cylinder of revolution, so that it shall contain one half the volume of a solid cylinder of the same dimensions ? 872. In order that a cylindrical tank with a depth of 12 ft. may contain 2000 gal., what should be its diameter ? 873. How many cubic inches of iron would be required to make that tank, its walls being one third of an inch thick ? 344 SOLID GEOMETRY. — BOOK X, SPHERES. Proposition VII. Theorem. 699. The area of the surface generated by a straight line revolving ohout an axis in its -plane, is measured by the product of the projection of that line upon the axis by the circumference of the circle whose radius is the perpendicular froin the axis to the mid point of the line. DO Given: ah, the projection upon XF of AB revolving about XY, and OP ± to ^i? at its mid point, and meeting xrin 0; To Prove : Area generated by ^5 is equal io ah x2 it - OP. Draw PD 1. to XY, and AC W to XY. - Since the surface generated hj AB is the lateral surface (576) (689) (292) (285) (237) of a frustum of a cone, area gen. hj AB = ab x 2 tt • PD. Now A ABC is similar to A POD; .: AB: OP = AC: PD; .'. AB X PD = AC X OP = ah X OP; .: AB X 2 TT • PD = ah X 2 77 • OP; i.e., area generated hy AB = ah x 2 tt - OP. q.e.d. If ^j5 meets XY, the surface generated is still a conical surface whose area=a6 x 2 tt • OP, as follows from Prop. III. If ^P is parallel to XY, the surface generated is a cylin- drical surface whose area = a6 x 2 tt • OP, as follows from Prop. I. SPHERES. 345 Proposition VIII. Theorem. 700. The area of the surface of a sphere is meas- ured hy the product of its diameter hy the circum- ference of a great circle. (699) Given : S, the surface of a sphere generated by the revolution of the semicircle ABODE about the diameter AOU, where OA equals R; To Prove : Area of s is equal to AR x 2 tt - R. Inscribe in the semicircle a regular semipolygon AB ... E, of any number of sides, and draw Bb, Co, Bd, Js to ^^. From draw OP _L to AB. Then OP bisects AB (172), and is equal to each of the Js drawn from to the equal chords BC, CD, DE (182). Now area AB = Ab. x 2 tt • OP, area BC = bc x 2 tt - OP, area CD = cd x 2 tt - OP, etc. ; .-. if S' denote the surface generated by the semipolygon, S' = {Ab -{-bc + cd-\- dE) x2Tr'OP = AEx2ir'OP. Conceive the number of sides of the semipolygon to be indefinitely increased. Then, as OP has for limit R, the semipolygon for limit the semicircle, and -S' for limit S, S= AE X2 TT' R. Q.E.D. (236) 701. Cor. 1. Tlie surface of a sphere is equivalent to four great circles. 'Fov 8=2 Rx2tt' R=4.Tv R\ (700) and TT • ii- is the area of a great circle. (398) 346 SOLID GEOMETRY. — BOOK X. 702. Cor. 2. The areas of the surfaces of two spheres are to each other as the squares of their radii or diameters. 703. Definitions. A zone is a portion of the surface of a sphere included between two parallel planes. The altitude of the zone is the perpendicular distance between the paral- lel planes. The bases of the zone are the circumferences of the bounding circles. The zone is called a zone of one base, if one of the parallel planes is tangent to the sphere ; that is, a zone of one base is the surface cut off by a plane. 704. Cor. 1. The area of a zone is measured by the product of its altitude by the circumference of a great circle. For (see diagram for Prop. VIII.) the area of the zone generated by the revolution of the arc 5(7.= 6c x 2 tt • i?. 705. Cor. 2. Zones on the same sphere, or on equal spheres, are to each other as their altitudes. 706. Cor. 3. The area of a zone of one base is meas- ured by the area of the circle whose radius is the chord of the generating arc. For the arc AB generates a zone of one base whose area is Ab X2 IT ' R = Tr ' Abx AE^TT'AB', (398) {^VHQQ AB X AE = AB .) Exercise 874. The diameter of a sphere being 1 ft., what is the area of a great circle of that sphere, and of the sphere itself ? 875. If the area of its surface is 400 sq. in., what is the diameter of the sphere ? 876. What is the ratio of the surfaces of two spheres whose radii are 10 in. and 12 in. respectively ? 877. What is the area of a zone of a sphere 12 in. in diameter, the altitude of the zone being 3 in. ? 878. What fraction of the diameter of a sphere should the altitude of a zone be so as to contain - th of the surface ? SPHERES. 347 Proposition TX. Theorem. 707. The volume of a sphere is measured hy one third the product of its surface hy its radius. Given : V, the volume, and R, the radius, of the sphere whose surface is S; To Prove : V is equal to ^ .s x i?. Conceive a great number of points to be taken on the surface of the sphere, and join them, two and two, by arcs of great circles, so as to form on the surface a network of triangles such as ABC. The planes of these arcs will, by their intersections, form the lateral faces of triangular pyramids, such as 0-ABC, having their vertices at 0, and their bases plane triangles whose vertices coincide with those of the spherical triangles. Since the volume of each pyramid is equal to one third the product of its base and altitude (555), if we denote by V' the sum of these volumes, by b, b', b", ••• the bases, and by h, h', h", ••• the altitudes, we find , V' = ^{b'h-\-b' 'h' + b" . h" -{- ...). Conceive the number of triangles to be indefinitely increased ; then, as V' has for limit V, the sum of the bases has for limit S, and each altitude has for limit R, V=}^S'R. Q.E.D. (236) 348 SOLID GEOMETRY. — BOOK X. 708. Cor. 1. Since s==4:7r'R^, (700) F= i X 47r . i^^ • i? = Itt • 7?