JKSfl ■■'■'■'- ¥l!\ ^ Cornell University Library arV19505 Treatise on plane and solid geometi 3 1924 031 307 733 olin,anx Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031307733 ICIEOTIG EDUCATIONAL SERIES. TREATISE PLANE AND SOLID GEOMETRY COLLEGES, SCHOOLS, AND PRIVATE STUDENTS. WEITTEN rOB TME MATHEMATICAl COURSE OF JOSEPH RAY, M.D., BY ELI T. TAPPAN, M.A., PBOFBSSOB OF MATHEMATICS, MT. AUBOBN INSTITUTE. VAN ANTWERP, BRAGG & CO., 137 WALNUT STREET, 28 BOND STREET, CINCINNATI. NEW YOKK. Ra k<- A Thorough and Progressive Course in Arithmetic, Algebra, and the Higher Mathematics. ithinetic. Higher Arithmetic. Test Examples in Arithmetic. Trimary Aritlimetic Intellectual Arithmetic. Rudiments of Arithmetic. Practical Aritlimetic. New Elementary Algebra. New Higher Algebra. Plane and Solid Geometry. By Eli T. Tappan, A.M., Pres't Kenyan College. 12mo, doth, 276 pp. Oeometry and Trigonometry. By Eli T. Tappan, A.M. Prei't Kenyan College. Sra, sheep, 420 pp. Analytic Geometry. By Geo. H. Howisok, A.M., Prof, in Mass. Institute of Technology. Treatise on Analytic Geometry, especially as applied to the Properties of • Conies ; including the Modern Methods of Abridged Notation. Elements of Astronomy. By S. H. Peaeody, A.M., Prof, of Physics and Civil Engineering, Amherst College. Handsomely and profusely illustrated. 8to, sheep, 336 pp. K E'VS. Ray's Arithmetical Hey ( To Intellectual and Practical) ; Hey to Ray's Higher Arithmetic ; Key to Ray's New Elementary and Higher Algebras. The Publishers furnish Descriptive Circulars of the ahove Mathe- tnaHcal TextSoohs, with Prices and other information concerning them. Entered according to Act of Congress, in the year 1868, by Sargent, Wilson & HlNKLE, in tlie Clerk's Ofiice of the District Court of the United States for the Sontheru District of Ohio. PREFACE. The science of Elementary Geometry, after remaining nearly stationary for two thousand years, has, for a century past, been making decided progress. This is owing, mainly, to two causes: discoveries in the higher mathematics have thrown new light upon the elements of the science ; and the demands of schools, in all enlightened nations, have called out many works by able mathematicians and skillful teachers. Professor Hayward, of Harvard University, as early as 1825, defined parallel lines as lines having the same direc- tion. Euclid's definitions of a straight line, of an angle, and of a plane, were based on the idea of direction, which is, indeed, the essence of form. This thought, employed in all these leading definitions, adds clearness to the science and simplicity to the study. In the present work, it is sought to combine these ideas with the best methods and latest discoveries in the science. By careful arrangement of topics, the theory of each class of figures is given in uninterrupted connection. No attempt is made to exclude any method of demonstration, but rather to present examples of all. The books most freely used are, "Cours de gfeomfetrie el6mentaire, par A. J. H. Vincent et M. Bourdon ; " " G6- om§trie th^orique et pratique, etc., par H. Sonnet;" "Die (iii) IV PEEFACE. reine elementar-mathematik, von Dr. Martin Ohm ; " and "Treatise on Geometry and its application to the Arts, by Rev. D. Lardner." The subject is divided into chapters, and the articles are numbered continuously through the entire work. The con- venience of this arrangement for purposes of reference, has caused it to be adopted by a large majority of writers upon Geometry, as it had been by writers on other scien- tific subjects. In the chapters on Trigonometry, this science is treated as a branch of Algebra applied to Geometry, and the trig- onometrical functions are defined as ratios. This method has the advantages of being more simple and more brief, yet more comprehensive/ than the ancient geometrical method. For many things in these chapters, credit is due to the works of Mr. I. Todhunter, M. A., St. John's College, Cam- bridge. The tables of logarithms of numbers and of sines and tangents have been carefully read with the corrected edi- tion of Callet, with the tables of Dr. Schron, and with those of Babbage. ELI T. TAPPAN. Ohio XJniveksitt, Jan. 1, 1868. CONTENTS. PAGE. PART FIRST.— mTEODUCTORY. CHAPTER I. PEELIMINARY. Logical Teems, 9 G-ENpRAL Axioms, 11 Ratio and Proportion, 12 CHAPTER II. THE SUBJECT STATED. Definitions, 17 Postulates of Extent and of Form, 19 Classification of Lines, 22 Axioms op Direction and of Distance, .... 23 Classification op Buepaces, 24 Division of the Subject, 26 PART SECOND.— PLANE GEOMETRY. CHAPTER III. STRAIGHT LINES. Problems, 28 Beoken Lines, 31 Angles, 32 VI CONTENTS. PAGE, Perpendicular and Oblique Lines 38 Parallel Lines, . 43 CHAPTEE IV. CIRCUMFERENCES. General Properties op Circumferences, ... 52 Arcs and Eadii, 53 Tangents, 58 Secants, . 59 Chords, 60 Angles at the Center, 64 Intercepted Arcs, 72 Positions op Two Circumperences, 78 CHAPTER V. TRIANGLES. General Properties of Triangles, .... 85 Equality of Triangles, 93 Similar Triangles, .101 CHAPTER VI. QUADRILATERALS. General Properties op Quadrilaterals, . . . 119 Trapezoids, . . . . ' 122 Parallelograms, . . . . 123 Measure op Area, 128 Equivalent Surfaces, I35 CHAPTER VII. POLYGONS. General Properties op Polygons, I43 Similar Polygons, I47 CONTENTS. vu Regular Polygons, 151 isopf.rimetry, 159 CHAPTER VIII. CIRCLES. Limit of Inscribed Polygons, 164 Rectification of the Circumference, 166 Quadrature of the Circle, 172 PART THIED.— GEOMETRY OF SPACE. CHAPTER IX. STRAIGHT LINES AND PLANES. Lines and Planes in Space, 177 Diedral Angles, 185 Parallel Planes, 190 Triedrals, 195 Polyedrals, 209 CHAPTER X. POLYEDRONS. Tetraedrons, 213 Pyramids, 222 Prisms, 226 Measure of Volume, 232 Similar Polyedrons, 239 Regular Polyedrons, 241 CHAPTER XI. SOLIDS OF REVOLUTION. Cones, 247 Cylinders, 249 Viu CONTENTS. PAGE. Spheres, 250 Spherical Areas, 261 Spherical Volumes, 270 Mensuration, 276 PART FOURTH.— TEIGONOMETRY. CHAPTEK XII. PLANE TRIGONOMETRY. Measure or Angles, 277 Functions op Angles, 279 Construction and Use op Tables, 296 Right angled Triangles, 302 Solution op Plane Triangles, 304 CHAPTEE XIII. SPHERICAL TRIGONOMETRY. Spherical Arcs and Angles, 314 Right angled Spherical Triangles, 324 Solution op Spherical Triangles, 329 CHAPTER XIV. LOGARITHMS. Use op Common Logarithms, 334 TABLES. Logarithmic and Trigonometric Tables, . . . 345 ELEMENTS GEOMETRY CHAPTER I.— PRELIMINARY. Article 1. Before the student begins the study of geometry, he should know certain principles and defini- tions, which are of frequent use, though they are not peculiar to this science. They are very briefly pre- sented in this chapter. LOGICAL TERMS. 3. Every statement of a principle is called a Propo- sition. Every proposition contains the subject of which the assertion is made, and the property or circumstance asserted. When the subject has some condition attached to it, the proposition is said to be conditional. The subject, with its condition, if it have any, is the Hypothesis of the proposition, and the thing asserted is the Conclusion. Each of two propositions is the Converse of the other, when the two are such that the hypothesis of either is the conclusion of the other. (9) 10 ELEMKNTS OF GEOMETKY. 3. A proposition is either theoretical, that is, it de- clares that a certain property belongs to a certain thing; or it is practical, that is, it declares that something can be done. Propositions are either demonstrable, that is, they may be established by the aid of reason; or they are indemon- strable, that is, so simple and evident that they can not be made more so by any course of reasoning. A Theorem is a demonstrable, theoretical proposition. A Problem is a demonstrable, practical proposition. An Axiom is an indemonstrable, theoretical propo- sition. A Postulate is an indemonstrable, practical propo- sition. A proposition which flows, without additional reason- ing, from previous principles, is called a Corollary. This term is also frequently applied to propositions, the demonstration of which is very brief and simple. 4. The reasoning by which a proposition is proved is called the Demonstration. The explanation how a thing is done constitutes the Solution of a problem. A Direct Demonstration proceeds from the premises by a regular deduction. An Indirect Demonstration attains its object by showing that any other hypothesis or supposition than the one advanced would involve a contradiction, or lead to an impossible conclusion. Such a conclusion may be called absurd, and hence the Latin name of this method of reasoning — reductio ad absurdum. A work on Geometry consists of definitions, proposi- tions, demonstrations, and solutions, with introductory or explanatory remarks. Such remarks sometimes have the name of scholia. GENERAL AXIOMS. IJ 5. Eemark. — The student should learn each proposition, so as to state separately the hypothesis and the conclusion, also the condition, if any. He should also learn, at each demonstration, whether it is direct or indirect; and if indirect, then what is the false hypothesis and what is the absurd conclusion. It is a good exercise to state the converse of a proposition. In this work the propositions are first enounced in general terms. This general enunciation is usually followed by a particu- lar statement of the principle, as a fact, referring to a diagram. Then follows the demonstration or solution. In the latter part of the work these steps ^re frequently shortened. The student is advised to conclude every demonstration with the general proposition which he has proved. The student meeting a reference, should be certain that he can state and apply the principle referred to. GENEKAL AXIOMS. 6. Quantities which are each equal to the same quarir tity, are equal to each other. 7. If the same operation he performed upon equal quantities, the results will be equal. For example, if the same quantity be separately added to two equal quantities, the sums will be equal. 8. If the same operation be performed upon unequal quantities, the results will be unequal. Thus, if the same quantity be subtracted from two unequal quantities, the remainder of the greater will exceed the remain'der of the less. 9. The whole is equal to the sum of all the parts. 10. The whole is greater than a part. EXERCISE. 11. What is the hypothesis of the first axiom ? Ans. If sev- eral quantities are each equal to the same quantity. 12 ELEMENTS OF GEOMETRY. What is the subject of the first axiom ? Ans. Several quan- tities. What is the condition of the first axiom ? Ans. That they are each equal to the same quantity. What is the conclusion of the first axiom ? Ans. Such quan- tities are equal to each other. Give an example of this axiom. RATIO AND PROPORTION 12. All mathematical investigations are conducted by comparing quantities, for we can form no conception of any quantity except by comparison. 13. In the comparison of one quantity with another, the relation may be noted in two ways : either, first, how much one exceeds the other; or, second, how many times one contains the other. The result of the first inethod is the difference be- tween the two quantities ; the result of the second is the Ratio of one to the other. Every ratio, as it expresses " how many times " one quantity contains another, is a number. That a ratio and a number are quantities of the same kind, is fur- ther shown by comparing them; for we can find their sum, their difference, or the ratio of one to the other. When the division can be exactly performed, the ratio is a whole number; but it may be a fraction, or a radical, or some other number incommensurable with unity. 14. The symbols of the quantities from whose com- parison a ratio is derived, are frequently retained in its expression. Thus, The ratio of a quantity represented by a to another represented by b, may be written , . A ratio is usually written a : b, and is read, a is to b. RATIO AND PROPORTION. 13 This retaining of the symbols is merely for conven- ience, and to show the derivation of the ratio; for a ratio may be expressed by a single figure, or by any other symbol, as 2, m, \/3, or tt. But since every ratio is a number, therefore, when a ratio is thus expressed by means of two terms, they must be understood to represent two numbers having the same relation as the given quantities. The second term is the standard or unit with which the first is compared. So, when the ratio is expressed in the form of a frac- tion, the first term, or Antecedent, becomes the numera- tor, and the second, or Consequent, the denominator. 15. A Proportion is the equality of two ratios, and is generally written, a : h : : c : d, and is read, a is to 6 as c is to d, but it is sometimes written, a : b^^c : d, or It may be, h^d' all of which express the same thing: that a contains h exactly as often as c contains d. The first and last terms are the Extremes, and the second and third are the Means of a proportion. The fourth term is called the FoURTH Proportional of the other three. A series of equal ratios is written, a : b :: c : d : : e -.f, etc. When a series of quantities is such that the ratio of each to the next following is the same, they are written, a : b : c : d, etc. I 4 ELEMENTS OF GEOMETRY. Here, each term, except the first and last, is both an- tecedent and consequent. When such a series consists of three terms, the second is the Mean Proportional of the other two. 16. Proposition. — The product of the extremes of any proportion is equal to the product of the means. For any proportion, as a : b : : c : d, is the equation of two fractions, and may be written, a c b~d- Multiplying these equals by the product of the denom- inators, we have (7) aXd=bXc, or the product of the extremes equal to the product of the means. 17. Corollary. — The square of a mean proportional is equal to the product of the extremes. A mean pro- portional of two quantities is the square root of their product. 18. Proposition. — When the product of two quanti- ties is equal to the product of two others, either two may be the extremes and the other two the means of a proportion. Let aXd = bXc represent the equal products. If we divide by b and d, we have j = ^; or, a:b::c:d. (1st.) If we divide by c and d, we have f=-^; or, a: c :: b : d. (2d.) If we arrange the equal products thus : bXc=aXd, RATIO AND PROPORTION. 15 and then divide by a and c, we have l:a::d:c. (Sd.'l By similar divisions, the student may produce five other arrangements of the same quantities in pro- portion. 19. Proposition. — The order of the terms may le changed without destroying the proportion, so long as the extremes remain extremes, or both become means. Let a : b : : : d represent the given proportion. Then (16), we have aXd=bXo. Therefore (18), a and d may be taken as either the extremes or the means of a new proportion. 20. When we say the first term is to the third as the second is to the fourth, the proportion is taken by alternation, as in the second case. Article 18. When we say the second term is to the first as the fourth is to the third, the proportion is taken inversely, as in the third case. 21. Proposition. — Ratios which are equal to the same ratio are equal to each other. This is a case of the first axiom (6). 21^. Proposition.; — If two quantities have the same multiplier, the multiples will have the same ratio as the given quantities. Let a and b represent any two quantities, and m any multiplier. Then the identical equation, wjXaX6 = JwXJX«, gives the proportion, jwXa : mXb : : a : b (18). 23. Proposition. — In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. IG ELEMENTS OF GEOMETRY. Let a:h::c:d::e:f::g:h, etc., represent the equal ratios. Therefore (16), aXd^hY^c aXf=bXe aXh = bXg To these add aXb = hXa aXib+d+f+h) = bX{a+c+e+g). ", Therefore (18), a^c^e^g : b+d+f+h :: a:b. This is called proportion by Composition. 24. Proposition. — The difference between the first and second terms of a proportion is to the second, as the dif- ference between the third and fourth is to the fourth. The given proportion, a -.b :: c : d, may be written, ^- — ^• Subtract the identical equation, b _d b^d The remaining equation, a^b c — d '~h~^~d~' may be written, a — b : 6 : : c — d : d. This is called proportion by Division. 25. Proposition — If four quantities are in proportion, their same powers are in proportion, also their same roots. Thus, if we have a:b: : e : d. then, a^ . j2 . : 0 ELEMENTS OF GEOMETRY. 166. The art of turning consists in the production of the cir- cular form by mechanical means. The substance to be turned is placed in a machine called a lathe, which gives it a, rotary mo- tion. The edge of a cutting tool is placed at a distance from the axis of revolution equal to the radius of the intended circle. As the substance revolves, the tool removes every part that is further from the axis than the radius, and thus gives a circular form to what remains. PROBLEMS IN DRAWING. 167. The compasses enable us to draw a circumference, or an arc of a given radius and given center. Open the instrument till the points are on the two ends of the given radius. Then fix one point on the given center, and the other point may be made to revolve around in contact with the surface, thus tracing out the circumference. The revolving leg may have a pen or pencil at the point In the operation, care should be taken not to vary the opening of the compasses. f 16S. It is evident that with the ruler and compasses (69), 1. A straight line can be drawn through two given points. 2. A given straight line can be producedf^any length. 3. A circumference can be described from any center, with any radius. I 169. The foregoing are the three pos^lates of Euclid. Since the straight line and the circumference are the only lines treated of in elementary geometry, these Euclidian postulates are a suf- ficient basis for all problems. Hence, the rule that no instruments shall be used except the ruler and the compasses (68). 1'70. In the Elements of Euclid, which, for many ages, was the only text-book on elementary geometry, the problems in drawing occupy the place of problems in geometry. At present, the mathe- maticians of Germany, France, and America put them aside as not forming a necessary part of the theory of the science. English writers, however, generally adhere to Euclid. 171. Problem. — To bisect a given straight line. With A and B as centers, and with a radius greater than the half of AB, describe arcs which intersect in the two points D PROBLEMS IN DRAWING. 57 and B. The straight line joining these two points will bisect AB at C. Let the demonstration be given ^D, by the student (109 and 151). X B y^ E^ A 172. Problem. — To erect a perpendicular on a given straight line at a given point. Take two points in the line, one on each side of the given point, at equal distances from it Describe arcs as in the last prob- lem, and their intersection gives one point of the perpendicular. Demonstration to be given by the student IT'S. ProWem. — To let fall a perpendicular from a given point on a given straight line. With the given point as a cen- ter, and a radius long enough, describe an arc cutting the given line BC in the points D and E. The line may be produced, if necessary, to be cut by the arc in two places. With D and E as centers, and with a radius greater than the half of DE, describe arcs cutting each other in F. The straight line joining A and F is perpendicular to DE. Let the student show why. 174. Problem. — To draw a line through a given point parallel to a given line. Let a perpendicular fall from the point on the line. Then, at the given point, erect a perpendicular to this last It will be par- allel to the given line. Let the student explain why (129). ITS. Problem To describe a circumference through three given points. The solution of this problem is evident, from Article 149. B .^i^ jl^ n Xp 58 ELEMENTS OF GEOMETRY. 176. Problem — To find the center of a given arc or circumference. Take any three points of the arc, and proceed as in the last problem. I'y'y. The student is advised to make a drawing of every prob- lem. First draw the parts given, then the construction requisite for solution. Afterward demonstrate its correctness. Endeavor to make the drawing as exact as possible. Let the lines be fine and even, as they better represent the abstract lines of geometry. A geometrical principle is more easily understood by the student, when he makes a neat diagram, than when his drawing is careless. TANGENT. ITS. Theorem — A straight line which is perpendicular to a radius at its extremity, touches the circumference in only one point. Let AD be perpendicular to the radius BC at its extremity B. Then it is to be proved that AD touches the circumference at B, and at no other point. If the center C be joined by straight lines with any points of AD, the perpendicular BC will be shorter than any such oblique line (104). Therefore (153), every point of the line AD, except B, is outside of the circumference. 179. A Tangent is a line touching a circumference in only one point. The circumference is also said to be tangent to the straight line. The common point is called the point of contact. SECANT. 59 APPLICATION. ISO. Tangent lines are frequently used in the arts. A com- mon example ia when a strap is carried round a part of the cir- cumference of a wheel, and extending to a distance, sufficient tenision is given to it to produce such a degree of friction between it and the wheel, that one can not move without the other. 181. Problem in Drawing — To draw a tangent at a given point of an arc. Draw a radius to the given point, and erect a perpendicular to the radius at that point. It will be necessary to produce the radius beyond the arc, as the student has not yet learned to erect a perpendicular at the extremity of a line without producing it. SECANT. 182. Theorem — A straight line which is oblique to a radius at its extremity, cuts the circumference in two points. Let AD be oblique to the radius CB at its extrem- ity B. Then it will cut the cir- cumference at B, and at some other point. From the center C, let CE fall perpendicularly on AD. On ED, take EF equal to EB. Then the distance from C to any point of the line AD be- tween B and F is less than the length of the radius CB (110), and to any point of the line be- yond B and F, it is greater than the length of CB. Therefore (153), that portion of the line AD between B and F is within, and the parts be- yond B and F are without the circumference. Hence, the oblique line cuts the circumference in two points. 60 ELEMENTS OF GEOMETEY. 183. Corollary — A tangent to the circumference is perpendicular to the radius which extends to the point of contact. For, if it were not perpendicular, it would be a secant. 184. Corollary At one point of a circumference, faere can be only one tangent (103). CHOKDS. 183. Theorem The radii being equal, if two arcs are equal their chords are also equal. If the arcs AOE and BCD are equal, and their radii are equal, then AE and BD are equal. For, since the radii are equal, the circumferences are equal (154) ; and the arcs may be placed one upon the other, and will coincide, so that A will be upon B, and E upon D. Then the two chords, being straight lines, must coincide (51), and are equal. 186. Every chord subtends two arcs, which together form the whole circumference. Thus th? chord AE sub- tends the arcs AOE and AIE. The arc of a chord always means thr smaller of the two, unless otherwise expressed. 187. Theorem. — The radius which is perpendicular to a chord bisects the chord and its arc. CHORDS. 61 Let CD be perpendicular at E to the chord AB, then Will AE be equal to EB, and the arc AD to the arc DB. Produce DC to the circum- ference at F, and let that part of the figure on one side of DF be turned upon DF as upon an axis. Then the semi-circum- ference DAF will coincide with DBF (157). Since the angles at E are right, the line EA will take the direction of EB, and the point A will fall on the point B. Therefore, EA and EB will coincide, and are equal; and the same is true of DA and DB, and of FA and FB. 188. Corollary — Since two conditions determine the position of a straight line (52), if it has any two of the four conditions mentioned in the theorem, it must have the other two. These four conditions are, 1. The line passes through the center of the circle, that is, it is a radius. 2. It passes through the center of the chord. 3. It passes through the center of the arc. 4. It is perpendicular to the chord. 