IN MEMORIAM FLOR1AN CAJORI PLANE AND SOLID GKOMBTRY SUGGESTIVE METHOD BY C. A. VAN VELZER Professor of Mathematics, University of Wisconsin, AND GEO. C. SHUTTS Professor of Mathematics and History, State Normal School, Whitewater, Wis. MADISON, WIS. TRACY, GIBBS & CO. COPYRIGHT, 1894. C. A. VAN VELZER AND GEO. C. SHUTTS. Tracy, Gibbs & Co., Printers and Stereotypers. PREFACE. An examination will show that this book is constructed upon a unique plan, and to give proper credit for this plan I have requested the privilege of writing this preface myself. The plan is due entirely to Professor Shutts who has used it for years from mimeograph notes with such excellent results that it was thought best to put the matter in permanent form for a text-book. Accordingly the mimeograph notes were carefully revised by Professor Shutts and myself and put into nearly the present form, when the work was again tested by Professor Shutts in his classes in the State Normal School at Whitewater, Wis. After this further trial the work was again revised and improved. The method of this book therefore is not untried, but has been sub- jected to a thorough test and found to give most excellent results. In putting the work into its present form the scientific classifica- tion of the subject-matter has been departed from when it was thought that by so doing the work could be better graded to the abil- ity of the average pupil. For this reason the subject of triangles has been introduced before the relations of lines and angles have been fully discussed. A departure from ordinary methods will be noticed in the treat- ment of proportion. It has not been thought wise to follow the or- dinary method in this country of limiting the subject to proportions whose terms are pure numbers, nor yet to follow the Euclidian method common in England, which indeed admits of proportions whose terms are concrete magnitudes, but which is so difficult that it can be understood by only the best students. The method in the text will be found to admit of proportions whose terms are concrete magnitudes and yet will do no violence to the fundamental ideas of arithmetic regarding operations upon concrete magnitudes. M306168 IV PREFACE. The subject of limits is treated in a manner so simple that begin- ners can grasp it. With an average high school class all the work, including exercises, can be done in one year. In preparing this book for the press the works which have been consulted most freely and have been most serviceable are Byerly's edition of Chauvenet's Geometry, and the geometry prepared by the English Society for the Improvement of Geometrical Teaching. The thanks of the authors are due to Professor A. J. Hutton, of the State Normal School, of Platteville, Wis., for valuable suggestions and criticisms, C. A. VAN VELZER. SUGGESTIONS TO TEACHERS. Geometry is essentially a disciplinary study, and benefit derived from its study depends upon the independent thought expended by the pupil. A geometry is in the nature of a "key" to the extent to which the demonstrations are written out for the pupil. That part of the work which a pupil can do for himself should not be done for him. The teacher and text-book should furnish the data and stimu- late thought rather than give him a set form of words which he may repeat entire with or without the ideas which those words should convey. In the following pages are suggestions arranged in logical order which are intended to direct and stimulate the thought of the pupil so that he may work out his own demonstrations. Model demonstrations are given of a few propositions to show the student how to work out his own demonstrations, and in what form they should be given. By comparison it will be seen that the answers to the suggestions logically linked together, constitute the demonstra- tion. The suggestions should be studied in their order, for usually each suggestion depends upon the preceding one. The answer to a suggestion should consist of a statement of the relations asked for, together with the authority in full for such statement. To be indif- ferent in regard to the authority in any instance is to encourage care- lessness, slovenliness and inaccuracy in demonstration. A common error is to apply authority that does not exactly fit the conditions un- der consideration. The pupil should be made to clearly understand that the authority should, without exception, be a definition, an axiom, or a previously proved proposition. "It seems so," or, "it looks reasonable," or any expression of judgment will not do. The vi SUGGESTIONS TO TEACHERS. pupil should be encouraged to search out his own authority, even when the authority is quoted for him in the suggestions, using the reference simply for verification. A pride in independent work is a most important factor in securing satisfactory results. In the preparation of the lesson the pupil should write out his demonstration, noting carefully the form of the "models." This will ensure correct form and avoid haziness of thought. During the first few weeks the teacher should scan carefully this written work as well as tests taken in the recitation. The exercises, or at least a part of them, should be taken along with the propositions as they occur, and not be studied all together at the end of a chapter. Time will be saved in the end by starting slowly but surely, pass- ing over nothing that is not clearly understood. Since each demonstration involves previous propositions and def- initions, facility in demonstration can be best secured by committing to memory each theorem and definition; for that authority cannot be readily recognized and applied which is only imperfectly in mind. CONTENTS. PLANE GEOMETRY. CHAPTER I. PAOE. RECTILINEAR FIGURES, ------ .-1 PROPOSITIONS IN CHAPTER I,- ------ 77 CHAPTER II. THE CIRCLE, ...-82 PROPOSITIONS IN CHAPTER II, ------- 133 CHAPTER III. PROPORTIONAL LINES AND SIMILAR POLYGONS, - 137 PROPOSITIONS IN CHAPTER III, - .... 173 CHAPTER IV. COMPARISON AND MEASUREMENT OF POLYGONS, ... 17G PROPOSITIONS IN CHAPTER IV, - - .... 206 CHAPTER V. REGULAR POLYGONS AND CIRCLES, --.... 209 PROPOSITIONS IN CHAPTER V, --.... 239 Vlll CONTENTS. GEOMETRY OF THREE DIMENSIONS. CHAPTER VI. PAGE. LINES AND PLANES, - --241 PROPOSITIONS IN CHAPTER VI, ------- 283 CHAPTER VII. POLYHEDRONS, 287 PROPOSITIONS IN CHAPTER VII, ...... 335 CHAPTER VIII. THE THREE. ROUND BODIES, PROPOSITIONS IN CHAPTER VIII, INDEX, CHAPTER I. RECTILINEAR FIGURES. DEFINITIONS. 1. The block represented in the figure occupies a lim- ited portion of space. Conceive it to be removed. Its form or shape may still be retained in mind. This is true of any object or body. The space conceived to be occupied by a body, apart from its substance, is called a geometrical solid. The matter or substance of which the body is composed constitutes a physical solid. Hence a geometrical solid is simply the shape or form of a physical solid, or some form or figure conceived by the mind. Definition. A geometrical solid is a limited portion of space and has length, breadth and thickness. The terra solid will hereafter be used for geometrical solid. 2. The boundary that separates a solid from the space outside of itself is called the surface of the solid. Dis- tinct flat portions of the bounding surface are called faces. As we can think of the surface of a body without in- cluding any of its substance, surface may be regarded as having no thickness. Definition. Surface is that which has length and breadth without thickness. 1 Geo. 2 PLANE GEOMETRY. 3. If a surface be divided into distinct portions, the boundaries of those portions are called lines. In the solid represented on page 1, the edges, or boundaries of the faces, are lines. These lines being the intersections of faces which have no thickness can themselves have neither breadth nor thickness. Definition. A line is that which has length but has neither breadth nor thickness. 4. If a line be divided into distinct portions the limits of these portions are called points. In the solid repre- sented on page 1, the corners, or limits of the edges, are points. These points being the intersections of lines which have neither breadth nor thickness can themselves have neither length, breadth nor thickness. Definition. A point is that which has simply posi- tion but neither length, breadth nor thickness. 5. A surface may be conceived apart from a solid, a line apart from a surface and a point apart from a line. If a point be conceived to move, the path in which it moves is a line, hence a line may be regarded as the path, or locus, of a moving point. When a line is regarded as formed by a moving point the point is said to describe the line. 6. A geometrical figure is a combination of points, lines, surfaces or solids in any given relations. Geometrical figures are ideal, i. e., mental conceptions, though none the less real, and can be represented to the eye only by some material substance; for instance, a line may be represented by a mark made by a pencil, crayon, etc. A solid may be represented by a drawing or a block of wood or some material in any given shape. DEFINITIONS. To avoid multiplying words, the representation of geometrical figures will generally be referred to for the concepts themselves. 7. A straight line is a line such that any part of it, however placed, will lie wholly in any other part, if its extremities be made to lie in that part. A line is read by naming the letters placed at its extremities, as line AB. 2. 8. A broken line is one made up of a succession of different straight lines, as A 'B CD E. 9. A curved line, or a curve, is a line no portion of which is straight, as C D. 10. A plane surface, or a plane, is a surface such that if any two of its points be joined by a straight line the line will lie wholly in the surface. 11. A plane figure is a figure which lies wholly in the same plane. 12. A plane figure which is made wholly of straight lines is called a rectilinear figure. 13. When reference is made simply to extent, figures are called magnitudes. 14. Geometry is the science which treats of points, lines, surfaces and solids, and is concerned with the con- struction and measurement of geometric figures. 15. Plane geometry treats of plane figures. PLANE GEOMETRY. 16. Solid geometry treats of figures which are not wholly in the same plane. ANGLES. 17. When two straight lines meet or intersect they are said to contain, or to make with each other, an angle. The two lines are the sides of the angle, and the point of meeting is the vertex of the angle. When a line, coincident with one side of an angle, re- volves about the vertex, always remaining in the same plane, until it arrives at the position of the other side of the angle, the line is said to turn. through the angle, and the greater the amount of turning the greater the angle. The magnitude of an angle may be made clear by means of a pair of dividers, the legs of the dividers representing the sides of the angle and the hinge representing the vertex. If the dividers be opened a given amount an angle is represented; if from that position they be closed to a certain extent a smaller angle is represented; if they be opened farther a larger angle is represented. A line may turn through an angle in two ways, hence there are two angles which have the same sides and the same vertex. For example, the side O B in the figure may turn through the angle m by a motion opposite to that of the hands of a watch to . the position O A, or it may turn through the angle n by a motion with that of the hands of a watch to the same position O A. Two angles which have the same vertex and the same sides are called conjugate angles. When an angle is spoken of, the smatterofihe two conjugate angles is al- ways meant unless the contrary is specifically mentioned. DEFINITIONS. 18. When the two sides of two conjugate angles lie in the same straight line, the angle formed is called a straight angle, and the conjugate angles are equal; for example the angle A O B is a straight angle. A o /? An angle may be read by reading the letter at the ver- tex of the angle between the letters upon the sides of the* angle, as angle A B C or angle CB A. When there is only one angle at a given A vertex it is sufficient to read the letter at ^/ the vertex alone, as angle B. When two ;S or more angles have a common vertex a ^C^^ letter is frequently placed near the vertex ^*^\^ between the sides of the angle, to desig- nate the angles, as m and n in the figure at the right. For example, we say angle m instead of angle C B A, and angle n instead of angle C B D. 19. Definition. Two angles which have a common vertex and one common side, and are on opposite sides of this common side, are called adjacent angles. Angles m and n in the second figure in article 18 are ad- jacent angles. 20. Definition. When one straight line meets another straight line so that the two adjacent angles formed are equal, the angles are right angles. If angle A O B equals angle A O C, the angles A O B and A O C are each right angles B O 6 PLANE GEOMETRY. A O 21. Definition. An angle less than a right angle is called an acute angle, and an angle greater than a right angle is called an obtuse angle. Angle B O C is an acute angle, and angle A O B is an obtuse angle. Acute and obtuse angles are called oblique angles. 22. Definition. If a line makes right angles with an- other line, the two lines are said to be perpendicular to each other. In the figure in article 20, A O is perpen- dicular to B C and B C is perpendicular to A O. 23. Definition. If a line makes an oblique angle with another line, the two lines are said to be oblique to each other. In the figure in article 21 B O is oblique to A C and A C is oblique to B O. 24. When two lines intersect, the opposite angles are called ver- tical angles. In the figure at the right, the angles A O D and COB are ver- tical angles; also the angles A O C and B O D are vertical angles. 25. If two adjacent angles are together equal to a right angle, the two angles are said to be com- plements of each other. If the angles M B D and MBA are each right angles, angles CBD A R D and M B C are complements of each other; also angles ABE and M B E are complements of each other. DEFINITIONS. 26. If two adjacent angles are together equal to two right angles, the two angles are said to be supplements of each other. In the figure in article 21, the angles B O C and A O B are supplements of each other. Two adjacent angles which are supplements of each other are called supplementary adjacent angles. LOGICAL TERMS. 27. A theorem is a truth which requires demonstra- tion. For example: If two straight lines intersect each other > the vertical angles are equal. 28. The statement of a theorem is called its enuncia- tion, or the general enunciation. V/hen a drawing is made to illustrate a theorem, the description of the drawing is called the special enuncia- tion. 29. A theorem consists of two parts, the conditions, or suppositions, and the conclusion. The conditional part of a theorem is called the hypothesis. The hy- pothesis is composed of at least two parts: for example, in the theorem stated in article 27, first> two lines are straight, second, they intersect. The two parts of the hypothesis are sometimes called premises. 30. The truth depending upon or following from the hypothesis is called the conclusion. In the theorem stated in article 27, the truth that the vertical angles are equal depends upon the hypothesis that two straight lines intersect each other. 31. The course of reasoning by which the truth or falsity of a statement is established is called the demon- stration, or proof. 8 PLANE GEOMETRY. 32. A problem is a question proposed for solution, or the statement of certain relations which are to be pro- duced. For instance: To construct a line perpendicular to a given line at a given point. 33. A proposition is a general term for a theorem 01 a problem. 34. A corollary is a proposition easily deduced from the proposition to which it is attached, with the aid, if necessary, of one or more previous propositions. 35. A scholium is a remark upon one or more propo- sitions with respect to their applications, limitations or connections. 36. An axiom is a truth which, from its simplicity, must be admitted without demonstration: as, The whole of anything is equal to the sum of its parts. 37. A postulate is a problem whose solution, on ac- count of its simplicity, is admitted to be possible: as, Let it be granted that a straight line can be drawn between two points. 38. AXIOMS. 1. Things which are equal to the same or equal things are equal to each other. 2. If equals be added to equals the sums will be equal. 3. If equals be subtracted from equals the differences will be equal. 4. If equals be multiplied by equals the products will be equal. 5. If equals be divided by equals the quotients will be equal. DEFINITIONS. 9 6. If equals be added to unequals the sums will be un- equal, and that sum will be greater which is obtained from the greater magnitude. 7. If equals be subtracted from unequals the differ- ences will be unequal, and that difference will be greater which is obtained from the greater magnitude. 8. The whole is greater than any of its parts. 9. The whole is equal to the sum of all its parts. 10. A straight line is the shortest distance between two- points. 11. If two straight lines coincide in two points they will coincide throughout their whole extent, and be one and the same straight line. From this axiom it follows first, that two straight lines can inter- sect in only one point, and second, but one straight line can be drawn between two points. 12. Magnitudes which occupy the same space are equal. 39. SYMBOLS AND ABBREVIATIONS. /_ angle. Ax. axiom. Z^s angles. Cons, construction. _L perpendicular. Cor. corollary. _Ls perpendiculars. Def. definition. || parallel. Ex. exercise. || s parallels. Give auth. give authority. A triangle. Hyp. hypothesis. As triangles. Rt. right. 7 parallelogram. Sch. scholium. 7s parallelograms. St. straight. O circle. Sug. suggestion. Os circles. Sugs. suggestions. 10 PLANE GEOMETRY. PROPOSITION I. THEOREM. 40. At a given point in a straight line, one perpen- dicular can be erected to that line f and but one. D Let A B represent any straight line, and O any point in the line. I. To prove that one perpendicular can be erected to A B at O. SUG. 1. Revolve a line, as O C, from the position O B through the st. Z A O B to the position O A. Art. 17. SUG. 2. In the first part of the revolution, how does /.COB compare with Z_ A O C? When the revolution is nearly completed, how does /.COB compare with Z. A O C ? SUG. 3. Is there a position of O C during the revolu- tion at which Zs C O B and A O C are equal, and hence rt. Zs? Why? II. To prove that only one perpendicular can be erected. SUG. 1. Let D O represent the position of C O, when the two adjacent angles are equal. How much can D O be revolved, either way, and keep the two adjacent angles equal ? RECTILINEAR FIGURES. 11 SuG. 2. Then how many _Ls can be drawn to the line A B at the point O ? and hence to any line at any point ? Therefore 41. COROLLARY I. Through the ver- tex of a given angle but one line bisect- ing the angle can be drawn. SuG. See method in Art. 40, I. 42. COROLLARY II. All right angles are equal. Let A O C and A' <7 C' represent two right angles. To prove that they are equal. SUG. 1. Place B C upon B' C with the point O upon O '. SUG. 2. Where will O A fall with re- spect toVA'l Why ? See Art. 40, II. Note. The pupil is expected to study the sug- gestions carefully, to follow the directions when directions are given, to answer the questions when questions are asked, and to give the authority on B o A' a C' which the answers are based; then to go through the whole demon- stration in a consecutive manner, without the aid of the suggestions. To illustrate what is expected of the pupil, a model demonstration is given of a few propositions, but this model should not be consulted until after the proposition has been studied by means of the suggevs tions. 12 PLANE GEOMETRY. MODEL. PROPOSITION I. THEOREM. 43. At a given point, in a straight line, one perpen- dicular can be erected to the line, and but one. o Let A B represent any straight line, and O any point in the line. I. To prove that one perpendicular can be erected to A B at O. Suppose O C to revolve from the position of O B through the st. Z. to the position of O A. In the first part of the revolution Z. B O C is smaller than Z. A O C. When the revolution is nearly completed, as at O C 1 , Z. B O C is greater than Z A O C' . Hence, as Z. B O C is at first small, and continually increases, while Z. A O C continually decreases, there must be a position in the revolution of O C from O B to O A, as O D, at which these Z$ are equal, and hence by definition are rt. Z&. Hence, by definition, O D is _L to A B. Therefore, a perpendicular can be erected to a given line, at a given point in the line. RECTILINEAR FIGURES. 13 II. To prove that only one perpendicular can be erected. O D is the only perpendicular, for if it be revolved ever so little, either way, one of the angles becomes smaller, and the other larger, and hence they are no longer equal. Therefore, there can be but one perpendicular erected to a given line, at a given point of that line. From parts I and II of this proposition, it follows that: At a given point, in a straight line, one perpendicular can be erected to that line, and but one. 44. COROLLARY I. Through the vertex of a given angle, one, and but one, straight line can be drawn which bisects the angle. Let the angle A O C represent the given angle. If a line, as O B, be revolved from the position O C, through the given Z. to the position O A, there is one position at which the adjacent Z!s are equal, accord- ing to the reasoning in Prop. I. And if from this position O D be revolved ever so little, in either direction, one of the Z. increases, and the other decreases, and hence they can no longer be equal. 45. COROLLARY II. All right angles Q are equal. Let the angles A O B and A' a ff represent two right angles. To prove that they are equal. Place Z.A O C upon Z A' a C', B C _ upon B' C', and O upon a. B> O A must fall upon <7 A'. Art. 40, II. C' 14 PLANE GEOMETRY. PROPOSITION II. THEOREM. 46. If one straight line, meets another straight line, the sum of the adjacent angles formed equals two right angles. A B Let C .B and A J> represent two straight lines, wJiich meet at a point, as B. To prove that the sum of the angles ABC and C B D equals two right angles. SUG. 1. Suppose, first, that Zs A B Cand C B D are equal. What do you conclude concerning the number of rt. Zs formed? Why? SUG. 2. Suppose that Zs A B C and C B D are un- equal, and let a JL, B M, be erected to the line A D, at the point B. Give authority for the construction. SUG. 3. How does the sum of Zs A B C and C B D compare with the sum of Zs ABM and M B D ? SUG. 4. What kind of Zs are A B M and MB Dt Why? What, then, does the sum of Zs A B C and CB D equal ? Therefore QUERY. In proposition II what is the hypothesis? What is the conclusion ? RECTILINEAR FIGURES. 15 MODEL. PROPOSITION II. THEOREM. 47. If one straight line meets another straight line, the sum of the two adjacent angles is equal to two right angles. M A B Let C B meet A D, at B. To prove that the sum of the angles A B Cand C B D equals two right angles. If A B C and CBD are equal, two rt. Zs are formed. ART. 20. If A B C and CBD are unequal, draw B M _L to A D, at B. PROP. I. ABC + Z CBD = Z A B M + Z MB D. ARTS. 17, 18. (Because each sum equals the st. ^ A B D.) But Z.A BM+ Z MBD = Zrt. Zs. ART. 22- Hence, Z A B C + Z CB D = 2 rt. Zs. Ax. 1. 48. COROLLARY I. The sum of all the angles on one side of a straight line, having a common vertex in the line, equals t\w> right angles. 49. COROLLARY II. The total angular magnitude, about a point, equals four right angles. 16 PLANE GEOMETRY. PROPOSITION III. THEOREM. 50. If two straight lines intersect, the vertical angles formed are equal. O Let A B and C Z> intersect at O, forming the vertical angles AO D and COB. To prove that angle A O D equals angle COB. SUG. 1. ^AOC+Z.COB= what ? Why? Sue. 2. Z.AOC+Z.AOD = what ? Why ? SUG. 3. Compare Z. A O C -f Z. COB with Z. A O C + Z. A O D. Give auth. SUG. 4. Compare Z. C O B with Z. A O D. Ax. 3. SUG. 5. In a similar manner, compare ^/s A O C and DOB. Therefore MODEL. PROPOSITION III. THEOREM. 51. If two straight lines intersect, the vertical angles formed are equal. A Let A B and C D intersect at O, forming the vertical angles A O D and COB. RECTILINEAR FIGURES. 17 To prove that angle A O D equals angle COB. ZAOC+ZCO=2rt.Zs. PROP. II. Also Z A O C + Z A O D = 2 rt. Zs. PROP. II. Hence, /.AOC+Z.COB = ^AOC+Z.AOD. Ax. 1. Hence, Z.COB=Z.AOD. Ax. 3. In a similar manner it may be proved that 4L A O C = /.DOB. Therefore, if two straight lines intersect, the vertical angles formed are equal. EXERCISES. 1. How many straight lines can be drawn through three points, not in the same straight line, if each line connects two points ? 2. How many straight lines can be drawn through four points, no three of which are in the same straight line, if each line connects two points ? 3. What is the greatest number of points in which three straight lines can intersect ? 4. What is the greatest number of points in which four straight lines can intersect ? 5. If an angle is a right angle, what is its supple- ment? 6. If an angle is two thirds of a right angle, what is its supplement ? 7. If an angle is three fourths of a right angle, what is its complement ? 8. In the figure in Art. 21, if the angle C O B is one half of a right angle, the angle A O B equals what ? 9. In the figure in Art. 25, if the angle D B C is two thirds of a right angle, and the angle A B E is three fourths of a right angle, the angle C B E equals what ? 2 Geo. 18 PLANE GEOMETRY. DEFINITIONS. 52. A plane triangle is a portion of a plane bounded by three straight lines. The points of intersection of the bounding lines are called the vertices of the triangle, and the portions of the bounding lines included between the vertices are called the sides of the triangle. 53. A right angled triangle, or simply a right tri- angle, is a triangle one of whose angles is a right angle. 54. An acute angled triangle, or simply an acute triangle, is a triangle all of whose angles are acute. 55. An obtuse angled triangle, or simply an obtuse triangle, is a triangle one of whose angles is obtuse. Obtuse angled triangles and acute angled triangles are often spoken of as oblique angled triangles, or simply oblique triangles. 56. A triangle, no two sides of which are equal, is called a scalene triangle. 57. A triangle, which has two equal sides, is called an isosceles triangle. 58. A triangle, which has all three sides equal, is called an equilateral triangle. 59. The base of a triangle is the side upon which it is supposed to stand. Generally, any side may be assumed as the base, but, in an isosceles triangle, that side which is not one of the two equal sides is always considered as the base. 60. The angle which is opposite the base, is called the vertical angle. RECTILINEAR FIGURES. 19 PROPOSITION IV. THEOREM. 61. If two triangles have two sides and the in- cluded angle of one, equal to two sides and the in- cluded angle of the other, each to each, the triangles are equal in all respects. Let ABC and D EF represent two triangles, in which AB=DE, BC=EF, and angle B = angle E. To prove that the triangles ABC and DBF are equal in all respects, i. e. , that A C = D F, angle A = angle D, and angle C = angle F. SUG. 1. Place A A B C upon A D E F, so that B C coincides with E F, B with E and C with F. SUG. 2. What direction will B A take with respect to E D? ART. 17. SUG. 3. Where will the point A lie ? Why ? SUG. 4. Having located the points C and A, where, with respect to D F, does the line A C lie ? Ax. 11. SUG. 5. What, now, is the position of A A B C with respect to A D E Ft SUG. 6. How, then, does A A B C compare with &DEFt Ax. 12. Therefore 20 PLANE GEOMETRY. MODEL. PROPOSITION IV. THEOREM. 62. If two triangles have two sides and the in- cluded angle of one t equal to two sides and the in- cluded angle of the other, each to each, the triangles are equal in all respects. A P Let ABC and & E F represent two triangles, in which = DE, BC=EF, and angle B = angle E. To prove that the triangles ABC and D E F are equal in all respects, i. and C upon F. Why can this be done ? SUG. 2. What direction will B A take ? Why? SUG. 3. Where will A fall ? Why ? SUG. 4. What direction will C A take? Why? SUG. 5. Where, now, will the point A fall ? Why ? SUG. 6. How, then, does the A A B C compare with the A D E Ft Therefore SCHOUUM. When two triangles are equal, equal sides lie opposite equal angles, and equal angles lie op- posite equal sides. 22 PLANE GEOMETRY. PROPOSITION VI. THEOREM. 64. The angles opposite the equal sides of an isos- celes triangle are equal. A BMC Let ABC represent an isosceles triangle, A JB "being equal to AC. To prove that angle B is equal to angle C. SUG. 1. Suppose A M drawn so as to bisect Z. A, and extended until it meets B C, at M. SUG. 2. Compare A A Mwiih A A CM. PROP. IV. SUG. 3. Compare Z. B with Z. C. Sen. ART. 63. Therefore Ex. 10. If a straight line bisects one of a pair of vertical angles, prove that it bisects the other also. If MN bisects Z.AOD, prove that , , \0> it bisects Z C O B\ that is, that' Z.COM=Z.BOM. ^ \ Ex. 11. If the equal sides of an isosceles triangle be extended beyond the base, prove that the exterior angles formed with the base are equal. Ex. 12. Points, in the sides of an isosceles triangle, equidistant from the base, are equidistant from the vertex. RECTILINEAR FIGURES. 23 MODEL. PROPOSITION VI. THEOREM. 65. The angles, opposite the equal sides of an isos- celes triangle, are equal. Let ABC represent an isosceles triangle, A B being eqiuil to A C. To prove that angle B is equal to angle C. Let A M be drawn, bisecting the Z. A y and meeting B C at M. In the &sABM and ACM, A B = A C. HYP. Z_BAM=Z_CAM. CONS. And A Mis common. Therefore, the As ABM and ACM are equal in all respects. PROP. IV. If two ^s have two sides, and the included ^/ of one equal to two sides, and the included ^_ of the other, each to each, the triangles are equal in all respects. Therefore, Z. B = Z. C, being corresponding ^s of equal As. Sen. ART. 63. Ex. 13. Prove that the line which bisects the vertical angle of an isosceles triangle, bisects the triangle. PLANE GEOMETRY. PROPOSITION VII. THEOREM. 66. If a perpendicular be erected at the middle point of a straight line, the distances from any point in the perpendicular to the extremities of the line are equal. Let A C represent any line, B its middle point, B O, a perpendicular to A C, at B, and O any point in the per- pendicular. To prove that A O and O C are equal. SUG. 1. In the As O B A and O B C, what parts are equal, each to each ? Why ? SuG. 2. How do the As O B A and O B C compare ? Why? ART. 61. SUG. 3. How, then, do O A and O C compare ? Therefore Ex. 14. Prove that the bisector of the vertical angle of an isosceles triangle bisects the base, and is perpendicu- lar to the base. Ex. 15. The base of an isosceles triangle, together with the bisectors of the angles at the base, form a second isosceles triangle. RECTILINEAR FIGURES. 25 PROPOSITION VIII. THEOREM. 67. Any side of a triangle is less than the sum of the other two. Let ABC represent any triangle. To prove that any side, as A B, is less than the sum of the other two. SUG. Which represents the shorter distance from A to B, that along the line A B, or along the lines A C and CBt Why ? Ax. 10. Therefore Ex. 16. If the angular magnitude about a point is divided into six equal angles, each angle is what part of a right angle ? Ex. 17. If the angular magnitude about a point is divided into three angles, the second of which is twice the first, and the third is three times the first, how many right angles in each of the three angles ? Ex. 18. A line which is perpendicular to the bisector of an angle makes equal angles with the sides of the angle. PLANE GEOMETRY. PROPOSITION IX. THEOREM. 68. If a perpendicular ~be erected at the middle point of a straight line, the distances, from a point not in the perpendicular, to the extremities of the line are unequal. Let B C represent a sfraiyht line, M its middle point f O M, a perpendicular to 1? C 9 at the point M, A, a point not in the perpendicular, and A JB and A C lines drawn from A to the extremities of the line It C. 7o prove that A B and A C are unequal. SuG. 1. Let O be the intersection of A B, and the per- pendicular. Draw the line O C. SuG. 2. Compare A C with A O + O C. Give auth. SUG. 3. Compare O C with O B. Give auth. SuG. 4. Compare A C with A O + O B, or with A B. Therefore QUERY. Why draw the line O C? Ex. 19. Prove that an equilateral triangle is also equiangular. SUG. See Prop. VI. RECTILINEAR FIGURES. 27 Locus OF A POINT. 69. To be able to fix, definitely, the position of a point in a plane, it is necessary to know two things about the point; or, as it is usually expressed, two con- ditions limiting the position of a point must be given. If only one condition is given, the point is, to some extent, fixed, but not completely fixed. For example, a point which is at a given distance from some fixed point, is it- self not completely fixed, but may move, provided that in its motion it always satisfies the requirement of being at the given distance from the fixed point. The moving point, then, may occupy any position whatever in a line which will be defined later as the circumference of a circle whose center is the fixed point. Definition. When the position of a point in a plane is restricted in such a way that the point cannot be exactly determined, but is limited to, and may be anywhere in, a line, or group of lines, the line, or group of lines, is called the locus of the point. The locus of a point is both inclusive and exclusive; that is, the line, or group of lines, must include all the points which satisfy the given condition, and exclude all points which do not satisfy that condition. Ex. 20. If two straight lines intersect, and one of the angles formed is a right angle, all the angles are right angles. Ex. 21. If the angles at the base of an isosceles tri- angle are bisected, the line which joins the intersection of the bisectors to the vertical angle bisects the vertical angle. 