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It is in successful use la all parts of the country, being in exclusive use in the public schools of Boston, Cambridge ; Newton, Mass. ; Washington, D. C, and many other representative cities. K3^~ In the /bllo-uing distinctive and important features, KERUS System of English Grammar claims superiority over other systems ; i. It teaches more that is of practical utility. 2. It contains a simpler, sounder, and more comprehensive article on the Analysis of Sentences. 3. It contains the true theory of Moods and Tanses, and shows better the nature of Participles and Infinitives. 4. It contains a much better article on Capital Letters. 5. It contains a much better article on Punctuation. 6. It contains a much better article on Versification, — probably the only set of principles that teach the true mechanism of English verse. 7. It contains a better article on Rhetorical Figures, and on the other devices which give beauty and vigor to style. 8. 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Principles are made plain by examples that show what is meant ; abstruse points are brought within easy reach, by familiar and striking explanations ; and all things are made practical by exercises. The typography is also superior. %W Full descriptive Circulars 0/ the series, giving titles and prices, will be sent by mail on application. 1 niL riiyutAi ikjl\s\lj js.rj,rwjt\. 1 ins. — Full of interesting and valuable Educational information, is published three times a year, bearing date respectively January, May and September, and will be sent to teachers and educationists, without charge, on application. Ivison, Blakeman, Taylor & Co., educational plblishers, 138 & 140 Grand St., New Vork. 133 & 135 State St., Chicago. GIFT OF Dr. Wttlliam H. Tvio ROBINSON'S MATHEMATICAL SERIES, ELEMENTS OF GEOMETRY, PLANE AND SPHERICAL; WITH NUMEROUS PRACTICAL PROBLEMS, BY HORATIO N. ROBINSON, LL.D., AUTHOR OP A FULL COURSE^OF MATHEMATICS. REWRITTEN BY I. F. QUINBY, A.M., LL.D., PROFESSOR OF MATHEMATICS AND NATURAL PHILOSOPHY, UNIVER8rTY OF EOCHE8TER ; AUTHOR OF DIFFERENTIAL AND INTEGRAL CALCULUS. IVISON, BLAKEMAN, TAYLOR & CO. NEW YORK: CHICAGO: • ptOBusrsoisrs SERIES OF MATHEMATICS. The most Complete, most Practical, and most Scientific Series of Mathematical Text-Books ever issued in this country. *♦* Robinson's Progressive Table HooJe, ....... Robinson's Progressive Primary Arithmetic, - - - - - Robinson's Progressive Intellectual Arithmetic, .... Robinson's Rudiments of Written Arithmetic, - Robinson's Progressive Practical Arithmetic, - Robinson's Key to Practical Arithmetic, ------ Robinson's Progressive Higher Arithmetic, - - - - - Robinson's Key to Higher Arithmetic, .---.- Robinson's Arithmetical Examples, ------- Robinson's New Elementary Algebra, Robinson's Key to Elementary Algebra, Robinson's "University Algebra, -------- Robinson's Key to University Algebra, ------ Robinson's New University Algebra, ------- Robinson's Key to New University Algebra, - - - - - Robinson's New Geometry and Trigonometry, Robinson's New Geometry only, -------- Robinson's New Trigonometry only, Robinson's Surveying and Navigation, ------ Robinson's Analytical Geometry and Conic Sections, - - - Robinson's Differential and Integral Calculus, - Kiddle's New Elementary Astronomy, ---.-. Robinson's University Astronomy, ..-.--- Robinson's Mathematical Operations, ------- Robinson's Key to Geometry and Trigonometry, Conic Sections, and Analytical Geometry, --------- ), according to Act of Congress, in the year 1868, by DANIEL W. FISH, A.M., in the Clerk's Office of the District Court of the United States for the Eastern District of New York. PEEFAOE, In the preparation of this work, the author's previous treatise, Elements of Geometry, has formed the ground- work of construction. But in adapting the work to the present advanced state of Mathematical education in our best Institutions, it was found necessary so to alter the plan, and the arrangement of subjects, as to make this essentially a new work. The demonstrations of proposi- tions have undergone radical changes, many new proposi- tions have been introduced, and the number of Practical Problems greatly increased, so that the work is now believed to be as full and complete as could be desired in an elemen- tary treatise. In view of the fact that the Seventh Book is so much larger than the others, it may be asked why it is not divided into two. We answer, that classifications and divisions are based upon differences, and that the differences seized upon for this purpose must be determined by the nature of the properties and relations we wish to investigate. There is such a close resemblance between the geometrical prop- erties of the polyedrons and the round bodies, and the demonstrations relating to the former require such slight modifications to become applicable to the latter, that there seems no sufficient reason for separating into two Books that part of Geometry which treats of them. M5498G7 i T PREFACE. Practical rules with applications will be found throughout the work, and in addition to these, there is a full collection of carefully selected Practical Problems. These are given to exercise the powers and test the proficiency of the pupil, and when he has mastered the most or all of them, it is not likely that he will rest satisfied with present acquisi- tion, but, conscious of augmented strength and certain of reward, he will enter new fields of investigation. The author has been aided, in the preparation of the present work, by I. F. Quinby, A.M., of the University of Eochester, N. Y., late Professor of Mathematics in the United States Military Academy at West Point. The thorough scholarship and long and successful experience of this gentleman in the class-room, eminently qualify him for such a task; and to him the public are indebted for* much that is valuable, both in the matter and arrangement of this treatise. October, 1860. CONTENTS. PLANE GEOMETRY. DEFINITIONS. Geometrical Magnitudes ... ... ....... . .. . Page 9 Plane Angles 10 Plane Figures of Three Sides w*.-~.U^ 12 Plane Figures of Four Sides. .....-^. ...... ....... 13 The Circle ......... ..... ... 14 Units of Measure .................. 15 Explanation of Terms. 16 Postulates.. —...-...-r. .. . 16 Axioms -. .. 17 Abbreviations 17 BOOK I. Of Straight Lines, Angles, and Polygons... ....... -^.. ... . 19 BOOK II. Proportion, and its Application to Geometrical Investigations. ... 59 BOOK III. Of the Circle, and the Investigation of Theorems dependent on its Properties 88 1* M ri CONTENTS. BOOK IV. Problems in the Construction of Figures in Plane Geometry.. ... Ill BOOK V. On the Proportionalities and Measurement of Polygons and Circles. 130 Practical Problems 142 BOOK VI. On the Intersections of Planes, the Kelative Positions of Planes, and of Planes and Lines 152 BOOK VII. Solid Geometry t 172 Practical Problems 229 BOOK VIII. Practical Geometry. — Application of Algebra to Geometry, and also Propositions for Original Investigation 231 Miscellaneous Propositions in Plane Geometry 238 BOOK IX. Spherical Geometry 243 Definitions 243 GEOMETRY. DEFINITIONS. 1. Geometry is the science which treats of position, and of the forms, measurements, mutual relations, and pro- perties of limited portions of space. Space extends without limit in all directions, and contains all bodies. 2. A Point is mere position, and has no magnitude. 3. Extension is a term employed to denote that pro- perty of bodies by virtue of which they occupy definite portions of space. The dimensions of extension are length, breadth, and thickness. 4. A Line is that which has extension in length only. The extremities of a line are points. 5. A Right or Straight Line is one all of whose parts lie in the same direction. 6. A Curved Line is one whose consecutive parts, how- ever small, do not lie in the same direction. 7. A Broken or Crooked Line is composed of several straight lines, joined one to another successively, and extending in different directions. When the word line is used, a straight line is to be understood, unless otherwise expressed. 8. A Surface or Superficies is that which has extension in length and breadth only. 9. A Plane Surface, or a Plane, is a surface such that (9) 10 GEOMETRY. if any two of its points be joined by a straight line, every' point of this line will lie in the surface. 10. A Curved Surface is one which is neither a plane, nor composed of plane surfaces. 11. A Plane Angle, or simply an Angle, is the difference in the direction of two lines proceeding from the same point. The other angles treated of in geometry will be named and defined in their proper connections. 12. A Volume, Solid, or Body, is that which has exten- sion in length, breadth, and thickness. These terms are used in a sense purely abstract, to denote mere space — whether occupied by matter or not, being a question with which geometry is not concerned. Lines, Surfaces, Angles, and Volumes constitute the different kinds of quantity called geometrical magnitudes. 13. Parallel Lines are lines which have the same direction. Hence parallel lines can never meet, however far they may be produced ; for two lines taking the same direction cannot approach or recede from each other. Two parallel lines cannot be drawn from the same point; for it parallel, they must coincide and form one line. PLANE ANGLES. To make an angle apparent, the two lines must meet in a point, as AB and A 0, which meet in the point A , Angles are measured by degrees. 14. A Degree is one of the three hundred and sixty equal parts of the space about a point in a plane. If, in the above figure, we suppose A C to coincide with AB, there will be but one line, and no angle ; but if AB retain its posi tion, and A G begin to revolve about the point A, an angle will be formed, and its magnitude will be expressed by that number of the fl. c DEFINITIONS. 11 860 equal spaces about the point A, which is contained between AB and A G. Angles are distinguished in respect to magnitude by the terms Right, Acute, and Obtuse Angles. J 15. A Right Angle is that formed by one line meeting another, so as to make equal angles with that other. The lines forming a right angle are perpendicular to *ach other. 16. An Acute Angle is less than a right angle. 17. An Obtuse Angle is greater than a right angle. Obtuse and acute angles are also called oblique angles; and lines which are neither parallel nor perpen- dicular to each other are called oblique lines. 18. The Vertex or Apex of an angle is the point in which the including lines meet. 19. An angle is commonly designated by a letter at its vertex; but when two or more angles have their vertices at the same point, they cannot be thus distinguished. For example, when the three lines AB y A C, and AD meet in the common point Aj we designate either of the an- gles formed, by three letters, placing that at the vertex between those at the opposite extremities of the including lines. Thus, we say, the angle BAG, etc. B 20. Complements. — Two angles are said to be comple ments of each other, when their sum is equal to one right angle. 21. Supplements. — Two angles are said to be supple- ments of each other, when their sum is equal to two n^ht angles. r 12 GEOMETRY. PLANE FIGURES. 22. A Plane Figure, in geometry, is a portion of a plane bounded by straight or curved lines, or by both combined. 23. A Polygon is a plane figure bounded by straight Hues, called the sides of the polygon. The least number of sides that can bound a polygon is three, and by the figure thus bounded all other polygons are analyzed. FIGURES OF THREE SIDES. 24. A Triangle is a polygon having three sides and three angles. Tri is a Latin prefix signifying three ; hence a Triangle is lite < /ally a figure containing three angled. Triangles are denominated from the relations both of their sides and angles. 25. A Scalene Triangle is one in which no two sides are equal. 26. An Isosceles Triangle is one in which two of the sides are equal. 27. An Equilateral Triangle is one in arhich the three sides are equal. 28. A Right -Angled Triangle is one which has one of the angles a right angle. 29. An Obtrse-Angled Triangle is one 1 aving an obtuse angle. DEFINITIONS. 13 30. An Acute- Angled Triangle is one in which each angle is acute. 31. An Equiangular Triangle is one having its three angles jqual. Equiangular triangles are dso equilateral, and vice versa. FIGURES OF FOUR SIDES. 32. A Quadrilateral is a polygon having four sides and four angles. 33. A Parallelogram is a quadrilateral 7 which has its opposite sides parallel. / / Parallelograms are denominated from the rela- _• tions both of their sides and angles. 34. A Rectangle is a parallelogram hav- ing its angles right angles. 35. A Square is an equilateral rectangle. 36. A Rhomboid is an oblique-angled parallelogram. 37. A Rhombus is an equilateral rhom- boid. 38. A Trapezium is a quadrilateral having 10 two sides parallel. 39. A Trapezoid is a quadrilateral in f hich two opposite s* J i le other two oblique. v/hich two opposite sides are parallel, and / 40. Polygous bounded by a greater number of sides 2 14 GEOMETRY. than four are denominated only by the number of sides. A polygon of five sides is called a Pentagon • of six, a Hexagon ; of seven, a Heptagon / of eight, an Octagon ; of nine, a JVonagon, etc. 4L Diagonals of a polygon are lines joining the vertices of angles not ad- jacent. 42. The Perimeter of a polygon is its boundary consid ered as a whole. 43. The Base of a polygon is the side upon which the polygon is supposed to stand. 44. The Altitude of a polygon is the perpendicular distance between the base and a side or angle opposite the base. 45. Equal Magnitudes are those which are not only equal in all their parts, but which also, when applied the one to the other, will coincide throughout their whole extent. 46. Equivalent Magnitudes are those which, though they do not admit of coincidence when applied the one to the other, still have common measures, and are therefore numerically equal. 47. Similar Figures have equal angles, and the same number of sides. Polygons may be similar without being equal ; that is, the angles and the number of sides may be equal, and the length of the sides and the size of the figures unequal. X THE CIRCLE. 48. A Circle is a plane figure bound- ed by one uniformly curved line, all of the points in which are at the same distance from a certain point within, called the Center, 49. The Circumference of a circle is the curved line that bounds it. \ DEFINITIONS. 15 50. The Diameter of a circle is a line passing througn its center, and terminating at both ends in the circum- ference. 51. The Radius of a circle is a line extending from its center to any point in the circumference. It is one half of the diameter. All the diameters of a circle are equal, as are also all the radii. 52. An Arc of a circle is any portion of the circum- ference. 53. An angle having its vertex at the center of a circle is measured by the arc intercepted by its sides. Thus, the arc AB measures the angle A OB ; and in gen- eral, to compare different angles, we have but to compare the arcs, included by their sides, of the equal circles having their centers at the vertices of the angles. UNITS OF MEASURE. 54. The Numerical Expression of a Magnitude is a number expressing how many times it contains a magnitude of the Bame kind, and of known value, assumed as a unit. For lines, the measuring unit is any straight line of fixed value, as an inch, a foot, a rod, etc. ; and for surfaces, the measuring unit is a square whose side may be any linear unit, as an inch, a foot, a mile, etc. The linear unit being arbitrary, the surface unit is equally so ; and its selection is determined by considerations of convenience and propriety. For example, the parallelogram A BBC is mea- c D sured by the number of linear units in CD, mul- tiplied by the number of linear units in AC ox BD; the product is the square units in ABDC. For, conceive CD to be composed of any number A B of equal parts — say five — and each part some unit of linear measure, and AC composed of three such units; from each point of divi- sion on CD draw lines parallel to A C, and from each point of divi- sion on A C draw lines parallel to CD or AB ; then it is as obvious 16 GEOMETRY. as an axiom that the parallelogram will contain 5 X 3 = 15 square units. Hence, to find the areas of right-angled parallelograms, mul- tiply the base by the altitude. EXPLANATION OF TERMS. 55. An Axiom is a self-evident truth, not only too sim- ple to require, but too simple to admit of, demonstration. 56. A Proposition is something which is either pro- posed to be done, or to be demonstrated, and is either a problem or a theorem. 57. A Problem is something proposed to be done. 58. A Theorem is something proposed to be demon- strated. 59. A Hypothesis is a supposition made with a view to draw from it some consequence which establishes the truth or falsehood of a proposition, or solves a problem. 60. A Lemma is something which is premised, or demon- strated, in order to render what follows more easy. 61. A Corollary is a consequent truth derived imme- diately from some preceding truth or demonstration. 62. A Scholium is a remark or observation made upon something going before it. 63. A Postulate is a problem, the solution of which is self-evident. POSTULATES. Let it be granted — I. That a straight line can be drawn from any one poirt to any other point ; IL. That a straight line can be produced to any distance, or terminated at any point ; III. That the circumference of a circle can be de- Bcrjbed about any center, at any distance from that center. DEFINITIONS. 17 AXIOMS. 1. Things which are equal to the same thing are equal U each other. 2. Wlien equals are added to equals the wholes are equal, 3. When equals are taken from equals the remainders are equal. 4. When equals are added to unequals the wholes are unequal. 5. Wlien equals are taken from unequals the remainders are unequal. 6. Things which are double of the same thing, or equal things, are equal to each other. 7. Things which are halves of the same thing, or of equal things, are equal to each other. 8. The whole is greater than any of its parts. 9. Every whole is equal to all its parts taken together. 10. Things which coincide, or fill the same space, are identical, or mutually equal in all their parts. 11. All right angles are equal to one another. 12. A straight line is the shortest distance between two points. 18. Two straight lines cannot inclose a space. ABBREVIATIONS. The common algebraic signs are used in this work, and demonstrations are sometimes made through the medium of equations ; and it is so necessary that the student in geometry should understand some of the more simple operations of algebra, that we assume that he is acquainted with the use of the signs. As the terms circle, angle, triangle, hypothesis, axiom, theorem, cor- ollary, and definition, are constantly occurring in a course of geometiy, we shall abbreviate them as shown in the following list : 20 GEOMETRY. By Th. 1, any two supplementary angles, as ABD^ ABO, are together equal to two right angles. And since the angular space about the point B is neither increased nor diminished by the number of lines drawn from that point, the sum of all the angles DBA, ABU, JEBH, HBQ, fills the same spaces as any two angles HBD, HBO. Hence the theorem ; from any point in a line, the sum of all the angles that can be formed on the same side of the line is equal to two right angles. Cor. 1. And, as the sum of all the angles that can be formed on the other side of the line, OB, is also equal to two right angles; therefore, all the angles that can be formed quite round a point, B, by any number of lines, are together equal to four right angles. Oor. 2. Hence, also, the whole circum- ference of a circle, being the sum of the measures of all the angles that can be made about the center F, (Def. 53), is the measure of four right angles; conse- quently, a semicircumference. is the mea- sure of two right angles ; and a quadrant, or 90°, is the measure of one right angle. THEOREM III. X If one straight line meets two other straight lines at a common point, forming two angles, which together are equal tc two right angles the 'two straight lines are one and the. tame line. Let the line AB meet the lines BD and BE at the com- mon point B, making the sum of tne two angles ABB, ABB, equal to two right angles; We j£ are to prove that DB and BE* are one straight line j^' BOOK I. 21 If I)B and BE are not in the same line, produce DB to 0, thus forming one line, BBC Now by Th. 1, ABB + ABO must be equal to two right angles. But by hypothesis, ABB 4- ABE is equal to two right angles. * Therefore, ABB + ABO is equal to ABB -f ABE, (Ax. 1). From each of these equals take away the com- mon angle ABB, and the angle ABO will be equal to ABE, (Ax. 3). That is, the line BE must coincide with BO, and they will be in fact one and the same line, and they cannot be separated as is represented in the figure. Hence the theorem ; if one line meets two other lines at a common point, forming two angles which together are equal to two right angles, the two lines are one and the same line. THEOKEM IV. If two straight lines intersect each other, the opposite or vertical angles must be equal. If AB and OB intersect each other at E, we are to demonstrate that the angle AEO is equal to the vertical angle BEB ; and the angle AEB, to the vertical angle OEB. As AB is one line met by BE, another line, the two angles AEB and BEB, on the same side of AB, are equal to two right angles, (Th. 1). Also, because OB is a right line, and AE meets it, the two angles AEO and AEB are together equal to two right angles. Therefore, AEB + BEB = AEO + AEB. (Ax. 1.) If from these equals we take away the common angle AEB, the remaining angle BEB must be equal to the remaining angle AEO, (Ax. 3). In like manner, we can prove that AEB is equal to OEB. Hence the theo**en. ; if the two lines intersect each other, the vertic %l angle* mu it be equal. 20 GEOMETRY. By Th. 1, any two supplementary angles, as ABD^ ABC, are together equal to two right angles. And since the angular space about the point B is neither increased nor diminished by the number of lines drawn from that point, the sum of all the angles DBA, ABU, EBH, HBO, fills the same spaces as any two angles HBD, HBQ. Hence the theorem ; from any point in a line, the sum of all the angles that can be formed on the same side of the line is equal to two right angles. Cor. 1. And, as the sum of all the angles that can be formed on the other side of the line, CD, is also equal to two right angles; therefore, all the angles that can be formed quite round a point, B, by any number of lines, are together equal to four right angles. Cor. 2. Hence, also, the whole circum- ference of a circle, being the sum of the measures of all the angles that can be made about the center F, (Def. 53), is the measure of four right angles; conse- quently, a semicircumference, is the mea- sure of two right angles ; and a quadrant, or 90°, is the measure of one right angle. THEOREM III. t If one straight line meets two other straight lines at a common point, forming two angles, which together are equal tc two right angles the 'two straight lines are one and the tame line. Let the line AB meet the lines BD and BE at the com- A mon point B, making the sum of tne two angles ABB, ABE, equal to two right angles ; We are to prove that DB and BlM are one straight line. BOOK I. 21 If 1)B and BE are not in the same line, produce BB to 0, thus forming one line, BBC. Now by Th. 1, ABB -f ABO must be equal to two right angles. But by hypothesis, ABB + ABB is equal to two right angles. * Therefore, ABB + ABO is equal to ABB + ABB, (Ax. 1). From each of these equals take away the com- mon angle ABB, and the angle ABO will be equal to ABB, (Ax. 3). That is, the line BE must coincide with BO, and they will be in fact one and the same line, and they cannot be separated as is represented in the figure. Hence the theorem ; if one line meets two other lines at a common point, forming two angles which together are equal to two right angles, the two lines are one and the same line. THEOREM IV. If two straight lines intersect each other, the opposite or vertical angles must be equal. If AB and OB intersect each other at E, we are to demonstrate that the angle AEO is equal to the vertical angle BEB ; and the angle AEB, to the vertical angle OEB. As AB is one line met by BE, another line, the two angles AEB and BEB, on the same side of AB, are equal to two right angles, (Th. 1). Also, because OB is a right line, and AE meets it, the two angles AEO and AEB are together equal to two right angles. Therefore, AEB + BEB = AEO + AEB. (Ax. 1.) If from these equals we take away the common angle AEB, the remaining angle BEB must be equal to the remaining angle AEO, (Ax. 3). In like manner, we can prove that AEB is equal to OEB. Hence the thecen. ; if the two lines intersect each other, the vertic il angle* km it be equal. 22 GEOMETRY. Second Demonstration. By Def. 11, the angle DEB is the difference in the direction of the lines ED and EB ; and the angle AEQ is the difference in the direction of the lines EC and EA. But ED is opposite in direction to EC; and EB is opposite in direction to EA. Hence, the difference in the direction of ED and EB is the same as that of EC and EA> as is obvious by in- spection. Therefore, the angle DEB is equal to its opposite AEC In like manner, we may prove AED = CEB. Hence the theorem ; if two lines intersect each other, ths vertical angles must be equal. THEOREM V. If a straight line intersects two parallel lines, the sum of the two interior angles on the same side of the intersecting line is equal to two right angles. [Note. — By interior angles, we mean angles which lie between the parallels ; the exterior angles are those not between the parallels.] Let the line EF intersect the parallels AB and CD \ then we are to demonstrate that the angles BGE + GHD = 2K.L Because GB and ED are C /H d parallel, they are equally in- <£ clined to the line EF, or have the same difference of direction from that line. There- fore, |_ FGB = [__ GHD. To each of these equals add the \_BGH, and we have FGB +BGH=GITD+BGB:. But by Th. 1, the first member of this equation is equal to two right angles ; and the second member is the sum of the two angles between the parallels. Hence the theo- rem ; if a line intersects two parallel lines, the sum of the two interior angles on the same side of the intersecting line must be equal to two right angle*. BOOK I. 23 Sctt&Ln* — As AB and CD are parallel lines, and EF is a line intersecting them, AB and EF must make angles equal to those made by CD and EF. That is, the angles about the point G must be equal to the corresponding angles about the point H. THEOREM VI. If a line intersects two parallel lines, the alternate interior angles are equal. Let AB and CD be paral- lels, intersected by EF at E and G. Then we are to prove that the angle A GE is equal to the alternate angle GED, and QEG = EGB. c ~7W " D By Th. 5, [_BGE + \__ / GED = two right angles. Al- so, by Th. 1, \_AGE + [_BGE = two right angles. From these equals take away the common angle BGE, and L GHD will be left, equal to \_AGE, (Ax. 3). In like manner, we can prove that the angle QEG is equal to the angle EGB. Hence the theorem ; if a line intersects two parallel lines, the alternate interior angles are equal. Cor. 1. Since [__ A GE = [__ FGB, and \_AGE=[__GED; Therefore, L FGB - |_ &HD (Ax. 1). Also, L A &F + L A GH = 2 R - L> (Th. 1), and L OEG + [_AGE = 2 K. L, (Th. 5); Therefore \_AGF + l_AGE= [_CEG + L_AGE,(Ax.l); and L A GF = L OEG, (Ax. 3). That is, the exterior angle is equal to the interior opposiU angle on the same side of the intersecting line. Cor. 2. Since \_AGE = [_FGB, and l_AGE=\_CEE; Therefore, [__FGB = |_ OEE. In the same manner it may be shown that [__AGF = \_EED. Hence, the alternate exterior angles are equal. v * GEOMETKY. THEOREM VII. If a line intersects two other lines, making the sum of the two interior angles on the same side of the intersecting line equal to two right angles, the two straight lines are parallel. Let the line EF intersect the lines AB and OB, making the two angles BQR + GHB A = to two right angles ; then we are to demonstrate that AB and OB are parallel. C /H "~ d As EF is a right line and E ' BGr meets it, the two angles FCrB and BGH are together equal to two right angles, (Th. 1). But by hypothesis, the angles, BGHnnd GHB, are together equal to two right angles. From these two equals take away the common angle BGrH, and the re- maining angles FGB and GHD must be equal, (Ax. 3). Now, because GB and HD make equal angles with the same line EF, they must extend in the same direction ; and lines having the same direction are parallel, (Def. 13). Hence the theorem ; if a line intersects two other lines, making the sum of the two interior angles on the same side of the in- tersecting line equal to two right angles, the two lines must b« 'parallel. Cor. 1. If a line intersects two other lines, making the alternate interior angles equal, the two lines intersected must be parallel. Suppose the \_ AGH = |__ & HI) - Adding L H& B to each, we have [_AGH + [_HGB = L_ GHB + [__HGB. but the first member of this equation, that is, [_AGfH-+ |__ HGB, is equal to two right angles ; hence the second member is also equal to the same ; and by the theorem, the lines AB and OB are parallel. Cor. 2. If a line intersects two other lines, making the BOOK I. 25 opposite exterior and interior angles equal, the two lines intersected must be parallel. Suppose the [_ FGB = [_ GRD. Adding the [__ HGB to each, we have [_FGB + \_EGB = L &HD + ##£. But the first member of this equation is equal to two right angles ; hence the second member is also equal to two right angles ; and by the theorem, the lines AB and OB are parallel. Oor. 3. If a line intersects two other lines, making tbo alternate exterior angles equal, the lines must be parallel: Suppose [_BGF=[_ORF, and [_AGF = [_DRF, ByTh.4, [_BGF^[__AGR,2ind[_ORF = [_DRG And since [_BGF = [_ORF, [__AGR=[_DRG . That is, the alternate interior angles are equal; an* J hence (by Cor. 1) the two lines are parallel. THEOREM VIII. If two angles have their sides parallel, the two angles will be either equal or supplementary. Let A be parallel to BD, and AH parallel to BF or to BG. Then we are to pror^ that the angle DBx is equal to the angle OAR, and that the angle DBG is supplementary to the angle A. The angle OAR is formed by the differ- ence in the direction of A and A R; and the angle DBF is formed by the differ- ence in the direction of BB and BF. But A and A R have the same direc- tions as BB and BF, because they are respectively paral- lel. Therefore, by Def. 11, L CAR= \__BBF. But the line BG has the same, direction as BF, and the angle DBG is supplementary to DBF. Hence the theorem; angles whose sides are parallel are either equal or supple* inentary. 3 X \ 26 GEOMETRY. v, • THEOREM IX. The opposite angles of any parallelogram we equal. Let AEBG be a parallel- ogram. Then we are to \ prove that the angle GBE 6 \] is equal to its opposite angle A. Produce EB to D, and GB to F; then, since BD is par- allel to AG, and BF to AE, the angle DBF is equal co the angle A, (Th. 8). But the angles GBE and DBF, being vertical, are equal, (Th. 4). Therefore, the opposite angles GBE and A, of the parallelogram AEBG, are equal. In like manner, we can prove the angle E equal to the angle G. Hence the theorem ; the opposite angles of any parallelogram are equal. THEOREM X. The sum of the angles of any parallelogram i» equal to four right angles. Let ABQD be a parallelo- gram. We are to prove that the sum of the angles A, B, and D, is equal to four right angles, or to 360°. Because AD and BO are parallel lines, and AB inter- sects them, the two interior angles A and B are together equal to two right angles, (Th. 5). And because CD in- tersects the same parallels, the two interior angles O and D are also together equal to two right angles. By addi- tion, we have the sum of the four interior angles of the parallelogram ABCD, equal to four right angles. Hence the theorem ; the sum of the angles of any parallelogram is taual tc four right angles. is. x A BOOK I. THEOREM XI. 27 V The sum of the three angles of any triangle is equal to two right angles. Let A B be a triangle, and through its vertex draw a line parallel to the base AB, and produce the sides A C and BO. Then the angles A and a, being exterior and in- terior opposite angles on the same side of the line AC, are equal to each other. For the same reason, [_ E = L b. The angles C and c, being vertical angles, are also equal, (Th. 4). Therefore, the angles A, B, C are equal to the angles a, b, c respect- ively. But the angles around the point C, on the upper side of the parallel CD, are equal to two right angles, (by Th. 2). Hence the theorem ; the sum of the three angles, etc. Second Demonstration. Let AEBG- be a parallelogram Draw the diagonal GE ; thus di- viding the parallelogram into two triangles, and the opposite angles G and E each into two angles. Because GB and AE are parallel, the alternate interior angles BGE and GEA are equal, (Th. 6). Designate each of these by b. In like manner, because EB and AG are parallel, the alternate interior angles, BEG and EGA, are equal. Designate each of these by a. Now we are to prove that the three angles B, b, and a, and also that the three angles A, a, and b, are equal to two right angles. 28 GEOMETRY. Because A and B are opposite angles of a parallelo- gram, they are equal, (Th. 9), and [_A -f [_B = 2 [__A. And all the interior angles of the parallelogram are equal to four right angles, (Th. 10). Therefore, 2 A + 2a -f- 2b = 4 right angles. Dividing by 2, and J. + a + & = 2 " That is, all the angles of the triangle AGrE are together equal to two right angles Hence the theorem ; the sum of the three angles, etc. Scholium. — Any triangle, as AGE, may be conceived to be part of a parallelogram. For, let A GE be drawn independently of the paral- lelogram ; then draw EB from the point E parallel to A G, and through the point G draw GB parallel to AE, and a parallelogram will be formed embracing the triangle ; and thus the sum of the three angles of any triangle is proved equal to two right angles. This truth is so fundamental, important, and practical, as to require special attention ; we therefore give a Third Demonstration. Let ABQ be a triangle. Then we are to show that the angles A, Q, and ABQ, are together equal *o two right angles. Let AB be produced to D, and from B draw BE parallel to A 0. Then, EBB and CAB being exterior and interior op- posite angles on the same side of the line AB, are equal, (Th. 6, Cor.1). Also, QBE and AQB, being alternate angles, are equal, (Th. 6). By addition, observing that [__ QBE, added to \__EBB, must make |__ QBD, we have l_CBB = l_A + \_Q. (1.) To each of these equals add the angle QBA, and we shall have L QBA + L OBB =L^ + L_tf+L OB A. But (by Th. 1), the sum of the first two is equal to two BOOK I. 29 rigltt angles; therefore, the three angles, A, C, and CBA f are together equal to two right angles. Hence the theorem ; the sum of the three angles, etc. y THEOREM XII. If any side of a triangle is produced, the exterior angle i* equal to the sum of the two interior opposite angles. Let ABO be a triangle. Pro- duce AB to B; and we are to prove that the angle CBB is equal to the sum of the two angles A and C. We establish this theorem by a course of reasoning in all respects the same as that hy which we obtained Eq. (1.), third demonstration, (Th. 11). Cor. 1. Since the exterior angle of any triangle is equal to the sum of the two interior opposite angles, therefore it is greater than either one of them. Cor. 2. If two angles in one triangle be equal to two angles in another triangle, the third angles will also be equal, each to each, (Ax. 3); that is, the two triangles will be mutually equiangular. Cor. 3. If one angle in a triangle be equal to one angle in another, the sum of the remaining angles in the one will also be equal to the sum of the remaining angles in the other, (Ax. 3). Cor. 4. If one angle of a triangle be a right' angle, the sum of the other two will be equal to a right angle, and each of them singly will be acute, or less than a right -. angle. Cor. 5. The two smaller angles of every triangle are acute, or each is less than a right angle. Cor. 6. All the angles of a triangle may be acute, but no triangl 3 can have more than one right or one obtuse angle. 