^ = i-TT • Z)l 709. Cor. 2. The volumes of spheres are to each other as the cubes of their radii or diameters. For since v^^tt- R^ = ^ttD^ and v' = ^ttR'^ = ^tt - D'% 710. Definitions. A spherical sector is a portion of a sphere generated by a sector of the semicircle that generates the sphere. Thus the revolution of the sector AOB generates the spherical sector 0-ABB'. The revolution of the sector OCD generates the spherical sector 0-GDD^C\ The base of the spherical sector is the zone gener- ated by the circular sector. 711. Cor. 1. The volume of a spherical sector is measured by one third the product of the zone that forms its base, by the radius of the sphere. In the case of the spherical sector 0-A B B', it is manifest, that whatever part the base bb' is of the surface of the sphere, that same part is 0-ABB' of the sphere. In the case of 0-CBD'g', we see that this sector consists of the difference of the spherical sectors 0-CA'c' and 0-DA'b', and its base consists of the difference of the portions of the surface of the sphere that are cut off by C& and DZ)' respec- tively. If V denote the volume of the sector, z the area of the zone, and H its "altitude, then 712. CoR. 2. The volumes of spherical sectors on equal spheres are as the zones that form their bases, or as their altitudes. SPHERES. 349 For since V=^Z' R^^tt- R^-H, Smd V' = l Z' • R' =z^7r' r'^-h, V: r' = Z: Z' = H: H'. 713. Definition. A spherical pyramid is a solid bounded by a spherical polygon as base, whose vertex is the center of the sphere, and whose sides are the sides of the polyhedral angle formed by the planes of the sides of the polygon. 714. CoR. 1. The volume of a spherical pyramid is meas- ured by one third the product of its base by the spherical radius. For it is evident that, whatever portion the base of the pyramid is of the surface of the sphere, the same portion will the volume of the pyramid be of the volume of the sphere. 715. CoR. 2. Spherical pyramids on equal spheres are to each other as their bases. 716. Definitions. A spherical segment is a portion of a sphere included between two parallel planes. The altitude of the segment is the perpendicular distance between the parallel planes ; the bases are the sections of the sphere made by those planes. If one of the bounding planes is tangent to the sphere, the segment is said to be a segment of one base. Exercise 879. What is the volume of a sphere one foot in diameter ? 880. Find the diameter of a sphere whose volume is one cubic foot. 881. Find the surface of a sphere whose volume is one cubic foot. 882. The diameters of two spheres are 10 in. and 12 in. respectively. Find the ratio (1) of their surfaces ; (2) of their volumes. 883. One sphere has twice the volume of another. Find the ratio of the radius of the first to the radius of the second. 884. The altitude of the base of a spherical sector of a sphere 10 in. in diameter is 2 in. ; find the volume of the sector. 885. A spherical pyramid has for base a trirectangular triangle. What fraction is the pyramid of the sphere ? 350 SOLID GEOMETRY. — BOOK X. Proposition X. Theorem. 717. The volume of a spherical segment is meas- ured hy one half the product of the sum of its bases by its altitude, plus the volume of a sphere of luhich that altitude is the diameter* Given: V, the volume of the segment generated by the revo- lution about il/OJV of ABCD, where AD = r, Ba = r', and CD= OC— OD = h; To Prove : V=^h {irr^ + tt?-'-) + i ttIA Join OA and OB. The solid generated hj ABCD consists of the spherical sector generated by GAB, plus the cone generated by OBC, minus the cone generated by GAB; .'. V=^7r7i ■ R' + ^TT • BC' ' GO — ^TT ' Iff • GD (711 ; 692) = i7rj2 7?--/i + (7?'— dc^)GC-{R'— Gb')GD\ (347) = \Tr\2R^'-h -\-R\GC - GD) - (GC^ - off) \ = 1 TT^fS R' — {0C~ 4- GC • GD -j- 01?)\. But since {GC -GD)- = h\ (Hyp.) OC' ~\- GC- GD-{- G~If = |( OC^ 4- cff) — — = \ ll{Tzr -f- 7rr'-) -|- \ irh^. Q.E.D. SPHERES. 351 718. Cor. The volume of a spherical segment of one base is measured by one half the volume of the cylinder haviyig the same base and altitude, plus the volume of the sphere having that altitude for diameter. For when r' = 0, i.e., when the segment has but one base, then F=i7rr7i+i7r/l'l Proposition XI. Theorem. 