28 PLANE GEOMETRY. PROPOSITION X. PROBLEM. 70, Determine the locus of a point at equal dis- tances from the extremities of a given line. Let A U represent the given line. To determine the locus of a point at equal distances from A and B. This problem means that we are to find the line, or lines, which pass through all those points which are the same distance from A as from B, and through no other points. SUG. 1. What line has all its points equally distant from A and #? Why? SUG. 2. Are any points, without said line, equally dis- tant from A and B ? Why ? SUG. 3. What, then, is the locus of a point, equally distant from the extremities of A B, and hence from the extremities of any line ? Ex. 22. The line joining the vertices of two isosceles triangles, on opposite sides of the same base, bisects the base and is perpendicular to it. Ex. 23. If the straight line which joins the vertex of a triangle with the middle point of the base is perpendic- ular to the base, the triangle is isosceles. Kx. 24. Prove that a line drawn from the vertex of an isosceles triangle to the middle point of the base, (1) bi- sects the triangle, (2) bisects the vertical angle, (3) is perpendicular to the base. RECTILINEAR FIGURES. 29 PROPOSITION XI. THEOREM. 71. Two triangles, having the three sides of one equal, respectively, to the three sides of the other, are equal in all respects. ADA A J$ C and D E F represent iwo triangles, having To prore that the triangles ABC and D E F are equal in all respects. SUG. 1. Place the A D E F upon the A A B C, so that the longest side, D F, of the A D E F, coincides with the longest side, A C, of the A A B C, D upon A, and F upon C, but the point E upon the opposite side of A C, from B. Draw B E. SUG. 2. In the A A B E, compare A B with A E. Give auth. SUG. 3. Compare Z.ABE with Z.AEB. Give auth. SUG. 4. In the A C B E, compare Z. C B E with Z C E B. Give auth. SUG. 5. Compare Z. A B C with Z. A E C. auth. SUG. 6. Compare A A B C with A A E C. auth. SUG. 7. Then how does 'A A B C compare with A D E Ft Why ? Therefore - Give Give 30 PLANE GEOMETRY. Ex. 25. In Prop. XI place two equal shorter sides upon each other, as A B upon D E, connect C and F, and demonstrate the pioposition. Ex. 26. If two straight lines bisect each other at right angles, any point of either is equidistant from the ex- tremities of the other. Ex. 27. The straight lines bisecting the equal angles of an isosceles triangle and terminating in the sides, are equal. 72. The premises of a proposition are the conditions given upon which the conclusion is based. The converse of a given proposition is a proposition which has the conclusion of the given proposition for one of its premises, and one of the premises of the given proposition for its conclusion. Proposition XII, separated into parts would read as follows: First premise There are two adjacent angles. Second premise Their sum equals two right angles. Conclusion Their exterior sides form a straight line. Proposition XII is the converse of Prop. II; which, separated into its parts, would read as follows: First premise There are two adjacent angles. Second premise Their exterior sides form a straight line. Conclusion Their sum equals two right angles. When a proposition is proved to be true, it does not necessarily follow that its converse is also true. RECTILINEAR FIGURES. 31 PROPOSITION XII. THEOREM. 73. If the sum of two adjacent angles equals two right angles, their exterior sides form a straight line. A~ O C" '"M Let OA, OB, and O C, be three lines, which meet to farm, two adjacent angles, A O B and B O C, whose sum equals two right angles. To prove that the exterior sides, O A and O C, form a straight line. SUG. 1. A O C is either a straight line or a broken line. To determine which of these suppositions is true, draw an extension of A O, as O M. SUG. 2. How many rt. Zs in Z A O B -f Z B O Ml Why? SUG. 3. How many rt. Zs in Z. A O B + Z B O C ? Why? SUG. 4. Compare the sum of Z A O B + Z. B O M> with the sum oiZ.AOB + BOC. Give auth. SUG. 5. Compare Z B O C with /.BOM. Give auth. SUG. 6. Since the Zs B O C and B O M are equal, and O C and O M are on the same side of O B, where does O C lie with respect to O Ml SUG. 7. But O Mis an extension of A O, by construc- tion. What relation does O C sustain to A Ol Therefore 32 PLANE GEOMETRY. MODEL. PROPOSITION XII. THEOREM. 74. If the sum of two adjacent angles equals two right angles, their exterior sides form a straight line. M Let O A, O B, and O C, be three lines which meet to form two adjacent angles, A O B, and B O C, whose sum equals two right angles. To prove that the exterior sides^ O A and O C, form a straight line. The line, A O C, is either a straight line or a broken line. Draw O M an extension of A O. Z A O B + Z B O M = 2 rt. Zs. PROP. II. Z A O B + Z B O C = 2 rt. Zs. HYP. Hence, /_AOB+^BOM = Z.AOB+Z_BOC. Ax. 1. Hence, Z.BOM=Z.BOC. Ax. 3. Since the Zs B O M and B O C are equal, and O C and (9 M are on the same side of O B, O C must lie upon O M. But A O M is a straight line by construction. Hence, as O C lies upon O M, A O C must be a straight line. Therefore, if the sum of two adjacent angles equals two right angles, their exterior sides form a straight line. RECTILINEAR FIGURES. 33 75. It is often more convenient to express the magni- tude of an angle in some other way than by stating how many right angles the given angle contains. To obtain another method, a right angle is divided into ninety equal parts, called degrees. The magnitude of an angle may, then, be expressed by stating how many degrees the given angle contains. EXERCISES. 28. How many degrees in a straight angle? In all the angular magnitude about a point ? 29. How many degrees in the supplement of two thirds of a right angle ? 30. How many degrees in an angle whose comple- ment equals one fourth of its supplement ? 31. The supplement of ten degrees is how much more than the complement of ten degrees ? 32. The supplement of any acute angle is how much more than the complement of the same acute angle ? Note. In elementary geometry it is only acute angles which have complements, but either acute or obtuse angles have supplements. 33. The supplement of the complement of any acute angle is how much more than the angle itself ? 34. The complement of the supplement of an obtuse angle is how much less than the given obtuse angle? 35. Two angles are complements of each other, and the greater exceeds the less by 38 degrees. What are the angles ? 36. If the bisectors of two adjacent angles are per- pendicular to each other, the angles are supplements of each other. 3 Geo. 34: PLANE GEOMETRY. PROPOSITION XIII. THEOREM. 76. From a point without a line, one, and only one, perpendicular can be drawn to the line. M O Let C D represent the line, and A the point. I. To prove that one perpendicular can be drawn from A to the line C D. SUG. 1. Draw any line, as M N, and at some point of this line, as O, erect the JL, O P. PROP. I. SUG. 2. Place the line, M N, upon the line, CD, and move it back and forth in C D. SUG. 3. Is it possible for O P to embrace the point A ? Therefore II- Only one perpendicular can lie drawn to the line. B Let A O represent a perpendicular from the point, A, to tJie line, C >. RECTILINEAR FIGURES. 35 To prove that no other perpendicular can be drawn from the point A to the line C D. SUG. 1. If another _L can be drawn, let it be repre- sented by A M. SUG. 2. Extend A O to D, making O B = A O, and connect M and B. SUG. 3. In the As A O M and BOM, compare A O with^? O, M O with M <9, and Z A OM\vithZ O M. SUG. 4. Now, compare Z A M O with Z_ B M O. Give auth. SUG. 5. Then, if by construction Z. A M O is a rt. Z., is line A MB a straight or a broken line? Why ? SUG. 6. Then, how many straight lines are drawn from A to B ? SUG. 7. What, then, do you conclude about the state- ment that A M B is a straight line? SUG. 8. Then, what do you conclude as to the possi- bility of A M being a _L from A to CD? SUG. 9. Then, how many _Ls can be drawn from a point to a straight line ? Therefore Ex. 37. If a perpendicular be dropped from the vertex of an isosceles triangle to the base, prove, first, that it bisects the base; second, that it bisects the vertical angle; third, that it bisects the triangle. Ex. 38. The triangles made with the base of an isos- celes triangle, and the lines bisecting the angles at the base, are equal. Ex. 39. A line drawn through the vertex of an angle, perpendicular to the bisector of the angle, makes equal angles with the sides of the given angle. PLANE GEOMETRY. MODEL. PROPOSITION XIII PART II. THEOREM. 77. Only one perpendicular can be drawn from a point to a line. M/ \ X H Let A O represent a perpendicular from the point A to the line C D. To prove that no other perpendicular can be drawn from the point A to the line C D. If another _L can be drawn, let it be represented by A M. Extend A O to D, so that A O = O B, and connect Mandfi. In the As A O M and B O M, A O = B O (by cons.), MO = MO (identical), and ZAOM= Z B O M (each being a rt. Z_ ). PROP. IV. If two As have two sides and the included ^/ of one, equal to two sides and the included ^/ of the other, each to each 4 , the triangles are equal in all respects. Therefore, the As A O M and O M are equal in all respects, and hence Z. A M O = Z. D M O. Since Z. A M O is a rt. ^, by construction, and Z. A M O equals Z. B M O, A M B is a straight line. PROP. XII. RECTILINEAR FIGURES. 37 If two adjacent angles are together equal to two right angles, their exterior sides form a straight line. But A O B is a straight line, hence A M B cannot be a straight line. Ax. 11. Therefore, only one J_ can be drawn from a point to a line. Ex. 40. If two vertical angles are bisected by two straight lines, prove that the bi- sectors together form one and the same straight line. Prove that tf- N O M is a straight line. SUG. See Prop. XII. Ex. 41. If A B Cbe an isosceles triangle, whose ver- tex is A, and whose base is B C, and if ^/be the middle point of A C and N the middle point of A B, and if the lines B M and C N intersect at O, prove (1) that B M equals C N, (2) that the triangles B C M and C B N art equal, (3) that the triangles C O M and B O N are equal, and (4) that the triangle B O C is isosceles. Ex. 42. If A B Cis an equilateral triangle, and D, E, and F are points in the sides A B, B C, and C A, re- spectively, such that A D = B E = C F, prove that the triangle D E F is equilateral. Ex. 43. If two lines, A B and CD, intersect in the point O, and if the lines A C and B D be drawn, A B -f C D is greater than A C + B D. Ex. 44. If A B C and A B D are two triangles on the same base, and on the same side of it, such that A C equals B D, and A D equals B C, and A D and B C in- tersect at 6>, prove (1) that the triangles A B Cand A B D are equal in all respects, (2) that the triangles A O C and BOD are equal in all respects, and (3) that the tri- angle A O B is isosceles. 38 PLANE GEOMETRY. 78. Definition. In a right triangle, the side which is opposite the right angle is called the hypotenuse. PROPOSITION XIV. THEOREM. 79. Two rijKt triangles which have the hypotenuse and a side of one, equal to the hypotenuse and a side of the other, are equal in all respects. C D F ^C M B Let ABC mid D E F represent two right triangles, having tJie hypotenuse A C equal to the hypotenuse D F 9 and the side A 15 equal to the side D E. To prove that the triangles ABC and D E F are equal in all respects. SuG. 1. Place the triangles so that A B coincides with D E, A upon D, and B upon E, with the vertices, C and /% on opposite sides of A B. SuG. 2. Is C B Fa. straight or a broken line ? Why ? SUG. 3. Connect A with the middle point, M, of C F. SUG. 4. Compare As A M C and A M F. Give auth. SuG. 5. Compare Zs C M A andFAfA. Give auth. SuG. 6. Then, are Zs C MA and F M A right or oblique Zs ? Why ? SUG. 7. Then, what relation of position do A M and A B sustain to each other ? And, hence, ^/and B ? Why ? SuG. 8. Compare As ABC and A B F, and hence As A B C and D E F, Therefore RECTILINEAR FIGURES. 39 PROPOSITION XV. THEOREM. 80. Two right triangles which have the hypotenuse and an acute angle of one, equal to the hypotenuse and an acute angle of the other, are equal in all respects. Let ABC and D EF represent two right triangles having tJie hypotenuse, A B, equal to tlie hypotenuse, D E, and the angle A equal to tin angle />. To prove that the triangles ABC and D E F are equal in all respects. SUG. 1. Place A A B C upon A D E F, so that A B coincides with D E, A upon D, and B upon E. SUG. 2. What direction will A C take ? Why ? SUG. 3. Where will the point C fall ? Why ? SUG. 4. Since B C and E F are both _L to the line D F, where will B C lie ? Why ? . PROP. XIII. SUG. 5. Where will the point C fall ? (Compare an- swers to sugs. 3 and 4.) SUG. 6. How, then, do the two As compare ? Therefore Note. In the answer to Sug. 3 it will be seen that C must lie somewhere in the line D F t and in the answer to Sug. 4 C must lie somewhere in the line E F. Hence, in the answer to Sug. 5, C can be exactly located. 4:0 PLANE GEOMETRY. 81, Straight lines, in the same plane, that do not and cannot meet, however far extended, are called parallel lines. PROPOSITION XVI. THEOREM. 82. Two lines, which are perpendicular to the same line, are parallel. D Let A C and B D represent two lines, each perpendicu- lar to the same line, A B, at the points A and B respec- tively. To prove that A C and B D are parallel. SUG. 1. The lines A C and B D will either meet or not meet. SUG. 2. If they meet, represent the point of meeting by O. SUG. 3. Compare the assumption that they meet, with Prop. XIII. SuG. 4. What, then, is your conclusion concerning their meeting ? What do you conclude with respect to their being || ? Give auth. Therefore RECTILINEAR FIGURES. 41 83- A transversal or secant line is a line which crosses two or more lines; as the line A B in the figure. When a transversal cuts two lines eight angles are formed, viz.; the angles 1-8 in the figure. C Jjb D T The interior angles are those within, or between, the lines; as 3, 4, 5, 6. The exterior angles are those without the lines; as 1, 2, 7, 8. Alternate interior angles are pairs of interior angles on opposite sides of the transversal, and not adjacent; as 3 and 6, 5 and 4. Alternate exterior angles are pairs of exterior angles on opposite sides of the transversal, and not adjacent; as 1 and 8, 2 and 7. Corresponding angles are pairs of angles on the same side of the transversal, one exterior and one interior but not adjacent; as 2 and 6, 4 and 8, etc. Note. The two lints which the transversal crosses mayor may not be parallel. If they are not parallel, and are extended until they meet, then the two lines, together with the transversal, form a tri- angle. Ex. 45. The perpendiculars from the extremities of the base of an isosceles triangle to the opposite sides, are equal, Ex. 46. If D & the middle point of the side B C, of a triangle ABC, and B E and C-Fare the perpendiculars from B and C to A D, prove that B E equals C F. 42 PLANE GEOMETRY. 84. Axiom 13. Only one line can be drawn through a given point parallel to a given line. PROPOSITION XVII. THEOREM. 85. If one of two parallel lines is perpendicular to a given line, the other one is perpendicular to the same line. Let E C and F D be two parallel lines, and let E C be perpendicular to A B. To prove that F D is perpendicular to A B. SUG. 1. From some point in F D, as O, draw O M _ to A B. SUG. 2. What relation does O M sustain to E Cl PROP. XVI. SUG. 3. Wh at relation does FD sustain to E t HYP. SUG. 4. What relation does MO sustain to F Dl Ax. 13. SUG. 5. Then what relation does F D sustain to A B ? Why? SeeSug. 1. Therefore RECTILINEAR FIGURES. 43 PROPOSITION XVIII. THEOREM. 86. If two parallel lines are cut by a transversal, the alternate interior angles are equal. U Let MNand R P represent two parallel lines cut by the secant A B. I. To prove that the alternate interior angles M E B and A F P are equal. SUG. 1. Through O, the middle point of E F, draw C D _L to R P. SUG. 2. What relation does CD sustain to M N? Why? SUG. 3. Compare As O F D and O E C. Give auth. SUG. 4. Then, how do Zs ME B and A FP com- pare ? Why ? II. To prove that the alternate interior angles NEB and R F A are equal. SUG. 1. /.MEB+ Z.NEB= Z.RFA + ^LPFA. Why? SUG. 2. Compare Zs NEB and R FA. (See Sug. 4, Part I, and Sug. 1, Part II.) Therefore 44 PLANE GEOMETRY. PROPOSITION XIX. THEOREM. 87. If two parallel lines are cut by a transversal, the corresponding angles are equal. A c D 77r B Let C D and E F represent tivo parallel lines, cut by the transversal A B. To prove thai the corresponding angles \ 2 -and 6, or 3 and j, etc. , are equal. SUG. See Prop. XVIII. PROPOSITION XX. THEOREM. 88. If two parallel lines are cut l>y a transversal, the interior angles on the same side of the transversal are supplements of each other. -D L represent two lines cut by the trans- versal E F, so that the corresponding angles EO A and E N C are equal. To prove that A B and C D are parallel. SuG. 1. Through O, the point of intersection of A B and E F, draw R S to represent a line || to C D. SuG. 2. Compare Z. E O R with Z. E N C. Give auth. SUG. 3. Compare Z. E O A with Z. E N C. Give auth. SUG. 4. Compare Z. E O R with Z. E O A. Give auth. SUG. 5. Since the side E O is the same in the two Z.s E O A and E O R, what relation of position must O A and O R sustain to each other ? SUG. 6. Since R S is, by construction, || to CD, what re- lation does A B sustain to C D ? Why ? Therefore 4:8 PLANE GEOMETRY. PROPOSITION XXIII. THEOREM. 92. // two straight lines are cut by a transversal so that the interior angles on the same side of the trans- versal are supplements of each other t the lines are parallel. ,B / -s OL. g R / / c OL D Let A B and C D represent two lines cut l>y the trans- versal E F 9 so that the angles A O F and E J\ C are sup- plements of each other. To prove ihat A B and C D are parallel. SUG. 1. Through O, the point of intersection of A D and E F, draw R S to represent a line || to C D. Suo.2. Compare Z.AOP with Z.ROF. (See method in Prop. XXI.) Complete the demonstration. There lore 93. An exterior angle of a triangle is an angle formed by one side of a triangle and an adjacent side extended. Angle n is an ex- terior angle of the triangle ABC. RECTILINEAR FIGURES. 49 94. When one side of a triangle has been extended to form an exterior angle, the two interior angles of the tri- angle which are not adjacent to the exterior angle are called opposite interior angles. In the figure in the previous article n is an exterior angle, and the angles A and C are the opposite interior angles of the tri- angle ABC. PROPOSITION XXIV. THEOREM. 95. An exterior angle of a triangle equals the sum of the opposite interior angles. BC represent a triangle, D AC an exterior angle t and B and C the opposite interior angles. To prove that the angle D A C equals the sum of the angles B and C. SUG. 1 . Through the point A draw a line M N || to B C. SUG. 2. Compare /.DAN with Z. B. Give auth. SUG. 3. Compare Z. C A A 7 ' with Z. C. Give auth. SUG 4. Compare Z D A C with Z B -f Z C. Therefore 4 Geo. 50 PLANE GEOMETRY. PROPOSITION XXV. THEOREM. 96. The sum of the interior angles of a triangle equals two right angles. '~ ~"D B Let ABD represent a triangle. To prove that the sum of the angles A, B and B D A equals two right angles. SuG. 1. Extend one of the sides, as B D. SuG. 2. Apply Prop. XXIV and complete the demon- stration. Therefore 97. COROLLARY I. A triangle cannot have more than one obtuse angle. 98. COROLLARY II. Every right triangle has two acute angles, each of which is the complement of the other. PROPOSITION XXVI. THEOREM. 99. If two triangles have two angles of one equal respectively to two angles of the other, the third angles are equal. SUG. Apply Prop. XXV to each A and complete the demonstration. 100. COROLLARY. If two right triangles have an acute angle of one equal to an acute angle of the other, the re- maining acute angles are equal. RECTILINEAR FIGURES. 51 PROPOSITION XXVII. THEOREM. 101. If two right triangles have a side and an acute angle of one equal to a side and an acute angle of the other, the triangles are equal in all respects. "Let ABC and D E F represent two right triangles in which A C equals D F, and ^_ B equals ^_ E. To prove that the triangles ABC and DBF are equal in all respects. Sue. Compare Zs A and D. Give auth. Complete the demonstration. Therefore Ex. 47. Prove Prop. XXVII by taking the side B C and the angle A, equal to the side E F and the angle D respectively; also, by taking the side A B and the angle B, equal to the side D E and the angle E respectively. PROPOSITION XXVIII. THEOREM. 102. If two triangles have two angles and a side of one equal respectively to two angles and a correspond- ing side of the other f the triangles are equal in all re- spects. Draw a figure and give the demonstration. 52 PLANE GEOMETRY. PROPOSITION XXIX. THEOREM. 103. If two angles of a triangle are equal, the sides opposite them cure equal and the triangle is isosceles. B M C Let ABC represent a triangle in which the angle B equals the angle C. To prove that the side A C equals the side A B. SUG. 1. Draw a line A M to represent a J_ from A ioC SUG. 2. Compare As^4 MB and A M C. Give auth. SUG. 3. Compare A C with A B. Therefore Ex. 48. If two parallel lines are cut by a transversal, prove that the alternate exterior angles are equal. Ex. 49. If the middle points of the three sides of an isosceles triangle be taken for the three vertices of a sec- ond triangle, prove that the second triangle is isosceles. Ex. 50. If from any point in the base of an isosceles triangle perpendiculars be drawn to the sides, these per- pendiculars make equal angles with the base. Ex. 51. In the figure for Prop. XX, compare angles 1 and 6, also angles 4 and 7. RECTILINEAR FIGURES. 53 PROPOSITION XXX. THEOREM. 104. // two angles of a triangle are unequal, the sides opposite them are unequal, the greater side being opposite the greater angle. Let ABC represent a triangle in which the angle C is greater than the angle B. To prove that the side B A is greater than the side C A. SUG. 1. Draw C M to represent a line making the Z B CM equal to the Z B. SUG. 2. Compare B Mand CM. Give auth. SUG. 3. Compare CM -f MA with C A. Give auth. SUG. 4. Compare B M + MA with C A. SUG. 5. Compare B A with C A. Therefore Ex. 52. Prove Prop. XXV by another method. SUG. Through A draw a line || to D B. Ex. 53. In Prop. XXV extend D A and B A through A, and draw through A, a line || to O B, and then prove the proposition in another way. 54 PLANE GEOMETRY. PROPOSITION XXXI. THEOREM. 105. If two sides of a triangle are unequal the angles opposite them are unequal, the greater angle being opposite the greater side. A jy c Let ABC represent a triangle in which the side A Bis greater than the side A C. To prove that the angle C is greater than the angle B. SUG. 1. The Z C equals Z B, or is less than Z B, or is greater than Z B. SUG. 2. If Z C were equal to Z B how would A B and A C compare ? Why ? Does this agree with the hyp . ? Is it possible for Z* C to equal Z B ? SUG. 3. If Z C were less than Z B how would A B and A C compare ? Why ? Does this agree with the hyp. ? Is it possible for Z C to be less than Z B ? SUG. 4. From the results reached in Sugs. 1,2,3, what must be the relation between Zs C and B ? Therefore - Note. If the student, in original investigation of the proposition just given, or any similar one, should chance to consider the supposi- tion which leads to the truth before one or more of the others, the re- maining suppositions should be investigated. When all possible sup- positions have been stated, one of which is true, and it has been shown that all but one are false, it is evident that the one remaining must be true. RECTILINEAR FIGURES. 55 106. Proposition XXXI is a good illustration of what is known as the indirect method, or the reductio adabsurdum method of reasoning. Its peculiarity consists in the fact that the statement of the proposition is not directly proved to be true, but that everything which contradicts the statement of the proposition is shown to lead to some manifest absurdity, and is therefore false. This method often presents difficulties to the beginner on account of the fact that he is obliged to admit, temporarily, and for argument's sake, something which the argument itself goes to destroy. But, as the reasoning is examined more closely, it will be found to be just as satisfactory as the direct demonstration, and in many cases (especially the converse of propositions which have been proved by di- rect demonstration), this method is the easiest and best one to apply. In applying this method care must be taken that every possible case which contradicts the propo- sition be considered and each one shown to lead to an ab- surdity. Then, and then only, is this method of demon- stration rigid. Ex. 54. If A B C be a right triangle with the right angle at C, and if through C a line meeting the hypote- nuse at D be drawn in such a manner that the angle A C D equals the angle B and the angle BCD equals the angle A, prove that CD is perpendicular to the hypotenuse A B. Ex. 55. If A B C be a right triangle with the right angle at C, and if through C a line meeting the hypote- nuse at D be drawn in such a manner that the angle A CD equals the angle A, and the angle BCD equals the angle B, prove that CD bisects the hypotenuse A B. 56 PLANE GEOMETRY. PROPOSITION XXXII. THEOREM. 107. If a perpendicular and oblique lines are drawn from a point to a given line: I- The perpendicular is shorter than any oblique line. II- Two oblique lines which meet the given line at equal distances from the foot of the perpendicular, are equal. III. Of two oblique lines meeting the given line at unequal distances from the foot of the perpendicular, the more remote is the greater. Let AD be perpendicular to a given line J5 E 9 and A &, A C and A E oblique lines, meeting the given line at J5, C and E respectively; and let DBbe greater tJian D E, and Z> C equal to D E. I. To prove that A D is the shortest line from A to B E. SuG. 1. If A C represents any oblique line, how does Z. A D C compare with Z A C D* Why? SUG. 2. How does A D compare with A C, and hence with any oblique line from A to B EJ II. To prove that A C equals A E. SuG. Compare the triangle ADC and A D E. RECTILINEAR FIGURES. 57 Complete the demonstration of this part of the propo- sition. III. To prove that A B is greater than A E. SUG. 1. Compare ^.s A B D and A CD with respect to magnitude. Give auth. SUG. 2. Compare Zs A CD and A C B with respect to magnitude. Give auth. SUG. 3. Compare Zs A C B and ABC with respect to magnitude. SUG. 4. Compare A B with A C, and hence with A E. Give auth. SUG. 5. Notice the conclusions in parts I, II and III. Therefore Ex. 56. A line drawn from one end of the base of an isosceles triangle, perpendicular to the opposite side, makes with the base, an angle equal to one half the ver- tical angle. Ex. 57. A straight line drawn from any point in the bisector of an angle to either side and parallel to the other side, makes, with the bisector and the side to which the line is drawn, an isosceles triangle. Ex. 58. The angle formed by the bisectors of the angles at the base of an isosceles triangle is equal to an exterior angle at the base of the triangle. Ex. 59. A C B and A D B are two triangles on the same side of A B, such that A C equals B D, and A D equals B C, and A D and B C intersect at O. Prove that A O B is an isosceles triangle. Ex. 60. The difference between any two sides of a tri- angle is less than the third side. 58 PLANE GEOMETRY. PROPOSITION XXXIII. THEOREM. 108. Converse of Prop.XXXII,!. The shortest line from a point to a line is a perpendicular to the line. M B "Let A B represent the shortest line from A to the line C D. To prove that A B is perpendicular to C D. SUG. 1. If A B is not _L to C D, draw A M to repre- sent a _L to C D. SUG. 2. If A M is J_ to C D, how does it compare in length with A B1 Prop. XXXII. SUG. 3. Compare your answer with the hypothesis. Therefore 109. The distance from a point to a line is the length of the perpendicular from the point to the line. Ex. 61. Lines which are perpendicular to parallel lines are parallel. Ex. 62. If the equal angles of an isosceles triangle be bisected, the bisectors and the base of the triangle form a new isosceles triangle whose vertical angle is equal to an exterior angle at the base of the original triangle. Ex. 63. The sum of the three sides of any triangle is greater than the double of any one side, but less than the sum of the doubles of any two sides. RECTILINEAR FIGURES. 59 PROPOSITION XXXIV. THEOREM. 110. Two equal oblique lines, drawn from the same point in the perpendicular to a given line f cut off equal distances from the foot of the perpendicular. C B D Let ABbe perpendicular to C D, and let AC and A be equal oblique lines drawn from A to CD. To prove that B C equals B D. SUG. Compare As A B C and A B D. Give auth. Therefore Ex. 64. Each angle of an equilateral triangle is one third of two right angles, or two thirds of one right angle. Ex. 65. The line joining the feet of the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides is parallel to the base. Ex. 66. The sum of the three straight lines drawn from any point within a triangle to the three vertices, is less than the sum and greater than the half sum of the sides. Ex. 67. If a line intersects the sides of an isosceles triangle at equal distances from the vertex, the line is parallel to the base. 60 PLANE GEOMETRY. PROPOSITION XXXV. THEOREM. 111. Two unequal oblique lines, drawn from the same point in the perpendicular to a given line, cut off unequal distances from the foot of the perpendicu- lar, the longer line cutting off the greater distance. Let ABbe perpendicular to C D, and let A C and A D be unequal oblique lines drawn from A to C D, A C being longer than A D. To prove that B C is greater than B D. SUG. 1. B Cis equal to B D, or is less than B D, or is greater than B D. SUG. 2. If B C were equal to B D, how would A C compare with A D ? Why ? SUG. 3. If B C were less than B D, how would A C compare with A D ? Give auth. SUG. 4. How, then, must B C compare with B Dl Therefore Ex. 68 If the vertical angle of an isosceles triangle is one half as great as an angle at the base, a bisector of' a base angle divides the given triangle into two isosceles triangles. RECTILINEAR FIGURES. 61 EXERCISES. 69. Prove that every point in the bisector angle is equally distant from the sides of the angle. SUG. O M and O N are J_ to A B and A C respectively. (See definition of dis- tance.) Prove O M equal to O N. 70. Prove that every point not in the bisector of an angle is unequally distant from the sides of the angle. Prove that O M and O N are unequal. SUG. Draw S R _L to A N. Connect R and O. Compare 5 O + S R with OR, O M with OR, O M with O N. 71. What is the locus of a point equally distant from the sides of an angle ? 72. What is the locus of a point at equal distances from the two intersecting lines ? 73. The middle point of the hypotenuse of a right tri- angle is at equal distances from the three vertices. 74. The sum of the exterior angles at the base of any triangle is equal to two right angles plus the vertical angle of the triangle. 75. In a triangle ABC, the angle C is twice the sum of the angles A and B, and the angle B is twice the angle A\ find all three angles of the triangle. 76. In a triangle ABC, the angle B is three times the angle A, and the angle C is five times the angle A\ find each angle of the triangle. 62 PLANE GEOMETRY. PROPOSITION XXXVI. THEOREM. 112. If two lines be drawn from a point in a tri- angle to the extremities of a side, the sum of the two lines so drawn is less than the sum of the ottier two sides of the triangle. Let ABC represent any triangle, O any point within the triangle, and O B and O C lines drawn from O to B and C, respectively. To prove that O B + O Cis less than A B + A C. SUG. 1. Extend one of the lines through O to meet the side of the triangle at M. SUG. 2. How does B A + A M compare with B Ml Why? SUG. 3. Then how does BA + AM+MC compare with^^/+ M C? SUG. 4. Compare O M + M C with O C. SUG. 5. Then how does B O + O M + M C compare with B O + O C? SUG. By referring to your answers to Sugs. 3 and 5, compare B A + A C with B O + O C. Therefore RECTILINEAR FIGURES. 