30 GEOMETRY. V THEOREM XIII. In any polygon, the sum of all the interior angles is equal to twice as many riglit angles, less four, as the figure has Let ABODE be any polygon ; we are to prove that the sum of all its interior angles, A+B + O + D+E, is equal to twice as many right angles, less four, as the figure has sides. From any point, p, within the figure, draw lines pA, pB, pG, etc., to all the angles, thus dividing the polygon into as many triangles as it has sides. Now, the sum of the three angles of each of these triangles is equal to two right angles, (Th. 11) j and the sum of the angles of all the triangles must be equal to twice as many right angles as the figure has sides. But the sum of these angles contains the sum of four right angles about the point p ; taking these away, and the remainder is the sum of the interior angles of the figure. Therefore, the sum must be equal to twice as many right angles, less four, as the figure has sides. Hence the theorem ; in any polygon, etc. From this Theorem is derived the rule for finding the sum of the interior angles of any right-lined figure : Subtract 2 from the number of sides, and multiply the re- mainder by 2 ; the product will be the number of right angles. Thus, if the number of sides be represented by S, the number of right angles will be represented by (2#— 4). The Theorem is not varied in case n of a re-entrant angle, as repre- ^/"^ sen ted at d, in the figure ABCdEF. -^^^^ L Draw lines from the angle d to \ "/( 7 E the several opposite angles, making \ / \ / as many triangles as the figure has \/ \y sides, less two, and the sum of the A F three angles of each triangle equals two right angles. X BOOK I. 31 THEOREM XIV. If the sides of one angle be respectively perpendicular to the sides of a second angle, these two angles will be either equal or supplementary. Let BAD be the first angle, and from any point within it, as C, draw / CB and CD, at right angles, the 3 T^>C first to AB } and the second to AD, I \ ^""-Jl and produce CD in the direction / j CD, thus forming at C the supple- A B entary angles BCE, BCD ; then will the angle BCD be equal to the angle A, and therefore BCD, which is the supplement of BCD, will also be the supplement of the angle A. For since ABCD is a quadrilateral, the sum of the four interior angles is four right angles (Prop. 13), and because the angles ABC and ADC are each right angles, the sum of the angles BAD, BCD is two right angles. But the sum of the adjacent angles BCD, BCD is also two right angles. Hence, if in these last two sums we omit the com- mon angle BCD, we have remaining the angle BCD, equal to the angle BAD, and consequently the angle BCD which is the supplement of the first of these equal angles is also the supplement of the other. Hence the Theorem. ScnoLiUM. — If the vertex of the second angle be without the first angle, we would draw through any assumed point within the first angle parallels to the sides of the second ; the above demonstration will then apply to the first angle, and the angle formed by the parallels. THEOREM XT. ,X From any point ivithout a straight line, but one perpen- dicular can be drawn to that line. From the point A let us suppose it possible that two perpendiculars, A B and A C, can be drawn. Now, be- cause AB is a supposed perpendicu- lar, the angle ABC is a right angle ; — and because AC is a supposed per- B C 32 GEOMETRY. pendicular, the angle ACB is also a right angle ; and if two angles of the triangle ABC are together equal to two right angles, the third angle, BAG, must be infinitely small, or zero ; that is, the two perpendiculars being drawn through the common point A, and including no angle, must necessarily coincide, and form one and the same per- pendicular. Hence the theorem ; from any point without a straight line, etc. Cor. At a given point in a straight line but one per- pendicular can be erected to that line ; for, if there could be two perpendiculars, we should have unequal righ angles, which is impossible. THEOREM XVI. Two triangles which have two sides and the included angle in the one, equal to two sides and the included angle in the other, each to each, are equal in all respects. In the two A's, ABC and BEF, on the supposition that AB = BE, AC=BF, and [_A = \__B, we are to prove that BC must = EF, the [__B = [__E, and the [_C = LJJ Conceive the A ABC cut out of the paper, taken up, and placed on the A BEF in such a manner that the point A shall fall on the point B, and the line AB on the line BE; then the point B will fall on the point E, because the lines are equal. Now, as the [__A = [__B, the line A C must take the same direction as BF, and fall on BF; and as AC = BF, the point C will fall on F, B being on E and C on F, BC must be exactly on EF, (otherwise, two straight lines would enclose a space, Ax. 13), and BC= EF, and the two magnitudes exactly fill the same space. Therefore, BC = EF, [_B = [__E, _C ~ [_F, and the two A's are equal, (Ax. 10). Hence the theorem ; two triangles which have two sides, evo. BOOK I. 33 THEOREM XVII. When two triangles have a side and two adjacent angles in the one, equal to a side and two adjacent angles in the other, each to each, the two triangles are equal in all respects. In two A's, as ABC and T)EF, on the supposition th&tBC = EF, \_B=\_E, and [_C — [_F, we are to prove that AB = BE, AC = DF, andL^. = L^>- Conceive the A ABC taken up and placed on the A DBF, so that the side BC shall exactly coincide with its equal side EF; now, because the angle B is equal to the angle E, the line BA will take the direction of ED, and will fall exactly upon : t ; and because the angle C is equal to the angle F, the line CA will take the direction of FD, and fall exactly upon it ; and the two lines BA and CA, exactly coinciding with the two lines EB and FD, the point A will fall on B, and the two magnitudes will exactly fill the same space ; therefore, by Ax. 10, they are equal, and AB » DE, AC=DF, and the [_A = \__D. Hence the theorem ; when two triangles have a side and two adjacent angles in the one, equal to, etc. /\ THEOREM XVIII. If two sides of a triangle are equal, the angles opposite to these sides are also equal. Let ABC be a triangle; and on the supposition that AC = BC, we are to prove that the l_A=the [_B. Conceive the angle C divided into two equal angles by the line CD; then we have two A's, ADC and BDC, which have the two sides, AC and CD of the one, equal to the two sides, CB and CD of the other ; and o 34 GEOMETRY. the invaded angle AOB, of the one, equa. to the in- cluded angle BOB of the other: therefore, (Th. 16), AB = BB, and the angle A, opposite to OB of the one tri- angle, is equal to the angle B, opposite to OB of the other triangle ; that is, |_ A = |__ B. Hence the theorem ; if two sides of a triangle are equal, the angles, etc. Cor, 1. Conversely : if two angles of a triangle are equal, the sides opposite to them are equal, and the triangle is isosceles. For, if A is not equal to BO, suppose BO to be the greater, and make BE— AE; then will A ABB be isos- celes, and [_EAB = l_EBA ; hence [_EAB = [_ CAB, or a part is equal to the whole, which is absurd ; therefore, OB cannot be greater than AO, that is, neither of the sides AO, BO, can be greater than the other, and conse- quently they are equal. Cor. 2. As the two triangles, A CB and BOB, are in all respects equal, the line which bisects the angle included between the equal sides of an isosceles A also bisects the base, and is perpendicular to the base. Scholium 1. — If in the perpendicular DC, any other point than C be taken, and lines be drawn to the extremities A and B, such lines will be equal, as is evident from Th. 16 ; hence, we may announce this truth: Any point in a perpendicular drawn from the middle of a line, is at equal distances from the two extremities of the line. ScnoLiuM 2. — Since two points determine the position of a line, it follows, that a line which joins two points each equally distant from the extremities of a given line, is perpendicular to this line at its middle point, < K r^ THEOREM XIX. The greater side of every trianjle has the greater angle opposite to it. Let ABC be a A ; and on the supposition that A C is greater than AB, we are to prove that the angle ABCi* BOOK I. 85 greater than the [_ 0. From AC, the greater of the two sides, take AB, equal ^ to the less side AB, and draw BB, thus /\ making two triangles of the original tri- / \ angle. As AB = AB, the [__ABB = / \ theL^^A (Th.18). B \r~\ But the |__ ABB is the exterior angle ^\\ of the A BBC, and is therefore greater c than C, (Th. 12, Cor. 1); that is, the L ABD is greater than the [_ C. Much more, then, is the angle ABC greater than the angle C. Hence the theorem ; the greater side of every triangle, etc. Cor, Conversely: the greater angle of any triangle has the greater side opposite to it. In the triangle ABC, let the angle B be greater than the angle A ; then is the side AC greater than the side BC For, if BC — A C, the angle A must be equal to the angle B, (Th. 18), which is contrary to the hypothesis ; and if BC^>AC, the angle A must be greater than the angle B, by what is above proved, which is also contrary to the hypothesis ; hence BC can be neither equal to, nor greater, than AC; it is therefore less than AC. THEOREM XX. The difference between any two sides of a triangle is less than thi third side. Let ABC he a A, in which A C is greater than AB ; then we are to prove that A C —AB is less than BC. On AC, the greater of the two sides, lay off AB equal to AB. Now, as a straight line is the shortest distance between two points, we have AB + BOAC. (1) 36 GEOMETRY. x From these unequals suotract the equals AB = AB, and we have BO>AO— AB. (Ax. 5). Hence the theorem ; the difference between any two Met of a triangle, etc. -^ \ THEOREM XXI. ff two triangles have the three sides of the one equal to the three sides of the other, each to each, the two triangles are equxl, and the equal angles are opposite the equal sides. In two triangles, as ABO and ABB, on the supposition that the side AB of the one = the side AB of the other, AO=AB, and BO=BJ), we are to demonstrate that [_AOB = \__ABB, L BAO = l_BAB, and [_ABO= [_ABD. Conceive the two triangles to t>e joined together by their long- est equal sides, and draw the line OB. Then, in the triangle AOD, because AO is equal to AB, the angle AOD is equal to the angle ABO, (Th. 18). In like manner, in the triangle BOB, because BO is equal to BB, the angle BOB is equal to the angle BBO. Now, the angle AOB being equal to the angle ABO, and the angle BOB to the angle BBO, [_AOB + [_BOB = L. ABO + [_BBO, (Ax. 2) ; that is, the whole angled OB ia equal to the whole angle ABB. Since the two sides A and OB are equal to the two sides AB and BB, each to each, and their included angles A OB, ABB, are also equal, the two triangles ABO, ABB, are equal, (Th. 16), and have their other angles equal ; that is, [_BAO= \_BAB, and [_ABO= [__ABB. Hence the theorem ; if two triangles have the three siie* of the one, etc. BOOK I 87 THEOREM XXII. If two triangles have two sides of the one equal to tw* tides of the other, each to each, and the included angles ui> equal, the third sides will be unequal, and the greater third side will belong to the triangle which has the greater included angle. In the two A's, ABO and ACB, let AB and AC of the one A be equal to AB and A of the other A, and the angle BAC greater than the angle BAG; we are to prove that the side BO is greater than the side OB. Conceive the two a's joined together by their shorter equal sides, and draw the line BB. Now, as AB = AB, ABB is an isosceles A. From the vertex A, draw a line bisecting the angle BAB. This line must be perpendic- ular to the base BB, (Th. 18, Cor. 2). Since the \_BAO is greater than the \_BAO, this line must meet BO, and will not meet OB. From the point E, where the per- pendicular meets BO, draw BB. Now BB - BB, (Th. 18, Scholium 1). Add BO to each ; then BO = BE + EC. But BE + EO is greater than BO. Therefore BOy BO. Hence the theorem ; if two triangles have two sides of one equal to two sides of the other, etc. Cor. Any point out of the perpendicular drawn from the middle point of a line, is unequally distant from the extremities of the line. -f y GEOMETRY. THEOREM XXIII, A perpendicular is the shortest line that can be drawn from any point to a straight line ; and if other lines be drawn from the same point to the same straight line, that which meets it farthest from the perpendicular will be longest ; and lines at equal distances from the perpendicular, on opposite sides, are equal. Let A be any point without the line BE ; let AB be the perpen- dicular; and AC, AB, and AE oblique lines: then, if BO is less than BB, and BO— BE, we are to show, 1st. That AB is less than A C 2d. That A is less than AB. 3d. That AC= AE. 1st. In the triangle ABO, as AB is perpendicular to BC, the angle ABC is a right angle; and, therefore (by Theorem 12, Cor. 4) ; the angle BCA is less than a right angle ; and, as the greater side is always opposite the greater angle, AB is less than AC; and AC may be any line not identical with AB ; therefore a perpen- dicular is the shortest line that can be drawn from A to the line DE. 2d. As the two angles, A OB and A CD, are together equal to two right angles, (Th. 1), and A OB is less than a right angle, AOB must be greater than a right angle ; consequently, the [__ B is less than a right angle ; and, in the A AOB, AB is greater than AC, or A is less than AB, (Th. 19 Cor). 3d. In the A's ABC and ABE, AB is common, CB=> BE, and the angles at B are right angles; therefore, AC — AE, (Th. 16). Hence the theorem; a perpendicular is the shortest line etc. Cor. Conversely : if two equal oblique lines be drawn BOOK I. 39 from the same point to a given straight line, they will meet the line at equal distances from the foot of the per- pendicular drawn from that point to the given line. THEOREM XXIV. The opposite sides, and also the opposite angles of any par- allelogram, are equal. Let ABOD be a parallelogram. Then we are to show that AB = 1)0, AD = BO, [__A - L °> and [_ADC « \__ABO. Draw a diagonal, as BD; now, be- cause AB and DC are parallel, the al- ternate angles ABB and BBC are equal, (Th. 6). Foi the same reason, as AB and BO are parallel, the angles ABB and BBO are equal. !Now, in the two triangles ABB and BOB, the side BD is common, the \_ABB = [_DBO (1) tmd[_BDO = \_ABD (2) Therefore, the angle A = the angle C, and the two tri- angles are equal in all respects, (Th. 17); that is, the sides opposite the equal angles are equal ; or, AB — DO, and AD = BO. By adding equations ( 1 ) and ( 2 ) ? we have the angle ADO= the angle ABO, (Ax. 2). Hence the theorem ; the opposite sides, and the opposite angles, etc. Cor. 1. As the sum of all the angles of a parallelogram is equal to four right angles, and the angle A is always equal to the opposite angle O; therefore, if A is a right angle, is also a right angle, and the figure is a rect- angle. Oor. 2. As the angle ABO, added to the angle A, gives the same sum as the angles of the /\ADB; therefore, the two adjacent angles of a parallelogram are together equal to two right angles. -f X 40 GEOMETRY. THEOREM XXV. If the opposite sides of a quadrilateral are equal, they are also parallel, and the figure is a parallelogram. Let A BCD be any quadrilateral; on the supposition that AD = BC, and AB =* DC. we are to prove that AD is parallel to BC, and AB parallel to DC Draw the diagonal BD; we now have two triangles, ABD and BOB, which have the side BD common, AD of the one =. BO of the other, and AB of the one = CD of the other ; therefore the two A's are equal, (Th. 21), and the angles opposite the equal sides are equal ; that is, the angle ADB = the angle CBD ; but these are alternate angles; hence, AD is parallel to BC, (Th. 7, Cor. 1); and because the angle ABD = the angle BDC, AB is parallel to CD, and the figure is a parallelogram. Hence the theorem ; if the opposite sides of a quadri- lateral, etc. Cor. This theorem, and also Th. 24, proves that the two A's which make up the parallelogram are equal; and the same would be true if we drew the diagonal from A to C; therefore, the diagonal of any parallelogram bisects the parallelogram. THEOREM XXVI. The lines which join the corresponding extremities of two equal and parallel straight lines, are themselves equal and parallel; and the figure thus formed is a parallelogram. On the supposition that AB is equal and parallel to DC, we are to prove that AD is equal and parallel to BC; and that the figure is a par- allelogram. Draw the diagonal BD ; now, since BOOK 1. 41 AB and DO are parallel, and BB joins them, the alter- nate angles ABB and BBO are equal ; and since the side AB = the side BO, and the side BB i« common to the two A's ABB and OBB, therefore the two triangles are eqnal, (Th. 16) ; that is, AB —BO, the angle A = 0, and the \__ABB = the [_BBO; also AB is parallel to BO; and the figure is a parallelogram. Hence the theorem ; the lines which join the corresponding extremities, etc. THEOREM XXVII. Parallelograms on the same base, and between the same parallels, are equivalent, or equal in respect to area or sur- face. Let ABEO and ABBF be two parallelograms on the same base AB, and between the same paral- lels AB and OB ; we are to prove that these two parallelograms are equal. Now, OB and FB are equal, be- cause they are each equal to AB, (Th. 24) ; and, if from the whole line OB we take, in succession, OB and FB, there will remain EB = OF, (Ax. 3) ; but BB = A 0, and AF= BB, (Th. 24); hence we have two A's, OAF and EBB, which have the three sides of the one equal to the three sides of the other, each to each ; therefore, the tv r o A's are equal, (Th. 21). If, from the whole figure A BBO, we take away the A OAF, the parallelogram ABBF will remain ; and if from the whole figure we take away the other A EBB, the parallelogram ABEO will remain. Therefore, (Ax. 3), the parallelogram ABBF = the parallelogram ABEO. Hence the theorem ; Parallelograms on the same base, etc. 4* 42 GEOMETRY. THEOREM XXVIII. Triangles on the same base and between the same parallel* are equivalent. Let the two A's ABE and ABF have the same base AB, and be be- tween the same parallels AB and EF', then we are to prove that they are equal in surface. From B draw the line BB, par- allel to AF; and from A draw the line AC, parallel to BE ; and produce EF, if necessary, to C and D ; now the parallelogram ABBF = the parallelogram ABEC, (Th. 27). But the A ABE is one half the parallelogram ABEC, and the A ABF is one half the parallelogram ABDF; and halves of equals are equal, (Ax. T); there-, fore the A ABE - the A ABF. Hence the theorem ; triangles on the same base, etc. THEOREM XXIX. Parallelograms on equal bases, and between the same par allels, are equal in area. Let ABCB and EFaH, be two parallelograms on equal bases, AB and EF, and between the same parallels, AF and BG ; then Ve are to prove that they are equal in area. AB = EF=HG; but lines which join equal and parallel lines, are themselves equal and parallel, (Th. 26) ; therefore, if AH said BGr be drawn, the figure ABGrlTis a parallelogram == to the parallelogram ABCB, (Th. 27) ; and if we turn the whole figure over, the two parallelo- grams, G-HEF and GRAB, will stand on the same base, GH, and between the same parallels ; therefore, GHEF - GHAB, and consequently ABCB = EFGff, (Ax. 1). Hence the theorem ; Parallelograms on equal bases, (tc. DC H G BOOK I. I — 43 Cor. Triangles on equal bases, and between the same parallels, are equal in area. For, draw BB and E Gr\ the A ABB is one half of the parallelogram AC, and the A EFGr is one half of the equivalent parallelogram FE; therefore, the A ABB - the A EFG, (Ax. 7). THEOREM XXX. If a triangle and a parallelogram are upon the same or equal bases, and between the same parallels, the triangle is equiva- lent to one half the parallelogram. Let ABC be a A, and ABBE a parallelogram, on the same base AB, and between the same parallels ; then we are to prove that the A ABC is equivalent to one half of the parallel- ogram ABBE. Draw EB the diagonal of the parallelogram ; now, because the two A's ABC and ABE are on the same base, and between the same parallels, they are equiva- lent, (Th. 28) ; but the A ABE is one half the parallel- ogram ABBE, (Th. 25, Cor.) ; therefore the A ABC is equivalent to one half of the same parallelogram, (Ax. 7). Hence the theorem ; if a triangle and a parallelogram, etc THEOREM XXXI. The complementary parallelograms described about any point in the diagonal of any parallelogram, are equivalent to each other. Let A C be a parallelogram, and BB its diagonal ; take any point, as E, in the diagonal, and through this point draw lines parallel to the sides of the parallelogram, thus forming four parallelograms. We are now to prove that the complementary paral- lelograms, AE and EC, are equivalent. 44 GEOMETRY. By (Th. 25, Cor.) we learn that the A ABB = A BBC. Also by the same Cor., A a = A b, and Ac = A<7* + "AS 2 +7RP But, four times the square of the half of a line is equiv- alent to the square of the whole line, (Th. 36, Corollary) ; therefore ±AW = AQ\ and 4EB 2 = BB 2 ; and by sub- stituting these values, we have AQ 2 + BB 2 =~AB 2 +~BQ % +~BQ 2 + ~AB 2 , which equation conforms to the enunciation of tfc* theorem. K H L BOOK I. 57 THEOREM XLIV. If a line be "bisected and produced, the rectangle contained by the whole line and the part produced, together with the square of one half the bisected line, will be equivalent to the square on a line made up of tlie part produced and one half the bisected line. Let AB be any line, bi- sected in and produced to D. On OB describe the square OF, and on BB describe the square BE. The sides of the square BE being produced, the G~~ m" square GrL will be form- ed. Also, complete the construction of the rectangle ABEK. Then we are to prove that the rectangle, AE, and the aquare, GrL, are together equivalent to the square, CBFa. The two complementary rectangles, CL and LF, are equal, (Th. 31). But OL=AH, the line AB being bisected at 0; therefore AL is equal to the sum of the two com- plementary rectangles of the square OF. To AL add the square BE, and the whole rectangle, AE, will be equal to the two rectangles CE and EM. To each of these equals add EM, or the square on HL or its equal OB, and we have rectangle AE -f square EM = OB 2 ; but rectangle AE = AB x BB, and square EM = CB 2 * Hence the theorem, etc. Scholium, — If we represent AB by 2a, and BD by x, then AD — 2a -f x, and AD X BD = 2ax + x*. But CB 1 = a' ; adding this equation to the preceding, member to member, we get AD X BD -f- CB 2 = a* -f- 2ax -f x 3 = a + x\ But CD — a + s; hence this equa- tion is equivalent to the equation AL X DB -f- CE? = CD , which is the algebraic proof of the theorem. 58 GEOMETRY THEOREM XLV. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the two un- equal parts together with the square of the line between the points of division, will be equivalent to the square on one half the line. Let AB be a line bisected in 0, and divided into two unequal parts in B, We are to prove that AB x BB + . p c Cff^AQ\ov~CB\ B A We see by inspection that AB = AC + CB, and BB m, AC— OB; therefore by (Th. 38), we have ABx BB = AC 2 —~CB\ By adding UB 2 to each of these equals, we obtain AB x BB + ~CD* =~AC % Hence the theorem. BOOK II. 59 BOOK II PROPORTION. DEFINITIONS AND EXPLANATIONS. The word Proportion, in its common meaning, de- notes that general relation or symmetry existing between the different parts of an object which renders it agree- able to our taste, and conformable to our ideas of beauty or utility ; but in a mathematical sense. 1. Proportion is the numerical relation which one quan- tity bears to another of the same kind. As the magnitudes compared must be of the same kind, proportion in geometry can be only that of a line to a line, a surface to a surface, an angle to an angle, or a volume to a volume. 2. Ratio is a term by which the number which meas- ures the proportion between two magnitudes is desig- nated, and is the quotient obtained by dividing the one by the other. Thus, the ratio of A to B is --, or A : B, in which A is called the antecedent, and B the consequent. If, therefore, the magnitude A be assumed as the unit or standard, this quotient is the numerical value of B ex- pressed in terms of this unit. It is to be remarked that this principle lies at the found- ation of the method of representing quantities by num- bers. For example, when we say that a body weighs twenty-five pounds, it is implied that the weight of this body has been compared, directly or indirectly, with that of the standard, one pound. And so of geometrica 60 GEOMETRY. magnitudes ; when a line, a surface, or a volume is said to be fifteen linear, superficial, or cubical feet, it is un- derstood that it has been referred to its particular unit, and found to contain it fifteen times ; that is, fifteen is the ratio of the unit to the magnitude. "When two magnitudes are referred to the same unit, the ratio of the numbers expressing them will be the ratio of the magnitudes themselves. Thus, if A and B have a common unit, a, which is contained in A, m times, and in B, n times, then A =• ma j t> , B na n and B = na. and -r = — = — . A ma m To illustrate, let the line A contain the line . A a six times, and let the , line B contain the same a line a five times : then i i i j j j A= 6a and B—5a, which B . B ba 5 glVe 2 = 6-a = 6- 3. A Proportion is a formal statement of the equality of two ratios. Thus, if we have the four magnitudes A, B, and D, fmch that -r = ^, this relation is expressed by the pro- portion A: B :: C: D, or A : B = : D, the first of which is read, A is to B as O is to D ; and the second, the ratio of A to B is equal to that of to D. 4. The Terms of a proportion are the magnitudes, or caore properly the representatives of the magnitudes compared. 5. The Extremes of a proport on are its first and fourth terms. 6. The Means of a proportion are its second and third terms. 7. A Conplet consists of the two terms of a ratio. The BOOK II. 61 first and second terms of a proportion are called the first couplet, and the third and fourth terms are called the second couplet. 8. The Antecedents of a proportion are its first and third terms. 9. The Consequents of a proportion are its second and fourth terms. In expressing the equality of ratios in the form of a proportion, we may make the denominators the ante- cedents, and the numerators the consequents, or the reverse, without affecting the relation "between the magni- tudes. It is, however, a matter of some little importance to the "beginner to adopt a uniform rule for writing the terms of the ratios in the proportion ; and we shall always, unless otherwise stated, make the denominators of the ratios the antecedents, and the numerators the conse- quents.* 10. Equimultiples of magnitudes are the products arising from multiplying the magnitudes by the same number. Thus, the products, Am and Bm, are equimultiples of A and B. U. A Mean Proportional between two magnitudes is a magnitude which will form with the two a proportion, when it is made a consequent in the first ratio, and an antecedent in the second. Thus, if we have three mag- nitudes A, B y and 0, such that A : B : : B : (7, B is a mean proportional between A and 0. 12. Two magnitudes are reciprocally, or inversely pro- portional when, in undergoing changes in value, one is multiplied and the other is divided by the same number. Thus, if A and B be two magnitudes, so related that when j> A becomes mA, B becomes — , A and B are said to be m inversely proportional. * For discussion of the two methods of expressing Ratio, see Uni vorsity Algebra. 6 62 GEOMETRY. 13. A Proportion is taken inversely when the ante- cedents are made the consequents and the consequents the antecedents. 14. A Proportion is taken alternately, or by alternation, when the antecedents are made one couplet and the con- sequents the other. 15. Mutually Equiangular Polygons have the same num- ber of angles, those of the one equal to those of the ethers, each to each, and the angles like placed. 16. Similar Polygons are such as are mutually equi- angular, and have the sides about the equal angles, taken in the same order, proportional. 17. Homologous Angles in similar polygons are those which are equal and like placed ; and 18. The Homologous Sides are those which are like dis- posed about the homologous angles. THEOREM I. If the first and second of four magnitudes are equal, and also the third and fourth, the four magnitudes may form a proportion. Let A, B, 0, and B represent four magnitudes, such that A — B and C—B; we are to prove that A : B : : C : B. Now, by hypothesis, A is equal to B, and their ratio is therefore 1 ; and since, by hypothesis, is equal to B, their ratio is also 1. Hence, the ratio of A to B is equal to that of C to B ; and, (by Def. 3), A : B : : : D. Therefore, four magnitudes which are equal, two and two, constitute a proportion. BOOK II. 63 THEOREM II. If four magnitudes constitute a proportion, the product of the extremes is equal to the product of the means. Let the four magnitudes A, B, C, and D form the pro- portion A : B : : : D ; we are to prove that A x D = Bx C. The ratio of A to B is expressed hy -j = r. The ratio of to B is expressed hy ^ = r. Hence, (Ax. 1), ^ - ~. Multiplying each of these equals hy A x C, we have B x C=Ax D. Hence the theorem ; if four magnitudes are in propor- Hon, etc. Cor. 1. Conversely; If we ha/ve the product of two mag- nitudes equal to the product of two other magnitudes, they will constitute a proportion of which either two may be made the extremes and the other two the means. Let the magnitudes B x C = A x D. Dividing both members of the equation by A x C, we obtain B _B A~ C Hence the proportion A : B : : (7:2). Cor. 2. If we divide both members of the equation Ax B = B x C by A, we have D = — - A — . A That is, to find the fourth term of a proportion, mul- tiply the second and third terms together and divide the pro- duct by the first term. This is the Rule of Three of Arithmetic- 64 . GEOMETRY. This equation shows that any one of the foui terms can be found by a like process, provided the other three are given. THEOREM III. If three magnitudes are continued proportionals, the product of the extremes is equal to the square of the mean. Let A, B, and represent the three magnitudes : Then A : B : : B : C, (by Def. 11). But, (by Th. 2), the product of the extremes is equal to the product of the means ; that is, A x (7= B*. Hence the theorem ; if three magnitudes, etc, THEOREM IV. Equimultiples of any two magnitudes have the same ratio as the magnitudes themselves ; and the magnitudes and their equimultiples may therefore form a proportion. Let A and B represent two magnitudes, and mA and mB their equimultiples. Then we are to prove that A : B : : mA • *nB, The ratio of A to B is ~, and of mA Us mB is mB B '. ,. — T = -r, the same ratio. mA A 7 Hence the theorem; equimultiples of any twc &4y*i tudes, etc, THEOREM V. If four magnitudes are proportional, they will be propor~ tional when taken inversely. If A : B : : mA : mB, then B : A : : mB : mA ; For in either case, the product of the extremes equals that of the means; or the ratio of the couplets is the same. Hence the theorem ; if four quantities are propor- tional, etc. BOOK II THEOREM VI. Magnitudes which are proportional to the same propor tionals, are proportional to each other. If A : B = P : Q 1 Then we are to prove that and a : b = P : Q) A : B = a : b. From the 1st proportion, — = ~ ; A. Jr From the 2d " - = -^; a P 7? h Therefore, by (Ax. 1), -j = -, or A : B = a : b. jA. a Hence the theorem ; magnitudes which are proportional to the same proportionals, etc. Cor. 1. This principle may be extended through any number of proportionals. Cor. 2. If the ratio of an antecedent and consequent of one proportion is equal to the ratio of an antecedent and conse- quent of another proportion, the remaining terms of the two proportions are proportional. For, if A : B : : C : D and M : N : : P : Q in which A= W then C=P> hence : D : : P : Q. THEOREM VII. If any number of magnitudes are proportional, any one of the antecedents will be to its consequent as the sum of all the antecedents is to the sum of all the consequents. Let A, B, C, B, E, etc., represent the several magni tudes whi 3h give the proportions 6* A : B :: : D A : B :: E : F A : B : : a : H, E etc., etc 66 GEOMETRY. To which we may annex the identical proportion, A : B : : A : B. Now, (by Th. 2), these proportions give the following equations, A x D - B x A x F = B x E A x H= B x a A x B = B x A, etc. etc. From which, by addition, there results the equation, A(B + D + F + H, etc.) = B(A+ C+F+ #, etc.) But the sums B + D + F, etc., and A -f C + U, etc., may be separately regarded as single magnitudes ; there- fore, (Th. 2, Cor. 1), A : B :: A+C+F+ G, etc. : .B + D -f #+ J7", etc. Hence the theorem ; if any number of magnitudes are pro- portional, etc. THEOREM VIII. If four magnitudes constitute a proportion, the first will be to the sum of the first and second as the third is to the sum of the third and fourth. By hypothesis, A : B :: C : D; then we are to prove that A : A + B :: : C + D. By the given proportion, — = — . Adding unity to both members, and reducing them to the form of a fraction, we have — - — = — ^ — . Chang- ing this equation into its equivalent proportion al form, we have A : A 4- B : : : C + B. Hence the theorem ; if four magnitudes constitute a pro* portion, etc. T> Cor. If we subtract each member of the equation -j = BOOK II. 67 D p from unity, and reduce as before, we shall have A : A — B :: : C — D. Hence also ; if four magnitudes constitute a proportion, the first is to the difference between the first and second, as the third is to the difference between the third and fourth. THEOREM IX. If four magnitudes are proportional, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. Let A, B, 0, and D be the four magnitudes which give the proportion A : B :: C : D; we are then to prove that they will also give the propor- tion A + B : A—B :: C + D : Q — D. By Th. 8 we have A : A + B - : (7+2). Aisoby Corollary, sameTh., A : A — B = C : C—D. Xow, if we change the order of the means in these pro- portions, which may be done, since the products of ex- tremes and means remain the same, we shall have A : C = A + B : C+ B. A : C = A — B : C—B. Hence, (Th. 6), we have A + B : 0+ D - A — B : C—D. Or, A + B : A — B - + D : Q—D. Ilence the theorem ; if four magnitudes are proportional, etc. THEOREM X. If four magnitudes are proportional, like powers or like roots of the same magnitudes are also proportional. If the four magnitudes, A, B, C, and D, give the pro- portion 68 GEOMETRY. A : B : : : D, we are to prove that A n : B n :: C n : B n . The hypothesis gives the equation — = — . Eaising A G both members of this equation to the nth power, we have B n D n -^ = -^, which, expressed in its equivalent proportional form, gives A n : B n :: C n : B n . If n is a whole number, the terms of the given propor- tion are each raised to a power ; but if n is a fraction having unity for its numerator, and a whole number for its denominator, like roots of each are taken. As the terms of the proportion may be first raised to like powers, and then like roots of the resulting propor- tion be taken, n may be any number whatever. Hence the theorem ; if four magnitudes, etc. THEOREM XI. If four magnitudes are proportional, and also four others, the products which arise from multiplying the first four by the second four, term by term, are also proportional. Admitting that A : B and X: Y We are to show that AX : BY : B, Mi N, QMiBK From the first proportion, — = — ; a. o Y N From the second, __ = -_. X M Multiply these equations, member by member, and AX CM' Or, AX i BY n CM: BN. The same would be true in any number oi proportions. Hence the theorem ; if four magnitudes are, etc. BOOK II. 69 THEOREM XII. If four magnitude* are 'proportional, and also four others, the quotients which arise from dividing the first four by the swond four, term by term, are proportional. By hypothesis, A : B : : : D, and X : Y ::M: K Multiply extremes and means, AD == OB, ( 1 ) and XN=MY. (2) Divide (1) hy (2), and ± x * ? JJ x * Convert these four factors, which make two equal pro- ducts, into a proportion, and we have X : Y' : M l JV" By comparing this with the given proportions, we find it is composed of the quotients of the several terms of the first proportion, divided by the corresponding terms of the second. Hence the theorem ; if four magnitudes are proportional, etc. THEOREM XIII. If four magnitudes are proportional, we may multiply the first couplet, the second couplet, the antecedents or the conse- quents, or divide them by the same quantity, and the results will be proportional in every case. Let the four magnitudes A, B, 0, and D give the pro- portion A: B :: 0: D. By multiplying the extremes and means we have A.D = B.O (1) Multiply both members of this equation by any num- ber, as a, and we have aA.D = aB.O By converting this equation into a proportion in four different ways, we have as follows : 70 GEOMETRK aA : aB :: : B A : B :: aO : aB aA : B : : aQ : B A : aB : : C: aB. Kesuming^the original equation, (1), and dividing both members by a, we have A.B _ B.Q a a This equation may also be converted into a proportion in four different ways, with the following results : ; x> : : : D a a A : B • : . a B a A . B . . (7. B a a A : B a : : C : B a Hence the theorem ; if four magnitudes are in proportion, etc, THEOREM XIV. If three magnitudes are in proportion, the first is to the third as the square of the first is to the square of the second. Let A, B, and C, be three proportionals. Then we are to prove that A : C— A 2 : B* By(Th. 3) AC=B> Multiply