719. T7^e total surface and the volume of a cylinder circumscribed about a sphere, are respectively to the surface and volume of the sphere as 3 is to 2. Given: A cylinder AD^, circumscribed about a sphere bece' ; To Prove : Surf. AI)' : surf. BECE'=yo\. AI)': vol. BECE' =3 : 2. For we may suppose the sphere and the cylinder to be generated respectively by the revolution about J5C of the semicircle BEC and the rectangle BD that is circumscribed about BEC. Then surf. AI)' = 27r' AB(AB -\- BC) = ^tt • AB^; surf. BECE' = 4 TT • ab' vol. AD' tT'AB 'BC = 4.7r'AB' = 2 TT • AB' YOl.BECE' = \Tr'BC AB . (669) (700) (674) (708) 352 SOLID GEOMETRY, — BOOK X. Hence, making the necessary simplifications, we have surf. AD^ : surf. BECE' = vol. AD' : vol. BECE' = 3:2. q.e.d. This interesting theorem is known as the Theorem of Archimedes, it having been discovered by that celebrated geometer. 720. Scholium. If we have a cone having the same base and altitude as the cylinder circumscribed about the sphere, then, since the volume of such a cone is ^ir - AB • BC = J IT • AB , we obtain the relation : cylinder : sphere : cone = 3:2:1. EXERCISES. NUMERICAL. 886. Find the area of the surface of a sphere whose radius is 3| in. 887. The surface of a sphere is to be 100 sq. in. What radius should be taken ? 888. Two spheres have radii of 9 in. and 5 in. respectively. What is the ratio of the surfaces of those spheres ? Of their volumes ? 889. The areas of the surfaces of two spheres are as 125 to 27. What is the ratio of their diameters ? Of their volumes ? 890. Two parallel planes intersect a sphere of 18 in. radius at dis- tances of 9 in. and 13 in., respectively, from the center. Find the area of the intercepted zone. 891. What is the volume of the spherical sector that has for base the zone just mentioned ? 892. Find the altitude of a zone whose area is 100 sq. in, on the surface of a sphere of 12 in. radius. 893. In the same sphere, what is the altitude of a zone that con- tains one fourth of the surface of the sphere ? 894. What is the volume, of a sphere whose diameter is (1) 1 ft. ; (2) 18 in. ? EXERCISES. 353 895. The surface of a sphere is 64 sq. in. Find its volume. 896. The volume of a sphere is 5 cu. ft. Find its diameter and surface. 897. Find the difference of the volumes of two spheres whose radii are 12 in. and 7 in. respectively. 898. The volumes of two spheres are as 27 to 8. Find the ratio (1) of their diameter ; (2) of their surfaces. 899. The radii of two spheres are as 4 to 5. Find the ratio (1) of their volumes ; (2) of their surfaces. 900. In a sphere whose radius is 6 in., find the altitude of a zone whose area shall be that of a great circle. 901. The area of a zone forming the base of a spherical sector is 60 sq. in. ; the radius of the sphere is 12 in. Find the altitude of the zone and the volume of the sector, 902. The volume of a spherical sector is 25 cu. in. ; the diameter of the sphere is 14 in. Find the area of the zone that forms the base of the sector. 903. The altitude of a cylinder circumscribing a sphere is 5 in. Find the surface and volume of the sphere, 904. The volume of a sphere is one cubic foot. Find the surface of the circumscribing cylinder. 905. The surface and volume of a sphere are expressed by the same number. Find its diameter. 906. Find the volume of a sphere inscribed in a cube whose volume is 1331 cu. in. 907. Find the surface of a cube circumscribed about a sphere whose surface is 150 sq. in, . 908. If a spherical shell have an exterior diameter of 12 in., what should be the thickness of its wall so that it may contain 696.9 cu, in.? 909. If an iron sphere, 6 in. in diameter, weigh n lbs., what will be the weight of an iron sphere whose diameter is 8 in. ? 910. In a sphere 10 in. in diameter, the radius of the lower base of a spherical segment is 8 in. Find the volume of the segment, its altitude being 2 in.' Geom, — 23 354 SOLID GEOMETRY. — BOOK X. THEOREMS. 911. The lateral area of a cylinder of revolution is equal to the area of a circle whose radius is a mean proportional between the altitude and diameter of the cylinder. 912. The lateral areas of the two cylinders generated by revolving a rectangle successively about each of its containing sides, are equal. 913. If the containing sides of the above rectangle are as m is to n, the total areas, and also the volumes, of the cylinders generated, will be as n is to m. 914. If the slant height of a cone of revolution is equal to the diameter of its base, its total area is to that of the inscribed sphere as 9 is to 4. 915. The arms of a right triangle are a and 6. Find the area of the surface generated by revolving the triangle about its hypotenuse. 916. An equilateral triangle revolves about one of its altitudes. What is the ratio of the lateral surface of the generated cone to that of the sphere generated by the circle inscribed in the triangle ? 917. An equilateral triangle revolves about one of its altitudes. Compare the volumes generated by the triangle, the inscribed circle, and the circumscribed circle respectively. 