63 PARTIAL DEMONSTRATION. Answer to Sug. 3. B A + A C is greater than B M + MC. (If equals be added to unequals the results are unequal, that being greater which is obtained by addition to the greater of the two given unequals.) Answer to Sug. 5. B M + M C is greater than B O f O C. (Same reason.) Answer to Sug. 6. Hence, much more must BA + A C be greater than B O + O C. Ex. 77. Prove that the three bisectors of the angles of a triangle meet in the same point. SUG. 1. Draw a A A B C. SUG. 2. The bisector of the Z A is the locus of a point equally distant from the sides A B and A C. See exercises 69 and 70. SUG. 3. The bisector of the Z B is the locus of what? SUG. 4. Prove, now, that the point of intersection of these two bisectors is on the bisector of the /_ C. Ex. 78. If a straight line intercepted between parallel lines is bisected, any other straight line drawn through the point of bisection to meet the parallel lines is also bi- sected at that point; and the two intersecting lines cut off equal portions on the parallel lines. Ex. 79. A B is the hypotenuse of a right triangle ABC, B D is drawn bisecting the angle B, meeting A C at D, and D E is drawn perpendicular to A C, meeting A B at E. Prove that E D B is an isosceles triangle. 64 PLANE GEOMETRY. PROPOSITION XXXVII. THEOREM. 113. If two triangles have two sides of one equal to two sides of the other and the included angles un- equal, the remaining sides are unequal, that "being greater which belongs to the triangle having the greater included angle. Let ABC and A' B' ' represent two triangles, in which A B equals A' B', A C equals A' C' t and ^_ A is greater tfian ^_A'. To prove that B C is greater than B' C'. SUG. 1. Place A A' B 1 C' upon A A B C so that A' & coincides with A B, A' upon A, and B 1 upon B. SUG. 2. Since ^ A' is less than ^ A, where, with re- spect to Z B A C, must A' C' fall ? CASE I. Suppose that C falls without the triangle ABC. SUG. 3. Bisect the Z. C A C', and extend the bisector to meet B C at O. Connect O and C'. SUG. 4. Compare As A O C and A O C'. Give auth. SUG. 5. Compare O C and O C. SUG. 6. Compare B O + O C' with B C, hence B C with B C', hence B C with B 1 C'. SUG. 7. What is your conclusion in this case ? RECTILINEAR FIGURES. 65 CASE II. Suppose that C' falls upon the line B C. SUG. 8. In the figure at the A right compare B C' with B C t hence B' C' with B C. (A part is less than the whole.) SUG. 9. What is your con- clusion in this case ? B C' ~"^ CASE III. Suppose that C f falls within the triangle ABC. SUG. 10. In the figure at the right compare A C + C^with A C+C B. SUG. 11. As^Cequals A C', how does B C' compare with B C, and hence B 1 C' with B C? SUG. 12. What is your conclusion in this case? Therefore Note. The pupil can read- ily see the possibility of the three cases in this proposition by an examination of the fig- ure at the right. B Ex. 80. Give another demonstration of Case I, of Proposition XXXVII, by using the following sug- gestions: SUG. 1. Connect Cand C'. B SUG. 2. Compare Z. A C Cwith A CC. SUG. 3. Compare ^LBC Cwith B CC. Therefore 5 Geo. 66 PLANE GEOMETRY. Ex. 81. Give another demonstration of Case III, of Prop. XXXVII, by us- ing the following sug- gestions: SUG. 1. Connect C and C. SUG. 2. Compare ZACCwithZACC SUG. 3. Compare sup- ^ plement of Z. A C 1 C with Z B C 1 C. SUG. 4. Compare supplement of Z. A C C' with Z.BCC . SUG. 5. Compare Z B C 1 C with Z B C C'. Therefore PROPOSITION XXXVIII. THEOREM. 114. If two triangles have two sides of one equal to two sides of the other and the third sides unequal the angles opposite the third sides are unequal, that being greater which is opposite the greater third side. Let ABC and A' B' C' represent two triangles having A B equal to A' B' and A C equal to A' C' and B C greater tJian B' C'. RECTILINEAR FIGURES. 67 To prove that the angle A is greater than the angle A' . SUG. 1. If Z A equals Z A', how do B C and B' C compare ? SUG. 2. If Z. A is less than Z A', how do B C and B' C' compare ? SUG. 3. How, then, must Z A compare with Z. A ' ? Therefore POLYGONS. 115. A polygon is a portion of a plane, bounded by straight lines; as A B C D E. The bounding lines are called the sides of the polygon, and their sum is called the perimeter of the polygon. The angles formed by the sides of the polygon on the side of the enclosed space, are called the in- terior angles of the polygon, as the angle B A E in the figure. An angle formed by one side of the polygon and an adjacent side extended, is called an exterior angle of the polygon; as the angle m in the figure. The vertices of the interior angles of a polygon are called the vertices of the polygon. A line joining any two vertices, not adjacent, is called a diagonal of a polygon. When no ambiguity arises, a polygon is frequently read by naming any two vertices, not adjacent; as A D, B D etc. 68 PLANE GEOMETRY. 116. Polygons are classified according to the number of their sides. The least number of sides a polygon can have is three. A polygon of three sides is a triangle. A polygon of four sides is called a quadrilateral. A polygon of five sides is called a pentagon. A polygon of six sides is called a hexagon. Etc. 117. An equilateral polygon is a polygon all of whose sides are equal. An equiangular polygon is one all of whose angles are equal. A convex polygon is a polygon no side of which, if extended, would enter the space enclosed by the perim- eter of the polygon; as ABCDE in the figure of Art. 113. A concave polygon is a polygon, two or more sides of which, if extended, would A enter the space enclosed by the perimeter of the polygon; as ^ \. A B C D E F,- in the figure at \ / the right. If either A B or B C V / r _ft I be extended through B it would enter the space enclosed by the perimeter of the polygon. The angle A B C in this figure is called a re-entrant angle. QUADRILATERAL. 118. Quadrilaterals are divided into classes as follows: The trapezium, which has no two of its sides parallel. The trapezoid, which has two sides parallel. The parallelogram, which has its pairs of opposite sides parallel. RECTILINEAR FIGURES. 69 C At 119. The parallel sides of a trapezoid are called bases, the non-parallel sides are called legs, and the perpendicular dis- tance between the bases is called the altitude of the trapezoid. In the figure at the right A B and C D are the bases, A C and B D are the legs, and A Mis the altitude of the trapezoid. 120. Either pair of opposite sides of a parallelogram may be called bases, but when one pair is selec- ted as bases the other pair must not be called bases. When a pair of parallel sides of a paral- lelogram is selected as bases one of these sides is called the primary base, and the other the secondary base. The perpendicular distance between the bases of a parallelogram is called the altitude of the parallelogram. Usually the words lower base and upper base are used instead of primary base and secondary base respectively, but as geometry does not take into account the idea of up and down, the terms primary and secondary are preferable. 121. Parallelograms are classified as follows: A parallelogram all of whose angles are oblique is called a rhomboid, and a paralello- gram all of whose angles are right angles is called a rect- angle. B PARALLELOGRAM. 70 PLANE GEOMETRY. The distinction between the different kinds of parallelograms should be kept clearly in mind. Students often make the mistake of drawing the figure of a rectangle when proving a proposition about a parallelogram. A rectangle is a parallelogram, but a parallelogram is not necessarily a rectangle. Every property of a parallelogram is also a property of a rectangle, but a property of a rectangle may not be a property of parallelograms in general, and, if a rectangle is drawn in connection with a proposition about a parallelogram, there is danger of being misled into applying to parallelograms some prop- erty which is true only of rectangles. RHOMBUS. SQUARE. 122. A rhomboid whose sides are all equal is called a rhombus, and a rectangle whose sides are all equal is called a square. PROPOSITION XXXIX. THEOREM. 123. The opposite sides of a parallelogram are equal. D Let AB C D represent a parallelogram. To prove that the opposite sides A C and B D are equal; also that the sides C B and A D are equal. RECTILINEAR FIGURES. 71 SUG. 1. Draw the diagonal A B. SUG. 2. Compare the As A B C and A B D. SUG. 3. How then does A C compare with D B, and C B with A D ? Therefore 124. COROLLARY. The diagonal of a parallelogram divides it into two equal triangles. PROPOSITION XI^. THEOREM. 125. If a quadrilateral have two of its sides equal and parallel, it is a parallelogram. Let A C B D represent a quadrilateral in which tJie side A C is equal tint! parallel to the side B D. To prove that A B C D is a parallelogram. SUG. 1. Draw the diagonal A B. SUG. 2. Compare the As A C B and A D B. SUG. 3. Then how does ^/ m compare with ^ n ? Why? SUG. 4. If Z. m equals ^ n, what relation does A D sustain to C B ? Does the quadrilateral A C B D agree with the defini- tion of a EU ? Therefore 72 PLANE GEOMETRY. PROPOSITION XU. THEOREM. 126. A quadrila teral whose opposite sides are equal is a parallelogram. Let ACB D represent a quadrilateral in which A C equals D B and A Z> equals C B. To prove that the quadrilateral A C B D is a parallelo- gram. SUG. 1. Draw the diagonal A B. SuG. 2. Compare As A C B and A D B. SUG. 3. How does Z! m compare with ^ n ? Why ? SUG. 4. See Prop. XL. Therefore - QUERY. How many and what statements have been made that may be used as definitions of a parallelogram ? Note. The pupil should carefully distinguish between a property of a figure and a statement that may be used as a definition. A definition of a figure is such a description of it as serves to dis- tinguish it from every other figure. A property of a figure is simply something which is true of the figure, and may also be true of other different figures. Ex. 82. In any triangle ABC, the angle A plus twice the angle B minus three times the angle C equals 110 degrees, and the angle A minus twice the angle B plus three time the angle C equals 90 degrees; find the angles A, B and C. RECTILINEAR FIGURES. 73 PROPOSITION XLH. THEOREM. 127. Two parallelograms which have two sides and the included angle of one equal to two sides and the included angle of the other, each to each, are equal in all respects. B Let AD and A' J>' represent two parallelograms in which A S equals A' B' 9 A C equals A' C' and the angle A equals the angle A ' . To prove that the parallelograms are equal in all re- spects. SUG. 1. Place the 7 A B D C upon 7 A' B' D' C so that A B coincides with A' B', A upon A ', and B upon B', and so that the two figures shall fall upon the same side of A' B'. SUG. 2. What diiection will A Ctake? Why? SUG. 3. Where will the point Cfall? Why? SUG. 4. What direction will C D take ? Why ? SUG. 5. Where will the point D fall ? Why ? Therefore Ex. 83. In proposition XLJI, suppose A C equals A' C', CD equals C' D ', and the angle C equals the angle C'. then draw the diagonals A D and A' D ', and prove the proposition in another way. 74 PLANE GEOMETRY. PROPOSITION XUII. THEOREM. 128. The diagonals of a parallelogram bisect each other. Let K TJ N M represent a parallelogram, and L M and K Nits diagonals, intersecting at O. To prove that K O equals O N and L O equals O M. SUG. Compare A K O L with A M O N, or ^ L O N with A 1C O M. Therefore PROPOSITION XUV. THEOREM. 129. The opposite angles of a parallelogram are equal. Let A C B D be a parallelogram. To prove that the angle CAD equals the angle C B or that the angle C equals the angle D. RECTILINEAR FIGURES. 75 SUG. 1. Draw the diagonal A B. SUG. 2. Compare ^s m and n. Give auth. SUG. 3. Compare Zs C A B and A B D. Give auth. SUG. 4. Compare Zs C A D and C B D. Give auth. SUG. 5. In a similar manner compare Z!s C and Z>. Therefore PROPOSITION XLV. THEOREM. 130. The sum of the interior angles of any convex polygon is equal to twice as many right angles as the polygon has sides, minus four right angles. B A E Let A B C D E F represent a convex polygon. To prove that the sum of the interior angles of the poly- gon equals twice as many fight angles as the polygon has sides, minus four right angles. SUG. 1 . Connect each vertex with O, any point within the polygon. SUG. 2. If the polygon has n sides, how many As are formed ? SUG. 3. How many rt. ^.s in the sum of the angles of each A? How many in the sum of the angles of all the As ? SUG. 4. What is the sum of the Zs about <9? SUG. 5. Then, how many rt. ^s in the sum of the in- terior Z-s of the polygon ? Therefore 76 PLANE GEOMETRY. PROPOSITION XI/VI. THEOREM. 131. The sum of the exterior angles of any convex polygon is equal to four right angles. L,et A B C D E represent a convex polygon. To prove that the sum of the exterior angles a, c, e, m and o, equals four right angles. SUG. 1. What is the sum of the Z.& a and ? Of c and dt Off and e ? Etc. SUG. 2. If the polygon has n sides, the sum of all the exterior and interior Z^s equals how many rt. ^.s ? SUG. 3. See Prop. XLV, and complete the demonstra- tion. Therefore PROPOSITIONS IN CHAPTER I. PROPOSITION I. At a given point, in a straight line, one perpendicular can be erec- ted to that line, and but one. PROPOSITION II. If one straight line meets another straight line, the sum of the ad- jacent angles formed equals two right angles. PROPOSITION III. If two straight lines intersect, the vertical angles formed are equal. PROPOSITION IV. If two triangles have two sides and the included angle of one equal to two sides and the included angle of the other, each to each, the tri- angles are equal in all respects. PROPOSITION V. If two triangles have two angles and the included side of one, equal to two angles and the included side of the other, each to each, the triangles are equal in all respects. PROPOSITION VI. The angles opposite the equal sides of an isosceles triangle are equal. PROPOSITION VII. If a perpendicular be erected at the middle point of a straight line, the distances from any point in the perpendicular to the extremities of the line are equal. PROPOSITION VIII. Any side of a triangle is less than the sum of the other two. 78 PLANE GEOMETRY. PROPOSITION IX. If a perpendicular be erected at the middle point of a straight line, the distances, from a point not in the perpendicular, to the extrem- ities of the line are unequal. PROPOSITION X. Determine the locus of a point at equal distances from the extrem- ities of a given line. PROPOSITION XI. Two triangles having the three sides of one equal, respectively, to the three sides of the other, are equal in all respects. PROPOSITION XII. If the sum of two adjacent angles equals two right angles, their ex- terior sides form a straight line. PROPOSITION XIII. From a point without a line, one, and only one, perpendicular can be drawn to the line. PROPOSITION XIV. Two right triangles which have the hypotenuse and a side of one equal to the hypotenuse and a side of the other, are equal in all re- spects. PROPOSITION XV. Two right triangles which have the hypotenuse and an acute angle of one equal to the hypotenuse and an acute angle of the other, are equal in all respects. PROPOSITION XVI. Two lines which are perpendicular to the same line, are parallel. PROPOSITION XVII. If one of two parallel lines is perpendicular to a given line the other one is perpendicular to the same line. PROPOSITION XVIII. If two parallel lines are cut by a transversal, the alternate interior angles are equal. RECTILINEAR FIGURES. 79 PROPOSITION XIX. If two parallel lines are cut by a transversal, the corresponding angles are equal. PROPOSITION XX. If two parallel lines are cut by a transversal, the interior angles on the same side of the transversal are supplements of each other. PROPOSITION XXI. If two straight lines are cut by a transversal, so that the alternate interior angles are equal, the lines are parallel. PROPOSITION XXII. If two straight lines are cut by a transversal so that the correspond- ing angles are equal, the lines are parallel. PROPOSITION XXIII. If two straight lines are cut by a transversal so that the interior angles on the same side of the transversal are supplements of each other, the lines are parallel. PROPOSITION XXIV. An exterior angle of a triangle equals the sum of the opposite in- terior angles. PROPOSITION XXV. The sum of the interior angles of a triangle equals two right angles. PROPOSITION XXVI. If two triangles have two angles of one equal respectively to two angles of the other, the third angles are equal. PROPOSITION XXVII. If two right triangles have a side and an acute angle of one equal to a side and an acute angle of the other, the triangles are equal in all respects. PROPOSITION XXVIII. If two triangles have two angles and aside of one equal respectively to two angles and a corresponding side of the other, the triangles are equal in all respects. 80 PLANE GEOMETRY. PROPOSITION XXIX. If two angles of a triangle are equal, the sides opposite them are equal, and the triangle is isosceles. PROPOSITION XXX. If two angles of a triangle are unequal, the sides opposite them are unequal, the greater side being opposite the greater angle. PROPOSITION XXXI. If two sides of a triangle are unequal, the angles opposite them are unequal, the greater angle being opposite the greater side. PROPOSITION XXXII. If a perpendicular and oblique lines are drawn from a point to a given line: I. The perpendicular is shorter than any oblique line. II. Two oblique lines which meet the given line at equal distances from the foot of the perpendicular, are equal. III. Of two oblique lines meeting the given line at unequal dis- tances from the foot of the perpendicular, the more remote is the greater. PROPOSITION XXXIII. CONVERSE OF PROP. XXXII, I. The shortest line from a point to a line is a perpendicular to the line. PROPOSITION XXXIV. Two equal oblique lines, drawn from the same point in the perpen- dicular to a given line, cut off equal distances from the foot of the perpendicular. PROPOSITION XXXV. Two unequal oblique lines, drawn from the same point in the per- pendicular to a given line, cut off unequal distances from the foot of the perpendicular, the longer line cutting off the greater distance. PROPOSITION XXXVI. If two lines be drawn from a point in a triangle to the extremities of a side, the sum of the two lines so drawn is less than the sum of the other two sides of the triangle. RECTILINEAR FIGURES. 81 PROPOSITION XXXVII. If two triangles have two sides of one equal to two sides of the other and the included angles unequal, the remaining sides are unequal, that being greater which belongs to the triangle having the greater in- cluded angle. PROPOSITION XXXVIII. If two triangles have two sides of one equal to two sides of the other and the third sides unequal, the angles opposite the third sides are unequal, that being greater which is opposite the greater third side. PROPOSITION XXXIX. The opposite sides of a parallelogram are equal. PROPOSITION XL. If a quadrilateral have two of its sides equal and parallel, it is a parallelogram. PROPOSITION XLI. A quadrilateral whose opposite sides are equal is a parallelogram. PROPOSITION XLII. Two parallelograms which have two sides and the included angle of one equal to two sides and the included angle of the other, each to each, are equal in all respects. PROPOSITION XLIII. The diagonals of a parallelogram bisect each other. PROPOSITION XLIV. The opposite angles of a parallelogram are equal. PROPOSITION XLV. The sum of the interior angles of any convex polygon is equal to twice as many right angles as the polygon has sides, minus four right angles. PROPOSITION XLVI. The sum of the exterior angles of any convex polygon is equal to four right angles. 6 Geo. CHAPTER II. THE CIRCLE. DEFINITIONS. 132. A circle is a portion of a plane bounded by a curved line all points of which are equally distant from a fixed point within it. The fixed point is called the center of the circle, and the bounding line is called the circumference of the circle. In higher branches of mathematics the word circle is used to denote what is here defined as the circumference, i. e., the curved line bounding a portion of a plane instead of that portion of a plane itself, but in this book the above definitions will be adhered to. 133. Any straight line drawn from the center to the circumference is called a radius. (O A is a radius.) Any straight line drawn through the center, terminated both ways by the cir- cumference, is called a diameter. (C B is a diameter.) From the definition of a circle all radii of the same circle are equal; also, all diameters of the same circle are equal, each diameter being twice the radius. 134. An arc of a circle is any por- tion of its circumference; as A M B. THE CIRCLE. 83 A chord of a circle is a straight line joining any two points on the circumference; as A B. A chord of a circle is said to subtend an arc whose extremities are the same as the extremities of the chord. The chord A B subtends the arc A M B, and also the arc A C D B. Thus, any chord subtends two arcs which together make up the whole circumference, but, when an arc and its chord are spoken of the smaller of the two arcs is always understood unless the contrary is specifically stated. 135. A secant of a circle is any straight line meeting the circumference in two points and passing through at least one of these points; as C D E. A secant may be considered as a chord extended. 136. A tangent to a circle is a straight line which has one and but one point in com- ... mon with the circle; as M T N. The line is said to be tangent to the circle, and the circle to the line. I The common point is called the point of contact or point of tan- gency. 137. A segment of a circle is a portion of a circle bounded by an arc and its subtending chord; as segment A M B in the figure in article 134, i. e., the part of the circle included between the arc A MB and the chord A B. 138. A sector of a circle is a portion of a circle bounded by two radii and the intercepted arc. In the figure in Art. 134, the sector C O D is bounded by the arc and the radii O C and O D. 84: PLANE GEOMETRY. The arc intercepted between two radii is said to sub- tend the angle made by the radii; as arc C D subtends the angle COD. 139. A circle is often read by naming the letter at the center of the circle. From the definition of a circumference it may be con- sidered as the locus of a point at a fixed distance from the center. Also the distance from the center to any point within the circle is less than the radius and the distance from the center to any point without the circle is greater than the radius. PROPOSITION I. THEOREM. 140. Two circles are equal if the radius of one equals the radius of the other. Let the figures O and S represent two circles having equal radii. To prove that the circles O and S are equal. SuG. 1. Place the O upon the O S, with the center O upon the center 5". THE CIRCLE. 85 SUG. 2. Where must the circumference of O fall with respect to the circumference of ,S ? (See definition of O.) Therefore INDIRECT METHOD. SUG. 1. Same as above. SUG. 2. Suppose some part of circumference of one O should fall outside of the circumference of the other O, how would the radii compare? PROPOSITION II. THEOREM. 141. A diameter divides a circle into two equal parts. Let A D B C represent a circle, and ABa diameter. To prove that A B divides the circle into two equal parts. SUG. 1. Revolve the segment A D B about the line A O B into the plane A C B. SUG. 2. Where will the arc A D B fall with respect to the arc A C B ? (See definition of O.) 142. A semicircle is a segment of a circle bounded by a diameter and the arc it subtends, as A D B in the above figure. 86 PLANE GEOMETRY. PROPOSITION III. THEOREM. 143. In the same circle, or in equal circles, equal angles at the center intercept equal arcs at the cir- cumference. Let O and S represent equal circles, and let A O S and C & X> represent equal angles at the centers O and S respe- ctively. To prove that the arc A B is equal to the arc C D. SUG. 1. Place O O upon O S, so that the lines O A and O B fall upon 5 C and 5 D, respectively. Why is this possible ? SUG. 2. Where will the arc A B fall ? Why ? Therefore Ex. 84. A parallelogram having one right angle is a rectangle. Ex. 85. Prove that a quadrilateral is a parallelogram if its opposite angles are equal. Ex. 86. A quadrilateral is a parallelogram if its diag- onals bisect each other. Ex. 87. The diagonals of a rhombus or a square bisect each other at right angles. THE CIRCLE. 87 PROPOSITION IV. THEOREM. 144. Converse of Prop. III. In the same circle, or in equal circles, equal arcs subtend equal angles at the center. Let O and S represent equal circles, and let A Band CD represent equal arcs. To prove that the angle A O B equals the angle C S D. SUG. 1. Place O O upon O 5 so that arcs A B and C D coincide. Why is this possible ? Complete the demonstration. Therefore Ex. 88. If two adjacent sides of a rectangle are equal the figure is a square. Ex. 89. If a line parallel to the base of a triangle bi- sects one side it bisects the other side also. A B is bisected at D, and D E is parallel Prove A E equals E C. SUG. Draw D M \\ to A C. Ex. 90. In the previous exercise instead of drawing D M parallel to A C, draw E M parallel to A B, and prove the exercise. 88 PLANE GEOMETRY. PROPOSITION V. THEOREM. 145. In the same circle, or in equal circles, chords which subtend equal arcs are equal. Let O and S represent equal circles, in which the arc A S equals the arc D. To prove that the chord A B is equal to the chord C D. SUG. 1. Draw the radii O A, OB, S C, and 5 D. SUG. 2. Compare ^/s O and ,S. Give auth. SUG. 3. Compare As A O B and C S D. Give auth. SUG. 4. Compare chord A B with chord C D. Therefore Ex. 91. With the same construction as in exercise 89, connect E and M, and prove the exercise without refer- ence to any proposition in quadrilaterals. Ex. 92. If a line bisects the two legs of a triangle, prove that it is parallel to the base. Ex. 93. In exercise 89, prove that D E equals one half the base B C. THE CIRCLE. 89 PROPOSITION VI. THEOREM. 146. Converse of Prop. V. In the same circle, or in equal circles, arcs which are subtended by equal chords are equal. Let O and S represent equal circles in which the chord A jB equals tJie chord C D. To prove that the arc A B is equal to the ate CD. SUG. 1. Draw radii O A, OB, S C, and 5 D. SUG. 2. Compare As A O B and C S D. Give auth. SUG. 3. Compare ^s O and 6". Give auth. SUG. 4. Compare arcs A B and C D. Give auth. Therefore Ex. 94. The diameter of a parallelogram divides it into two equal parts. A diameter of a quadrilateral is a line which joins the middle points of two opposite sides. Ex. 95. The two diameters of a parallelogram bisect each other. 90 PLANE GEOMETRY. PROPOSITION VII. THEOREM. 147. In the same circle, or in equal circles, two chords which subtend unequal arcs are unequal, that chord ~being greater which subtends the greater arc. Let O and S represent equal circles, and let the arc A B be less than the arc C J>. To prove that the chord A B is less than the chord C D. SUG. 1. Connect the extremities of the arcs with their respective centers. SUG. 2. Place O upon O S, O upon S, and A upon C. SUG. 3. As arc A B is less than arc C D, where will the point B fall with respect to the arc C D ? SUG. 4. Compare Z C S E with Z C S D. Give auth. SUG. 5. How, then, does chord C E compare with chord C D, and hence A B with C D ? Why ? Therefore Ex. 96. Draw two parallel lines and a transversal. Bisect each of the two interior angles on the same side of the transversal, and prove that the bisectors are per- pendicular to each other. THE CIRCLE. 91 PROPOSITION VIII. THEOREM. 148, Converse of Prop. VII. In the same circle, or in equal circles, two arcs which are subtended by un- equal chords are unequal, that arc being greater which is subtended by the greater chord. Let O and S represent equal circles, and let the chard A -B be less than tlie cliord C D. To prove that the arc A B is less than the arc C D. SUG. 1. Draw radii A O, B O, C S, and D S. SUG. 2. Compare Zs A O B and C S D. Give auth. SUG. 3. Place sector A O B upon sector C S D, so that A O falls upon C S. SUG. 4. Where will O B fall with respect to the sec- tor C S D ? SUG. 5. Where will B fall ? SUG. 6. With A upon C, and B located with respect to D, how does arc A B compare with arc C D ? Therefore Bx. 97. A B C is an isosceles triangle whose base is B C and whose vertex is A: Extend B A through A to Q, making A O equal to B A. Connect O and C. Prove that O C is perpendicular to B C. 92 PLANE GEOMETRY. PROPOSITION IX. THEOREM. A radius which is perpendicular to a chord bisects the chord and its subtended arc. Let O Dbea radius perpendicular to the chord L M. To prove that L E equals E M, and that arc L D equals arc D M. SuG. Compare the As L O and M E O. Complete the demonstration. Therefore Ex. 98. If A is the vertex and B C the base of an isos- celes triangle ABC, and if from any point D in the side A B a line be drawn perpendicular to the base and meet- ing C A extended at E, prove that the triangle A D E is isosceles. Ex. 99. If a perpendicular be drawn from the vertex of the right angle of a right-angled tri- ^ angle to the hypotenuse, prove that the two triangles formed are mutually equi- angular to the original triangle. Prove that As A B M and A C M are mutually equiangular to A A B C. THE CIRCLE. PROPOSITION X. THEOREM. 150. A line perpendicular to a chord at its middle point passes through the center. Let A B represent a cliord, and C D a perpendicular t& A B at its middle point C. To prove that C D passes through the center of the circle. SUG. 1. From 0, the center, of the circle, drop a J_ to the chord A />. SUG. 2. Where will C O lie with respect to C D ? Why? Therefore 151. COROLLARY I. If two circles intersect, the line joining their centers is perpendicular to their common chord at its middle point. SUG. Erect a _L to the common chord at its middle point and extend both ways. 152. COROLLARY II. If two equal circles intersect, the common chord bisects at right angles the line join- ing their centers. PLANE GEOMETRY. PROPOSITION XI. THKORKM. 153. In the same circle, or in equal circles, equal chords are equally distant from the center, and of two unequal chords the greater is nearer the center. I. Let A B and C D represent equal chords in the equal circles O and S, and O M and S N their respective distances from the centers O and S. To prove that O M equals S N. SUG. 1. Draw the radii O B and 5 D. SUG. 2. Compare A O MB with A S N D. Give auth. SUG. 3. What is your conclusion about the equality of II. In the circle O let A B and A E represent unequal chords, A B being greater tlian A E, and O M and O P their respective distances from the center O. To prove that O M is less than O P. SUG. 1. Compare arc A E with arc A B. Prop. VIII. SUG. 2. Where does E lie with respect toA? Why ? SUG. 3. L,et F represent the intersection of P O and A B. Why will they intersect ? THE CIRCLE. 95 SUG. 4. Compare O M with O F, O F with O P, and, finally, O M with O P. SUG. 5. Compare 5* N with O P. Therefore PROPOSITION XII. THEOREM. 1 54. A straight line perpendicular to the radius of a circle at its extremity is tangent to the circle. Let O .B represent a straight line perpendicular to the radius O S at its extremity O. To prove that O B is tangent to the circle. SUG. 1. The point O is common to the circle 5 and the line OB. SUG. 2. I^et M represent any other point in O B. Draw S M. SUG. 3. Compare 5* O and S Mm respect to length. SUG. 4. Where, then, is M with respect to the circle ? See Art. 139. Therefore 96 PLANE GEOMETRY. PROPOSITION XIII. THEOREM. 155. Converse of Prop . XII If a straight line is tan- gent to a circle the radius meeting it at the point of contact is perpendicular to it. Let O B represent a tangent to the circle 8 9 O the point of contact, and S O the radius drawn to the point of con- tact. To prove that S O is perpendicular to O B. SUG. 1. Draw 6* M, any other line from 6* to.O B. SUG. 2. Compare 6* O and S M in respect to length. SUG. 3. See article 108. Therefore 156. CORONARY. At any point in a circumference, one, and only one, tangent can be drawn. Ex. 100. In the figure at the right, A E and B C are paral- lel, and M O is a transversal. B O bisects the angle A O M, and CO bisects the angled O M. B ' JM C Prove that B M equals M C. THE CIRCLE. 97 PROPOSITION XIV. THEOREM. 157. Arcs of a circle intercepted by parallel chords are equal. Let A B and C Z> represent two parallel lines intercept- ing the arcs A C and B D. To prove that arc A C equals arc B D. SUG. 1. Drop a J_ from O to CD, and extend it to meet the circumference at M. SUG. 2. How is O M related to A B ? See Art. 85. SUG. 3. See Prop. IX, and complete the demonstra- tion. Therefore Ex. 101. If one of the equal sides of an isosceles tri- angle be extended through the vertex and the exterior angle formed be bisected, prove that the bisector is par- allel to the base. Ex. 102. If the diagonals of a parallelogram are equal, prove that the parallelogram is a rectangle. Ex. 103. The bisectors of the interior angles of a par- allelogram form a rectangle. 7-Geo. 98 PLANE GEOMETRY. PROPOSITION XV. THEOREM. 158. Through three points, not in the same straight line, one circumference, and only one, is possible. A* B Let A, B and C represent three given points not in the same straight line. To prove that through A, B and C one circumference, and but one^ is possible. SUG. 1. What is the locus of a point equally distant from A and C? Give auth. SUG. 2. What is the locus of A point equally distant from A and B ? Give auth. SUG. 3. Will these two loci intersect ? Why ? SUG. 4. Is there a point equally distant from .A, B, and C? SUG. 5. Can a circumference pass through the points A, #and C? SUG. 6. Can there be more than one such circumfer- ence ? Why ? Therefore THE CIRCLE. 99 MEASUREMENT. 