this equation by the numeral value of A, and we have % A*C=A& This equation gives the following proportion : Ax C=A 2 :B\ Hence the theorem. Remark. — It is now proposed to make an application of the pre- ceding abstract principles of proportion, in geometrical investigations BOOK II 71 D THEOREM XV. If two parallelograms are equal in area, the base and per- pendicular of either may be made the extremes of a propor- tion, of which the base and perpendicular of the other are the means. Let ABQD, and HLNM, be two paral- lelograms hav- ing equal areas, A by hypothesis ; then we are to prove that AB : LN : : MK : BF, in which MK and BF are the altitudes or perpendiculars of the parallelograms. This proportion is true, if the product of the extremes is equal. to the product of the means; that is, if the equation AB.BF = LN.MKis true. But AB.BF is the measure of the rectangle ABFE, by (Definition 54, B. I.), and this rectangle is equal in area to the parallelogram ABOD, (B. I., Th. 27). In the same manner, we may prove that LN.MK is the measure of the parallelogram NLHM. But these two parallelograms have equal areas by hypothesis. Therefore, AB.BF = LN.MK is a true equation, and Th. 2, Cor. 1), gives the proportion AB : LN : : MK : BF. Hence the theorem ; if two parallelograms are equal in area, etc. THEOREM XVI. Parallelograms having equal altitudes are to each other as their bases. Since parallelograms having equal bases and equal altitudes are equal in area, however much their angle* 72 GEOMETRY. may differ, we can suppose the two parallelograms under consideration to be mutually equiangular, without in the least impairing the generality of this theorem. There- fore, let ABOB and AEFB be two parallelograms having equal alti- tudes, and let them be placed with their bases on the same line AE, and let the side, AB t be common. First suppose their bases commensurable, and that AE being divided into nine equal parts, AB contains five of those parts. If, through the points of division, lines be drawn paral- lel to AB, it is obvious that the whole figure, or the parallelogram, AEFB, will be divided into nine equal parts, and that the parallelogram, ABOB, will be com- posed of five of those parts. Therefore, ABCB : AEFB : : AB : AE : : *5 : 9. Whatever be the whole numbers having to each other the ratio of the lines AB and AE, the reasoning would remain the same, and the proportion is established when the bases are commensurable. But if the bases are not to each other in the ratio of any two whole numbers, it remains still to be shown that AEFB : ABOB :: AE : AB (1) If this propor- tion is not true, there must be a line greater or less than AB, to which AE will have the A B L same ratio that AEFB has to ABOB. Suppose the fourth proportional greater than AB, as AK, then, AEFB : ABOB :: AE : AK (2). BOOK II. 78 If we now divide the line AE into equal parts, each less than the line BK, one point of division, at least, will fall between B and K. Let L be such point, and draw LM parallel to BO. This construction makes AE and AL commensura- ble; and by what has been already demonstrated, we have AEFD : ALMD :: AE : AL. (3) Inverting the means in proportions ( 2 ) and ( 3 ), they become AEFD : AE :: ABQD : AK; and AEFD : AE : : ALMD : AL. Hence, (Th. 6), ABOD : AK : : ALMD : AL. By inverting the means in this last proportion, we have ABOD : ALMD : : AK : AL. But AK is, by hypothesis, greater than AL; hence, if this proportion is true, ABOD must be greater than ALMD; but on the contrary it is less. We therefore conclude that the supposition, that the fourth propor- tional, AK, is greater than AB, from which alone this absurd proportion results, is itself absurd. In a similar manner it can be proved absurd to sup- pose the fourth proportional less than AB. Therefore the fourth term of the proportion ( 1 ) can be neither less nor greater than AB ; it is then AB itself, and parallelograms having equal altitudes are to each other as their bases, whether these bases are commensur- able or not. Hence the theorem ; Parallelograms having equal alti- tudes, etc. Oor. 1. Since a triangle is one half of a parallelogram having the same base as the triangle and an equal alti- tude, and as the halves of magnitudes have the same ratio as their wholes ; therefore, 7 74 GEOMETRY. Triangles having the same or equal altitudes are to each other as their bases. Cor. 2. Any triangle has the same area as a right- angled triangle having the same hase and an equal alti- tude ; and as either side about the right angle of aright- angled triangle may be taken as the base, it follows that Two triangles having the same or equal bases are to each other as their altitudes. Cor. 3. Since either side of a parallelogram may be taken as its base, it follows from this theorem that Parallelograms having equal bases are to each other as their altitudes. THEOREM XVII. If lines are drawn cutting the sides, or the sides 'produced, of a triangle proportionally, such secant lines are parallel to the base of the triangle ; and conversely, lines drawn parallel to the base of a triangle cut the sides, or the sides produced, proportionally. Let ABC be any triangle, and draw the line BE dividing the sides AB and AC into parts wmich give the proportion AB : BB : : AE : EC. We are to prove that BE is parallel to BC. If BE is not a parallel through the point B to the line BC, suppose Bm to be that parallel ; and draw the lines BC and Bm. Now, the two triangles ABm and mBC, have the same altitude, since they have a common vertex, B, and their bases in the same line, AC; hence, they are to each other as their bases, Am and mC, (Th. 16, Cor. 1). BOOK II. 75 That is, A ADm : A mDQ : : Am : mO, Also, A AmD : £ J>mB :: AD : DB. But, since Dm is supposed parallel to BO, the triangles DBm and DCm have equal areas, because they are on the same base and between the same parallels, (Th. 28, B.I). Therefore the terms of the first couplets in the two preceding proportions are equal each to each, and conse- quently the terms of the second couplets are proportional, (Theorem 6). That is, AD : DB : : Am : mO But AD : DB :: AE : EQ by hypothesis. Hence we again have two proportions having the first couplets, the same in both, and we therefore have AE : EC : : Am : mO By alternation this becomes AE : Am :: EC : mO That is, AE is to Am, a greater magnitude is to a less, as EQ is to mO, a less to a greater, which is absurd. Had we supposed the point m to fall between E and 0, our conclusion would have been equally absurd ; hence the suppositions which have led to these absurd results are themselves absurd, and the line drawn through the point D parallel to BO must intersect AO in the point E. Therefore the parallel and the line DE are one and the same line. Conversely : JfDE be drawn parallel to the base of the triangle, then will AD i DB :: AE i EQ For as before, A ADE : a EDO : : AE : EQ and A DEB : A ADE x; DB i AD Multiplying the corresponding terms of these propor- 76 GEOMETRY. tions, and omitting the common factor, a ADE, in the first couplet, we have A DEB : A EDO : : AE x DB : EC x AD. But the a's DEB and EDO have equal areas, (Th. 28, B. I) ; hence AE x DB = EC x AD, which in the form of a proportion is AE : EC : : AD : DB or, AD : DB : : AE : EC and therefore the line parallel to the base of the triangle, divides the sides proportionally. It is evident that the reasoning would remain the same, had we conceived ADE to be the triangle and the sides to be produced to the points B and 0. Hence the theorem; if lines are drawn cutting the sides, etc. Cor. 1. Because DE is parallel to BO, and intersects the sides AB and A 0, the angles ADE and ABO are equal. For the same reason the angles AED and AOB are equal, and the A's ADE and ABO are equiangular. Let us now take up the triangle ADE, and place it on ABO; the angle ADE falling on |__ B, the side AD on the side AB, and the side DE on the side BO Now, since the angle A is common, and the angles AED and A OB are equal, the side AE of the A ADE, in its new position, will be parallel to the side AO of the A ABO. The last proportion of this Th. gives (Th. 8 and Th. 5), AD : AE w AB : AC From the above construction we obtain, by a similar course of reasoning, the proportion AD : DE : : AB : BO And in like manner it may be shown that AE : ED :: AO : OB That is, the sides about the equal angles of equiangular triangles, taken in the same order, are proportional, and the triangles are similar, (Def. 16). BOOK II 77 Cor. 2. Two triangles having an angle in one equal to an angle in the other, and the sides about these equal angles pro* portional, are equiangular and similar. For, if the smaller triangle be placed on the larger, the equal angles of the triangles coinciding, then will the sides opposite these angles be parallel, and the triau- gles will therefore be equiangular and similar. THEOREM XVIII. If any triangle have its sides respectively proportional to the like or homologous sides of another triangle, each to each, then the two triangles will be equiangular and similar. Let the triangle abc have its sides pro- portional to the triangle ABO ; that is, ac to A as cb to OB, and ac to AC as ab to AB ; then we are to prove that the a's, abc and ABO, are equi- angular and similar. On the other side of the base, AB, and from A, conceive the angle BAB to be drawn = to the |__ a ; and from the point B, conceive the angle ABB to be drawn = to the [_ b. Then the third [__ D must be =a to the third [_ o, (B. I, Th. 12, Cor. 2) ; and the A ABB will be equiangular to the A abc by construction. Therefore, ac : ab = AB : AB By hypothesis, ac : ab = A O : AB Hence, AB : AB = A : A B, (Th. 6). In this last proportion the consequents are equal; therefore, the antecedents are equal : that is, AB - AO In the sa me manner we may prove that BB = OB 78 GEOMETRY. But AB is common to the two triangles ; therefore, the three iides of the A ABB are respectively equal to the three sides of the A ABO, and the two a's are equal, (B. I, Th. 21). But the A's ABB, and abc, are equiangular by con- struction; therefore, the A's, ABO, and abc, are also equiangular and similar. Hence the theorem ; if any triangle have its sides, etc, Second Demonstration. Let abc and ABO be two triangles whose sides are respectively propor- tional, then will the triangles be equi- angular and similar. That is, L« = LA L& = L 2> and If the [_ c be in fact equal to the [__ 0, the tri- angle abc can be placed on the triangle ABO, ca taking the direction of OA and cb of OB. The line ab will then divide the sides OA and OB proportionally, and will therefore be parallel to AB, and the triangles will be equiangular and similar, (Th. 17). But if the [_c be not equal to the [_ 0, then place ae on i(J as before, the point c falling on 0. Under the present supposition cb will not fall on OB, but will take another direction, OV, on one side or the other of OB Make OV equal to cb and draw aV. Now, the A abc is represented in magnitude and posi- tion by the A a VO; and if, through the point a, the line ab be drawn parallel to AB, we shall have Oa : OA n ab : AB; hut by (Hy.) Oa : OA : : aV : AB. BOOK II. 79 Hence, (Th. d), ab : AB :: aV : XB; which requires that ab = aV, but (Th. 22, B. 1) ab can not be equal to aV; hence the last proportion is absurd, and the supposition that the [__ c is not equal to the [__ (7, which leads to this result, is also absurd. Therefore, the [_c is equal to the [__ (7, and the triangles are equi- angular and similar. Hence the theorem ; if any triangle have its sides, etc. THEOREM XIX. If four straight lines are in proportion, the rectangle con- tained by the lines which constitute the extremes, is equivalent to that contained by those which constitute the means of the proportion. Let A, B, C, D, represent the four A '~ lines ; then T> we are to show, geo- c , metrically, that A x JD — B x Q. D j Place A and B at right angles to each other, and draw the hypotenuse. Also place and D at right angles to each other, and draw the hypotenuse. Then bring the two triangles together, so that shall be at right angles to B, as represented in the figure. Now, these two A's have each a E. [_, and the sides about the equal angles are pro- portional ; that is, A : B : : C : D ; hence, (Th. 17, Cor. 2), the two A's are equiangular, and the acute angles which meet at the extremities of B and (7, are together equal to one right angle, and the lines B and G are so placed as to make another right angle; therefore, also, the extremities of A, B, 0, and D, are in one right line, (Th. 3, B. I), and that line is the diag- l\ B \ BC \ C \» ■ AD \ 80 GEOMETRY. onal of the parallelogram be. By Th. 31, B. I, the complementary parallelograms about this diagonal are equal ; but, one of these parallelograms is B in length, and in width, and the other is D in length and A in width; therefore, B x C = A x B. Hence the theorem; if four straight lines are in propor- tion, etc. Cor. When B = (7, then A X B = B\ and B is the mean proportional between A and D. That is, if three straight lines are in proportion, the rectangle contained by the first and third lines is equivalent to the square described on the second line. THEOREM XX. Similar triangles are to one another as the squares of their homologous sides. Let ABO and DBF be two similar triangles, and LQ and MF perpendiculars to the sides A B and BE respectively. Then we are to prove that aABC:aBEF = AB*:BE\ By the similarity of the tri- angles, we have, AB :BE = LC : MF But, AB ; BE = AB : BE Hence, AB 2 : JJW = AB x LQ : BE x MF. But, (by Th. 30, B. I), AB x LQ is double the area of the A ABC, and BE x MF is double the area of the A BEF. Therefore, A ABO: ABEFnAB x LQ :BExMF And, (Th. 6), A ABC: A BEF= AW : BE\ Hence the theorem ; similar triangles are to one another, etc. BOOK II. 81 The following illustration will enable the learner fully to comprehend this important theorem, and it will also serve to impress it upon his memory. Let abc and ABO represent two equiangular triangles. Suppose the length of the side ac to be two units, and the length of the corresponding side A to be three units. Now, drawing lines through the points of division of the sides ac and A 0, parallel to the other sides of the triangles, we see that the smaller triangle is com- posed of four equal triangles, while the larger contains nine such triangles. That is, the sides of the triangles are as 2 : 3, and their areas are as 4 : 9 = 2 2 : 3'. THEOREM XXI Similar polygons may be divided into the same number of triangles; and to each triangle in one of the polygons there will be a corresponding triangle in the other polygon, these triangles being similar and similarly situated. JjQtABCDUtmd. abcde be two similar polygons. Now it is obvious that we can divide each polygon into as many triangles as the figure has sides, less two; and as the polygons have the same number of sides, the diagonals drawn from the vertices of the homologous angles will divide them into the same number of tri- angles. 82 GEOMETRY. Since the polygons are similar, the angles EAB and cab, are equal, and EA : AB : : ea : ab. Hence the two triangles, EAB and eab, having an angle in the one equal to an angle in the other, and the sides about these angles proportional, are equiangular and similar, and the angles ABE and abe are equal. But the angles ABO and abc are equal, because the polygons are similar. Hence, [_ABO— [_ABE= \__abc — [__abe; that is, [___EBO = [__ebc. The triangles, EAB and eab, being similar, their ho- mologous sides give the proportion, AB : BE : : ab : be; ( 1 ) and since the polygons are similar, the sides about the equal angles B and b are proportional, and we have AB : BO :: ab : be; or, BO : AB :: be : ab. (2) Multiplying proportions (1) and (2), term by term, and omitting in the result the factor AB common to the terms of the first couplet, and the factor ab common to the terms of the second, we have BO : BE n be i be. Hence the A's EBO and ebe are equiangular and similar; and thus we may compare all of the triangles of one polygon with those like placed in the other. Hence the theorem ; similar polygons may be divided^ etc THEOREM XXII. The perimeter 8 of similar polygons are to one another as their homologous sides ; and their areas are to one another as the squares of their homologous sides. Let ABODE and abode be two similar polygons ; then we are to prove that AB is to the sum of all the sides BOOK II. of the polygon A BOB, as ah \% to the sum of all the sides of the polygon abed. E We have the identical proportion AB : ah : : AB : ah; and since the polygons are similar, we may write the following : AB : ah : 1 BO : he AB : ah : : OB : cd AB : ah : : BE : de 9 etc. etc. Hence, (Th. 7), AB : ah :: AB+BO+OD+DE, etc.: ab+bc+cd+de, etc. Therefore, the perimeters of similar polygons are to one another as their homologous sides. This is the first part of the theorem. Since the polygons are similar, the triangles EAB, eah, are similar, and if the triangle EAB is a part expressed by the traction -, of the polygon to which it belongs, n the triangle eah is a like part of the other polygon. Therefore, EAB : eah : : ABOBEA : ahedea. But, (Th. 20), EAB : eah : : AB 2 : ah\ Therefore, (Th. 6), ABOBEA : ahedea : : AB 2 : ah\ Therefore, the similar polygons are to one another as the squares on their homologous sides. This is the second part of the theorem. Hence the theorem ; the perimeters of similar polygon* are to one another, ete, THEOREM XXIII. Two triangles which have an angle in the one equal tc an angle in the other, are to each other as the rectangle of thu aides about the equal angles. 84 GEOMETRY. Let ABC and def be two triangles having the angles A and d equal. It is to be proved that the areas ABC and def are to each other as AB.AC is to de.df. Conceive the triangle def placed on the tri- angle ABC, so that d shall fall on A, and de on AB ; then df will fall on AC, because the L'si and d are equal. On AB, lay off Ae, equal to de ; and on AC, lay off Af, equal to df, and draw ef The tri- angle Aef will then be equal to the triangle def Join B and/. Now, as triangles having the same altitude are to each other as their bases, (Th. 16, Cor. 1), we have Aef : ABf :: Ae : AB also, ABf : ABC : : Af : AC Multiplying these proportions together, term by term, omitting from the result ABf, a factor common to the terms of the first couplet, we have Aef : ABC : : Ae . Af : AB . AC But Aef is equal to def, Ae to de, and Af to df; therefoie, def : ABC :: de . df : AB . AC Hence the theorem ; two triangles which have an angle, ttc. Scholium. — If we suppose that AB : AC :: de : df the two triangles will be similar ; and if we multiply the terms ot the first couplet of this proportion by AC, and the terms of the second ccmplet by df, we shall have AB . AC : AC % : : de ^ : dj* U, AB . AC : de . df :: AC 2 : df BOOK II. 85 Comparing this with the last proportion in this theorem, and we have, (Th.6); _ _ def: ABC xx df : AC 9 Remark. — This scholium is therefore another demonstration of Theorem 20, and hence that theorem need not necessarily have been made a distinct proposition. We require no stronger proof of the cer- tainty of geometrical truth, than the fact that, however different the processes by which we arrive at these truths, we are never led into inconsistencies ; but whenever our conclusions can be compared, they will harmonize with each other completely, provided our premises are true and our reasoning logical. It is hoped that the student will lose no opportunity to exercise his powers, and test his skill and knowledge, in seeking original demonstrations of theorems, and in deducing consequences and conclusions from those already established. THEOREM XXIV. If the vertical angle of a triangle be bisected, the bisecting line will cut the base into segments proportional to the adja- cent sides of the triangle. Let ABO be any triangle, and the vertical angle, 0, be bi- *s1 sected by tbe straight line CD. Then we are to prove that AB : BB = AC : OB. Produce A O to E, making A — D r OE = OB, and draw EB. The exterior angle A OB, of the A CEB, is equal to the two angles E, and CBE; but the angle E = QBE, because OB — OE, and the tri- angle is isosceles; therefore the angle AOB, the half of the angle A OB, i3 equal to the angle E, and BO and BE are parallel, (Cor.2,Th. 7,B. I). Now, as ABE is a triangle, and OB is parallel to BE, we have AD : BB = AQ : OE or OB, (Th. 17). Hence the theorem ; if the vertical angle of a triangle be bisected, etc. 8 86 GEOMETRY. THEOREM XXV. If from the right angle of a right-angled triangle, a per pendicular is drawn to the hypotenuse ; 1. The perpendicular divides the triangle into two similar triangles, each of which is similar to the whole triangle, 2. The perpendicular is a mean proportional between the segments of the hypotenuse. 3. The segments of the hypotenuse are in proportion to the squares on the adjacent sides of the triangle. 4. The sum of the squares on the two sides is equivalent to the square on the hypotenuse. Let BAG be a triangle, right an- gled at A ; and draw AD perpendicu- lar to BO. 1. The two A 's, ABO and ABB, B DC have the common angle, B, and the right angle BAQ = the right angle BDA ; therefore, the third |_ 's are equal, and the two A's are similar by Th. 17, Cor. 1. In the same manner we prove the A ADC similar to the A ABC; and the two triangles, ABB, ABC, being similar to the same A ABO, are similar to each other. 2. As similar triangles have the sides about the equal angles proportional, (Def. 16), we have BB : AD :: AD : OD; or, the perpendicular is a mean proportional between the seg- ments of the hypotenuse. 3. Again, BO j_BA :: BA : BD hence, BA* - BO.BD (1) also, BQjJJA : : OA : OD hence, OA 2 = BOOD (2) Dividing Eq. (1) by Eq. (2), member by member, wo obtain ~BA* , BD ~OA 2 "" OD BOOK II. 87 which, in the form of a proportion, is CA* :~BA* :: CD : BD; that is, the segments of the hypotenuse are proportional to the squares on the adjacent sides. 4. By the addition of (1) and (2), we have BJl + CA* m BQ(BD + CD) =BC\ that is, the sum of the squares on the sides about the right angle is equivalent to the square on the hypotenuse. This ia another demonstration of Theorem 39, B. I. Hence the theorem , if from the right angle of a right* angled triangle, etc. 88 GEOMETRY. BOOK III. OF THE CIRCLE, AND THE INVESTIGATION OF THEtt REMS DEPENDENT ON ITS PROPERTIES. DEFINITIONS. 1. * A Curved Line is one whose consecutive parts, how- ever small, do not lie in the same direction. 2. A Circle is a plane figure bounded by one uniformly curved line, all of the points of which are at the same distance from a certain point within, called the center 3. The Circumference of a cir- cle is the curved line that bounds it. 4. The Diameter of a circle is a line passing through the center, and terminating at both extremities in the circumfer- ence. Thus, in the figure, is the center of the circle, the curved line ACrBD is the cir- cumference, and AB is a diameter. 5. The Radius of a circle is a line extending from the center to any point in the circumference. Thus, CD is a radius of the circle. 6. An Arc of a circle is any portion of the circum- ference. * The first six of the above definitions have been before given among the general definitions of Geometry, but it was deemed advisable to reinsert them here. BOOK III. 89 7. A Chord of a circle is the line connecting the ex- tremities of an arc. 8. A Segment of a circle is the portion of the circle on either side of a chord. Tims, in the last figure, EGrF is an arc, and EF is a chord of the circle, and the spaces bounded by the chord EF, and the two arcs EGrF and EDF, into which it divides the circumference, are segments. 9. A Tangent to a circle is a line which, meeting the circumference at any point, will not cut it on being produced. The point in which the tangent meets the circumference is called the point of tangency. 10. A Secant to a circle is a line which meets the cir- cumference in two points, and lies a part within and a part without the circumference. 11. A Sector of a circle is a portion of the circle included between any two radii and their intercepted arc. Thus, in the last figure, the line HL, which meets the circumference at the point D, but does not cut it, is a tangent, D being the point of tangency; and the line MN, which meets the circumference at the points P and Q, and lies a portion within and a portion without the circle, is a secant. The area bounded by the arc BD, and the two radii OB, CD, is a sector of the circle. 12. A Circumscribed Polygon is one all of whose sides are tangent to the circumference of the circle ; and conversely, the circle is then said to be inscribed in the polygon. 13. An Inscribed Polygon is one the vertices of whose angles are all found in the circumference of the circle ; and conversely, the circle is then said to be circumscribed about the polygon. 14. A Regular Polygon is one which is both equiangu- lar and equilateral. 8* 90 GEOMETRY. The last three definitions are illustrated by the last figure. THEOREM I. Any radius perpendicular to a chord, bisects the chord, and also the arc of the chord. Let AB be a chord, the center of the circle, and OE & radius perpen- dicular to AB ; then we are to prove that AB = BB, and AE = EB. Since is the center of the circle, AC— BO, CB is common to the two A's A OB and BOB, and the angles at B are right angles ; therefore the two A's ABO and BBO are equal, and AB = BB, which proves the first part of the theorem. Now, as AB — BB, and BE is common to the two spaces, ABE and BBE, and the angles at B are right angles, if we conceive the sector OBE turned over and placed on CAE, OE retaining its position, the point B will fall on the point A, because AB = BB and AO = BO; then the arc BE will fall on the arc AE; otherwise there would be points in one or the other arc unequally distant from the center, which is impossible ; therefore, the arc AE — the arc EB, which proves the second part of the theorem. Hence the theorem. Cor. The center of the circle, the middle point of the chord AB, and of the subtended arc AEB, are three points in the same straight line perpendicular to the chord at its middle point. Now as but one perpen- dicular can be drawn to a line from a given point in that line, it follows : 1st. That the radius drawn to the middle point 3f any arc bisects, and is perpendicular to, the chord of the arc. BOOK III 91 2d. That the perpendicular to the cl ord at its middle point passes through the center of the circle and the middle of the subtended arc. THEOREM II. Equal angles at the center of a circle are subtended by equal chords. Let the angle A CE = the angle ECB; then the two isosceles triangles, ACE, and ECB, are equal in all re- spects, and AE = EB. Hence the theorem. THEOREM III. In the same circle, or in equal circles, equal chords are equally distant from the center. Let AB and EF be equal chords, and C the center of the circle. From C, draw CG and CH, perpendicular to the respective chords. These perpendiculars will bisect the chords, (Th. 1), and we shall have A G = EH. "We are now to prove that CG = CH. Since the A's ECH and AQG are right-angled, we have, (Th. 39, B. I), EH 2 +~WC 7 =-EC 2 and, AG 2 + GO 2 = 'AC 2 . By subtracting these equations, member from mem- ber, we find that EH 2 — AG 2 + 7W 2 — ~GC 2 =~EC 2 — AC* (1) But the chords are equal by hypothesis, hence their halves, EH and AG, are equal; also EC = AC, being radii of the circle. Wherefore, 92 GEOMETRY. EH 2 — AG 2 « and, W — ~A0 2 - 0. These values in Equation ( 1 ) reduce it to HC? - GO 2 = or, H0 2 =G0 2 and, #(7 - #(7. Hence the theorem. (7 EH 2 , and A G > EH. Therefore, AB, the double of AG, is greater than EF, the double of EH; that is, of two chords in the same or equal circles, the one nearer the center is the greater. The equation, EH 2 + HO 2 = AG 2 -f ~G0 2 , being true, whatever be the position of the chords, we may suppose GO to have any value between and A 0, the radius of the circle. When GO becomes zero, the equation reduces to EH 2 + ~H0 2 - AG* - B 2 ; that is, under this supposition, AG coincides with A 0, and AB becomes the diameter of the circle, the greatest chord that can he drawn in it. BOOK 111. 93 THEOREM IV A line tangent to the circumference of a circle is at right angles with the radius drawn to the point of contact. Let A be a line tangent to the circle at the point B, and draw the radius, EB, and the lines, AE and OE. Now, we are to prove that EB is per- pendicular to AC. Because B is the only point in the line A which meets the circle, (Def. 9, B. Ill), any other line, as AE or CE, must be greater than EB ; therefore, EB is the shortest line that can be drawn from the point E to the line A 0; and EB is the perpendicu- lar to AC, (Th. 23, B. I). Hence the theorem. THEOREM V. In the same circle, or in equal circles, equal chords subtend or stand on equal portions of the circumference. Conceive two equal circles, and two equal chords drawn within them. Then, conceive one circle taken up and placed upon the other, center upon center, in such a po- sition that the two equal chords will fall on, and exactly coincide with, each other; the circles must also coin- ciie, because they are equal; and the two arcs of the two circles on either side of the equal chords must also coin- cide, or the circles could not coincide; and magnitudes which coincide, or exactly fill the same space, are in all respects equal, (Ax. 10). Hence the theorem. 94 GEOMETRY. THEOREM VI. Through three given points, not in the same straight line, one circumference can be made to pass, and but one. Let A, B, and be three given points, not in the same straight line, and draw the lines AB and BO. If a circumference is made to pass through the two points A and B, the line AB will be a chord to such a circle ; and if a chord is bisected by a line at right angles, the bisecting line will pass through the center of the circle, (Cor., Th. 1) ; therefore, if we bisect the line AB, and draw DF, perpendicular to AB, at the point of bisection, any circumference that can pass through the points, A and B, must have its center somewhere in the line DF. And if we draw FIG- at right angles to BO at its middle point, any circumference that can pass through the points B and must have its center somewhere in the line EO. Now, if the two lines, DF and EGr, meet in a common point, that point will be a center, about which a circumference can be drawn to pass through the three points, A, B, and 0, and DF and EO will meet in every case, unless they are parallel ; but they are not parallel, for if they were, it would follow (Th. 5, B. I) that, since DF is intersected at right angles by the line AB, it must also be intersected at right angles by the line BO, having a direction different from that of AB ; which is impossible, (Th. 7, B. I). Therefore the two lines will meet ; and, with the point H, at which they meet, as a center, and HB — HA = HQ as a radius, one circumference, and but one, can be made to pass through the three given points. Hence the theorem. BOCK III. 95 THEOREM VII. If two rircles touch each other, either internally or exter- nally, the two centers and the point of contact will be in one right line. Let two circles touch each other internally, as represented at A, and conceive AB to be a tangent at the common point A. Now, if a line, perpendicular to AB, be drawn from the point A, it must pass through the center of each circle, (Th. 4) ; and as but one perpendicular can be drawn to a line at a given point in it, A, 0, and B, the point of contact and the two centers must be in one and the same line. Next, let two circles touch each other externally, and from the point of contact conceive the common tangent, AB, to be drawn. Then a line, A 0, perpendicular to AB, will pass through the center of one circle, (Th. 4), and a per- pendicular, AB, from the same point, A, will pass through the center of the other circle ; hence, BAG and BAB are together equal to two right angles ; therefore CAB is one continued straight line, (Th. 3, B. I). Cor. "When two circles touch each other internally, the distance between their centers is equal to the difference of their radii ; and when they touch each other extern- ally, the distance between their centers is equal to the sum of their radii. THEOREM VIII. An angle at the circumference of any circle is measured by one half the arj on which it stands. In this work it is taken as an axiom that any angle whose vertex is at the center of a circle, is measured by 96 GEOMETRY. the arc on which it stands ; and we now proceed to prove that when the arcs are equal, the angle at the circumference is equal to one half the angle at the center. Let ACB be an angle at the center, and D an angle at the circumference, and at first suppose D in a line with AC. We are now to prove that the angle ACB is double the angle D. The A DCB is an isosceles triangle, because CD = CB ; and its exterior angle, ACB, is equal to the two interior angles, D, and CBD, (Th. 12, B. I), and since these two angles are equal to each other, the angle ACB is double the angle at D. But ACB is measured by the arc AB ; therefore the angle D is measured by one half the arc AB. Next, suppose D not in a line with AC, but at any point in the circum- ference, except on AB ; produce DC to E. Now, by the first part of this theorem, the angle ECB = 2EDB, also, EC A = 2EDA, by subtraction, ACB = 2 ADB. But ACB is measured by the arc AB; therefore ADB or the angle D, is measured by one half of the same arc Hence the theorem. THEOREM IX. An angle in a semicircle is a right angle ; an angle in a (segment greater than a semicircle is less than a right angle ; and an angle in a segment less than a semicircle is greater than a right angle. If the angle ACB is in a semicircle, the opposite seg- ment, ADB, on which it stands, is also a semicircle ; and the angle ACB is measured by one half the arc ADB BOOK III. b« (Th. 8) ; that is, one half of 180°, or 90°, which is the measure of a right angle. If the angle A CB is in a segment greater than a semicircle, then the opposite segment is less than a semi- circle, and the measure of the angle is less than one half of 180°, or less than a right angle. If the angle ACB is in a segment less than a semicircle, then the opposite segment, ABB, on whiefc the angle stands, is greater than a semicircle, and its half is greater than 90°; and, consequently, the angle is greater than a right angle. Hence the theorem. Cor. Angles at the circumference, and standing on the same arc of a circle, are equal to one another ; for all angles, as BAG, BBC, BBC, are equal, because each is measured by one half of the arc BC. Also, if the angle BEC is equal to CEG-, then the arcs Bd and CG- are equal, be- cause their halves are the measures of equal angles. THEOREM X. The sum of two opposite angles of any quadrilateral in* scribed in a circle, is equal to two right angles. Let ACBD represent any quadri- lateral inscribed in a circle. The angle ACB has for its measure, one half of the arc ABB, and the angle ABB has for its measure, one half of the arc ACB; therefore, by addition, the sum of the two opposite angles at C and B, are together measured by one half of the whole circumference, or by 180 degrees, «= two right angles. Hence the theorem 98 GEOMETRY. THEOREM XI. An angle formed by a tangent and a chord is measured ty one half of the intercepted arc. Let AB be a tangent, and AD a chord, and A the point of contact ; then we are to prove that the angle BAD is measured by one half of the arc AED. From A draw the radius AC; and from the center, C, draw CE per- pendicular to AD. The [_BAD + [__DAC = 90°, (Th. 4). Also, [_C+l_DAC= 90°, (Cor. 4, Th. 12, B. 1), Therefore, by subtraction, BAD — (7=0; by transposition, the angle BAD = C. But the angle C, at the center of the circle, is measured by the arc AE, the half of AED ; therefore, the equal angle, BAD, is also measured by the arc AE, the half of AED. Hence the theorem. THEOREM XII. An angle formed by a tangent and a chord, is equal to an angle in the opposite segment of the circle. Let AB be a tangent, and AD a chord, and from the point of contact, A, draw any angles, as AOD, and AED, in the segments. Then we are to prove that [__ BAD = [__ A CD, and L GAD = L AED- By Th. 11, the angle BAD is meas- ured by one half the arc AED ; and as the angle A CD is measured by one half of the same arc, (Th. 8), we have |_ BAD = [_ ACD. BOOK III. 90 Again, as AEBO is a quadrilateral, inscribed in a circle, the sum of the opposite angles, AOB + AEB - 2 right angles. (Th. 10). Also, the sum of the angles BAD + BAG = 2 right angles. (Th. 1, B. I). By subtraction (and observing that BAB has just been proved equal to AOB), we have, AEB — BAG = 0. Or, by transposition, AEB = BAG* Hence the theorem. THEOREM XIII. Arcs of the circumference of a circle intercepted by paral- lel chords, or by a tangent and a parallel chord, are equal. Let AB and OB be parallel chords, and draw the diagonal, AB ; now, be- cause AB and CB are parallel, the angle BAB - the angle ABO (Th. 6, B. I) ; but the angle BAB has for its meas- ure, one half of the arc BB\ and the angle ABO has for its measure, one half of the arc AO, (Th. 8) ; and because the angles are equal, the arcs are equal ; that is, the arc BB = the arc A 0. Next, let EF be a tangent, parallel to a chord, OB, and from the point of contact, G, draw GB. Since EF and OB are parallel, the angle OBG = the angle BGF. But the angle OBGr has for its measure, one-half of the arc 00-, (Th. 8) ; and the angle BGF has for its measure, one half of the arc GB, (Th. 11) ; therefore, these measures of equals must be equal ; that is, the arc CG=the arc GD. Hence, the theorem. 