918. A circle of cardboard being given, what is the angle of the sector that must be cut from it so that with the remainder, a cone with a vertical angle of 90° may be formed ? 919. If the diameter of a sphere be divided by a perpendicular plane in the ratio m to /^, the zones thus formed will also be as m to n. 920. The volume of a cylinder of revolution is equal to one half the product of its lateral area by the radius of its base. 921. If the altitude of a cylinder of revolution is equal to the diameter of its base, its volume is equal to one third the product of its total area by the radius of its base. 922. The base of a cone is equal to a great circle of a sphere, and the altitude is equal to a diameter of the sphere. What is the ratio of their volumes ? 923. The volume of a sphere is to that of the inscribed cube as; TT is to 2 -=- Vs. 924. A sphere is to the circumscribed cube as tt is to 6. APPENDIX >>*ic SYMMETRY. p' I. SYMMETRY WITH RESPECT TO A CENTER. 721. Definition. Two points are said to be symmetrical with resj^ect to a third point, if this point bisects the line joining the two points. p. Thus the points P and P' are ^~ symmetrical with respect to O, if the line PP^ is bisected in O. The point is then called the center of symmetry. 722. Definition. Two figures are said to be symmetrical with respect to a point, called their center of symmetry, if every point in the one has its symmetrical point in the other. Thus the figures ABC, A'b'c', are symmetrical with respect to the center 0, if every point in ABC has its symmetrical point in A'b'c'. 723. Definition. In symmetrical figures, sides whose extremities are mutually symmetrical are said to be homolo- gous. Thus ^ C is homologous to A'c', since A is symmetrical with A', and C with c'. Exercise 925. The opposite vertices of a regular polygon of an even number of sides have a common center of symmetry. 926. The opposite vertices of a parallelopiped have a common cen- ter of symmetry. 365 356 GEOMETRY. — APPENDIX, Pkoposition I. Theorem. 724. If tivo polygons are syinjnetrical tuith respect to a center, any two homologoivs sides are equal and parallel, and drawn in opposite directions. :.0..- Given: Two polygons AB"-E, A'b' " e', symmetrical with respect to o ; To Prove: AB, BC, etc., are resp. = and II to A'b', b'c', etc. Since OA = OA', OB = OB', (Hyp.) and Zaob = Z a'ob', (50) Aaob = Aa'ob'; (66) .*. AB = A'B', and Z GAB = Z oa'b' ; (70) .-. AB is II to A'B'-; (110) also AB, a'b', are drawn in opposite directions ; (115) and similarly for BC and b'c', CD and c'd', etc. q.e.d. 725. Cor. 1. Any line mm', intercepted between two homolo- gous sides, AE, A'e', and passing through 0, is bisected in 0. For since AE is II to A'e', the triangles AOM, A'om', are equiangular; and OA=OA' ; .-. 0M= OM' (70). 726. Cor. 2. If two polygons have their sides respectively equal and parallel, and drawn in opposite directions, they have a center of symmetry. For \i AB and A'B' are equal and parallel, and drawn on opposite sides of A A', BB', then A A', BB', are diagonals of SYMMETRY. 357 what could be made a parallelogram (142) ; hence A A', bb', bisect each other in (146). - 727. Scholium. When two polygons are symmetrical with respect to a center, one can be made to coincide with the other by revolving it about the center through two right angles in their common plane. 728. Definition. A figure is symmetrical with respect to a point, if every intercept that passes through the point is bisected there. Thus the figure AB '•• cd is symmetrical with regard to 6, if every intercept, as MM', that passes through 0, is bisected in 0. II. SYMMETRY WITH RESPECT TO AN AXIS. 729. Definition. Two points are said to be symmetrical with respect to a straight line, if this line bisects at right angles the straight line joining the two points. Thus the points P and P' are symmetrical with respect to X Y, if XF bisects PP' at right angles. The line xr is then called the aons of symmetry as regards P and P'. Exercise 927. A circle is symmetrical with respect to what point ? 928. A parallelogram is symmetrical with respect to what point ? 929. A trapezium has no center of symmetry. 930. Every regular polygon of an even number of sides has a cen- ter of symmetry. 931. The axis of symmetry of the extremities of a chord is what line? 932. The axis of symmetry of opposite vertices of a square is what line ? 358 GEO ME TR Y. — APPENDIX. A B 730. Definition. Any two figures are said to be symmet- rical with respect to an axis, if every point in the one has a point in the other symmetrical with respect to that axis. Thus the figures ABC, A'b'c', are symmetrical with respect to ^ XY, if corresponding to every point in ^5 C there is a point in A'b'c' symmetrical with respect to X Y. 731. Scholium. It is obvious that, if the portion of the plane above XT be revolved about XF as an axis, till it coin- cides with the portion of the plane below X Y, the figure ABC will coincide with a'b'c', since the homologous points are at equal distances from X F. 732. Definition. A plane fig- ure is symmetrical loith respect to an axis, if the axis divides the figure into two symmetrical fig- ures. Thus the figure AB --- B' is symmetrical with regard to X F if its homologous points are symmetrical with respect to XY. Exercise 933. How many axes of symmetry may a circle have ? What common axis of symmetry have two circles ? 