159. Anything which can be measured by a unit of the same kind is called quantity. The quantities used in geometry are the geometric magnitudes, viz. , lines, surfaces and solids. To measure a quantity is to find out how many times it contains another selected quantity of the same kind, called the unit of measure. In every day experience, the unit of measure is a standard ac- cepted by general consent; as a foot, a square yard, a ton, a cord, etc. 160. The measure of a quantity is the number which expresses how many times the unit of measure is con- tained in the given quantity. A quantity is completely expressed when the measure and unit of measure are both expressed; as three yards. In this example three is the measure, a yard is the unit of measure, and three yards is the quantity completely expressed. 161. The ratio of two quantities is the number of times the first contains the second; i. e., the measure of the first regarding the second as the unit of measure; or, having measured both quantities by the same unit, the quotient of the measure of the first divided by the meas- ure of the second. The ratio of line A to line B is the A number of times A contains B, which may^be determined by laying off B B upon A. Or, if A and B b^ meas- ured by same unit m t the ratio of m A to B is the number of times the measure of A contains the measure of B. Suppose m 100 PLANE GEOMETRY. is contained c times in A, and d times in B, then the ratio of A to B equals c divided by d> which may be ex- c pressed as -7- As ratio plays an important part in subsequent geometry, it is im- portant that the pupil have clear and definite ideas of what a ratio actually is, and the relation it sustains to the subject of division. From the definition of the ratio of two quantities as the number of times the first contains the second, it follows immediately that a ratio can exist only between quantities of the same kind, and also that the ratio of two quantities is always a pure number'. For example, 6 ft. contains 3 ft. twice, hence the ratio of 6 ft. to 3 ft. is 2, not 2 ft. but simply 2. Again, division has been defined as the process of finding how many times one quantity is contained in another. Now, it is clear that, with this conception of division, the quotient is identical with the ratio of two quantities, as above defined. But there are two kinds of division; the one just described maybe called the division of meas- urement; the other is the division of separation, and means the proc- ess of dividing or separating a quantity into a certain number of equal parts. For example, 6 ft. may be divided into three equal parts, and each part will be 2 ft. In the division of measurement the divisor is a quantity of the same kind as the dividend, and the quotient is a pure number; but, in the division of separation, the divisor is a pure number, and the quotient is a quantity of the same kind as the divi- dend. Now, the ratio of two quantities identifies itself with the quo- tient only in the division of measurement. As the ratio of two quantities is the quotient, in one kind of divis- A ion, the ratio of A to B is often written in the fractional form, . It B is very important to keep constantly in mind the fact that A and B are quantities of the same kind, and that the ratio -^ is zpure number. The case considered above, in which A and B are lines, and each is measured by the the same third line m, gives ^=^- Jt wil1 be seen here that A and B are geometric quantities, and c and and hence can be no greater than E B. If E B is exactly contained in C D, then E B is the greatest common unit; for if it is contained in C D it is contained in any number of times C D, and hence in A E and also in A E plus E B. I^ay off E B upon C D as many times as possible. If there is a remainder proceed as before. When a remainder is obtained which is exactly contained in the last preceding divisor, it is the greatest common unit. The student should give the reason for each statement in the above process. 102 PLANE GEOMETRY. PROPOSITION XVII. PROBLEM. 1 64. Given the greatest common unit of measure of two quantities to find their ratio. SUG. See definition of ratio. 165. Two quantities are said to be incommensurable when it is impossible to exactly measure both of them by the same unit, i. *?., to express their ratio as a frac- tion whose numerator and denominator are whole num- bers. The side and diagonal of a square may be shown to be incommensurable, as follows: B Let A C represent a square, and B D the diagonal. To find whether there is a common unit of measure of the side and diagonal. Upon the diagonal B D lay off D E equal to the side DC, as in Prop. XVI. There is a remainder B E, and the greatest common unit of measure of D C and D B is the greatest common unit of measure of B E and DC, or of B E and B C. At E, draw E F, _L to the diagonal B D. In the rt. &BEF\hzZ_FBE= half a rt. Z, and hence the Z BFE = half art. Z. Hence B E = E F. THE CIRCLE. 103 Again, if a line D F were drawn, the two rt. As D C F and DBF would have the same hypotenuse, and the sides D C and D E equal. Hence E F = F C. Our problem now is to find the greatest common unit of measure, if one exists, of B E and B C. Upon the diagonal F B lay off F M = B E. Then as C Fand F M are each equal to B E, it is clear that B E is contained twice in B C with the remainder B M. The greatest common unit of measure, if one exists, of B E and B C is the greatest common unit of measure of B M and B E. At M draw M R, _L to the diagonal B F. In the same manner that it was proved that B E =* E F ' = F C it may also be shown that B M M R = RE. Then B M is the side of a square of which B R is the diagonal. If we now lay off on B E a line equal to B M, the con- ditions are the same as when B E was laid off on B C. However far the process is continued it is evident that each remainder becomes the side of a new square, and the last preceding divisor is divided into two parts, one of which is the diagonal of the square, and the other a side. Hence, at no time in the process will the division be exact, and therefore no common unit of measure exists for the diagonal and side of a square. Hence they are incommensurable. 166. The ratio of two incommensurable quantities is called an incommensurable ratio, or an incommensurable number. The ratio of the diagonal and side of a square has been computed to equal - 104 PLANE GEOMETRY. Note. An incommensurable number must not be looked upon as an inexact number. If the side of a square is one foot the diagonal is ]/ 2 feet, but this diagonal is a perfectly definite length. 167- A constant is a quantity whose value is fixed. A variable is a quantity which, under the conditions imposed upon it, may assume an indefinite number of values. For example, the distance from a railway sta- tion to a moving train of cars is a quantity which has one value at one time, another value a minute later, and still another value another minute later, and so on. 168, When a variable so changes that it continually approaches sonfe fixed quantity which it cannot reach, but from which it may be made to differ by an amount less than any assigned quantity, however small, the fixed quantity is called the limit of the variable, and the va- riable is said to approach its limit. Note. By the above definitions it will be noticed that not all va- riables have limits; but those considered in the following pages do have limits. 169. ILLUSTRATIONS. One mih of a given line is a va- riable which approaches zero as a limit if m is made to increase indefinitely, and a minus one mth of a is a vari- able which approaches a as a limit if m is made to in- crease indefinitely. If a point be made to move along a given line, the dis- tance from a fixed point to the moving point is a vari- able. This variable may or may not have a limit. If such conditions be imposed upon the moving point that during a second it moves along half the length of the line, the next second half the remaining distance, the next, half the remaining distance, etc., the length of the given line is the limit of the variable. THE CIRCLE. 105 The pupil should remember that the limit and the va- riable are always the same kind of quantities; as in the example just given, the limit and variable are both dis- tances. PROPOSITION XVIII. THEOREM. 170. // two variables are always equal as they ap- proach tlieir limits, their limits are equal. O O D Let A m and C n represent two variables, and A JB and C D tlieir respective limits. To prove that A B equals C D. Lay off A B upon C D. Let A B equal C O. By hypothesis A m and C n are always equal, i. e. : A m = C n, A m' = Cn', etc. But as A m approaches A B, or its equal C O, as its limit, and C n approaches C D as its limit, there must come a time, if C O is greater or less than C D, when the value A m is not equal to C n, which is contrary to the hypothesis. Hence C O, or its equal A B, must equal CD. Therefore, if two variables are always equal as they approach their limits, their limits are equal. Ex. 104. If the middle points of the sides of a quad- . rilateral be joined in order, the figure formed is a paral- lelogram. SuG. Draw the diagonals of the given quadrilateral. 106 PLANE GEOMETRY. PROPOSITION XIX. THEOREM. 171. In the same circle, or in equal circles, two angles at the center have the same ratio as the arcs which they intercept at the circumference. D Let O and S represent two equal circles, A O U and C S -D two angles at their centers 9 and A JB and C 2> the arcs which they intercept at the circumference. ., . Z. A O B cere A B To prove that ^r-^ equals -^^. There are two cases. CASE I. When the angles A O B and C S D are com- mensurable. SuG. 1. I/et m represent the common unit of meas- ure for the two angles. SuG. 2. If Z. m is contained 5 times in Z! A O B and 4 times in Z! C S D, what is the ratio . r ? n e( l ua l to? SuG. 3. Extend the lines of division of the ^/s to the arcs. How do arcs A E, E F, C I, I K, etc. , com- pare ? Why ? SUG. 4. How does the number of arcs in arc A B com- pare with the number of Z.$> in ^ A O B ? THE CIRCLE. 107 How does the number of arcs in arc C D compare with the number of ^.s in Z. C S Z?? Sue. 5. Then the ratio -^-= equals what? arc CD SUG. 6. Compare the ratio of the ^s with the ratio of the arcs. Therefore CASE II. When the angles A O B and C S D are in- commensurable. D Let the angles A O B and C 8 D be incommensurable. SUG. 1. Take any unit of measure of one of the ^.s, as of Z. C S D, and apply this unit to the Z A O B. There mnst be a remainder. Why ? Let this remainder be represented by /.MOB. SUG. 2. The Zs A O M and COD are commensur- able. Why ? SUG. 3. Compare the ratio > ^r- with the ratio . C o U arc AM ^. ,~. Give auth. arc CD SUG. 4. By taking the unit of measure of the Z. C SD smaller and smaller continually, the remainder, viz., the Z. M O B, may be made as small as we please, i. e., it may be made to approach zero at its limit. Can it be made to disappear entirely ? Why? (See Sug. 1.) 108 PLANE GEOMETRY. SUG. 5. The Z A O M is a variable. Why ? (See definition of variable.) SUG. 6. Is Z. C S D a variable or a constant ? ^ A. O M SUG. 7. Is the ratio -, _ ? _ a constant or a vari- able ? Why ? SUG. 8. What is the limit of ^ ^ c ff * * toward ^L. C o Xx , ,. . Z.A O M what ratio is . approaching ? SUG. 9. Is the ratio ^-=j a constant or a variable ? arc CD Why? fly/* ^\ H/F SUG. 10. What is the limit of - ^^ z. . arc A M what ratio is 7=r^r approaching r arc CD ^. A O M SUG. 11. Compare the two variables en an< ^ ^_ C o D /y y/" /4 1W _ as they approach their limits. (See Sug. 3.) arc CD SUG. 12. Compare their limits. (Prop. XVIII.) Therefore Ex. 105. The diameters of any quadrilateral bisect each other. SUG. See definition of diameter in exercise 94. Ex. 106. The sum of the angles at the vertices of a five-pointed star is equal to two right angles. See figure at the right. Ex. 107. If a line be drawn in a trape- zoid, bisecting one of the legs and parallel to the bases, prove that it bisects the other leg also. THE CIRCLE. 109 MODEL. PROPOSITION XIX. THEOREM. 172. In the same circle, or in equal circles, two angles at the center have the same ratio as the arcs which they intercept at the circumference. Let O and S represent two equal circles, A O B and C S D two angles at tJieir centers, and A B ana C I> the arcs which they int< rccpt at the circumference. ., . Z- A O B , arc A B To brove that , ^ ^ equals CS D There are two cases. arc C CASE I. When the angles A O B and C S D are com- mensurable, Let /. m represent the common measure of Z.S A O B and C S D. If m is contained 5 times in Z. A O B and 4 times in = - (See definition of Z. CS D, the ratio / (- O - ratio.) Extend the sides of the ^s to the arcs; then arcs A E > E F,C I, I K, etc., are all equal. 110 PLANE GEOMETRY. Hence, the ratio (In the same circle, or in equal circles, equal angles at the center in- tercept equal arcs.) There are 5 equal arcs in arc A B, and 4 in arc C D t for each angle at the center intercepts an arc. arc A B _ 5 arc CD == 4' Z A O B arc A B Therefore, . _ = - -. Z. C S D arc C D Therefore, when the angles at the centers are commen- surable they have the same ratio as their intercepted arcs. CASE II. When the angles A O B and C S D are in- commensurable. B Let the angles A O S and C S D be incommensurable. Take a unit of measure of one of the Zs, as of /. C S D, and apply this unit to the Z. A O B. Since the Zs are incommensurable, this unit will be contained in Z_ A OB a certain number of times with a remainder, as Z MOB. Since the Zs A O M and C S D have a common unit of measure, they are commensurable, also the arcs A M and CD are commensurable, and hence, by case I, Z A O M _ arc A M Z CS D " S arcCD ' THE CIRCLE. Ill Now, by taking the unit of measure of the ,/ C S D smaller and smaller continually, the remainder, viz., the ^L M O B, which is always less than the unit, may be made as small as we please, but cannot be made to dis- appear entirely, for then the Zs A O B and C S D would be commensurable. Hence, the remainder, MOB, ap- proaches zero as a limit, and, therefore, the Z A O M approaches the /LA O B as a limit. But the Z. C S D is a constant, and hence the ratio Z A O M / r c ri~ 1S a variable which approaches the ratio 2 C o JJ AAOB arc A M Again, the ratio -- ^ _ is a variable which ap- arc CD . arc A B ,. . proaches the ratio - F-^T as a limit. arc CD ^AOM . arc A M Hence, as ..- and - _ _ are two variables Z C S D arc C D which are always equal as they approach their limits, their limits must be equal (by Prop. XVIII). That is, Z A OB arc A B Z CS D~ != arc C D' Therefore, in the same circle, or in equal circles, two angles at the center have the same ratio as the arcs which they intercept at the circumference. 173, SCHOLIUM I. A degree, which has already been denned as ^ of a right angle, or, what is the same thing, 5^-ff of the whole angular magnitude about a point, is often taken as a standard unit angle; and the arc it in- tercepts on the circumference, also called a degree, or -gfa of the circumference, is often taken as a standard unit arc. 112 PLANE GEOMETRY. 174. SCHOLIUM II. By the proposition it will be seen that the number of times the unit angle is contained in the given angle at the center is the same as the num- ber of times the unit arc is contained in the arc inter- cepted by the given angle. Hence, we say, briefly, that an angle at the center is measured by its intercepted arc. 175. SCHOLIUM III. A right angle which contains 90 degrees (written 90), is also often used as a unit angle in estimating the size of other angles. 176. An inscribed angle is one whose vertex is in the circumference, and whose sides are chords of the circle. The angle D E F is an inscribed angle. An angle is inscribed in a segment of a circle when its vertex is in the arc of the segment and its sides meet the ex- tremities of the chord of the arc. The angle D F is inscribed in the segment DBF. PROPOSITION XX. THEOREM. 177. An inscribed angle is measured by one half its intercepted arc. Let A be an inscribed angle, whose sides intercept the arc B C. THE CIRCLE. 113 To prove that the angle A is measured by one half the arc B C. There are three cases. CASE I. When one side of the angle passes through the center of the circle. SUG. 1. Connect O, the center of the circle, with C. SUG. 2. Compare Z A with Z. O C A. Give auth. SUG. 3. Compare /.BO Cwith Z A. Give auth. SUG. 4. By what arc is Z B O C measured ? Why ? SUG. 5. Then, by what part of the arc B C is the Z A measured ? CASK II. angle. When the center lies between the sides of the SUG. 1. Through the vertex A draw a diameter A M. SUG. 2. By what arc is the /.BAM measured? Why? SUG. 3. By what arc is the /.CAM measured? Why? SUG. 4. Then, by what arc is the Z. B A C measured ? 8 Geo. 114 PLANE GEOMETRY. CASE III. When both sides of the angle are on the same side of the center. SUG. 1. Through the vertex A draw a diameter A M. SUG. 2. By what arc is the /.MAC measured? SuG. 3. By what arc is the /_ M A B measured ? SUG. 4. Then, by what arc is the ^ B A C measured? SuG. 5. Compare the last suggestion with Sug. 4, case II; and Sug. 5, case I. Therefore PROPOSITION XXI. THEOREM. 178. An angle formed by a tangent and a chord is measured by one half the intercepted arc. A B Af Let A B represent a tangent, and AC a chord. To prove that the angle B A C is measured by one half the arc ADC. THE CIRCLE. 115 SUG. 1. Draw, through A, the diameter A M. SUG. 2. By what arc is the Z MA C measured? Why? SUG. 3. By what arc is the Z. B A M measured ? Why? SUG. 4. Then, by what arc is the Z B A C measured ? Therefore PROPOSITION XXII. THEOREM. 179, The angle formed by two secants, meeting without the circle, is measured by one half the differ- ence of the intercepted arcs. .A L,et A B and A C repi'esent two secants, meeting at A 9 without the circle, and forming trie angle SAC. To prove that the angle A is measured by one half the difference of the arcs B C and E D. SUG. 1. Through E, draw a chord E M\\ to A B. SUG. 2. By what arc is the Z. ME C measured? Why? SUG. 3. Compare the Z. A with the Z ME C. SUG. 4. Then, by what arc may Z. A be said to be measured ? SUG. 5. Express the arc M C in terms of B C and D E. SUG. 6. Then, by what arc may Z. A be said to be measured ? Therefore 116 PLANE GEOMETRY. PROPOSITION XXIII. THEOREM. 180. An angle formed by two intersecting chords is measured by one half the sum of the intercepted arcs. Let A JB and C D represent two chords intersecting at X. To prove that the angle A X D is measured by one half the sum of the arcs C B and A D. SUG. 1. Through D, the extremity of the chord CD, draw D M\\ to A B, the other chord. SUG. 2. Compare Z CXB with Z C D M. Give auth. SUG. 3. Then, by what arc is /.CXB measured? Why? SUG. 4. Express the arc CM in terms of B C and B M, and then in terms of B C and A D. SUG. 5. Then, by what arc may the Z. C X B be said to be measured. Therefore 181. A polygon is said to be inscribed in a circle when each angle of the polygon is an in- scribed angle; as polygon A B C D. When a polygon is inscribed in a circle, the circle is said to be circumscribed \ about the polygon. THE CIRCLE. 117 A polygon is circumscribed about a circle when each of its sides is tangent to the circle; as polygon A B C D. When a polygon is circumscribed about a circle, .the circle is said to be inscribed in the polygon. PROPOSITION XXIV. THEOREM. 1 82. If two circles are tangent to each other the line joining their centers passes through the point of con- tact* Let the circles O and S be tangent to each other, X the point of contact, and O S the line joining ttie centers. To prove that O S, extended if necessary, passes through the point of contact X. SUG. 1. Draw a common tangent, A B, to the two circles. SUG. 2. At the point of contact, X, erect a J_ to A B, and extend it through both circles. SUG. 3. Why will the _L pass through the centers ? SUG. 4. What relation will this line sustain to O S ? Give auth. Therefore 118 PLANE GEOMETRY. PROBLEMS OF CONSTRUCTION. 183. In the previous work the constructions have been represented, instead of being actually performed. When it is known that a certain relation of points, lines, or surfaces is possible, it has been considered sufficient to represent that relation rather than to actually con- struct it. Since elementary geometry deals only with figures which can be made from straight lines and circumfer- ences of circles, the constructions of elementary geometry are those which it is possible to effect by means of a straight edge and dividers, which are necessitated by the postulates named below. Problems of construction belong in no sense to pure geometry, .but are simply applications of the principles demonstrated in pure geometry. POSTULATES OF CONSTRUCTION. 184. Let it be granted: 1. That a straight line may be drawn between any two points and may be extended to any length through either extremity. 2. That a circle may be drawn with any point as a center, and with any straight line as a radius. Note. It will be observed that the first postulate necessitates the straight edge, and the second the dividers. Ex. 108. If the middle points of the three sides of a triangle be joined by straight lines, the triangle is divided into four triangles which are equal in all respects. Ex. 109. A diameter of a circle is greater than any other chord. THE CIRCLE. 119 PROPOSITION XXV. PROBLEM. 185. To bisect a given straight line. Let ABbe given^a straight line. To bisect A B. SUG. 1. Use the^truth that if two equal circles inter- sect, the line joining their centers is bisected at right angles by their common chord. SUG. 2. To use this truth, A and B must be the cen- ters of two equal intersecting circles. Hence, construct two equal intersecting circles, with A and B as centers, by postulate 2, and draw their common chord by postu- late 1. Note. A little experience will suggest how to omit the unessential parts of lines. For instance, in the above construction all the cir- cumferences may be omitted, except short arcs near the points of in- tersection, M and N. Ex. 110. The sum of the bases of a trapezoid is equal to twice the \"~~7 \ diameter joining the middle point \ / \ d\Z- A of the legs. Prove A B equals one half the sum of C D and E G. EN SUG. Draw M N || to D G, through A. 120 PLANE GEOMETRY. PROPOSITION XXVI. PROBLEM. 186. To erect a perpendicular to a given line, at a given point in that line. 'T-. A Let A B be the given line, and M the given point. To erect a perpendicular to A B, at~M. SUG. Use the same principle as in the preceding prob- lem. In order to use the principle, M must be the middle point of the line joining the centers of the Os. Hence, find points on the line A B equally distant from M (post- ulate 2), and proceed as in the preceding problem. SECOND METHOD. SUG. 1. Use the two principles that only one straight line can be drawn through two points (Ax. 11), and that all the points in the _L, at the middle point of a straight line, are equally distant from the extremities of the line. Hence, if two points of the _L can be found, the line drawn through them must be the J_. SUG. 2. What point of the J_ is given and may be used as one of the two points ? SUG. 3. To find another point see Prop. XXV. THE CIRCLE. 121 THIRD METHOD. SUG. 1. Use the truth that an /_ inscribed in a semi- circle is a rt. < and the side included between these angles equal to the line C D. SUG. Apply Prop. XXXI, so as to make C and D each a vertex of the required A. Ex. 113. Prove that the line which bisects a chord and its subtended arc passes, if extended, through the center of the circle. THE CIRCLE. 127 PROPOSITION XXXV. PROBLEM. 195. Given two sides and ihe included angle of a triangle, to construct the triangle. M Let M and N represent the two sides, and O the included angle. To construct a triangle having M and N for iwo of its sides, and the angle foimed by those sides equal to the angle O. This problem is so simple that the student can solve it without the aid of any suggestions. Ex. 114. Parallel strright lines, included between par- allel straight lines, are equal. Ex. 115. What is the locus of the middle points of all the chords of a circle which are parallel to a given line. Ex. 116. A radius drawn to the middle point of an arc will bisect its subtended chord. Ex. 117. A straight line cannot cut the circumference of a circle in more than two points. SUG. If it cut in three points, how many equal lines could there be drawn from a point to a straight line ? How many straight lines can be drawn from a point to a straight line ? 128 PLANE GEOMETRY. PROPOSITION XXXVI. PROBLEM. 1 96. To construct a triangle whose sides are three given lines. A B r. D Let A B, C D and E F represent three given lines. To construct a triangle whose sides are A B> C D and E F respectively. SUG. 1. Draw the line A B. How many vertices of the required A are now located ? SUG. 2. With the extremity C, of the line C D, fixed at A) what is the locus of the point D? SUG. 3. From the answer to Sug. 2, where does the third vertex of the required A lie ? SUG. 4. With the extremity E, of the line E F, fixed at B, what is the locus of the point F1 SUG. 5. From the answer to Sug. 4, where does the third vertex of the required A lie ? , SUG. 6. Now, give complete directions for construct- ing the A. Can three such lines be given that the problem is im- possible ? Ex. 118. Prove that the two tangents from a point to a circle are equal. SUG. Connect the center of the circle with the given point. Complete the construction. THE CIRCLE. 129 EXERCISES. 119. If a tangent and a chord of a circle are parallel, prove that the arcs intercepted between the tangent and chord are equal. 120. Upon a given base, construct an isosceles triangle, in which the sum of two equal sides shall equal a given line. 121. All angles inscribed in the same segment are equal. 122. Prove that an angle inscribad in a semicircle is a right angle. To prove that the angle B A C is a right angle. 123. Demonstrate Prop. XXII by an- other method. Sue. Draw B E. By what arc s D E meas- Then, by ured ? By what arc is Z. B E C measured ? what arc must Z. A be measured ? 124. In the same circle, or in equal circles, an angle inscribed in the smaller ^ of two segments is larger than an angle ^ inscribed in the larger segment. Prove that the angle A E B is larger than the angle COD. 125. An angle formed by a tangent and a secant is measured by one half the difference of the intercepted arcs. 126. The segments of a straight line intercepted by concentric circles are equal. Prove A B equals CD. Concentric circles are circles having the same center. 9-Geo. 130 PLANE GEOMETRY. PROPOSITION XXXVII. PROBLEM. 197. Through a given point, to draw a tangent to a given circle. CASE I. When the point is on the circumference. Let A represent the given paint in tlie circumference of the given circle. To draw a tangent to the circle O, through the point A. SUG. 1. What relation does a tangent bear to the ra- dius drawn to the point of contact ? SUG. 2. Give the method of making the construction in this case. II. When the given point is without the circle. A Let A represent the given point without the circle O. To draw a tangent to the circle O, through the point A. THE CIRCLE. 131 SUG. 1. Connect O and A. SUG. 2. If the required point of contact were joined both to O and A, what kind of an Z. would be formed ? SUG. 3. What is the locus of the vertex of the Z in a A whose base is O A ? See Art. 177, and Bxs. 121, 122. SUG. 4. Give, now, complete directions for finding the point of contact, and hence for constructing the tangent. PROPOSITION XXXVIII. PROBLEM. 198. To circumscribe a circle about a given tri- angle. Let ABC represent the given triangle. To circumscribe a circle about the triangle ABC. SUG. 1. The problem is to find the center of a O whose circumference passes through A, B and C; i. e., to find a point equally distant from A, B and C. SUG. 2. See Art. 158, and then give complete direc- tions for circumscribing a O about the A A B C. QUERY. How many circles can be circumscribed about a triangle ? Why ? 132 PLANE GEOMETRY. PROPOSITION XXXIX. PROBLEM. 199. To inscribe a circle in a given triangle* Let ABC represent the triangle. To inscribe a circle in the triangle ABC. SUG. 1. If a O can be inscribed in the A, the center of the O must be equally distant from the three sides. SUG. 2. Is there such a point? See exercise 71. SUG. 3. How may this point be found ? SUG. 4. Give complete directions for inscribing a circle in a given A. QUERY. How many circles can be inscribed in a given triangle? Why? PROPOSITIONS IN CHAPTER II. PROPOSITION I. Two circles are equal if the radius of one equals the radius of the other. PROPOSITION II. A diameter divides a circle into two equal parts. PROPOSITION III. In the same circle, or in equal circles, equal angles at the center intercept equal arcs at the circumference. PROPOSITION IV. CONVERSE OF PROP. III. In the same circle, or in equal circles, equal arcs subtend equal angles at the center. PROPOSITION V. In the same circle, or in equal circles, chords which subtend equal arcs are equal. PROPOSITION VI. CONVERSE OF PROP. V. In the same circle, or in equal circles, arcs which are subtended by equal chords are equal. PROPOSITION VII. In the same circle, or in equal circles, two chords which subtend unequal arcs are unequal, that chord being greater which subtends the greater arc. PROPOSITION VIII. CONVERSE OF PROP. VII. In the same circle, or in equal circles, two arcs which are subtended by unequal chords are unequal, that arc being greater which is subtended by the greater chord. PROPOSITION IX. A radius which is perpendicular to a chord bisects the chord and its subtended arc. 134 PLANE GEOMETRY. PROPOSITION X. A line perpendicular to a chord at its middle point passes through the center. PROPOSITION XI. In the same circle, or in equal circles, equal chords are equally distant from the center, and of two unequal chords the greater is nearer the center. PROPOSITION XII. A straight line perpendicular to the radius of a circle at its extrem- ity is tangent to the circle. PROPOSITION XIII. CONVERSE OF PROP. XII. If a straight line is tangent to a circle the radius meeting it at the point of contact is perpendicular to it. PROPOSITION XIV. Arcs of a circle intercepted by parallel chords are equal. PROPOSITION XV. Through three points, not in the same straight line, one circumfer- ence, and only one, is possible. PROPOSITION XVI. To find the greatest common unit of measure of two given lines. PROPOSITION XVII. Given the greatest common unit of measure of two quantities to find their ratio. PROPOSITION XVIII. If two variables are always equal as they approach their limits, their limits are equal. PROPOSITION XIX. In the same circle, or in equal circles, two angles at the center have the same ratio as the arcs which they intercept at the circumfer- ence. THE CIRCLE. 135 PROPOSITION XX. An inscribed angle is measured by one half its intercepted arc. PROPOSITION XXI. An angle formed by a tangent and a chord is measured by one half the intercepted arc. PROPOSITION XXII. The angle formed by two secants, meeting without the circle, is measured by one half the difference of the intercepted arcs. PROPOSITION XXIII. An angle formed by two intersecting chords is measured by one half the sum of the intercepted arcs. PROPOSITION XXIV. If two circles are tangent to each other the line joining their cen- ters passes through the point of contact. PROPOSITION XXV. To bisect a given straight line. PROPOSITION XXVI. To erect a perpendicular to a given line, at a given point in that line. PROPOSITION XXVII. To draw a perpendicular to a given line from a given point with- out that line. PROPOSITION XXVIII. To bisect a given arc. PROPOSITION XXIX. To bisect a given angle. PROPOSITION XXX. To find the center of a circle, when any arc of the circumference is given. 136 PLANE GEOMETRY. PROPOSITION XXXI. At a given point in a given line, to construct an angle equal to a given angle, with the given line as one side. PROPOSITION XXXII. Through a given point, to construct a straight line parallel to a given straight line. PROPOSITION XXXIII. To construct (1) the complement of a given angle; (2) the supple- ment of a given angle; (3) the third angle of a triangle, having given the other two angles. PROPOSITION XXXIV. Given two angles and the included side of a triangle, to construct the triangle. PROPOSITION XXXV. Given two sides and the included angle of a triangle, to construct the triangle. PROPOSITION XXXVI. To construct a triangle whose sides are three given lines. PROPOSITION XXXVII. Through a given point, to draw a tangent to a given circle. PROPOSITION XXXVIII. To circumscribe a circle about a given triangle. PROPOSITION XXXIX. To inscribe a circle in a given triangle CHAPTER III. PROPORTIONAL LINES, AND SIMILAR POLYGONS. THE THEORY OF PROPORTION. 200. Proportion is an equality between two ratios. If A and B form one ratio, and C and D an equal ratio, a proportion is formed which may be written in either of the forms: Either of these is read: "the ratio of A to B equals the ratio of C to D ; " or, " A is to B as C is to Z>." 201. The terms of a proportion are the four numbers, or quantities, compared. In any ratio the first term is called the antecedent, and the second term the consequent. Hence, in any proportion the first and third terms are called anteced- ents, and the second and fourth terms are called conse- quents. In any proportion, the first and fourth terms are called extremes, and the second and third terms are called means. A C In the proportion - - = , the first term is A, the j JLS second B, the third C, and the fourth D. A and C are the antecedents, B and D are the consequents, A and D are the extremes, and B and C are the means. 138 PLANE GEOMETRY. PROPOSITION I. THEOREM. 