100 GEOMETRY. THEOREM XIV. When two chords intersect each other within a circle, the angle thus formed is measured by one half the sum of the twa intercepted arcs. Let AB and CD intersect each other within the circle, forming the two angles, E and E f , with their equal vertical angles. Then, we are to prove that the angle E is measured by one half the sum of the arcs AC and BD; and the angle E' is measured by one half the sum of the arcs AD and CB. First, draw AF parallel to CD, and FD will be equal to AC, (Th. 13); then, by reason of the parallels, [__ BAF = |__ E. But the angle BAF is measured by one half of the arc BDF; that is, one half of the arc BD plus one half of the arc AC. Now, as the sum of the angles E and E f is equal to two right angles, that sum is measured by one half the whole circumference. But the angle E, alone, as we have just proved, is measured by one half the sum of the arcs BD and AC; therefore, the other angle, E', is measured by one half the sum of the other parts of the circumference, AD + CB. Hence the theorem. THEOREM XV. When two secants intersect, or meet each other without a circle, the angle thus formed is measured by one half the dif ference of the intercepted arcs. ^ BOOK III. 101 Let DE and BE be two secants meeting at E ; and draw A F parallel to CD. Then, by reason of the parallels, the angle E, made by the intersection of the two secants, is equal to the angle BAF. But the angle BAF is measured by one half the arc BF; that is, by one half the difference be- tween the arcs BD and A C. Hence the theorem. THEOREM XVI. The angle formed by a secant and a tangent is measured by one half the difference of the intercepted arcs. Let BQ be a secant, and CD a tan- gent, meeting at C. We are to prove that the angle formed at C, is meas- ured by one half the difference of the arcs BD and DA. From A, draw AE parallel to CD ; then the arc AD = the arc DE; BD --DE = BE; and the [__ BAE = L O. But the angle BAE is measured by one half the arc BE, (Th. 8,) that is, by one half the difference between the arcs BD and AD; there- fore, the equal angle, C, is measured by one half the arc BE. Hence the theorem. THEOREM XVII. When two chords intersect each other in a circle, the rect- angle contained by the segments of the one, will be equivahrt to the rectangle contained by the segments of the other. 9* 102 GEOMETRY. Let AB and CB be two chords inter- Becting each other in E. Then we are to prove that the rectangle AE x EB = the rectangle CE X EB. Draw the lines AB and. CB, forming the two triangles AEB and CEB. The angles B and B are equal, because they are each measured by one half the arc, AG. Also the angles A and C are equal, because each is measured by one half the arc, BB ; and L AEB = |__ CEB, because they are vertical angles ; hence, the triangles, AEB and CEB, are equiangular and similar. But equiangular tri- angles have their sides about the equal angles propor- tional, (Cor. 1, Th. 17, B. II); therefore, AE and EB, about the angle E, are proportional to CE and EB, about the same or equal angle. That is, AE : EB :: CE : EB; Or, (Th. 19, B. II), AE x EB = CE x EB. Hence the theorem. Cor. When one chord is a diameter, and the other at right angles to it, the rectangle contained by the segments of the diameter is equal to the square of one half the other chord ; or one half of the bisected chord is a mean proportional be- tween the segments of the diameter. For, ABxBB*=FBx BE. But, if AB passes through the center, C, at right angles to FE, then FB = BE (Th. 1) ; and in the place of FB, write its equal, BE, in the last equation, and we have ABxBB = EE\ or, (Th. 3, B II), AB : BE : : BE : BB. Put, BE = x, CB == y, and CE = B, the radius of the circle. BOOK III. 103 Then AD ~ B - -y, and DB - B -f #. With this nota- tion, AD x DB = DE % becomes, (B -y)(R + y) = x* or, B % — y 2 = x* or, B 2 = x*+y* That is, f A« square of the hypotenuse of the right-angled triangle, DCE, is equal to the sum of the squares of the other two sic THEOREM XVIII. If from a point without a cirde, a tangent line be drawn to the circumference, and also any secant line terminating in tht- concave arc, the square of the tangent will be equivalent to b*. rectangle contained by the whole secant and its external sea ment. Let A be a point without the circle DUG, and let AD be a tangent and AE any secant line. Then we are to prove that AOxAE = AD\ In the two triangles, ADE and ADC, the angles ADO and AED are equal, since each is meas- ured by one half of the same arc, DO; the angle A is com- mon to the two triangles ; their third angles are therefore equal, and the triangles are equiangular and similar. Their homologous sides give the proportion AE : AD i :_AZ) : AC whence, AE xAC = AD 2 Hence the theorem. Cor. If AE and AF are two secant lines drawn from %\ e same point without the circumference, we shall have 104 GEOMETRY. AOx AF=AB* and, ABxAF^AB* hence, AOx AF = AB x AF, which, in the form of a proportion, gives AC : AF ::AB : AF. That is, the secants are reciprocally proportional to their ex- ternal segments. Scholium. — By means of this theorem we can determine the diam- eter of a circle, when we know the length of a tangent drawn from a point without, and the external segment of the secant, which, drawn from the same point, pasees through the center of the circle. Let Am be a secant passing through the center, and suppose the tangent AB to he 20, and the external seg- ment, An, of the secant to be 2. Then, if D denote the diameter, we shall have im=2+i), whence, Am x An = 2 (2 + B) = 4 -f 2B = (20) 2 = 400, 22) = 396, and B = 198. Ifiw, the height of a mountain on the earth, and AB, the distance of the visible sea horizon, be given, we may determine the diameter of the earth. For example ; the perpendicular height of a mountain on the island of Teneriffe is about 3 miles, and its summit can be seen from ships when they are known to be 154 or 155 miles distant ; what then is the diameter of the earth ? Designate, as before, the diameter by B. Then A m «■ 3 + 2), and Am x An = 9 + SB. AB = 154. 5 ; hence, 9 + 32) = (154. 5) 2 = 23870. 25, from which we find 2) = 7953.75, which differs but little from the true diameter* of the earth. One source of error, in this mode of computing the diameter of the earth, is atmospheric refraction, the ex planation of which does not belong here. HOOK III. 105 THEOREM XIX. If a circle be described about a triangle, the rectangle con- tained by two sides of the triangle is equivalent to the rectangle contained by the perpendicular let fall on the third side, and the diameter of the circumscribing circle. Let ABO be a triangle, AC and OB, the side3, OB the perpendicular let fall on the base AB, and OB the diameter of the circumscribing circle. Then we are to prove that AO x OB= OEx OB. The two A's, A OB and OEB, are equiangular, because \__A=\__E, both being measured by the half of the arc OB; also, ABO is a right angle, and is equal to OBB, an angle in a semi- circle, and therefore a right angle ; hence, the third angle, AOB = [_BOE, (Th. 12, Cor. 2, B. I). Therefore, (Cor. 1, Th. 17, B. II), AO x OB :: OE : OB and, AOxBO=OEx OB. Hence the theorem ; if a circle, etc. Oor. The continued product of three sides of a triangle is equal to twice the area of the triangle into the diameter of its tircumscribing circle. Multiplying both members of the last equation by A B, we have, AOx BOxAB=OEx {AB x OB). But OB is the diameter of the circle, and (AB x OB) «= twice the area of the triangle ; Therefore, AOx OB x AB= diameter multiplied by twice the area of the triangle. 106 GEOMETRY. THEOREM XX. The square of a line bisecting any angle of a triangle, to* gether with the rectangle of the segments into which it cuts the opposite side, is equivalent to the rectangle of the two sides including the bisected angle. Let ABO be a triangle, and CD a Line bisecting the angle C. Then we are to prove that CD* + (AD x DB) = ACx CB. The two A's, ACE and CDB, are equiangular, because the angles E and B are equal, both being in the same segment, and the [__ ACE = BCD, by hypothesis. Therefore, (Th. 17, Cor. 1, B. H), AC : CE :: CD : CB. But it is obvious that CE = CD -f DE, and by substi- tuting this value of CE, in the proportion, we have, AC i CD + DE :: CD : CB. By multiplying extremes and means, CD 2 + (DE x CD) = ACx CB. But by (Th. 17), DE x CD = ADx DB, and substituting, we have, HB* + (AD x DB) = ACx CB. Hence the theorem. THEOREM XXI. The rectangle contained by the two diagonals of any quad- rilateral inscribed in a circle, is equivalent to the sum of the two rectangles contained by the opposite sides of the quadri- lateral. Let ABCD be a quadrilateral inscribed in. a circle; then we are to prove that ACx BD = (AB x DC) + (AD x BC). From 0, draw CE, making the angle DOE equal to BOOK III. 107 the angle A OB ; and as the angle BAO is equal to th* angle ODE, both being in the same seg- ment, therefore, the two triangles, BEO and ABO, are equiangular, and we have (Th. IT, Cor. 1, B. II), AB : AC :: BE : BO (1) The two A's, ABO and BEO, are equiangular; for the [_BAO= \_EBO, both being in the same segment; and the L BOA — \_EOB, for BOE = BOA; to each of these add the ai gle EOA, smdL BOA = EOB; therefore, (Th. 17, Cor. 1, B.II), AB : AO :: BE : BO (2). By multiplying the extremes and means in proportions (1) and (2), and adding the resulting equations, we have, (AB x BO) + (AB x BO) = (BE + BE) x AO. But, BE + BE = BB; therefore, (AB x 2)(7) -f (AB x 5(7) = i(Jx BB, Oor. When two adjacent sides of the quadrilateral are equal, as AB and BO, then the resulting equation is, (AB x BO) + (AB x AB) = AO x 5D; or, -15 x (2)<7 -f AB) = AO x BB; or, AB : AO : : BB : BO + AB. That is, one of the two equal sides of the quadrilateral is to the adjoining diagonal, as the transverse diagonal is to the sum of the two unequal sides. THEOREM XXII. If two chords intersect each other at right angles in a cir- cle, the sum of the squares of the four segments thus formed is equivalent to the square of the diameter of the circle. Let AB and OB be two chords, intersecting eacb other at right angles. Draw BE parallel to EB, and draw BE and AF. Now, we are to prove that IE 1 +^F +~W % f EB 2 =~AF\ 108 GEOMETRY. As BF is parallel to ED, ABF is a right angle, and therefore AF is a diam- eter, (Th. 9). Also, because BF is parallel to CD, OB = DF, (Th. 13). Because CEB is a right angle, CE 2 + EB 2 =CB 2 = DF\ Because AED is a right angle, • AE 2 +~ED 2 = AD\ Adding these two equations, we have, W 2 + EB 2 + AE 2 +~ED 2 = DF 2 + ~AD\ But, as AF is a diameter, and ADF a right ang\e, (Th. 9), _ _ DT+AD 2 = AF 2 ; therefore, CE 2 + EB 2 -f AE 2 + Ji) 2 = ZF 2 . Hence the theorem. Scholium. — If two chords intersect each other at right angles, in a circle, and their opposite extremities be joined, the two chords thus formed may make two sides of a right-angled triangle, of which tho diameter of the circle is the hypotenuse. For, AD is one of these chords, and CB is the other ; and we have shown that CB = DF; and AD and DF are two sides of a right- angled triangle, of which AF is the hypotenuse ; therefore, AD and CB may be considered the two sides of .a right-angled triangle, and AF its hypotenuse. THEOREM XXIII. If two secants intersect each other at right angles, the sum of their squares, increased by the sum of the squares of the two segments without the circle, will be equivalent to the square of the diameter of the circle. Let AE and ED be two secants in- tersecting at right angles at the point E. From B, draw BF parallel to CD, and draw AF and AD. Now we are to prove that f3M ED 2 -r EB 2 +~Wf T IF. BOOK III. 109 Because BF is parallel to CD, ABF is a light angle, and consequently AF is a diameter, and BC = BF; and because AF is a diameter, J.2>^ is a right angle. As ABB is a right angle, AF'+FD^AD' Also, FB'+EC^BC^DF 2 Byadditio^XI^^SV^'+^^ZS'+SF^AF 1 Hence the theorem. THEOREM XXIV. If perpendiculars be drawn bisecting the three sides of a triangle, they will, when sufficiently produced, meet in a com- mon point. The three angular points of a triangle are not in the same straight line; consequently one circumference, and but one, may be made to pass through them. Conceive a triangle to be thus circumscribed. The sides of the triangle then become chords of the circum- scribing circle. Now if these sides be bisected, and at the points of bisection perpendiculars be drawn to the sides, each of these perpendiculars will pass through the center of the circle (Th. 1, Cor.) ; and the perpendiculars will therefore meet in a common point. Hence the theorem. THEOREM XXV. The sums of the opposite sides of a quadrilateral circum- scribing a circle are equal. Let ABOB be a quadrilateral circumscribed about a circle, whose center is 0. Then we are to prove that AB + BC=AB + BC. From the center of the circle draw OF and OF to the points of contact of the sides AB and BO. Then, 10 ttO GEOMETRY. the two right-angled triangles, OEB and OFB, are equal, because they have the hypotenuse OB common, and the side OF = OE; therefore, BE = BF, (Cor., Th. 39, B. I). In like manner we can prove that AE= AH, CF= Ca, nn&DG=DK Now, taking the equation BE = BF, and adding to its first mem- ber CG-, and to its second the equal line CF. we have, BE+ CG = BF+ OF (1) The equation AE=AH, by adding to its first member DCr, and to the second the equal line, DH, gives AE+DG=AH+DH (2) By the addition of (1) and (2) ? we find that BE + AE+CG + DG = BF + CF+AK+DIT. That is, AB + CD = BC + AD. Hence the theorem. aooK IV. Ill BOOK IV. PROBLEMS In this section, we have, in most instances, merely shown the construction of the prohlem, and referred to the theorem or theorems that the student may use, to prove that the object is attained by the construction. In obscure and difficult problems, however, we have gone through the demonstration as though it were a theorem. PROBLEM I. To bisect a given finite straight line. Let AB be the given line, and from its extremities, A and B, with any radius greater than one half of AB, (Postulate 3), describe arcs, cutting A — each other in n and m. Draw the line nm ; and 0, where it cuts AB, will be the middle of the given line. Proof, (B. I, Th. 18, Sch. 2). PROBLEM II. To bisect a given angle. Let ABO be the given angle. With any radius, and B as a center, describe the arc AC. From A and 0, as centers, with a radius greater than one half of AC, de- scribe arcs, intersecting in n; join B and n; the joining line will bisect the given angle. Proof, (Th. 21, B. I). X X 112 GEOMETRY. Proof, X-tf PROBLEM III. From a given point in a given line, to draw a perpendicular to that line. Let AB be the given line, and the given point. Take n and m, at equal distances on opposite sides of 0; and with the points m and n, as centers, and any radius greater than nO or mO, describe arcs cutting each other in & Draw SO, and it will be the perpendicular required. (B. I, Th. 18, Sch. 2). The following is another method, which is preferable, when the given point, 0, is at or near the end of the line. Take any point, 0, which is mani- festly one side of the perpendicular, as a center, and with 00 as a radius, describe a circum- ference, cutting AB in m and 0. Draw mn through the points m and 0, and meeting the arc again in n ; mn is then a diameter to the circle. Draw On, and it will be the perpendicular required. Proof, (Th. 9, B. III). PROBLEM IV. From a given point without a line, to draw a 'perpendicular to that line. Let AB be the given line, and the given point. From draw any oblique line, as On. Find the mid- dle point of On by Problem 1, and with that point, as a center, describe a semicircle, having On as a diam- eter. From m, where this semi-cir- cumference cuts AB, draw Om, and it will be the pcrpcn dicular required. Proof, (Th. 9, B. III). BOOK IV. 113 PROBLEM V. At a given point in a line, to construct an angle equal to a given angle. Let A be the point given in the line AB, and DOE the given angle. With as a center, and any radius, OE, draw the are ED. With i as a center, and the radius AF=OE, describe an indefinite arc; and with J 7 as a center, and FCr as a radius, equal to ED, describe an arc, cutting the other arc in Q-, and draw A G\ QAF will be the angle required. Proof, (Th. 2, B. III). PROBLEM VI. From a given point, to draw a line parallel to a given line. Let A be the given point, and BO the given line. Draw A 0, making an angle, AOB; and from the given point, A, in the line A 0, draw the angle OAD m AOB, by Problem 5. Since AD and BO make the same angle with A 0, they are, therefore, parallel, (B. I, Th. 7, Cor. 1). PROBLEM VII. To divide a given line into any number of equal parte. Let AB represent the given line, and let it be required to di- vide it into any number of equal parts, say £ve. Prom one end of the line A, draw AD, indefinite in both length and position. Take any convenient distance in the di- 10* h 114 GEOMETRY. viders, as A a, and set it off on the line AD, thus making the parts Aa, ab, be, etc., equal. Through the last point, e, draw EB, and through the points a, b, c, and d, draw parallels to eB, by Problem 6 ; these parallels will divide the line a3 required. Proof, (Th. 17, Book II). PROBLEM VIII. To find a third proportional to two given lines. Let AB and A be any two lines. Place them at any angle, and draw OB. On the greater line, AB, take AD mm AC, and through D, draw DE parallel to BO', AE is the third proportional required. Proof, (Th. 17, B. II). PROBLEM IX, To find a fourth proportional to three given lines. L^t AB, AC, AD, represent the ihrca given lines. Place the first two at any angle, as BAO, and draw BO On AB place AD, and from the point D, draw DE parallel to BO, by Problem 6 ; AE will be the fourth proportional required. Proof, (Th. 17, B. H). PROBLEM X. Tt find the middle, m mean proportional, between two given lines BOOK IV. 115 Place AB and BC in one right jne, and on A C, as a diameter, de- scribe a semicircle, (Postulate 3), and from the point B, draw BD at right angles to AC, (Problem 3); BD is the mean proportional re- quired. Proof, (B. m, Th. 17, Cor.). PROBLEM XI. To find the center of a given circle. Draw any two chords in the given cir- cle, as AB and CD, and from the middle points, m and n, draw perpendiculars to AB and CD ; the point at which these two perpendiculars intersect will be the center of the circle. Proof, (B. m, Th. 1, Cor.). PROBLEM XII. To draw a tangent to a given circle, from a given either in or without the circumference of the circle* When the given point is in the cir- cumference, as A, draw the radius A C, and from the point A, draw AB per- pendicular to AC; AB is the tangent required. Proof, (Th. 4, B. III). When the given point is without the circle, as A, draw AC to the center of the circle ; on A C, as a diameter, describe a semicircle ; and from B, where the semi-ciruumfer- ence cuts the given circumference, draw AB, and it will be tangent to the circle. Proof, (Th. 9, B. Ill), and, (Th. 4, B. III). point, 116 GEOMETRY. PROBLEM XIII. On a given line, to describe a segment of a circle, that shall contain an angle equal to a given angle. Let AB be the given line, and the given angle. At the ends of the given line, form angles DAB, DBA, each equal to the given angle, C. Then draw AH and BE perpendiculars to AD and BD ; and with E as a center, and EA, or EB, as a radius, describe a circle; then AFB will be the segment required, as any angle F, made in it, will be equal to the given angle, 0. Proof, (Th. 11, B. Ill), and (Th. 8, B. III). PROBLEM XIV. From any given circle to cut a segment, that shall contain a given angle. Let be the given angle. Take any point, as A, in the circumfer- ence, and from that point draw the tangent AB ; and from the point A, in the line AB, construct the angle BAD = 0, (Problem 5), and O AED is the segment required. Proof, (Th. 11, B. HI), and (Th. 8, B. HI). PROBLEM XV. To construct an equilateral triangle on a given straight line. Let AB be the given line; from the extremities A and B, as centers, with a radius equal to AB, describe arcs cutting each other at 0. From 0, the point of intersection, draw QA and CB; ABO will be the triangle required. The construction is a sufficient demonstration. Or, (Ax. 1\ BOOK IV. 117 PROBLEM XVI. To construct a triangle, having its three sides equal to three jiven lines, any two of which shall be greater than the third. Let AB, OB, and EF, represent the E p three lines. Take any one of them, as c D AB, to be one side of the triangle. From B, as a center, with a radius equal to OB, describe an arc ; and from A, as a center, with a radius equal to EF, describe an- other arc, cutting the former in n. Draw An and Bn, and AnB will be the A re- quired. Proof, (Ax. 1). PROBLEM XVII. To describe a square on a given line. Let AB be the given line ; and from the extremities, A and B, draw A and BB per- 9 pendicular to AB. (Problem 3.) From A, as a center, with AB as radius, strike an arc across the perpendicular at 0; i and from draw OB parallel to AB ; AOBB is the square required. Proof, (Th. 26, B. I). PROBLEM XVIII. To construct a rectangle, or a parallelogram, whose adia .ent sides are equal to two given lines. Let AB and A be the two given A c lines. From the extremities of one A n line, draw perpendiculars to that line, as in the last prob- lem; and from these perpendiculars, cut off portions equal to the other line ; and, by a parallel, complete the figure. 118 GEOMETRY. When the figure is to be a parallelogram, with oblique angles, describe the angles by Problem 5. Proof, (Th 26, B. I). PROBLEM XIX. To describe a rectangle that shall be equivalent to a given square, and have a side equal to a given line. Let AB be a side of the given square, c D and CD one side of the required rect- A B angle. E p Find the third proportional, FF, to CD and AB, (Prob- lem 8). Then we shall have CD : AB : : AB : FF, Construct a rectangle with the two given lines, CD and FF, (Problem 18), and it will be equal to the given square, (Th. 3, B. II). PROBLEM XX. To construct a square that shall be equivalent to the differ ence of two given squares. Let A represent a side of the greater of two given squares, and B a side of the less square. On A, as a diameter, describe a semicircle, and from one extremity, n, as a center, with a radius equal to B, describe an arc. and, from the point where it cuts the circumference, — - — draw mp and np ; mp is the side of a square, which, when constructed, (Problem 17), will be equal to the difference of the two given squares. Proof (Th. 9, B. Ill, and Th. 39, B. I.) To construct a square equivalent to the sum of two given squares, we have only to draw through any point two lines at right angles, and lay off on one a distance equal to the side of one of the squares, and on the other BOOK IV. 119 a distance equal to the side of the other. The straight line connecting the extremities of these lines will be the side of the required square, (Th. 39, B. I). PROBLEM XXI. To divide a given line into two parts, which shall be in the ratic of two other given lines. M^ NH Let AB be the line A — ~ PROBLEM XXII. To divide a given line into any number of parts, having to each other the ratios of other given lines* Let AB be the given M»- line to be divided, and Nl M, It, P, etc., the lines to which the parts of AB are to be propor- tional. Through the point A draw an indefinite line, making, with AB, any conve- nient angle, and on this line lay off from A the lines M, N, P, etc., successively. Join the extremity of the last line to the point B by a straight line, parallel to which draw other lines through the points of division of the indefinite line, and they will divide the line AB at the points 0, D, etc., into the required parts. (Proof, Th. 17, B. II). PROBLEM XXIII. To construct a square that shall be to a given square, as a line, M, to a line, N. Place M and N in a line, and on the sum describe a semicir- cle. From the point where the two lines meet, draw a perpen- dicular to meet the circumfer- ence in A. Draw Am and An, and produce them indefinitely. On An or An produced, take AG = to the side of the given square ; and from C, draw CB parallel to mn ; AB is a side of the required square. For, Also, Am An* An Am* Therefore, ~A& :A0 2 ::M AB 2 : AC\ (Th. 17, B. II). M :JST, (Th.25,B.II). JST, (Th. 6, B. II). BOOK IV. 121 PROBLEM XXIV. To cut a line into extreme and mean ratio ; that is, sj that the whole line shall be to the greater part, as that greater part is to the less. Remark. — The geometrical solution of this problem is not imme- diately apparent, but it is at once suggested by the form of the equa- tion, which a simple algebraic analysis of its conditions leads to. Represent the line to be divided by 2a, the greater part by x, and consequently the other, or less part, by 2a — x. Now, the given line and its two parts are required, to satisfy the following proportion : 2a : x : : x : 2a — x whence, x % = 4a 2 — 2ax By transposition, x 2 -f 2ax = 4a 2 = (2a) 2 If we add a 2 to both members of this equation, we shall have, x 2 + 2ax + a 2 = (2a) 2 -f a 2 or, (x -fa) 2 = (2a) 2 -f a 2 This last equation indicates that the lines represented by (x + a), 2a, and a, are the three sides of a right- angled triangle, of which (x + a) is the hypotenuse, the given line, 2a, one of the sides, and its half, a, the other. Therefore, let AB represent the given line, and from the extremity, B, draw BO at right angles to AB, and make it equal to one half of AB. With C, as a center, and radius OB, describe a circle. Draw A C and pro- duce it to F. With A as a center and AD as a radius, describe the arc D.E; this arc will divide the line AB, as required. We are now to prove that AB : AE :: AE : EB 11 122 GEOMETRY. By Th. 18, B. Ill, we have, AF x AD = AB 2 or, AF : AB : : AB : AD Then, (by Cor., Th. 8, Book II), we may have, (AF— AB) : AB :: (AB — AD) : AD Since OB = \AB = \DF\ therefore, AB = LI Hence, AF — AB = AF — DF — AD = A#. Therefore, AF : AB :: FB : AF By taking the extremes for the means, we have, AB i AF ii AF : FB. PROBLEM XXV. To describe an isosceles triangle, having its two equal angles each double the third angle, and the equal sides of any given length. Let AB be one of the equal sides of the required triangle ; and from the point A, with the radius AB, describe an arc, BD. Divide the line AB into extreme acd mean ratio by the last problem, and sup- pose the point of division, and A the greater segment. From the point B, with AC, the greater segment, as a radius, describe another arc, cutting the arc BD in D. Draw BD, DO, and DA. The triangle ABD is the tri- angle required. As AC = BD, by construction ; and as AB is to A as A C is to B C, by the division of AB ; therefore AB : BD : : BD : BC Now, as the terms of this proportion are the sides ot" the two triangles about the common angle, B, it follows, (Cor. 2, Th. 17, B. II), that the two triangles, ABD and BOOK IV. 123 BBC, are equiangular; but the triangle ABB is isos- celes; therefore, BBQ is isosceles also, and BB == BC\ but BB = AC: hence, BC = AC, (Ax. 1), and the tri- angle ACB is isosceles, and the [__ CBA = | -^L. But the exterior angle, BCB = CBA -f A, (Th. 12, B. I), Therefore, \__BCB, or its equal [__B = L CBA + [__A ; or the angle B = 2[__A. Hence, the triangle ABB has each of its angles, at the base, double of the third angle. Scholium. — As the two angles, at the base of the triangle ABD, are equal, and each is double the angle A, it follows that the sum of the three angles is Jive times the angle A. But, as the three angles of every triangle are always equal to two right angles, or 180°, the angle A must be one fifth of two right angles, or 36° ; therefore, BD is a chord of 36°, when AB is a radius to the circle ; and ten such chords would extend exactly round the circle, or would form a decagon. PROBLEM XXVI, Within a given circle to inscribe a triangle, equiangular to a given triangle. Let ABC be the circle, and abc the given triangle. From any point, as A, draw EB tan- gent to the given circle at A 9 (Problem 12). From the point A, in the line AB, lay off the angle BAC — the angle b, (Problem 5), and the angle BAB — the angle c, and draw BC. The triangle ABC is inscribed in the circle; It 13 equi- angular to the triangle abc, and hence it is the triangle required. Proof, (Th. 12, B. HI). 124 GEOMETRY. PROBLEM XXVII, To inscribe a regular pentagon in a given circle. 1st. Describe an isosceles tri- angle, abc, having each of the equal angles, b and c, double the third angle, a, by Problem 25. 2d. Inscribe the triangle, ABO, in the given circle, equi- angular to the triangle abc, by Problem 26 ; then each of the angles, B and 0, is double the angle A. 3d. Bisect the angles B and 0, by the lines BD and OE, (Problem 2), and draw AE, EB, OB, BA; and the figure AEBOB is the pentagon required. By construction, the angles BAO, ABB, BBO, BOE, EOA, are all equal ; therefore, (B. in, Th. 9, Cor.), the arcs, BO, AB, BO, AE, and EB, are all equal; and if the arcs are equal, the chords AE, EB, etc., are equal. Scholium. — The arc subtended by one of the sides of a regular pen- 360° tagon, being one fifth of the whole circumference, is equal to — — ==72°' u PROBLEM XXVIII. To inscribe a regular hexagon in a circle. D aw any diameter of the circle, as AB. and from one extremity, B, draw BB equal to BO, the radius of the circle. The arc, BB, will be one sixth part of the whole circumference, and the chord BB will be a side of the regu- lar polygon of six sides. In the A OBB, as OB = OB, and BB= OB by con- struction, the A is equilateral, and of course equiangular. Since the sum of the three angles of every A is equal to two right angles, or to 180 degrees, when the BOOK IV. 125 three angles are equal to one another, each one of them must be 60 degrees ; but 60 degrees is a sixth part of 360 degrees, the whole number of degrees in a circle ; therefore, the arc whose chord is equal to the radius, is a sixth part of the circumference ; and, if a polygon of six equal sides be inscribed in a circle, each side will be equal to the radius. Scholium. — Hence, as BD is the chord of 60°, and equal to BC or CD, we say generally, that the chord of 60° is equal to radius. PROBLEM XXIX. To find the side of a regular polygon of fifteen sides, which may be inscribed in any given circle. Let CB be the radius of the given circle; divide it into extreme and mean ratio, (Problem 24), and make BD equal to CB, the greater part; then BD will be a side of a regular polygon of ten sides, (Scholium to Problem 25). Draw BA = to CB, and it will be a side of a polygon of six sides. Draw DA, and that line must be the side of a polygon which cor- responds to the arc of the circle expressed by \ less j\, of the whole circumference ; or J — j^ = g % = T ^ ; that is, one-fifteenth of the whole circumference ; or, DA is a side of a regular polygon of 15 sides. But the 15th part of 360° is 24° ; hence the side of a regular inscrioed polygon of fifteen sides is the chord of an arc of 24°. PROBLEM XXX. In a given circle to inscribe a regular polygon of any num her of sides, and then to circumscribe the circle by a similar polygon. 11* 126 GEOMETKY. Let the circumference of the circle, whose cetter is C, be divided into any number of equal arcs, as amb, bnc, cod, etc. ; then will the polygon abode, etc., bounded by the chords of these arcs, be regu- lar and inscribed ; and the poly- gon ABODE, etc., bounded by the tangents to these arcs at their middle points m, n, o, etc, be a 3imilar circumscribed polygon. First. — The polygon abode, etc., is equilateral, because its sides are the chords of equal arcs of the same circle, (Th. 5, B. Ill) ; and it is equi- angular, because its angles are inscribed in equal segments of the same circle, (Th. 8, B. III). Therefore the poly- gon is regular, (Def. 14, B. Ill), and it is inscribed, since the vertices of all its angles are in the circumference of the circle, (Def. 13, B. III). Second. — If we draw the radius to the point of tangency of the side AB of the circumscribed polygon, this radius is perpendicular to AB, (Th. 4, B. Ill), and also to the chord ab, (B. Ill, Th. 1, Cor.) ; hence AB is parallel to a5, and for the same reason BO is parallel to be ; therefore the angle ABO is equal to the angle abc, (Th. 8, B. I). In like manner we may prove the other angles of the circumscribed polygon, each equal to the corresponding angle of the inscribed polygon. These polygons are therefore mutually equiangular. Again, if we draw the radii Om and On, and the line 0B> the two A's thus formed are right-angled, the one at n and the other at n, the side OB is common and Om is equal to On ; hence the difference of the squares descril^ed on OB and Om is equivalent to the difference of the squares described on OB and On. But the first difference is equivalent to the square described on Bm, and the second diffeience is equivalent to the square described BOOK IV. 127 on Bn ; hence Brn is equal to Bn, and the two right- angled triangles are equal, (Th. 21, B. I), the angle BOm opposite the side Bm being equal to the angle BOn, op- posite the equal side Bn. The line OB therefore passea through the middle point of the arc mbn ; hut because m and n are the middle points of the equal arcs amb and bnc, the vertex of the angle abc is also at the middle point of the arc mbn. Hence the line OB, drawn from the center of the circle to the vertex of the angle ABC, also passes through the vertex of the angle abc. By pre- cisely the same process of reasoning, we may prove that 00 passes through the point e, OB through the point d, etc. ; hence the lines joining the center with the vertices of the angles of the circumscribed polygon, pass through the vertices of the corresponding angles of the inscribed polygon ; and conversely, the radii drawn to the vertices of the angles of the inscribed polygon, when produced, pass through the vertices of the corresponding angles of the circumscribed polygon. Now, since ab is parallel to AB, the similar A's abO and ABO, give the proportion Ob : OB :: ab : AB, and the A'a,bcO and BOO, give the proportion Ob : OB : : be : BO. As these two proportions have an antecedent and con- sequeut, the same in both, we have, (Th. 6, B. II), ab : AB : : bo : BO. In like manner we may prove that be : BO : : cd : OB, etc., etc. The two polygons are therefore not only equiangular, but the sides about the equal angles, taken in the same order, are proportional ; they are therefore similar, (Def 16. B. IT). J28 GEOMETRY. Cor. 1. To inscribe any regular polygon in a circle, we have only to divide the circumference into as many equal parts as the polygon is to have sides, and to draw the chords of the arcs; hence, in a given circle, it is possible to inscribe regular polygons of any number of sides whatever. Having constructed any such polygon in a given circle, it is evident, that by changing the radius of the circle without changing the number of sides of the polygon, it may be made to represent any regular poly- gon of the same name, and it will still be inscribed in a circle. As this reasoning is applicable to regular poly- gons of whatever number of sides, it follows, that any regular polygon may be circumscribed by the circumference of a circle. Cor. 2. Since ab, be, cd, etc., are equal chords of the same circle, they are at the same distance from the center, (Th. 3, B. Ill) ; hence, if with as a center, and Ot, the distance of one of these chords from that point, as a radius, a circumference be described, it will touch all of these chords at their middle points. It follows, therefore, that a circle may be inscribed within any regular polygon. Scholium. — The center, 0, of the circle, may be taken as the center of both the inscribed and circumscribed polygons ; and the angle A OB, included between lines drawn from the center to the extremities of one of the sides A B, is called the angle at the center." The perpen- dicular drawn from the center to one of the sides is called the Ajpothem of the polygon. Cor. 3. The angle at the center of any regular polygon is equal to four right angles divided by the number of sides of the polygon. Thus, if n be the number of sides of the polygon, the angle at the center will be expressed v 360° n Cor. 4. If the arcs subtended by the sides of any regular inscribed polygon be bisected, and the chords of these semi-arcs be drawn, we shall have a regular BOOK IV. 129 inscribed polygon of double the number of sides. Thus, from the square we may pass successively to regular inscribed polygons of 8, 16, 32, etc., sides. To get the corresponding circumscribed polygons, we have merely to draw tangents at the middle points of the arcs sub- tended by the sides of the inscribed polygons. Cor, 5. It is plain that each inscribed polygon is but a part of one having twice the number of sides, while