934. An isosceles triangle is symmetrical with respect to which altitude ? 935. An equilateral triangle has how many axes of symmetry ? 936. How many axes of symmetry may be drawn for (1) a square ? (2) A rhombus ? (3) A regular pentagon? (4) A regular hexa- gon ? (5) A regular polygon of 2 w sides ? (6) Of 2 w + 1 sides ? 937. In a quadrilateral ABCD, AB=AD, and GB = CD. Show that AC is an axis of symmetry, and is perpendicular to BD. SYMMETRY. 359 Proposition II. Theorem. 733. If a figure is symmetrical with respect to tivo axes at right angles to each other, it is also syimnet- rical with respect to their intersection as center. F 1 E G ,J^. Q \^ '■■■■.. ;s D M O '>y C A 7^ Given: AB •" II, symmetrical with respect to axes XX', YY', intersecting in 0; To Prove : AB •-• iiis symmetrical with respect to as center. From P, any point in the perimeter of the figure, draw PQIi ± to YY', and through E draw RSP' ± to XX'. Join OP, OP', and QS. Since p Q = Q r, (Hyp.) SiiidOS=QR, (13G) PQ=OS. (Ax. 1) Also PQ is II to OS; • (106) .-. OP is II and = to QS. (13G) In the same way it may be proved that OP' is II and = to QS; .'. pop' is 3i straight line, and is bisected in o; i.e., is the center of symmetry of AB •" ii. q.e.d. 734. Scholium. The axes xx', YY', evidently divide the figure into four equal parts. Any one of these parts may be made to coincide with either of the adjacent parts by revolving it about one of the axes, or may be made to coin- cide with the opposite part by revolving it, in the plane of the figure, through two right angles. 360 GEOMETRY.— APPENDIX. SYMMETRICAL POLYHEDRONS. I. SYMMETRY WITH RESPECT TO A CENTER. 735. Definition. Two polyhedrons are said to be symmet- rical with respect to a center, when each vertex of the one has its symmetrical vertex on the other polyhedron. Thus in the polyhedrons S-A B c, s'-A'b'c', if the lines join- ing the vertices A and A', B and J5', etc., all pass through the same point 0, and are bisected in that point, the polyhedrons are said to be symmetrical with respect to 0, which is called their center of symmetry. 736. Cor. If two polyhedrons are symmetrical with respect to a center, their homologous faces are severally equal, their dihedral angles are severally equal, their polyhedral angles are symmetrical, and the polyhedrons are equivalent. 737. Definition. A polyhedron A B cn-A' is said to be symmetrical tvith respect to a center O, if its vertices, taken two and two, are symmetrical with regard to 0; i.e., if A A', BB', etc., are each bisected in the same point 0. 738. Cor. A polyhedron, in order to have a center of symmetry, must have an even number of edges, must have its homologous edges equal and parallel, its homologous plane angles and dihedral angles equal, and its homologous polyliedral angles symmetrical. SYMMETRICAL POLYHEDRONS. 361 II. SYMMETRY WITH RESPECT TO AN AXIS. 739. Definition. Two polyhedrons are said to be sym- metrical ivith respect to an axis, when this axis is an axis of symmetry for the corresponding vertices of the two polyhedrons. 740. Definition. A polyhedron is said to be symmetrical tuith respect to an axis, when this line is an axis of sym- metry for the corresponding vertices of the polyhedron, taken two and two. Thus the polyhedron AB'-'E' is sym- metrical with respect to the axis 00 when this axis is an axis of symmetry for each of the pairs of vertices, A and D, B' and E', etc. o D' III. SYMMETRY WITH RESPECT TO A PLANE. 741. Definition. Two points are said to be symmetrical with respect to a plane, when the plane is perpendicular to, and bisects, the straight line joining the two points. Exercise 938. Every intercept between two opposite faces of a polyhedron having a center of symmetry, and passing through that center, is bisected there. 939. Every prism whose bases are polygons symmetrical with respect to a point, has a center of symmetry. 940. Where is the center of symmetry of a parallelopiped ? 941. A right prism whose bases are symmetrical with respect to a center, has an axis of symmetry. 942. A rectangular parallelopiped has three axes of symmetry. 943. How many axes of symmetry has a cube ? 944. A regular pyramid having an even number of lateral faces, has an axis of symmetry. 362 GEOME TRY. — A PPENDIX. 742. Definition. Two polyhedrons are said to be sym- metrical ivitli respect to a plane, when this plane is a plane of symmetry for the corresponding vertices of the polyliedrons, taken two and two. 743. Cor. hi order that two polyhedrons may he symmetri- cal with respect to a plane, their homologous faces must he severally equal, their dihedral angles must he equal, their polyhedral angles must he symmetrical, and the ijolyhedrons must he equivalent. 