201. In a proportion, all of whose terms are num- bers, the product of the means equals the product of the extremes. Let -^ = -=- be a proportion in tvhich A, B, C and D Jo JLJ are numbers. To prove A D = B C. SUG. 1. What process is performed upon the expression A TT to produce the product ADI > Q SUG. 2. What, then, must the expression -=: be changed into ? Why ? Therefore MODEL. PROPOSITION I. THEOREM. 202. In a proportion, all of whose terms are num- bers, the product of the means equals the product of the extremes. Let -^ = -=r be a proportion in which A, B, C and D j> u are numbers. To prove A D = B C. ~ X B D = A D and ~ X B D = B C. Hence, by Ax. 4, A D = B C. Therefore, in a proportion, all of whose terms are num- bers, the product of the means equals the product of the extremes PROPORTION SIMILAR POLYGONS. 139 PROPOSITION II. THEOREM. 203. If the product of two numbers equals the product of two other nuiribers, the factors of one product may be made the means, and the factors of the other product the extremes, of a proportion. Let AB= C D be an equation, in which A, B, C and D are tlie numbers. To prove -. SUG. 1. What process must be performed upon A B to ^ produce the fraction ^_- ? SUG. 2. By the same process the second member, C D, is changed into what? Therefore Note. It frequently occurs that expressions are used which are perfectly intelligible in some cases and have no meaning in other cases. For example, the expression A B (that is, A times B} has a perfectly definite meaning, if A and B are both numbers; it also has a definite meaning when A is a pure number and B a quantity, provided A be con- sidered as the multiplier; but if A and B are both quantities, A B has no meaning. Likewise, the expression -= has a definite meaning if A and B are both numbers; it also has a definite meaning if A is a quantity and B a number; also, if A and B are both quantities of the same kind; but has no meaning if A is a number and B is a quantity, or if A and B are quantities unlike in kind. See Art. 161. In all the following operations upon ratio, and applications of it, care should be taken to interpret the symbols of number or quantity, and see if the principles of the fundamental rules of arithmetic and al- gebra can be applied. The scholia following the theorems in the the- ory of proportion will suggest the limitations which the principles un- derlying the operations impose upon the propositions. 140 PLANE GEOMETRY. 204. A mean proportional between two given num- bers, or quantities, is the second or third term of a pro- portion in which the means are alike, and the extremes are the given numbers or quantities. In the proportion A. B -n = -f? i B is a mean proportional between A and C. PROPOSITION III. PROBLEM. 205. To determine a mean proportional between two given numbers. Liet A and C be the given nutribers, and B the required mean proportional, To determine B. SUG. In the equation -=- = -= find the value of B. 13 C Note. In propositions I, II and III, the terms of the propor- tions must be regarded not as geometric magnitudes, but as numbers or the measures of magnitudes. These propositions, however, afford aid in investigating propositions whose terms are geometric magni- tudes instead of numbers. For example, suppose it is required to determine the mean proportional between two lines, A and C, in which a rod is taken as the unit of measure. If B is the required mean pro- A /? portional, the equation -= = -= must hold. Since A and C are each some number of rods, B must also be some number of rods. Suppose A a rods, B = b rods, C = c rods. Now, B is determined as soon as b is determined. From the pro- portion -= = it follows that = is true, and from this last jD C O C equation b may be found. Ex. 127. An angle formed by two tangents is meas- ured by one half the difference of the intercepted arcs. PROPORTION SIMILAR POLYGONS. 141 PROPOSITION IV. THHOREM. 206. If, in several successive ratios, the consequent of the first equals the antecedent of the second, the con- sequent of the second equals the antecedent of the third, etc. , the ratio of the antecedent of the first to the consequent of the last equals the product of tlie ratios. Let -g-, ^-, -=j, -= and -^ represent the ratios. To prove that -^ equals the product of the ratios. SUG. 1. If A, B, C, D, E and F are numbers, apply the algebraic rule for multiplying fractions. SUG. 2. If A, B, C, D, E and F are geometric magni- tudes, they are all magnitudes of the same kind, and by definition each ratio is a number. ^ = w ^ w C_ r D_ = <.,. m, n, r, s and / b2ing numbers. SUG. 3. Then, A = m B, B = n C, C = r D, D = s E and E = / F. Why ? SUG. 4. Since A m B and B n C, A = m n C. Since A = m n C and C = r D, A = m n r D. In the same way A = m n r s E. Finally, A = m n r s t F. SUG. 5. Since A^mnrstF, -= mnrst. Why> r SUG. 6. Notice, now, that m n r 1 1 is the product of the given ratios. Therefore 142 PLANE GEOMETRY. Note j. A careful interpretation of the symbols should be made at each step of the preceding proposition. Note 2. The preceding proposition may be illustrated as follows: 1 bushel _ 1 peck _ 1 quart _ 1 peck " ' 1 quart "" ' 1 pint Therefore, - US = 64. the product of the ratios. 1 pint To say that = 4, is only another form of saying that 1 bushel 4 pecks. -= = m is only another form of A = m B. This should be illustrated until the pupil sees that they are only different forms of expressing the same thought, viz., that 1 peck is the unit by which 1 bushel is measured, or that B is the unit by which A is measured, in whichever way it is expressed. QUERY. In the equation A m n r D, which sym- bols represent numbers and which represent quantities ? PROPOSITION V. THEOREM. 207. Both terms of a ratio may be multiplied by any number without changing the value of the ratio. Let T represent a ratio. ~ A m A . To prove -=? = ^-, m being a number. ti m B A SUG. 1. I^et the ratio -^ equal the number r. r> ^ SUG. 2. In the equation -=- = r, find what A equals. B SUG. 3. Multiply both members of the equation last found by m } and from the result find the value of the ra- m A tio m B ' SUG. 4. How, then, does -jc compare with ~, Therefore PROPORTION SIMILAR POLYGONS. 14.3 208. COROLLARY. Both terms of a ratio may be divided by the same number; also, by the same quantity, provided it [is a quantity of the same kind as the terms of the given ratio. 209. SCHOLIUM. In the foregoing proposition A and B may both be numbers, or both like quantities. PROPOSITION VI. THEOREM. 210. If four numbers, or like quantities, are in pro- portion, the ratio of the first to the third equals the ratio of the second to the fourth. . A C To prove ^ = - A C SUG. 1. If 2j- = the number m, what does -= equal? SUG. 2. Find A and C in terms of B and D, respec- tively. Express the ratio of A to C, and reduce by Prop. V, Cor. Therefore -- 211. SCHOLIUM. Proposition VI practically states that the second and third terms, of a proportion may be interchanged. This is true if the terms of the propor- tion are numbers, or quantities of the same kind, but not true if the terms are quantities unlike in kind. For ex- . ., . 16 rods 1 peck . ample, the proportion -^ - j- = 7-^ is true, since 2 rods 1 pint each ratio equals 8, but if the second and third terms be interchanged the result would be -= -- = - : , an 1 peck 1 pint absolutely meaningless expression. PLANE GEOMETRY. 212. When from any proportion a new proportion is obtained by taking the antecedents for one ratio, and the consequents, in the same order, for the other ratio, the second proportion is said to be deduced from the first by alternation. MR... . . M N -- = is deduced by alternation from -5- = -^-. YV o /to PROPOSITION VII. THEOREM. 213. If four numbers, or quantities, are in propor- tion, the ratio of the second to the first equals the ra- tio of the fourth to the third. A (^ Let ^- = -=r be the given proportion. Jt> u B D fo prove --=-. A C SUG. 1. L/et -=r = m. Then - = m. JD JJ SUG. 2. From the equations in Sug. 1, find the values of A and C respectively. SUG. 3. Divide by A and C, respectively, the two equations obtained in Sug. 2. Compare the results and reduce. Therefore - 214. SCHOiyiUM. Proposition VII states that in a proportion the terms of one ratio may be interchanged, provided the terms of the other ratio are interchanged also. In this proposition the terms of one ratio may be different in kind from the terms of the other ratio. For 16 rods 1 peck example, the proportion -^ == -i t 1S true > an( i the proportion obtained by applying the theorem, viz., 2 rods 1 pint . . r-z T- = T-Z =- is also true. 16 rods 1 peck PROPORTION SIMILAR POLYGONS. 145 215. When, from any proportion, a new proportion is formed by inverting the ratios, the second proportion is said to be deduced ironi the first by inversion. * PROPOSITION VIII. THEOREM. 216. If four numbers, or quantities, are in propor- tion, the ratio of the first plus the second, to the second, equals the ratio of the third plus the fourth, to the fourth. Let jj = jrj *> e M ie ffiven proportion. A + B C+ D Toprove _____ = ______ SUG. 1. Let -= = m. Then = m. > D SUG. 2. From -- = m, what does -7- equal ? SUG. 3. From -^ = m, what does -= equal ? Compare answers to Sugs. 3 and 4. Therefore 217. SCHOLIUM. In proposition VIII, the terms of one ratio may be different in kind from the terms of the other ratio and the proposition still be true. For ex- ample the proportion - = - * is true and 6 rods 5 quarts the proportion obtained by applying the theorem, viz., 18 rods 15 quarts . - - = g - is also true. 6 rods 5 quarts A C 218. When from the proportion -^ = -^ the pro- portion - - = - - is obtained, the second is Jo D said to be deduced from the first by composition. 10 Geo. 146 PLANE GEOMETRY. PROPOSITION IX. THEOREM. 219. If four numbers, or quantities, are in propor- tion, the ratio of the first minus the second to the sec- ond, equals the ratio of the third minus the fourth, to the fourth. Let ]g j) &e fhe given proportion. To prove that - = . J3 JD A C SUG. 1. Let -=,- = m. Then -^- = m. r> U ^ ^ g SUG. 2. From -=- = m, what does - - equal ? J ) SUG. 3. From -=r = m, what does - '== equal ? SUG. 4. Compare answers to Sugs. 2 and 3. Therefore 220. SCHOLIUM. In Proposition IX, the terms of one ratio may be different in kind from the terms of the other ratio and the proposition still be true. Illustrate this fact. A C 221. When from the proportion -=r = -=r , the pro- h JJ A B C- D . ,..'- portion = = =r is obtained, the second is B L) said to be deduced from the first by division. PROPOSITION X. THEOREM. 222. In a series of equal ratios, all of whose terms are numbers or quantities of the same Tcind, the ratio of the sum of the antecedents to the sum of the conse- quents equals any one of the ratios. PROPORTION SIMILAR POLYGONS. 147 B~ D ~ F ~ H* A + C+ E + G A C Toprove B + D + F+f/ == ^ == ^ etc. SuG. 1. Let each of the given ratios = x. SuG. 2. Put each ratio separately equal to x and find each antecedent. SUG. 3. Find the sum of all the antecedents. SUG. 4. From the result of Sug. 3, find the value of x. SuG. 5. Equate the value of x just found to any of the given ratios. Therefore - 223. In proposition X, the terms of the given ratios must be numbers or quantities of the same kind. Illus- trate the absurdity of using quantities of different kinds. Ex. 128. If J - , Prove that -^ = -^ A + B C + D and also that ^ = -= . A L, In deducing the second proportion from the first, are A, B, C and D restricted to being numbers, or may they be quantities as well, and, if quantities, are they all quantities of the same kind, or may they differ in kind ? Ex. 129. If J = g, prove that -^ = ^-, A - B C - D and also that = . In deducing the second proportion from the first, are A, B t Cand D restricted in any way, and, if so, how ? 148 PLANE GEOMETRY. PROPOSITION XI. THEOREM. 224. If three terms of one proportion are respec- tively equal to the three corresponding terms of an- other proportion, the fourth terms are equal. Let =% an* = %. To prove that C = M. SUG. 1. If ^ = r, then ~ = r and ^ = r. Why? SUG. 2. From -=? = r, find C, and from = r find M. SUG. 3. Compare C and M. Therefore Note. In all the preceding propositions in the theory of propor- tion, the theorems are true if the terms of the proportions are num- bers. If the terms are geometric magnitudes, the limitations noted should be carefully studied and applied. Read Arts. 159, 160 and 161 on measurement, and note their bearing on the theory of proportion. Ex. 129. Prove that an angle formed by a tangent and a chord, is measured by one half the intercepted Afy arc, using the following construction: Drop a _L from the center of the O to the chord and extend to the arc. Con- nect O and the point of contact A. SUG Compare Z. M O A with Z. between tangent and chord. Ex. 130. In the figure of Prop. XXIII, Chap. II, con- nect B and D, and prove the proposition in another manner. PROPORTION SIMILAR POLYGONS. 149 PROPOSITION XII. THEOREM. 225. If parallel lines intercept equal segments on one transversal, they intercept equal segments on all transversals. Go / Let E F, G H, etc., represent parallel lines intercepting equal segments E G, G I, etc., on the transversal A B, and V.w segments F H, H M, etc., on the transversal C D. To prove F H = HM= M N, etc. SUG. 1. From E, G, /, etc., draw lines E O, G P t IS, etc., || to CD. SUG. 2. In the As E O G, G P /, etc., compare the lines E O, G P, IS, etc. SUG. 3. Compare E O with F H, G P with H M, I S with M N, etc. Therefore Ex. 131. If a quadrilateral is inscribed in a circle the sum of the opposite angles equals two right angles. Prove that Z A plus Z. C equals two rt. Zs. 150 PLANE GEOMETRY. 132. An angle formed by a tangent and a secant is measured by one half the difference of the intercepted arcs. PROPOSITION XIII. THEOREM. 226. If a line is parallel to the base of a triangle, the ratio of the segments of one side equals the ratio of the segments of the other side. Let DEbe a line parallel to the base B C of the tri- angle ABC. AD A E To prove -^2=-. CASE I. When A D and D B are commensurable. SUG. 1. Since A D and D B are commensurable, sup- pose each to be measured, and express their ratio. SUG. 2. Through the points of division in A B, draw lines || to the base and intersecting A C. SUG. 3. Compare the segments of A and E C with respect to size. How many are there as compared with the number of segments in A B ? SUG. 4. Compare the ratio -^-^ with the ratio Tr ^ s , c, L JJ Jo PROPORTION SIMILAR POLYGONS. 151 CASE II. When A B and D B are incommensurable. B SUG. 1. Divide A D into any number of equal parts, and lay off the unit of measure upon D B as many times as possible. There will be a remainder, MB, less than the unit of measure. Why ? SUG. 2. Through M, draw M N || to B C A D A F SUG. 3. Compare -prT> with -p-T>- Give auth. D M EN SUG. 4. If the unit of measure be continually dimin- A D ished, the ratio ^ is a variable. Why ? SUG. 5. What is its limit ? Why ? A E SUG. 6. The ratio -^.y is also a variable. What is its limit ? SUG. 7. How do the limits of -^^ and T^> com- D M EN pare ? (See Art. 170.) Therefore Note. Compare this demonstration with that of Prop. XIX, Chap. II. 227. COROU,ARY. If a line be drawn parallel to the base of a triangle, one side is to either of its segments as the other side is to its corresponding segment. SUG. Take the conclusion of the proposition by com- position. 152 PLANE GEOMETRY. PROPOSITION XIV. THEORKM. 228. If a line divides two sides of a triangle pro- portionally, it is parallel to the base. -W V Let A B C be a triangle, and let D E divide tlie sides A B and A C so that 44^ - 4|?* A Jo AC To prove that D E is parallel to the base B C. SUG. 1. From D, draw D M \\ to B C. A. D A. M .SuG. 2. Compare the ratios -7-5 and r . ./i Jo A. C- SuG. 3. See hypothesis, and compare A R and A M. Ax. 1. SUG. 4. Complete the demonstration. Therefore SIMILAR POLYGONS . 229. Similar polygons are those which have their corresponding angles equal, and corresponding sides proportional. Points, lines, or angles, which are similarly situated in similar polygons, are called homologous. PROPORTION SIMILAR POLYGONS. 153 In similar polygons, the ratio of two homologous sides is called the ratio of similitude of the polygons. B '0' The polygons A B C D E and A' I? C D' E' are similar if Z. A = Z ^', Z /? = Z /?', etc., and ^^ C CZ) ^' ff ~'~ B' C ~~ ~~ C U ' The ratio ^. is the ratio of similitude of the poly- A' B' gons. Ex. 133. If two opposite sides of a quadrilateral in- scribed in a circle be equal, ihe other two sides are par- allel. Ex. 134. The locus of the middle points of all chords which pass through a given point, is a circle whose diameter is the line connecting the given point and the center of the circle. Prove that the circle described on O S is the required locus. Ex. 135. Through one of the points common to two intersecting circumferences, draw the diameters of the circles and prove that the line connecting the other ex- tremities of the diameters passes through the other point of intersection of the circumferences. 154* PLANE GEOMETRY. PROPOSITION XV. THEOREM. 230. Two triangles which, cure mutually equian- gular are similar. A 1 Let ABC and A B' C' 1>e two triangles in which the angle A = the angle A', tJie angle B = toe angle B' and the angle C = the angle C' . To prove that the triangles ABC and A' B' C' are similar. . SUG. 1. Place the A A' B' C upon the A A B C, so that A' B' lies upon A B with A' upon A and B' upon M. SUG. 2. What direction will A' C' take ? Why ? SUG. 3. Where will the point C' fall ? SUG. 4. What relation of position does M N sustain to BC1 Why? K AM AN _ SUG. 5. r-fi = -A~T' ^ lve autn - A > A C A SUG. 6. In the same way compare the ratios Jj A BC and -p-g.. SUG. 7. Apply definition of similar polygons and draw conclusion. Therefore - PROPORTION SIMILAR POLYGONS. 155 231. COROLLARY I. If two triangles have two angles of one respectively equal to two angles of the other, the triangles are similar. 232. COROLLARY II. Two right triangles which have an acute angle of one equal to an acute angle of the other, are similar. Ex. 136. If the sum of the opposite angles of a quadrilateral equals two right angles, prove that a circumference is possible through the four vertices. SUG. A circumference may be de- scribed through three vertices, as A, B and C. If it does hot pass through >, it must cut the line C D, or C D extended, as at M. Compare the sum of Zs B and C M A with two rt. Zs. Also, the sum of Zs B and C D A with two rt. Zs. Ex. 137. If a quadrilateral be circum- scribed about a circle, the sum of one pair of opposite sides is equal to the sum of the other pair. Ex. 138. If two Circles are tangent, and two secants be drawn through the point of con- tact, the chords joining the inter- sections of the secants and the cir- cumferences are parallel. Prove A D parallel to B C. SUG. Draw the common tangent, M N. Compare Z. A O M with Z C O N. Compare Z A OMwith Z. ADO\ also, ZCONvtiih 156 PIWVNE GEOMETRY. PROPOSITION XVI. THEOREM. 233. If two triangles have an angle of one equal to an angle of the other, and the sides including the equal angles proportional, the triangles are similar. In the triangles ABC and A' B' C' let the angle A A T> equal the angle A' and let the ratio ^ r equal the ra- AC tio ., , . A C To prove that the triangles ABC and A' B' C' are similar. SuG. 1. Place the A A B C upon the A A' B' C', with A upon A' t and A B and A C upon A' B' and A' C' re- spectively. Why is this possible ? * SuG. 2. Where will B and C fall ? SUG. 3. What relation of position will the line B C sustain to B' C ? Why ? SUG. 4. How do the ^s B and C compare with the ^s B' and C' respectively ? SUG. 5. See Prop. XV. Therefore PROPORTION SIMILAR POLYGONS. 157 PROPOSITION XVII. THKOREM. 234. Two triangles whose sides are proportional are similar. * r^ ~ BO ^O C ~AJW ~ W~W ~ ATV' To prove that the triangles ABC and A' & C' are similar. SUG. 1 On A B, lay off A M equal to A' B' and on A C, lay off A N equal to A' C'. Connect Maud N. A B A' /?' SUG. 2. Compare the ratios and p , . Give yy -* ^ auth. SUG. 3. Compare -^-^ with ; /. *., compare -^-^ with -, . Give auth. SUG. 4. Compare B' C and M N. SUG. 5. Then, how do As A' B' C' and A M N compare ? Why ? SUG. 6. The line M N divides A B and A C propor- tionally. Why ? See hyp. SUG. 7. How is the A A MN related to the A A B C? SUG. 8. Then, how are the As A' B' C' and A B C related ? Therefore 158 PLANE GEOMETRY. PROPOSITION XVIII. THEOREM. 235. The ratio of homologous altitudes of similar triangles equals the ratio of similitude of the tri- angles. B Let A Mand E N be homologous altitudes in tJie similar triangles ABC and E FG 9 and let -^-~, represent the ratio of similitude of the triangles. . AM AB To prove that Ar = -=r-^. jc, jtv E, r SuG. 1. The As A MB and E NF are rt. As. Why ? SuG. 2. In the rt. As A MB and E N F, compare Z B with Z F. SuG. 3. Then, how are the As A M B and E NF re- lated? See Art. 232. SuG. 4. Compare the ratio . with the ratio Give auth. Therefore Ex. 139. Through a given point, to draw a straight line so that the portion of it intercepted between two given inter- secting lines shall be divided at the point in a given ratio. SUG. Through the point draw a line parallel to one of the given lines. PROPORTION SIMILAR POLYGONS. 159 PROPOSITION XIX. THEOREM. 236. // two polygons are composed, of the same number of triangles, similar each to each and simi- larly placed, the polygons are similar. C' Let the polygons A B C D E F and A' B' C' D' E' F 1 be composed of the same number of triangles, similar each to each and similarly placed, ABC being similar to A' B' C', and so on for the other triangles. To prove that the polygons are similar. SUG. 1. Compare the Z A with the Z A', Z. B with Z.B' , etc. SUG. 2. Compare the ratio , , with SUG. 3. Compare the ratio B C B 1 C with A'C also the ratio with A C A'C' B C Now, compare , , with CD CD" SUG. 4. In a similar manner compare the ratios of other corresponding sides in 'the two polygons. SUG. 5. Apply the definition of similar polygons, and draw the conclusion. Therefore 160 PLANE GEOMETRY. PROPOSITION XX. THEOREM. 237. Converse of Prop. XIX. Two similar polygons may be divided into the same number of triangles, similar each to each and similarly placed. Let the polygons AB C D EF and A' B' ' D' E' F' be similar, and let all possible diagonals be drawn from corresponding vertices A and A'. To prove that the triangles in one polygon are similar to those in the other polygon. SUG. 1. Compare the A A B C and A' B' C' (See Prop. XVI. AC B C SUG. 2. Compare the ratios CD , B C CD' CD and also, the ratios A'C B'C A C Compare the ratios , _, ^i C SUG. 3. Compare Z. A CD with Z. A 1 CD 1 . SUG. 4. Compare the As A CD and A' C D' . SUG. 5. In a similar way compare the other corre- sponding As in the two polygons. Therefore PROPORTION SIMILAR POLYGONS. 161 PROPOSITION XXI. THEOREM. 238. The ratio of the perimeters of two similar polygons equals the ratio of similitude of the polygons. B> perimeter ABODE To prove that the ratio - r- -- equals perimeter A B' C D E * A ~fi the ratio of similitude . SUG. See proposition X, and give the demonstration. Therefore Ex. 140. Straight lines drawn through an) 7 point will intercept proportional segments upon two parallel lines. Prove that - _ - /B ~ FN >e AC C P SUG. Compare each of the ratios - ir -= and -= =, with Jo D D f 1 the ratio Complete the demonstration. 11 Geo. 162 PLANE GEOMETRY. PROPOSITION XXII. THEOREM. 239. If a perpendicular be dropped to the hypote- nuse from the vertex of a right angle in a right tri- angle: ! The triangles thus formed are similar to each other and to the whole triangle. II. The perpendicular is a mean proportional be- tween the segments of the hypotenuse. III. Each side is a mean proportional between the hypotenuse and the segment adjacent to that side. Let ABC represent a right triangle, right angled at~A 9 and let AM represent a perpendicular drawn from A to tlie hypotenuse B C. I. To prove that the triangles A M C, A MB and B A C are similar. SUG. 1. In the rt. As A M C and A MB, compare Z CA MvfiihZB. SUG. 2. See Art. 232. , SUG. 3. In the rt. As A M C and B A C, notice that Z C is common. See Art. 232. TT T ,/ , CM AM II. To prove that -^ = ^-^. SUG. Deduce this proportion from the similar AsAMC zn&AMB. PROPORTION SIMILAR POLYGONS. 163 ., . CB AB CB CA III. To prove that B = SUG. Deduce the first proportion from the similar As ABC and ABM. Deduce the second proportion from the similar As A B Cand ACM. Therefore - * PROPOSITION XXIII. THEOREM. 240. If two chords of a circle intersect, the ratio of either segment of the first to either segment of the sec- ond equals the ratio of the remaining segment of the second to the remaining segment of the first. B Let the chords A B and C D intersect at X. AX CX To prove that D X ~ B X' SUG. 1. Connect A and C\ also, B and D. SUG. 2. Compare Z. A with Z D\ also, Z. C with Z B. (Art. 177.) SUG. 3. Compare As A X C and D XB with respect to form. (Art. 230.) SUG. 4. Complete the demonstration. Therefore 164 PLANE GEOMETRY. PROPOSITION XXIV. THEOREM. 241. If two secants intersect without a circle, the ratio of the first secant to the second equals the ratio of the external segment of the second to the external segment of the first. Let A B and A C represent two secants meeting at A. To prove that c =lf D . SuG. 1. Connect D and C\ also, B and E. SUG. 2. Compare the As A D C and A E B. SUG. 3. Complete the demonstration. Therefore ^ n/M T , A C ... A + B C+D Ex. 141. If ^ = -^ , prove that - =j ^ = ^ D . Are A, B, C and D restricted to being numbers, or may they be quantities ? If quantities, are they all of the same kind, or may they differ in kind ? Ex. 142. Construct a triangle having a given base, a given altitude and a given vertical angle. SUG. Solve by intersection of two loci. PROPORTION SIMILAR POLYGONS. 165 PROPOSITION XXV. THEOREM. 242. //' a secant and a tangent meet without a circle, the tangent is a mean proportional between the secant and its external segment. Let A C represent a tangent, and O C a secant meeting the tangent at tfie point C. O C AC To prove that ^-j^. SUG. 1. Connect A and O\ also, A and B. SUG. 2. Compare the As A B C and O A C with re- spect to form. SUG. 3. Complete the demonstration. Therefore Ex. 143. What is the locus of the vertex of the right angle of a right-angled triangle constructed upon a given line as a hypotenuse? (See Ex. 122.) Ex. 144. Construct a triangle whose base and vertical angle are given, and whose vertex is at a given distance from the middle point of the base. SUG. Solve by intersection of two loci. 166 PLANE GEOMETRY. PROPOSITION XXVI. PROBLEM. 243. To divide a given straight line into any num- ber of equal parts. K Let A B be the given straight line. To divide A B into any number of equal parts for ex- ample, four. SUG. 1. Through A, draw an indefinite straight line making any convenient Z. with A B. SUG. 2. Take any convenient unit of length and lay it off four times on the indefinite line. SUG. 3. Connect the last point of division, F, with the point B, and draw || s to FB through the other points of division, C, D and E. SUG. 4. Why will these || s divide A B into equal parts ? Ex. 145. With a given fixed base, and a given verti- cal angle of a triangle, find the locus of the vertex. See exercise 121. Ex. 146. Construct upon a given line, as a chord, a segment of a circle which shall contain a given angle. SuG. Construct at the ends of the given line, and upon it, angles whose sum is the supplement of the given angle. PROPORTION SIMILAR POLYGONS. 167 PROPOSITION XXVII. THEOREM. 244. To divide a given straight line into parts pro- portional to any given lines. F B Let A B be the given straight line. To divide A B into parts such that ^r = ry- , etc. M N SUG. 1. Through A, draw an indefinite straight line making any convenient ^ with A B. SUG. 2. On this indefinite line lay off, in succession, the lines M, N, R, etc. SUG. 3 Connect the last point of division with the point , and draw || s from the other points of division, C, D t etc. SUG. 4. Why will these || s divide A B into parts pro- portional to M, N, R, etc.? 245. A fourth proportional to three given quanti- ties is the fourth term of a proportion whose first three terms are the three given quantities taken in order. If the three given quantities are 6 ft., 12 ft. and 18 ft., the fourth proportional is 36 ft. ; if the given quantities are 12 ft., 6 ft. and 18 ft., the fourth proportional is 9 ft.; if the given quantities are 18 ft., 6 ft. and 12 ft., the fourth proportional is 4 ft. 168 PLANE GEOMETRY. PROPOSITION XXVIII. PROBLEM. 246. To find a fourth proportional to three given lines. B C D B P Let A B, C D and E F be three given lines. To find a fourth proportional to these three lines. SUG. 1. In the figure in Art. 226, notice that E C is a fourth proportional to A D, D B and A E. SUG. 2. Construct a figure in which one of the lines shall be a fourth proportional to A B, CD and E F. SUG. 3. Work out at least three solutions of this prob- lem. 247. A third proportional to two given quantities is the fourth term of a proportion whose first term is one of the given quantities, and whose second and third terms are each the second of the given quantities. If 7- = , c is a third proportional to a and b. Ex. 147. Give a solution of Prop. XXXII, Chap. II, by using the truth that if two lines are perpendicular to the same line they are parallel. PROPORTION SIMILAR POLYGONS. 169 PROPOSITION XXIX. PROBLEM. 248. To find a third proportional to two given lines. A. C D Let A J5 and C D be two given lines. To find a third proportional to A B and C D. SUG. 1. See suggestions to the preceding proposition, and make the construction. SUG. 2. Make out at least two solutions. PROPOSITION XXX. PRJDBLEM. 249. To find a mean proportional between two given lines. A. C D Let A B and C D be two given lines. To find a mean proportional between A B and C D. SUG. 1. See suggestions to Prop. XXII, II and make the construction. SUG. 2. Make out two or more solutions. Every prop- osition which involves a mean proportional should sug- gest a solution to the problem. Ex. 148. Prove Prop. XVIII, Chap. Ill, by taking f~* -~ as the ratio of similitude of the triangles. 170 PLANE GEOMETRY. PROPOSITION XXXI. PROBLEM. 250. To construct, upon a given line as a ~base, a triangle similar to a given triangle. B C Let A B C be a given triangle, and E Ga given line. To construct, upon E G as a base, a triangle similar to the triangle ABC. SUG. 1. If B C be regarded as the base of the A A B C, how must the /_ at E, in the required A , compare with the Z B ? SUG. 2. How must the ^ at G compare with the Z. C? SUG. 3. Make the required construction. Ex. 149. If two parallel lines are cut proportionally by a set of secant lines, prove that the secant lines all pass through a common point. _- AC CE that the lines B A, D C, F E, etc., intersect at a common point. SUG. Extend two of them, B A and D C, until they meet at O. Connect O and E, and extend O E to the other of the parallels. Determine the relative position of the lines O M and E F. ~~ = - Why? J3 D AC CE why? PROPORTION SIMILAR POLYGONS. 171 PROPOSITION XXXII. PROBLEM. 251. To construct a polygon similar to a given poly- gon and liaving a given line as one side. Let MN P 8 K be a given polygon, and M.' XT a given straight line. To construct \ with M' N' as one side, a polygon similar toMNPSR. SUG. 1. Let M' N' be homologous to the side M N, and from M draw all possible diagonals of the given polygon. SUG. 2. On M' N' construct a A, M' N 1 P* , similar to the A MNP. Give auth. SUG. 3. On M' P' construct a A similar to the A M P S. Continue the construction. PROPOSITIONS IN CHAPTER III. PROPOSITION I. In a proportion, all of whose terms are numbers, the product of the means equals the product of the extremes. PROPOSITION II. If the product of two numbers equals the product of two other numbers, the factors of one product may be made the means, and the factors of the other product the extremes, of a proportion. PROPOSITION III. To determine a mean proportional between two given numbers. PROPOSITION IV. If, in several successive ratios, the consequent of the first equals the antecedent of the second, the consequent of the second equals the antecedent of the third, etc. , the ratio of the antecedent of the first to the consequent of the last equals the product of the ratios. PROPOSITION V. Both terms of a ratio may be multiplied by any number without changing the value of the ratio. PROPOSITION VI. If four numbers, or like quantities, are in proportion, the ratio of the first to the third equals the ratio of the second to the fourth. PROPOSITION VII. If four numbers, or quantities, are in proportion, the ratio of the second to the first equals the ratio of the fourth to the third. PROPOSITION VIII. If four numbers, or quantities, are in proportion, the ratio of the first plus the second, to the second, equals the ratio of the third plus the fourth, to the fourth. PROPORTION SIMILAR POLYGONS. 173 PROPOSITION IX. If four numbers, or quantities, are in proportion, the ratio of the first minus the second, to the second, equals the ratio of the third minus the fourth, to the fourth. PROPOSITION X. In a series of equal ratios, all of whose terms are numbers or quantities of the same kind, the ratio of the sum of the antecedents to the sum of the consequents equals any one of the ratios. PROPOSITION XI. If three terms of one proportion are respectively equal to the three corresponding terms of another proportion, the fourth terms are equal. PROPOSITION XII. If parallel lines intercept equal segments on one transversal, they intercept equal segments on all transversals. PROPOSITION XIII. If a line is parallel to the base of a triangle, the ratio of the seg- ments of one side equals the ratio of the segments of the other side. PROPOSITION XIV. If a line divides two sides of a triangle proportionally, it is parallel to the base. PROPOSITION XV. Two triangles which are mutually equiangular are similar. PROPOSITION XVI. If two triangles have an angle of one equal to an angle of the other, and the sides including the equal angles proportional, the triangles are similar. PROPOSITION XVII. Two triangles whose sides are proportional are similar. 