each circumscribed polygon is but a part of one having o^p hqlf thp. number of sides 130 GEOMETRY. BOOK V. ON THE PROPORTIONALITIES AND MEASUREMENT OF POLYGONS AND CIRCLES. PROPOSITION I. — THEOREM. The area of any circle is equal to the product of its radius by one half of its circumference. Let CA be the radius of a circle, and AB a very small portion of its circumference; then AOB will be a sector. We may conceive the whole circle made up of a great number of such sectors; and when each sector is very small, the arcs AB, BB, etc., each one taken separately, may be regarded as nght lines ; and the sectors CAB, CBD, etc., will be triangles. The triangle, AOB, is measured by the product of the base, A 0, multiplied into one half the altitude, AB, (Th. 33, Book I) ; and the triangle BCD is measured by the pro- duct of BC, or its equal, AC, into one half BB ; then the area, or measure of the two triangles, or sectors, is the product of A C, multiplied by one half of AB plus one half of BB, and so on for all the sectors that compose the circle ; therefore, the area of the circle is measured by th g product of the radius into one half the circumference. BOOK V. 131 PROPOSITION II. — THEOREM. Circumferences of circles are to one another as their radii, and their areas are to one another as the squares of their radii. Let CA be the radius of a circle, and Ca the radius of another circle. Conceive the two circles be so placed upon each other so as to have a common center. Let AB be such a certain definite portion of the circumference of the larger circle, that m times AB will represent that cir- cumference. But whatever part AB is of the greater circumference, the same part ab is of the smaller; for the two circles have the same number of degrees, and are of course sus- ceptible of division into the same number of sectors. But by proportional triangles we have, CA : Ca : : AB : ab Multiply the last couplet by m, (Th. 4, B. II), and we have CA : Ca :: m.AB : m.ab. That is, the radius of one circle is to the radius of another, as the circumference of the one is to the circumference of the other. To prove the second part of the theorem, let C repre- sent the area of the larger circle, and c that of the smaller ; now, whatever part the sector CAB is of the circle C, the sector Cab is the corresponding part of the circle c. That is, C : c but, CAB : Cab Therefore, C : c CAB : Cab, {CAf:(Ca) 2 , (Th.20,B.H). (CAY : (Caf, (Th. 6, B. II). That is, the area of one circle is to the area of another, as 132 GEOMETRY. the square of the radius of the one is to the square of the radius of the other. Hence the theorem. Cor. If : c :: (OAf : (Ca) 2 , then, C : c ::4(£M) 2 : 4 ((7a) 2 . But 4 (QAf is the square of the diameter of the larger circle, and 4 (Oaf is the square of the diameter of the smaller. Denoting these diameters respectively by B and d, we have, : c : : D 2 : d\ That is, the areas of any two circles are to each other, a% the squares of their diameters. Scholium. — As the circumference of every circle, great or small, is assumed to be the measure of 360 degrees, if we conceive the circum- ference to be divided into 360 equal parts, and one such part repre- sented by AB on one circle, or ab on the other, AB and ab will be very near straight lines, and the length of such a line as ^LBwill be greater or less, according to the radius of the circle ; but its absolute length cannot be determined until we know the absolute relation between th« diameter of a circle and its circumference. PROPOSITION III. — THEOREM. When the radius of a circle is unity, its area and semi- circumference are numerically equal. Let B represent the radius of any circle, and the Greek letter, *, the half circumference of a circle whose radius is unity. Since circumferences are to each other as their radii, when the radius is B, the semi-circumference will be expressed by *B. Let m denote the area of the circle of which B is the radius ; then, by Theorem 1, we shall have, for the area of this circle, icB 2 = m y which, when B = 1, reduces to «r = m. This equation is to be interpreted as meaning that the semi-circumference contains its unit, the radius, as many BOOK V. 133 times as the area of the circle contains its unit, the square of the radius. Remark. — The celebrated problem of squaring the circle has for its object to find a line, the square on which will be equivalent to the area of a circle of a given diameter ; or, in other words, it proposes to find the ratio between the area of a circle and the square of its radius. An approximate solution only of this problem has been as yet dis- covered, but the approximation is so close that the exact solution ii no longer a question of any practical importance. PROPOSITION IV. — PROBLEM. Given, the radius of a circle unity, to find the areas of regular inscribed and circumscribed hexagons. Conceive a circle described with the radius CA, and in this circle inscribe a regular polygon of six sides (Prob. 28, B. IV), and each side will be equal to the radius CA ; hence, the whole perimeter of this poly- gon must be six times the ra- dius of the circle, or three times the. diameter. The chord bd is Ft bisected by CA. Produce Cb and Cd, and through the point A, draw BD parallel to bd ; BD will then be a 3ide of a regular polygon of six sides, circumscribed about the circle, and we can compute the length of this line, BD, as follows : The two triangles, Cbd and CBD, are equiangular, by construction ; therefore, Ca : bd : : CA : BD. Now, let us assume CA = Cd = the radius of the circle, equal unity; then bd = l, and the preceding pro- portion becomes Ca : 1 :: 1 : BD (1) In the right-angled triangle Cad, we have, (Cay + (ad)* = (C'dy, (Th. 39, B. I). That is, (Ca) 2 + i = l, because Cd=l, and ad— \. 12 134 GEOMETRY. "Whence, Ca = J >/3. This value of Ca, substituted in proportion ( 1 ), gives |^3 : 1 : : 1 : BD; hence, BD= JL But the area of the triangle Cbd is equal to bd(= 1,) multiplied by J Ca = \ ^3 ; and the area of the triangle CBD is equal to BD multiplied by \CA. "Whence, area, Cbd = J v/3, and, area, CBD ■» ,i But the area of the inscribed polygon is six times that of the triangle Cbd, and the area of the circumscribed polygon is six times that of the triangle CBD. Let the area of the inscribed polygon be represented by p, and that of the circumscribed polygon by P. Then^ = ?V3, andP ?« ^J - 2^3. 2 •a v/3 3 3 Whence^ : P : : ^3 : 2%/3 ::£:2::3:4::9 : 12 ^ = 2^3 = 2.59807621. P = 2^3 = 3.46410161. Now, it is obvious that the area of the circle must be included between the areas of these two polygons, and not far from, but somewhat greater than, their half sum, which is 3.03 -f ; and this may 'be regarded as the first approximate value of the area of the circle to the radius unity. PROPOSITION Y. — PROBLEM. (riven, the areas of two regular polygons of the same nvm her of sides, the one inscribed in and the other circumscribed about, the same circle, to find the areas of regular inscribed and circumscribed polygons of double the number of sides. Let j? represent the area of the given inscribed polygon, and P that of the circumscribed polygon of the same BOOK V. 135 nunibei of sides. Also denote by p' the area of the inscribed polygon of double the number of sides, and by P' that of the corresponding circumscribed polygon. Now, if the arc KAL be some exact part, as one-fourth, one fifth, etc., of the circumference of the circle, of which is the center and CA the radius, then will KL be the side of a regular inscribed polygon, and the triangle KCL will be the same part of the whole polygon that the arc KAL is of the whole circumference, and the triangle CDB will be a like part of the circumscribed polygon. Draw QA to the point of tangency, and bisect the angles ACB and ACD, by the lines CGr and CLJ, and draw KA. It is plain that the triangle ACK is an exact part of the inscribed polygon of double the number of sides, and that the A ECG- is a like part of the cir- cumscribed polygon of double the number of sides. Repre- sent the area of the A LCKbj a, and the area of the A BCD by b, that of the A ACK by x, and that of the A ECG- by y, and suppose the A's, KCL and BBC, to be each the nth. part of their respective polygons. Then, na—p, nb = P, 2nx = p', and, 2ny = P f ; But, by (Th. 33, B. I), we have CM. MK=a (1) CA . AD =b (2) QA . MK= 2x (3) Multiplying equations (1) and (2), member by member, we have (CM . AD) x (CA . MK) =- ab (4 ) 136 GEOMETRY. From the similar A's OMK and CAD, we have CM : MK :: CA : AD whence CM . AD = CA . MK But from equation ( 3 ) we see that each memher of this last equation is equal to 2x; hence equation (4) becomes 2x . 2x = <*5 If we multiply both members of this by n 2 ■■ n h, we shall have 4t& 2 .z 2 = wa.w5 = p.P or, taking the square root of both members, 2nx = s/p^P That is, the area of the inscribed polygon of double the number of sides is a mean proportional between the areas of the given inscribed and circumscribed polygons p and P. Again, since CE bisects the angle ACD, we have, by, (Th. 24, B. II), AE : ED CA : CD CM: CK CM: CA CM: CM+ CA. hence, AE : AE +ED Multiplying the first couplet of this proportion by CA, and the second by MK, observing that AE -f ED = AD, we shall have AE.CA : AD.CA :: CM.MK : (CM + CA) MK. But AE. CA measures the area of the A CEG-, whi 3h we have called y, AD.CA = A CBD — b, CM.MK - A CKL = a, and (CM 4- CA)MK= a CKL + 2 a CAK = a -h 2x, as is seen from equations (1) and (3). Therefore, the above proportion becomes y : b :: a : a + 2x. Multiplying the first couplet by 2n, and the second by w, we shall have BOOK V. 137 2m/ : 2nb : : na : na -f 2wa: That is, P' : 2P :: p : p + / whence, P' = — -±- and as the value of p r has been previously fourd equal to *SPp, the value of P' is known from this last equation, and the problem is completely solved. PROPOSITION VI. — PROBLEM. To determine the approximate numerical value of the area of a circle, when the radius is unity. "We have now found, (Prob. 4), the areas of regular inscribed and circumscribed hexagons, when the radius of the circle is taken as the unit ; and Prob. 5 gives us formulae for computing from these the areas of regular inscribed and circumscribed polygons of twelve sides, and from these last we may pass to polygons of twenty-four sides, and so on, without limit. Now, it is evident that, as the number of sides of the inscribed polygon is increased, the polygon itself will increase, gradually approaching the circle, which it can never sur- pass. And it is equally evident that, as the number of sides of the circumscribed polygon is increased, the poly- gon itself will decrease, gradually approaching the circle, less than which it can never become. The circle being included between any two corres- ponding inscribed and circumscribed polygons, it will differ from either less than they differ from each other ; and the area of either polygon may then be taken as tne area of the circle, from which it will differ by an amount less than the difference between the polygons. It is also plain that, as the areas of the polygons ap- proach equality, their perimeters will approach coinci- dence with each other, and with the circumference of the circle. 12* 138 GEOMETRY. Assuming the areas already found for the inscribed and circumscribed hexagons, and applying the formulae of Prob. 5 to them and to the successive results ob. tained, we may construct the following table : NUMBER OB 1 SIDES. n INSCRIBED POLYGONS. CIRCUMSCRIBED POLYGONS. 6 1^ - 2.59807621 2^3= =3.46410161 12 24 3 6 = 3.0000000 = 3.1058286 12 2+/3 -3.2153904 3.1596602 ^2+x/3 48 3.1326287 3.1460863 96 3.1393554 3.1427106 192 3.1410328 3.1418712 384 3.1414519 3.1416616 768 3.1415568 3.1416092 1536 3.1415829 3.1415963 3072 3.1415895 3.1415929 6144 3.1415912 3.1415927 Thus we have found, that when the radius of a circle is 1, the semi-circumference must be more than 3.1415912, and less than 3.1415927; and this is as accurate as can be determined with the small number of decimals here used. To be more accurate we must have more decimal places, and go through a very tedious mechanical opera- tion; but this is not necessary, for the result is well known, and is 3.1415926535897, plus other decimal places to the 100th, without termination. This result was dis- covered through the aid of an infinite series in the Dif- ferential and Integral Calculus. The number, 3.1416, is the one generally usod in prac- tice, as it is much more convenient than a greater num- ber of decimals, and it is sufficiently accurate for all ordinary purposes. In analytical expressions it has become a general cus- tom with mathematicians to represent this number by BOOK V. 139 the Greek letter #, and, therefore, when any diameter of a circle is represented by D, the circumference of the same circle must be *D. If the radius of a circle is re- presented by B, the circumference must be represented by 2«B. Scholium. — The side of a regular inscribed hexagon subtends an arc of 60°, and the side of a regular polygon of twelve sides subtends an arc of 30° ; and so on, the length of the arc subtended by the sides of the polygons, varying inversely with the number of sides. Angles are measured by the arcs of circles included between their sides ; they may also be measured by the chords of these arcs, or rather by the half chords called sines in Trigonometry. For this purpose, it becomes necessary to know the length of the chord of every possible arc of a circle. PROPOSITION VII. — PROBLEM. Given, the chord of any arc, to find the chord of one half that arc, the radius of the circle being unity. Let FE be the given chord, and draw the radii QA and QE, the first perpen- dicular to FE, and the second to its ex- tremity, E. Denote FE by 2c, and the chord of the half arc AE by. a;. Then, in the right-angled triangle, DOE, we have ~ DQ 2 = QE 2 — DE 2 . Whence, since QE = 1, D<7 = ^l — c\ If from QA =■ 1 we subtract DO, we shall have AD. That is, AD = 1 — and EF— c — x, and the proportion becomes, 1 : x : : x : BF, Hence, BF= x\ Also, Fa = 2 — x\ As AE and i?6r are two chords intersecting each other at the point F, we have, GFx FB = AFx FE, (Th. 17, B. HI). That is, (2 — x 2 ) x* = x (c — x) ; or, # 3 — 3a; = — - site angle is 60° ; required the length of each of the other two equal sides, and the number of degrees in each of the other angles. 2. One angle of a right-angled triangle is 30° ; what is the other angle ? Also, the least side is 12, what is the hypotenuse ? A J The hypotenuse is 24, the double of the least ***' \ side. Why? 3. The perpendicular distance between two parallel lines is 10 ; what angles must a line of 20 make with these parallels to extend exactly from the one to the other ? Arts. The angles must be 30° and 150°. 4. The perpendicular distance between two parallels is 20 feet, and a line is drawn across them at an angle of 45° ; what is its length between the parallels ? Ans. 20^2. 5. Two parallels are 8 feet asunder, and from a point in one of the parallels two lines are drawn to meet the other ; the length of one of these lines is 10 feet, and that of the other 15 feet ; what is the distance between the points at which they meet the other parallel ? Ans. 6.69 ft., or 18.69 ft. (See Th. 39, B. I). 6. Two parallels are 12 feet asunder, and, from a point on one of them, two lines, the one 20 feet and the other 18 feet in length, are drawn to the other parallel ; what is the distance between the two lines on the other parallel, and what is the area of the triangle so formed ? c The distance on the other parallel is 29.416 Ans. < feet, or 2.584 feet; and the area of the tri 1 angle is 176.496, or 15.504 square feet. 7. The diameter of a circle is 12, and a chord of the 144 GEOMETRY. circle is 4; what is the length of the perpendicular drawn from the center to this chord ? (See Th. 3, B. III). Ans. 4^2. 8. Two parallel chords in a circle were measured and found to be 8 feet each, and their distance asunder was 6 feet ; what was the radius of the circle ? Ans. 5 feet. 9. Two chords on opposite sides of the center of a cirsle are parallel, and one of them has a length of It) and the other of 12 feet, the distance between them being 14 feet. What is the diameter of the circle ? Ans. 20 feet. 10. An isosceles triangle has its two equal sides, 15 each, and its base 10. What must be the altitude of a right-angled triangle on the same base, and having an equal area? 11. From the extremities of the base of any triangle, draw lines bisecting the other sides ; these two lines in- tersecting within the triangle, will form another triangle on the same base. How will the area of this new tri- angle compare with that of the whole triangle ? Ans. Their areas will be as 3 to 1. 12. Two parallel chords on the same side of the center of a circle, whose diameter is 32, are measured and found to be, the one 20, and the other 8. How far are they asunder? Ans. ^240"— ^156"= 3 +. If we suppose the two chords to be on opposite sides of the center, their distance apart will then be v'240 -j- v/156 = ] 5.49 -f- 12.49 = 27.98. 13. The longer of the two parallel sides of a trapezoid is 12, the shorter 8, and their distance asunder 5. What is the area of the trapezoid ? and if we produce the two inclined sides until they meet, what will be the area of the triangle so formed ? Ans. Area of trapezoid, 50 ; area of triangle, 40 ; area of triangle and trapezoid, 90. BOOK V. 145 14. The base of a triangle is 697, one of the sides is 534, and the other 813. If a line be drawn bisecting the angle opposite the base, into what two parts will the bisecting line divide the base ? (See Th. 24, B. II). A ( The greater part will be 420.684 ; m ' \ The less " " 276.316. 15. Draw three horizontal parallels, making the dis- tance between the two upper parallels 7, and that be- tween the middle and lower parallels 9 ; then place be- tween the upper parallels a line equal to 10, and from the point in which it meets the middle parallel draw to the lower a line equal to 11, and join the point in which this last line meets the lower parallel, with the point in the upper parallel, from which the line 10 was drawn. Required the length of this line, and the area of the triangle formed by it and the two lines 10 and 11. The adjoining figure will illustrate. Let A be the point on the upper parallel from which the line 10 is drawn. Then, AF = 7, AB = 10, FB = \/l00 — 19 = •51. BI1 = FD = 9, B C = U,HC= BE. For, the triangles BAG, BAB, and BAE are all right- angled at J., the side BA is common, and AC= AD=AE by construction, hence, (Th. 16, B. I), BC=BD = BE. Moreover, since AF> AE, the oblique line BF> BE. Cor. If any number of equal oblique lines be drawn from the point B to the plane, they will all meet the plane in the circumference of a circle having the foot of the perpendicular for its center. It follows from this, that, if three points be taken in a plane equally distant from a point out of it, the center of the circle whose cir* cumference passes through these points will be the foot of *he ]aerx>endicular drawn from the point to the plane. BOOK VI. 157 THEOREM V. The line which joins any point of a perpendicular to a plane, with the point in which a line in the plane is inter- sected, at right angles, by a line through the foot of the per- pendicular, will be at right angles to the line in the plane Let AB be perpendic- ular to the plane ST, and AD a line through its foot at right angles to EF, a line in the plane. Connect D with any point, as B, of the perpendicular; and BD will be perpendicular to EF. Make DF '= DE, and join B to the points E and F. Since DE = DF, and the angles at D are right angles, the oblique lines, AE and AF, are equal ; and, since AE= AF, we have, (Th. 4), BE=BF; therefore the line BD has two points, B and D, each equally distant from the extremities E and F of the line EF, and hence BD is perpendicular to EF at its middle point D. Cor. Since FD is perpendicular to the two lines AD und BD at their intersection, it is perpendicular to their plane ADB, (Th. 3). ScnoLiuM. — The inclination of a line to a plane is measured by the angle included between the given line and the line which joins the point in which it meets the plane and the foot of the perpendicular drawn from any point of the line to the plane ; thus, the angle BFA ia the inclination of the line BF to the plane ST. THEOREM VI. If either of two parallels is perpendicular to a plane, the other is also perpendicular to the plane. Let BA and ED be two parallels, of which one, BA, is perpendicular to the plane ST; then will the other also be. perpendicular to the same plane. 14 158 GEOMETRY. The two parallels de- termine a plane which intersects the given plane in AD; through D draw MN perpendicular to AD; then, (Cor., Th. 5,) will MN be perpendicu- lar to the plane BAD, and the angle MDE is therefore a right angle ; but ED A is also a right angle, since BA and ED are parallel, and BAD is a right angle by hypothesis; hence, ED is perpendicular to the two lines MD and AD in the plane ST; it is therefore perpen- dicular to the plane, (Th. 3). Cor. 1. The converse of this proposition is also true , that is, if two straight lines are both perpendicular to the same plane, the lines are parallel. For, suppose BA and ED to be two perpendiculars ; if not parallel, draw through D a parallel to BA, and this last line will be perpendicular to the plane ; but ED is a perpendicular by hypothesis, and we should have two perpendiculars erected to the plane at the same point, which is impossible, (Cor. 2, Th. 3). Cor. 2. If two lines lying in the same plane are each parallel to a third line not in the same plane, the two lines are parallel. For, pass a plane perpendicular to the third line, and it will be perpendicular to each of tlva others ; hence they are parallel , THEOREM VII, A straight line is parallel to a plane, when it is parallel to a line in the plane. Suppose the line MX to be parallel to the line CD, in the plane ST; then will Jf^be parallel to the plane ST BOOK VI. 159 For, OB being in the plane ST, and at the same time g parallel to MN, it must be the intersection of the pl°.ne of these parallels with the plane ST; hence, if MN meet the plane ST, it must do so in the T line OB, or OB produced; but MN and OB are parallel, and cannot meet; therefore MN, nowever far produced, can have no point in the plane ST, and hence, (Def. 5), it is parallel to this plane. THEOREM VIII. If two lines are parallel, they will be equally inclined to any given plane. Let AB and OB be two parallels, and ST any plane met by them in the points A and 0; then will the lines AB and OB be equally inclined to the plane ST. For, take any distance, AB, on one of these parallels, and make OB = AB, and draw A and BB. From the points B and B let fall the perpendiculars, BE and BF, on the plane ; join their feet by the line BF, and draw AB and OF. Now, since AB is equal and parallel to OB, ABBOh ft parallelogram, and BB is equal and parallel to A 0, and BB is parallel to the plane ST, (Th. 7) ; and, since BE and BF are both perpendicular to this plane, they are parallel ; but BB and EF are in the plane of these parallels; and as EF is in the plane ST, and BB is parallel to this plane, these two lines must be parallel and equal, and BBFE is also a parallelogram Now, 160 GEOMETRY. we have shown that BD is equal and parallel to AC, and EF equal and parallel to BD; hence, (Cor. 2, Th. 6), EFis equal and parallel to AC, and ACFE is a parallel- ogram, and AE — CF. The triangles J.^J/ and CDF have, then, the sides of the one equal to the sides of the other, each to each, and their angles are consequently equal; that is, the angle BAE is equal to the angle DCF; but these angles measure the inclination of the lines AB and CD to the plane ST, (Scholium, Th. 5). Scholium. — The converse of this proposition is not generally true ; that is, straight lines equally inclined to the same plane are not neces- sarily parallel. THEOREM IX. The intersections of two parallel planes by a third plane, are parallel. Let the planes QR and ST he intersected by the third plane, AD : then will the intersections, AB and CD, be parallel. Since the lines AB and CD are in the same plane, if they are not parallel, they will meet if sufficiently produced; but they cannot meet out of the planes QR and ST, in which they are respectively found; therefore, any point common to the lines, must be at the same time common to the planes ; and since the planes are parallel, (hey have no common point, and the lines, therefore, do not intersect ; hence they are parallel. THEOREM X. If two planes are perpendicular to the same straight line, they are parallel to each other. Let QR and ST be two planes, perpendicular to the line AB; then will these planes be parallel. '- — \ s ] 1— i" n "~"~ — ■ . \ BOOK VI. 161 For, if not parallel, suppose M to be a point in their line of intersection, and from this point draw lines to the extremities of the perpendicular m" AB, thus forming a tri- angle, MAB. Now, since the line AB is perpendicular to both planes, it is perpendicular to each of the lines MA and MB, drawn through its feet in the planes, (Def. 2); hence, the triangle has two right angles, which is impos- sible ; the planes cannot therefore meet in any point as M, and are consequently parallel. Cor. Conversely : The straight line which is perpendicu- lar to one of two parallel planes, is also perpendicular to the other. For, if AB be perpendicular to the plane QB, draw in the other plane, through the point in which the perpendicular meets it, any line, as AC. The plane of the lines AB and A C will intersect the plane QB in the line BD ; and since the planes are parallel by hypothesis, the lines AC and BD must be parallel, (Th. 9) ; but the angle DBA is a right angle ; hence, BA C must be a right angle, and the line BA is perpendicular to any line what; ever drawn in the plane through the point A ; BA is therefore perpendicular to the plane ST. THEOREM XI. If two straight lines be drawn in any direction through parallel planes, the planes will cut the lines proportionally. Conceive three planes to be parallel, as represented in the figure, and take any points, A and B, in the first and third planes, and draw AB, the line passing through the second plane at E. 14* L 162 GEOMETRY. Also, take any other two points, as and D, in the first and third planes, and draw OB, the line passing through the second plane at F. Join the two lines by the diagonal AB, which passes through the second plane at G. Draw BB, EG, GF, and AC. We are now to prove that, AE : EB : : OF : FB. For the sake of brevity, put AG=X, and GD= Y. As the planes are parallel, BD is parallel to EG ; from the two triangles ABD and AEG, we have, (Th. 17, B.II); A E : EB : : X : Y. Also, as the planes are parallel, GF is parallel to A 0, and we have, OF : FD : : X : Y. By comparing the proportions, and applying Th. 6, B. II, we have AE : EB : : OF : FB. THEOREM XII. If a straight line is perpendicular to a plane, all planes passing through that line will be perpendicular to the plane. Let MNbe a plane, and AB a per- pendicular to it. Let BO be any other plane, passing through AB ; this plane will be perpendicular to MM Let BB be the common intersec- tion of the two planes, and from the point B, draw in MN BE at right angles to DB. Then, as AB is perpendicular to the plane MN, it is perpendicular to every line in that plane, passing through M D ^ BOOK VI. 163 B ; (Def. 2,) ; therefore, ABE is a right angle. But the angle ABE, (Def. 3), measures the inclination of the two planes ; therefore, the plane CB is perpendicular to the plane MN; and thus we can show that any other plane, passing through AB, will be perpendicular to MN. Hence the theorem. THEOREM XIII. If two planes are perpendicular to each other, and a line be drawn in one of them perpendicular to their common in- tersection, it will be perpendicular to the other plane. Let the two planes, QR and ST, be perpendicular to each other, and draw in QR the line CD at right angles to their common intersection, R V; then will this line be perpendicular to the plane ST. In the plane ST draw ED, perpen- dicular to VR at the point D. Then, since the planes QR and ST are perpendicular to each other, the angle ODE is a right angle, and CD is perpendicular to the two lines, ED and VR, passing through its foot in the plane ST. CD is therefore perpendicular to the plane ST, (Th. 3). Cor. Conversely: if we erect a perpendicular to the plane ST, at any point, D, of its intersection with the plane QR, this perpendicular will lie in the plane QR. For, if it be not in this plane, we can draw in the plane the line CD, at right angles to VR', and, from what has been shown above, CD is perpendicular to the^lane ST, and we should thus have two perpendiculars erected to the plane, ST, at the same point, which is impossible, (Cor. 2, Th. 8V 164 GEOMETRY. THEOREM XIV. The common intersection of two planes, 'both of which are perpendicular to a third plane, will also be perpendicular to the third plane. Let MN be the common intersection of the two planes, QR and VX, both of which are perpendicular to the plane ST; then will MZ^be perpendicular to the plane ST For, if we erect a perpendicular to the plane ST, at the point M, it will lie in both planes at the same time, (Cor. Th. 13); and this perpendicular must therefore be their intersection. Hence the theorem. THEOREM XV. Parallel straight lines included between parallel planes, are equal. Let AB and 2) (7 be two parallel lines, included by the two parallel planes, QR and ST; then will AB = BO. For, the plane AC, of the parallel lines, intersects the planes, QR and ST, in the parallel lines, AB and BO, (Th. 9) ; hence ABGD is a parallelogram, and its oppo- site sides, AB and BO, are equal. Oot. It follows from this proposition, that parallel planes are everywhere equally distant ; for, two perpendiculars drawn at pleasure between the two planes are parallel lines, (Cor. 1, Th. 6), and hence are equal ; but these per- pendiculars measure the distance between the planes. BOOK Vl 165 THEOREM XVI. Two planes are parallel when two lines not parallel, lying in the one, are respectively parallel to two lines lying in the other. Let QR and ST be two planes, the first containing the two lines AB and CD which intersect each other at E, and the second the two lines LM and NO, respect- ively parallel to AB and CD-, then will these planes be par- allel. For, if the two planes are not parallel, they must intersect when sufficiently produced; and their common section lying in both planes at the same time, would be a line of the plane QR. Now, the lines AB and CD intersect each other by hypothesis ; hence one or both of them must meet the common sec- tion of the two planes. Suppose AB to meet this com- mon section ; then, since AB and LM are parallel, they determine a plane, and AB cannot meet the plane ST in a point out of the line LM; but AB and LM being par- allel, have no common point. Hence, neither AB nor CD can meet the common section of the two planes ; that », they have no common section, and are therefore par- allel. Cor. Since two lines which intersect each other, deter- mine a plane, it follows from this proposition, that the plane of two intersecting lines is parallel to the plane of two other intersecting lines respectively parallel to the first lines 166 GEOMETRS*. THEOREM XVII. When two intersecting lines are respectively parallel to two other intersecting lines lying in a different plane, the angles formed by the last two lines will be equal to those formed by the first two, each to each, and the planes of the angles will be parallel. Let QB be the plane of the two lines AB and CD, which inter- sect each other at the point E, and ST the plane of the two lines LM and NO, respect- ively parallel to AB and QB ; then will the [__BED = \_MPO, and L EEC - L MPN, etc., and the planes QB and ST will be parallel. That the plane of one set of angles is parallel to that of the other, follows from the Corollary to Theorem 16 ; we have then only to show that the angles are equal, each to each. Take any points, B and B, on the lines AB and CD, and draw BD. Lay off PM, equal to and in the same direction with JEB, and PO, equal to and in the same direction with BD, and draw MO. Now, since the planes QR and ST are parallel, and BD is equal and parallel to PO, BDOP is a parallelogram, and DO is equal and par allel to EP. For the same reason, BM is equal and parallel to EP; therefore, BDOM is a parallelogram, and MO is equal and parallel to BD. Hence the A's, EBD and PMO, have the sides of the one equal to the sidea of the other, each to each; they are therefore equal, and BOOK VI. 167 the I MPO — the [ BED. In the same manner it can be proved that \_BEC = \_MPN, etc. Cor. 1. The plane of the parallels AB and LM is in- tersected by the plane of the parallels CD and NO, in the line EP. Now, EB and ED are the intersections of these two planes with the plane QR, and PM and PO are the intersections of the same planes with the parallel plane ST. It has just been proved that the |__ BED = [_MPO. Hence, if the diedral angle formed by two planes, be cut by two parallel planes, the intersections of the faces of the diedral angle with one of these planes will include an angle equal to that included by the intersections of the faces with the other plane. Cor. 2. The opposite triangles formed by joining the cor- responding extremities of three equal and parallel straight lines lying in different planes, will be equal and the plane* of the triangles will be parallel. Let EP, BM, and DO, be three equal and parallel straight lines lying in different planes. By joining their corresponding extremities, we have the triangles EBD and PMO. Now, since EP and BM are equal and parallel, EBMP is a parallelogram, and EB is equal and parallel to PM; in the same manner, we show that ED is equal and parallel to PO, and BD to MO; hence the triangles are equal, ha ring the three sides of the one, respectively, equal to the three sides of the other. That their planes are parallel, follows from Cor., Theo rem 16. THEOREM XVIII. Any one of the three plane angles bounding a triedral angle, is less than the sum of the other two. Let A be the vertex of a solid angle, bounded by the three plane angles, BAG, BAD, and DAC; then will any one of these three angles be less than the sum of the 168 GEOMETRY. other two. To establish this proposition, we have only to compare the greatest of the three angles with the sum of the other two. Suppose, then, BAG to be the greatest angle, and draw in its plane B« the line AE, making the angle CAE equal to the angle CAD. On "iT AE, take any point, E, and through it draw the line CEB. Take AD, equal to AE, and draw BD and DC. Now, the two triangles, CAD and CAE, having two sides and the included angle of the one equal to the two sides and included angle of the other, each to each, are equal, and CE = CD; but in the triangle, BDC, BC< BD + DC. Taking EC from the first member of this inequality, and its equal, DC, from the second, we have, BE < BD. In the triangles, BAE and BAD, BA is common, and AE = AD by construction ; but the third side, BD, in the one, is greater than the third side, BE, in the other ; hence, the angle BAD is greater than the angle BAE, (Th. 22, B. I); that is, [_BAE < [_BAD; adding the \_EAC to the first member of this inequality, and its equal, th.e.