189. Theorem. — The radii being equal, when two arcs are each less than a semi-circumference, the greater arc has Ihe greater chord. If the arc AMB is greater than CND, and the radii of the circles are equal, then AB is greater than CD. Take AME equal to CND. Join AE, OE, and OB. Then AE is equal to CD (185). Since the arc AMB is less than a semi-circumference, the chord AB will pass between the arc and the center 0, Hence, it cuts the radius OE at some point I. 62 ELEMENTS OF GEOMETRY. Now, the broken line OIB is greater than OB (54), or its equal OE. Subtracting 01 from each (8), the remainder IB is greater than the remainder IE. Add- ing AI to each of these, we have AB greater than AIE. But AIE is greater than AE. Therefore, AB, the chord of the greater arc, is greater than AE, or its equal CD, the chord of the less. 190. Corollary. — When the arcs are both greater than a semi-circumference, the greater arc has the less chord. DISTANCE FROM THE CENTER. 191. Theorem — When the radii are equal, equal chords are equally distant from the center. Let the chords AB and CD be equal, and in the equal o circles ABG and CDF; then the distances of these chords from the centers E and H will also be equal. CHORDS. 63 Let fall the perpendiculars EK and HL from the centers upon the chords. Now, since the chords AB and CD are equal, the arcs AB and CD are also equal (185) ; and we may apply the circle ABG to its equal CDF, so that they will coincide, and the arc AB coincide with its equal CD. Therefore, the chords will coincide. Since K and L are the mid- dle points of these coinciding chords (187), K will fall upon L. Therefore, the lines EK and HL coincide and are equal. But these equal perpendiculars measure the distance of the chords from the centers (105). If the equal chords, as MO and AB, are in the same circle, each may be compared with the equal chord CD of the equal circle CDF. Thus it may be proved that the distances NE and EK are each equal to HL, and therefore equal to each other. 193. Theorem. — When the radii are equal, the less of two unequal chords is the farther from the center. Let AB be the greater of two chords, and FG the less, in the same or an equal circle. Then FG is farther from the center than AB. Take the arc AC equal to FG. Join AC, and from the center D let fall the perpendiculars DE and DN upon AB and AC. Since the arc AC is less than AB, the chord AB will be between AC and the center D, and will cut the perpendicular DN. Then DN, the whole, is greater than DH, the part cut off; and DH is greater than DE (104). So much the 61 ELEMENTS OP GEOMETRY. more is DN greater than DE. Therefore, AC and its equal FG are farther from the center than AB. 193. Corollary. — Conversely of these two theorems, when the radii are equal, chords which are equally dis- tant from the center are equal ; and of two chords which are unequally distant from the center, the one nearer to the center is longer than the other. 104. Problem in Drawing. — To bisect a given arc. Draw the chord of the arc, and erect a perpendicular at its center. State the theorem and the problems in drawing here used. 195. "The most simple case of the division of an arc, after its bisection, is its trisection, or its division into three equal parts. This problem accordingly exercised, at an early epoch in the prog- ress of geometrical science, the ingenuity of mathematicians, and has become memorable in the history of geometrical discovery, for having baffled the skill of the most illustrious geometers. "Its object was to determine means of dividing any given arc into three equal parts, without any other instruments than the rule and compasses permitted by the postulates prefixed to Euclid's Elements. Simple as the problem appears to be, it never has been solved, and probably never will be, under the above conditions." — Lardner's Treatise. ANGLES AT THE CENTER. 19G. Angles which have their vertex at the center of a circle are called, for this reason, angles at the center. The arc between the sides of an angle is called the in- tercepted arc of the angle. 197. Theorem. — The radii being equal, any two angles at the center have the same ratio as their intercepted arcs. This theorem presents the three following cases: 1st. If the arcs are equal, the angles are equal. ANGLES AT THE CENTER. 65 For the arcs may be placed one upon the other, and will coincide. Then BC will coincide with AO, and DC with EO. Thus the angles may coincide, and are equal. The conyerse is proved in the same manner. 2d. If the arcs have the ratio of two whole numbers, the angles have the same ratio. Suppose, for example, the arc BD : arc AE : : 13 : 5. Then, if the arc BD be divided into thirteen equal parts, and the arc AE into five equal parts, these small arcs will all be equal. Let radii join to their respective cen- ters all the points of division. The small angles at the center thus formed are all equal, because their intercepted arcs are equal. But BCD is the sum of thirteen, and AOE of five of these equal angles. Therefore, angle BCD : angle AOE : : 13 : 5 ; that is, the angles have the same ratio as the arcs. Geom.— 6 66 ELEMENTS OF GEOMETKY. 3d. It remains to be proved, that, if the ratio of the arcs can not be expressed by two whole numbers, the angles have still the same ratio as the arcs ; or, that the radius being the same, the arc BD : arc AE : : angle BCD : angle AOE. If this proportion is not true, then the first, second, A and third terms being unchanged, the fourth term is either too large or too small. We will prove that it is neither. If it were too large, then some smaller angle, as AOI, would verify the proportion, and arc BD : arc AE : : angle BCD : angle AOI. Let the arc BD be divided into equal parts, so small that each of them shall be less than EI. Let one of these parts be applied to the arc AE, beginning at A, and marking the points of division. One of those points must necessarily fall between I and E, say at the point U. Join OU. Now, by this construction, the arcs BD and AU have the ratio of two whole numbers. Therefore, arc BD : arc AU : : angle BCD : angle AOU, These last two proportions may be written thus (19) : arc BD : angle BCD : : arc AE : angle AOI ; arc BD : angle BCD : : arc AU ; angle AOU. METHOD OF LIMITS. 67 Therefore (21), arc AE : angle AOI : : arc AU : angle AOU; or (19), arc AE : arc AU : ; angle AOI : angle AOU. But this last proportion is impossible, for the first antecjedent is greater than its consequent, while the second antecedent is less than its consequent. There- fore, the supposition which led to this conclusion is false, and the fourth term of the proportion, first stated, is not too large. It. may be shown, in the same way, that it is not too small. Therefore, the angle AOE is the true fourth term of the proportion, and it is proved that the arc BD is to the arc AE as the angle BCD is to the angle AOE. DEMONSTEATION BY LIMITS. 198. The third case of the above proposition may be demonstrated in a different manner, which requires some explanation. We have this definition of a limit: Let a magnitude vary according to a certain law which causes it to ap- proximate some determinate magnitude. Suppose the first magnitude can, by this law, approach the second indefinitely, but can never quite reach it. Then the second, or invariable magnitude, is said to be the limit of the first, or variable one. 199. Any curve may be treated as a limit. The straight parts of a broken line, having all its vertices in the curve, may be diminished at will, and the broken line made to approximate the curve indefinitely. Hence, a curve is the limit of those broken lines which have all their vertices in the curve. 68 ELEMENTS OF GEOMETRY. 300. The arc BC, which is cut oflF by the secant AD, may be diminished by successive bisections, keeping the remain- ders next to B. Thus AD, re- volving on the point B, may approach indefinitely the tan- gent EF. Hence, the tangent at any point of a curve is the limit of the secants which may cut the curve at that point. SOI. The principle upon which all reasoning by the method of limits is governed, is that, whatever is true up io the limit is true at the limit. We admit this as an axiom of reasoning, because we can not conceive it to be otherwise. Whatever is true of every broken line having its vertices in a curve, is true of that curve also. What- ever is true of every secant passing through a point of a curve, is true of the tangent at that point. We do not say that the arc is a broken line, nor that the tangent is a secant, nor that an arc can be without extent; but that the curve and the tangent are limits toward which variable magnitudes may tend, and that whatever is true all the way to within the least pos- sible distance of a certain point, is true at that point. SOS. Having proved (first and second parts, 197) that, when two arcs have the ratio of two whole numbers, the angles at the center have the same ratio, we may then suppose that the ra- tio of BD to BF can not be ex- pressed by whole numbers. Now, if we divide BF into two equal parts, the point of division will be at a certain METHOD OP INFINITES. 69 distance from D. We may conceive tlie arc BF to be divided into any number of equal parts, and by in- creasing this number, the point 0, the point of division nearest to D, may be made to approach within any con- ceivable distance of D. By the second part of the theorem (197), it is proved that arc BO : arc BF : : angle BCO : angle BCF. Now, although the arc BD is itself incommensurable with BF, yet it is the limit of the arcs BO, and the angle BCD is the limit of the angles BCO. Therefore, since whatever is true up to the limit is true at the limit, arc BD : arc BF : : angle BCD : angle BCF. That is, the intercepted arcs have the same r^tio as their angles at the center. METHOD OF INFINITES. SOS. Modern geometers have made much uso of a kind of reasoning which may be called the method of infinites. It consists in supposing that any line cf def- inite extent and form is composed of an infinite num- ber of infinitely small straight lines. A surface is supposed to consist of an infinite number of infinitely narrow surfaces, and a solid of an infinite number of infinitely thin solids. These thin solids, nar- row surfaces, and small lines, are called infinitesimals. 304. The reasoning of the method of infinites is substantially the same in its logical rigor as of the method of limits. The method of infinites is a p»uch abbreviated form of the method of limits. The student must be careful how he adopts it. For when the infinite is brought into an argument bj the unskillful, the conclusion is very apt to be absurd. It 70 ELEMENTS OF GEOMETRY. is sufficient to say, that where the method of limits can be used, the method of infinites may also be used with- out error. The method of infinites has also been called the method of indivisibles. Some examples of its use will be given in the course of the work. ARCS AND ANGLES. We return to the subject of angles at the center. The theorem last given (197) has the following 205. Corollary. — If two diameters are perpendicular to each other, they divide the whole circumference into four equal parts. 206. A Quadrant is the fourth part of a circumference. 207. Since the angle at the cen- ter varies as the intercepted arc, mathematicians have adopted the same method of measuring both an- gles and arcs. As a right angle is the unit of angles, so a quadrant of a certain radius may be taken as the standard for the measurement of arcs that have the same radiuS. For the same reason, we usually say that the inter- cepted arc measures the angle at the center. Thus, the right angle is said to be measured by the quadrant; half a right angle, by one-eighth of a circumference; and so on. APPLICATIONS, 20S. In the applicatioos of geometry to practical purposes, the quadrant and the right angle are divided into ninety equa^i parts, each of which is called a degree. Each degree is marked ARCS AND ANGLES. 71 thus °, and is divided into sixty minutes, marked thus '; and each minute is divided into sixty seconds, marked thus ". Hence, it appears that there are in an entire circumference, or in the sum of all the successive angles about a point, 360°, or 2160(y, or 1296 000" Some astronomers, mostly the French, divide the right angle and the quadrant into one hundred parts, each of these into one hundred; and so on. 309. Instruments for measuring angles are founded upon the principle that arcs are proportional to angles. Such instruments usually consist either of a part or an entire circle of metal, on the surface of which is accurately engraven its divisions into de- grees, etc. Many kinds of instruments used by surveyors, navi- gators, and astronomers, are constructed upon this principle. 210. An instrument called a protractor is used, in drawing, for measuring angles, and for laying down, on paper, angles of any required size. It consists of a semicircle of brass or mica, the circumference of which is divided into degrees and parts of a degree. PROBLEMS IN DRAWING. 211. Problem — To draw an angle equal to a given angle. Let it be required to draw a line making, with the given line BC, an angle at B equal to the given angle A. With A as a center, and any as- sumed radius AD, draw the arc DB cutting tlie sides of the angle A. With B as a center, and the same radius as before, draw an arc FG. Join DE. With F as a center, and a radius equal to DE, draw an arc cut- ting FG at the point G. Join BG. Then GBF is the required angle. For, joining FG, the arcs DE and FG have equal radii and equal chords, and therefore are equal (185). Hence, they sub- tend equal angles (197). 212. Corollary. — An arc equal to a given arc may be drawn in the same way. 72 ELEMENTS OF GEOMETRY. 313. Problem — To draw an angle equal to the sum of two given angles. Let A and B be the given an- gles. First, make the angle DCE equal to A, and then at C, on the line CE, draw the angle EOF equal to .B. The angle FCD is equal to the sum of A and B (9). 514. CoroUary. — In a similar manner, draw an angle equal to the sum of several given angles; also, an angle equal to the dif- ference of two given angles ; or, an angle equal to the supplement, or to the complement of a given angle. 515. Corollary. — By the same methods, an arc may be drawn equal to the difference of two arcs having equal radii, or equal to the sum of several arcs. 516. Problem. — To erect a perpendicular to a given line at its extreme point, without producing the given line. A right angle may be made separately, and then, at the end of the given line, an angle be made equal to the given angle. This is the method universally employed by mechanics and draughtsmen to construct right angles and perpendiculars by the use of the square. 317. Problem — To draw a line through a given point parallel to a given line. This has been done by means of perpendiculars (174). It may be done with an oblique secant, by making the alternate or the corresponding angles equal. ARCS INTERCEPTED BY PARALLELS. 318. An arc which is included between two parallel lines, or between the sides of an angle, is called inier- cepted. 319. Theorem — Two parallel lines intercept equal arcs of a circumference. INTERCEPTED ARCS. 73 The two lines may be both secants, or both tangents, 3T one a secant and one a tangent. 1st. When both are secants. The arcs AC and BD inter- cepted by the parallels AB and CD are equal. For, let fall from the center a perpendicular upon CD, and produce it to the circumference at E. Then OE is also perpendicular to AB (127). Therefore, the arcs EA and EB are equal (187) ; and the arcs EC and ED are equal. Subtracting the first from the second, there remains the arc AC equal to the arc BD. 2d. When one is a tangent. Extend the radius OE to the point of contact. This radius is perpendicular to the tangent AB (183). Hence, it is perpen- dicular to the secant CD (127), and therefore it bisects the arc CED at the point E (187). That is, the intercepted arcs EC and ED are equal. 3d. When both are tangents. Extend the radii OE aind 01 to the points of contact. These radii being perpendicular ^ -p u (183) to the parallels, must (103 and 127) form one straight line. Therefore, EI is a diameter, and divides (157) the circumference into equal parts. But these equal parts are the arcs intercepted by the para-llel tangents. G l u Therefore, in every case, the arcs intercepted by two parallels are equal. Geom. — 7 74 ELEMENTS OF GEOMETRY. ARCS INTERCEPTED BY ANGLES. 230. An Inscribed Angle is one whose sides are chords or secants, and whose vertex is on the circum- ference. An angle is said to be inscribed in an arc, when its vertex is on the arc and its sides extend to or through the ends of the arc. In such a case the arc is said to contain the angle. Thus, the angle AEI is inscribed in the arc AEI, and the arc AEI con- tains the angle AEI. An angle is said to stand upon the arc intercepted between its sides. Thus, the angle AEI stands upon the arc AOL 331. Corollary. — The arc in which an angle is in- scribed, and the arc intercepted between its sides, com- pose the whole circumference. 333. Theorem — An inscribed angle is measured by half of the intercepted arc. This demonstration also presents three cases. The center of the circle may be on one of the sides of the angle, or it may be inside, or it may be outside of the angle. 1st. One side of the angle, as AB, may be a diameter. Make the diameter DE, paral- lel to BC, the other side of the angle. Then the angle B is equal to its alternate angle BOD (125), which is measured by the arc BD (207). This arc is equal to CE (219), and also to EA (197). Therefore, the arc INTEKCEPTED ARCS. 75 BD is equal to the half of AO, and the inscribed angle B is measured by half of its intercepted arc. 2d. The center of the circle may be within the angle. From the vertex B extend a diameter to the opposite side of the circumference at D. As just proved, the angle ABD is measured by half of the arc AD, and the angle DBG by half of the arc DC. Therefore, the sum of the two angles, or ABC, is measured by half of the sum of the two arcs, or half of the arc ADC. 3d. The center of the circle maybe outside of the angle. Extend a diameter from the vertex as before. The angle ABC is equal to ABD diminished by DBC, and is, therefore, meas- ured by half of the arc DA di- minished by half of DC; that is, by the half of AC. 323. Corollary When an inscribed angle and an angle at the center have the same intercepted arc, the inscribed angle is half of the angle at the center. 2!S4. Corollary — All angles in- scribed in the same arc are equal, for they have the same measure. 3S5. Corollary. — Every angle inscribed in a semi- circumference is a right angle. If the arc is less than a semi-circumference, the angle is obtuse. If the arc is greater, the angle is acute. 76 ELEMENTS OF GEOMETRY. SS6. Theorem — The angle formed by a tangent and a chord is measured by half (he intercepted arc. The angle CEI, formed by the tangent AC and the chord EI, is measured by half the intercepted arc IDE. Through I, miike the chord 10 parallel to the tangent AC. The angle CEI is equal to its alternate EIO (125), which is measured by half the arc OME (222), which is equal to the arc IDE (219). Therefore, the angle CEI is measured by half the arc IDE. The sum of the angles AEI and CEI is two right angles, and is therefore measured by half the whole cir- cumference (207). Hence, the angle AEI is equal to two right angles diminished by .the angle CEI, and is measured by half the whole circumference diminished by half the arc IDE ; that is, by half the arc lOME. Thus it is proved that each of the angles formed at E, is measured by half the arc intercepted between its sides. 227. This theorem may be demonstrated very ele- gantly by the method of limits (200). 228. Theorem — Every angle whose vertex is within the circumference, is measured by half the sum of the arcs intercepted between its sides and its sides pro- dvjced. Thus, the angle OAE is meas- ured by half the sum of the arcs OE and lU. To be demonstrated by the student, using the previous theorems (219 and 222). INTERCEPTED ARCS. 77 S39. Theorem. — Every angle whose vertex is outside of a circumference, and whose sides are either tangent or secant, is measured by half the difference of the inter- cepted arcs. Thus, the angle ACF is measured by half the dif- ference of the arcs AF and AB ; the angle FCG, by half the difference of the arcs FG and BI; and the angle ACE, by half the difference of the arcs AFGB and ABIE. This, also, may be demon- strated by the student, by the aid of the previous theo- rems on intercepted arcs. PROBLEMS IN DRAWING. S30. Problem — Through a given point out of a cir- cumference, to draw a tangent to the circumference. Let A be the given point, and C tlie center of the given circle. Join AC. Bisect AC at the point B (171). With B as a center and BC as a radius, describe a circumference. It will pass through C and A (153), and will cut the cir- cumference in two points, D and E. Draw straight lines from A through D and E. AD and AE are both tangent to the given circumference. Join CD and CE. The angle CDA is inscribed in a semi- circumference, and is therefore a right angle (225). Since AD is perpendicular to the radius CD, it is tangent to the circumference (178). AE is tangent for the same reasons. 78 ELEMENTS OF GEOMETRY. Make the angle DAB 331. Problem Upon a given chord to describe an arc which shall contain a given angle. Let AB be tlie chord, and C the angle, equal to C. At A erect a perpendicular to AD, and erect a perpendicu- lar to AB at its center (172). Produce these till they meet at the point F (137). With F as a center, and FA as a ra- / dius, describe a circum- Z C ference. Any angle in- Kcribed in the arc BGHA will be equal to the given angle C. For AD, being perpendicular to the radius FA, is a tangent (178). Therefore, the angle BAD is measured by half of the arc AIB (226). But any angle contained in the arc AHGB is also measured by half of the same arc (222), and is therefore equal to BAD, which was made equal to C. POSITIONS OF TWO CIECUMFERENCES. 23!3. Theorem — Two circumferences can not cut each other in more than two points. For three points determine the position and extent of a circumference (149). Therefore, if two circumfer- ences have three points common, they must coincide throughout. 333. Let us investigate the various positions which two circumferences may have with reference to each other. Let A and B be the centers of two circles, and let these points be joined by a straight line, which there- fore measures the distance between the centers. First, suppose the sum of the radii to be less than AB. TWO CIRCUMFERENCES. 79 Then AC and BD, the radii, can not reach each other. At C and D, where the curves cut tlie line AB, let perpendic- ulars to that line be erected. These perpendiculars are paral- lel to each other (129). They are also tangent respectively to the two circumferences (178). It follows, therefore, that CD, the distance between these parallels, is also the least distance between the two curves. 234. Next, let the sum of the radii AC and BC be equal to AB, the distance between the centers. Then both curves will pass through the point C (153). At this point let a perpendicular be erected as before. This per- pendicular CG is tangent to both the curves (178); that is, it is cut by neither of them. Therefore, the curves have only one common point C. 235. Next, let AB be less than the sum, but greater than the difference, of the radii AC and BD. Then the point C will fall within the circum- ference DF. For if it fell on or outside of it, on the side toward A, then AB would be equal to or greater than the sum of the radii ; and if the point fell on or out- side of the curve in the direction toward B, then AB would be equal to or less than the difference between the radii. Each of these is contrary to the hypothesis. For the same reasons, the point D will fall within tha 80 ELEMENTS OF GEOMETRY circumference CG. Therefore, these circumferences cut each other, and have two points common (232). 236. Next, let the difference between the two radii AC and BC be equal to the distance AB. A perpendic- ular to this line at the point C will be a tangent to both curves, and they have a com- mon point at C. They have no other common point, for the two curves are both symmetrical about the line AC (158), and, therefore, if they had a common point on one side of that line, they would have a corresponding com- mon point on the other side; but this can not be, for they would then have three points common (232). 