174 PLANE GEOMETRY. PROPOSITION XVIII. The ratio of homologous altitudes of similar triangles equals the ratio of similitude of the triangles. PROPOSITION XIX. If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar. PROPOSITION XX. CONVERSE OF PROP. XIX. Two similar polygons may be divided into the same number of triangles, similar each to each, and similarly placed. PROPOSITION XXI. The ratio of the perimeters of two similar polygons equals the ra- tio of similitude of the polygons. PROPOSITION XXII. If a perpendicular be dropped to the hypotenuse from the vertex of a right angle in a right triangle: I. The triangles thus formed are similar to each other and to the whole triangle. II. The perpendicular is a mean proportional between the seg- ments of the hypotenuse. III. Each side is a mean proportional between the hypotenuse and the segment adjacent to that side. PROPOSITION XXIII. If two chords of a circle intersect, the ratio of either segment of the first to either segment of the second equals the ratio of the re- maining segment of the second to the remaining segment of the first. PROPOSITION XXIV. If two secants intersect without a circle, the ratio of the first secant to the second equals the ratio of the external segment of the second to the external segment of the first. PROPOSITION XXV. If a secant and a tangent meet without a circle, the tangent is a mean proportional between the secant and its external segment. PROPORTION SIMILAR POLYGONS. 175 PROPOSITION XXVI. To divide a given straight line into any number of equal parts. PROPOSITION XXVII. To divide a given straight line into parts proportional to any given lines. PROPOSITION XXVIII. To find a fourth proportional to three given lines. PROPOSITION XXIX. To find a third proportional to two given lines. PROPOSITION XXX. To find a mean proportional between two given lines. PROPOSITION XXXI. To construct, upon a given line as a base, a triangle similar to a given triangle. PROPOSITION XXXII, To construct a polygon similar to a given polygon and having a given line as one side. CHAPTER IV. COMPARISON AND MEASUREMENT OF POLYGONS. DEFINITION. 252. The area of a surface is its ratio to some selec- ted unit of measure, coupled with the unit of measure. (See Arts. 159 and 160.) If m denotes the unit of measure for determining the A area of a surface A, and if the ratio -- equals b (article 161), the area of the surface A is bm ; i. e., b times the unit m. The unit ef measure used to express the area of sur- faces is a square, whose side is a given linear unit; as a square inch, a square foot, etc. PROPOSITION I. THEOREM. 253. Two parallelograms having equal bases and equal altitudes are equal in area* B A' -m* B'A' Let A B CD and A' B' ' D' be two parallelograms having D C and D' C' equal, and the altitudes m n and m' n' equal. To prove that the area of A C equals the area of A' C f . MEASUREMENT OF POLYGONS. 177 SUG. Place the EJ A C upon the 7 A' C', D C upon U C 1 , and compare the parts external to each other. Therefore PROPOSITION II. THEOREM. 254. If two rectangles have equal altitudes, the ra- tio of their areas equals the ratio of their bases. A RE F Let A C and E G be two rectangles having equal alti- tudes, D A and HE. A area of A C DC To prove that - J ' _ _ =- ^^ . area of EG HG CASE I. When D C and H G are commensurable. SUG. 1. Measure D Cand H G by some common unit of measure. Let the unit be contained m times in D C, and n times in H G. SUG. 2. What is the ratio ofDCto&G* SUG. 3. Through the points of division erect J_s to D C and H G, and extend the JLs to the secondary bases, thus dividing A C and E G into rectangles. SUG. 4. How many rectangles in A C? How many in H G ? How do these smaller rectangles compare in size ? Why? SUG. 5. What does the ratio of the areas of the rect- angles A C and E G equal ? SUG. 6. Compare the ratio of the areas with the ratio of the bases. 12 Geo. 178 PLANE GEOMETRY. CASK II. When the bases are incommensurable. ARM BE G SUG. 1. Take any measure of K /and lay it off on D C as many times as possible. There must be a remainder less than one of these parts. Why ? Suppose the meas- ure of K /is contained in D C a certain number of times with a remainder D N. Erect N M J_ to D C, at the point N. SUG. 2. Compare the ratio of the areas ofMC and E I with the ratio of N C and K I. SUG. 3. Use, now, a unit of measure smaller than D N, and let the remainder now be D S. Erect S R _L to D C at S, and compare the ratio of the areas of R C and E I with the ratio of 5 C and K I. SUG. 4. By continually decreasing the unit of measure the two ratios compared are variables. The limit of the area of first ratio is T . area of E I ond ratio is -=^rj . Why ? Why ? The limit of the sec- SUG. 5. What relation exists between the variables as they approach their limits ? (See Siig. 3 ) SUG. 6. What relation exists between the limits ? Therefore - 255. COROLLARY. If two rectangles have equal bases the ratio of their areas equals the ratio of their altitudes. Note i. Compare the demonstration of Prop. II with that of Prop. XIX, Chap. II. What other proposition is demonstrated in a similar manner ? MEASUREMENT OF POLYGONS. 179 Note 2. Hereafter a ratio like -^r will be written in the area & I A C form -F-r. PROPOSITION III THEOREM. 256. The number of units of area in any rectangle is equal to the product of the number of linear units in the base and altitude. "Let A represent a rectangle, U a unit of measure for area, and u the linear unit, viz., a side of the square U. Let u be contained a times in m and b times in n. To prove that A contains the unit U y a X b times. SUG. 1. Construct a rectangle B whose altitude is n and whose base is u. SUG. 2. What is the ratio of A to B1 (Prop. II.) Re- duce the answer to the simplest form. SUG. 3. What is the ratio of B to /? Reduce as be- fore. SUG. 4. What, then, is the ratio of A to Ut (Art. 206.) Therefore Ex. 150. In a given line determine a point which is equally distant from two given points not in the line. Note. In this, and as many other problems of construction as pos- sible, employ the intersection of loci. 180 PLANE GEOMETRY. MODEL. PROPOSITION III. THEOREM. 257. The number of units of area in any rectangle is equal to the product of the number of linear units in the base and altitude. A n u. B m. n. Let A represent a rectangle, V a unit of measure for area, and u the linear unit, viz., a side of the square V. Let u be contained a times in m and b times in n. To prove that A contains the unit U, a x b times. Construct the rectangle B, whose altitude is n and whose base is u. Since A and B have equal altitudes -=--. (Prop- osition II.) But since u is contained a times in m, - = a. u Hence, =- = a. n Ft n, Since B and U have equal bases, jj . But since u is contained b times in n, = b. n Hence, -~ b. Since -^ a, and -=- T = b > U MEASUREMENT OF POLYGONS. 181 Therefore, ^ = a x b (Art. 206). That is, A contains /, a X b times. Therefore, the number of units of area in any rectangle is equal to the product of the number of linear units in the base and altitude. 258. SCHOLIUM I. In the applications of this theorem the base and altitude must be expressed in terms of the same unit, and the unit of area must be the square whose side is the linear unit. 259. SCHOLIUM II. By comparison of the theorem with the definition of area (Art. 252), it will be observed that the area of a rectangle is the product of the measures of the base and altitude joined to the name of the unit of area. ILLUSTRATIONS. If the base of a rectangle is 10 ft. and the altitude 6 ft., the area of the rectangle is (6 X 10) sq. ft. If the base of a rectangle is 10 ft. and the altitude 6 in. (or \ ft.), the area of the rectangle is ( X 10) sq. ft. (Sch. I.) 260. SCHOLIUM III. The expression " the product of the base and altitude," is a common abbreviation for "the product of the measures of the base and altitude joined to the name of the unit of area." ' ' The product of two lines, * ' or ' ' the square of a line, ' ' must not be interpreted in any other sense than that just stated. With this interpretation, Prop. Ill is usually stated as follows: The area of a rectangle is equal to the product of the base and altitude. The numbers which represent the measures of the lines may be integral, fractional, or incommensurable. 182 PLANE GEOMETRY. Note. It may make the thought in the foregoing demonstration clearer to.illustrate it by the arithmetical form of analysis. U is contained in B the same number of times that u is in n, i. *., b times; also, B is contained in A the same number of times that u is in m, i. e. t a times. Now, as U is contained b times in B t and B is contained a times in A, it follows that U is contained a X b times in A. To take a definite case, let the base and altitude be measured by some linear unit, as 1 inch; and suppose this unit is contained 5 times in the altitude and 8 times in the base, then, as seen in the figure at the right, there are 8 columns, with 5 squares in each column, and hence, in all, 8 X 5 squares; i. e. t 40 square inches. It is readily seen, by constructing a figure, that if the unit is con- tained a fractional number of times in the base and altitude the same rule is reached. QUERY. Can the conclusion of the proposition, in all its generality, be drawn from the method in common use in arithmetic? PROPOSITION IV. THEOREM. 26 1 . The area of a parallelogram equals the product of its base and altitude. SUG. See Props. I and III, and demonstrate. Ex. 151. Determine a point equally distant from two given points, and at a given distance from a given line. How many points answer the conditions of the prob- lem? Ex. 152. Determine a point equally distant from two given parallel lines, and equally distant from two given points. MEASUREMENT OF POLYGONS. 183 Ex. 153. Determine a point at given distances from each of two given intersecting lines. How many points answer the conditions of the prob- lem ? Ex. 154. Determine a point at given distances from two given points. PROPOSITION V. THEOREM. 262. The area of a triangle equals one half the product of its base and altitude. Let ABC represent a triangle, AB its base and M C its altitude. To prove that the area of the triangle ABC equals one half of A B times M C. SUG. 1. From C, draw CD equal and || to A B, and connect B with D. What kind of a figure is A D ? Why? SUG. 2. Compare the A A B C with the A CD B. SUG. 3. Compare the A A B C with the EH A Z>, in respect to magnitude; also, compare bases and altitudes. SUG. 4. What, then, is the area of the A A B C in terms of A B and Therefore 184 PLANE GEOMETRY. 263, COROLLARY. If two triangles have equal alti- tudes their areas have the same ratio as their bases. Then Give In the triangles A C D and A 1 C D f , let the altitudes A B and A 1 B' be equal. A CD \A B X CD CD A' CD' '' \A' B' x C' D' '''' CD'' auth. In a similar manner, prove that if two triangles have equal bases their areas have the same ratio as their alti- tudes. 264, SCHOUUM. For the interpretation of the ex- pression, " the product of the base and altitude," in prop- ositions IV and V, see Art. 260. PROPOSITION VI. THEOREM. 265. The area of a trapezoid equals one half the product of its altitude by the swm of its 'bases. Let A B CD represent a trapezoid, and A Mits altitude. MEASUREMENT OF POLYGONS. 185 To prove that the area of the trapezoid, A B C D, equals one half of A M times the sum of A B and D C. SUG. 1. Draw the diagonal, B D. SUG. 2. In the A D C B, if D C is taken as the base, what is the altitude? Then, what is the area of the A D C B equal to ? SUG. 3. In the A D A B, if A B is taken as the base, what is the altitude? Then, what is the area of the A D A B equal to ? SUG. 4. What, then, is the area of the trapezoid A BCD equal to ? Therefore Ex. 155. The ratio of the squares of the legs of a right triangle is equal to the ratio of the segments formed by dropping a perpendicular from the vertex of the right angle upon the hypotenuse. SUG. 1. See figure in Art. 239. By use of part III of the proposition in Art. 239, find an expression for the square of each leg of the triangle. SUG. 2. What is the ratio of the squares found ? Ex. 156. The sum of the squares of the legs of a right triangle equals the square of the hypotenuse. SUG. 1. See Sug. 1, of Ex. 155. SUG. 2. Add the squares. Ex. 157. Determine a point at a given distance from a given point, and equally distant from two parallel lines. How many points answer the conditions of the problem ? Is this problem always possible ? Ex. 158. Determine a point at a given distance from a given point, and equally distant from two given intersect- ing lines. Show when there are four points, when three, when two, when one, and when not any. 186 PLANE GEOMETRY. AREA OF A POLYGON. 266. Various methods have been used to find the area of poly- gons. Among the methods used the following may be noticed: From any vertex of the polygon draw all possible diagonals, as in fig. 1, at the right. The polygon is, by this means, divided into triangles, and, if the bases and altitudes of these triangles can be measured, their areas can be computed, and then, by addition, the area of the polygon can be found. Another method is to draw the longest diagonal of the polygon, and, from the ver- tices, drop perpendiculars upon this diag- onal, as in fig. 2, at the right. The poly- gon is, in this way, divided into triangles and trapezoids, and, if the bases and alti- tudes of these triangles and trapezoids can be measured, their areas can be computed, and then, by addition, the area of the poly- gon can be obtained. Still another method is to draw, through any vertex of the polygon, a straight line, exterior to the polygon, and, from the ver- tices, drop perpendiculars upon this line, as in fig. 3. In this way, triangles aud trapezoids are formed, and, if the various bases and altitudes can be meas- ured, the areas of the triangles and trapezoids can be computed, and, if the areas of the parts exterior to the polygon be subtracted from the sum of the other areas, the difference gives the area of the polygon. FIQ 3 The method just described is the one by which surveyors sometimes compute irregular areas bounded by straight lines. In considering the area of a field, or any large irregular polygon, it is often convenient to make a map or outline of the polygon accu- rately drawn to a scale as large as convenient. Then, any line con- nected with the polygon considered, can be determined by means of the homologous line in the constructed polygon. MEASUREMENT OF POLYGONS. 187 PROPOSITION VII. THEOREM. 267. The ratio of the areas of two similar triangles is equal to the ratio of the squares of their homologous edges, or homologous altitudes. B^ -sr*c F Let ABC and EFO represent two similar triangles, and A M and E N Iwmologous altitudes. ,L * A B C B C AM To brove that =-= = x or EN 2 SUG. 1. What is the area of each A? What is the ratio of their areas in terms of their bases and altitudes ? Sue. 2. What is the ratio of the bases in terms of the altitudes ? SUG. 3. What, then, is the ratio of the areas of the As in terms of their bases ? What in terms of their alti- tudes ? Therefore Ex. 159. Draw the common chord of two intersecting circles. From any point in the chord extended draw tangents to the two circles. Prove that the tangents are equal. Ex. 160. To construct a triangle similar to a given tri- angle, having a given perimeter. 188 PLANE GEOMETRY. PROPOSITION VIII THEOREM. 268. The ratio of the areas of two similar polygons equals the ratio of the squares of two Jwmologous sides. C' Let O and O' represent two similar polygon* To prove that -^ = =^ . SUG. 1. Divide the polygons into similar As; ABC similar to A' B' C' , etc. SUG. 2. What is the ratio of the areas of As A B C and A' B' C' in terms of A B and A' B' ? What is the ratio of the areas of As A CD and A' C' H in terms of CD and C' ZX ? Art. 267. SUG. 3. Compare the ratios A APE etc. O SUG. 4. Compare the ratio -=,- with either of the ratios in Sug. 3. (Art. 222.) SUG. 5. What is the ratio of the areas of the polygons in terms of the sides ? (See Sug. 2.) Therefore 269. COROLLARY. The ratio of the areas of two sim- ilar polygons equals the ratio of the squares of any two homologous lines. MEASUREMENT OF POLYGONS. PROPOSITION IX. THEOREM. 270. The square described upon the hypotenuse of a right triangle is equal to the sum of the squares de- scribed upon the other two sides. N Let ABC represent a right triangle, whose hypotenuse isAC, and let A E be the square upon the hypotenuse, and B S and B Mthe squares on the other two sides. To prove that A E = B S plus B M. SUG. 1. Draw B G \\ to C E, and extend to meet N E, at O. Draw B N, B E, C S and A M. SUG. 2. Compare As B C E and A CM. SuG. 3. Compare the area of A A CM with the area of the square B M, and the area of A B C E with the area of the rectangle C O. SUG. 4. Compare the area of the square B M with the area of the rectangle C O. SUG. 5. Compare As C A S and NAB. SUG. 6. Compare the square B S with the rectangle A O. SUG. 7. Compare the sum of the areas of the squares B M and B S, with the sum of the areas of the rectangles C O and A O\i. e., with the area of the square C N. Therefore 190 PLANE GEOMETRY. 271. SCHOLIUM I. This proposition is known as the Pythagorean proposition. It is so named in honor of Pythagoras, who is supposed to have given the first demonstration of it. 272, SCHOLIUM II. The "square of a line'* is an expression often used instead of the square described upon a line. But if the former expression is used it must be interpreted as already explained in Art. 260. Using this language, the Pythagorean proposition is often stated as follows: The square of the hypotenuse of a right tri- angle equals the sum of the squares of the other two sides. The proposition is often put in the form of an equation, thus: A~C 2 = ^4~2? 2 + B~C* . From this equation two others are readily obtained by mere trans- position, viz.: A~B* =^T 2 -B~C* and 'B~C* = A~C* - ~AB'\ When any two of the three sides of a right triangle are given, the remaining side may be found from one of the three equations just written. Ex. 161. If two triangles have an angle of one equal to an angle of the other, the ratio of their areas equals the ratio of the products of the sides including the equal angles. A A B C A B X A C Ve A A B' C == AB'xA~C T ' Sue. Connect B and C'. Compare each A with A A B C' (Prop. V, Cor.). MEASUREMENT OF POLYGONS. 191 PROPOSITION X. THEOREM. 273. The square described on the sum of two lines equals the sum of the squares described on the two lines plus twice the rectangle whose sides are the two lines. M /V Let A ~B and B C be two given lines, and A C their sum. Let AN be tJie square described on A C. To prove that A N equals the sum of the squares de- scribed on A B and B C plus twice the rectangle whose sides are A B and B C. SUG. 1. Lay off A E equal to A B, and draw E O || to A C; also draw B M\\ to C N. SUG. 2. What kind of a figure is E Bl SUG. 3. Compare C O with A B '; also O N and M N with B C. SUG. 4. What kind of a figure is M Ol SUG. 5. Compare the rectangles B O and E M. SUG. 6. Notice that the square A N, is divided into four parts, and then note the answers to Sugs. 2, 4 and 5. Therefore - 274. SCHOLIUM. With the interpretation given in Art. 260 for the square of a line and the product of two lines, this proposition may be expressed thus: If A C = AB + B C 2 + ~B~C* + 2 A B X B C. then C* = 192 PLANE GEOMETRY. EXERCISES. 162. Prove that the area of a rhombus equals one half the product of its diagonals. 163. Given the area of a trapezoid equal to 104 sq. ft., the altitude equal to 6 ft. and the difference between the bases equal to 2 ft. Find the two bases of the trapezoid. 164. A B is a chord of a circle. C D is a diameter perpendicular to A B^ and intersecting A B at E. C E is 5 in. and A C 10 in. Find the diameter of the circle. 165. Find the dimensions of a rectangle whose perim- eter is 16 in. and whose area is 15 sq. in. 166. Find the dimensions of a rectangle whose perim- eter is 28 ft. and whose diagonal is 10 ft. 167. If the hypotenuse of a right triangle is 15 ft. and the ratio of the legs is f-, what is the area ? 168. Prove that the area of an equilateral triangle con- structed on the hypotenuse of a given right triangle, is equal to the sum of the areas of the equilateral triangles constructed on the other two sides of the given right tri- angle. 169. If one acute angle of a right triangle is 60, prove that the area of the equilateral triangle constructed on the hypotenuse is equal to the area of a rectangle whose sides are the two legs of the right triangle. 170. The altitude of a trapezoid is 3 ft. and the bases are 8 and 12 ft. respectively. Extend the non-parallel sides until they meet, and find the areas of the two tri- angles of which the trapezoid is the difference. 171. If two triangles have two sides of one equal re- spectively to two sides of another and the included angles supplementary, the triangles are equal in area. MEASUREMENT OF POLYGONS. 193 PROPOSITION XI. THEOREM. 275. The square described on the difference of two lines equals the sum of the squares described on the two lines minus twice the rectangle whose sides are the two lines. c G Let A "B and B C be two given lines, and A C their dif- ference. Let A F be the square described on A C. To prove that A F equals the sum of the squares de- scribed on A B and B C, minus twice the rectangle whose sides are A B and B C. STJG. 1. Let A D be the square described on A B and B K the square described on B C. SUG. 2. Extend K C to Z,, and HFioG. SuG. 3. Compare E B with B C, and B G with C A. Compare E G with A B. SuG. 4. Compare the rectangle E F with the rect- angle D H. SUG. 5. Notice that the square A F equals the sum of the squares A D and B K> minus the sum of the rect- angles E /^and D H. Therefore - 13 -Geo. 194: PLANE GEOMETRY. 276. SCHOLIUM. With the interpretation given in Art. 260 for the square of a line and the product of two lines, this proposition may be expressed thus: If AC=AB-BC then AC* **AB* +3 C' ZABx BC. 277. The projection of a point upon a straight line is the foot of the perpendicular from the point to the line. The projection of a given straight line upon another straight line is that part of the second line included be- tween the projections of the extrem- ities of the given line. The projection ofAJB upon C D is M N. B M N PROPOSITION XII. THEOREM. 278. The square on the side opposite an acute angle of a triangle equals the sum of the squares of the other two sides minus twice the product of one side by the projection of the other side upon that side. G Let ABCbea triangle of wJiich the angle A is acute, and let B L, A D and A K 1>e the squares described on ttie sides B C, AC and A B respectively. MEASUREMENT OF POLYGONS. 195 To prove that the square B L equals the sum of the squares A D and A K, minus twice A B times the projec- tion of A C upon A B. SUG. 1. From A, draw A O _L to B C, and extend it to R. From B, draw Bm _L to A C, and extend it to E. From C, draw Cn _L to A B and extend it to H. SUG. 2. Draw A /, A L, B D, B F, C^and C G. SUG. 3. Compare A A B /with A C B K. Compare A A B I with rectangle B R, and A C B I with rect- angle B H. Compare rectangle B H with rectangle B R. SUG. 4. In a similar manner compare rectangle C E with rectangle C R. SUG. 5. Compare A C A G with A B A F. Compare A C A G with rectangle A H. Compare A B A /'with rectangle A E. Compare the rectangle A H with rect- angle A E. SUG. 6. The square B L = the sum of the rectangles B H and C E. Why ? SUG. 7. The square B L = the sum of the squares A K and A D, minus the sum of what two rectangles ? SUG. 8. The square B L = the sum of the squares A K and A D, minus twice the rectangle A H. Why ? (That is, the square B L = the sum of the squares A K and A D, minus 2 x A G X An.) SUG. 9. Compare A G X An with A B X An. SUG. 10. Notice that An is the projection of A C upon A B. Therefore Ex. 172. The non-parallel sides of a trapezoid are each 15 inches, and the bases are 12 inches and 30 inches re- spectively. Find the area of the trapezoid. 196 PLANE GEOMETRY. PROPOSITION XIII. THEOREM. 279. The square.on the side opposite an obtuse angle of a triangle equals the sum of the squares of the other two sides plus twice the product of one of the sides "by the projection of the other side upon that side. Let A B Cbea triangle, of which tlie angle A is obtuse; and let JB L, A K and A Dbe squares on the sides JB C,AB and A C respectively. To prove thai the square B L equals the sum of tht squares A K and A D plus twice A B times the projection of A C upon A -B. SUG. 1. From A, draw A O J_ to B C, and extend to R. From B, draw Bm _L to C A extended, and extend Bm to E. From C, draw Cn _L to B A extended, and extend Cn to H. SUG. 2. Draw A 7, A L, B D, B F, CJ^smdC G. SUG. 3. Compare As A B I and C B K. Compare A A B I with rectangle B R, and A C B K with rect- angle B H. Compare rectangle B R with rectangle B H. MEASUREMENT OF POLYGONS. 197 SUG. 4. In a similar manner compare rectangle C R with rectangle C E. SUG. 5. Compare As B A F and C A G. Compare A B A F with rectangle A E, and A C A G with rect- angle A H. Compare rectangle A E with rectangle A H. SUG. 6. The square B L equals the sum of the rect- angles B H and C E. Why ? SUG. 7. The square B L equals the sum of the squares A K and A D plus twice the rectangle A H. Why ? (That is, the square B L = the sum of the squares A K and A D plus 2 x A G X An.*) SUG. 8. Compare A G X An with A B X An. SUG. 9. Notice that An is the projection of A C upon A B. Therefore Ex. 173. If one circle is inscribed in a right triangle, and another circle circumscribed about the same right triangle, the sum of the diameters of the circles is equal to the sum of the two legs of the right triangle. Ex. 174. If a circle is inscribed in a right triangle the sum of the two legs of the triangle exceeds the hypote- nuse by an amount equal to the diameter of the circle. Ex. 175. The circle inscribed in an equilateral triangle has the same center as the circle circumscribed about the same triangle, and the radius of the circumscribed circle is double that of the inscribed circle. Ex. 176. The ratio of similitude of two similar poly- gons is fy, and the sum of their areas is 518 sq. in. Find the area of each polygon. Ex. 177. Upon a given base construct a right triangle having given the perpendicular from the right angle to the hypotenuse. 198 PLANE GEOMETRY. PROPOSITION XIV. PROBLEM. 280. Upon a given line as base, to construct a rect- angle equal in area to a given square. A Jjet CD be the given line, and A E the given square. To construct upon C D as a base, a rectangle equal in area to the square A R. SUG. 1. lyet X stand for the required altitude. Suice the area of the rectangle is to be equal to the area of the square, the product of the measures of C D and X must equal the square of the measure of A B. Hence, ABxAB=CDxX. SUG. 2. Make a proportion from the equation in Sug. 1. Art. 203. SUG. 3. All the terms of the proportion, except X, be- ing given, construct X by a method based upon a propo- sition involving proportion. SuG. 4. Having now constructed X, construct the rect- angle. Therefore Ex. 178. Find the base and altitude of a rectangle whose diagonal is 15 ft. and whose area is 108 sq. ft. MEASUREMENT OF POLYGONS. 199 PROPOSITION XV. PROBLEM. 281. To constmct a square equal in area to a given rectangle* B Let A B C D be a given rectangle* To construct a square equal in area to rectangle A C. SUG. 1. What is the area of A C? SUG. 2. If X represents the side of the square equal in area to A C make a proportion of D C, B C and X. SUG. 3. Construct the line represented by X in the proportion. Note . There are at least three previous propositions by which the above problem can be worked. It may be time well used by the pu pil to work more than one solution. Ex. 179. Find the area of a right triangle whose hypotenuse is 1 ft. 8 in., and one of whose legs is 1 ft. Ex. 180. Construct a square equal in area to a given trapezoid. Ex. 181. Construct a square equal in area to a given triangle. Ex. 182. Construct a square equal in area to a given parallelogram. 200 PLANE GEOMETRY. PROPOSITION XVI. PROBLEM. 282. To construct a square whose area is equal to the sum of the areas of two given squares. Let A and U be two given squares. To construct a square whose area equals the sum of the areas of A and B. SUG. Apply Art. 270. PROPOSITION XVII. PROBLEM. 283. To construct a square whose area is equal to the difference of the areas of two given squares. SUG. Apply Art. 272. PROPOSITION XVIII. PROBLEM. 284. To construct a rectangle in which the sum of the base and altitude equals a given line f and the area equals the area of a given square. A B F E Let C D be the given line, and ABEF the given squcvre. MEASUREMENT OF POLYGONS. 201 To construct a rectangle in which the sum of the base and altitude equals C D, and the area equals the area of the square A B E F. SUG. 1. The side A B of the given square must be a mean proportional between the two segments into which C D must be divided. SUG. 2. See some former proposition about a mean proportional: for example, Art. 239, II. QUERY. Can only one, or can an indefinite number, of rectangles be constructed whose area equals the area of the given square ? Ex. 183. Prove that the product of the sum of two lines by their difference equals the difference of their squares. Ex. 184. Construct a square whose area is equal to the sum of the areas of three given squares. Ex. 185. Upon a given line as a base, construct a rect- angle whose area is equal to the sum of the areas of a given square, a given trapezoid and a given triangle. Ex. 186. Construct a parallelogram having a given angle, whose base and altitude taken together equal a given line, and whose area equals the area of a given square. Ex. 187. Two parallel chords of a circle are, respect- ively, 36 in. and 48 in. long; the radius of the circle is 30 in. What is the distance between the chords ? Ex. 188. The area of a square inscribed in a circle is one half the area of a square circumscribed about the same circle. 202 PLANE GEOMETRY. PROPOSITION XIX. THEOREM. 285. To construct a triangle equal in area to a given polygon. Let A B C D E Fbea given polygon. To construct a triangle equal in area to the polygon A B CD EF. SuG. 1. Draw a diagonal cutting off one vertex; as B D. SUG. 2. Through C, the vertex cut off, draw a line || to the diagonal B D. SUG. 3. Extend E D until it meets the line || to the diagonal at O. Connect O with B. SUG. 4. Compare the areas of the AsCD and BOD. SUG. 5. Compare the areas of the polygons A B CD EF m&ABOEF. SUG. 6. Notice that the number of sides of the last polygon is one less than of the original polygon. Continue the process until the polygon is reduced to a triangle. Ex. 189. The area of a triangle is equal to one half the product of its perimeter by the radius of the inscribed circle. MEASUREMENT OF POLYGONS. 203 PROPOSITION XX. PROBLEM. 286. To find two straight lines which have the same ratio as two given similar polygons. Let A and B be two given similar polygons. To find two lines whose ratio equals the ratio -^. SuG. 1. Construct a right triangle, ABC, having for its legs A B and B C, any two homologous lines in the two polygons, and from B, the vertex of the rt. Z., drop a _L B D upon the hypotenuse A C. SuG. 2. Compare the ratio Py on w jth the ratio polygon N A n 2 AD . Compare the last ratio with the ratio ^-^ . SuG. 3. What, then, are the required lines? Ex. 190. If tangents to a circle be drawn at the ex- tremities of any chord, these tangents make, with each other, an angle which is twice the angle between the chord and the diameter of the circle drawn through the extremity of the chord. Ex. 191. Any straight line drawn through the center of a parallelogram divides the parallelogram into two parts which are equal in all respects. 204: PLANE GEOMETRY. PROPOSITION XXI. PROBLEM. 287. To construct a polygon similar to a given poly- gon and equal in area to another given polygon. Let A and B be two given polygons. To construct a polygon similar to A, and equal in area to B. SUG. 1. Construct squares equal in area to the polygons A and B respectively, and let m and n represent the sides of the squares. SUG. 2. L,et a represent one side of the polygon A, and find a fourth proportional to m> n and a. SUG. 3. Let p represent the fourth proportional just found, and construct a polygon P similar to A, having the side p in the polygon P homologous to the side a in the polygon A. SUG. 4. Compare the ratio -p- with the ratio j. A 2 SUG. 5. Compare the ratios -^ and ^ . Compare -^ with 3- . (See Sugs. 2 and 3.) Compare the ratios -=- and 2- . Compare B with P. Jr 288. A line is divided into mean and extreme ratio when the ratio of the whole line to the greater segment equals the ratio of the greater to the lesser segment. MEASUREMENT OF POLYGONS. 205 PROPOSITION XXII. PROBLEM. 