[_DAC, to the other, we. have [_BAE+l_EAC< [_BAD + [_DAC. And, as the [_ BAG is made up of the angles BAE and EAC, we have, as enunciated, ]_BAC< [_BAD + [_DAC. THEOREM XIX. The sum of the plane angles forming any solid angle, is always less than four right angles. Let the planes which form the solid angle at A, be cut by another plane, which we may call the plane of the base, BCDE. Take any point, a, in this plane, and draw aB y aC, aD, aE, etc., thus making as many triangles on BOOK VI. 169 the plane of the base as there are tri- angular planes forming the solid angle A. Now, since the sum of the angles of every A is two right angles, the sum of all the angles of the A's which have their vertex in A, is equal to the Bum of all angles of the A's which have their vertex in a. But, the angles BOA + AOD, are, together, greater than the angles BOa + aOD, or BCD, by the last proposition. That is, the sum of all the angles at the bases of the A's which have their vertex in A, is greater than the sum of all the angles at the bases of the A's which have their vertex in a. Therefore, the sum of all the angles at a is greater than the sum of all the angles at A ; but the sum of all the angles at a is equal to four right angles ; there- fore, the sum of all the angles at A is less than four right angles. THEOREM XX. If two solid angles are formed by three plane angles respect- ively equal to each other, the planes which contain the equal angles will be equally inclined to each other. Letthe [_ASC=the [_DTF, the |_ ASB = the [_ J> TE, and the [_BSO= the [_ETF; then will the inclination of the planes, A SO, ASB, be equal to that of the planes, DTF, DTE. Having taken SB at pleas- are, draw BO perpendicular to the plane ASO; from the point 0, at which that perpen- dicular meets the plane, draw OA and 00, perpendicular to SA and SO; draw AB and BO; next take TE = SB, and draw EP perpendicular to the plane DTF; from the 15 170 GEOMETRY. point P, draw PB and PF, perpendicular to TB and TF; lastly, draw BE and FF. The triangle SAB, is right-angled at A, and the tri- angle TBF, at B, (Th. 5) ; and since the |__ ASB = the L J> TF, we have [_ SB A - L 2^ J likewise, SB=TF; therefore, the triangle SAB is equal to the triangle TBF) hence, SA = TB, and AB = Di?. In like manner it may be shown that SO = TF, and BO = J£F. That granted, the quadrilateral SAOO is equal to the quadri- lateral TBPF; for, place the angle A SO upon its equal, 2)2^, and because SA = 2'i), and SO = TF, the point A will fall on i), and the point on F; and, at the same time, A 0, which is perpendicular to SA, will fall on PB, which is perpendicular to TB, and, in like manner, 00 on PF; wherefore, the point will fall on the point P, and A will be equal to BP. But the triangles, A OB, BPF, are right angled at and P; the hypotenuse AB = BF, and the side AO = BP; hence, those triangles are equal, (Cor, Th. 39, B. I), and \_OAB=[_PBF. The angle OAB is the inclination of the two planes, ASB, ASO; the angle PBF is that of the two planes, BTF, BTF; conse- quently, those two inclinations are equal to each other. Hence the theorem. Scholium 1. — The angles which form the solid angles at S and T, may be of such relative magnitudes, that the perpendiculars, BO and EP, may not fall within the bases, ASC and DTF; but they will always either fall on the bases, or on the planes of the bases produced, and will have the same relative situation to A, S, and C, as P has to D, T, and F. In case that and P fall on the planes of the bases produced, the angles BCO and EFP, would be obtuse angles ; but the demonstration of the problem would not be varied in the least. Scholium 2. — If the plane angles bounding one of the triedral angles be equal to those of the other, each to each, and also be simi larly arranged about the triedral angles, these solid angles will be ab- solutely equal. For it was shewn, in the course of the above demon- stration, that the quadrilaterals, SAOC and TDPF, were equal; and on being applied, the point falls on the point P; and since the trian- gles A OB and DPE are equal, the perpendiculars OB and PE ar« BOOK VI. 171 ilsc equal. Now, because the plane angles are like arranged about the triedral angles, these perpendiculars lie in the same direction ; hence the point B will fall on the point \E, and the solid angles will exactly coincide. Scholium 3. — When the planes of the equal angles are not like dis- posed about the triedral angles, it would not be possible to make these triedral angles coincide ; and still it would be true that the planes of the equal angles are equally inclined to each other. Hence, these triedral angles have the plane and diedral angles of the one, equal to the plane and diedral angles of the other, each to each, without having of themselves that absolute equality which admits of superposition. Magnitudes which are thus equal in all their component parts, but will not coincide, when applied the one to the other, are said to be symmetrically equal. Thus, two triedral angles, bounded by plane fcngbs equal each to each, but not like placed, are symmetrical triedral 172 GEOMETRY. BOOK VII, SOLID GEOMETRY. DEFINITIONS. 1. A Polyedron is a solid, or volume, bounded on all sides by planes. The bounding planes are called the faces of the polyedron, and then intersections are its edges, 2. A Prism is a polyedron, having two of its faces, called bases, equal polygons, whose planes and homolo- gous sides are parallel. The other, or lateral faces, are parallelograms, and constitute the convex surface of the prism. The bases of a prism are distinguished by the terms, upper and lower; and the altitude of the prism is the per pendicular distance between its bases. Prisms are denominated triangular, quadrangular, pent angular, etc., according as their bases are triangles, quad- rilaterals, pentagons, etc. 3. A Right Prism is one in which the planes of the lateral faces are perpendicular to the planes of the bases. 4. A Parallelopipedon is a prism whose bases are parallelograms. 5. A Rectangular Parallelopipedon is a right parallelopipedon, with rectangular bases. BOOK VII. 173 6. A Cube or Hexaedron is a rectangu- lar parallelopipedon, whose faces are all equal squares. 7. A Diagonal of a Polyedron is a straight line joining the vertices of two solid angles not adjacent. 8. Similar Polyedrons are those which are bounded by the same number of similar polygons like placed, and whose homologous solid angles are equal. Similar parts, whether faces, edges, diagonals, or angles, similarly placed in similar polyedrons, are termed homologous. 9. A Pyramid is a polyedron, having for one of its faces, called the base, any polygon whatever, and for its other faces triangles having a common vertex, the sides opposite which, in the several trian- gles, being the sides of the base of the pyramid. 10. The Vertex of a pyramid is the common vertex of the triangular faces. U. The Altitude of a pyramid is the perpendicular distance from its vertex to the plane of its base. 12. A Right Pyramid is one whose base is a regular polygon, and whose vertex is in the perpendicular to the base at its center. This perpendicular is called the axis of the pyramid. 13. The Slant Height of a right pyramid is the perpen- dicular distance from the vertex to one of the sides of the base. 14. The Frustum of a Pyramid is a portion of the pyr- amid included between its base and a section made by a plane parallel to the base. Pyramids, lik* prisms, are named from the forms of their bases. 15* 174 GEOMETRY. 15. A Cylinder is a body, having for its ends, or bases, two equal circles, the planes of which are perpendicular to the line joining their centers ; the remainder of its surface may be con- ceived as formed by the motion of a line, which constantly touches the cir- cumferences of the bases, while it remains parallel to the line which joins their centers. We may otherwise define the cylinder as a body gen- erated by the revolution of a rectangle about one of its sides as an immovable axis. The sides of the rectangle perpendicular to the axis generate the bases of the cylinder; and the side opposite the axis generates its convex surface. The line joining the centers of the bases of the cylinder is its axis, and is also its altitude. If, within the base of a cylinder, any polygon be in- scribed, and on it, as a base, a right prism be con- structed, having for its altitude that of the cylinder, such prism is said to be inscribed in the cylinder, and the cylin- der is said to circumscribe the prism. Thus, in the last figure, ABODEc is an inscribed prism, and it is plain that all its lateral edges are con- tained in the convex surface of the cylinder. d If, about the base of a cylinder, any polygon be circumscribed, and on it, as a base, a right prism be con- structed, having for its altitude that of the cylinder, such prism is said to be circumscribed about the cylinder, and the cylinder is said to be inscribed in the prism. Thus, ABCLEFc is a circum- scribed prism; and it is plain that BOOK Vll. 175 the lino, mn, which joins the points of tangency of the sides, EF and ef, with the circumferences of the bases of the cylinder, is common to the convex sur- faces of the cylinder and prism. 16. A Cone is a body bounded by a circle and the surface generated by the motion of a straight line, which con- stantly passes through a point in the perpendicular to the plane of the circle at its center, and the different points in its circumference. The cone may be otherwise' defined as a body gene rated by the revolution of a right-angled triangle about one of its sides as an immovable axis. The other side of the triangle will generate the base of the cone, while the hypotenuse generates the convex surface. The side about which the generating triangle revolves is the axis of the cone, and is at the same time its altitude. If, within the base of the cone, any polygon be inscribed, and on it, as a base, a pyramid be constructed, having for its vertex that of the cone, such pyramid is said to be inscribed in the cone, and the cone is said to circumscribe the pyramid. Thus, in the accompanying figure, V — ABODE, is an inscribed pyramid, and it is plain that all its lateral edges are contained in the convex surface of the cone. If, about the base of a cone, any poly- gon be circumscribed, and on it, as a base, a pyramid be constructed, having for its vertex that of the cone, such pyramid is said to be circumscribed about the cone, and the cone is said to be inscribed in the pyramid. 176 GEOMETRY. 17. The Frnstnm of a Cone is the portion of the cone that is included between its base and a section made by a plane parallel to the base. 18. Similar Cylinders, and also Similar Cones, are such as have their axes proportional to the radii of their bases. 19. A Sphere is a body bounded by one uniformly-curved surface, all the points of which are at the same distance from a certain point within, called the center. We may otherwise define the sphere as a body gene- rated by the revolution of a semicircle about its diameter as an immovable axis. 20. A Spherical Sector is that portion of a sphere which is in- cluded between the surfaces of two cones having a common axis, and their vertices at the center of the sphere. Or, it is that portion of the sphere which is generated by a sector of the generating semicircle. 21. The Radius of a Sphere is a straight line drawn from the center to any point in the surface ; and the diameter is a straight line drawn through the center, and limited on both sides by the surface. All the diameters of a sphere are equal, each being twice the radius. 22. A Tangent Plane to a sphere is one which has a single point in the surface of the sphere, all the others being without it. 23. A Secant Plane to a sphere is one which has more than one point in the surface of the sphere, and lies partly within and partly without it. Assuming, what will presently be proved, that the in- tersection of a sphere by a plane is a circle, 24. A Small Circle of a sphere is one whose plane does not pass through its center; and BOOK VII 177 25. A Great Circle of a sphere is one whose plane passes through the center of the sphere. 26. A Zone of a sphere is the portion of its surface in- cluded between the circumferences of any two of its paral- lel circles, called the bases of the zone. "When the plane of one of these circles becomes tangent to the sphere, the zone has a single base. 27. A Spherical Segment is a portion of the volume of a sphere included between any two of its parallel circles, called the bases of the segment. The altitude of a zone, or of a segment, of a sphere, is the perpendicular distance between the planes of its bases. \ 28. The area of a surface is measured by the product of its length and breadth, and these dimensions are always conceived to be exactly at right angles to each other. 29. In a similar manner, solids are measured by the product of their length, breadth, and height, when all their dimensions are at right angles to each other. The product of the length and breadth ©f a solid, is the measure of the surface of its base. Let P, in the annexed fig- ure, represent the measuring unit, and AF the rectangular solid to be measured. A side of P is one unit in length, one in breadth, and one in height; one inch, one foot, one yard, or any other unit that may be taken. Then, lxlxl 1, the unit cube. Now, if the base of the solid, AC, is, as here repr< • sented, 5 units in length and 2 in breadth, it is obvious that (5x2 = 10), 10 units, each equal to P, can be placed on the base of AC, and no more; and as each of these onits will occupy a unit of altitude, therefore, 2 units of M 178 GEOMETRY. altitude will contain 20 solid units, 3 units of altitude, 30 solid units, and so on ; or, in general terms, the num- ber of square units in the base multiplied by the linear units in perpendicular altitude, will give the solid units in any rect- angular solid. THEOREM I. If the three plane faces bounding a solid angle of one prism be equal to the three plane faces bounding a solid angle of another, each to each, and similarly disposed, the prisms will be equal. Suppose A and a to be the vertices of two solid angles, bounded by equal and similarly placed faces; then will the prisms, ABODE— -i^and abcde — n, be equal. For, if we place the base, abcde, upon its equal, the base ABODE, they will coincide; and since the solid angles, whose vertices are A and a, are equal, the lines ab, ae, and ap, respectively coincide with AB, AE, and AP ; but the faces, al and ao, of the one prism, are equal, each to each, to the faces, AL and AO, of the other; therefore pi and po coincide with PL and PO, and the upper bases of the prisms also coincide : hence, not only the bases, but all the lateral faces of the two prisms coincide, and the prisms are equal. Oor. If the two prisms are right, and have equal bases and altitudes, they are equal. For, in this case, the rect- angular faces, al and ao, of the one, are respectively equal to the rectangular faces, AL and AO, of the other; and hence the three faces bounding a triedral angle in the one, are equal and like placed, to the faces bounding a triedral angle in the othei BOOK VII. 179 THEOREM II. The opposite faces of any parattelopipedon axe equal, and their planes are parallel. Let ABOD — E be any parallelopipedon ; then will its opposite faces be equal, and their planes will be parallel. The bases ABOD and FEGH are equal, and their planes are parallel, by definitions 2 and 4 of this Book ; it remains tor us, therefore, only to show that any two of the opposite lateral faces are equal and parallel. Since all the faces of the parallelopipedon are parallel- ograms, AB is equal and parallel to DC, and AH is also equal and parallel to DF; hence the angles HAB and FDO are equal, and their planes are parallel, (Th. 17, B. VI), and the two parallelograms, HABGr and FDCE, having two adjacent sides and the included angle of the one equal to the two adjacent sides and included angle of the other, are equal. Oor. 1 Hence, of the six faces of the parallelopipedon, any two lying opposite may be taken as the bases. Cor. 2. The four diagonals of a parallelopipedon mutu- ally bisect each other. For, if we draw A and HE, we shall form the parallelogram A OEH, of which the diago- nals are AE and HO, and these diagonals are at the same time diagonals of the parallelopipedon; but the diagonals of a parallelogram mutually bisect each other. Now, if the diagonal FB be drawn, it and HO will bisect each other, since they are diagonals 'of the parallelogram FHBO. In like manner we can show that if DGr be drawn, it will be bisected by AE. Hence, the four diag- onals have a common point within the parallelopipedon. Scholium. — It is seen at once that the six faces of a parallelopipe- don intersect each other in twelve edges, four of which are equal to HA, four to AB, and four to AD. Now, we may conceive the parallel- opipedon to be bounded by the planes determined by the three lines 180 GEOMETR AH, AB, and AD, and the three planes passed through the extremi* ties, H, B, and JD, of these lines, parallel to the first three planes. THEOREM III. The convex surface of a right prism is measured by the perimeter of its base multiplied by its altitude. Let ABODE — iVbe a right prism, of which AP is the altitude ; then will its convex surface he measured by {AB + BO + OB + DE + EA)xAP. For, its convex surface is made up of the rectangles AL, BM, ON, etc., and each rectangle is measured by the product of its base by its altitude ; but the altitude of each rectangle is equal to AP, the alti- tude of the prism ; hence the convex sur- face of the prism is measured by the pro- duct of the sum of the bases of the rectangles, or the perimeter of the base of the prism, by the common alti- tude, AP. Oor. Kight prisms will have equivalent convex surfaces, when the products of the perimeters of their bases by their altitudes are respectively equal ; and, generally, their convex surfaces will be to each other as the products of the perimeters of their bases by their altitudes. Hence, if the altitudes are equal, their convex surfaces will be as the perimeters of their bases ; and if the perimeters of their bases are equal, the^r convex surfaces will be as their altitudes. THEOREM IV. The two sections of a prism made by parallel planes between its bases are equal polygons. Let the prism ABODE — N be cut between its bases by two parallel planes, making the sections QBS, etc.. BOOK VII, 181 and TVX, etc. ; then will these sections be equal polygons. For, since the secant planes are paral- lel, their intersections, QR and TV, by the plane of the face JEAPO are parallel, (Th. 9, B. VI) ; and being included be- tween the parallel lines, AP and EO, they are also equal. In the same manner we may prove that BS is equal and parallel to VX, and so on for the intersections of the secant planes by the other faces of the prism. Hence, these polygonal sections have the sides of the one equal to the sides of the other, each to each. The angles QBS and TVX are equal, because their sides are parallel and lie in the same direction ; and in like manner we prove [__ RSY = [_ VXZ, and so on for the other corresponding angles of the polygons. Therefore, these polygons are both mutually equilateral and mutually equiangular, and consequently are equal. Cor. A section of a prism made by a plane parallel to the base of the prism, is a polygon equal to the base. THEOKEM V. Two parallelopipedons, the one rectangular and the other oblique, will be equal in volume when, having the same base and altitude, two opposite lateral faces of the one are m the 'planes of the corresponding lateral faces of the other. Designating the parallelo- pipedons by their opposite diagonal letters, let AG be the rectangular, and AL the obnque, parallelopipedon, hav- ing the same base, A C, and the same altitude, namely, the perpendicular distance be- 16 182 GEOMETRY. tween the parallel pranes, AC and EL. Also let the fa^e, AK, be in the plane of the face, AF, and the face, BL, m the plane of the face, BG. We are now to prove that the oblique parallelopipedon is equivalent to tne rectangular parallelopipedon. As the faces, AF and AK, are in the same plane, and the parallelopipedons have the same altitude, EFK is a straight line, and EF— IK, because each is equal to AB. If from the whole line, FK, we take FF, and then from the same line we take IK= FF, we shall have the re- mainders, FI and FK, equal ; and since AF and BF are parallel, \_AFI = [_BFK ; hence the A's, AFI and BFK, are equal. Since HE and MI are both parallel to DA, they are parallel to each other, and FIMH is a par- allelogram ; for like reasons, FKLG is a parallelogram, and these parallelograms are equal, because two adjacent sides and the included angle of the one are equal to two adjacent sides and the included angle of the other. The parallelograms, BE and CF, being the opposite faces of the parallelopipedon, AG-, are equal. Hence, the three plane faces bounding the triedral angle, E, of the triaii* gular prism, EAI — E, are equal, each to each, and like placed, to the three plane faces bounding the triedral angle F, of the triangular prism, FBK — G, and these prisms are therefore equal, (Th. 1). Now, if from the whole solid, EABK — H, we take the prism, EAI — H, there will remain the parallelopipedon, AL; and, if from the srme solid, we take the prism, FBK—G, there will remain the rectangular parallelopipedon, AG. Therefore, the oblique and the rectangular parallelopipedons are equiva- lent. Cor. The volume of the rectangular parallelopipedon, AG, is measured by the base, ABCB, multiplied by the altitude, AF, (Def. 29) ; consequently, the oblique paral- lelopipedon is measured by the product of the same base by the same altitude. BOOK VII. 183 Scholium.- -If neither of the parallelopipedons is rectangular, but they still have the same base and the same altitude, and two opposite lateral faces of the one are in the planes of the corresponding lateral faces of the other, by precisely the same reasoning we could prove the parallelopipedons equivalent. Hence, in general, any two parallelo- pipedons will be equal in volume when, having the same base and aUtiPude, two opposite lateral faces of the one are in the planes of the correspond- ing lateral faces of the other, THEOREM VI. Two parallelopipedons having equal bases and equal alti- tudes^ are equivalent. Let A G- and AL be two paral- lelopipedons, having a common lower base, and their upper bases in the same plane, HF. Then will these parallelopipedons be equivalent. Since their upper bases are in the same plane, and the lines IM and KL are par- allel, and also EF and HG, these lines will intersect, when produced, and form the parallelogram NOPQ, which will be equal to the common lower base of the two parallelopipedons. Now, if a third parallelo- pipedon be constructed, having BD for its lower base, and OQ for its upper base, it will be equivalent to the par- allelopipedon AG, and also to the parallelopipedon AL, (Th. 5, Scholium) ; hence, the two given parallelopipe- dons, being each equivalent to the third parallelopipe- don, are equivalent to each other. Hence, two parallelopipedons having equal bases, etc. THEOREM VII. The volume of any parallelopipedon is measured by the product of its base and altitude, or the product of its three dimensions. 184 GEOMETRY. Let ABCD—Gr be any parallelopipedon ; tl en will its volume be expressed by the product ^_h of tbe area of its base and altitude. hj If the parallelopipedon is oblique, l/ we may construct on its base a right parallelopipedon, by erecting perpen- diculars at the points A, B, 0, and D, and making them each equal to the altitude of the given parallelopipedon ; and the right parallelopipedon, thus constructed, will be equivalent to the given parallelopip- edon, (Th. 6). Eow, if the base, ABOD, is a rectangle, the new parallelopipedon will be rectangular, and meas- ured by the product of its base and altitude, (Def. 29). But if the base is not rectangular, let fall the perpen- diculars, Be and Ad, on CD and CD produced, and take the rectangle ABcd for the base of a rectangular paral- lelopipedon, having for its altitude that of the given parallelopipedon. "We may now regard the rectangular face, ABFE, as the common base of the two parallelo- pipedons, Ag and AG-', and, as they have a common base, and equal altitude, they are equivalent. Thus we have reduced the oblique parallelopipedon, first to an equivalent right parallelopipedon on the same base, and then the right to an equivalent rectangular parallelopip- edon on an equivalent base, all having the same alti- tude. But the rectangular parallelopipedon, Ag, is measured by product of its base, ABcd, and its altitude; hence, the given and equivalent oblique parallelopipedon is measured by the product of its equivalent base and equal altitude. Hence, the volume of any parallelopipedon, etc. Cor. Since a parallelopipedon is measured by the pro- duct of its base by its altitude, it follows that parallelo pipedons of equivalent bases, and equal altitude?, are equiva lent, or equal in volume. BOOK VII 185 THEOREM VIII. Parallelopipedons on the same, or equivalent hoses, are to each other as their altitudes ; and parallelopipedons having equal altitudes, are to each other as their bases. Let P and p represent two parallelopipedons, whose bases are denoted by B and b, and altitudes by A and a, respectively. Now, P = B x A, and p = b x a, (Th. 7). But magnitudes are proportional to their numerical measures ; that is, P : p :: B x A : b X a. If the bases of the parallelopipedons are equivalent, we have B — b; and if the altitudes are equal, we have A = a. Introducing these suppositions, in succession, in the above proportion, we get P : p :: A : a, and P : p : : B : b. Hence the theorem ; Parallelopipedons on the same, etc, THEOREM IX. Similar parallelopipedons are to each other as the cubes of their like dimensions. Let P and p represent any two similar parallelopipe- dons, the altitude of the first being denoted by h, and the length and breadth of its base by I and n, respect- ively; and let h', V, and n r , in order, denote the correeh ponding dimensions of the second. Then we are to prove that P : p :: n 8 : n" :: I* : V* :: h n : h'\ We have P = Inh, and p - Vn'K* (Th. 7) ; and by dividing the first of these equations by the second, member by member, we get 16* 186 GEOMETRY. P _ Inh J~~ I'n'h'' which, reduced to a proportion, gives P : p :: Inh : Vn'Jt. But, by reason of the similarity of the patallelopipo dons, we have the proportions I if : ; n : n' h : h 1 : : n : n' ; we have also the identical proportion, n : n' : : n : n' . By the multiplication of these proportions, term by term, we get, (Th. 11, B. II), Inh : l'n f h f : : n* : n n . That is, P : p : : n* : ri\ By treating in the same manner the three proportions, I : V : : h : h' n : n r : : h : h r h : h' : : h : h f , we should obtain the proportion P : p :: h* : h"; and, by a like process, the three proportions, h : h' : : I : V n : n' : : I : V I : V :: I : I', will give us the proportion P : p :: I* : V\ Hence the theorem ; similar parallelopipedons are to each other, etc. THEOREM X. The two triangular prisms into which any parallelopipedon is divided, by a plane passing through its opposite diagonal edges, are equivalent. Let ABCD — F be a parallelopipedon, and through the diagonal edges, BF and DH, pass the plane BH. divi- ding the parallelopipedon into the two triangular prisms. BOOK VII. 187 ABB-E and BCB—G ; then we are to prove that these prisms an equivalent. Letus divide the diagonal, BB, in which the se- cant plane intersects the base of the parallelopipedon, into three equal parts, a and c being the points of division. In the base, AB CD, con- st ruct the complementary paral- lelograms, aC and a A, and in the parallelogram, badD, construct the complementary parallelograms, cd and cb, and conceive these, to- gether with the parallelograms, Ba, ac, cB, to be the bases of smaller parallelopipedons, having their lateral faces parallel to the lateral faces of, and their altitude equal to the altitude of, the given parallelopipedon, A Gr. Now it is evident that the triangular prism, BCD — Gr, is composed of the parallelopipedons on the bases, a and cd, and the triangular prisms, on the side of the secant plane with this prism, into which this plane divides the parallelopipedons on the bases, Ba, ac, and cB. The triangular prism, ABB — E, is also composed of the par- allelopipedons on the bases, Aa and be, together with the triangular prisms on the side of the secant plane with this prism, into which this plane divides the parallelopip- edons on the bases, Ba, ac, and cD. But the parallelograms, a and a A, being complement- ary, are equivalent, (Th. 31, B. I) ; and for the same reason the parallelograms, cd and cb, are equivalent ; and since parallelopipedons on equivalent bases and of equal altitudes, are equivalent, (Cor., Th. 7), we have the sum of parallelopipedons on bases a and cd, equivalent to the sum of parallelopipedons on the bases, aA and cb. Hence, the triangular prisms, ABB — E and BOB — Gr, 188 GEOMETRY. differ in volume only by the difference which may exist between the sums of the triangular prisms on the two Bides of the secant plane into which this plane divides the parallelopipedons on the bases, Ba, ac, and cd. Now, if the number of equal parts into which the diag- onal is divided, be indefinitely multiplied, it still holds true that the triangular prisms, ABB — B and BOB — 6r, differ in volume only by the difference between the sums of the triangular prisms on the two sides of the secaLt plane into which this plane divides the parallelopipedons constructed on the bases whose diagonals are the equal portions of the diagonal, BB. But in this case the sum of these parallelopipedons themselves becomes an indefi- nitely small part of the whole parallelopipedon, A Cr, and the difference between the parts of an indefinitely small quantity must itself be indefinitely small, or less than any assignable quantity. Therefore, the triangular prisms, ABB — B and BOB — Gr, differ in volume by less than any assignable volume, and are consequently equiv- alent. Hence the theorem ; the two triangular prisms into which, etc. Cor. 1. Any triangular prism, as ABB — B, is one half the parallelopipedon having the same triedral angle, A, and the same edges, AB, AB, and AB. Cor. 2. Since the volume of a parallelopipedon is meas- ured by the product of its base and altitude, and the tri- angular prisms into which it is divided by the diagonal plane, have bases equivalent to one half the base of the parallelopipedon, and the same altitude, it follows that, the volume of a triangular prism is measured by the product of its base and altitude. The above demonstration is less direct, but is thought to be more simple, than that generally found in authors, and which is here given aa a BOOK VII 189 Second Demonstration Let ABCD — F be a parallelo- pipedon, divided by the diagonal plane, BH, passing through the edges, BF and BR; then we are to prove that the triangular prisms, ABB—E and BQD—G, thus formed, are equivalent. Through the points B and F, pass planes perpendicular to the edge, BF, and produce the late- ral faces of the parallelopipedon to intersect the plane through B ; then the sections Bcda and Fghe are equal parallelograms. For, since the cutting planea are both perpendicular to BF, they are parallel, (Th. 10, B. VI) ; and because the opposite faces of a parallelo- pipedon are in parallel planes, (Th. 2), and the intersec- tions of two parallel planes by a third plane are parallel, (Th. 9, B. VI), the sections, Bcda and Fghe, are equal parallelograms, and may be taken as the bases of the right parallelopipedon, Bcda — h. But the diagonal plane divides the right parallelopipedon into the two equal tri- angular prisms, aBd — e and Bed — g, (Th. 1). We will now compare the right prism with the oblique triangular prism on the same side of the diagonal plane. The volume ABD — e is common to the two prisms, ABD — F xnd aBd — e ; and the volume eFh — F, which, added to this common part, forms the oblique triangular prism, is equal to the volume aBd — A, which, added to the common part, forms the right triangular prism. For, since ABFF and aBFe are parallelograms, AF = ae, and taking away the common part Ae, we have aA—eF; and since BFHD and BFhd are parallelograms, we have DE = dh ; and from these equals taking away the common part. Dh, we have dD = hE. Now, if the volume eFh — H 190 GEOMETRY. be applied to the volume aBd — B, the base eFh falling on the equal base aBd, the edges eE and hH will fall upon aA and dB respectively, because they are perpen- dicular to the base aBd, (Cor. 2, Th. 3, B. VI), and the point E will fall upon the point A, and the point H upon the point B ; hence the volume eFh — H exactly coincides with the volume aBd — B, and the oblique triangular prisrr. ABB — E is equivalent to the right triangular prism aBd — e. In the same manner, it may be proved that the oblique triangular prism, BQBOr, is equivalent to the right tri- angular prism, Bcdg. The oblique triangular prism on either side of the diagonal plane is, therefore, equivalent to the corresponding right triangular prism ; and, as the two right triangular prisms are equal, the oblique trian- gular prisms are equivalent. Hence the theorem ; the two triangular prisms, etc. THEOREM XI. The volume of any prism whatever is measured by the prod* net of the area of its base and altitude. For, by passing planes through the homologous diag- onals of the upper and lower bases of the prism, it will be divided into a number of triangular prisms, each of which is measured by the product of the area of its base and altitude. Now, as these triangular prisms all have, for their common altitude, the altitude of the given prism, when we add the measures of the triangular prisms, to get that of the whole prism, we shall have, for this measure, the common altitude multiplied by the sum of the areas of the bases of the triangular prisms : that is, the product of the area of the polygonal base and the altitude of the prism. Hence the theorem ; the volume of any prism, etc. Cor. If A denote the area of the base, and H the alti» BOOK VII. 191 hide of a prism, its volume will be expressed by A X 11. Calling this volume V, we have Denoting by A', W, and V, in order, the area of the base, altitude, and volume of another prism, we have V = A 1 x H f . Dividing the first of these equations by the second, u ember by member, we have YL A x H V ' A' x R' y which gives the proportion, V : V : : A x H : A 1 x H'. If the bases are equivalent, this proportion becomes V : V : : H : H' ; anl if the altitudes are equal, it reduces to V : V : : A : A'. Hence, prisms of equivalent bases are to each other as their altitudes; and prisms of equal altitudes are to each other as their bases. THEOREM XII. A plane passed through a pyramid parallel to its base, divides its edges and altitude proportionally, and makes a section, which is a polygon similar to the base. Let ABODE — V be any pyramid, whose base is in the plane, MN, and vertex in the parallel plane, mn ; and let a plane be passed through the pyramid, parallel to its base, cutting its edges at the points, a b, c } d, e, and the altitude, EF, at the point I. By joining the points, a, J, c, etc., we have the polygon formed by the intersection of the plane and the sides of the pyramid. Eow, we are to prove that the edges, VA, VB, etc., and the altitude, FE, are divided proportionally at the points, a, b, etc., and I; and that the polygon, a, b, c, d, e, is similar to the base of the pyramid. 192 GEOMETRY. bince the cutting plane is parallel to the base of the pyramid, ab is parallel to AB, (Th. 9, B. VI) ; for the same reason, bo is parallel to BO, cd to OB, etc. Now, in the triangle VAB, because ab is parallel to the base AB, we have, (Th. 17, B. II), the proportion, VA : Va : : VB : Vb. In like manner, it may be shown that VB : Vb : : VO : Vc, and so on for the other lateral edges of the pyramid. F being the point in which the perpendicular from F pierces tbe plane mn, and I the point in which the parallel secant plane cuts the perpendicular, if we join the points F and V, and also the points I and e by straight lines, we have in the triangle FFV, the line le parallel to the base FV; hence the proportion VF : Ve :: FF : Fl Therefore, the plane passed through the pyramid par- allel to its base, divides the altitude into parts which have BOOK VII. 193 to each other the same ratio as the parts into which it divides the edges. Again, since ab is parallel to AB, and be to BO, the angle abc is equal to the angle ABC, (Th. 17, B. VI.) ; in the same manner we may show that each angle in the polygon, abode, is equal to the corresponding angle in the polygon, ABODE-, therefore these polygons are mutually equiangular. But, because the triangles VBA and Vba are similar, their homologous sides give the proportion Vb : VB :: ah : AB; and because the triangles Vbc and VBO are similar, we also have the proportion Vb : VB :: be : BO. Since the first couplets in these two proportions are the same, the second couplets are proportional, and give ab : AB : : be : BO. By a like process, we can prove that be : BO :: ed : OD, and that cd : CD :: de : DE, and so on, for the other homologous sides of the two polygons. Hence, the two polygons are not only mutually equi- angular, but the sides about the equal angles taken in the same order are proportional, and the polygons are there- fore similar, (Def. 16, B. II). Hence the theorem; a plane passed through a pyramid, etc. Cor, 1. Since the areas of similar polygons are to each other as the squares of their homologous sides, (Th. 22, B. H), we have area abode : area ABODE i ab* i AB*. But, ab : AB :: Va : VA :: Fl : FE; hence, ab* : AB 2 :: jf : FE*: therefore, area abode : area ABODE : Fl* : FE*. 17 N 194 GEOMETRY. That is, the area of a section parallel to the base of a pyramid, is to the area of the base, as the square of the perpendicular distance from the vertex of the pyramid to the section, is to the square of the altitude of the pyramid. Cor. 2. Let V— ABODE and X—RST be two pyra- mids, having their bases in the plane MN, and their ver- tices in the parallel plane mn ; and suppose a plane to be passed through the two pyramids parallel to the common plane of their bases, making in the one the section abode, and in the other the section rsL Kow,area ABODE: area abode : :AB 2 : ab\ (Th.22,B.II), and " EST: " rst ::~M 2 :r~s~\ But, AB : ab : : VB : Vb, and BS : rs : : XR : Xr. Because the plane which makes the sections is parallel to the planes MN and mn, we have, (Th. 11, B. YI), VB : Vb :: XR : Xr; therefore, (Cor. 2, Th. 6, B. II), AB :ab::RS: rs. By squaring, AB 2 : ab 2 : RS 2 : rs 2 ; hence, area ABODE : area abode : : area RST : area* rst. That is, if two pyramids having equal altitudes, and their bases in the same plane, be cut by a plane parallel to the com* mon plane of their bases, the areas of the sections will be proportional to the areas of the bases ; and if the bases are equivalent, the sections will also be equivalent. THEOREM XIII. If two triangular pyramids have equivalent bases and equal altitudes, they are equal in volume. Let V— ABO and v—abc be two triangular pyramids, having the equivalent bases, ABO and abc, and let the altitude of each be equal to OX; then will these two pyramids be equivalent. BOOK VII 195 -i i t/ AFT"'? 