337. Lastly, suppose the distance AB less than the difference of the radii AC and BD, by the line CD. That is, AB + BD + DC = AC. Join A, the center of the larger circle, with F, any point of the smaller cir- cumference, and join BF. Then AB and BD are to- gether equal to AB and BF, which are together greater than AF. Therefore, AD is greater than AF. Hence, the point D is farther from A than any other point of the circumference DF. It follows that CD is the least distance between the two curves. The above course of reasoning develops the follow- ing principles: 238. Theorem — Two circumferenees may have, with reference to each other, five positions : TWO CIROUMFERENCES. g^ 1st. Each may he entirely exterior to the other, when the distance between their centers is greater than the sum of their radii. 2d. They may touch each other exteriorly, having one point common, wlien the distance between the centers is equal to the sum of the radii. Zd. They may cut each other, having two points com- mon, when the distance between the centers is less than the sum and greater than the difference of the radii. Ath. One may be within the other and tangent, having one point common, when the distance between the centers is equal to the difference of the radii. bih. One may be entirely withiyi the other, when the distance between the centers is less than the difference of the radii. 339. Corollary.^Two circumferences can not have more than one chord common to both. 240. Corollary — The common chord of two circum- ferences is perpendicular to the straight line which joins their centers and is bisected by it. For the ends of the chords are equidistant from each of the centers, the ends of the other line (109). 341. Corollary, — When two circumferences are tan-- gent to each other, the two centers and the point of contact are in one straight line. 343. Corollary — When two circumferences have no common point, the least distance between the curves is measured along the line which joins the centers. 343. Corollary — ^When the distance between the cen- ters is zero, that is, when they coincide, a straight line through this point may have any direction in the plane; and the two curves are equidistant at all points. Such circles are called Concentric. S2 ELEMENTS OF GCOIIETRY. S44. A Locus is a line or a surface all the points of which have some common property, which does not belong to any other points. This is also frequently called a geometrical locus. Thus, The circumference of a circle is the locus of all those points in the plane, which are at a given distance from a given point. A straight line perpendicular to another at its center is the locus of all those points in the plane, which are at the same distance from both ends of the second line. The geometrical locus of the centers of those circles which have a given radius, and are tangent to a given straight line, is a line parallel to the former, and at a distance from it equal to the radius. 245. The student will find an excellent review of the preceding pages, in demonstrating the theorems, and solving the problems in drawing which follow. In his eflforts to discover the solutions of the more difficult problems in drawing, the student will be much assisted by the following Suggestions. — 1. Suppose the problem solved, and the figure completed. 2. Find the geometrical relations of the difi'erent parts of the figure thus formed, drawing auxiliary lines, if necessary. 3. From the principles thus developed, make a rule for the solution of a problem. This is the analytic method of solving problems. EXERCISES. 1. Take two straight lines at random, and find their ratio. Make examples in this way for all the problems in drawing. 2. liisect a quadrant, also its half, its fourth ; and so on. EXERCISES. 83 3. From a given point, to draw the shortest line possible to a given straight line. 4. With a given length of radius, to draw a circumference through two given points. 5. From two given points, to draw two equal straight lines which shall end in the same point of a given line. 6. From a point out of a straight line, to draw a second lii e making a required angle with the first. 7. If from a point without a circle two straight lines extend to the concave part of the circumference, making equal angles with the line joining the same point and the center of the circle, then the parts of the first two lines which are within the circumfer- ence are equal. 8. To draw a line through a point such that the perpendicu- lars upon this line, from two other points, may be equal. 9. From two points on the same side of a straight line, to draw two other straight lines which shall meet in the first, and make equal angles with it. 10. In each of the five cases of the last theorem (238), how many straight lines can be tangent to both circumferences'? The number is different for each case. 11. On any two circumferences, the two points which are at the greatest distance apart are in the prolongation of the line which joins the centers. 12. To draw a circumference with a given radius, through a given point, and tangent to a given straight line. 13. With a given radius, to draw a circumference tangent to two given circumferences. 14. What is the locus of the centers of those circles which have a given radius, and are tangent to a given circle? 15. Of all straight lines which can be drawn from two given points to meet on the convex circumference of a circle, the sum of those two is the least which make equal angles with the tan- gent to the circle at the point of concourse. 16. If two circumferences be such that the radius of one is the diameter of the other, any straight line extending from their point of contact to the outer circumference is bisected by the inner one. 84 ELEMENTS OF GEOMETRY. ] 7. If two circumferences cut each other, and from either point of intersection a diameter be made in each, the extremities of these diameters and the other point of intersection are in the same straight line. 18. If any straight line joining two parallel lines be bisected, any other line through the point of bisection and joining the two parallels, is also bisected at that point. .19. If two circumferences are concentric, a line which is a chord of the one and a tangent of the other, is bisected at the point of contact. 20. If a circle have any number of equal chords, what is the locus of their points of bisection? 21. If any point, not the center, be taken in a diameter of a circle, of all the chords which can pass through that point, that one is the least which is at right angles to the diameter. 22. If from any point there extend two lines tangent to a circumference, the angle contained by the tangents is double the angle contained by the line joining the points of contact and the radius extending to one of them. 23. If from the ends of a diameter perpendiculars be let fall on any line cutting the circumference, the parts intercepted be- tween those perpendiculars and the curve are equal. 24. To draw a circumference with a given radius, so that the eiaes of a given angle shall be tangents to it. 25. To draw a circumference through two given points., with the center ia a given line. 26. Through a given point, to draw a straigiit line, making equal angles with the two sides of a given angle. PKOPEKTIES OF TRIANGLES. 85 CHAPTER V. TKIANGLES. S46. Next in regular order is the consideration of those plane figures which inclose an area; and, first, of those whose boundaries are straight lines. A Polygon is a portion of a plane bounded by straight lines. The straight lines are the sides of the polygon. The Perimeter of a polygon is its boundary, or the sum of all the sides. Sometimes this word is used to designate the boundary of any plane figure. S^T. A Triangle is a polygon of three sides. Less than three straight lines can not inclose a sur- face, for two straight lines can have only one common point (51). Therefore, the triangle is the simplest polygon. From a consideration of its properties, those of all other polygons may be derived. 248. Problem. — Any three points not in the same straight line may be made the vertices of the three angles of a triangle. For these points determine the plane (60), and straight lines may join them two and two (47), thus forming the required figure. INSCRIBED AND CIRC UM SCRIBED. 349. Corollary Any three points of a circumference may be made the vertices of a triangle. A circumfer- 86 ELEMENTS OF GEOMETRY. ence may pass through the vertices of any triangle, for it may pass through any three points not in the same straight line (149). 250. Theorem. — Within every triangle there is a point equally distant from the three sides. In the triangle ABC, let lines bisecting the angles A a ad B be produced until they meet. The point D, where the two bisecting lines meet, is equally distant from the two sides AB and BC, since it is a point of the line which bisects the angle B (113). For a similar reason, the point D is equally distant from the two sides AB and AC. Therefore, it is equally distant from the three sides of the triangle. 331. Corollary. — The three lines which bisect the sev- eral angles of a triangle meet at one point. For the point D must be in the line which bisects the angle C (113). S3S. Corollary — With D as a center, and a radius equal to the distance of D from either side, a circum- ference may be described, to which every side of the triangle will be a tangent. 233. When a circumference passes through the ver- tices of all the angles of a polygon, the circle is said to be circumscribed about the polygon, and the polygon to 1)6 inscribed in the circle. When every side of a polygon is tangent to a circumference, the circle is inscribed and the polygon circumscribed. 334. The angles at the ends of ^^ one side of a triangle are said to be adjacent to that side. Thus, the PROPERTIES OF TRIANGLES. SI angles A and B are adjacent to the side AB. The angle formed by the other two sides is opposite. Thus, the angle A and the side BC are opposite to each other. SUM OF THE ANGLES. 255. Theorem — The sum of the angles of a triangle is equal to two right angles. Let the line DE pass through the vertex of one an- gle, B, parallel to the op- p BE posite side, AC. —^ Then the angle A is equal ^/^ \ to its alternate angle DBA A''=— -^C (125). For the same rea- son, the angle C is equal to the angle EBC. Hence, the three angles of the triangle are equal to the three consecutive angles at the point B, which are equal to two right angles (91). Therefore, the sum of the three angles of the triangle is equal to two right angles. 256. Corollary. — Each angle of a triangle is the sup- plement of the sum of the other two. 257. Corollary At least two of the angles of a tri- angle are acute. 258. Corollary If two angles of a triangle are equal, they are both acute. If the three are equal, they are all acute, and each is two-thirds of a right angle. 259. An Acute Angled triangle is one which has all its angles acute, as a. A Right Angled triangle has one of the angles right, as h. 88 ELEMENTS OF GEOMETRY. An Obtuse Angled triangle has one of the angles obtuse, as c. S60. Corollary — In a right angled triangle, the two acute angles are complementary (94). 361. Corollary — If one side of a triangle be pro- duced, the exterior angle thus B formed, as BCD, is equal to the sum of the two interior angles not adjacent to it, as A A'= and B (256). So much the more, the exterior angle is greater than either one of the interior angles not adja- cent to it. 1362. Corollary. — If two angles of a triangle are re- spectively equal to two angles of another, then the third angles are also equal. 363. Either side of a triangle may be taken as the hase. Then the vertex of the angle opposite the base is the vertex of the triangle. The Altitude of the triangle is the distance from the vertex to the base, which is measured by a perpen- dicular let fall on the base produced, if necessary. 364. Corollary. — The altitude of a triangle is equal to the distance between the base and a line through the vertex parallel to the base. 365. When one of the angles at the base is obtuse, the perpendicular falls outside of the triangle. When one of the angles at the base is right, the alti- tude coincides with the perpendicular side. When both the angles at the base are acute, the alti- tude falls within the triangle. Let the student give the reason for each case, and illustrate it with a diagram. PROPERTIES OF TRIANGLES. 89 LIMITS OF SIDES. S66. Theorem — Each Me of a triangle is smaller than the sum of the other two, and greater than their dif- ference. The first part of this theorem is an immediate conse- quence of the Axiom of Distance „ (54) ; that is, AC gram having all its sides equal. -.^_^--^^^^ ^^^^^^^ The rhombus has the follow- ing peculiarities, which may be demonstrated by the student. 186 ELEMENTS OF GEOMETRY. 366. Theorem. — The diagonals of a rhombus are per- pendicular to each other. 367. Theorem. — The diagonals of a rhombut bisect its angles. SQUARE. 368. A Square is a quadrilateral having its sides equal, and its angles right angles. The square may be shown to have all the properties of the parallelogram (359), of the rectangle, and of the rhombus. 369. Corollary — The rectangle and the square are the only parallelograms which can be inscribed in a circle (348). EQUALITY. 370. Theorem. — Two parallelograms are equal when two adjacent sides and the included angle in the one, are respectively equal to those parts in the other. For the remaining sides must be equal (356), and this becomes a case of Article 345. 371. Corollary — Two rectangles are equal when two adjacent sides of the one, are respectively equal to those parts of the other. 373. Corollary — Two squares are equal when a side of the one is equal to a side of the other. APPLICATIONS. 373. The rectangle is the most frequently used of all quadri- laterals. The walls and floors of our apartments, doors and win- dows, books, paper, and many other articles, have this form. Carpenters make an ingenious use of a geometrical principle in order to make their door and window-frames exactly rectangular. Having made the frame, with its sides equal and its ends equal, PARALLELOGRAMS. 127 they measure the two diagonals, and make the frame take such a shape that these also will be equal. In tlii.s operation, what principle is applied? 3*74. A rhombus inscribed in a rectangle is the basis of many orna- ments used in architecture and other work. Sfd. An instrument called parallel rulers, used in drawing parallel lines, consists of two rulers, connected by cross pieces a B \/ith pins in their ends. The ) ^ ■ g^ | rulers may turn upon the pins, varying their distance. The dis- i — tances between the pins along C D the rulers, that is, AB and CD, must be equal; also, along the cross pieces, that is, AC and BD. Then the rulers will always be parallel to each other. If one ruler be held fast while the other is moved, lines drawn along the edge of the other ruler, at different positions, will be parallel to each other. What geometrical principles are involved in the use of this instrument? EXERCISES. 370. — 1. State the converse of each theprem that has been given in this chapter, and determine whether each of these con- verse propositions is true. 2. To construct a parallelogram when two adjacent sides and an angle are given. 3. What parts need be given for the construction of a rect- angle? 4. What must be given for the construction of a square ? 5. If the four middle points of the sides of any quadrilateral be joined by straight lines, those lines form a parallelogram. 6. If four points be taken, one in each side of a square, at equal distances from the four vertices, the figure formed by join- ing these successive points is a square. 12S ELEMENTS OF GEOMETRY. 7. Two parallelograms are similar when they have an angle in the one equal to an angle in the other, and these equal angles included between proportional sides. MEASURE OF AREA. 377. The standard figure for the measure of surfaces is a square. That is, the unit of area is a square, the side of which is the unit of length, whatever be the ex- tent of the latter. Other figures might be, and sometimes are, used for this purpose; but the square has been almost univers- ally adopted, because 1. Its form is regular and simple; 2. The two dimensions of the square, its length and breadth, are the same; and, 3. A plane surface can be entirely covered with equal squares. The truth of the first two reasons is already known to the student : that of the last will appear in the fol- lowing theorem. 378. Any side of a polygon may be taken as the base. The Altitude of a parallelogram is the distance be- tAveen the base and the opposite side. Hence, the alti- tude of a parallelogram may be taken in either of two ways. AREA OF RECTANGLES. 379. Theorem. — The area of a rectangle is measured hy the product of its hose hy its altitude. That is, if we multiply the number of units of length contained in the base, by the number of those units F — -i- — + MEASURE OP AREA. 129 contained in the altitude, the product is the number of units of area contained in the surface. Suppose that the base AB and the altitude AD are multiples of the same unit of length, for example, four and three. Di- vide AB into four equal parts, and through all the points of divi- sion extend lines parallel to AD. Divide AD into three equal parts, " " and through the points of division extend lines paral- lel to AB. All the intercepted parts of these two sets of parallels must be equal (357); and all the angles, right angles (124). Thus, the whole rectangle is divided into equal squares (372). The number of these squares is equal to the number in one row multiplied by the number of rows ; that is, the number of units of length in the base multiplied by the number in the altitude. In the exam- ple taken, this is three times four, or twelve. The result would be the same, whatever the number of divisions in the base and altitude. If the base and altitude have no common measure, then we may assume the unit of length as small as we please. By taking for the unit a less and less part of the altitude, the base will be made the limit of the lines commensurable with the altitude. Thus, the demonstra- tion is made general. 380. Corollary The area of a square is expressed by the second power of the length of its side. An- ciently the principles of arithmetic were taught and il- lustrated by geometry, and we still find the word square in common use for the second power of a number. 381. By the method of infinites (203), the latter part of the above demonstration would consist in supposing 130 ELEMENTS OF GEOMETRY. the base and altitude of the rectangle divided into infi- nitely small and equal parts ; and then proceeding to form infinitesimal squares, as in the former part of the demonstration. If a straight line move in a direction perpendicular to itself, it describes a rectangle, one of whose dimen- sions is the given line, and the other is the distance which it has moved. Thus, it appears that the two di- mensions which every surface has (33), are combined in the simplest manner in the rectangle. A rectangle is said to be contained by its base and altitude. Thus, also, the area of any figure is called its superficial contents. APPLICATION. 3S2. All enlightened nations attach great importance to exact and uniform standard measures. In this country the standard of length is a yard measure, carefully preserved by the National Government, at Washington City. By it all the yard measures are regulated. The standards generally used for the measure of surface, are the square described upon a yard, a foot, a mile, or some other cer- tain length; but the acre, one of the most common measures of surface, is an exception. The number of feet, yards, or rods in one side of a square acre, can only be expressed by the aid of a radical sign. The public lands belonging to the United States are divided into square townships, each containing thirty-six square miles, called sections. AREA OF PARALLELOGRAMS. 383. The area of a parallelogram is measured hy the product of its hose hy its altitude. At the ends of the base AB erect perpendiculars, and MEASURE OF AREA. 131 produce them till they meet the opposite side, in the points E and P. Now the right angled triangles AED and BFC are equal, having the side BF equal to AE, since they are perpendiculars between par- allels (133); and the side EC equal to AD, by hypoth- esis (288). If each of these equal triangles be subtracted from the entire figure, ABCE, the remainders ABFE and ABCD must be equivalent. But ABFE is a i ectangle having the same base and altitude as the parallelogram ABCD. Hence, the area of the parallelogram is measured by the same product as that which measures t'' e area of the rect- angle. 3S4. Corollary. — Any two parallelograms have their areas in the same ratio as the products of their bases by their altitudes. Parallelograms of equal altitudes have the same ratio as their bases, and parallelograms of equal bases have the same ratio as their altitudes. 385. Corollary Two parallelograms are equivalent when they have equal bases and altitudes ; or, when the two dimensions of the one are the extremes, and the two dimensions of the other are the means, of a pro- portion. AREA OF TRIANGL*ES. 386. Theorem. — The area of a triangle is measured hj half the product of its base hy its altitude. For any triangle is one-half of a parallelogram having the same base and altitude (358). 132 ELEMENTS OF GEOMETRY. 3S7. Corollary, — The areas of triangles are in the ratio of the products of their bases by their altitudes. 388. Corollary. — Two triangles are equivalent when they have equal bases and altitudes. 389. Corollary — If a parallelogram and a triangle have equal bases and altitudes, the area of the paral- lelogram is double that of the triangle. 390. Theorem — If from half the sum of the three sides of a triangle each side be subtracted, and if these remain- ders and the half sum be multi-plied together, then the square root of the product will be the area of the triangle. Let DEF be any triangle, DF being the base and EG the altitude. Let the E extent of the several lines be represented by letters; that is, let ^ ^ DF = a, EF = 6, DE = c, EG=A, GiF^m, DG = «. and DE + EF + FD=s. Then (328), m + n : b-\-c : : b—c : m-^n. Therefore, m — n= — r^. ' m-tn By hypothesis, m-\-n=a. Adding, 2m = a +^^. a^+b^-c^ Then, , m= ^^—. Again (327), m^+h^=b\ Substituting for m^ its value, and transposing, Therefore, ^^V^'—i'^^^^)' MEASURE OF AREA. 1 33 But the area of the triangle is half the product of the base a by the altitude h. Hence, In this expression, we have the area of the triangle in terms of the three sides. For greater facility of cal- culation it is reduced to the following: area = J |/ (a -\-h-\- c){a -\-h — e)(a — h + c)( — a-\- b + c). The exact equality of these two expressions is shown by performing as far as is possible the operations indi- cated in each. But, by hypothesis, {a-\-b -\-c)^8 = 2{%\. Therefore, (a + 6 — c)=2(| — el, (a — 6 + c) = 2(|— ft), and, (— a + 6 + e) = 2/|— «\. Substituting these in the equation of the area, it be- comes, area =V(i)(l-Hi-')(i-")- 391. Theorem The areas of similar triangles are 10 each other as the squares of their homologous lines. Let AEI and BCD be similar triangles, and 10 and DH homologous altitudes. 134 ELEMENTS OF GEOMETRY. Then (310), 10 : DH : : AE ; ;BC. Multiply by AE : BC : : AE ; ;BC. Then, AB X 10 ; ; BC X DH : : AE ; 2 ;BC. But (387), AE X 10 ; : BC X DH : : area AEI: area BCD. Therefore (21), 2 2 area AEI : area BCD : : AE :B0. In a similar manner, prove that the areas have the same ratio as the squares of the altitudes 10 and DH, or as the squares of any homologous lines. AREA OF TRAPEZOIDS. 39S. Theorem — The area of a trapezoid is equal to half the j^-7oduct of its altitude by the sum of its parallel sides. The trapezoid may be divided by a diagonal into two triangles, having for their bases the parallel sides. Thf altitude of each of these triangles is equal to that of the trapezoid (264). The area of each triangle being half the product of the common altitude by its base, the area of their sum, or of the whole trapezoid, is half the product of the altitude by the sum of the bases. EXERCISES. 393. — 1. Measure the length and breadth, and find the area of the blackboard; of the floor. 2. To divide a given triangle into any number of equivalent triangles. 3. To divide a given parallelogram into any number of equiva- lent parallelograms. EQUIVALENT SURFACES. I35 4. To divide a given trapezoid into any number of equivalent trapezoids. 5. The area of a triangle is equal to half the product of the perimeter by the radius of the inscribed circle. 6. What is the radius of the circle inscribed in the triangle whose sides are 8, 10, and 12? EQUIVALENT SUEPACES. 