744. Definition. A polyhedron is said to be symmetri- cal with respect to a plane, when the plane divides it into two polyhedrons symmetrical with respect to that plane. MAXIMA AND MINIMA. 745. Definition. A magnitude is said to be a maximum or minimum according as it is the greatest or least of a given class. Thus the diameter of a circle is a maximum among all inscribed straight lines; and, among all the straight lines drawn from a given point to a given straight line, the per- pendicular is the minimum. 746. Definition. Isoperimetric figures are those which have equal perimeters. Exercise 945. How many planes of symmetry has any right prism ? 946. How many has a parallelopiped ? 947. Has a cylinder a plane of symmetry ? A sphere ? 948. Show that, if a polyhedron has two planes of symmetry that are perpendicular to each other, their common intersection is an axis of symmetry. 949. Show that, if a polyhedron has three planes of symmetry, the point common to the three planes is a center of symmetry. MAXIMA AND MINIMA 363 Proposition I. Theorem. 747. Of all triangles having two given sides, that in which tlxese sides are perpendicular to each other has the maximum area. Given: In triangles ABC, A'BC, AB eqnal to A^B, BC equal to B c, but only A B perpendicular to BC ; To Prove : Triangle ABC is greater tlian triangle a'b c. Draw A' I) ± to BC. Since AB =z A'B, (Hyp.) but A'B > A'B, (93) the altitude AB ^ the altitude A'D; .: Aabc> AA'BC. Q.E.D. (333) Exercise 950. Two lines, whose lengths are a and b respectively, being given, find the length of the third line that will form with them the maximum triangle. 951. Of all triangles of given base and area, the isosceles has the greatest vertical angle. 952. Of all triangles of given base and vertical angle, the isosceles is the greatest. 953. Of all triangles of given altitude and vertical angle, the isos- celes is the least. 954. Of all triangles of given base and vertical angle, the isosceles has the greatest perimeter. 955. Divide a given arc into two parts such that the sum of the chords subtending them shall be a maximum. 364 GEOMETRY. — APPENDIX. Proposition II. Theorem. 748. Of all equivalent triangles on the same base, the isosceles triangle has the Tuinimum perimeter. Given : In ^ J5 C, A^B C, equivalent triangles on the base B C, AB equal to AC ; To Prove : AB -\- AC -\- BC < A^B + A'C-\- BC. Produce B A ^o that AD = AB= AC; join DC and AA\ and produce ^ J.' to ^. Since ABC and A^BC are equivalent, and on the same base, (Const.) ABC and A^B C have the same altitude, (332) (since otherwise they would not be equivalent;) .-. AA^El^ II to BC; .'. EB = EC; (147) .-. ^^is± to DC; (97) .-. A^D=.A'C. (96) But BD < A'B-{- A'D; i.e., AB -}-AC< A'B-{- a'c. .'. AB -{- AC -{- BC < A'B -\- A'C-\- BC. Q.E.D. 749. CoR. Of all equivalent triangles, that which is equi- lateral has the least perimeter. MAXIMA AND MINIMA. 365 Proposition III. Theorem. 750. Of all isoperiinetric triangles on the same base, that which is isosceles has the maximuTn area. Given: Two isoperimetric triangles ABC, A'BC, and AB equal to A c; To Proves Triangle ABCi^ greater than triangle A^BC. Draw AD 1. to BC, and A^E II to BC, and meeting AD pro- duced if necessary in E; also join EB, EC. Since A ^'^C, z 5 (7, have the same altitude, (Const.) A A'BC=o=AEBC. (332) But Z 5 (7 is an isosceles A ; (96) .-. A'B-\-A'C>EB-]-EC; (748) .'. AB-^AC>EB,-\-EC, (Hyp.) (since AB -{- AC= a'b -f A'c;) .'. AB>EB; (Ax. 7) .: AD>ED; (99) .'. A ABOAebcov AA'BC. Q.E.D. (333) 751. CoR. Of all isoperiinetric triangles, that which is equilateral has the maximum area. Exercise 956. Find the area of the maximum triangle whose perim- eter is n feet. 957. Of all triangles inscribed in a given circle, the equilateral has the greatest perimeter. 366 GEOMETR Y. — APPENDIX. Proposition IV. Theorem. 752. Of all polygons formed of sides all given hut one, the majciinum can be inscribed in a semicircle having the undetermined side as diameter. Given: ABCDEF, the maximum polygon, having the given sides AB, BC, CD, DE, EF, and an undetermined side AF ; To Prove : ABCDEF can be inscribed in a senricircle. Join any vertex, as C, with A and F. Then the triangle A CF must be the maximum of all triangles formed with the given sides A G and CF. For otherwise, by changing /. ACF, leaving AC and CF unchanged, we could increase A A CF, leaving the rest of the polygon unchanged. That is, the whole polygon AB--- F would be increased, which it cannot be, since AB •" F is a maximum. Hence A ACF must be the maximum triangle formed with the given sides A C, CF; .'. Z ACF is a rt. Z, (747) (since otherwise A ACF would not be a maximum ;) .-. C is on the semicircumf. whose diam. is AF; .: each vertex of ABCDEF is on that semicircumf. q.e.d. Exercise 958. Of all parallelograms of a given base and area, the rectangle has the least perimeter. 