289. To divide a straight line into mean and exr treme ratio. Let ABbea given straight line. To divide A B into mean and extreme ratio. SUG. 1. At B, erect a _L to A B, and on this _L lay off B O = \A B. With O as a center, and O B as a radius, describe a circumference. Draw a line through A and O, meeting the circumference at N and M. On A lay off a distance A X equal to A N. A M A B . 2. AN SUG. 3. By division -- -. Giveauth. BX _. Explain. SUG. 4. Therefore, by inversion and substitution AB _ AX AX~'' XB' PROPOSITIONS IN CHAPTER IV. PROPOSITION I. Two parallelograms having equal bases and equal altitudes are equal in area. PROPOSITION II. If two rectangles have equal altitudes, the ratio of their areas equals the ratio of their bases. PROPOSITION III. The number of units of area in any rectangle is equal to the prod- uct of the number of linear units in the base and altitude. PROPOSITION IV. The area of a parallelogram equals the product of its base and altitude. PROPOSITION V. The area of a triangle equals one half the product of its base and altitude. PROPOSITION VI. The area of a trapezoid equals one half the product of its altitude by the sum of its bases. PROPOSITION VII. The ratio of the areas of two similar triangles is equal to the ratio of the squares of their homologous sides, or homologous altitudes. PROPOSITION VIII. The ratio of the areas of two similar polygons equals the ratio of the squares of two homologous sides. PROPOSITION IX. The squnre described upon the hypotenuse of a right triangle is equal to the sum of the squares described upon the other two sides. MEASUREMENT OF POLYGONS. 207 PROPOSITION X. The square described on the sum of two lines equals the sum of the squares described on the two lines plus twice the rectangle whose sides are the two lines. PROPOSITION XI. The squares described on the difference of two lines equals the sum of the squares described on the two lines minus twice the rectangle whose sides are the two lines. PROPOSITION XII. The square on the side opposite an acute angle of a triangle equals the sum of the squares oc the other two sides minus twice the product of one side by the projection of the other side upon that side. PROPOSITION XIII. The square on the side opposite an obtuse angle of a triangle equals the sum of the squares on the other two sides plus twice the product of one of the sides by the projection of the other side upon that side- PROPOSITION XIV. Upon a given line as base, to construct a rectangle equal in area to a given square. PROPOSITION XV. To construct a square equal in area to a given rectangle. PROPOSITION XVI. To construct a square whose area is equal to the sum of the areas of two given squares. PROPOSITION XVII. To construct a square whose area is equal to the difference of the areas of two given squares. PROPOSITION XVIII. To construct a rectangle in which the sum of the base and altitude equals a given line and the area equals the area of a given square. 208 PLANE GEOMETRY. PROPOSITION XIX. To construct a triangle equal in area to a given polygon. PROPOSITION XX. To find two straight lines which have the same ratio as two given similar polygons. PROPOSITION XXI. To construct a polygon similar to a given polygon and equal in area to another given polygon. PROPOSITION XXII. To divide a straight line into mean and extreme ratio. CHAPTER V. REGULAR POLYGONS AND CIRCLES. DEFINITION. 290. A regular polygon is a polygon which is both equilateral and equiangular. The equilateral triangle and square are examples of regular polygons. PROPOSITION I. THEOREM. 291. An equilateral polygon inscribed in a circle is a regular polygon. Let ABC D EF represent an equilateral polygon in- scribed in a circle. To prove that A B C D E F is a regular polygon. SUG. 1. Compare the arcs A B, B C, etc. Give auth. SUG. 2. Compare Zs A B C, B C D, etc. Give auth. SUG. 3. Apply the definition of a regular polygon to A B CD EF. Therefore 14-Geo. 210 PLANE GEOMETRY. 292. COROI/LARY. A regular polygon may have any number of sides. For, if the circumference of a circle be divided into any number of equal parts, the lines joining the points of division will form an inscribed equilateral polygon. How does the number of sides of the potygon compare with the number of parts into which the circum- ference is divided ? PROPOSITION II. THEOREM. 293. A. circle can be circumscribed about a regular polygon, and a circle can be inscribed in a regular polygon. A M Let A B C D E F represent a regular polygon. I. To prove that a circle can be circumscribed about A B CDEF. SUG. 1. At M and N, the middle points of two adja- cent sides, erect JLs and extend them until they meet at O ; then O is the center of a circumference which passes through A, B and C. Why ? SUG. 2. Join O with each vertex of the polygon. SUG. 3. Compare O C with O B. SUG. 4. Compare Z O B C with Z. O C B. SUG. 5. Compare Z A B C with Z B C D. SUG. 6. Compare Z. O B A with Z O CD. REGULAR POLYGONS CIRCLES. 211 SUG. 7. Compare A O B A with A O C D, and hence, compare O A with O D. SUG. 8. The circumference through A, B and C, also passes through D. Why ? SUG. 9. In a similar manner show that the same cir- cumference passes through E and F. SUG. 10. Can a O be circumscribed about the polygon ABCDEFJ II. To prove that a circle can be inscribed in the poly- gon A B C D E F. SUG. 1. Compare the distances of the various sides of the polygon from O. (See Art. 153.) SUG. 2. Can a circle be inscribed in the polygon ABCDEF1 Therefore 294. The radius of a regular polygon is the radius of the circumscribed circle. The apothem of a regular polygon is the radius of the inscribed circle. The center of a regular polygon is the center of the inscribed or circumscribed circle. The angle at the center of a regular polygon is the angle formed by two radii drawn to two adjacent vertices of the polygon. 295. From the definitions just given three inferences may be immediately drawn: (1) The angle at the center of a regular polygon equals four right angles divided by the number of sides of the polygon. (2) The radius of a regular polygon bisects the angle of the polygon to which it is drawn. (3) The angle at the center of a regular polygon is the supplement of the interior angle of the polygon. 212 PLANE GEOMETRY. PROPOSITION III. THEOREM. 296. If a circumference is divided into any num- ber of equal parts, the tangents drawn through the points of division form a regular circumscribed poly- gon. C K Let the circumference ABCDE Foe divided into equal parts, AB,BC, etc. 9 and at the points A, B, etc., let tan- gents be drawn forming the polygon G H K L M N. To prove that G H K L M N is a regular polygon. SuG. 1. By what is the /_ G measured ? By what is the Z. H measured ? Compare ^/s G and H. Compare ^s H and K, etc. SUG. 2. Compare the As A G B and B H C. Com- pare the As B H C and C K D, etc. SUG. 3. Compare GH and HK\ also, H XT and KL, etc. SUG. 4. Apply the definition of a regular polygon to GHK LM N. Therefore 297. COROU,ARY I. If the vertices of a regular in- scribed polygon be connected with the middle of the arcs subtended by the sides, a regular inscribed polygon of double the number of sides is formed. (See Art. 145.) REGULAR POLYGONS CIRCLES. 213 298. COROLLARY II. The perimeter of a regular in- scribed polygon is less than the perimeter of a regular inscribed polygon of double the number of sides. 299. COROLLARY III. If a regular polygon is circumscribed about a circle, and if tangents are drawn at the middle points of the arcs between the original points of contact, a regular circumscribed polygon of double the number of sides is formed. 300. COROLLARY IV. The perimeter of a regular cir- cumscribed polygon is greater than that of a regular cir- cumscribed polygon of double the number of sides. 301. COROLLARY V. If a regular polygon is inscribed in a circle, tangents drawn at the middle points of the arcs subtended by the sides of the inscribed polygon, form a regular circumscribed polygon whose sides are parallel to the sides of the inscribed polygon, and whose vertices lie in the radii extended of the inscribed polygon. SUG. 1. Draw radii of the O to the points of contact of tangents; also radii to the vertices of the inscribed poly- gon, and extend these latter radii. SUG. 2. Show that each radius of the latter set bisects the Z formed by two adjacent radii of the former set. Ex. 192. The line joining the middle points of the bases of a trapezoid divides the trapezoid into two parts equal in area. 214 PLANE GEOMETRY. PROPOSITION IV. THEOREM. 302. Regular polygons of the same number of sides are similar H K Let A J> and HP represent two regular polygons of the same number of sides. To prove thai A D and HP are similar polygons. SUG. 1. Compare the sum of the interior Z!s of A D with the sum of the interior ^s of HP. SUG. 2. Compare Zs A and H, B and K, etc. SuG.. 3. Compare the ratios - ^ and -= - , etc. Jj C K. L, A B B C* SuG. 4. Compare the ratios -77-^= and -==-j , etc. JTl 4\. Therefore Ex. 193. Given the three angles of a triangle, and the area equal to the area of a given polygon; construct the triangle. SUG. Construct any triangle having the three given angles, and then consult Art. 287. Bx. 194. What is the locus of the center of a circle which is tangent to a given line at a given point ? REGULAR POLYGONS CIRCLES. 215 PROPOSITION V. THEOREM. 303, The perimeters of two regular polygons of the same number of sides have the same ratio as their radii, or their apothems. H N K Let ABC I> E F and HK LPRT represent two regu- lar polyyoiitt of the same number of sides. . AB+ BC+CD+DE+EF+FA To prove that HK + KL +LP+PR +R T~+Ttf O A OM ~~ Stf == SJV' SUG. 1. Compare the ratio of the perimeters with the A B ratio of any two homologous sides; as -77^- SUG. 2. Compare ^j^ with -^~F>, or with -=-T>. -T> IV O A STJG. 3. Compare the ratio of the perimeters with -prr>, o ft . , OM orwlth __ r . Therefore - 304. COROLLARY. The areas of two regular polygons of the same number of sides have the same ratio as the squares of their radii, or the squares of their apothems. 216 PLANE GEOMETRY. PROPOSITION VI. THEOREM. 305. If the number of sides of a regular inscribed polygon be increased indefinitely the apothem is a variable which approaches the radius as a limit. Let AB be a side of a regular polygon, and O H its apothem. To prove that if the number of sides of the polygon be increased indefinitely, the apothem is a variable which ap- proaches O A as a limit. SuG. 1. In the A A O H, compare O A and O H. SuG. 2. Compare O A O H with A H\ with A B. SuG. 3. Since, by sufficiently increasing the number of sides of the polygon, one side, A B, may be made shorter than any assumed line, however short, what re- lation does O A sustain to O HJ That is, what relation does the radius sustain to the apothem ? Therefore Ex. 195. If the middle points of two adjacent sides of a parallelogram be joined, a triangle is formed, equal in area to one eighth of the area of the parallelogram. REGULAR POLYGONS CIRCLES. 217 EXERCISES. 196. Draw a circle through a given point tangent to a given line at a given point. 197. The diagonals of a parallelogram divide it into four triangles equal in area. 198. What is the locus of the vertex of an isosceles tri- angle upon a given base ? 199. A B C is a triangle, and D any point in B C ex- tended. Find a point E in A B, such that the area of the triangle E B D will be equal to the area of the tri- angle ABC. 200. Draw a line through a given point in a side of a triangle so as to divide the triangle into two parts equal in area. 201. If the diagonals of a quadrilateral intersect at right angles, show that the area of the quadrilateral is one half the area of a rectangle whose sides are equal to the diagonals of the quadrilateral. 202. The hypotenuses of three isosceles right triangles form a right triangle. Prove that one of the isosceles triangles is equal to the sum of the other two. 203. If the diagonals of a quadrilateral intersect at right angles, prove that the sum of the squares on one pair of opposite sides equals the sum of t-ie squ'ares on the other pair. 204. If a parallelogram be inscribed in, or circum- scribed about, a circle, the diagonals pass through the center. 205. A circle circumscribes an isosceles triangle, and tangents are drawn to the circle through the vertices of the triangle. Prove that these tangents form a second isosceles triangle, and that the two triangles cannot have equal vertical angles unless both are equilateral. 218 PLANE GEOMETRY. PROPOSITION VII. THEOREM. 306. If the number of sides of a regular inscribed polygon be increased indefinitely, the perimeter of the polygon is a variable which approaches the circum- ference of the circle as a limit. Let CD represent the side of a regular inscribed poly- gon, and p its perimeter. To prove that as the number of the sides of the polygon increases^ p is a variable which approaches the circumference of the circle as a limit. SUG. 1. L,et E B be the side of a circumscribed poly- gon similar to the inscribed polygon, and let P denote the perimeter of the circumscribed polygon. L,et r de- note the radius and a the apothem of the inscribed poly- gon. P r SUG. 2. Compare the ratios - and . f _ P a SUG. 3. The ratio ^ = what? See Ex. 128. SUG. 4. From the answer to Sug. 3, find the value SUG. 5. Since P is greater, and p less, than the cir- cumference of the 0, the difference P p must be greater than the difference between the circumference and either P or/. I^et c represent the circumference of the O, then c p is less than P p. REGULAR POLYGONS CIRCLES. 219 SuG. 6. Show that by increasing the number of sides of the polygon, the value of P p (Sug. 4), may be made to decrease indefinitely, yet cannot be made equal to zero. SUG. 7. Hence, as c p is less than P ' p and still c p cannot be made equal to zero, therefore, c is the limit of p. That is, the circumference of the O is the limit of the perimeter of the inscribed polygon when the number of sides of the polygon is made to increase in- definitely. Therefore ADDITIONAL HELPS. The numbers attached to the following additional helps indicate that they are answers to suggestions of corresponding numbers. ' >-.- 8 . fji---^. 4. P-p=(r~a) x - r . p 6. - continually decreases, since P decreases and r remains unchanged; also, r a diminishes indefinitely p (Prop. VI). Hence (r a) X diminishes indefin- itely, but cannot be made absolutely zero. Ex. 206. To draw a common tangent to two given circles. Ex. 207. If three equal circles are tangents to one an- other, the lines joining their centers form an equilateral triangle. 220 PLANE GEOMETRY. PROPOSITION VIII. THEOREM. 307. If tine number of sides of a regular inscribed polygon be increased indefinitely, the area of the poly- gon is a variable which approaches the area of the circle as a limit. Let C J> represent a side of a regular inscribed polygon, and let A denote the wrea of the inscribed polygon and A' the area of the circle. To prove that if the number of sides of the polygon be in- definitely increased, A is a variable whose limit is A'. SUG. 1. Let r denote the radius, and a the apothem, of the inscribed polygon; also, let/> denote the perimeter of the inscribed polygon, and P the perimeter of a similar circumscribed polygon. Let E B be a side of the cir- cumscribed polygon homologous to the side C D of the inscribed polygon. SUG. 2. What is the area of the trapezoid E B C Dm terms of E B, CD and F G ? That is, in terms of E B> CZ>and (r a). SUG. 3. What is the difference of the areas of the cir- cumscribed and inscribed polygons in terms of P, p and (r a) ? REGULAR POLYGONS CIRCLES. 221 SUG. 4. Let A" represent the area of the circumscribed polygon. Then A" A equals what in terms of P. f> and (r a) ? SUG. 5. Compare the answer to Sug. 4 with Px (ra). SUG. 6. As the number of sides of the polygon is in- creased, P evidently diminishes, and (r a) approaches zero as a limit. Therefore, P X (r a) approaches zero as a limit. What, then, may be said of the difference A" A as the number of sides of the polygons is in- creased ? Therefore ADDITIONAL HELPS. The numbers attached to the following additional helps indicate that they are answers to suggestions of corre- sponding numbers. 2. Trapezoid ABCD=(EB+ CD)X (r- a). Why? 4. A" A = (P+fi)x(ra). Why? 5. (P+/)X(r )X(r--). Why? 6. A" A continually decreases and approaches zera as a limit. 308. COROLLARY. The area of the circle is the limit of the area of a regular circumscribed polygon as the number of sides is indefinitely increased. Ex. 208. If the radius of one circle is the diameter of another, the circles are tangent to each other, and any line drawn from the point of contact to the outer circum- ference is bisected by the inner one. Ex. 209. Draw a circle of a given radius, through a given point, tangent to a given circle. When will there be two solutions, when one solution and when no solu- tion. 222 PLANE GEOMETRY. PROPOSITION IX. THKOREM. 309. The area of a regular polygon is equal to one half the product of its perimeter and apothem. -D Let AB C D EF be a regular polygon, and O N its apothem. To prove that the area of the polygon equals one half the product of its perimeter and apothem. SUG. 1. Circumscribe a circle, and draw radii to the vertices of the polygon. SuG. 2. Compare the distances from the center to the sides of the polygon. SuG. 3. The area of the A A O B equals what ? Give an expression for the area of each of the other As into which the polygon is divided. The sum of the areas of all the As equals what? Therefore Ex. 210. Draw a circle with its center at a given point tangent to a given circle. When is there only one solu- tion? REGULAR POLYGONS CIRCLES. 223 EXERCISES. 211. Construct a triangle equal in area to a given tri- angle, two sides of tne required triangle being given. Show when there are two solutions, when one solution and when no solution. 212. Construct a line that shall equal 1^3. SUG. Let x = 1/3". Then x* = 3; or, - = ^ . See X JL Art. 249. 213. Construct a line that shall equal l/S in. Give two solutions. 214. Construct three equal circles which shall be tan- gent to one another and to a given circle. Consider two cases: (1) that in which the three circles are within the given circle, and (2) that in which the three circles are without the given circle. 215. If the diameter of a circle be divided into two parts, and on these parts, as diameters, circles be de- scribed, the sum of the circumferences of the two circles equals the circumference of the given circle. 216. Find the side of a square equal in area to a rect- angle whose sides are 6 and 9. 217. Find the length of the chord joining the points of contact of two tangents to a circle whose radius is 8 in. drawn from a point 12 in. from the center. 218. Construct a triangle whose area is equal to 9 times the area of a triangle whose sides are 6, 7 and 9. 219. What is the ratio of the areas of two similar tri- angles whose homologous sides have the ratio f . 220. Give.i two similar triangles, construct a third tri- angle similar to the other two whose area shall be equal to the sum of the areas of the other two. 224 PLANE GEOMETRY. PROPOSITION X . THEOREM. 310. The circumferences of two circles have the same ratio as their radii. Let O and S represent two circles, JB and H' their radii and C and C' their circumferences. To prove that ^ = -~. SUG. 1. Inscribe in the two circles similar regular poly- gons. I,et P and P' represent their perimeters. Then -p = -^7- . Give auth. r> SUG. 2. From the equation in Sug. 1, P = Px-=- . K SuG. 3. Now, let the number of sides of each polygon continually increase, always, however, keeping the num- ber of sides of one polygon the same as the number of sides of the other polygon. During this change the r> variable P is always equal to the variable P x -^ . Why? K SuG. 4. Since P and P are variables which approach C and C' respectively, as their limits (Prop. VII), there- r> fore P and P X -^- are variables which approach C and TO C'x-^7- respectively, as their limits. Why ? REGULAR POLYGONS CIRCLES. 225 SUG. 5. C= C'xr. Why? /t SUG. 6. Hence ^ = ~ . Why ? Therefore - 311. COROU,ARY I. The circumferences of two circles have the same ratio as their diameters. If D and D' represent their diameters, C D_ C ~ = >' ' 312. CORONARY II. The ratio of the circumference of a circle to its diameter is the same for all circles. For, by Cor. 1, ^ = -jy , C C' and by alternation -~ = . This ratio is represented by the Greek letter n (pi). Hence, ^ = n ; or, ^ = n . Hence, C = n D ; or, C = 2 n R. Ex. 221. Draw a line through the vertex of a triangle so as to divide the triangle into parts whose areas have the ratio f. Ex. 222. Two circles are tangent externally; locate a line of a given length so that it shall pass through the point of contact and have its extremities in the circum- ferences of the circles. Ex. 223. Given one leg of a right triangle and the ra- dius of the inscribed circle, construct the triangle. Ex. 224. Inscribed a circle in a given rhombus. 15 Geo. 226 PLANE GEOMETRY. PROPOSITION XI. THEOREM. . 313, The areas of two circles have the same ratio as the squares of the radii. Let A and A' represent the areas, and M and H' the radii of the circles O and O' respectively. To prove that -*r =* -fsr A J\. SUG. 1. Inscribe in the two circles regular polygons of the same number of sides, and let M and M 1 represent their respective areas. M SUG. 2. What does the ratio -^ equal in terms of R M' u and R ? (Art. 268.) SUG. 3. Proceed as in Prop. X. Therefore 314. COROLLARY. The areas of circles have the same ratio as the squares of their diameters. Ex. 225. Construct a triangle equal in area to a given triangle, the base and the distance from the vertex to the middle point of the base being given. REGULAR POLYGONS CIRCLES. 227 PROPOSITION XII. THEOREM. 315. The area of a circle equals one half the product of its circumference and radius. Let O represent a circle, H its radius, C its circumfer- ence and A its area. To prove that A equals one half of C X R. SUG. 1. Inscribe in the circle a regular polygon, and let P denote its perimeter, Afits area, and r its apothem. SUG. 2 Express M in terms of P and r. Give auth. SUG. 3. If the number of sides of the polygon be in- definitely increased, express the value of M. SUG. 4. Express the limit of M t the area of the vary- ing polygon. SUG. 5. Express the limit of the varying value of M. SUG. 6. What is the area of the O equal to ? Why ? Therefore ADDITIONAL HELPS. The numbers attached to the following helps indicate that they are answers to the above suggestions of the same numbers. 2. Af=$Pxr. 3. M = \ Px r however great the number of sides. 4." Limit of M = A. (Art. 307.) 5. Since the limit of P is C (Art. 306), the limit of \P*R is \ CxR. 6. Therefore A = %CxR. (Art. 170.) 228 PLANE GEOMETRY. "* 316. COROLLARY I. The area of a circle equals n times the square of the radius. If the value of C given in article 312 be substituted in the expression for the area, the above equation becomes A = \x2nRxR>, that is, A = TT R*. 317. COROLLARY II. The area of a sector equals one half the product of radius and arc. SUG. Compare the ratio of the sector to the circle, with the ratio of the arc of the sector to the circumference of the circle. PROPOSITION XIII. THEOREM. 318. One side of a regular hexagon equals the ra- dius of the circumscribed circle. Let ABC represent a regular hexagon inscribed in a cvrcle whose radius is O A. To prove A B = O A. . SUG. 1. Draw the radius O B. SUG. 2. The Z. A O B is what part of four rt. Zs ? Then, how many degrees are there in Z. A O B1 SuG. 3. How many degrees in the sum of the Z!s O A B and O B A ? SUG. 4. Compare Zs O A B and O B A. How many degrees in each of these REGULAR POLYGONS CIRCLES. 229 SUG. 5. Compare the three Zs of the A A O B. SUG. 6. Compare the sides O A and A B. Therefore 319. Similar arcs, similar sectors and similar seg- ments in different circles are those which correspond to equal angles at the center. PROPOSITION XIV. THEOREM. 320. The areas of tivo similar segments have tlie same ratio us the squares of their radii. Let BCD nnd F HK represent two similar segments of circles tvhose centers are A and E respectively. seg. SUG. 1. Provethatthe As^ #C and ^^7 are similar. SUG. 2. Express the ratio of the areas of As A B C and E F H in terms of A B and EF\ also the ratio of the areas of sectors A B D C and EFKH in terms of A B and E F. SUG. 3. Compare the ratio of the areas of the sectors with the ratio of the areas of the As. SUG. 4. Take the proportion of Sug. 3, by alternation and then by division. Compare the ratio of the areas of segments with the ratio of the areas of As or of sectors. SUG. 5. Compare the ratios ^. and seg.FHK Therefore - 230 PLANE GEOMETRY. PROPOSITION XV. PROBLEM. 321. Given the radius of a circle and the side of a regular inscribed polygon, required to find the side of a regular inscribed polygon of double the number of sides. C Let O represent a circle, H its radius, A B the side of a regular inscribed polygon, and A C the side of a regular inscribed polygon of double the number of sides. To find the length of A C in terms of the known quanti- ties, R and A B. SUG. 1. Connect O and A. SuG. 2. In the rt. A A S O, express the value of 5 O in terms of O A (= R} andAS(=$A B}. (Art. 270.) SUG. 3. Express C S in terms of R and O, and then in terms of R and A B. SUG. 4. In the rt. A A S C, A S and C S have both been expresssed in terms of R and A B. Hence, express A C in terms of R and A B, and reduce the answer to its simplest form. The result is: A c = V' If R equals unity A C = AB *-V-A~B*. REGULAR POLYGONS CIRCLES. 231 EXERCISES. 226. If, from a point without a circle, two secants be drawn whose external segments are 8 in. and 3 in., while the internal segment of the latter is 17 in., what is the internal segment of the former ? 227. The sides of a triangle are 5, 6 and 7, and the side corresponding to 6, in a similar triangle, is 36; find the other two sides of the triangle. 228. Find the altitude of an equilateral triangle whose side is 10 in. Find the side when the altitude is 10 in. 229. If three -similar polygons be constructed on the three sides of a right triangle, prove that the area of the polygon constructed on the hypotenuse equals the sum of the areas of the polygons constructed on the other two sides. SUG. See Art. 268. 230. All equal chords of any circle are tangents to some other circle. 231. In A B, the diameter of a circle, or, in A B ex- tended, take any point C, and draw CD perpendicular to A B ; if A be joined with any point P in CD, and A P meet the circumference at Q\ then A PxA Q=*ACx AB. That is, A P X A Q = a constant. 232. If A is a given point, and P any point in a given straight line, and if a point Q be taken in the line joining A and P, so that A P x A Q is constant, then, as Amoves along the given line, Q will move on the circumference of a circle which passes through A. (See Ex. 231.) 233. If A B be the diameter of a circle, and if a point P be taken on any chord A Q, or A Q extended, so that A PxA Q is constant, the locus of P will be a straight line perpendicular to A B. (See Ex. 231.) 232 PLANE GEOMETRY. PROPOSITION XVI. PROBLEM. 322. To compute, approximately, the ratio of the circumference of a circle to its diameter. Consider the circle whose radius is unity. By Prop. XV, A C = J 2 - 1/4 -A~*. If A Cis one of the equal sides of a regular polygon of n sides, n X A C = the perimeter, which may be re- garded as an approximation to the circumference of the circle; the greater the number of sides the closer the ap- proximation. If A B equals one side of a regular inscribed hexagon, A B = the radius, which = 1. The formula then gives -l/~3 = .51763809. A C= A/2-- 1/3T- (I) 2 = Hence, the perimeter of a regular twelve-sided poly- gon = 12 x. 51763809 = 6.21165708. The result found for a side of a regular twelve-sided polygon substituted for A B in the above formula, gives the side of a regular twenty-four-sided polygon. In this way the sides of regular inscribed polygons of greater and greater number of sides may be found. Some of the results are given in the following table: NO. SIDES. ONE SIDE. PERIMETER. 6 1$ 24 48 96 192 1/2 - >/3 1 '=.51763809 = .26105238 = .13080626 = .06533817 - .03271346 6 6.21165708 6.26525722 6.27870041 6.28206396 6.28290510 1/2 -N/4 -(.51763809) 2 V2 - x/4 - (.26105238) 2 1/2 -v/4 -(.13080626)2 1/2 -v/4 - (.06533817) 2 REGULAR POLYGONS CIRCLES. 233 In this way, the perimeter of a polygon of 768 sides has been computed to be 6. 283169 + . Dividing this result by the diameter, i. e., by 2, gives 3. 141584 + , as an approximate value of the ratio of the circumference of a circle to its diameter, i. e. , an approx- imate value of n. The approximate value usually used is 3.1416. Therefore, n = 3.1416 approximately. 323. COROLLARY. As an approximate value of n has been found, the area of a circle may be found approx- imately in terms of its radius. The approximate value is found by multiplying the square of the radius by 3*14.16. (Art. 315.) 324. SCHOLIUM. Archimedes (born 287 B. C), found an approximate value of n. He proved that its value is between 3| and 3fJ. The smaller of these two values is often used as an approximate value of n when great ac- curacy is not required. In modern times the value of n has been computed to a large number of decimal places. Clausen and Dase, in- dependently of each other, computed the value to the two hundredth decimal place. Other computers have given the value to over five hundred decimal places, but their results have not been verified. The number is in- commensurable, and cannot be expressed exactly by any number of decimal places. Ex. 234. The altitude of an equilateral triangle equals one and one half times the radius of the circumscribed circle. Ex. 235. If the center of each of two equal circles lies on the circumference of the other, the square on the common chord is three times the square on the radius. 234 PLANE GEOMETRY. PROPOSITION XVII. PROBLEM. 325. To inscribe a square in a given circle. Let Obea given circle. To inscribe a square in the circle O. ' SuG. In any square, at what /_ do the diagonals in- tersect ? How, then, can a square be inscribed in a circle ? 326. COROLLARY. By bisecting the arcs subtended by the sides of the square, and joining each point of di- vision with the two adjacent vertices, a regular inscribed octagon is formed. PROPOSITION XVIII. PROBLEM. 327. To inscribe a regular hexagon in a given circle. O Let O be a given circle. To inscribe a regular hexagon in the circle O. REGULAR POLYGONS CIRCLES. 235 SUG. Lay off six times as a chord, the line which equals one side of the hexagon. How can this line be obtained ? 328. COROLLARY I. By joining the alternate vertices of the regular inscribed hexagon, an equilateral triangle is inscribed in the circle. 329. COROLLARY II. By bisecting the arcs subtended by the sides of a regular inscribed hexagon, and joining the points of division with the adjacent vertices of the hexagon, a regular dodecagon (i. e., a regular polygon of twelve sides), is inscribed in the circle. Ex/236. Divide a circle into segments such that an angle inscribed in one segment shall be three times an angle inscribed in the other segment. Ex. 237. From a point without a circle two tangents are drawn which, with the chord of contact, from an equi- lateral triangle whose side is 18 in. Find the diameter of the circle. Ex. 238. Find the locus of the center of a circle which passes through two fixed points. Ex. 239. Find the locus of the point of intersection of tangents to a fixed circle which intersect at a given angle. Ex. 240. Find the locus of a point equally distant from the circumferences of two eqaal non-intersecting circles. Ex. 241. Find the locus of the center of a circle tan- gent to two intersecting straight line. Ex. 242. Find the locus of a point equally distant from the circumferences of two concentric circles. Ex. 243. Prove that the apothem of an inscribed equi- lateral triangle equals one half of the radius of the circle. 236 PLANE GEOMETRY. PROPOSITION XIX. PROBLEM. 330. To inscribe a regular decagon in a given circle. Let Obea given circle. To inscribe a regular decagon in the circle O. SUG. 1. Divide the radius O A into mean and extreme ratio, and let O C be the greater segment. SUG. 2. Draw the chord A B equal to to O C, and con- nect O with B, and C with B. SUG. 3. See the definition of mean and extreme ratio and form a proportion with the lines O A, O C and C A. SUG. 4. In the proportion just formed substitute for O C its equal, A B, thus forming a new proportion. SUG. 5. By means of the last proportion, the As O A B and B A C in respect to similarity. In this comparison notice that the Z. A is common to the two As. SUG. 6. Compare the lines A B and B C ; also the lines O C andC. SUG. 7. Compare /.ABC with Z O ; also, Z CB O with Z. O. Compare Z. A B O with Z. O. Compare Z. B A O with Z O. SUG. 8. The sum of the three Zs of the A OB A con- tains the Z O how many times ? REGULAR POLYGONS CIRCLES. 237 SUG. 9. The Z O is what part of two rt. Zs ? What part of four rt. Z!s ? Hence, the chord, A B, subtends what part of the circumference ? SUG. 10. Now, give the method of inscribing a regular decagon in a circle. 331. COROLLARY. By joining the alternate vertices of a regular inscribed decagon r a regular inscribed penta- gon is formed. 