394. IsoPERiMBTEiCAL figures are those whose perim- eters have the same extent. S93. Theorem — Of all equivalent triangles of a given lose, the one having the least perimeter is isosceles. The equivalent triangles having the same base, AE, have also the same altitude (388). Hence, their vertices are in the same line parallel to the base, that is, in DB. Now, the shortest line that a E can be made from A to E through some point of DB, will constitute the other two sides of the triangle of least perimeter. This shortest line is the one making equal angles with DB, as ACE, that is, making ACD and ECB equal (115). The angle ACD is equal to its alternate A, and the angle ECB to its alternate E. Therefore, the angles at the base are equal, and the tri- angle is isosceles. 39G. Corollary — Of all isoperimetrical triangles of a given base, the one having the greatest area is isosceles. 397. To draw a square equivalent to a given fig- ure, is called the squaring, or quadrature of the figure. How this can be done for any rectilinear figure, is shown in the following. 136 ELEMENTS OF GEOMETRY. PROBLEMS IN DRAWING. 398. Problem. — To draw a rectangle with a given base, equivalent to a given parallelogram. With the given base as a first term, and the base and altitude of the given figure as the second and third terms, find a fourth proportional (319). This is the required altitude (385). 399. Problem — To draw a square equivalent to a given parallelogram. Find a mean proportional between the base and altitude of the given figure (330). This is the side of the square (385). 400. Problem., — To draw a triangle equivalent to a given polygon. Let ABCDB be the given polygon. Join DA. Produce BA, and through E draw EF parallel to DA. Join DF. ^2^ Now, the triangles DAF fC^"^ xl V\\ and DAE are equivalent, for / \ / / \ \\,,^p they have the same base DA, / y / \ \ / \ and equal altitudes, since / / \ / '-A "i their vertices are in the line // \ / \ / \\ EF parallel to the base (264). f- ^ ' g- -^ To each of these equal8,,add the figure ABCD, and we have the quadrilateral FBCD equiva- lent to the polygon ABODE. In this manner, the number of sides may be diminished till a triangle is formed equivalent to the given polygon. In this diagram it is the triangle FDGr. 401. Problem. — To draw a square equivalent to a given triangle. Find a mean proportional between the altitude and half the base of the triangle. This will be the side of the required square. EQUIVALENT SQUARES. 402. Having shown (379) how an area is expressed by the product of two lengths, it follows that an equa- EQUIVALENT SURFACES. 137 tion will represent equivalent surfaces, if each of its terms is composed of two factors which represent lengths. For example, let a and h represent the lengths of two straight lines. Now we know, from algebra, that what- ^ ever be the value of a and h, This formula, therefore, includes the following geomet- rical 403. Theorem — The square described upon the sum. of two lines is equivalent to the sum of the squares described on the two lines, increased hy twice the rectangle contained hy these two lines. Since the truths of algebra are universal in their application, this theorem is demon- strated by the truth of the above equation. Such a proof is called algebraic. It is also called analytical, but with doubtful propriety. Let the student demonstrate the theorem geometrically, by the aid of this diagram. 404. Theorem The square described on the difference of two straight lines is equivalent to the sum of the squares described on the two lines, diminished by twice the rectan- gle contained by those lines. This is a consequence of the truth of the equation, {a—lf=a^—2ab-\-h^. 405. Theorem. — The rectangle contained by the sum and the difference of two straight lines is equivalent to the difference of the squares of those lines. Geom.— ]2 ab ¥ a? ab 138 ELEMENTS OP GEOMETRY. This, again, is proved by the principle expressed in the equation, (a + b) {a—h)^a^—b\ 406. These two theorems may also be demonstrated by purely geometrical reasoning. The algebraic method is sometimes called the modern, while the other is called the ancient geometry. The algebraic method was invented by Descartes, in the seventeenth century, while the other ia twenty centuries older. THE PYTHAGOREAN THEOREM. 407. Since numerical equations represent geomet- rical truths, the following theorem might be inferred from Article 327. This is called the Pythagorean Theorem, because it was discovered by Pythagoras. It is also known as the Forty -seventh Proposition, that being its number in the First Book of Euclid's Elements. It has been demonstrated in a great variety of ways. One is by dividing the three squares into parts, so that the several parts of the large square are respectively equal to the several parts of the two others. The fame of this theorem makes it proper to give here the demonstration from Euclid. 408. Theorem — The square described on the hypote- nuse of a right angled triangle is equivalent to the sum of the squares described on the tivo legs. Let ABO be a right angled triangle, having the right angle BAG. The square described on the side BO is equivalent to the sum of the two squares described on BA and AO. Through A make AL parallel to BD, and join AD and FC. EQUlVALJiNT SURFACES. 139 Then, because each of the angles BAG and BAG is a right angle, the line GAG is one straight line (100). For the same reason, BAH is one straight line. The angles FBC and DBA are equal, since each is the sum of a right angle and the an- gle ABG. The two tri- angles FBG and DBA are equal, for the side FB in the one is equal to BA in the other, and the side BG in the one is equal to BD in the other, and the included angles are equal, as just proved. Now, the area of the parallelogram BL is double that of the triangle DBA, because they have the same base BD, and the same altitude DL (389). And the area of the square BG is double that of the triangle FBC, be- cause these also have the same base BF, and the same altitude FG. But doubles of equals are equal (7). Therefore, the parallelogram BL and the square BG are equivalent. In the same manner, by joining AE and BK, it is demonstrated that the parallelogram GL and the square GH are equivalent. Therefore, the whole Square BE, described on the hypotenuse, is equivalent to the two squares BG and GH, described on the legs of the right angled triangle. 409. Corollary. — The square described on one leg is equivalent to the difference of the squares on the hypot- enuse and the other leg. 140 ELEMENTS OF GEOMETRY. 410. If from the extremities of one line perpen- diculars be let fall upon another, then the part of the second line between the perpendiculars is called the projection of the first line on the second. If one end of the first line is in the second, then only one perpen- dicular is necessary;. 411. Theorem — The square described on the side oppo- site to an acute angle of a triangle, is equivalent to the sum of the squares described on the other two sides, diminished by twice the rectangle contained by one of these sides and the projection of the other on that side. Let A be the acute angle, and from B let a perpen- dicular fall upon AC, produced if necessary. Then, AD is the projection of AB upon AC. And it is to be proved that the square on BC is equivalent to the sum of the squares on AB and on AC, diminished by twice the rectangle contained by AC and AD. 2 BD = AB — AD; 2 2 2 CD = AC + AD— 2AC X AD. For (409), and (404), 2 2 2 2 By addition, BD + CD =:: AB + AC— 2AC X AD. But the square on BC is equivalent to BD + CD (408). Therefore, it is also equivalent to AB-f AC— 2ACXAD. EQUIVALENT SURFACES. 141 413. Theorem. — The square described on the side oppo- site an obtuse angle of a triangle, is equivalent to the sum of the squares described on the other two sides, increased by twice the rectangle of one of those sides and the pro- jection of the other on that side. In the triangle ABC, the square on BC which is op- posite the obtuse angle at B A, is equivalent to the sum of the squares on AB and on AC, and twice the rect- angle contained by CA and AD. For, BD = AB— AD; 2 2 2 and (403), CD = AC + AD + 2AC X AD. By addition, BD-i-CD=AB + AC-i- 2AC X AD. But, BC=BD + CD. Therefore, BC is equivalent to AB + AC + 2ACXAD. 413. Corollary. — If the square described on one side of a triangle is equivalent to the sum of the squares described on the other two sides, then the opposite an- gle is a right angle. For the last two theorems show that it can be neither acute nor obtuse. EXERCISES. 4J4. 1. When ii quadrilateral has its opposite angles supple- mentary, a circle can be circumscribed about it. 2. From a, given isosceles triangle, to cut off a trapezoid which 142 ELEMENTS OF GEOMETRY. shall have the same base as the triangle, and the remaining three sides equal to each other. 3. The lines which bisect the angles of a parallelogram, form a rectangle whose diagonals are parallel to the sides of the paral- lelogram. 4. In any parallelogram, th5 distance of one vertex from ^ straight line passing througli the opposite vertex, is equal to tliy sum or difference of the distances of the line from the other two vertices, according as the line is without or within the paral- lelogram. 5. When one diagonal of a quadrilateral divides the figure into equal triangles, is the figure necessarily a parallelogram? 6. Demonstrate the theorem, Article 329, by Articles 113 and 387. 7. What is the area of a lot, which has the shape of a right angled triangle, the longest side being 100 yards, and one of the other sides 36 yards. 8. Can every triangle be divided into two equal parts? Into three? Into nine? 9. Two parallelograms having the same base and altitnle are equivalent. To be demonstrated without using Articles 379 or 383. 10. A triangle is divided into two equivalent parts, by a line from the vertex to the middle of the base. To be demonstrated without the aid of the principles of this chapter. 11. To divide a triangle into two equivalent parts, by a line drawn from a given point in one of the sides. 12. Of all equivalent parallelograms having equal bases, what one has the minimum perimeter? 13. Find the locus of the points such that the sum of the squares of the distances of each from two given points, shall be equivalent to the square of the line joining the given points. POLYGONS. 143 CHAPTER VII. POLYGONS. 415. Hitherto the student's attention has been given to polygons of three and of four sides only. He has seen how the theories of similarity and of linear ratio have grown out of the consideration of triangles; and how the study of quadrilaterals gives us the principles for the measure of surfaces, and the theory of equiva- lent figures. In the present chapter, some principles of polygons of any number of sides will be established. A Pentagon is a polygon of five sides ; a Hexagon has six sides ; an Octagon, eight ; a Decagon, ten ; a Dodecagon, twelve ; and a Pentedecagon, fifteen. The following propositions on diagonals, and on the sum of the angles, are more general statements of those in Articles 340 to 346, DIAGONALS. 416. Theorem The number of diagonals from any vertex of a polygon, is three less than the number of sides. For, from each vertex a diagonal may extend to every other vertex except itself, and the one adjacent on each side. Thus, the number is three less than the number of vertices, or of sides. 417. Corollary The diagonals from one vertex di- 144 ELEMENTS OF GEOMETRY. vide a polygon into as many triangles as the polygon has sides, less two. Polygons may be divided into this number, or into a greater number of triangles, in various ways; but a polygon can not be divided into a less number of tri- angles than here stated. 418. Corollary The whole number of diagonals pos- sible in a polygon of n sides, is ^ w (ra — 3). For, if we count the diagonals at all the n vertices, we have n {n — 3), but this is counting each diagonal at both ends. This last product must therefore be divided by two. EQUAL POLYGONS. 419. Theorem. — Two polygons are equal when they are composed of the same number of triangles respectively equal and similarly arranged. This is an immediate consequence of the definition of equality (40). 430. Corollary. — Conversely, two equal polygons may be divided into the same number of triangles respect- ively equal and similarly arranged. 431. Theorem — Two polygons are equal when all the sides and all the diagonals from one vertex of the one, are respectively equal to the same lines in the other, and are similarly arranged. For each triangle in the one would have its three sides equal to the similarly situated triangle in the other, and would be equal to it (282). Therefore, the polygons would be equal (419). 432. Theorem — Two polygons are equal when aU the sides and the angles of the one are respectively equal to the same parts of the other, and are similarly arranged. POLYGONS. 145 For each triangle in the one is equal to its homolo- gous triangle in the other, since they have two sides and the included angle equal. It is enough for the hypothesis of this theorem, that all the angles except three be among the equal parts. SUM OF THE ANGLES. 423. Theorem — The sum of all the angles of a poly- gon is equal to twice as many right angles as the polygon has sides, less two. Eor the polygon may be divided into as many trian- gles as it has sides, less two (417); and the angles of these triangles coincide altogether with those of the polygon. The sum of the angles of each triangle is two right angles. Therefore, the sum of the angles of the poly- gon is equal to twice as many right angles as it has sides, less two. The remark in Article 346 applies as well to this theorem. 434. Let R represent a right angle; then the sum of the angles of a polygon of n sides is 2 (n — 2) R; or, it may be written thus, (2m — 4) R. The student should illustrate each of the last five theorems with one or more diagrams. 425. Theorem. — If each side of a convex polygon be produced, the sum of all the exterior angles is equal to four right angles. Let the sides be produced all in one way; that is, all to the right or all to the left. Then, from any point in the plane, extend lines parallel to the sides thus pro- duced, and in the same directions. Geom. — 13 146 ELEMENTS OF GEOMETRY. The angles thus formed are equal in number to the exterior angles of the polygon, and are re- spectively equal to them (138). But the sum of \ / those formed about the "-*',« point is equal to four \ right angles (92). Therefore, the sum of the exterior angles of the poly- gin is equal to four right angles. 4S6. This theorem will also be true of concave poly- gons, if the angle formed by producing one side of the reentrant angle is considered as a negative quantity. Thus, the remainder, after subtracting the angle formed at A by producing GA, from the sum of the angles formed at B, C, D, E, F, and G, is four right angles. This may be demonstrated by the aid of the previous theorem (423). EXERCISES. A^T.— 1. What is the number of diagonals that can be in a pentagon? In a decagon? 2. What is the sum of the angles of a hexagon? Of a dodec- agon ? 3. What is the greatest number of acute angles which a con- vex polygon can have? 4. Join any point within a given polygon with every vertex of the polygon, and with the figure thus formed, demonstrate the theorem, Article 423. 5. Demonstrate the theorem, Article 425, by means of Article 23, and without using Article 92. POLYGONS. 147 PROBLEMS IN DRAWING. 498. Problem — To draw a polygon equal to a given polygon. By diagonals divide the given polygon into triangles. The prob- lem then consists in drawing triangles equal to given triangles. 4S9. Problem — To draw a polygon when all its sides and all the diagonals from one vertex, are given in their proper order. This consists in drawing triangles with sides equal to three given lines (295). 430. Problem — To draw a polygon when the sides and angles are given in their order. It is enough for this problem if all the angles except three be given. For, suppose first that the an- gles not given are consecutive, as at D, ^^-'^7 ~~»D B, and C. Then, draw the triangles a, e, i, and o (297). Then, having DC, com- plete the polygon by drawing the trian- ^ " / ,-' / tt ^g gle DBC from its three known sides (295). Suppose the angles not given were D, C, and F. Then, draw the tri- angles a, e, and i, and separately, the triangle u. Then, having the three sides of the triangle o, it may be drawn, and the poly- gon completed. SIMILAR POLYGONS. 431. Theorem Similar polygons are composed of the same number of triangles, respectively similar and simi- larly arranged. Since the figures are similar, every angle in one has 148 ELEMENTS OF GEOMETUr. its corresponding equal angle in the other (303). If, then, diagonals be made to divide one of the polygons into triangles, every angle thus formed may have its corresponding equal angle in the other. Therefore, the triangles of one polygon are respectively similar to those of the other, and are similarly arranged. 433. Theorem — If two polygons are composed of the same number of triangles which are respectively similar and are similarly arranged, the polygons are similar. By the hypothesis, all the angles formed by the given lines in one polygon have their corresponding equal angles in the other. It remains to be proved that an- gles formed by any other lines in the one have their corresponding equal angles in the other polygon. This may be shown by reasoning, in the same man- ner as in the case of triangles (304). Let the student make the diagrams and complete the demonstration. 433. Theorem — Two polygons are similar when the angles formed hy the sides are respectively equal, and there is the same ratio between each side of the one and its homologous side of the other. Let all the diagonals possible extend from a vertex A BC N OBLIQUE LINES AND PLANES. 529. Theorem If from a point without a plane, a perpendicular and oblique lines he extended to the plans, then two oblique lines which meet the plane at equal dis' ianoes from the foot of the perpendicular, are equal. Let AB be perpendicular, and AC and AD oblique to the plane MN, and the dis- tances BC and BD equal. Then the triangles ABC and ABD are equal (284), and AC is equal to AD. 530. Corollary — A perpendicular is the shortest line from a point to a plane. Hence, the distance from a point to a plane is measured by a perpendicular line. 531. Corollary — All points of the circumference of a circle are equidistant from any point of its axis. 533. If from all points of a line perpendiculars be let fall upon a plane, the line thus described upon the plane 's the projection of the given line upon the given plane. 533. Theorem. — The projection of a straight line upon a plane is a straight line. Let AB be the given line, and MN the given plane. Then, from the points A and B, let the perpen- diculars, AC and BD, fall upon the plane MN. Join CD. AC and BD, being per- pendicular to the same plane, are parallel (526), and lie in cne plane (121). Now, every perpendicular PLANE AND LINES. 183 to MN let fall from a point of AB, must be parallel to ED, and must therefore lie in the plane AD, and meet the plane MN in some point of CD. Hence, the straight line CD is the projection of the straight line AB on the plane MN. y There is one exception to this proposition. When the given line is perpendicular to the plane, its projection is a point. 534. Corollary. — A straight line and its projection on a plane, both lie in one plane. 933. Theorem The angle which a straight line makes with its projection on a plane, is smaller than the angle it maJces with any other line in the plane. Let AC be the given line, and BC its projection on the plane MN. Then the angle ACB is less than the angle made by AC with any other line in the plane, as CD. With C as a center and BC as a radius, de- scribe a circumference in the plane MN, cut- ting CD at D. Then the triangles ACD and ACB have two sides of the one respectively equal to two sides of the other. But the third side AD is longer than the third side AB '530). Therefore, the angle ACD is greater than the angle ACB (294). 536. Corollary — The angle ACE, which a line makes with its projection produced, is larger than the angle made with any other line in the plane. S3!7. The angle which a line makes with its projec 184 ELEMENTS OP GEOMETRY. tion in a plane, is called the Angle of Inclination of the line and the plane. PARALLEL LINES AND PLANE. 53S. Theorem If a straight line in a plane is paral- lel to a straight line not in the plane, then the second line and the plane can not have a common point. For if any line is parallel to a given line in a plane, and passes through any point of the plane, it will lie ■wholly in the plane (121). But, by hypothesis, the sec- ond line does not lie wholly in the plane. Therefore, it can not pass through any point of the plane, to what- ever extent the two may be produced. 539. Such a line and plane, having the same direc- tion, are called parallel. 510. Corollary. — If one of two parallel lines is par- allel to a plane, the other is also. 511. Corollary — A line which is parallel to a plane is parallel to its projection on that plane. 512. Corollary A line parallel to a plane is every- where equally distant from it. APPLICATIONS. 543. Three points, however placed, must always be in the same plane. It is on this principle that stability is more readily obtained by three supports than by a greater number. A three- logged stool must be steady, but if there be four legs, their ends should be in one plane, and the floor should be level. Many sur- veying and astronomical instruments are made with three legs. 544. The use of lines perpendicular to planes is very frequent in the mechanic arte. A ready way of constructing a line perpen- dicular to a plane is by the use of two squares (114). Place the luigle of each at the foot of the desired perpendicular, one side of DIEDRAL ANGLES. 185 each square resting on the plane surface. Bring their perpendic- ular sides together. Their position must then be that of a per- pendicular to the plane, for it is perpendicular to two lines in the plane. 545. When a circle revolves round its axis, the figure under- goes no real change of position, each point of the circumference taking successively the position deserted by another point. On this principle is founded the operation of millstones. Two circular stones are placed so as to have the same axis, to which their faces are perpendicular, being, therefore, parallel to each other. The lower stone is fixed, while the upper one is made to revolve. The relative position of the faces of the stones under- goes no change during the revolution, and their distance being properly regulated, all the grain which passes between them will be ground with the same degree of fineness. 546. In the turning lathe, the axis round which the body to be turned is made to revolve, is the axis of the circles formed by the cutting tool, which removes the matter projecting beyond a proper distance from the axis. The cross section of every part of the thing turned is a circle, all the circles having the same axis. DIEDKAL ANGLES. 54'7. A DiEDRAL Angle is formed by two planes meeting at a common line. This figure is also called simply a diedral. The planes are its faces, and the in- tersection is its edge. In naming a diedral, four letters are used, one in each face, and two on the edge, the letters on the edge being between the other two. This figure is called a diedral angle, because it is simi- lar in many respects to an angle formed by two lines. MEASURE OF DIEDRALS. 548. The quantity of a diedral, as is the case with a linear angle, depends on the diflference in the directions Gconi. — 16 180 ELEMENTS OF GEOMETRY. of the faces from the edge, without regard to the extent of the planes. Hence, two diedrals are equal when they can be so placed that their planes will coincide. 549. Problem, — One diedral may be added to another. In the diagram, AB, AC, and AD represent three planes having the common intersection AE. Evidently the sum of BEAC and CEAD is equal to BEAD. 530. Corollary Diedrals may be subtracted one from another. A diedral may be bisected or divided in any required ratio by a plane pass- ing through its edge. 551. But there are in each of these planes any num- ber of directions. Hence, it is necessary to determine which of these is properly the direction of the face from the edge. For this purpose, let us first establish the following principle: 553. Theorem. — One diedral is to another as the plane angle, formed in the first hy a line in each face perpen- dicular to the edge, is to the similarly formed angle in the other. Thus, if FO, GO, and HO are each perpendicu- lar to AE, then the die- dral CEAD is to the die- dral BEAD as the angle GOH is to the angle FOH. This may be de- monstrated in the same manner as the proposi- tion in Article 197. DIEDRAL ANGLES. 