1 // y 1/ -i'H m // /A /I F ~^f ^B Place tlie bases of the pyramids on the same plane, with their vertices in the same direction, and divide the altitude into any number of equal parts. Through the points of division pass planes^ parallel to the plane of the bases ; the corresponding sections made in the pyramids by these planes are equivalent, (Th. 12, Cor. 2) ; that is, the triangle DJEF is equivalent to the triangle def, the triangle Or HI to the triangle glii, etc. JSTow, let triangular prisms be constructed on the tri- angles ABO, DEF, etc., of the pyramid V— ABO, these prisms having their lateral edges parallel to the edge, VO, of the pyramid, and the equal parts of the altitude, OX, for their altitudes. Portions of these prisms will be exterior to the pyramid V — ABO, and the sum of their volumes will exceed the volume of the pyramid. On the bases clef, ghi, etc., in the other pyramid, con- struct interior prisms, as represented in the figure, their lateral edges being parallel to vc, and their alti- tudes also the equal parts of the altitude, OX. Portions of the pyramid, v — dbc, will be exterior to these prisms, 196 GEOMETRY. and the volume of the pyramid will exceed the sum of the volumes of the prisms. Since the sum of the exterior prisms, constructed in connection with the pyramid V — ABO, is greater than the pyramid, and the sum of the interior prisms, con- structed in connection with the pyramid v — abc, is less than this pyramid, it follows that the difference of these eums is greater than the difference of the pyramids them- selves. But the second exterior prism, or that on the base DEF, is equivalent to the first interior prism, or that on the base def, and the third exterior prism is equivalent to the second interior prism, (Th. 10, Cor. 2), and so on. That is, beginning with the second prism from the base of the pyramid, V — ABO, and taking these prisms in order towards the vertex of the pyramid, and comparing them with the prisms in the pyramid, v — abc, beginning with the lowest, and taking them in order toward the vertex of this pyramid, we find that to each exterior prism of the pyramid, V — ABO, exclusive of the first or lowest, there is a corresponding equivalent interior prism in the pyramid, v — abc. Hence the prism, ABODFF, is the difference between the sum of the prisms constructed in connection with the pyramid, V— ABO, and the sum of the interior prisms constructed in the pyramid, v — abc. But the, first sum being a volume greater than the pyramid, V— ABO, and the second sum a volume less than the pyramid, v — abc, it follows that the volumes of the pyramids differ by less than the prism, ABOBEF. Now, however great the number of equal parts into which the altitude, OX, be divided, and the correspond- ing number of prisms constructed in connection with each pyramid, it would still be true that the difference between the volumes of the pyramids would be less than the volume of the lowest prism of the pyramid V— ABO; Dutwhenwe make the number of equal parts into which BOOK VII. 19T the altitude is divided indefinitely great, the vok me ot this prism becomes indefinitely small : that is, the differ- ence between the volumes of the pyramids is less thau an indefinitely small volume ; or, in other words, there is no assignable difference between the two pyramids, and they are, therefore, equivalent. Ilence the theorem ; if two triangular pyramids, etc. THEOREM XIV. Any triangular pyramid is one third of the triangul xr prism having the same base and equal altitude. Let F — ABO be a triangular pyramid, and througa F pass a plane parallel to the plane of the base, ABO. Iu this plane, through F, construct the triangle, FDE, having its sides, FD, E DE, and EF, parallel and equal to BO, /^T ~~7\ OA, and AB, respectively. The tri- / 0>\ / angle, FDE, may be taken as the / /h\ upper base of a triangular prism of 'I 'V which the lower base is ABO. "\. ~\ / Now, this triangular prism is com- b posed of the given triangular pyramid, F — ABO, and of the quadrangular p/ramid, F — A ODE. This last pyramid may be divided by a plane through the three points, 0, F, and F, into tha two triangular pyra- mids, F—DEO and F—AOE. But the pyramid, J 7 — DEO, may be regarded as h^ing the triangle, EFD, equal to the triangle, ABO, for its base, and the point, 0, for its vertex. The two pyramids, F — ABO and — DEF, have equal bases and equal altitudes ; they are therefore equivalent, (Th. 13). Again, the two pyramids, F — DEO and F — AOE } have a common vertex, and equivalent bases in the same plane, and they are also equivalent. There- fore, the triangular prism, ABODEF, is composed of n* 198 GEOMETRY. three *tpiv«dent triangular pyramids, one of which is the given triangular pyramid, F — ABC. Hetice the theorem ; any triangular pyramid is one third of the triangular prism, etc. Cor. The volume of the triangular prism heing meas- ured hy the product of its hase and altitude, the volume of a triangular pyramid is measured by one third of the product of its base and altitude. THEOREM XV. The volume of any pyramid whatever is measured by one third of the product of its base and altitude. Let V — ABCDE be any pyramid ; then will its volume be measured by one third of the product of its base and altitude. In the base of the pyramid, draw the diagonals, AD and A C, and through its vertex and these diagonals, pass planes, thus dividing the pyramid into a number of triangular pyramids having the common vertex F", and the altitude of the given pyramid for their common altitude. Now, each of these triangular pyra- mids is measured by one third of the product of its base and altitude, (Cor., Th. 14), and their sum, which constitutes the polygonal pyramid, is therefore measured by one third of the product of the sum of the trian- gular bases and the common altitude ; but the sum of the triangular bases constitutes the polygonal base, ABCDE. Hence the theorem ; the volume of any pyramid what- ever, etc. Cor. 1. Denote, by B, ff, and V, respectively, the base, altitude, and volume of one pyramid, and by B' f IF, and BOOK VII. 199 P, the base, altitude, and volume of another ; then we Bhall have and V = i'B f x W. Dividing the first of these equations by the second, member by member, we have V _ B x E V B f x E' 9 which, in the form of a proportion, gives V : V : : B X E : B' x E>. From this proportion we deduce the following conse- quences : 1st. Pyramids are to each other as the products of their bases and altitudes. 2d. Pyramids having equivalent bases are to each other a* their altitudes. 3d. Pyramids having equal altitudes are to each other as their bases. Cor. 2. Since a prism is measured by the product of its base and altitude, and a pyramid by one third of the product of its base and altitude, we conclude that any pyramid is one third of a prism having an equivalent base and equal altitude THEOREM XVI. The volume of the frustum of a pyramid is equivalent to the sum of the volumes of three pyramids, each of which has an altitude equal to that of the frustum, and whose bases are, respectively, the lower base of the, frustum, the upper base of the frustum, and a mean proportional between these bases. Let V— ABODE and X—RST be two pyramids, the one polygonal and the other triangular, having equiva- lent bases and equal altitudes ; and let their bases be placed on the plane MN, their vertices falling on the parallel plane ran. Pass through the pyramids a plane 200 GEOMETRY. parallel to the common plane of their bases, cutting o .it the sections abode and rst ; these sections are equivalent, (Th. 12, Cor. 2), and the pyramids, V— abode and X—rst, are equivalent, (Th. 13). Now, since the pyramids, V— ABODE and X—BST, are equivalent, if from the first we take the pyramid, V — abode, and from the second, the pyramid, X — rst, the remainders, or the frusta, ABODE— a and BST—r, will be equivalent. If, then, we prove the theorem in the case of the frus- tum of a triangular pyramid, it will be proved for the frustum of any pyramid whatever. Let ABO — D be the frustum of a triangular pyramid. Through the points D, B, and 0, pass a plane, and through the points D, O, and E, pass another, thus dividing the frustum into three triangular pyra- mids, viz., D—ABO, O—DEF, and D—BEO. Now, the first of these has, for its BOOK VII. 201 base, the lower base of the frustum, and for its altitude the altitude of the frustum, since its vertex is in the upper base ; the second has, for its base, the upper base of the frustum, and for its altitude the altitude of the frustum, since its vertex is in the lower base. Hence, these are two of the three pyramids required by the enunciation of the theorem ; and we have now only to prove that the third is equivalent to one having, for its basa, a mean proportional between the bases of the frus- tum, and an altitude equal to that of the frustum. In the face ABED, draw HB parallel to BE, and draw HE and HO. The two pyramids, D — BEO and H—BEO, are equivalent, since they have a common base and equal altitudes, their vertices being in the line BH, which is parallel to the plane of their common base, (Th. 7, B. VI). "We may, therefore, substitute the pyramid, H—BEO, for the pyramid, D—BEC. But the triangle, BOH, may be taken as the base, and E as the vertex of this new pyramid ; hence, it has the required altitude, and we must now prove that it has the required base. The triangles, ABO and HBO, have a common vertex, and their bases in the same line ; hence, (Th. 16, B. II), A ABO : A HBO n AB : HB :: AB : BE. (1) In the triangles, BEE and HBO, \_E^\_B, and BE=HB; hence, if BEE be applied to HBO, L E fall- ing on L B, an( i tne s id e BE on HB, the point B will fall on H, and the triangles, in this position, will have a common vertex, H, and their bases in the same line ; hence, A HBO : A BEE : : BO : EF. (2) But, because the triangles, ABO and BEE, are similar, we nave AB . BE ii BO i EF. (3) From proportions (1), (2), and (3), we have, (Th. 6, b n), 202 GEOMETRY. A ABO : A HBO : : A HBO : A DEF; that is, the base, HBO, is a mean proportional between the lower and upper bases of the frustum. Hence the theorem ; the volume of the frustum of a pyra- mid, etc, THEOREM XVII. The convex surface of any right 'pyramid is measured by the perimeter of its base, multiplied by one half its slant height. Let S—ABOBEF be a right pyramid, s of which SH is the slant height ; then will k its convex surface have, for its measure, /M ±SH(AB + BO+OB + BE+EF+FA). Mm Since the base is a regular polygon, and /e/LjJa\ the perpendicular, drawn to its plane from ^1/ I \ \\ S, passes through its center, the edges, \ / i / SA, SB, SO, etc., are equal, (Th. 4, B. YI), ahb and the triangles SAB, SBC, etc., are equal, and isosceles, each having an altitude equal to SH. Now, AB x \SH measures the area of the triangle, SAB ; and BO X %SH measures the area of the triangle, SBO; and so on, for the other triangular faces of the pyramid. By the addition of these different measures, we get %SH(AB + BO + OB + BE + EF + FA), as the measure of the total convex surface of the pyramid. Hence the theorem ; the convex surface of any right pyramid, etc, THEOREM XVIII. The convex surface of the frustum of any right pyramid is measured by the sum of the perimeters of the two bases, mul- tiplied by one half the slant height of the frustum. Let ABOBEF — d be the frustum of a right pyramid; then will its convex surface be measured by $Ilh(AB-+ BC4-CD+DE-\ EF+FA+ab+bc+cd+de+ef+fa), book vir 203 For, the upper base, abcdef, of the frustum is a section of a pyramid by a plane parallel to the lower base, (Def. 14), and is, therefore, similar to the lower base, (Th. 12). But the lower base is a regular polygon, (Def. 12); hence, the up- per base is also a regular polygon, of the same name; and as ab and AB are intersections of a face of the pyramid by two parallel planes, A H B they are parallel. For the same reason, be is parallel to BO, cd to OB, etc., and the lateral faces of the frustum are all equal trapezoids, each having an altitude equal to Hh, the slant height of the frustum. The trapezoid ABba has, for its measure, \Hh(AB+ab), (Th. 34, Book I) ; the trapezoid BOcb has, for its meas- ure, \Hh{BO + be), and so on, for the other lateral faces of the frustum. Adding all these measures, we find, for their sum, which is the whole convex surface of the frustum, IHh {AB+ B C+ CD+DE+EF+ FA+ab+bc+cd+de+ef+fa). Hence the theorem ; the convex surface of the frustum. THEOREM XIX. The volumes of similar triangular prisms are to each other as the cubes constructed on their homologous edges. Let ABO— F ana abc— /be two similar triangular prisms ; then will their vol- umes be to each other as the cubes, whose edges are the homologous edges 204 GEOMETRY. AB and a b, or as the cubes, whose edges are the homol- ogous edges BE and be, etc. Since the prisms are similar, the solid angles, whose vertices are B and b, are equal ; and the smaller prism, when so applied to the larger that these solid angles coincide, will take, within the larger, the position represented by the dotted lines. In this position of the prisms, draw EH perpendicular to the plane of the base ABO, and join the foot of the perpen- dicular to the point B, and in the triangle BEIT draw, through e, the line eh, parallel to EH', then will EH represent the altitude of the larger prism, and eh that of the smaller. Now, as the bases ABO and aBc, are homologous faces, they are similar, and we have, (Th. 20, Book II), A ABO : A aBc : : AB* CoE* ( 1 ) But the A's BEH and Beh are equiangular, and there fore similar, and their homologous sides give the propor- tion BE : Be :: EH : eh (2) and from the homologous sides of the similar faces, ABED and aBed, we also have BE : Be :: AB : aB (3) Proportions (2) and (3 ), having an antecedent and con sequent the same in both, we have, (Th. 6, B. II), EH : eh w AB \ aB (4) By the multiplication of proportions (1) and (•*), term Dy term, we get A ABO X EH : A aBc X eu : : AB* : aB* But A ABO X EH measures the volume of the larger prism, and A aBc x eh measures the volume of the smaller. Hence the theorem; the volumes of similar triangular prisms^ etc. BOOK VII. 205 Cor. 1. The volumes of two similar prisms having any bases whatever, are to each other as the cubes constructed on their homologous edges. For, if planes be passed through any one of the late<-al edges, and the several diagonal edges, of one of these prisms, this prism will be divided into a number of smaller triangular prisms. Taking the homologous edge of the other prism, and passing planes through it and the seve- ral diagonal edges, this prism will also be divided into the same number of smaller triangular prisms, similar to those of the first, each to each, and similarly placed. ]STow, the similar smaller prisms, being triangular, are to each other as the cubes of their homologous edges ; and being like parts of the larger prisms, it follows that the larger prisms are to each other as the cubes of the homologous edges of any two similar smaller prisms. But the homologous edges of the similar smaller prisms are to each other as the homologous edges of the given prisms ; hence we conclude that the given prisms are to each other as the cubes of their homologous edges. Cor. 2. The volumes of two similar pyramids having any bases whatever, are to each other as the cubes constructed on their homologous edges. For, since the pyramids are similar, their bases are similar polygons ; and upon them, as bases, two similar prisms may be constructed, having for their altitudes, the altitudes of their respective pyramids, and their lateral edges parallel to any two homologous lateral edges of the pyramids. Now, these similar prisms are to each other as the cubes of their homologous edges, which may be taken as the homologous sides of their bases, or as their lateral edges, which were taken equal and parallel to any two arbitrarily assumed homologous lateral edges of the two pyramids ; hence the pyramids which are thirds of their respective prisms, are to each other as the cubes constructed on any two homologous edges. 18 206 GEOMETRY. Cor. 3. The volumes of any two similar polyedrons arz U each other as the cubes constructed on their homologous edges. "For, by passing planes through, the vertices of the homologous solid angles of such polyedrons, they may both be divided into the same number of triangular pyramids, those of the one similar to those of the other, each to each, and similarly placed. Now, any two of these similar triangular pyramids are to each other as the cubes of their homologous edges ; and being like parts of their respective polyedrons, it follows that the polyedrons are to each other as the cubes of the homologous edges of any two of the similar tri- angular pyramids into which they may be divided. But the homologous edges of the similar triangular pyramids are to each other as the homologous edges of the poly- edrons ; hence the polyedrons are to each other as the cubes of their homologous edges. THEOREM XX. The convex surface of the frustum of a cone is measured by the product of the slant height and one half the sum of the circumferences of the bases of the frustum. Let ABCD — abed be the frustum of a cone ; then will its convex surface be , , . (circ. 00 + circ. oc) measured by Aa x i — '-, in which the expression, circ. 00, de- notes the circumference of the circle of which 00 is the radius. Inscribe in the lower base of the frustum, a regu- lar polygon having any number of sides, and in the upper base a similar polygon, having its sides parallel to those of the polygon in the lower base. These polygons BOOK VII. 207 may be taken as the bases of the Irustum of a right pyramid inscribed in the frustum of the cone. Now, however great the number of sides of the in- scribed polygons, the convex surface of the frustum of the pyramid is measured by its slant height multiplied by one half the sum of the perimeters of its two bases, (Th. 18) ; but when we reach the limit, by making the number of sides of the polygon indefinitely great, the slant height, perimeters of the bases, and convex surfaco of the frustum of the pyramid become, severally, the slant height, circumferences of the bases, and convex sur- face of the frustum of the cone. Hence the theorem ; the convex surface of the frustum, etc. Cor. 1. If we make oc = OC, and, consequently, circ. oc = circ. OC, the frustum of the cone becomes, a cylin- der, and the half sum of the circumferences of the bases becomes the circumference of either base of the cylinder, and the slant height of the frustum, the altitude of the cylinder. Hence, the convex surface of a cylinder is meas- ured by the circumference of the base multiplied by the alti- tude of the cylinder. Cor. 2. If we make oc = 0, the frustum of the cone becomes a cone. Hence, the convex surface of a cone is measured by the circumference of the base multiplied by one half the slant height of the cone. Cor. 3. If through E, the middle point of Co, the line Ff be drawn parallel to Oo, and Em perpendicular to Oo, the line oc being produced, to meet Ff at/, we have, because the A's EFC and Efc are equal, Em = O0 + M . 2S if we multiply both members of this equation by 2*, we have 208 GEOMETRY. that is, circ. Em is equal to one half the sum of the cir cumferences of the two hases of the frustum. Hence, the convex surface of the frustum of a cone is measured by the. circumference of the section made by a plane half way between the two bases, and parallel to them, multiplied by the slant height of the frustum. Cor. 4. If the trapezoid, OCeo, he revolved ahout Oo as an axis, the inclined side, Cc, will generate the con- vex surface of the frustum of a cone, of which the slant height is Cc, and the circumferences of the hases are circ. OC and circ. oc. Hence, if a trapezoid, one of whose sides is perpendicular to the two parallel sides, be revolved about the perpendicular side as an axis, it will generate the frustum of a cone, the inclined side opposite the axis generating tho convex surface, and the parallel sides the bases of the frustum. THEOREM XXI. The volume of a cone is measured by the area of its base multiplied by one third of its altitude. Let V— ABC, etc., he a cone; then will its volume he measured hy area ABC, etc., multiplied hy iVO. Inscribe, in the base of the cone, any regular polygon, as ABCDEF, which may he taken as the base of a right pyra- mid, of which V is the vertex. The volume of this inscribed pyramid will have, for its measure, (Th. 15), polygon ABCDEF x iVO. Now, however great the number of sides of the pol f- gon inscribed in the base of the cone, it will still ho d true that the pyramid of which it is the base, and who-e vertex is V, will be measured by the area of the poly- gon, multiplied by one third of VO; but when we reach the limit, by making the number of sides indefi- BOOK VII. 209 uitely great, the polygon becomes the ^cle in which it is inscribed, and the pyramid become +he cone. Hence the theorem ; the volume of a cone, etc. Cor. 1. If R denote the radius of the base of a cone, and ff its altitude, or axis, its volume will be expressed by iff x *i2 2 ; hence, if V and V designate the volume ~> of two cones, of which R and R' are the radii of the bases, and H and ff' the altitudes, we have V: V :: J#x *R 2 : \H' x «R' 7 ;: Bx«& : ff' X «R'\ From this proportion we conclude, First. That cones having equal altitudes are to each other as their bases. Second. That cones having equal bases are to each other as their altitudes. Cor. 2. Retaining the notation above, we have YL El ^1 m V " H x R*' l J and, if the two cones are similar, ff : ff ! :: R : i2'; ff 1 R' , ff" R'* or > ~ff = R'> hence >ir = -&' By substituting for the factors, in the second member of eq. ( 1 ), their values successively, and resolving into a proportion, we get V : V :: R* : R"; and V : V :: ff 3 : H". Hence, similar cones are to each other as the cubes of the radii of their bases, and also as the cubes of their altitudes. Cor. 3. A cone is equivalent to a pyramid having an equiv> alent base and an equal altitude. 18* 210 GEOMETRY. THEOREM XXII. 'The volume of the frustum of a cone is equivalent to the sum of the volumes of three cones, having for their common altitude the altitude of the frustum, and for their several bases, the bases of the frustum and a mean proportional be- tween them. Let ABQD — abed be the frustum of a cone ; then will its volume be equiva- lent to the sum of the volumes, having Oo for their common altitude, and for their bases, the circles of which, OO, oc, and a mean proportional between OO and oc, are the respective radii. Inscribe in the lower base of the frus- tum any regular polygon, and in the upper base a similar polygon, having its sides parallel to those of the first. These polygons may be taken as the bases of the frustum of a right pyra- mid inscribed in the frustum of the cone. The volume of the frustum of the pyramid is equiva- lent to the sum of the volumes of three pyramids, having for their common altitude the altitude of the frustum, and for their several bases the bases of the frustum, and a mean proportional between them, (Th. 16). Eow, however great the number of sides of the poly- gons inscribed in the bases of the frustum of the cone, this measure for the volume of the frustum of the pyr* mid, of which they are the bases, still holds true ; biu when we reach the limit, by making the number of the sides of the polygon indefinitely great, the polygons be- come the circles, the frustum of the pyramid becomes the frustum of the cone, and the three partial pyramids, whose sum is equivalent to the frustum of the pyramid, become three partial cones, whose sum is equivalent to the frustum of the cone. BOOK VII. 211 Hence the theorem ; the volume of the frustum of a cone, etc. Cor. 1. Let R denote the radius of the lower base, R f that of the upper base, and IT the altitude of the frustum of a cone; then will its volume be measured, (Th. 21), by IE x *i2 2 + iH x «R n + iffx *R x R', since *R x R' expresses the area of a circle which is a mean proportional between the two circles, whose radii arc R andi2'. Now, if the bases of the frustum become equal, or 7? — R ', the frustum becomes a cylinder, and each of the last two terms in the above expression for the volume of the frustum of a cone will be equal to the first ; hence, the volume of a cylinder, of which iZ"is the altitude, and It the radius of the base, is measured by H X *R 2 . Therefore, the volume of a cylinder is measured by the area of its base multiplied by its altitude. Cor. 2. By a process in all respects similar to that pur- sued in the case of cones, it may be shown that similar cylinders are to each other as the cubes of the radii of their bases, and also as the cubes of their altitudes. Cor. 3. A cylinder is equivalent to a prism having an equivalent base and an equal altitude. THEOREM XXIII. If a plane be passed through a sphere, the section will be a circle. Let be the center of a sphere through which a plane is passed, making the section AmBn ; then will this section be a circle. From let fall the perpendic- ular Oo upon the secant plane, and draw the radii OA, OB, and Om, to different points in the intersection of the plane with the surface of the sphere. Now, 212 GEOMETRY. the oblique lines OA, OB, Om, are all equal, being ladii of the sphere; they therefore meet the plane at equal dis tances from the foot of the perpendicular Oo, (Cor., Th. 4, B.VI); hence oA, oB, om, etc., are equal: that is, all the points in the intersection of the plane with the surface of the sphere are equally distant from the point 0. This intersection is therefore the circumference of a circle of which o is the center. Hence the theorem; if a plane be passed thrcugh a sphere, etc. Cor. 1. Since AB, the diameter of the section, is a chord of the sphere, it is less than the diameter of the sphere ; except when the plane of the section passes through the center of the sphere, and then its diameter becomes the diameter of the sphere. Hence, 1. All great circles of a sphere are equal. 2. Of two small circles of a sphere, that is the greater whose plane is the less distant from the center of the sphere. 3. All the small circles of a sphere whose planes are at the same distance from the center, are equal. Cor. 2. Since the planes of all great circles of a sphere pass through its center, the intersection of two great circles will be both a diameter of the sphere and a com- mon diameter of the two circles. Hence, two great circles of a sphere bisect each other. Cor. 3. A great circle divides the volume of a sphere, and also its surface, equally. For, the two parts into which a sphere is divided by any of its great circles, on being applied the one to the other, will exactly coincide, otherwise all the points in their convex surfaces would not be equally distant from the center. Cor. 4. The radius of the sphere which is perpendicular to the plane of a small circle, passes through the center of th* circ le. BOOK VII. 213 Cor. 5. A plane passing through the extremity of a radius of a sphere, and perpendicular to it, is tangent to the sphere. For, if the plane intersect the sphere, the section is a circle, and all the lines drawn from the center of the sphere to points in the circumference are radii of the sphere, and are therefore equal to the radius which is per- pendicular to the plane, which is impossible, (Cor. 1, Th. 3, B. VI). Hence the plane does not intersect the sphere, and has no point in its surface except the extremity of the perpendicular radius. The plane is therefore tangent to the sphere by Def 22. THEOREM XXIV. If the line drawn through the center and vertices of two opposite angles of a regular polygon of an even number of sides, be taken as an axis of revolution, the perimeter of either semi-polygon thus formed will generate a surface whose measure is the axis multiplied by the circumference of the inscribed circle. Let ABODEF be a semi-polygon cut off from a regular polygon of an even number of sides by drawing the line AF through the center 0, and the vertices A and F, of two opposite angles of the poly- gon ; then will the surface generated by the perimeter of this semi-polygon re- volving about AF as an axis, be meas- ured by AF X circumference of the in- scribed circle. From m, the middle point, and the extremities B and C of the side B 0, draw mn, BK, and OL, perpendicular to AF; join also m and 0, and draw BIT perpendicular to QL. The surface of the frustum of the cone generated by the trapezoid BKLO, has for its measure circ. mn X BO, (Cor. 3, Th. 20). Since mO is perpendicular to BO, and mn to BIT, the two A's, BOH and mnO, are similar, and their homologous sides give the proportion 214 GEOMETRY. mn : mO :: BE (= KL) : BO and as circumferences are to each other as their radii, we have circ. mn : circ. mO :: KL : BO Hence, circ. mn x BQ = circ. mO X KL. But mO is the radius of the circle inscribed in tne polygon. Hence, the surface generated by BQ during the revolution of the semi-polygon, is measured by the cir- cumference of the inscribed circle multiplied by KL, the part of the axis included between the two perpendicu- lars let fall upon it from the extremities B and 0. The surface generated by any other side of the semi-polygon will be measured, in like manner, by the circumference of the inscribed circle multiplied by the corresponding part of the axis. By adding the measures of the surfaces generated by the several sides of the semi-polygon, we get Circ. mO x (AK + KL + LN+ JSTM+ MF) for the measure of the whole surface. Hence the theorem ; if the line drawn through the cm Mr, etc. Cor. It is evident that the surface generated by any portion, as CD and DK, of the perimeter, is measured by circ. mO x LM. THEOREM XXV. The surface of a sphere is measured by the circumference of one of its great circles multiplied by its diameter. Let a sphere be generated by the revolution of the Bemi-circle, AKF, about its diameter, AF; then will the surface of the sphere be measured by Circ. AOxAF. Inscribe in the semi-circle any regular semi-polygon, and let it be revolved, with the semi-circle, about the axii BOOK VII. 215 AF; the surface generated by its perim- eter will be measured by Circ. mO x AF, (Th. 24), and this measure will hold true, how- ever great the number of sides of the in- Hj scribed semi-polygon. But as the num- ber of these sides is increased, the radius mO, of the inscribed semi-circle, increases and approaches equality with the radius, AO; and when we reach the limit, by making the number of sides indefinitely great, the radii and semi-circles become equal, and the surface generated by the perimeter of the inscribed semi-polygon becomes the surface of the sphere. Therefore, the surface of the sphere has, for its measure, Circ. A x AF. Hence the theorem ; the surface of a sphere is meas- ured, etc. Cor. 1. A zone of a sphere is measured by the circumfer- ence of a great circle of the sphere multiplied by the altitude of the zone. For, the surface generated by any portion, as CD and DE, of the perimeter of the inscribed semi-polygon has, for its measure, circ. mO X LM, (Cor. Th. 24) ; and as the number of the sides of the semi-polygon increases, LM remains the same, the radius mO alone changing, and becoming, when we reach the limit, equal to AO: hence, the surface of the zone is expressed by Circ. Ad X LM, whether the zone have two bases, or but one. Cor. 2. Let H and H r denote the altitudes of two zones of spheres, whose radii are R and R ' ; then these zones will be expressed by 2<>rR x H and 2*R ' x R r ; and if tht surfaces of the zones be denoted by Z and Z' f we have 216 GEOMETRY. Z : Z 1 : 2«RxH : 2«R' x H f i: Rx E : R' x H'> •Hence, 1. Zones in different spheres are to each other at their altitudes multiplied by the radii of the spheres. 2. Zones of equal altitudes are to each other as the radii of the spheres, 3. Zones in the same, or equal spheres, are to each other as their altitudes. Cor. 3. Let R denote the radius of a sphere; then will its diameter be expressed by 222, and the circumference of a great circle by 2irR; hence its surface will be ex pressed by 2«R x 2R = 4iri2 2 . That is, the surface of a sphere is equivalent to the area of four of its great circles. Cor. 4. The surfaces of spheres are to each other as the squares of their radii. THEOREM XXVI. Tf a triangle be revolved about either of its sides as an axis, the volume generated will be measured by one third of the prod- uct of the axis and the area of a circle, having for its radius the perpendicular let fall from the vertex of the opposite angle on the axis, or on the axis produced. First. Let the triangle ABC, in which the perpendicular from C falls on the opposite side, AB, be revolved about AB as an axis ; then will *Vol. A ABChuve, for its measure, \AB x *CD . The two A's into which A ABC is divided by the perpendicular DC, are right-angled, and during the rev- olution they will generate two cones, having for their * Vol. A ABC, cone A ADC, are abbreviations for volume gener- ated by A ABC, cone generated by A ADC; and surfaces of revolu- sien generated by lines will hereafter be denoted by like abbreviations. BOOK VII. 217 common base the circle, of which D is the radius, and for their axes the parts DA and DB, into which AB is divided. Now, *Cone a ADO is measured by \AD x *DO\ (Th. 21), and cone A BDO, by %BD x «DO % \ but these two cones compose Yol. A ABO; and by adding their measures, we have, for that of Yol. A ABO, IAD x *D0 2 + iBD x *~D(7 2 = J AB x <^DO\ Second. Let the trian- 9 gle FFGr, in which the perpendicular from G- falls on the opposite side FF produced, be revolved about FF as an axis ; then will Yol. A FFG E F_ "h have, for its measure, %EF x *GfH\ CrH being the per- pendicular on FF produced. For, in this case it is appa- rent, that Yol. A FFGr is the difference between the cone A FHCr and the cone a FHCr. The first cone has, for its measure, \FH x *GH\ and the second, for its measure, %FH x wGrJI 2 ; hence, by subtraction, we have Vol. A FFG = IEH X hGH 2 — iFH X 7t~GH 2 = IEF X 7t~GH 2 . Hence the theorem ; if a triangle be revolved about either of its sides, etc. Scholium. — If we take either of the above expressions for the meas- ure of the volume generated by the revolution of a triangle about one of its sides, for example the last, and factor it otherwise, we have \EF X *GH* = EFX $GHxU*2GH= FFx $GHX — 3 Now, EF X %GH expresses the area of the triangle EFG; and 2x X GH , one third of the circumference described by the point Q o during the revolution. The expression, \AB X rtDC 2 , maybe factored and interpreted in the * See note on the preceding page. 19 218 GEOMETRY. same manner. Hence, we conclude that the volume generated by th* revolution of a triangle about either of its sides, is measured by the area of the triangle multiplied by one third of the circumference described in the revolution by the vertex of the angle opposite the axis. THEOREM XXVII. The volume generated by the revolution of a triangle about any line lying in its plane, and passing through the vertex of one of its angles, is measured by the area of the triangle mul- tiplied by two thirds of the circumference described, in the revolution, by the middle 'point of the side opposite the vertex through which the axis passes. Let the triangle ABCbe revolved about the line AG, drawn through the vertex A, and lying in the plane of the triangle, and let HE be the perpendicu- lar let fall from H, the middle point of BO, upon the axis AG- ; then will Vol. a ABO have, for its meas ure, A ABO x § circ. HE. From the extremities of BO, let fall the perpendicu- lars BE and OB, on the axis; and from A draw AST per pendicular to BO, or BO produced, and produce OB, until it meets the axis in G-. Now, it is evident that Yol. A ABO is the difference between Yol. A AGO and Yol. A AGB. But Yol. A AGO is expressed by a AGO x J circ. (7i>;and Yol. A AGB, by A AG B x J circ. BE, (Scholium, Th. 26). Hence, Vol. A ABC = ^ AGO X I circ. CD — A AGB X i circ. BF. Substituting for areas of A's, and for circumferences, their measures, we have BOOK VII. 219 Vol. A A£C= GO X IAK X ?^2— GB x UK X ^^ o o = GCx \AKx 2 J^R—{GC—BC) X UKX ^^ o • ~GCXIAKX^^— GCXIARX^^+BCXUK*^ 1 ^ o 8 8 = GO X \AK X ^(CD — BF) + BC X $AR X — „— . o o But i?iV being drawn parallel to J.#, we have GN = CD — BF; hence, substituting this value for CD — BF, in the first term of the second member of the last equation, we have o o = GCx CNx ±AK x 2 ^ + BC x %AK x ?