959. Of all rectangles of given area, the square has the least perim- eter. MAXIMA AND MINIMA. 367 Proposition Y. Theorem. 753. Of all polygons forfned with given sides, that which can he inscribed in a circle is the maximum. Given: Two mutually equilateral polygons, ABODE and A'b'c'd'e', of which only ABODE can be inscribed in a circle; To Prove : ABODE is greater than A'b'o'd'e'. Draw the diameter AF, and join FO, FD. Upon 0'd'{= OD) construct Af'o'd'= AFOD, and join^'i^'. Since polygon ABOF is inscribed in a semicircle, ABOF> A'B'O'F'; similarly AEDF> a'e'd'f' ; .'. ABCFDE> A'B'O'F'd'e'; (Ax. 4) .-. ABOFDE-FOD> A'b'O'F'D'e'-F'O'D' ; (Ax. 5) Q.E.D. (752) .e., ABODE> a'b'o'd'e'. 754. Scholium. The area of the inscribed polygon will be the same in whatever order the sides are arranged. For these sides are chords which cut off equal segments, in whatever order they occur; and the polygon is the difference between the circle and these segments. Exercise 960, Of all rectangles that can be inscribed in a given circle, the greatest is a square. 368 GEOMETRY. — APPENDIX. Proposition VI. Theorem. 755. Of isoperimetric polygons of the same numher of sides, the maxiTnum is a regular polygon. Given: A BCD, the maximum of isoperimetric polygons of n To Prove : ABCD is si, regular polygon. The polygon ABCD must be equilateral. For if any two of its sides, BA,BC, were unequal, then upon ^(7 as base we could construct an isosceles triangle B'AC, having the sum of its sides, B'A, B'c, equal to BA-\-BC. The triangle ^Uc would be greater than BAC (748), and therefore the polygon AB'CJ) would be greater than the maximum polygon ABCD. But this is impossible; hence ABCD must be equilateral. It can also be inscribed in a circle (753) ; hence it is a regular polygon (380). Exercise 961. Of all triangles having the same vertical angle, and whose bases pass through a given point, that whose base is bisected by the given point is least. 962. The rectangle contained by the segments of a line is a maxi- mum when the segments are equal. 963. Through a given point within a given circle, draw the maxi- mum and minimum chords that pass through that point. 964. From a given point without a given circle, draw the secant whose outer segment is a minimum. What about its inner segment ? MAXIMA AND MINIMA. 369 Proposition VII. Theorem. 756. Of isoperimetric regular polygons, that having the greatest nufnher of sides is the maximum. Given : A regular polygon ABC, of n sides ; To Prove : ABC is less than an isoperimetric regular polygon of n + 1 sides. In one of the sides, as AB, take any point P. We may now regard the given polygon as an irregular polygon of n -4-1 sides, in which the sides AP, PB, make with each other an angle equal to two right angles. This irregular polygon is less than the regular polygon of the same perimeter and having n-{-l sides (755) ; that is, a regular polygon of n sides is less than the isoperimetric regular polygon of n -{-1 sides, q.e.d. 757. Cor. The circle contains a maximum area ivithin a given perimeter. Exercise 9()o. On the circumference of a given circle find the point such that the sum of the squares of its distances from two given points without the circle shall be a minimum. 966. Of two given circles, one lies wholly within the other. Find the maximum and minimum chords of the outer that are tangent to the inner circle. 967. A line ABC is perpendicular to an indefinite line CM. Find the point P in CiHf at which the angle APE is a maximum. 968. Find a point within a quadrilateral such that the sum of the lines drawn from that point to the vertices shall be a minimum. 370 GEOMETR Y. — APPENDIX. Proposition YIII. Theorem. 758. Of two regular -polygons having equal areas, that having the greater nujnher of sides has the less perimeter. Given : P and Q, regular polygons of tlie same area, but Q with the greater number of sides ; To Prove : Perimeter of Q is less than perimeter of P. Let i? be a regular polygon having the same perimeter as Q and the same number of sides as P. Then since Q> E, (756) but Q = P, (Hyp.) P>Ji; .'. the perimeter ofR< the perimeter of P; (346) .-. the perimeter of Q < the perimeter of P. q.e.d. 759. Cor. TTie circumference of a circle is less than the perimeter of any polygon of equal area. Exercise 969. Through a given point within a given angle draw the intercept that cuts off the triangle of maximum area. 970. Through a point of intersection of two circles draw that inter- cept between the two circumferences which is a maximum. 971. In a given line find a point such that the sum of its distances from two given points without the line, and on the same side of it, shall be a minimum. 972. In a given line find the point such that the tangents drawn from it to a given circle contain the maximum angle. INDEX OF DEFINITIONS. PAGE PAGE Alternation, 122 Altitude of prism, 262 Angle, 15 of pyramid. 275 acute, 17 Analysis, 71, 105 adjacent, 16 Antecedent, 113 alternate, 51 Apothem, 199 at center of circle, 128 Arc, 78 of regular polygon. 