187 553. Corollary A diedral is said to be measured by the plane angle formed by a line in each of its faces perpendicular to the edge. 554. Corollary — Accordingly, a diedral angle may be acute, obtuse, or right. In the last case, the planes are perpendicular to each other. 555. Many of the principles of plane angles may be applied to diedrals, without further demonstration. All right diedral angles are equal (90). When the sum of several diedrals is measured by two right angles, the outer faces form one plane (100). When two planes cut each other, the opposite or ver- tical diedrals are equal (99). PEEPENDICULAR PLANES. 556. Theorem — If a line is perpendicular to a plane, then any plane passing through this line is perpendicular to the other plane. If AB in the plane PQ is perpendicular to the piano MN, then AB must be perpen- dicular to every line in MN which passes through the point B (519); that is, to RQ, the intersection of the two planes, and to BC, which is made perpendicular to the in- tersection RQ. Then, the an- gle ABC measures the inclina- tion of the two planes (553), and is a right angle. There- fore, the planes are perpendicular. 557. Corollary — Conversely, if a plane is perpen- dicular to another, a straight line, which is perpendicu- M N 188 ELEMENTS OF GEOMETRY. lar to one of them, at some point of their intersection, must lie wholly in the other plane (524). 558. Corollary If two planes are perpendicular to a third, then the intersection of the first two is a line perpendicular to the third plane. OBLIQUE PLANES. 559. Theorem. — If from a point within a diedral, per- pendicular lines be made to the two faces, the^ angle of these lines is supplementary to 'the angle which measures the diedral. Let M and N be two planes whose intersection is AB, and OF and CE perpendicu- lars let fall upon them from the point C ; and DF and DE the intersections of the plane FOE with the two planes M and N. Then the plane FOE mast be perpendicular to each of the planes M and N (556). Hence, the line AB is perpendicular to the plane FCE (558), and the angles ADF and ADE are right angles. Then the angle FDE measures the diedral. But in the quadrilateral FDEC, the two angles F and E are right angles. Therefore, the other two angles at C and D are supplementary. 560. Theorem — Every point of a plane which bisects a diedral is equally distant from its two faces. Let the plane FC bisect the diedral DBCE. Then it is to be proved that every point of this plane, as A, for example, is equally distant from the planes DC and EC. From A let the perpendiculars AH and AI fall upon the faces DC and EC, and let 10, AG, and HO be the DIEDRAL ANGLES. 189 intersections of the plane of the angle I AH with the three given planes. Then it may be shown, as in the last theorem, that the angle HOA measures the diedral FBCD, and the angle lOA the diedral FBCE. But these diedrals are equal, by hypothesis. Therefore, the line AO bisects the angle lOH, B B and the point A is equally distant from the lines OH and 01 (113). But the distance of A from these lines is measured by the same perpendiculars, AH and AI, which measure its distance from the two faces DC and EC. Therefore, any point of the bisecting plane is equally distant from the two faces of the given diedral. APPLICATIONS. 561. Articles 548 to 554 are illustrated by a door turning on its hinges. In every position it is perpendicular to the floor and ceiling. As it turns, it changes its inclination to the wall, in which it is. constructed, the angle of inclination being that which is formed by the upper edge of the door and the lintel. 563. The theory of diedrals is as important in the study of magnitudes bounded by planes, as is the theory of angles in the study of polygons. This is most striking in the science of crystallography, which teaches us how to classify mineral substances according to their geometrical forms. Crystals of one kind have edges of which the diedral angles measure a certain number of degrees, and crystals of another kind have edges of a different number of degrees. Crystals of many species may be thus classified, by measuring their diedrals. 563. The plane of the surface of a liquid at rest is called hori- zontal, or the plane of the horizon. The direction of a plumb- 190 ELEMENTS OF GEOMETRY. line when the weight is at rest, is a vertical line. The vertical line is perpendicular to the horizon, the positions of both being governed by the same causes. Every line in the plane of the horizon, or parallel to it, ia called a horizontal line, and every plane passing through a vertical line is called a vertical plane. Every vertical plane is perpendicular to the horizon. Horizontal and vertical planes are in most frequent use. Floors, ceilings, etc., are examples of the former, and walls of the latter. The methods of using the builder's level and plummet to determ- ine the position of these, are among the simplest applications of geometrical principles. Civil engineers have constantly to observe and calculate the position of horizontal and vertical planes, as all objects are re- ferred to these. The astronomer and the navigator, at every step, refer to the horizon, or to a vertical plane. EXERCISES. 564. — 1. If, from a point without a plane, several equal oblique lines extend to it, they make equal angles with the plane. 2. If a line is perpendicular to a plane, and if from its foot a perpendicular be let fall on some other line which lies in the plane, then this last line is perpendicular to the plane of the other two. 3. What is the locus of those points in space, «ach of which is equally distant from two given points? PARALLEL PLANES. S6S. Two planes ■which are perpendicular to the same straight line, at different points of it, are both fixed in po- sition (525), and they have the same directions. If the parallel lines AB and CD revolve about the line EF, to -which they are both perpendicular, then each of the revolving lines describes a plane. -1> Every direction assumed by one line is the same as PARALLEL PLANES. 191 that of the other, and, in the course of a complete revo- lution, they take all the possible directions of the two planes. Two planes which have the same directions are called parallel planes. Parallelism consists in having the same direction, whether it be of two lines, of two planes, or of a line and a plane. 560. Corollary — Two planes parallel to a third are parallel to each other. SST. Corollary. — Two planes perpendicular to the same straight line are parallel to each other. 568. Corollary — A straight line perpendicular to one of two parallel planes is perpendicular to the other. 569. Corollary. — Every straight line in one of two parallel planes has its parallel line in the other plane. Therefore, every straight line in one of the planes is parallel to the other plane. 570. Corollary — Since through any point in a piano there may be a line parallel to any line in the same plane (121), therefore, in one of two parallel planes, and through any point of it, there may be a straight line parallel to any straight line in the other plane. 571. Theorem. — Two parallel planes can not meet. For, if they had a common point, being parallel, they would have the same directions from that point, and therefore would coincide throughout, and be only one plane. STS. Theorem. — The intersections of two parallel planes hy a third plane are parallel lines. Let AB and CD be the intersections of the two par- allel planes M and N, with the plane P. Now, if through C there be a line parallel to AB, i4 102 ELEMENTS OF GEOMETRY. M must lie in the plane P (121), and also in the plane N (570). Therefore, it is the in- tersection CD, and the two in- tersections are parallel lines. When two parallel planes are cut by a third plane, eight diedrals are formed, which have properties similar to those of Articles 124 to 128. 573. Theorem The parts of two parallel lines inter- cepted between parallel planes are equal. For, if the lines AB and CD are parallel, they lie in one plane. Then AC and BD are the intersections of thi^ plane with the two parallel planes M and P. Hence, AC is parallel to BD, and AD is a parallelogram. Therefore, AB is equal to the opposite side CD. 574. Theorem. — Two parallel planes are everywhere equally distant. For the shortest distance from any point of one plane to the other, is measured by a perpendicular. But these perpendiculars are all parallel (526), and therefore equal to each other. 575. Theorem — If the two sides of an angle are each parallel to a given plane, then the plane of that angle is parallel to the given plane. If AB and AC are each parallel to the plane M, then the plane of BAG is parallel to the plane M. From A let the perpendicular AD fall upon the plane M, and let the projections of AB and AC on the plane ' M be respectively DE and DF, PARALLEL PLANES. 193 Since DE is parallel to AB (541), DA is perpendic- ular to AB (127). For a like reason, DA is per- pendicular to AC. There- fore, DA is perpendicular to the plane of BAG (517), and the two planes being perpendicular to the same line are parallel to each other (567). 576. Theorem — If two straight lines which cut each other are respectively parallel to two other straight lines which cut each other, then the plane of the first two is parallel to the plane of the second two. Let AB be parallel to EF, and CD parallel to GH. Then the planes M and P are parallel. For AB being parallel to EF, is parallel to the plane P in which it lies (538). Also, CD is par- allel to the plane P, for the same reason. There- fore, the plane M is par- allel to the plane P (575). 577. Corollary. — The angles made by the first two lines are respectively the same as those made by the sec- ond two. For they are the differences between the same directions. This includes ' the corresponding principle of Plane Geometry. 578. Theorem. — Straight lines cut by three parallel planes are divided proportionally. If the line AB is cut at the points A, E, and B, and Geom. — 17 194 ELEMENTS OP GEOMETRY. the line CD at the points C, F, and D, by the parallel planes M, N, and P, then AE : EB : : CF : FD. Join AC, AD, and BD. AD pierces the plane N in the point G. Join EG and GF. Now, EG and BD are par- allel, being the intersections of the parallel planes N and P by the third plane ABD (572). Hence (313), AE EB : : AG : GD For a like reason, AG GD : : CF FD. Therefore, AE EB : : CF FD. APPLICATION. 5f9. The general problem oi perspective in drawing, consista in representing upon a plane surface the apparent form of ob- jects in sight. This plane, the plane of the picture, is supposed to be between the eye and the objects to be drawn. Then each object is to be represented upon the plane, at the point where it would be pierced by the visual ray from the object to the eye. All the visual rays from one straight object, such as the top of a wall, or one corner of a house, lie in one plane (60). Their intersection with the plane of the picture must be a straight line (516). Therefore, all straight objects, whatever their posijon, must be drawn as straight lines. Two parallel straight objects, if they are also parallel to the plane of the picture, will remain parallel in the perspective. For the lines drawn must be parallel to the objects (572), and there- fore to each other. Two parallel lines, which are not parallel to the plane of the picture, will meet in the perspective. They will meet, if produced. TJilEDRALS. 195 at that point where the plane of the pictur6 is pierced by a Hne from the eye parallel to the given lines. EXERCISES. 5SO. — 1. A straight line makes equal angles with two paral- lel planea 2. Two parallel lines make the same angle of inclination with a given plane. 3. The projections of two parallel lines on a plane are parallel. 4. When two planes are each perpendicular to a third, and their intersections with the third plane are parallel lines, then the two planes are parallel to each other. 5. If two straight lines be not in the same plane, one straight line, and only one, may be perpendicular to both of them, 6. Demonstrate the last sentence of Article 579. TRIEDRALS. 581. When three planes cut each other, three cases are possible. 1st. The intersections may coincide. Then six diedrals are formed, having for their common edge the intersection of the three planes. 2d. The three intersections may be parallel lines. Then one plane is parallel to the intersection of the other two. 190 ELEMENTS OF GEOMETRY. 3d. The three intersections may meet at one point. Then the space about the point is divided by the three planes into eight parts. The student will apprehend this better when he reflects that two intersecting planes make four di- edrals. Now, if a third plane cut through the intersection of the first two, it will divide each of the diedrals into two parts, making eight in all. Each of these parts is called a triedral, because it has three faces. A fourth case is impossible. For, since any two of the intersections lie in one plane, they must either be parallel, or they meet. If two of the intersections meet, the point of meeting must be common to the three planes, and must therefore be common to all the in- tersections. Hence, the three intersections either have more than one point common, only one point common, or no point common. But these are the three cases just considered. 58S. A Teiedkal is the figure formed by three planes meeting at one point. The point where the planes and intersections all meet, is called the vertex of the trie- dral. The intersections are its edges, and the planes are its faces. The corners of a room, or of a chest, are illustrations of triedrals with rectangular faces. The point of a tri- angular file, or of a small-sword, has the form of a triedral with acute faces. TRIEDRALS. 197 The triedral has many things analogous to the plane triangle. It has been called a solid triangle ; and more frequently, but with less propriety, a solid angle. The three faces, combined two and two, make three diedrals, and the three intersections, combined two and two, make three plane angles. These six are the six elements or principal parts of a triedral. Each face is the plane of one of the plane angles, and two faces are said to be equal when these angles are equal. Two triedrals are said to be equal when their several planes may coincide, without regard to the extent of the planes. Since each plane is determined by two lines, it is evident that two triedrals are equal when their several edges respectively coincide. 583. A triedral which has one rectangular diedral, that is, whose measure is a right angle, is called a rect- angular triedral. If it has two, it is hiredangvlar ; if it has three, it is trireetangidar. A triedral which has two of its faces equal, is called isosceles; if all three are equal, it is equilateral. SYMMETRICAL TRIEDRALS. 584. If the edges of a triedral be produced beyond the vertex, they form the edges of a new triedral. The faces of these two triedrals are respect- ively equal, for the angles are vertical. Thus, the angles ASC and ESD are equal ; also, the angles BSC and FSE are equal, and the an- gles ASB and DSF. The diedrals whose edges are FS and BS are also 198 ELEMENTS OF GEOMETRY. equal, since, being formed by the same planes, EFSBC and DPSBA, they are vertically opposite diedrals (555). The same is true of the diedrals whose edges are DS and SA, and of the diedrals whose edges are ES and SO. In the diagram, suppose ASB to be the plane of the paper, C being above and E below that plane. But the two triedrals are not equal, for they can not be made to coincide, although composed of parts which are respectively equal. This will be more evident if the student will imagine himself within the first triedral, his head toward the vertex, and his back to the plane ASB. Then the plane ASC will be on the right hand, and BSC on the left. Then let him imagine himself in the other triedral, his head toward the vertex, and his back to the plane FSD, which is equal to ASB. Then the plane on the right will be FSE, which is equal to BSC, the one that had been on the left; and the plane now on the left will be DSE, equal to the one that had been on the right. Now, since the equal parts are not similarly situated, the two figures can not coincide. Then the difference between these two triedrals con- sists in the opposite order in which the parts are ar- ranged. This may be illustrated by two gloves, which we may suppose to be composed of exactly equal parts. But they are arranged in reverse order. The right hand glove will not fit the left hand. The two hands themselves are examples of the same kind. 585. When two magnitudes are composed of parts respectively equal, but arranged in reverse order, they are said to be symmetrical magnitudes. The werd symmetrical, as here used, has essentially the same meaning as that given in Plane Geometry (158). Two symmetrical plane figures, or parts of a figure, are TRIEDRALS. 199 divided by a straight line, while two such figures in space are divided by a plane. When two plane figures are symmetrical, they are also equal, for one can be turned over to coincide with the other, as with the figures m and n in Article 282. But this is not possible, as just shown, with figures that are not in one plane. ANGLES OF A TEIEDEAL. 586. Theorem — Each plane angle of a triedral is less than the sum of the other two. The theorem is demonstrated, when it is shown that the greatest angle is less than the sum of the other two. Let ASB be the largest of the three angles of the triedral S. Then, from the angle ASB take the part ASD, equal to the angle ASC. Join the edges SA and SB by any straight line AB. Take SO equal A, to SD, and join AC and BC. Since the triangles ASD and ASC are equal (284), AD is equal to AC. But AB is less than the sum of AC and BC, anJ from these, subtracting the equals AD and AC, we have BD less than BC. Hence, the trian- gles BSD and BSC have two sides of the one equal to two sides of the other, and the third side BD less than t6e third side BC. Therefore, the included angle BSD is less than the angle BSC. Adding to these the equal angles ASD and ASC, we have the angle ASB less than the sum of the angles ASC and BSC. 587. Theorem. — The sum of the plane angles which form a triedral is always less than four right angles. 200 ELEMENTS OF GEOMETRY. Through any three points, one in each edge of the triedral, let the plane ABO pass, making the intersec- tions AB, BC, and AC, with the faces. There is thus formed a triedral at each of the points A, B, and 0. Then the angle BAG is less than the sum of BAS and CAS (586). The angle ABC is less than the sum of ABS and CBS. The angle BCA is less than the sum of ACS and BCS. Adding together these inequalities, we find that the sum of the angles of the triangle ABC, which is two right angles, is less than the sum of the six angles at the bases of the tri- angles on the faces of the triedral S. The sum of all the angles of these three triangles is six right angles. Therefore, since the sum of those at the bases is more than two right angles, the sum of those at the vertex S must be less than four right angles. 58S. To assist the student to understand this theo- rem, let him take any three points on the paper or blackboard for A, B, and C. Take S at some distance from the surface, so that the plane angles formed at S will be quite acute. Then let S approach the surface of the triangle ABC. Evidently the angles at S be- come larger and larger, until the point S touches the surface of the triangle, when the sum of the angles becomes four right angles, and, at the same time, the triedral becomes one plane. SUPPLEMENTARY TEIEDBALS. 589. Theorem. — If, from a point within a triedral, perpendicular lines fall on the several faces, these lines TRIEDRALS. 201 will be the edges of a second triedral, whose faces will he supplements respectively of the diedrals of the first; and the faces of the first will he respectively supplements of the diedrals of the second triedral. A plane angle is not strictly the supplement of a die- dral, but we understand, by this abridged expression, that the plane angle is the supplement of that which measures the diedral. If from the point E, within the triedral ABCD, the perpendiculars EF, EG, and EH fall on the several faces, then these lines form a second trie- dral, whose faces are FEH, FEGr, and GEH. Then the angle FEH is the supplement of the diedral whose edge is BA, for the sides of the angle are perpendicular to the faces of the diedral (559). For the same reason, the angle FEG is the supplement of the diedral whose edge is CA, and the angle GEH is the supplement of the diedral whose edge is DA. But it may be shown that these two triedrals have a reciprocal relation; that is, that the property just proved of the second toward the first, is also true of the first toward the second. Let BF and BH be the intersections of the face FEH with the faces BAG and BAD ; OF and CG be the inter- sections of the face FEG with the faces BAG and GAD ; and DG and DH be the intersections of the face GEH with the faces CAD and BAD. Now, since the plane FBH is perpendicular to each of the planes BAG and BAD (556), their intersection AB is perpendicular to the plane FBH (558). For a 202 ELEMENTS OF GEOMETRY. like reason, AC is perpendicular to the plane FCG and AD is perpendicular to the plane GDH. Then, reason- ing as above, ■we prove that the angle BAG is the sup- plement of the diedral whose edge is FE ; and that each of the other faces of the first triedral is a supplement of a diedral of the second. S90. Two triedrals, in which the faces and diedral angles of the one are respectively the supplements of the diedral angles and faces of the other, are called supplementary triedrals. Instead of placing supplementary triedrals each within the other, as above, they may be supposed to have their vertices at the same point. Thus, at the point A, erect a perpendicular to each of the three faces of the trie- dral ABCD, and on the side of the face toward the triedral. A second triedral is thus formed, which is supplementary to the triedral ABCD, and is symmet- rical to the one formed within. SUM OF THE DIEDEALS. 501. Theorem. — In every triedral the sum of the three diedral angles is greater than two right angles, and less than six. Consider the supplementary triedral, with the given one. Now, the sum of the three diedrals of the given triedral, and of the three faces of its supplementary tri- edral, must be six right angles; for the sum of each pair is two right angles. But the sum of the faces of the supplementary triedral is less than four right angles (587), and is greater than zero. Subtracting this sum from the former, the remainder, being the sum of the three diedrals of the given triedral, is greater than two ml less than six right angles. TRIEDRALS. 203 EQUALITY OF TKIEDEALS. 59S. Theorem — When two triedrals hqve two faces, and the included diedral of the one respectively equal to the corresponding parts of the other, then the remaining face and diedrals of the first are respectively equal to the corresponding parts of the other. There are two cases to be considered. 1st. Suppose the angles AEO and BOG equal, and the angles AEI and BCD equal, jilso the included diedrals whoso edges are AE and BC. Let the ar- rangement be the same in both, so that, if we go around one triedral in the order 0, A, I, 0, and around the other in the order G, B, D, G, in both cases the triedral will be on the right. Then it may be proved that the two triedrals are equal. Place the angle BCD directly upon its equal, AEI. Since the diedrals are equal, and are on the same side of the plane AEI, the planes BCG and AEO will coin- cide. Since the angles BCG and AEO are equal, the lines CG and EO will coincide. Thus, the angles DCG and lEO coincide, and the two triedrals coincide throughout. 2d. Let the angles AEO and DCG be equal, and the aiigles AEI and BCD, also the included diedrals, whose edges are AE and DC. But let the arrangement be re- verse ; that is, if we go around one triedral in the order 0, A, 1, 0, and around the other in the order G, D, B, G, 20 t ELEMENTS OF GEOMETKY. in one case the triedral will be to the right, and in the other it will be to the left of us. Then it may be proved that the two triedrals are symmetrical. One of the triedrals can be made to coincide with the symmetrical of the other ; for if the edges EC, GO, and DC be produced beyond C, the triedral CFHK will have two faces and the included .