^^, by changing the order of factors in the first term of the second member. The homologous sides of the similar triangles, GOD and BCN, give the proportion GO : CD : : BO : ON whence, GO x ON = CD x BO Substituting this value for GO X CJSF, in the last equa- tion above, and arranging the factors as before, it becomes Vol. A ABC= BO x lAKx ^^ + BC x \AK x ^F - BO x UK x g^V BF >. a But CD + BF= 2HF; hence Vol, A ABC=--BCx \AKx —~=BCx ±AKx f .2*. J7#j o and since BO x \AK= A ABC, and § x 2«.HE - § circ. ##, this measure conforms to the enunciation. It only remains for us to consider the case in which the axis is parallel to the base BC of the triangle The 220 GEOMETKY. precedi :g demonstration will not now apply, because it supposes BO, or B produced, to intersect the axis. Let the axis AE, be parallel to the base BO, of the A ABO. From B and let fall on the axis the perpen- diculars BE and OB. Now it is plain that Vol. A ABO= cylinder rectangle BODE -f cone a ADO — cone A AEB. Substituting in second member, for cylinder and cones, their measures, we have Vol. AABO= BE x *£Z) 2 + IAD x ^OD 2 — \AE x «BE* =$DEx *7JD 2 +iDEx «OD 2 +§ADx*OD*—iAEx *BE\ But BE = OD, and ^DE + %AD = \AE. Reducing by these relations, we have Vol. A ABO= %DE x «OD 3 = %DE x \OD x 4*.OD = DEx lODx %. + EA x ' 2 Hence, A spherical segment having but one base, is equiva- lent to a sphere whose diameter is the altitude of the segment, plus one half of a cylinder having for base and altitude the base and altitude of the segment. Scholium. — "When the spherical segment has a single base, we may put the expression, \hEA + EA X — — , under a form to indicate a convenient practical rule for computing the volume of the segment. Thus, since the triangle DEO is right-angled, and 0j£= OA — EA, we have DE 2 =TDO a — ~OE 2 = O? — Ol 2 + 20A X EA—~EA 2 = 20Ax EA—~EA\ By substituting this value for DE 2 in the expression for the volume of the segment, we find UEA? + EAX?X(20AXEA — ~EA*) 2 Bsi*1Z f -f EA 2 X % [20A — EA) 2 = }rtEA 9 +U.ZEA i {20A — EA) = \hEA\EA + Q.OA — 3 EA) = lrtEA\6.0A — 2EA) = l7tEA\%OA — EA) Hence, the volume of a spherical segment, having a single base, is measured by one third of n times the square of the altitude of the seg- ment, multiplied by the difference between three times the radius of lfa& sphere and this altitude. RECAPITULATION Of some of the principles demonstrated in this and the pre- ceding Boohs. Let R denote the radius, and D the diameter of any circle or sphere, and H the altitude of a cone, or of a segment of a sphere ; then, -i*r + *&*" + **"•> BOOK VII. 229 Circuinforence of a circle == 2*JK. Surface of a sphere = 4*J2 2 , or nD\ Zone forming the base of a 1 _ ~ ^ „. segment of a sphere, / Volume or solidity of a sphere = {*B*, or £*!)•. Volume of a spherical sector = %nR 2 x JZ". Volume of a cone, of which ^ Jit is the radius of the V = \*IP x H. base ) Volume of a spherical seg-' ment, of which R' is the radius of one base, and R" the radius of the other, and whose altitude is#, If the so ir ment has but one^ . „, , ^hR'* ;,„ -, ,! = \*H* + H.—^ ; or, base, R" = zero, and the > 2 volume of the segment, J = J*JP(3i2 — H). PRACTICAL PROBLEMS. 1. The diameter of a sphere is 12 inches ; how many cubic inches does it contain? Arts. 904.78 cu. in. 2. What is the solidity of the segment of a single base that is cut from a sphere 12 inches in diameter, the altitude of the segment being 3 inches? Arts. 141.372 cu. in. 3. The surface of a sphere is 68 square feet ; what is its diameter ? Ans. D = 4.652 feet. 4. If from a sphere, whose surface is 68 square feet, a segment be cut, having a depth of two feet and a single base, what is the convex surface of the segment ? Ans. 29.229+ sq. ft. 5. What is the solidity of the sphere mentioned in the two {receding examples, and what is the solidity of the segment, having a depth of two feet, and but one base ? A ( Solidity of sphere, 52.71 cu. ft. t " " segment, 20.85 " 20 230 GEOMETRY. 6. In a sphere whose diameter is 20 feet, what is the solidity of a segment, the bases of which are on the same side of the center, the first at the distance of 3 feet from it, and the second of 5 feet; and what is the solidity of a second segment of the same sphere, whose bases are also on the same side of the center, and at distances from it, the first of 5 and the second of 7 feet ? a ( Solidity of first segment, 525.7 cu. ft US I " " second " 400.03 " 7. If the diameter of the single base of a spherical segment be 16 inches, and the altitude of the segment 4 inches, what is its solidity ? * Arts. 435.6352 cubic inches. 8. The diameter of one base of a spherical segment is 18 inches, and that of the other base 14 inches, these bases being on opposite sides of the center of the sphere, and the distance between them 9 inches ; what is the volume of the segment, and the radius of the sphere ? a ( Vol. seg., 2219.5 cubic inches. \ Bad. of sphere, 9.4027 inches. 9. The radius of a sphere is 20, the distance from the center to the greater base of a segment is 10, and the distance from the same point to the lesser base is 16 ; what is the volume of the segment, the bases being on the same side of the center? Ans. 4297.7088. 10. If the diameter of one base of a spherical segment be 20 miles, and the diameter of the other base 12 miles* and the altitude of the segment 2 miles, what is its solidity, and what is the diameter of the sphere ? * First find the radius of the sphere. Note. — The Key to this work contains full solutions to all the problems in the Geometry and Trigonometry, and the necessary diagrams for illustration. BOOK VIII. 231 BOOK VIII. PRACTICAL GEOMETRY. APPLICATION OF ALGEBRA TO GEOMETRY, AND ALSO PROPOSITIONS FOR ORIGINAL INVESTIGATION. No definite rules can be given for the algebraic solu- tion of geometrical problems. The student must, in a a great measure, depend on his own natural tact, and Lis power of making a skillful application of the geomet- rical and analytical knowledge he has thus far obtained. The known quantities of the problem should be repre- sented by the first letters of the alphabet, and the un- known by the final letters ; and the relations between these quantities must be expressed by as many inde- pendent equations as there are unknown quantities. To obtain the equations of the problem, we draw a figure, the parts of which represent the known and unknown magnitudes, and very frequently it will be found neces- sary to draw auxiliary lines, by means of which we can deduce, from the conditions enunciated, others that can be more conveniently expressed by equations. In many cases the principal difficulty consists in finding, from the relations directly given in the statement, those which are ultimately expressed by the equations of the problem. Having found these equations, they are treated by the known rules of algebra, and the values of the required magnitudes determined in terms of those given. 23:2 GEOMETRY. PROBLEM I. Given, the hypotenuse, and the sum of the other two sides of a right-angled triangle, to determine the triangle. Let ABO be the A. Put OB = y, AB = x, AO= h, and OB + AB = s. Then, by a given condition, we have x + y = s; and, x*+ y*= h\ (Th. 39, B. I). lleducing these two equations, and we have x = \s =b i^W^7; y mm \s =fc Jv^tf — * 2 . If A = 5 and * = 7, a; = 4 or 3, and y = 3 or 4. Remark. — In place of putting x to represent one side, and y the other, we might put [x -f- y) to represent the greater side, and (x — y) the less side ; then, h 2 x* + y* = -, and 2x = s, etc. PROBLEM II. (riven, the base and perpendicular of a triangle, to find the side of its inscribed square. Let ABO be the A. Put AB mm b, the base, OB — p, the perpendicular. Draw EF parallel to AB, and suppose it equal to EG, a side of the required square ; and put EF = x. Then, by similar A's, we have 01 : EF : : OB : AB. That is, Hence, p — x P b. bp — bx = px ; or, x m t f . r - b + p That is, the side of the inscribed square is equal to the product of the base and altitude, divided by their sum. BOOK VIII 233 PROBLEM III. In a triangle, having given the sides about the vertical angle, and the line hisecting that angle and terminating in the base, to find the base. Let ABO be the a, and let a cir- cle be circumscribed about it. Di- vide the arc AEB into two equal partu at the point E, and draw EO. This line bisects the vertical angle, (Cor., Th. 9, B. HI). Draw BE. Put AD = x, DB = y, AQ = a, OB = b, OD = c, and BE = w. The two A's, ADO and EBO, are equiangular; from which we have w + o : b : : a : c ; or, cw + \_EGF =\_AGC. Hence, the angles B GD and A G C are equal, and the lines AG and BG meet in a common point in the line CD, and made equal angles witk that line. 240 GEOMETRY. 3. If, from a point without a circle, two straight lines be drawn to the concave part of the circumference, making equal angles with the line joining the same point and the center, the parts of these lines which are intercepted within the circle, are equal. 4. If a circle be described on the radius of another circle, any straight line drawn from the point where they meet, to the outer circumference, is bisected by the interior one. 5. From two given points on the same side of a line given in position, to draw two straight lines which shall contain a given angle, and be terminated in that line. 6. If, from any point without a circle, lines be drawn touching the circle, the angle contained by the tangents is double the angle contained by the line joining the points of contact and the diameter drawn through one of them. 7. If, from any two points in the circumference of a circle, there be drawn two straight lines to a point in a tangent to that circle, they will make the greatest angle when drawn to the point of contact. 8. From a given point within a giv^n circle, to draw a straight line which shall make, with the circumference, an angle, less than any angle made by any other line drawn from that point. 9. If two circles cut each other, the greatest line that can be drawn through either point of intersection, is that which is parallel to the line joining their centers. 10. If, from any point within an equilateral triangle, perpendiculars be drawn to the sides, their sum is equal to a perpendicular drawn from any of the angles to the opposite side. 11. If the points of bisection of the sides of a given tri- angle be joined, the triangle so formed will be one fourth of the given triangle. 12. The difference of the angles at the base of any tri- angle, is double the angle contained by a line drawn from the vertex perpendicular to the base, and another bisect- ing the angle at the vertex. BOOK VIII. 241 13. If, from the three angles of a triangle, lines be *rawn to the points of bisection of the opposite sides, tfiese lines intersect each other in the same point. 14. The three straight lines which bisect the three ! angles of a triangle, meet in the same point. 15. The two triangles, formed by drawing straight lines from any point within a parallelogram to the ex- tremities of two opposite sides, are, together, one half the parallelogram. 16. The figure formed by joining the points of bisection of the sides of a trapezium, is a parallelogram. 17. If squares be described on three sides of a right- angled triangle, and the extremities of the adjacent sides be joined, the triangles so formed are equivalent to the given triangle, and to each other. 18. If squares be described on the hypotenuse and sides of a right-angled triangle, and the extremities of the sides of the former, and the adjacent sides of the others, be joined, the sum of the squares of the lines joining them will be equal to five times the square of the hypotenuse. 19. The vertical angle of an oblique-angled triangle inscribed in a circle, is greater or less than a right angle, by the angle contained between the base and the diam- eter drawn from the extremity of .the base. 20. If the base of any triangle be bisected by the diam- eter of its circumscribing circle, and, from the extremity of that diameter, a perpendicular be let fall upon the longer side, it will divide that side into segments, one of which will be. equal to one half the sum, and the other to one half the difference, of the sides. 21. A straight line drawn from the vertex of an equi- lateral triangle inscribed in a circle, to any point in the opposite circumference, is equal to the sum of the two lines which are drawn from the extremities of the base to the same point. 22. The straight line bisecting any angle of a triangle 21 Q 242 GEOMETRY. inscribed in a given circle, cuts the circumference in a point which is equi-distant from the extremities of the side opposite to the bisected angle, and from the center of a circle inscribed in the triangle. 23. If, from the center of a circle, a line be drawn to any point in the chord of an arc, the square of that line, together with the rectangle contained by the segments of the chord, will be equal to the square described on the radius. 24. If two points be taken in the diameter of a circle, equidistant from the center, the sum of the squares of the two lines drawn from these points to any point in the cir- cumference, will be always the same. 25. If, on the diameter of a semicircle, two equal circles be described, and in the space included by the three cir- cumferences, a circle be inscribed, its diameter will be % the diameter of either of the equal circles. 26. If a perpendicular be drawn from the vertical angle of any triangle to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base. 27. The square described on the side of an equilateral triangle, is equal to three times the square of the radius of the circumscribing circle. 28. The sum of the sides of an isosceles triangle is less than the sum of the sides of any other triangie on the same base and between the same parallels. 29. In any triangle, given one angle, a side adjacent to the given angle, and the difference of the other two sides, to construct the triangle. 30. In any triangle, given the base, the sum of the other two sides, and the angle opposite the base, to con- struct the triangle. 31. In any triangle, given the base, the angle opposite io the base, and the difference of the other two sides, to instruct the triangle. BOOK IX. 243 BOOK IX. SPHERICAL GEOMETRY. DEFINITIONS. 1. Spherical Geometry has for its object the investiga- tion of the properties, and of the relations to each other, of the portions of the surface of a sphere which are bounded by the arcs of its great circles. 2. A Spherical Polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles, called the sides of the polygon. 3. The Angles of a spherical polygon are the angles formed by the bounding arcs, and are the same as the angles formed by the planes of these arcs. 4. A Spherical Triangle is a spherical polygon having but three sides, each of which is less than a semi-circum- ference. 5. A Lime is a portion of the surface of a sphere in- cluded between two great semi-circumferences having a common diameter. 6. A Spherical Wedge, or TTngula, is a portion of the Bolid sphere included between two great semi-circles having a common diameter 244 GEOMETRY. 7. A Spherical Pyramid is a portion of a sphere bounded by the faces of a solid angle having its vertex at the center, and the spherical polygon which these faces inter- cept on the surface. This spherical polygon is called the base of the pyramid. 8. The Axis of a great circle of a sphere is that diameter of the sphere which is perpendicular to the plane of the circle. This diameter is also the axis of all small circles parallel to the great circle. 9. A Pole of a circle of a sphere is a point on the sur- face of the sphere equally distant from every point in the circumference of the circle. 10. Supplemental, or Polar Triangles, are two triangles on a sphere, so related that the vertices of the angles of either triangle are the poles of the sides of the other. PKOPOSITION I. Any two sides of a spherical triangle are together greater than the third side. Let AB, AC, and BC, be the three sides of the triangle, and D the center of the sphere. The angles of the planes that form the solid angle at D, are measured by the arcs AB, AG, snidBC. But any two of these angles are together greater than the third angle, (Th. 18, B. VI). Therefore, any two sides of the triangle are, together, greater than * ue third side. Hence the proposition. PROPOSITION II. The sum of the three sides of any spherical triangle is less than the circumference of a great circle. Let ABC be a spherical triangle ; the two sides, AB and AC, produced, will meet at the point which is diame- trically opposite to A f and the arcs, ABB and ACB are BOOK IX 245 together equal to a great circle. But, by the last proposition, BC is less than the two arcs, BD and DC There- fore, AB + BO + AC, is less than ABD + ACD; that is, less than a great circle. Hence the proposition. PROPOSITION III. The extremities of the axis of a great circle of a sphere are the poles of the great circle, and these points are also the poles of all small circles parallel to the great circle. Let be the center of the sphere, and BD the axis of the great circle, Cm Am" ; then will B and D, the extremities of the axis, be the poles of the circle, and also the poles of any parallel small cir- cle, as FnE. For, since BD is per- pendicular to the plane of the circle, Cm Am", it is perpendicular to the lines OA, 0m', Om", etc., passing through its foot in the plane, (Def. 2, B. VI); hence, all the arcs, Bm, Bm', etc., are quadrants, as are also the arcs Dm, Dm', etc. The points B and D are, therefore, each equally distant from all the points in the circumfer- ence, Cm Am" ; hence, (Def. 9), they are its poles. Again, since the radius, OB, is perpendicular to the plane of the circle, Cm Am", it is also perpendicular to the plane of the parallel small circle, FnE, and passes through its center, 0'. Now, the chords of the arcs, BF, Bn, BE, etc., being obiique lines, meeting the plane of the small circle a 4 - oqual distances from the foot of the o" -- \e / IS — * — Z^- hi m» [.-""'-""/ / ■"---», \ iJtr \j)f 246 GEOMETRY. perpendicular, BO', are all equal, (Th. 4, B. VI); hence, the arcs themselves are equal, and B is one pole of the circle, FnE. In like manner we prove the arcs, BF, Dn, BE, etc., equal, and therefore D is the other pole of the same circle. Hence the proposition, etc. Cor. 1. A point on the surface of a sphere at the distance of a quadrant from two points in the arc of a great circle, not at the extremities of a diameter, is a pole of that arc. For, if the arcs, Bm, Bm r , are each quadrants, the angles, BOm and BOm f , are each right angles; and hence, BO is perpendicular to the plane of the lines, Om and 0m f , which is the plane of the arc, mm'; B is therefore the pole of this arc. Cor. 2. The angle included between the arc of a great circle and the arc of another great circle, connecting any of its points with the pole, is a right angle. For, since the radius, BO, is perpendicular to the plane of the circle, Cm Am", every plane passed through this radius is perpendicular to the plane of the circle ; hence, the plane of the arc Bm is perpendicular to that of the arc (7m; and the angle of the arcs is that of their planes. PKOPOSITION IV. The angle formed by two arcs of great circles which inter- sect each other, is equal to the angle included between the tan- gents to these arcs at their point of intersection, and is meas- ured by that arc of a great circle whose pole is the vertex of the angle, and which is limited by the sides of the angle or the sides produced. Let AM and AN be two arcs intersecting at the point A, and let AE and AF be the tangents to these arcs at this point. Take AC and AD, each quadrants, and draw the arc CD, of which A is the pole, and OQ and OB are the radii. BOOK IX, 247 Now, since the planes of the arcs intersect in the radius OA, and AE is a tangent to one arc, and AF a tangent to the other, at the common point 'A, A these tangents form with each other an angle which is the measure of the angle of the planes of the arcs ; but the angle of the planes of the arcs is taken as the angle included by the arcs, (Def. 3). Again, because the arcs, AQ and AD, are each quadrants, the angles, A 00, AOD, are right angles ; hence the radii, OC and OD, which lie, one in one face, and the other in the other face, of the diedral angle formed by the planes of the arcs, are perpendicular to the common intersection of these faces at the same point. The angle, OOD, is therefore the angle of the planes, and consequently the angle of the arcs ; but the angle OOD is measured by the arc OD. Hence the proposition. Oor. 1. Since the angles included between the arcs of great circles on a sphere, are measured by other arcs of great circles of the same sphere, we may compare such angles with each other, and construct angles equal to other angles, by processes which do not differ in principle from those by which plane angles are compared and con- structed. Oor. 2. Two arcs of great circles will form, by their in- tersection, four angles, the opposite or vertical ones of which will be equal, as in the case of the angles formed by the intersection of straight lines, (Th. 4, B. I). PROPOSITION V. The surface of a hemisphere may be divided into three right- angled and four quadrantal triangles, and one of these right- angled triangles will be so related to the other two, that two of its sides and one of its angles will be complemental to the 248 GEOMETRY. Bides of one of them, and two of its sides supplemental to two of the side.s of the other. Let ABO be a right-angled spherical triangle, right angled at B. Produce the sides, AB and AC, and they will meet at A', the opposite point on the sphere. Produce BO, both ways, 90° from the point B, to P and P', which are, therefore, poles to the arc AB, (Prop. 3). Through A, P, and the center of the sphere, pass a plane, cutting the sphere into two equal parts, forming a great circle on the sphere, which great circle will be represented by the circle PAP' A' in the figure. At right angles to this plane, pass another plane, cutting the sphere into two equal parts ; this great circle is represented in the figure by the straight line, POP'. A and A' are the poles to the great circle, POP' ; and P and P f are the poles to the great circle, ABA f . !N"ow, OPB is a spherical triangle, right-angled at B, and its sides OP and OB are complemental respectively to the sides BO and A of the A ABO, and its side PD is complemental to the arc DO, which measures the [_BAO of the same triangle. Again, the A A' BO is right- angled at B, and its sides A'O, A'B, are supplemental respectively to the sides AO, AB, of the A ABO. There- fore, the three right-angled A's, ABO, OPB, and A'BO, have the required relations. In the A AOP, the side AP is a quadrant, and for this reason the A is called a quad- ran tal triangle. So also, are the A's A' OP, AOP', and P'OA', quadrantal triangles. Hence the proposition. Scholium. — In every triangle there are six elements, three sides and three angles, called the parts of the triangle. Now, if ail the parts of the triangle ABC are known, the parts of each of the A' s > PCD and A'BC, are as completely known. And when the parts of the A PCD are known, the parts of the A ' 8 A C2 book ix. 249 and A'CP are also known ; for, the side PD measures each of the | 's P^lCand PA'C, and the angle CPD, added* to the right angle A' PD t gives the | A' PC, and the | CPA is supplemental to this. Hence, the solution of the A ABC is a solution of the two right-angled and four quadrantal A's? which together with it make up the surface of the hemisphere. PROPOSITION VI. If there be three arcs of great circles whose poles are the angular points of a spherical triangle, such arcs, if produced, will form another triangle, whose sides will be supplemental to the angles of the first triangle, and the sides of the first triangle will be supplemental to the angles of the second. Let the arcs of the three great cir- cles be GH, PQ, KL, whose poles are respectively A, B, and 0. Produce the three arcs until they meet in D, E, and F. We are now to prove that E is the pole of the arc AO; D the pole of the arc BO; F the pole to the arc AB. Also, that the side EF, is supplemental to the angle A; EB to the angle 0; and BF to the angle B; and also, that the side A is supplemental to the angle E, etc. A pole is 90° from any point in the circumference of Us great circle ; and, therefore, as A is the pole of the arc Gff, the point A is 90° from the point E. As is the pole of the arc LK, is 90° from any point in that arc ; therefore, is 90° from the point E ; and E being fcO° from both A and 0, it is the pole of the arc AC. In the same manner, we may prove that B is the pole of BO, and F the pole of AB. Because A is the pole of the arc Git, the arc GH measures the angle A, (Prop. 4); for a similar reason, PQ measures the angle B, and LK measures the angle 0, Because E is the pole of the arc AO, EH = 90° Or, EG +GH= 90° For a like reason, FH + GH = 90° 250 GEOMETRY. Adding these two equations, and observing that QJf = A, and afterward transposing one A, we have, EG + GIT + FIT = 180° — A. Or, UF=1S0°— A } In like manner, jF2> = 180° — B \ («) And, BE = 180° — J But the arc (180° — A), is a supplemental arc to A, by the definition of arcs; therefore, the three sides of the triangle BEF, are supplements of the angles A, B y C, of the triangle ABO. Again, as E is the pole of the arc A C, the whole angle E is measured by the whole arc LH. But, AC + CE = 90° Also, AQ + AL = 90° By addition, AQ+AC+QH + AL = 180° By transposition, ^(7+ (7^+^^ = 180° —J. O That is, £#, or E= 180°— ^L<7 } In the same manner, F =■ 180°— ^ > (&) And, i)=180° — J5(7 J That is, the sides of the first triangle are supplemental to the angles of the second triangle. PROPOSITION Til. The sum of the three angles of any spherical triangle, is greater than two right angles, and less than six right angles. Add equations («), of the last proposition. The first member of the equation so formed will be the sum of the three sides of a spherical triangle, which sum we may designate by S. The second member will be 6 right angles (there being 2 right angles in each 180°) less the three angles A, B, and O. That is, S = 6 right angles — (A + B + C) By Prop. 2, the sum S is less than 4 right angles; BOOK IX. 251 therefore, to it add s, a sufficient quantity to make 4 right angles. Then, 4 right angles = 6 right angles — (A + B -f C) -f * Drop or cancel 4 right angles from both members, and transpose (A + B + O). Then, A + B + = 2 right angles + a. That is, the three angles of a spherical triangle make a greater sum than two right angles by the indefinite quantity *, which quantity is called the spherical excess, and is greater or less according to the size of the triangle. Again, the sum of the angles is less than 6 right angles. There are but three angles in any triangle, and each one of them must be less than 180°, or 2 right angles. For, an angle is the inclination of two lines or two planes ; and when two planes incline by 180°, the planes are parallel, or are in one and the same plane ; therefore, as neither angle can be equal to 2 right angles, the three can never be equal to 6 right angles. PROPOSITION VIII. On the same sphere, or on equal spheres, triangles which are mutually equilateral are also mutually equiangular ; and, conversely, triangles which are mutually equiangular are also mutually equilateral, equal sides lying opposite equal angles. First— -Let ABO and DEF, in which AB = BE, AO= DF, and BO = EF, be two triangles on the sphere whose center is 0; then will the [_ A, opposite the side BO, in the first triangle, be equal the [_D, opposite the equal side EF, in the second; also L# = l E, andl_AC, (Prop. 1.); or, substituting for AD its equal DB, we have, CD + DB> AC. If in the above inequality we now substitute CB for CD-\-DB, it becomes CB > CA. Conversely ; if the side CB be greater than the side CA, then is the \_A > th« [__i?. For, if the [_A is not greater than the [_B, it is either equal to it, or less than it. The [__A is not equal to the [_B ; for if it were, the triangle would be isosceles, and CB would be equal to CA, which is contrary to the hypothesis. The [_A is not less than the [_B; for if it were, the side CB would be less than the side CA, by the first part of the proposition, which is also contrary to the hypothesis ; hence, the [_A must be greater than the L^- book ix. 257 PROPOSITION XIII. Two symmetrical spherical triangles are equal in area Let ABO and BEF be two A's on the same sphere, having the sides and angles of the one equal to the sides and angles of the other, each to each, the triangles themselves not admitting of superposition. It is to be proved that these A's have equal areas. Let P be the pole of a small circle passing through the three points, ABO, and connect P with each of the points, A, B, and 0, by arcs of great circles. Next, through E draw the arc of a great circle, EP', making the angle DEP' equal to the angle ABP. Take EP' = BP, and draw the arcs of great circles, P'B, P'F. The A's, ABP and BEP', are equal in all their parts, because AB=BE, BP=EP', and the \__ABP=[_BEP', (Prop. 9). Taking from the \__ABO the \__ABP, and from the \__DEF the \_BEP' , we have the remaining angles, PBO and P'EF, equal; and therefore the A's, BOP and EFP 1 , are also equal in all their parts. Now, since the a's, ABP and DEP', are isosceles, they will coincide when applied, as will also the A's, BOP and EFP 1 , for the same reason. The polygonal areas, ABQP and BEFP 1 , are therefore equivalent. If from the first we take the isosceles triangle, PAO, and from the second the equal isosceles triangle, P'BF, the remainders, or the triangle* ABO and BEF, will be equivalent. Remark. — It is assumed in this demonstration that the pole P falls without the triangle. Were it to fall within, instead of without, no other change in the above process would be required than to add the isosceles triangles, PAC, P / DF, to the polygonal areas, to get the areas of the triangles, ABC, DEF. 258 GEOMETBY. Cor. Two spherical triangles on the same sphere, or on equal spheres, will be equivalent — 1st, when they are mutually equilateral; — 2d, when they are mutually equi- angular ; — 3d, when two sides of the one are equal to two sideb of the other, each to each, and the included angles are equal ; — 4th, when two angles of the one are equal to two angles of the other, each to each, and the included aides are equal. PKOPOSITION XIV. If two arcs of great circles intersect each otlier on the sur- face of a hemisphere, the sum of either two of the opposite tri- angles thus formed will he equivalent to a lune whose angle is the corresponding angle formed by the arcs. Let the great circle, AEBC, be the base of a hemi- sphere, on the surface of which the great semi-circumfer- ences, BBA and CBE, inter- sect each other at B ; then will the sum of the opposite tri- angles, BBC and BAB, be equivalent to the lune whose angle is BBC; and the sum of the opposite triangles, CBA and BBJE, will be equiv- alent to the lune whose angle is CBA. Produce the arcs, BB A and CBE, until they intersect on the opposite hemisphere at H\ then, since CBE and BEE are both semi-circumferences of a great circle, they are equal*. Taking from each the common part BE, we have CB = HE. In the same way we prove BB = HA, and AE = BC. The two triangles, BBC and HAE, are therefore mutually equilateral, and hence they are equivalent, (Prop. 13). But the two tri- angles, BAE and ABE f together, make up the lune BOOK IX. 259 BEHAB\ hence the sum of the a's, BBO and ABE, is equivalent to the same lune. By the same course of reasoning, we prove that the sum of the opposite A's, BAQ and BBE, is equivalent to the lune BOHAB, whose angle is ABO. PROPOSITION XV. The surface of a lune is to the whole surface of the sphere, as the angle of the lune is to four right angles ; or, as the arc which measures that angle is to the circumference of a great circle. I^ztABFCA be a lune on the surface of a sphere, and BCE an arc of a great circle, whose poles are A and F, the vertices of the angles of the lune. The arc, BO, will then measure the angles of the lune. Take any arc, as BB, that will be con- tained an exact number of times in BO, and in the whole circum- ference, BOEB, and, beginning at B, divide the arc and the circumference into parts equal to BB, and join the points of division and the poles, by arcs of great circles. We shall thus divide the whole surface of the sphere into a number of equal lunes. Now, if the arc BO con- tains the arc BB m times, and the whole circumference contains this arc n times, the surface of the lune will oontain m of these partial lunes, and the surface of the sphere will contain n of the same ; and we shall have, Surf, lune : surf, sphere : : m : n. But, m : n :: BO : circumference great circle ; hence, surf, lune : surf sphere : : BO : cir. great circle; or, surf, lune : surf, sphere :: [_BOO : 4 right angles. 260 GFOMETRY. This demonstration assumes that BD is a common measure of the arc, BC, and the whole circumference. It may happen that no finite common measure can be found ; but our reasoning would remain the same, even though this common measure were to become indefinitely small. Hence the proposition. Cor. 1. Any two lunes on the same sphere, or on equal spheres, are to each other as their respective angles. Scholium. — Spherical triangles, formed by joining the pole of an arc of a great circle with the extremities of this arc by the arcs of great circles, are isosceles, and contain two right angles. For this reason they are called bi-rectangular. If the base is also a quadrant, the vertex of either angle becomes the pole of the opposite side, and each angle is measured by its opposite side. The three angles are then right angles, and the triangle is for this reason called tri-rectangular. It is evident that the surface of a sphere contains eight of its tri- rectangular triangles. Cor. 2. Taking the right angle as the unit of angles, and denoting the angle of a lune by A, and the surface of a tri-rectangular triangle by T, we have, surf, of lune : 8^ :: A : 4; whence, surf, of lune = 2 A X T. Cor. 3. A spherical ungula bears the same relation to the entire sphere, that the lune, which is the base of the ungula, bears to the surface of the sphere ; and hence, any two spherical ungulas in the same sphere, or in equal spheres, are to each other as the angles of their re- spective lunes. PKOPOSITION XVI. The area cf a spherical triangle is measured by the excess of the sum of its angles over two right angles, multiplied by the tri-rectangular triangle. Let ABC be a spherical triangle, and DJEFLK the cir- cumference of the base of the hemisphere on which this triangle is situated. BOOK IX. 2G1 Produce the sides of the tri- angle until they meet this cir- cumference in the points, D, U, F, L, K, and P, thus forming the sets of opposite triangles, DAE, AKL ; BEF, BPK; CFL, CDP. Now, the triangles of each of these sets are together equal to a lune, whose angle is the cor- responding angle of the triangle, (Prop. 14) ; hence we have, ADAE + AAKL = 2 A x T % (Prop. 15, Cor. 2). ABEF + ABPK = 2B x T. A CFL + A CDP = 2(7 x T. If the first members of these equations be added, it is evident that their sum will exceed the surface of the hemisphere by twice the triangle ABC; hence, adding these equations member to member, and substituting for the first member of the result its value, 4T + 2 A ABC, we have 4I 7 + 2aABC = 2A.T + 2B.T+ 2CT or, 2T+ AABC=> A.T + B.T + C.T whence, A ABC = A.T + B.T + C.T—2T. That is, AABC - (A + B + C— 2) T. But A + B + C — 2 is the excess of the sum of the angles of the triangle over two right angles, and T de- notes the area of a tri-rectangular triangle. Hence the proposition ; the area, etc. 262 GEOMETRY. PROPOSITION XVII. 77ie area of any spherical polygon is measured by the excess of the sum of all its angles over two right angles, taken as many times, less two, as the polygon has sides, multiplied by the tri-rectangular triangle. Let AB CDE be a spherical poly- ^-—"7 gon; then will its area be meas- b^"^ / ured by the excess of the sum of /\ / / \ / the angles, A, B, 0, D, and E, over / two right angles taken a number / \ of times which is two less than J -~J*Je the number of sides, multiplied by \ / T, the tri-rectangular triangle. \. / Through the vertex of any of the p angles, as E, and the vertices of the opposite angles, pass arcs of great circles, thus divi- ding the polygon into as many triangles, less two, as the polygon has sides. The sum of the angles of the several triangles will be equal to the sum of the angles of the polygon. Now, the area of each triangle is measured by the excess of the sum of its angles over two right angles, multiplied by the tri-rectangular triangle. Hence the sum of the areas of all the triangles, or the area of the polygon, is measured by the excess of the sum of all the angles of the triangles over two right angles, taken as many times as there are triangles, multiplied by the tri- rectangular triangle. But there are as many triangles as the polygon has sides, less two. Hence the proposition ; the area of any spherical voly- gon, etc. Cor. If S denote the sum of the angles of any spherical polygon, n the number of sides, and T the tri-rectan- gular triangle, the right angle being the unit of angles ; the area of the polygon will be expressed by [£_ 2 (n — 2)] x T= (#— 2n + 4) T. 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