200 major and minor, 85 complementary, 17 Area, 165 corresponding. 51 Axiom, 19 difference of, 16 Axis of circle, 301 dihedral, 240 of cone, 297 equal, 16 of cylinder. 294 exterior, 51 of symmetry. 357 face. 251 Base of triangle. 28 including. 16 of polygon. 165 inscribed. 131 Bisector, 38 interior. 61 Center of polyhedrons, 360 oblique, 17 of regular polygon, 199 obtuse. 17 of symmetry of plane figures, 355 of the line and plane, 249 Chord, 78 plane, 15 Circle, 78 polyhedral, 251 of sphere. 301 right. 17 great or small, 301 spherical. 309 sector of, 210 straight, 16 segment of. 210 sum of. 16 Circumference, 15, 78 supplementary. 17 Commensurable, 112 trihedral, 251 Common measure. 112 vertical. 16 Complement, 17 Altitude of triangle. 74 CONCYCLIC, 160 of cone. 297 Cone, 297 of cylinder. 294 altitude of. 297, 33(> of frustum of pyramid, 275 base of. 297 of polygon. 165 circular, 297 371 372 INDEX OF DEFINITIONS. PAGE PAGE Cone, axis of circular, 297 Enunciation, 20 frustum of, 337 Figure, 13 lateral area of, 330 equal, 29 lateral surface of, 297 equivalent, 165 of revolution, 297 geometrical, 13 similar. 337 isoperimetric. 362 slant height of, 336 plane. 13 truncated. 337 Frustum of cone, 337 vertex of, 297 altitude of, 337 Conical surface, 297 of pyramid. 275 Consequent, 113 altitude of, 275 Constant, 117 Generatrix, 293, 297 Construction, 21 Geometrical concepts, 12 Corollary, 18 Geometry, 12 Cube, 273 of space, 225 Curve, 12 of three dimensions, 225 Cylinder, 294 plane, 13 axis of, 294 solid. 225 circular. 294 Hexahedron, 261 lateral area of, 332 Homologous, 144 oblique, 294 Hypotenuse, 35 of revolution, 294 Hypothesis, 20 similar, 333 Icosahedron, 261 radius of, 294 Incommensurable, 112 right, 294 Intercept, 62 Cylindrical surface, 293 Inversion, 122 Demonstration, 21 Limit, 117 Diagonal, 61 Line, 12 Diagram, 13 broken. 12 Diameter, 78 curved. 12 Dihedral angle, 240 of centers. 92 equal. 240 parallel. 51 plane angle of, 240 parallel to plane. 235 right. ^0 perpendicular to plane. 227 Direction, 14 straight. 12 of parallel lines, 57 Locus, 102 Directrix, 293, 297 LUNE, 324 Distance, central. 92 Magnitude, 13 equal, 15 equal, 13 of point from line. 47 Maximum, 362 of point from plane. 230 Median, 74 on sphere, 303 Minimum, 362 polar. 303,304 Nappe, upper and lower 297 Dodecahedron, 261 Numerical measure. 112 Element of surface, 293 Octahedron, 261 INDEX OF DEFINITIONS, 373 Parallel lines, PAGE 51 Proportion, continued, PAGE 124 planes, 235 Proportional, 114 Parallelogram, 61 fourth, 115 Parallelopiped, 2(!7 inversely. 152 rectangular, 267 mean. 119 right, 267 third, 119 Perimeter, 74 Pyramid, 275 Plane, 13 altitude of. 275 Point, 12 frustum of. 275 Polar distance, 303 regular, 275 triangle. 313 slant height of, 275 Pole, 301 triangular, etc., 275 Polygon, 59 truncated. 275 center of regular, 1<)9 Quadrilateral, 61 circumscribed. 101 Quantity, 111 convex, 59 Radius of circle, 15 inscribed. 101 of regular polygon, 199 radius of regular. 199 of sphere, 300 regular, 198 Ratio, 113 similar, 144 extreme and mean, 158 spherical. 310 inverse or reciprocal, 152 diagonal of. 311 Rectangle, 61 stellate. 215 Rhombus, 61 Polyhedral angle, 251 Scholium, 19 edge, face, A^ertex, 251 Secant, 90 equal or symmetrical, 255 Sector, 210 face angle of. 251 similar. 210 Polyhedron, 201 Segment, harmonic. 145 convex, 261 of circle. 210 equal, equivalent, 262 of a line. 140 regular. 286 of sphere, 349 similar. 262 similar, 210 volume of, 261 Semicircle, 80 Postulate, 19 Sbmicircumferbnce, 80 Prism, 262 Solid, 11 altitude of. 262 Space-concepts, 12 edge, faces. 262 Sphere, 300 lateral or convex surface, 2()3 tangent to plane, "307 regular, 262 tangent to sphere. 307 right. 262 Spherical angle. 309 right section of, 262 degree. 325 truncated, 2(56 excess. 316 Problem, 18 polygon. 310 Projection, 179, 247 pyramid, 349 Proportion, 114 sector. 348 374 INDEX OF DEFINITIONS. PAGE PAGE Spherical segment, 349 Triangle, obtuse, 29 triangle, 311 right, 29 Square, 61 scalene, 28 Supplement, 17 spherical, 311 Surface, 11 birectangular, 316 Symmetry, 355-362 symmetrical, 317 Synthesis, 71 trirectangular, 316 Tangent, 90 supplemental, 315 circles. 92 Trihedral angle, 261 spheres, 307 Ungula, 324 Tetrahedron, 261 Unit of area, 168 Theorem, 18 of length, 111 converse, 24 of volume. 273 Transversal, 51 Variable, 117 Trapezium, 61 Vertex, 15 Trapezoid, 61 Zone, 346 Triangle, 28 altitude oi, 346 acute, 29 bases of. 346 equilateral, 28 of one base, 346 isosceles. 28 PRIMERS IN SCIENCE, HISTORY, AND LITERATURE. Flexible cloth, ISino. 35 cents each. SCIENCE PRIMERS. Edited by Professors HUXLEY, ROSCOE, and BALFOUR STEWART. Introductory. Prof. T. H. Huxley, F.K.S. Chemistry. Prof. H. E. Roscoe, P.R.S. Physics. Prof. Balfour Stewart, F.R.S. Physical Geography. Prof. A. Gbikie, F.R.S. Geology. Prof. A. Geikie, P.R.S. Physiol og-y. M. Foster, M. D., F.R.S. Hygiene. R. S. 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