^ JK diedral respect- flf— T --'-■•■tF ively e q u a 1 to \ i -'' those parts of the \ \ triedral EAOI, \i / and arranged in the same order; that is, the re- verse of the tri- edral CDGB. Hence, as just shown, the trie- drals CFHK and EAOI are equal. Therefore, EAOI and CDGB are symmetrical triedrals. In both cases, all the parts of each triedral are re- spectively equal to those of the other. 593. Theorem — When two triedrals have one face and the two adjacent diedrals of the one respectively equal to the corresponding parts of the other, then the remaining faces and diedral of the first are respectively equal to the corresponding parts of the other. Suppose that the faces AEI and BCD are equal, that the diedrals whose edges are AE and BC are equal, that the diedrals whose edges are IE and DC are equal, and that these parts are similarly arranged in the two trie- drals. Then the one may coincide with the other. TRIEDRALS. 205 For BCD may coincide with its equal AEI, BC fall- ing on AE. Then the plane of BCG must coincide ■with that of AEO, since the diedrals are equal ; and the line CGr will fall in the plane of AEO. For a similar reason CG will fall on the plane of lEO. Therefore, it must coincide with their intersection EO, and the two triedrals coincide throughout. When the equal parts are in re- verse order in the two triedrals, the arrangement in one must be the same as in the sym- metrical of the other. Therefore, in that case, the two triedrals would be symmetrical. In both cases, all the parts of each triedral are re- spectively equal to those of the other. S94. Theorem — An isosceles triedral and its symmet- rical are equal. Let ABCD be an isosceles triedral, having the faces BAG and DAG equal, and let AEFG be its symmetrical triedral. Now, the faces BAG, DAG, FAG, and FAE, are all equal to each other. The diedrals whose edges are AG and AF being vertical, are also equal. Hence, the faces mentioned being all equal, corresponding equal parts may be taken in the same order in both triedrals ; that is, the face EAF equal to the face BAG, and the face FAG equal to GAD. Therefore, the two triedrals are equal. 206 ELEMENTS OF GEOMETRY. S95. Corollary In an isosceles triedral, the diedrala opposite the equal faces are equal. Fpr the diedrals whose edges are AB and AD, are each equal to the diedral whose edge is AE. 396. Corollary. — Conversely, if in any triedral two of the diedral angles are equal, then the faces opposite these diedrals are equal, and the triedral is isosceles. For, as in the above theorem, the given triedral can be shown to be equal to its symmetrical. 597. Theorem — When two triedrals have two faces of the one respectively equal to two faces of the other, and the included diedrals unequal, then the third faces are unequal, and that face is greater which is opposite the greater diedral. Suppose that the faces CBD and EAI are equal, and that the faces CBF and EAO are also equal, but that the diedral whose edge is CB is greater than the die- dral whose edge is EA. Then the face FBD will be greater than the face OAI. Through the line BC, let a plane GBC pass, making with the plane DBC a diedral equal to that whose edge is AB. In this plane, make the angle CBG equal to EAO. Let the diedral FBCG be bisected by the plane TRIEDRALS. 207 EEC, BH being the intersection of this plane with the plane FED. Then the two triedrals BCDGr and AEIO, having two faces and the included diedral in the one equal to the corresponding parts in the other, will have the remain- ing parts equal. Hence, the faces DEG and lAO aie equal. Again, the two triedrals BCFH and ECGH have the faces CEF and CEG equal, by construction, the face CEH common, and the included diedrals equal, by con- struction. Therefore, the third faces FBH and GBH are equal. To each of these equals add the face HBD, and we have the face FED equal to the sum of GEH and HED. Eut in the triedral EDGH, the face DEG is less thar the sum of the other two faces, GEH and HBD (586) Hence, the face DBG is less than FED. Therefore, the face OAI, equal to DBG, is less than FED. 598. Corollary — Conversely, when two triedrals have two faces of the one respectively equal to two faces of the other, and the third faces are unequal, then the die- dral opposite the greater face is greater than the diedral opposite the less. 599. Theorem When two triedrals have their three faces respectively equal, their diedrals will he respectively equal; and the two triedrals are either equal, or they are symmetrical. When two faces of one triedral are respectively equal to those of another, if the included diedrals are une- qual, then the opposite faces are unequal (597). Eut, by the hypothesis of this theorem, the third faces are equal. Therefore, the diedrals opposite to those faces must be equal. In the same manner, it may be shown that the other 208 ELEMENTS OF GEOMETRY. diedral angles of the one, are equal to the corresponding diedral angles of the other triedral. Therefore, the trie- drals are either equal or symmetrical, according to the arrangement of their parts. GOO. Theorem Two friedrals which have their die- drals respectively equal, have also their faces respectively equal; and the two triedrals are either equal, or they are symmetrical. Consider the supplementary triedrals of the two given triedrals. These ■will have their faces respectively equal, because they are the supplements of equal diedral an- gles (589). Since their faces are equal, their diedrals are equal (599). Then the two given triedrals, having iheir faces the supplements of these equal diedrals, will nave those faces equal ; and the triedrals are either equal or symmetrical, according to the arrangement of their parts. 601. The student may notice, in every other case of equal triedrals, the analogy to a case of equality of tri- angles ; but the theorem just demonstrated has nothing analogous in plane geometry. 602. Corollary. — All trirectangular triedrals are equal. 603. Corollary — In all cases where two triedrals are either equal or supplementary, equal faces are opposite equal diedral angles. EXERCISES. 604. — 1. In any triedral, the greater of two faces is opposite to the greater diedral angle; and conversely. 2. Demonstrate the principles stated in the last sentence of Article 590. 3. If a triedral have one right diedral angle, then an adjacent POLYEDRALS. 209 face and its opposite diedral are either both, acute, both right, or \)Oth obtuse. POLYEDKALS. 605. A PoLYBDRAL is the figure formed by several planes which meet at one point. Thus, a polyedral is composed of several angles having their vertices at a common point, every edge being a side of two of the angular faces. The triedral is a polyedral of three faces. 606. Problem — Any polyedral of more than three faces may he divided into triedrals For a plane may pass through any two edges which are not adjacent. Thus, a polyedral of four faces may be divided into two triedrals ; one of five faces, into three ; and so on. 607. This is like the division of a polygon into tri- angles. The plane passing through two edges not adja- cent is called a diag- onal plane. . A polyedral is //\ called convex, when / / i \\ / every possible diag- />Lj " \\ /<.' onal plane lies within / / ^"^^Vx / y'^ \\ the figure; otherwise ' ' it is called concave. 60S. Corollary If the plane of one face of a con- vex polyedral be produced, it can not cut the polyedral. 609. Corollary A plane may pass through the ver- tex of a convex polyedral, without cutting any face of the polyedral. 610. Corollary. — A plane may cut all the edges of a convex polyedral. The section is a convex polygon. de.nm. — 18 210 ELEMENTS OF GEOMETRY. 611. When any figure is cut by a plane, the figure that is defined on the plane by the limits of the figure so cut, is called a plane section. Several properties of triedrals are common to other polyedrals. 613. Theorem The sum of all the angles of a convex polyedral is less than four right angles. For, suppose the polyedral to be cut by a plane, then the section is a polygon of as many sides as the polye- dral has faces. Let n represent the number of sides of the polygon. The plane cuts off a triangle on each face of the polyedral, making n triangles. Now, the sum of the angles of this polygon is 2n — 4 right angles (424), and the sum of the angles of all these triangles is 2n right angles. Let v right angles represent the sum of the angles at the vertex of the polyedral ; then, 2n right angles being the sum of all the angles of the triangles, 2n — V is the sum of the angles at their bases. Now, at each vertex of the polygon is a triedral hav- ing an angle of the polygon for one face, and angles at the bases of the triangles for the other two faces. Then, since two faces of a triedral are greater than the third, the sum of all the angles at the bases of the tri- angles is greater than the sum of the angles of the polygon. That is, 2w — ■y>2w — 4. Adding to both members of this inequality, v + 4, and subtracting 2n, we have 4 > ti. That is, the sum of the angles at the vertex is less than four right angles. This demonstration is a generalization of that of Article 587. The student should make a diagram and special demonstration for a polyedral of five or six faces. DESCRIPTIVE GEOMETRY. 211 613. Theorem — In any convex polyedral, the sum of fJie diedrals is greater than the sum of the angles of a polygon having the same number of sides that the poly- edral has faces. Let the given polyedral be divided by diagonal planes into triedrals. Then this theorem may be demonstrated like the analogous proposition on polygons (423). The remark made in Article 346 is also applicable here. DESCRIPTIVE GEOMETRY. 614. In the former part of this work, vre have found problems in drawing to be the best exercises on the principles of Plane Geometry. At first it appears im- possible to adapt such problems to the Geometry of Space ; for a drawing is made on a plane surface, while the figures here investigated are not plane figures. This object, however, has been accomplished by the most ingenious methods, invented, in great part, by Monge, one of the founders of the Polytechnic School at Paris, the first who reduced to a system the elements of this science, called Descriptive Geometry. Dbsceiptivb Geometry is that branch of mathemat- ics which teaches how to represent and determine, by means of drawings on a plane surface, the absolute or relative position of points or magnitudes in space. It is beyond the design of the present work to do more than allude to this interesting and very useful science. EXBECISES. 615. — 1. What is the locus of those points in space, each of which is equally distant from three given points? 2. What is the locus of those points in space, each of which is equal!)' distant from two given planes? 212 ELEMENTS OF GEOMETRY. 3. What is the locus of those points in space, each of which is equally distant from three given planes ? 4. What is the locus of those points in space, each of which is equally distant from two given straight lines which lie in the same plane ? 5. What is the locus of those points in space, each of which is equally distant from three given straight lines which lie in the same plane ? 6. What is the locus of those points in space, such that the sum of the distances of each from two given planes is equal to a . given straight line ? 7. If each diedral of a triedral be bisected, the three planes have one common intersection. 8. If a straight line is perpendicular to a plane, every plane parallel to the given line is perpendicular to the given plane. 9. Given any two straight lines in space; either one plane may pass through both, or two parallel planes may pass through Ahem respectively. 10. In the second case of the preceding exercise, a line which is perpendicular to both the given lines is also perpendicular to the two planes. 11. If one face of a triedral is rectangular, then an adjacent diedral angle and its opposite face are eitlier both acute, both right, or both obtuse. 12. Apply to planes, diedrals, and triedrals, respectively, such properties of straight lines, angles, and triangles, as have not already been stated in this chapter, determining, in each case, whether the principle is true when so applied. TETRAEDRONS. 313 CHAPTER X. POLYEDRONS. 616. A PoLTEDKON is a solid, or portion of space, bounded by plane surfaces. Each of these surfaces is a face, their several intersections are edges, and the points of meeting of the edges are vertices of the poly- edron. 617. Corollary. — The edges being intersections of planes, must be straight lines. It follows that the faces of a polyedron are polygons. 618. A Diagonal of a polyedron is a straight line joining two vertices which are not in the same face. A Diagonal Plane is a plane passing through three vertices which are not in the same face. TETRAEDRONS. 619. We have seen that three planes can not inclose a space (581). But if any point be taken on each edge of a triedral, a plane passing through these three points /;\ would, with the three faces of >.;,; the triedral, cut oif a portion ^ :Si.;] of space, which would be in- ■"*■■■-• closed by four triangular faces. A Tetraedron is a polyedron having four faces. 214 ELEMENTS OF GEOMETRY. 630. Problem. — Any four points whatever, which do not all lie in one plane, may he taken as the four vertices of a tetraedron. For they may be joined two and two, by straight lines, thus forming the six edges ; and these bound the four triangular faces of the figure. 631. Either face of the tetraedron may be taken a;j the base. Then the other faces are called the sides, the vertex opposite the base is called the vertex of the tetraedron, and the altitude is the perpendicular distance from the vertex to the plane of the base. In some cases, the perpendicular falls on the plane of the base produced, as in triangles. 633. Corollary — If a plane parallel to the base of a tetraedron pass through the vertex, the distance between this plane and the base is the altitude of the tetrae- dron (574). 633. Theorem. — There is a point equally distant from the four vertices of any tetraedron. In the plane of the face BCF, suppose a circle whose circumference passes through the three points B, C, and F. At the center of this circle, erect a line perpendicular to the plane of BCF. Every point of this per- pendicular is equally distant from the three points B, C, and F (531). In the same manner, let a line perpendicular to the plane of BDF be erected, so that every point shall be equally distant from the points B, D, and F. These two perpendiculars both lie in one plane, the plane which bisects the edge BF perpendicularly at its TETRAEDRONS. 215 center (520). These two perpendiculars to two oblique planes, being therefore oblique to each other, will meet at some point. This point is equally distant from the four vertices B, C, D, and F. 634. Corollary. — The six planes which bisect perpen- dicularly the several edges of a tetraedron all meet in one point. But this point is not necessarily within the tetraedron. 655. Theorem. — There is a point within every tetrae- dron which is equally distant from the several faees. Let AEIO be any tetraedron, and let OB be the straight line formed by the intersection of two planes, A one of which bisects the /^/ \ diedral angle whose edge is ^^ / \ AO, and the other the die- ^^>^g--/-g \ dral whose edge is EO. ~\,^^ / ^^^-^-^ Now, every point of the I first bisecting plane is equally distant from the faces lAO and EAO (560) ; and every point of the second bisecting plane is equally distant from the faces EAO and BIO. Therefore, every point of the line BO, which is the intersection of those bisect- ing planes, is equally distant from those three faces. Then let a plane bisect the diedral whose edge is EI, and let C be the point where this plane cuts the line BO. Since every point of this last bisecting plane is equally distant from the faces EAI and EOI, it follows that the point C is equally distant from the four faces of the tet- raedron. Since all the bisecting planes are interior, the point found is within the tetraedron. 656. Corollary The six planes which bisect the several diedral angles of a tetraedron all meet at one point. 216 ELEMENTS OF GEOMETEY, EQUALITY OF TETEAEDRONS. 637. Theorem. — Two tetraedrons are equal when three faces of the one ara respectively equal to three faces of the other, and they are similarly arranged. For the three sides of the fourth face, in one, must be equal to the same lines in the other. Hence, the fourth faces are equal. Then each diedral angle in the one is equal to its corresponding diedral angle in the other (599). In a word, every part of the one figure is equal to the corresponding part of the other, and the equal parts are similarly arranged. Therefore, the two tetraedrons are equal. 638. Corollary. — Two tetraedrons are equal when the six edges of the one are respectively equal to those of the other, and they are similarly arranged. 6S9. Corollary — Two tetraedrons are equal when two faces and the included diedral of the one are respect- ively equal to those parts of the other, and they are similarly arranged. 630. Corollary — Two tetraedrons are equal when one face and the adjacent diedrals of the one are respect- ively equal to those parts of the other, and they are similarly arranged. 631. When tetraedrons are composed of equal parts in reverse order, they are symmetrical. MODEL TETRAEDRON. 633. The student may easily construct a model of a tetrae- dron when the six edges are given. First, with three of the edges which are sides of one face, draw the triangle, as ABC. Then, on each side of this first triangle, as a base, draw a triangle equal to the corresponding face; all of which can be done, for tho TETRAEDRONS. 217 edges, that is, the sides of these triangles, are given. Then, cut out the whole figure from the pa- per and carefully fold it at the lines AB, EC, and CA. Since BF is equal to BD, CF to CE, and AD to AE, the points F, D, and E may be united to form a vertex. In this vfay models of various forms may be made with more accuracy than in wood, and the student may derive much help from the work. But he must never forget that the geometrical figure exists only as an intellectual conception. To assist him in this, he should strive to generalize every demonstration, stating the argu- ment without either model or diagram, as in the demonstration last given. To construct models of symmetrical tetraedrons, the drawings may be equal, but the folding should, in the one case, be up, and in the other, down. SIMILAR TETRAEDRONS. 633. Since similarity consists in having the same form, so that every difference of direction in one of two similar figures has its corresponding equal differ- ence of direction in the other, it follows that when two polyedrons are similar, their homologous faces are simi- lar polygons, their homologous edges are of equal die- dral angles, and their homologous vertices are of equal polyedrals. 634. Theorem. — When two tetraedrons are similar, any edge or other line in the one is to the homologous line in the second, as any other line in the first is to its homolo- gous line in the second. If the proportion to he proved is between sides of homologous triangles, it follows at once from the simi- larity of the triangles. Geom. — 19 218 ELEMENTS OF GEOMETEY. When the edges taken in one of the tetraedrons are not sides of one face ; as, AE : BC : : 10 : DF, then, AE BC : : IE : CD, as just proved, and 10 DF: : IE : CD. Therefore, AE :BC: : 10 : DF. Again, suppose it is to be pro.ved that the altitudes AK and BH have the same ratios as two homologous edges. AK and BH are perpendicular lines let fall .from the homologous points A and B on the opposite faces. From K let the perpendicular KN fall upon the edge 10. Join AN, and from H let the perpendicular HG fall upon DF, which is homologous to 10. Join BGr. Now, the planes AKN and EIO are perpendicular to each other (556), and the line IN m one of them is, by construction, perpendicular to their intersection KN. Hence, IN is perpendicular to the plane AKN (557). Therefore, the line AN is perpendicular to IN, and the diedral whose edge is 10 is measured by the angle ANK. In the same way, it is proved that the diedral whose edge is DF, is measured by the angle BGH. But these two diedrals, being homologous, are equal, the angles ANK and BGH are equal, and the right an- gled triangles AKN and BHG are similar. Therefore, AK : BH : : AN : EG. TETEAEDRONS. 219 Also, tie right angled triangles ANI and BGD are similar, since, by hypothesis, the angles AEN and BDG are equal. Hence, AI : BD : : AN : BG. Therefore, " AE : BH : : AI : BD. Thus, by the aid of similar triangles, it may be proved that any two homologous lines, in two similar tetrae- drons, have the same ratio as two homologous edges. 635. Theorem — Two tetraedrons are similar when their faces are respectively similar triangles, and are simi- larly arranged. For we know, from the similarity of the triangles, that every line made on the surface of one may have its homologous line in the second, making angles equal to those made by the irst line. If lines be made through the figure, it may be shown, by the aid of auxiliary lines, as in the corresponding proposition of similar triangles, that every possible an- gle in the one figure has its homologous equal angle in the other. The student may draw the diagrams, and go through the details of the demonstration. 636. If the similar faces were not arranged similarly, but in reverse order, the tetraedrons would be symmet- rically similar. 637. Corollary Two tetraedrons are similar when three faces of the one are respectively similar to those of the other, and they are similarly arranged. For the fourth faces, having their sides proportional, are simi- lar also. 638. Corollary. — Two tetraedrons are similar when two triedral vertices of the one are respectively equal to two of the other, and they are similarly arranged. 220 ELEMENTS OF GEOMETRY. 639. Corollary Two tetraedrons are similar when the edges of one are respectively proportional to those of the other, and they are similarly arranged. 640. Theorem. — The areas of homologous faces of similar tetraedrons are to each other as the squares of their ^ edges. This is only a corollary of the theorem that the areas of similar triangles are to each other as the squares of their sides. 641. Corollary. — The areas of homologous faces of similar tetraedrons are to each other as the squares of any homologous lines. 643. Corollary — The area of any face of one tetrae- dron is to the area of a homologous face of a similar tetraedron, as the area of any other face of the first is to the area of the homologous face of the second. 643. Corollary — The area of the entire surface of one tetraedron is to that of a similar tetraedron as the squares of homologous lines. TETEAEDRONS CUT BY A PLANE. (>44. Theorem — If a plane cut a tetraedron parallel to the base, the tetraedron cut off is similar to the whole. For each triangular side is cut by a line parallel to its base (572), thus making all the edges of the two tetraedrons respectively proportional. 645. Theorem — If two tetraedrons, having the same altitude and their bases on the same plane, are cut by a plane parallel to their bases, the areas of the sections will have the same ratio as the areas of the bases. If a plane parallel to the bases pass through the ver- tex A, it will also pass through the vertex B (622). But TETBAEDKONS. 221 such a plane is parallel to the cutting plane GHP (566). A B D I Therefore, the tetraedrons AGHK and BLNP have equal altitudes. The tetraedrons AEIO and AGHK are similar (644). Therefore, EIO, the base of the first, is to GHK, the base of the second, as the square of the altitude of the first is to the square of the altitude of the second (641). For a like reason, the base CDF is to the base LNP as the square of the greater altitude is to the square of the less. Therefore, EIO : GHK : : CDF : LNP. By alt3rnation, EIO : CDF : : GHK : LNP. 046. Corollary — When the bases are equivalent the sections are equivalent. G47. Corollary — When the bases are equal the sec- tions are equal. For they are similar and equivalent. REGULAR TETRAEDRON. 648. There is one form of the tetraedron which de- serves particular notice. It has all its faces equilateral. This is called a regular tetraedron. 649. Corollary It follows, from the definition, that 222 ELEMENTS OF GEOMETRY. the faces are equal triangles, the vertices are of equal triedrals, and the edges are of equal diedral angles. 650. The area of the surface of a tetraedron is found by taking the sum of the areas of the four faces. When two or more of them are equal, the process is shortened by multiplication. But the discussion of this matter will be included in the subject of the areas of pyra- mids. The investigation of the measures of volumes will be given in another connection. EXERCISES. 651. — 1. State other cases, when two tetraedrons are similar, in addition to those given. Articles 635 to 639. 2. In any tetraedron, the lines which join the centers of the opposite edges bisect each other. 3. If one of the vertices of a tetraedron is a trirectangular trj- edral, the square of the area of the opposite face is equal to the sum of the squares of the areas of the other three faces. PYRAMIDS. 633. If a polyedral is cut by a plane which cuts its several edges, the section is a polygon, and a portion of space is cut off, which is called a pyramid. A Pyramid is a polyedron having for one face any polygon, and for its other faces, triangles whose vertices meet at one point. PYRAMIDS. . 223 The polygon is the base of the pyramid, the triangles are its sides, and their intersections are the lateral edges of the pyramid. The vertex of the polyedral is the vertex of the pyramid, and the perpendicular distance from that point to the plane of the base is its altitude. Pyramids are called triangular, quadrangular, pentag- onal, etc., according to the polygon which forms the base. The tetraedron is a triangular pyramid. 633. Problem — Every pyramid can le divided into the same number of tetraedrons as its base can be into triangles. Let a diagonal plane pass through the vertei of the pyramid and each diagonal of the base, and the solu- tion is evident. EQUAL PYRAMIDS. 654. Theorem. — Two pyramids are equal when ihe base and two adjacent sides of the one are respectively equal to the corresponding parts of the other, and they are simi- larly arranged. For the triedrals formed by the given faces in the two must be equal, and may therefore coincide; and the given faces will also coincide, being equal. But now the vertices and bases of the two pyramids coin- cide. These include the extremities of every edge. Therefore, the edges coincide; also the faces, and the figures throughout. SIMILAR PYRAMIDS. 633. Theorem. — Two similar pyramids are composed of tetraedrons respectively similar, and similarly arranged ; and, conversely, two pyramids are similar when com- posed of similar tetraedrons, similarly arranged. 224 ELEMENTS OF GEOMETRY. 636. Theorem. — When a pyramid is cut by a plane parallel to the base, the pyramid cut off is similar to the whole. These theorems may be demonstrated by the student. Their demonstration is like that of analogous proposi- tions in triangles and tetraedrons. REGULAR PYRAMIDS. 657. A Regular Pyramid is one whose base is a regular polygon, and whose vertex is in the line perpen- dicular to the base at its center. 638. Corollary. — The lateral edges of a regular pyra- mid are all equal (529), and the sides are equal isosce- les triangles. 639. The Slant Hight of a regular pyramid is the perpendicular let fall from the vertex upon one side of the base. It is therefore the altitude of one of the sides of the pyramid. 660. Theorem — The area of the lateral surface of a regular pyramid is equal to half the product of the pe- rimeter of the base by the slant hight. The area of each side is equal to half the product of its base by its altitude (386). But the altitude of each of the sides is the slant hight of the pyramid, and the sum of all the bases of the sides is the perimeter of the base of the pyramid. Therefore, the area of the lateral surface of the pyr- amid, which is the sum of all the sides, is equal to half the product of the perimeter of the base by the slant hight. 661. When a pyramid is cut by a plane parallel to the base, that part of the figure between this plane and PYRAMIDS. 225 the base is called a frustum of a pyramid, or a trwnc- ated pyramid. 662. Corollary — The sides of a frustum of a pyra- mid are trapezoids (572); and the sides of the frustum of a regular pyramid are equal trapezoids. 663. The section made by the cutting plane is called the upper base of the frustum. The slant hight of the frustum of a regular pyramid is that part of the slant hight of the original pyramid which lies between the bases of the frustum. It is therefore the altitude of one of the lateral sides. 664. Theorem. — The area of the lateral surface of the frustum of a regular pyramid, is equal to half the prod- uct of the sum of the perimeters of the bases by the slant hight. The area of each trapezoidal side is equal to half the product of the sum of its parallel bases by its altitude (392), which is the slant hight of the frustum. There- fore, the area of the lateral surface, which is the sum of all these equal trapezoids, is equal to the product of half the sum of the perimeters of the bases of the frustum, multiplied by the slant hight. 663. Corollary. — The area of the lateral surface of a frustum of a regular pyramid is equal to the product of the perimeter of a section midway between the two bases, multiplied by the slant hight. For the perimeter of a section, midway between the two bases, is equal to half the sum of the perimeters of the bases. 666. Corollary — The area of the lateral surface of a regular pyramid is equal to the product of the slant hight by the perimeter of a section, midway between the vertex and the base. For the perimeter of the middle section is one-half the perimeter of the base. 226 ELEMENTS OF GEOMETRY. MODEL PYRAMIDS. 66'?'. The student may construct a model of a regular pyra- mid. First, draw a regular polygon of any number of sides. Upon these sides, as bases, draw equal isosceles triangles, taking care that their altitude be greater than the apothem of the base. The figure may then be cut but and folded. EXERCISES. 66S. — 1. Find the area of the surface of a regular octagonal pyramid whose slant hight is 5 inches, and a side of whose base IS 2 inches. 2. What is the area in square inches of the entire surface of a regular tetraedron, the edge being one inch ? Ans. |/3. 3. A pyramid is regular when its sides are equal isosceles triangles, whose bases form the perimeter of the base of the pyramid. 4. State other cases of equal pyramids, in addition to those given. Article 654. 5. When two_ pyramids of equal altitude have their bases in the same plane, and are cut by a plane parallel to their bases, the areas of the sections are proportional to the areas of the bases. PKISMS. 669. A Prism is a polyedron -which has two of its faces equal polygons lying in par- allel planes, and the other faces parallelograms. Its possibility is shown by supposing two equal and parallel polygons lying in two par- allel planes (569). The equal sides being parallel, let planes unite them. The figure thus formed on each plane is a parallelogram, for it has tvv'o opposite sides equal and parallel. PRISMS. 227 The parallel polygons are called the hases, the paral- lelograms the sides of the prism, and the intersections of the sides are its lateral edges. The altitude of a prism is the perpendicular distance between the planes of its bases. GTO. Corollary. — The lateral edges of a prism are all parallel to each other, and therefore equal to each other (573). 671. A Right Pmsm is one whose lateral edges are perpendicular to the bases. A Regular Peism is a right prism whose base is a regular polygon. 672. Corollary — The altitude of a right prism is equal to one of its lateral edges ; and the sides of a right prism are rectangles. The sides of a regular prism are equal. 673. Theorem — If two parallel planes pass tlirough a prism, so that each plane cuts every lateral edge, the sections made hy the two planes are equal polygons. Each side of one of the sections is parallel to the corresponding side of the other section, since they are the intersections of two parallel planes by a third. Hence, that portion of each side of the prism which is between the secant planes, is a parallelogram. Since the sections have their sides respectively equal and parallel, their angles are respectively equal. There- fore, the polygons are equal. 674. Corollary — The section of a prism made by a plane parallel to the base is equal to the base, and the given prism is divided into two prisms. If two paral- lel planes cut a prism, as stated in the above theorem, that part of the solid -between the two secant planes is also a prism. 228 ELEMENTS OF GEOMETRY. HOW DIVISIBLE. 675. Problem — Every prism can he divided into the same nunAer of triangidar prisms as its base can he into triangles. If homologous diagonals be made in the two bases, as EO and CP, they will lie in one plane. For CE and OP being parallel to each other (670), lie in one plane. There- fore, through each pair of these homologous diagonals a plane may pass, and these diagonal planes divide the prisms injto triangular prisms. 676. Problem. — A triangular prism may be divided into three tetraedrons, which, taken two and two, have equal bases and equal altitudes. Let a diagonal plane pass through the points B, C, and H, making the intersections BH and CH, in the sides DP and DGr. This plane cuts off the tet- raedron BCDH, which has for one of its faces the base BCD of the prism; for a second face, the triangle BCH, being the sec- tion made by the diagonal plane; and for its other two faces, the triangles BDH and CDH, each being half of one of the sides of the prism. The remainder of the prism is a quadrangular pyra- mid, having the parallelogram BCGF for its base, and H for its vertex. Let it be cut by a diagonal plane through the points H, G, and B. PRISMS. 229 This plane separates two tetraedrons, HBCG and HBFG. The two faces, HBO and HBG, of the tetrae- dron HBCG, are sections made by the diagonal planes; and the two faces, HCG and BOG, are each half of one side of the prism. The tetraedron HBFG has for one of its faces the base HFG of the prism; for a second liice, the triangle HBG, being the section made by the diagonal plane; and, for the other two, the triangles HBF and GBF, each being half of one of the sides of the prism. Now, consider these two tetraedrons as having their bases BOG and BFG. These are equal triangles lying in one plane. The point H is the common vertex, and therefore they have the same altitude ; that is, a perpen- dicular from H to the plane BCGF. Next, consider the first and last tetraedrons described, HBCD and BFGH, the former as having BCD for its base, and H for its vertex; the latter as having FGH for its base, and B for its vertex. These bases are equal, being the bases of the given prism. The vertex of each is in the plane of the base of the other. Therefore, the altitudes are equal, being the, distance between these two planes. Lastly, consider the tetraedrons BCDH and BCGH as having their bases CDH and CGH. These are equal triangles lying in one plane. The tetraedrons have the common vertex B, and hence have the same altitude. 6T7. Corollary Any prism may be divided into tetraedrons in several ways ; but the methods above ex- plained are the simplest. 6f8. Eemakk. — On account of the importance of the above problem in future demonstrations, tlie student is advised to make a model triangular prism, and divide it into tetraedrons. A po tato may be used for this purpose. The student will derive most benefit from those models and diagrams which he makes himself 230 ELEMENTS OF GEOMETRY. EQUAL PKISMS. 679. Theorem. — Two prisms are equal, when a base and two adjacent sides of the one are respectively equal to the corresponding parts of the other, and they are simi- larly arranged. For the triedrals formed by the given faces in the two prisms must be equal (599), and may therefore be made to coincide. Then the given faces will also coin- cide, being equal. These coincident points include all of one base, and several points in the second. But the second bases have their sides respectively equal, and parallel to those of the first. Therefore, they also coin- cide, and the two prisms having both bases coincident, must coincide throughout. 6SO. Corollary — Two right prisms are equal when they have equal bases and the same altitude. 681. The theory of similar prisms presents nothing difficult or peculiar. The same is true of symmetrical prisms, and of symmetrically similar prisms. AREA OF THE SURFACE. 682. Theorem — The area of the lateral surface of a prism is equal to the product of one of the lateral edges hy the perimeter of a section, made hy a plane perpen- dicular to those edges. Since the lateral edges are parallel, the plane HN, perpendicular to one, is perpendicular to all of them. Therefore, the sides of the polygon, HK, KL, etc., are severally perpendicular to the edges of the prism which they unite (519). Then, in order to measure the area of each face of the prism, we take one edge of the prism as the base PRISMS. 231 of the parallelogram, and one side of the polygon HN as its altitude. Thus, area AG = ABxHP, area EB = EC X HK, etc. By addition, the sum of the areas of these parallelograms is the lateral surface of the prism, and the sum of the altitudes of the parallelograms is the perim- eter of the polygon HN. Then, since the edges are equal, the area of all the sides is equal to the product of one edge, multi- plieJ by the perimeter of the polygon. 683. Corollary — The area of the lateral surface of a right prism is equal to the product of the altitude by the perimeter of the base. 684. Corollary — The area of the entire surface of a regular prism is equal to the product of the perime- ter of the base by the sum of the altitude of the prism and the apothem of the base. EXERCISES. 6S5. — 1. A right prism has less surface than any other prism of equal base and equal altitude; and a regular prism has less surface than any other right prism of equivalent base and equal altitude. 2. A regular pyramid and a regular prism have equal hexag- onal bases, and altitudes equal to three times the radius of the base; required the ratio of the areas of their lateral surfaces. 3. Demonstrate the principle stated in Article 683, without the aid of Article 682. 232 ELEMENTS OF GEOMETRY. MEASURE OF VOLUME. 686. A Parallelopiped is a prism whose bases are parallelograms. Hence, a parallelopiped is a solid in- closed by six parallelograms. 687. Theorem. — The opposite sides of a parallelopiped are equal. For example, the faces AI and BD are equal. For 10 and DF are equal, being opposite sides of the parallelogram IF. For a like reason, EI is equal to CD. But, since these equal sides are also par- allel, the included angles BIO and CDF are equal. Hence, the parallelograms are equal. 688. Corollary. — Any two opposite faces of a paral- lelopiped may be assumed as the bases of the figure. 689. A parallelopiped is called right in the same case as any other prism. When the bases also are rectangles, it is called rectangular. Then all the faces are rectangles. 699. A Cube is a rectangular parallelopiped whose length, breadth, and altitude are equal. Then a cube is a solid, bounded by six equal squares. All its verti- ces, being trirectangular triedrals, are equal (602). All its edges are of right diedral angles, and therefore equal (555). The cube has the simplest form of all geometrical solids. It holds the same rank among them that the square does among plane figures, and the straight line among lines. MEASURE OP VOLUME. 233 The cube is taken, therefore, as the unit of measure of volume. That is, whatever straight line is taken as the unit of length, the cube whose edge is of that length is the unit of volume, as the square whose side is of that length is the measure of area. VOLUME OF PAEALLELOPIPEDS. 691. Theorem — The volume of a rectangular paral- lelepiped is equal to the product of its length, breadth, and altitude. In the measure of the rectangle, the product of one line by another was ex- plained. Here we have three lines used with a similar meaning. That is, the number of cu- bical units contained in a rectangular parallele- piped is equal to the product of the numbers of linear units in the length, the breadth, and the alti- tude. If the altitude AE, the length EI, and the breadth 10, have a common measure, let each be divided by it ; and let planes, parallel to the faces of the prism, pass through all the points of division, B, C, D, etc. By this construction, all the angles formed by these planes and their intersections are right angles, and each of the intercepted lines is equal to the linear unit used in dividing the edges of the prism. Therefore, the prism is divided into equal cubes. The number of these at the base is equal to the number of rows, mul- tiplied by the number in each row; that is, the product G-eom.— 20 234 ELEMENTS OF GEOMETRY. of the length by the breadth. There are as many layers of cubes as there are linear units of altitude. Therefore, the whole number is equal to the product of the length, breadth, and altitude. In the diagram, the dimensions being four, three, and two, the volume is twenty-four. But if the length, breadth, and altitude have no com- mon measure, a linear unit may be taken, successively smaller and smaller. In this, we would not take the whole of the linear dimensions, nor would we measure the whole of the prism. But the remainder of both would grow less and less. The part of the prism meas- ured at each step, would be measured exactly by the principle just demonstrated. By these successive diminutions of the unit, we can make the part measured approach to the whole prism as nearly as we please. In a word, the whole is the limit of the parts measured; and since the principle demon- strated is true up to the limit, it must be true at the limit. Therefore, the rectangular parallelopiped is meas- ured by the product of its length, breadth, and altitude. 692. Theorem. — The volume of any parallelopiped is equal to the product of its length, breadth, and altitude. Inasmuch as this has just been demonstrated for the rectangular parallelopiped, it will be sufficient to show that any parallelopiped is equivalent to a rectangular one having the same linear dimensions. Suppose the lower bases of the two prisms to be placed on the same plane. Then their upper bases must also be in one plane, since they have the same altitude. Let the altitude AB be divided into an infinite number of equal parts, and through each point of division pass a plane parallel to the base AI. Now, every section in either prism is equal to the MEASURE OF VOLUME. 235 base ; but the bases of the two prisms, having the same length and breadth, are equivalent. The several par- tial infinitesimal prisms are reduced to equivalent fig- ures. Although they are not, strictly speaking, paral- lelograms, yet their altitudes being infinitesimal, there can be no error in considering them as plane figures ; which, being equal to their respective bases, are equiva- lent. Then, the number of these is the same in each prism. Therefore, the sum of the whole, in one, is equivalent to the sum of the whole, in the other; that is, the two parallelepipeds are equivalent. Besides the above demonstration by the method of infinites, the theorem may be demonstrated by the or- dinary method of reasoning, which is deduced from principles that depend upon the superposition and co- incidence of equal figures, as follows ; Let AF be any oblique parallelepiped. It may be shown to be equivalent to the parallelepiped AL, which has a rectangular base, AH, since the prism LHEO is equal to the prism DGAI. But the paxallelopipeds AF and AL have the same length, breadth, and altitude. 230 ELEMENTS OF GEOMETRY. By similar reasoning, the prism AL may be shown to be equivalent to a prism of the same base and alti- tude, but with two of its opposite sides rectangular. This third prism may then be shown to be equivalent to a fourth, which is rectangular, and has the same dimen- sions as the others. 693. Corollary. — The volume of a cube is equal to the third power of its edge. Thence comes the name of cube, to designate the third power of a number. MODEL CUBES. 694. Draw six equal squares, as in the diagram. Cut out the figure, fold at the dividing lines, and glue the edges. It is well to have at least eight of one size. 695. Corollary. — The volume of any parallelopiped is equal to the product of its base by its altitude. 696. Corollary — The volumes of any two parallele- pipeds are to each other as the products of their threa dimensions. VOLUME OF PKISMS. 697. Theorem — The volume of any triangular prism is equal to the product of its base hy its altitude. The base of any right triangular prism may be con- sidered as one-half of the base of a right parallelopiped. Then the whole parallelopiped is double the given prism, for it is composed of two right prisms having equal bases and the same altitude, of which the given prism MEASURE OF VOLUME. 237 is one. Therefore, the given prism is measured by half the product of its altitude by the base of the parallel- epiped ; that is, by the product of its own base and altitude. If the given prism be oblique, it may be shown, by demonstrations similar to the first of those in Article 692, to be equivalent to a right prism having the same base and altitude. 698. Corollary — The volume of any prism is equal to the product of its base by its altitude. For any prism is composed of triangular prisms, having the com- mon altitude of the given prism, and the sum of their bases forming the given base. 699. Corollary The volume of a triangular prism is equal to the product of one of its lateral edges mul- tiplied by the area of a section perpendicular to that edge. VOLUME OF TETRAEDBONS. '700. Theorem. — Two tetraedrons of equivalent bases and of the same altitude are equivalent. Suppose the bases of the two tetraedrons to be in the O F same plane. Then their vertices lie in a plane parallel to the bases, since the altitudes are equal. Let the edge AB be divi^ded into an infinite number of parts, 238 ELEMENTS OF GEOMETRY. and through each point of division pass a plane parallel to the base AIO. Now, the several infinitesimal frustums into which the two figures are divided may, without error, be consid- ered as plane figures, since their altitudes are infinitesi- mal. But each section of one tetraedron is equivalent to the section made by the same plane in the other tet- raedron. Therefore, the sum of all the infinitesimal frustums in the one figure is equivalent to the sum of all in the other; that is, the two tetraedrons are equiv- alent. 701. Theorem — The volume of a tetraedron is equal to one-third of the product of the base hy the altitude. Upon the base of any given tetraedron, a triangular prism may be erected, which shall have the same alti- tude, and one edge coincident with an edge of the tet- raedron. This prism may be divided into three tetrae- drons, the given one and two others, which, taken two and two, have equal bases and altitudes (676). Then, these three tetraedrons are equivalent (700); and the volume of the given tetraedron is one-third of the volume of the prism ; that is, one-third of the prod- uct of its base by its altitude. VOLUME OF PYRAMIDS. 703. Corollary. — The volumn of any pyramid is equal to one-third of the product of its base by its altitude. For any pyramid is composed of triangular pyramids; that is, of tetraedrons having the common altitude of the given pyramid, and the sum of their bases forming the given base (653). 703. Corollary. — The volumes of two prisms of equiv- alent bases are to each other as their altitudes, and the SIMILAR POLYEDEONS. 239 volumes of two prisms of equal altitudes are to each other as their bases. The same is true of pyramids. '?04. Corollary. — Symmetrical prisms are equivalent. The same is true of symmetrical pyramids. 705. The volume of a frustum of a pyramid is found by subtracting the volume of the pyramid cut off from the volume of the whole. When the altitude of the whole is not given, it may be found by this proportion ; the area of the lower base of the frustum is to the area of its upper base, which is the base of the part cut off, as the square of the whole altitude is to the square of the altitude of the part cut off. EXERCISES. fOG. — I. What is the ratio of the volumea of a pyramid and prism having the same base and altitude? 2. If two tetraedrons have a triedral vertex in each equal, their volumes are in the ratio of the products of the edges which contain the equal vertices. 3. The plane which bisects a diedral angle of a tetraedron, divides the opposite edge in the ratio of the areas of the adjacent faces. SIMILAR POLYEDRONS. TOT. The propositions (640 to 643) upon the ratios of the areas of the surfaces of similar tetraedrons, may be applied by the student to any similar polyedrons. These propositions and the following are equally appli- cable to polyedrons that are symmetrically similar. TOS. Problem. — Any two similar polyedrons may be divided into the same number of similar tetraedrons, which shall be respectively similar, and similarly arranged. For, after dividing one into tetraedrons, the construe 240 ELEMENTS OF GEOMETRY. tion of the homologous lines in the other will divide it in the same manner. Then the similarity of the re- spective tetraedrons follows from the proportionality of the lines. TOO. Theorem The volumes of similar polyedrons are proportional to the cities of homologoics lines. First, suppose the figures to be tetraedrons. Let- AH and BGr be the altitudes. Then (641), EIO : CDF : : EF : CF^ : : AH^ : BG' By the proportionality of homologous lines, (634), i AH : i BG : : EI : CF : : AH : BG. Multiplying these proportions (701), we have AEIO : BCFD : : EP : CF^ : : AH^ : BG^ or, as the cubes of any other homologous lines. Next, let any two similar polyedrons be divided into the same number of tetraedrons. Then, as just proved, the volumes of the homologous parts are proportional to the cubes of the homologous lines. By arranging these in a continued proportion, as in Article 436, we may show that the volume of either polyedron is to the vol- ume of the other as the cube of any line of the first is to the cube of the homologous line of the second. EEGULAE POLYEDRONS. 241 TIO. Notice that in the measure of every area there are two linear dimensions ; and in the measure of every volume, three linear, or one linear and one superficial. EXERCISE. "Ml. What is the ratio between the' edges of two cubes, one of which has twice the volume of the other? This problem of the duplication of the cube was one of the celebrated problems of ancient times. It is said that the oracle of Apollo at Delphos, demanded of the Athenians a new altar, of the same shape, but of twice the volume of the old one. The efforts of the Greek geometers were chiefly aimed at a graphic so- lution; that is, the edge of one cube being given, to draw a line equal to the edge of the other, using no instruments but the rule and compasses. In this they failed. The student will find no difficulty in making an arithmetical solution, within any desired degree of approximation. KEGULAK POLYEDRONS. 712. A Regular Polyedbon is one whose faces are equal and regular polygons, and whose vertices are equal polyedrals. % The regular tetraedron and the cube, or regular hexa- edron, have been described. The regular oetaedron has eight, the dodecaedron twelve, and the ieosaedron twenty faces. Geom.— 21 £42 ELEMENTS OF GEOMETET. The class of figures here defined must not be con- founded with regular pyramids or prisms. 713. Problem — It is not possible to make more than five regular polyedrons. First, consider those whose faces are triangles. Each angle of a regular triangle is one-third of two right angles. Either three, four, or five of these may bo joined to form one polyedral vertex, the sum being, in each case, less than four right angles (612). But the sum of six such angles is not less than four right angles. Therefore, there can not be more than three kinds of regular polyedrons whose faces are triangles, viz. : the tetraedron, where three plane angles form a vertex; the octaedron, where four, and the icosaedron, where five angles form a vertex. The same kind of reasoning shows that only one regular polyedron is possible with square faces, the cub6; and only one with pentagonal faces, the dode- caedron. Regular hexagons can not form the faces of a regular polyedron, for three of the angles of a regular hexagon are together not less than four right angles ; and there- fore they can not form a vertex. So much the more, if the polygon has a greater num- ber of sides, it will be impossible for its angles to be the faces of a polyedral. Therefore, no polyedron is possible, except the five that have been described. MODEL REGULAR POLYEDRONS. 714. The possibility of regular polyedrons of eight, of twelve, and of twenty sides is here assumed, as the demonstration would occupy more space than the principle is worth. However, the student may construct models of these as follows. Plans for tix". regular tetraedron and the cube have already been given. REGULAR POLYEDRONS. 213 For the octaedron, draw eight equal regular trian- gles, as in the diagram. For the dodecaedron, draw twelve equal regular penta- gons, as in the diagram. For the icosaedron, draw twenty equal regular trian- gles, as in the diagram. There are many crystals, which, though not regular, in the geometrical rigor of the word, yet present a certain regularity of shape. EXERCISES. "715. — 1. How many edges and how many vertices has each of the regular polyedrons? 2. Calling that point the center of a triangle which is the inter- section of straight lines from each vertex to the center of the opposite side; then, demonstrate that the four lines which join the vertices of a tetraedron to the centers of the opposite faces, inter sect each other in one point. 1 3. In what ratio do the lines just described in the tetraedron divide each other? 4. The opposite vertices of a parallelopiped are symmetrical tricdrals. 5 The diagonals of a parallelopiped bisect each other; the lines which join the centers of the opposite edges bisect eacJi other; the lines which join the centers of the opposite faces bi- 244 ELEMENTS OF GEOMETKY. sect each o'iher; and the point of intersection is the same for all these lines. 6. The diagonals of a rectangular parallelepiped are equal. 7. The square of the diagonal of a rectangular parallelepiped is equivalent to the sum of the squares of its length, breadth, and altitude. 8. A cube is the largest parallelepiped of the same extent of surface. 9. If a right prism is symmetrical to another, they are equal. 10. Within any regular polyedron there is a point equally distant from all the faces, and also from all the vertices. 11. Two regular polyedrons of the same number of faces are similar. 12. Any regular polyedron may be divided into as many regu- lar and equal pyramids as it has faces. 13. Two different tetraedrons, and only two, may be formed with the same four triangular faces; and these two tetraedrons are symmetrical. 14. The area of the lower base of a frustum of a pyramid is five square feet, of the upper base one and four-fifths square feet, and the altitude iu two feet; required the volume. SOLIDS OF KEVOLUTION. 245 CHAPTER XI. SOLIDS OF REVOLUTION. 716. Op the infinite variety of forms there remain but three to be considered in this elementary work. These are formed or generated by the revolution of a plane figure about one of its lines as an axis. Figures formed in this way are called solids of revolution. TIT. A Cone is a solid formed by the revolution of a right angled triangle about one of its legs as an axis. The other leg revolv- Jk ing describes a plane surface (521). /Jl'lk This surface is also a circle, having for Ji i||. its radius the leg by which it is de- Jljl' 1|l scribed. The hypotenuse describes a fjl fjip-'^"-^ curved surface. The plane surface of a cone is called its base. The opposite extremity of the axis is the vertex. The alii- tude is the distance from the vertex to the base, and the slant hight is the distance from the vertex to the cir- cumference of the base. 718. A Cylinder is a solid described -.«« by the revolution of a rectangle about one of its sides as an axis. As in the cone, the sides adjacent to the axis de- scribe circles, while the opposite side describes a curved surface. "~ — ^ The plane surfaces of a cylinder are called its lases, 2i6 ELEMENTS OF GEOMETRY. and the perpendicular distance between them is its altitude. These figures are strictly a regular cone and a regular cylinder, yet but one word is used to denote the figures defined, since other cones and cylinders are not usually discussed in Elementary Geometry. The sphere, which is described by the revolution of a semicircle about the diameter, will be considered separately. 719. As the curved surfaces of the cone and of the cylinder are generated by the motion of a straight line, it follows that each of these surfaces is straight in one direction. A straight line from the vertex of the cone to the circumference of the base, must lie wholly in the sur- face. So a straight line, perpendicular to the base of a cylinder at its circumference, must lie wholly in the surface. For, in each case, these positions had been occupied by the generating lines. One surface is tangent to another when it meets, but being produced does not cut it. The place of contact of a plane with a conical or cylindrical surface, must be a straight line ; since, from any point of one of those surfaces, it is straight in one direction. CONIC SECTIOJfS. 720. Every point of the line which describes the curved surface of a cone, or of a cylinder, moves in a plane parallel to the base (565). Therefore, if a cone or a cylinder be cut by a plane parallel to the base, the section is a circle. If we conceive a cone to be cut by a plane, the curve formed by the intersection will be different according to the position of the cutting plane. There are three dif- CONES. 247 ferent modes in which it is possible for the intersection to take place. The curves thus formed are the ellipse, parabola, and hyperbola. These Oonio Sections are not usually considered in Elementary Geometry, as their properties can be better investigated by the application of algebra. CONES. •ySl. A cone is said to be inscribed in a pyramid, when their bases lie in one plane, and the sides of the pyramid are tangent to the curved surface of the cone. The pyramid is said to be circumscribed about the cone. A cone is said to be circumscribed about a pyramid, when their bases lie in one plane, and the lateral edges of the pyramid lie in the curved surface of the cone. Then the pyramid is inscribed in the cone. 722. Theorem — A cone is the limit of the pyramids which can be circumscribed about it; also of the pyramids which can be inscribed in it. Let ABODE be any pyramid circumscribed about a cone. The base of the cone is a circle inscribed in the base of the pyramid. The sides of the pyramid are tangent to the surface of the cone. Now, about the base of the cone there may be described •a polygon of double^the num- ber of sides of the first, each alternate side of the second polygon coinciding with a side of the first. This second polygon may be the base of a pyramid, having its vertex at A. Since the sides of its bases are tangent to the base of the cone, every 248 ELEMENTS OF GEOMETRY. side of the pyramid is tangent to the curved surface of the cone. Thus the second pyramid is circumscribed about the cone, but is itself within the first pyramid. By increasing the number of sides of the pyramid, it can be made to approximate to the cone within less than any appreciable diiFerence. Then, as the base of the cone is the limit of the bases of the pyramids, the cone itself is also the limit of the pyramids. Again, let a polygon be inscribed in the base of the cone. Then, straight lines joining its vertices with the vertex of the cone form the lateral edges of an inscribed pyramid. The number of sides of the base of the pyr- amid, and of the pyramid also, may be increased at will. It is evident, therefore, that the cone is the limit of pyramids, either circumscribed or inscribed. 733. Corollary. — The area of the curved surface of a cone is equal to one-half the product of the slant hight by the circumference of the base (660). Also, it is equal to the product of the slant hight by the circumfer- ence of a section midway between the vertex and the base (666). • 734. Corollary — The area of the entire surface of a cone is equal to half of the product of the circumfer- ence of the base by the sum of the slant hight and the radius of the base (499). 735. Corollary — The volume of a cone is equal to one-third of the product of the base by the altitude. 726. The frustum of a cone is defined in the same way as the frustum of a pyramid. 727. Corollary — The area of the curved surface of the frustum of a cone is equal to half the product of its slant hight by the sum of the circumferences of its bases (664). Also, it is equal to the product of its slant CYLINDERS. 249 hight by the circumference of a section midway between the two bases (665). TSS. Corollary — ^If a cone be cut by a plane paral- lel to the bq.se, the cone cut oif is similar to the whole (656). EXEKCISES. 739. — 1. Two cones are similar when they are generated by similar triangles, homologous sides being used for the axes. 2. A section of a cone by a plane passing through the vertex, is an isosceles triangle. CYLINDERS. 730. A cylinder is said to be in- scribed in a prism, when their bases lie in the same planes, and the sides of the prism are tangent to the curved surface of the cylinder. The prism is then said to be circumscribed about the cylinder. A cylinder is said to be circum- scribed about a prism, when their bases lie in the same planes, and the lat- eral edges of the prism lie in the curved surface of the cylinder; and the prism is then said to be inscribed in the cylinder. 731. Theorem. — A cylinder is the limit of the prisms which can be circumscribed about it; also of those which can be inscribed in it. The demonstration of this theorem is so similar to that of the last, that it need not be repeated. TaO ELEMBNTS OF GEOMETRY. 73S. Corollary. — The area of the curved surface of a Bylinder is equal to the product of the altitude by the ^ircumference of the base (683). 733. Corollary. — The area of the entire , surface of a eylinder is equal to the product of the circumference of ihe base by the sum of the altitude and the radius of the base (684). 734. Corollary The volume of a cylinder is equal to the product of the base by the altitude (698). MODEL CONES AND CYLINDERS. TfSa. Models of cones and cylinders may be made from paper, taking a sector of a circle for the curved surface of a cone, and a rectangle for the curved surface of a cylinder. Make the bases separately. EXERCISES. 736. — 1. Apply to cones and cylinders the principles demon- strated of similar polyedrons. 2. A section of a cylinder made by a plane perpendicular to the base is a rectangle. 3. The axis of a cone or of a cylinder is equal to its altitude. SPHERES. 737. A Sphere is a solid de- scribed by the revolution of a semicircle about its diameter as an axis. The center, radius, and diame- ter of the sphere are the same as those of the generating circle. The spherical surface is described by the circumference. SPHERES. 251 738. Corollary — Every point on the surface of the sphere is equally distant from the center. This property of the sphere is frequently given as its definition. 739. Corollary. — All radii of the same sphere are equal. The same is true of the diameters. 740. Corollary. — Spheres having equal radii are equal. 741. Corollary. — ^A plane passing through the center of a sphere divides it into equal parts. The halves of a sphere are called hemispheres. 742. Theorem. — A plane which is perpendicular to a radius of a sphere at its extremity is tangent to the sphere- For if straight lines extend from the center of the sphere to any other point of the plane, they are oblique and longer than the radius, ■which is perpendicular (530). There- fore, every point of the plane except one is beyond the surface of the sphere, and the plane is tangent. 743. Corollary. — The spherical surface is curved in every direction. Unlike those surfaces which are gen- erated by the motion of a straight line, every possible section of it is a curve. SECANT PLANES. 744. Theorem.- Every section of a sphere made by a plane is a circle. If the plane pass through the center of the sphere, every point in the perimeter of the section is equally distant from the center, and therefore the section is a circle. 252 ELEMENTS OF GEOMETRY But if the section do not pass through the center, as DGF, then from the center C let CI fall perpendicu- larly on the cutting plane. Let radii of the sphere, as CD and CG, extend to differ- ent points of the boundary of the section, and join ID and IG. Now the oblique lines CD and CG being equal, the points D and G must be equally distant from I, the foot of the perpendicular (529). The same is true of all the points of the .pe- rimeter DGF. Therefore, DGF is the circumference of a circle of which I is the center. '745. Corollary. — The circle formed by the section through the center is larger than one formed by any plane not through the center. For the radius BC is equal to GC, and longer than GI (104). 746. When the plane passes through the center of a sphere, the section is called a great circle; otherwise it is called a small circle. 747. Corollary — All great circles of the same sphere are equal. 748. Corollary — Two great circles bisect each other, and their intersection is a diameter of the sphere. 749. Corollary — If a perpendicular be let fall from the center of a sphere on the plane of a small circle, the foot of the perpendicular is the center of the cir- cle; and conversely, the axis of any circle is a diame- ter of the sphere. The two points where the axis of a circle pierces the spherical surface, are the poles of the circle. Thus, SPHERES. 253 N and S are the poles of both the sections in the last diagram. TSO. Corollary — Circles whose planes are parallel to each other have the same axis and the same poles. ARC OF A GREAT CIRCLE. 751. Theorem — The shortest line which can extend from one point to another along the surface of a sphere, is the arc of a great circle, passing tJirough the two points'. Only one great circle can pass through two given points on the surface of a sphere ; for these two points and the center determine the position of the plane of the circle. Let ABCDEFG be any curve whatever on the sur- face of a sphere from G to A. Let AKG be the arc of a great circle joining these points, and also AD and DG arcs of great cir- cles joining those points with the point D of the given curve. Then the sum of AD and DG is greater than AKG. For the planes of these arcs form a triedral whose vertex is at the center of the sphere. These arcs have the same ratios to each other as the plane angles which compose this triedral, for the arcs are intercepted by the sides of the angles, and they have the same radius. But any one of these angles is less than the sum of the other two (586). Therefore, any one of the arcs is less than the sum of the other two. Again, let AH and HD be arcs of great circles join- ing A and D with some point H of the given curve ; also let DI and IG be arcs of great circles. In the 254 ELEMENTS OF GEOMETRY. same manner as above, it may be sbown that AH and HD are greater than AD, and that the sum of DI and IG is greater than DGr. Therefore, the sum of AH, HD, DI, and IG is still greater than AKG. By continuing to take intermediate points and join- ing them to the preceding, a series of lines is formed, each greater than the preceding, and each approaching nearer to the given curve. Evidently, this approach can be made as nearly as we choose. Therefore, the curve is the limit of these lines, and partakes of their common character, in being greater than the arc of a great circle which joins its extremities. 733. Theorem — Every plane passing through the axis of a circle is perpendicular to the plane of that circle, and its section is a great circle. The first part of this theorem is a corollary of Arti- cle 556. The second part is proved "by the fact that every axis passes through the center of a sphere (749). 753. Corollary — The distances on the spherical sur- face from any points of a circumference to its pole, are the same. For the arcs of great circles which mark these distances are equal, since all their chords are equal oblique lines (529). 754. Corollary — The distance of the pole of a great circle from any point of the circumference is a quad- rant. APPLICATIONS. 755. The Btudent of geography will recognize the equator a8 a great circle of the earth, which is nearly a sphere. The paral- lels of latitude are small circles, all having the same poles as the equator. The meridians are great circles perpendicular to the equator. The application of the principle of Article 751 to navigation SPHERES. 255 has been one of the greatest reforms in that art. A vessel cross- ing the ocean from a port in a certain latitude to a port in the same latitude, should not sail along a parallel of latitude, for that is the arc of a small circle. ■ySB. The curvature of the sphere in every direction, renders it impossible to construct an exact model with plane paper. But ' the student is advised to procure or make a globe, upon which he can draw the diagrams of all the figures. This is the more im- portant on account of the difficulty of clearly representing these figures by diagrams on a plane surface. SPHERICAL ANGLES. '757. A Spherical Angle is the difference in the directions of two arcs of great cir- cles at their point of meeting. To obtain a more exact idea of this angle, notice that the direction of an arc at a given point is the same as the direction of a straight line tangent to the arc at that point. Thus, the direction of the arc BDF at the point B, is the same as the direction of the tangent BH. 758. Corollary. — A spherical angle is the same as the plane angle formed by lines tangent to the given arcs at their point of meeting. Thus, the spherical angle DBG is the same as the plane angle HBK, the lines HB and BK being severally tangent to the arcs BD and BG. 759. Corollary, — A spherical angle is the same as the diedral angle formed by the planes of the two arcs. For, since the intersection BF of the planes of the arcs is a diameter (748), the tangents HB and KB are both perpendicular to it, and their angle measures the diedral. II 7 1 \ v"^ "^^^v 256 ELEMENTS OF GEOMETRY. 760. Corollary — A spherical an- gle is measured by the arc of a cir- cle included between the sides of the angle, the pole of the arc being at the vertex. Thus, if DG- is an arc of a great circle whose pole is at B, then the spherical angle DBGr is measured by the arc DG. 761. A LuNB is that portion of the surface of a sphere included between two halves of great circles. That portion of the sphere included between the two planes is called a spherical wedge. Hence, two great circles divide the surface into four lunes, and the sphere into four wedges. SPHERICAL POLYGONS. 762. A Spherical Polygon is that portion of the surface of a sphere included between three or more arcs of great circles. Let C be the center of a sphere, and also the vertex of a convex polyedral. Then, the planes of the faces of this polyedral will cut the surface of the sphere in arcs of great circles, which form the poly- gon BDFGH. We say con- vex, for only those polygons which have all the angles convex are considered among spherical polygons. Conversely, if a spherical polygon have the planes of its several sides produced, they form a polyedral whose vertex is at the center of the sphere. SPHERES. 257 The angles of the polygon are the same as the die- dral angles of the polyedral (759). 763. Theorem. — The sum of all the sides of a spher- ical polygon is less than a circumference of a great circle. The arcs which form the sides of the polygon measure the angles which form the faces of the corresponding polyedral, for all the arcs have the same radius. But the sum of all the faces of the polyedral being less than four right angles, the sum of the sides must be less than a circumference. 761. Theorem. — A spherical polygon is always within the surface of a hemisphere. For a plane may pass through the vertex of the cor- responding polyedral, having all of the polyedral on one side of it (609). The section formed by this plane produced is a great circle, as KLM. But since the polyedral is on one side of this plane, the corres- ponding polygon must be con- tained within the surface on one side of it. 763. That portion of a sphere which is included be- tween a spherical polygon and its corresponding polye- dral is called a spherical pyramid, the polygon being its base. SPHERICAL TRIANGLES. 768. If the three planes which form a triedral at the center of a sphere be produced, they divide the sphere into eight parts or spherical pyramids, each hav- ing its triedral at the center, and its spherical triangle Geom.— 22 25S ELEMENTS OF GEOMETRY. at the surface. Thus, for every spherical triangle, there are seven others whose sides are respectively either equal or supplementary to those of fhe given triangle. / ^--^^^'>S Of these seven spherical tri- /, F/' / ;\ angles, that which lies vertically f'""-/ \ /^^•C \ opposite the given triangle, as I \/ ",^