332. GENERAL SCHOLIUM. Methods have already been explained of inscribing in a circle a regular polygon of 3, 4, 5, 6, 8 or 10 sides. Any regular inscribed polygon being given a regular inscribed polygon of double the number of sides, can be formed by bisecting the arcs sub- tended by the sides and joining the points of division to the adjacent vertices of the given polygon. Hence, by means of the inscribrd square regular polygons can be inscribed of 8, 16, 32, etc., sides; by means of the regular inscribed hexagon regular polygons can be inscribed of 12, 24, 48, etc., sides; by means of the regular inscribed decagon regular polygons can be inscribed of 20, 40, 80, etc. , sides. There is still one more set of polygons which can be inscribed in a circle. For, if from any point on the circumference a chord be drawn eqnal to one side of a regular inscribed hexagon, and from the same point an- other chord be drawn equal to the side of a regular in- scribed decagon, then the first chord subtends an arc which is one sixth of the circumference, and the second chord an arc which is one tenth of a circumference, and the difference between these two arcs is one fifteenth of the circumference. Hence, the chord joining the ex- tremities of the two chords previously drawn is one side of a regular inscribed polygon of fifteen sides, and from this regular polygons can be inscribed of 30, 60, etc., sides. 238 PLANE GEOMETRY. Until the beginning of the present century it was sup- posed that the polygons already enumerated were the only ones which could be inscribed by elementary geome- try, but, in a work published in 1801, Gauss proved by means of the ruler and dividers only, it is possible to in- scribe regular polygons of 17 sides, of 257 sides, and, in general, of any number of sides which can be expressed by 2*-f 1, n being an integer provided that 2"-f 1 is a prime number. PROPOSITIONS IN CHAPTER V. PROPOSITION I. An equilateral polygon inscribed in a circle is a regular polygon. PROPOSITION II. A circle can be circumscribed about a regular polygon, and a circle can be inscribed in a regular polygon. PROPOSITION III. If a circumference is divided into any number of equal parts, the tangents drawn through the points of division form a regular circum- scribed polygon. PROPOSITION IV. Regular polygons of the same number of sides are similar. PROPOSITION V. The perimeters of two regular polygons of the same cumber of sides have the same ratio as their radii, or their apothems. PROPOSITION VI. If the number of sides of a regular inscribed polygon be increased indefinitely the apothem is a variable which approaches the radius as a limit. PROPOSITION VII. Ii the number of sides of a regular inscribed polygon be increased indefinitely the perimeter of the polygon is a variable which ap- proaches the circumference of the circle as a limit. PROPOSITION VIII. If the number of sides of a regular inscribed polygon be increased indefinitely, the area of the polygon is a variable which approaches the area of the circle as a limit. 240 PLANE GEOMETRY. PROPOSITION IX. The area of a regular polygon is equal to one half the product of its perimeter and apothem. PROPOSITION X. The circumferences of two circles have the same ratio as their radii. PROPOSITION XI. The areas of two circles have the same ratio as the squares of their radii. PROPOSITION XII. The area of a circle equals one half the product of its circumfer- ence and radius. PROPOSITION XIII. One side of a regular hexagon equals the radius of the circum- scribed circle. PROPOSITION XIV. The areas of two similar segments have the same ratio as the squares of their radii. PROPOSITION XV. Given the radius of a circle and the side of a regular inscribed polygon, required to find the side of a regular inscribed polygon of double the number of sides. PROPOSITION XVI. To compute, approximately, the ratio of the circumference of a circle to its diameter. PROPOSITION XVII. To inscribe a square in a given circle. PROPOSITION XVIIL To inscribe a regular hexagon in a given circle PROPOSITION XIX. To inscribe a regular decagon in a given circle. GEOMETRY OF THREE DIMENSIONS. CHAPTER VI. LINES AND PLANES. DEFINITIONS. 333. Geometry of three dimensions treats of fig- ures whose parts are not confined to a single plane. Geometry of three dimensions is also called solid geometry and geometry of space. The results of plane geometry furnish the basis for the investigations of geometry of three dimensions, but it must be remembered that the statements of plane geome- try are made with respect to figures which are entirely in one plane, and, in the form stated, are not necessarily true in geometry of three dimensions. For example, in plane geometry it is stated that from a given point in a straight line only one perpendicular can be drawn to that line. This statement is true only of figures confined to a single plane, and hence is not true in geometry of three dimensions. In reasoning from plane geometry to geometry of three dimensions, the following axiom is used. 334. Axiom 14. The relations of the parts of a figure, or figures, in one plane are not changed by moving the plane containing the figure from one position to an- other. 16 Geo. 242 SOLID GEOMETRY. 335. A plane has been defined as a surface such that a straight line joining any two of its points will lie wholly in the surface. A plane is considered to be indefinite in extent, so that if any straight line of the plane be extended indefinitely it will continue to lie wholly in the surface. 336. A plane is said to be determined when it is ex- actly located, or is distinguished from every other plane. A plane is determined by lines or points when it is the only plane which contains those lines or points. Ex. 244. The apothem of a regular inscribed hexagon equals ^ R 1/3 ; R representing the radius of the circle. Ex. 245. One side of an inscribed equilateral triangle equals R 1/3 ; R being the radius of the circle. Ex. 246. One side of an inscribed square equals R 1/2 , and its apothem equals \ R 1/2; R being the radius of the circle. Ex. 247. Show how to cut off the corners of an equi- lateral triangle in such a way that the remaining figure will be a regular hexagon. Ex. 248. If any two polygons whatever be circum- scribed about the same circle, prove that their areas have the same ratio as their perimeters. Ex. 249. In a given circle inscribe an isosceles triangle in which each base angle is twice the vertical angle. Ex. 250. In a given circle inscribe a triangle whose angles are proportional to the numbers 3, 4 and 5. Ex. 251. What is the area of a sector whose arc is one sixth of the circumference, in a circle whose radius is 15 in. LINES AND PLANES. 243 POPOSITION I. THEOREM. 337. A plane may be passed through any straight line, but the line does not determine the plane. .D Let A I? represent any straight line. To prove that a plane may be passed through the line A B, but that the line does not determine the plane. Suo. 1. Consider any plane whatever, and draw any straight line in the plane. SUG. 2. Move the line and plane until the line of the plane coincides with the given line, A B. SUG. 3. Is it possible, then, to pass a plane through the line A B1 SUG. 4. Must the plane always occupy exactly the same position when the line of the plane coincides with A fit SUG. 5. Does the line determine the plane? Therefore 338. When a plane takes in succession the various possible positions with the line of the plane coinciding with the line A B, the plane is said to revolve about A B. 244: SOLID GEOMETRY. PROPOSITION II. THEOREM. 339. A plane is determined: I. By a straight line and a point without the line* II. By two intersecting straight lines. III. By two parallel straight lines. IV. By three points not in a straight line. C * l\r w Let A B represent a straight line, and C a point not in the line A B. To prove that the line A B and the point C determine a plane. SUG. 1. Through the line A B pass a plane, and let it revolve about A B until it contains the point C. SUG. 2. How much can the plane be revolved, either way, about B, and still contain the point C ? Why ? SUG. 3. What is your conclusion about part I of the proposition ? M / Let A B and C D represent two intersecting lines. To prove that A B and C D determine a plane. SUG. 1. Pass a plane, M ' N, through the line A B and some point of C D. SUG. 2. How many planes can occupy this position ? Why? LINES AND PLANES. 245 SuG. 3. Where does the line C D lie with respect to the plane M Nt (Art. 335.) How many points of C D are in the plane M Nt SuG. 4. Do the lines A B and C D determine a plane ? -.' Let A B and C D represent two parallel lines. To prove that A B and C D determine a plane. SuG. 1. By definition of || lines, in how many planes do they lie ? SuG. 2. Pass a plane through A B, and one point of C D. How many planes occupy this position ? Where is the plane of || lines located with respect to this plane ? Then, how many planes can pass through the lines A B andC>? "Let A f B and C represent three points not in one straight line. To prove that the points A, B and C determine a plane. SUG. Connect two of the points. What conditions have you now ? Therefore 246 SOLID GEOMETRY. PROPOSITION III. THEOREM. 340. The intersection of two planes is a straight line. \/ \/ J\ Let MN and C D represent two planes, and A and B two points of their intersection. To prove that M N and CD intersect in a straight line. SuG. 1. Connect A and B. Where is the line A B with respect to each of the planes ? SuG. 2. Take any point of the plane M N, without the line A B. Can this point be in the plane C D also ? Why ? (See Prop. II, part I.) Therefore QUERY. What is the locus of a point common to two planes ? 341. The point in which a straight line intersects a plane is called the foot of the line. 342. A line is perpendicular to a plane when it is per- pendicular to every line in the plane drawn through its foot. The plane is then said to be perpendicular to the line. LINES AND PLANES. 247 PROPOSITION IV. THEOREM. 343. Fj'om a given point without a plane, one, and only one, perpendicular can be dropped to the plane, and the perpendicular is the shortest line from the point to the plane. Let M N represent the given plane, and A the given point ivithout the plane. To prove that y from A y one, and only one, perpendicular can be drawn to the plane M JV t and that the perpendicular is shorter than any other line from A to the plane M N. SUG. 1. Of all lines from A to the plane M N, either there is one shortest line or a group of equal shortest lines. Suppose A C and A D are two of a group of shortest lines. Connect C and D. SUG. 2. If^Cand^Z?are=,whatkindof A is ^ CD? SuG. 3. From A, draw A B _L to CD. Compare A B and A C in respect to length. SUG. 4. If any two lines from A to the plane M N, are equal in length, can they be shortest fines to the plane ? SUG. 5. Then, how many shortest lines are there from A to the plane M N? SUG. 6. Let A O represent the shortest line from A to the plane M N, and draw E G, any line in the plane M N, through the point O. SUG. 7. Since A O is the shortest line from A to the line E G, what relation does A O sustain to E G ? Why ? SuG. 8. Since E G is any line drawn through the foot of A O, what relation must A O sustain to the plane M N1 SUG. 9. How many J_s can be drawn from A to the plane M N1 Why? Therefore SOLID GEOMETRY. PROPOSITION V. THEOREM. 344. At a given point in a plane, one perpendicu- lar, and only one, can be erected to the plane. .C Let C D represent a plane, and O a given point in the plane. To prove that one and only one perpendicular can be erected to the plane at the point- O. SUG. 1. L,et M N represent another plane, and A B a _L from A to the plane M N. SUG. 2. Place the plane M N on the plane C D, so that B coincides with O, When so placed, what rela- tion does AB sustain to the plane CD? (Ax. 14.) Can a _L be erected to the plane C D, at the point O ? SuG. 3. If more than one J_ can be erected, let O E and O G represent two J_s to C D, at O. I^et R S be the intersection of the plane BOG with the plane C D. What relation do both O E and O G sustain to RSI Why ? Can a second J_ be erected to a plane at a given point of the plane ? Therefore LINES AND PLANES. 249 PROPOSITION VI. THEOREM. 345. If, *from any point in the perpendicular to a plane t oblique -lines be drawn, those which meet the plane at equal distances from the foot of the perpen- dicular are equal ; and, of two unequal oblique lines, that which meets the plane at the greater distance from the foot of the perpendicular is greater. Let M N represent a plane, and A B a perpendicular to the plane ; also, let A D, A C and A E represent oblique lines meeting tfie plane MN at the points J>, C and E re- spectively, and let B C equal B E, and B D be greater tftan B E. To prove that the oblique lines A C and A E are equal \ and that the line A D is longer than the line A E. SUG. 1. What relation does A B sustain to B C, and also to B E ? Why ? (Art. 342.) SUG. 2. Compare As A B C and ABE. SUG. 3. Compare A C and A E. SUG. 4. Revolve the plane of the A A B E upon A B as an axis, into the plane of A B D, and let A B F be the position which ABE takes after the revolution. SUG. 5. Since Zs A B F and A B D are rt. Zs (why are they rt. Zs ?), where will B F lie with respect to B D ? SUG. 6. Compare A F with A D t and hence A E with A D. Therefore 250 SOLID GEOMETRY. PROPOSITION VII. PROBLEM. 346. If ou straight line is perpendicular to two lines of a plane at their point of intersection it is perpen- dicular to the plane. Let My represent a plane, A U and C J> two lines of the plane, and E O a perpendicular to the two lines at their point of intersection, O. To prove that O E is perpendicular to the plane M N. SuG. 1. Extend O E to S, making O S = E O, and let O R represent any line of the plane through the point O. Draw the line B Z>, intersecting O R at R. Con- nect both E and S with the points B, R and D. SuG. 2. In As E O B and SOB, compare E B and SB\ also, m&sEOD and SOD compare ED and SD. SuG. 3. In As E B D and S B D, compare Z. E B D with Z.SB D. SuG. 4. In As^^^andS^^compare.E^with.S^. SUG. 5. In As E O R and S O R, compare Z_$ E O R and SO R. SuG. 6. What relation does E O sustain to O R ? SUG. 7. Since O R is any line of the plane through the point O, what relation does E O sustain to the plane ? (Art. 342.) Therefore LINES AND PLANES. 251 PROPOSITION VIII. THEOREM. 347. All the perpendiculars to a given line at the same point lie in the same plane t and that plane is perpendicular to the given line. B Let BO be a given line, and O A, O C, O J>, etc., lines peri>emlicular to & O, at O. To prove that OA, O C, O D, etc., are in the same plane, and that the plane is perpendicular to B O, at O. SUG. 1. The lines O D and O C determine a plane. The lines O B and O A determine another plane. Let these two planes intersect in the line O M. SuG. 2. What relation does B O sustain to the plane determined by O D and O C? Why ? (Art. 346.) SUG. 3. What relation does OB sustain to Why? SUG. 4. What relation does O B sustain to O A ? SuG. 5. What relation then must O A sustain to SUG. 6. Then, where must O A lie with respect to the plane determined by O- D and O C? Therefore 252 SOLID GEOMETRY. 348. COROLLARY I. Through a given point in a straight line only one plane can be drawn perpendicular to the line. Suppose one plane drawn through O, _L to O B. Then, if a second plane can be drawn through O J_ to O B, all the lines in that second plane through O would be _L to O B, at O. But all the lines _L to OB at O are in the same plane. Hence, the second plane must coin- cide with the first. That is, only one plane can be drawn through O _L to B O. 349. COROLLARY II. From a given point without a line only one plane can be drawn perpendicular to the given line. X~ -*B the plane A B, through the point N y be _L to M O at O. If another plane can be drawn through N J_ to M O, represent it by C B, and suppose C B is _L to .M O at 5. Connect N and O ; also N and S. What relation must ON and 5 N each sustain to MS? Why? Ex. 252. Find the area of a circular ring included be- tween the circumferences of two concentric circles, whose radii are 3 and 6 respectively. LINES AND PLANES. 253 PROPOSITION IX. THEOREM. 350. If, from the foot of a perpendicular to a plane, a line be drawn perpendicular to any line of tJie plane, and, from the point of intersection, a line be drawn to any point of the perpendicular, the last line is perpendicular to the line of the plane. S^} D Ssr / --4L Let M N represent a plane, and A B a perpendicular to tJie plane. Let C D represent a line of the plane MN, and let B O be perpendicular to I) C, at O, and let the point of intersection O, be joined to F, a point in A B. To prove that F O is perpendicular to C D. SUG. 1. On the line D C, take O D = O C Connect F with C and D ; also connect B with C and D. SUG. 2. Compare As B O D and O C. SUG. 3. Compare B D with B C. SUG. 4. Compare As FB D and F B C. SUG. 5. Compare F C and F D. SUG. 6. Compare As F C O and F D O. SUG. 7. Compare Zs D O /^and C O F. Therefore 254 SOLID GEOMETRY. 351. A straight line is parallel to a plane when the line and plane cannot meet however far they may both be extended. In that case, the plane is also parallel to the line. 352. Two planes are parallel to each other when they cannot meet however far both may be extended. PROPOSITION X. THEOREM. 353. If one of two parallel straight lines is perpen- dicular to a plane, the other is perpendicular to the same plane. Let A J5 and CD be two parallel lines, meeting the plane MN in the points B and D respectively, and let CD be perpendicular to the plane MN. To prove that A B is perpendicular to the plane M N. SUG. 1. Connect B and Z>, and, through B, draw a line E F t in the plane M N, JL to B D. Connect B with any point O in CD. SUG. 2. What relation does OB sustain to EF, or EF to OBt What relation does EF sustain to B Dt SUG. 3. What relation does E F sustain to the plane ABDC1 (Art. 346.) What relation does E F sustain to A Bt LINES AND PLANES. 255 SUG. 4. What relation does A B sustain to E F* What relation does A B sustain to B Z>? Why ? What rela- tion does A B sustain to the plane M N1 Why ? Therefore PROPOSITION XI. THEOREM. 354. If two lines are perpendicular to the same plane, they are parallel. Lit the two straight lines A B and CD be perpendicular to tJie plane M N. To prove that A B and C D are parallel. SUG. 1. If A B is not || to CD, draw, through any point <9, of A B, a line O E || to C D. SUG. 2. What relation does O E sustain to the plane Mm (Art. 253.) SuG. 3. Compare the answer to Sug. 2 with the hy- pothesis, and finish the demonstration. Therefore Ex. 253. Find the line on which a paper triangle must be folded in order that the vertex may fall upon a given point of the base. 256 SOLID GEOMETRY. PROPOSITION XII. THEOREM. 355. If two straight lines are each parallel to a third straight line they are parallel to each other. D B Let A B and C J> each be parallel to E F. To prove that A B and C D are parallel to each other. SUG. 1. Draw a plane, M N, _L to E P. SUG. 2. What relation does A B sustain to the plane M N ? What relation does CD sustain to the plane MNt SUG. 3. What relation do A B and C D sustain to each other ? Therefore Ex. 254. Equal oblique lines, drawn from a point in the perpendicular to a plane, meet the plane at equal dis- tances from the foot of the perpendicular, and of two un- equal oblique lines the longer meets the plane at the greater distance from the foot of the perpendicular. Ex. 255. What is the locus of the foot of an oblique line of constant length drawn from a point in a perpen- dicular to a plane ? LINES AND PLANES. 257 PROPOSITION XIII. THEOREM. 356. Every plane containing one, and only one, of two parallel lines, is parallel to the other line. C tJ, Let A B and CD be two parallel lines, and MNa plane <-<>ni(ihtin, draw two lines || to A B and CD re- spectively. PROPOSITION XIV. THEOREM. 359. If a line and a plane are parallel, the inter- section of the plane with any plane containing the line is parallel to the line. { - 1, Let the line CD be parallel to the plane MN, and let the plane C B, containing the line C D, intersect M N in the line A B. To prove that A B and C D are parallel. QUERY. Of what proposition is this the converse ? SUG. Give two demonstrations, a direct and an iiidi- direct. Therefore LINES AND PLANES. 259 PROPOSITION XV. THEOREM. 360. Planes perpendicular to the same straight line cure parallel. J* M >j -IT 7 o / Let the planes MN and O P be perpendicular to the To prove that the planes M N and O P are parallel. SUG. 1. Let C and D be the points of intersection of the line A B with the planes M N and O P respectively. SUG. 2. If the planes M N and O P intersect, connect any point of their intersection with the points C and D. SUG. 3. What relation do these lines sustain to A B ? Why? SUG. 4. Is this possible ? Therefore - Ex. 256. If, from a point in a perpendicular to a plane, a line be drawn perpendicular to any line of the plane, a line joining the point of intersection to the foot of the perpendicular to the plane is perpendicular to the line of the plane. SUG. Use the indirect method. QUERY. Of what proposition is this exercise the con- verse ? 260 SOLID GEOMETRY. PROPOSITION XVI. THEOREM. 361. The intersections of two parallel planes with a third plane are parallel lines. Let MN and O P represent two parallel planes, and A B a plane intersecting them in the lines C Z> and E F. To prove that C D and E F are parallel lines, SUG. C D and E F are in same plane A B. Can they meet ? Why ? Therefore - PROPOSITION XVII. THEOREM. 362. If two angles not in the same plane have their sides parallel and lying in the same direction from their vertices, they are equal and their planes are par- allel. AL Let B AC and E D F represent two equal angles in the planes M N and O F respectively, having their sides parallel and lying in the same direction from their vertices. LINES AND PLANES. 261 To prove that the angles BAG and E D F are equal, and that the planes M N and O P are parallel. SUG. 1 . Take D E = A B, and D F = A C. Con- nect A and D, C and F, B and E, C and B, F and E. SUG. 2. What kind of a quadrilateral is A C F Dl Why ? What kind of a quadrilateral is A B E D ? SUG. 3. Compare ^ Z? with C F, also ^ Z> with B E. Give auth. Compare C F with ^ E. (Art. 355, and Ax. 1.) SUG. 4. Compare C/? with F E. Give auth. Sue. 5. Compare As B A C and E D F. Compare Z B A C with Z.E D F. SUG. 6. What relation does the plane (9 /> sustain to the line A B ? Also to the line A C ? Also to the line B C? Why? (Art. 356.) SUG. 7. If the planes M N and OP intersect, their line of intersection is || to A B. Why ? (See Prop. XIV.) Also, their line of intersection is || to A C. Why ? Is it possible for the planes M N and O P to intersect ? Therefore Ex. 257. Find the locus of a point equidistant from two given parallel planes. Ex. 258. What is the locus of a point 'in space equi- distant from two given points ? Ex. 259. What is the locus of a point equidistant from two given parallel planes and at the same time equidis- tant from two given points ? Ex. 260. What is the locus of a point equidistant from two given points, and at the same time equidistant from two other given points ? 262 SOLID GEOMETRY. PROPOSITION XVIII. THEOREM. 363. A straight line perpendicular to one of two parallel planes is perpendicular to the other also. f s^ C- --# / M ' A' 77 ^I^^'^^C^ 0'~ Let MNand O P represent two parallel planes, and let the line ABbe perpendicular to the plane MN. > To prove that the line A B is perpendicular to the plane O P. SuG. 1. Through A, in the plane MN, draw two lines CD and G H ' ; and, through B, in the plane O P, draw the lines E F and K L || to C D and G H respectively. Is this construction possible ? Why ? SuG. 2. What relation does A B sustain to E F? What relation does A B sustain to K LI Give auth. SUG. 3. What relation does A B sustain to the plane OP? Give auth. Therefore Ex. 261. What is the locus of a point equidistant from two given parallel planes, and at the same time equidis- tant from two other parallel planes ? Ex. 262. What is the locus of a point equally distant from three given points which are not in a straight line ? LINES AND PLANES. 263 PROPOSITION XIX. THEOREM. 364. If three parallel planes are intersected by two lines the segments of the lines are proportional. v fc K A., ^r ^A / ~V~ \ ~l s ** iir ~^\~ Let, M N, OP and R S represent three parallel planes, and A B and C D two lines intersectiny them, making senments E I and I L on A Bf and K H and H F on CD. To prove that -=-=- = . / L, ri r SUG. 1. Connect E and F, and let the line E F inter- sect the plane O Pat G. SuG. 2. The plane E F K intersects the planes M N and O P in what lines ? SuG. 3. The plane E F L intersects the planes OP and R S in what lines ? SUG. 4. In the A E F K, the lines G H and E K are how related? (Prop. XVI.) F (^ PC f-f SUG. 5. Compare the ratios and - (JT r 1 Mr SUG. 6. In a similar manner compare the ratios EG , El -= =c and GF IL SUG. 7. C Therefore F T ff f-f SUG. 7. Compare the ratios -^-f- and -Y 1 L, 264 SOLID GEOMETRY. DIHEDRAL ANGLES. 365. Two planes which meet and are terminated by their line of intersection are said to form a dihedral angle. The planes A B and C D, in the figure at the right, meet in the line A C and form a dihedral angle. The planes A B and C D are called the faces of the angle, and the line A C is called the edge of the angle. A dihedral angle is read by reading first, one face; second, the edge; third, the other face. This can be done by reading four letters, if none are re- peated; as B A CD. When but one dihedral angle is formed at an edge it may be read by reading the edge, as dihedral angle A C in the above figure. 366. The plane angle of a dihedral angle is an angle formed by drawing two lines, one in each face, perpendicular to the edge at the same point. In the above figure, if F E and F G are perpendicular to the edge A C, the angle E F G is the plane angle of the dihedral angle. 367. A plane is perpendicular to another plane if it forms with the other plane a right dihedral angle, z. , and let A B be a line in C D perpendicular to B D. To prove that A B is perpendicular to the plane M N. SUG. 1. In the plane M N, draw B E _L to B D, at B. SUG. 2. The Z. A B E is the plane Z. of the dihe- dral Z.. Why? How many degrees in Z. A B E ? Why? SUG. 3. What relation does A B sustain to the plane AfN? Why? (Art. 346.) Therefore 378. COROLLARY I. If two planes are perpendicular to each other, a perpendicular to one of them at any point of their intersection lies in the other. SuG. 1. A B, in the plane C D, is _L to M N by the proposition. SUG. 2. How many _Ls can be erected to M N at B ? 379. COROLLARY II. If two planes are perpendicular to each other, a perpendicular to one of them from any point in the other lies in the other. SUG. See Sug. 1 of Cor. I, and Art. 343. 272 SOLID GEOMETRY. PROPOSITION XXVI. THEOREM. 380. If two planes are perpendicular to a third plane, their intersection is perpendicular to that plane. Let C D and E F be two planes, each perpendicular to the plane M JV, and let A O be the line of intersection of C D and E F. To prove that A O is perpendicular to the plane M N. SUG. 1. From the point O, which is common to all three planes, erect a J_ to the plane M N. SUG. 2. Where will this J_ lie with respect to each of the planes C D and E Ft (Art. 378.) SUG. 3. What relation does the J_ erected sustain to the intersection of the planes C D and E Ft Therefore Ex. 275. If a plane intersects two parallel planes the alternate interior dihedral angles are equal. Kx. 276. If one plane intersects another, the sum of the two adjacent dihedral angles on the same side of either plane is equal to two right dihedral angles. Ex. 277. If a plane intersects two parallel planes the corresponding dihedral angles are equal. LINKS AND PLANES. 273 PROPOSITION XXVII. THEOREM. 381. Through a given straight line, one, and but one, plane can be passed perpendicular to a given plane. Af Let ABbea given straight line, and MNa given plane. To prove that one, and but one, plane can be passed through A B perpendicular to M N. SUG. 1. From some point, as A, in A B, drop a _L A Cto the plane M N. SUG. 2. A B and A C determine a plane B C. Give auth. SUG. 3. What relation does the plane B C sustain to the plane M Nt (Art. 376.) SUG. 4. Can a plane be drawn through A B J_ to M ' Nt SUG. 5. How many planes can be drawn through A B J_ to M Nt (Art. 339.) Therefore Ex. 278. If the sum of two adjacent dihedral angles is equal to two right dihedral angles, their exterior faces are in the same plane. . Ex. 279. If two planes intersect each other, their ver- tical dihedral angles are equal. 18 Geo. 274 SOLID GEOMETRY. PROPOSITION XXVIII. THKOREM. 382, Every point in a plane which bisects a dihe- dral angle is equidistant from the faces of the dihe- dral angle. Mite a plane bisecting the dihedral angle formed by the planes A C and A D , and let P be any point in the plane A M, and P E and P F perpendiculars from P to the faces A C and A D of the dihedral angle. To prove that P E and P F are equal. SuG. 1. Pass a plane through the lines P E and P F and let it intersect the planes A C, A M and A D in the lines O E, O P and O F. SuG. 2. What relation does the plane E F sustain to each of the planes A C and A D ? Give auth. SUG. 3. What relation does the plane E F sustain to A B, the edge of the dihedral Z. ? Give auth. SUG. 4. Compare Zs P O E and P O F. (Art. 371.) SUG. 5. Compare As P O E and POP. Give auth. SUG. 6. Compare the lines P E and P F. Therefore LINES AND PLANES. 275 383. The projection of a point upon a plane is the foot of the perpendicular from the point to the plane. The projection of a line /? upon a plane is the line in the plane which contains the pro- jections of all the points of the line. In the figure at the right, the point C is the pro- jection of the point A, and the 7 /N line C D is the projection of the line A B. PROPOSITION XXIX. THEOREM. 384. The projection of a straight line upon a plane is a straight line. Let ABbea given straight line and MN a given plane. To prove that the projection of A B upon the pla?ie M N is a straight line. Sue. 1. Let A D represent the plane which contains A B and is _L to the plane M N. This plane contains CD, the projection of A B upon the plane M N. Why ? SUG. 2. The two planes A D and M N intersect in what kind of a line ? Therefore 276 SOLID GEOMETRY. PROPOSITION XXX. THEOREM. 385. The angle ivhich a straight line makes with its own projection upon a plane is the least angle the line malces with any line of the plane. Let A JB represent any straight line, and 13 C its projec- tion upon the plane MN. To prove that the angle A B C is the least angle the line A B can make with any line of the plane M N. SUG. 1. Through B, draw any other line of the plane, as B D. From A, drop a _L A C to the plane M N. Ivay off B D = B C and draw A D. SUG. 2. Compare A C and A D in respect to length. Give auth. SUG. 3. Compare /.ABC with Z. A B D. Give auth. Therefore 386. The angle which a line makes with its projec- tion on a plane is called the angle of the line and the plane, of the inclination of the line to the plane. LINES AND PLANES. 277 POLYHEDRAL ANGLES. 387. The figure formed by three or more plaues meet- ing at a common point is called a polyhe- dral angle. In the figure at the right, A-B CD represents a polyhedral angle. The common point A is called the vertex of the angle; A ' B, A C, etc. , are the edges. The portions of planes intercepted by the edges; as B A C, CAD, etc., are the faces of the angle. The plane angles formed by the edges, as B A C, CAD, etc., are the face angles of the polyhe- dral angle. In a polyhedral angle each pair of adjacent faces forms a dihedral angle; as B C A D, etc. A polyhedral angle of three faces is called a trihedral angle, one of four faces is called a tetrahedral angle, etc. Two polyhedral angles are equal when they can be made to coincide. Two polyhedral angles are symmetrical when the parts of one one are equal respectively to the parts of the other but arranged in reverse order. D ^ ^C' A-B CD and A'- B' CD are symmetrical if Z. B A C : Z B' A' C, Z C A D = Z. C A' U and Z B A D . B' A' D'. 278 SOLID GEOMETRY. A polyhedral angle cannot be made to coincide with its symmetrical polyhedral angle. Note. The two hands, or the two sides of the face, of the human body will serve to illustrate symmetrical solids. To convince one that the two hands are symmetrical rather than equal, it might be suggested to try to put the left glove on the right hand. See defini- tion of equal polyhedral angles. PROPOSITION XXXI. THEOREM. 388. The swm of any two face angles of a trihedral angle is greater than the third face angle. B "Let ABC represent a trihedral angle in which each of the face angles DAB and B A C is smaller than the face angle D A C. To prove that the sum of the angles DAB and B A C is greater than the angle D A C. SUG. 1. In the face DA C draw A M, making Z DA M = Z.DAB. SuG. 2. Cut the edges by a plane D B C, so that A M shall = A B. SUG. 3. Compare &$DAB and DAM. Compare D M and B D. Give auth. Compare B CandMC. Give auth. SuG. 4. Compare /_BAC with Z MA C. Give auth. SUG. 5. Compare the sum of Zs DAB and B A C with Z. DA C. Therefore LINES AND PLANES. 279 PROPOSITION XXXII. THEOREM. 389. The sui) i of the face angles of any convex poly- hrdral angle is less than four right angles. Let A-BCDEF represent a convex polyhedral angle. To prove that the sum of the face angles B A C, C A D y etc. , is less than four right angles. SUG. 1. Pass a plane cutting the edges of the polyhe- dral Z in the points B, C, D, E and F. SUG. 2. Connect O, any point within the polygon B C D E F, with each of the vertices. SUG. 3. In the trihedral Z B compare the sum of the face Zs A B C and A B F with the face Z C B F. SUG. 4. In the trihedral Z C, compare the sum of the face Zs A C B and A CD with the face Z. B C D. SUG. 5. Compare the sum of the base Zs of the As whose vertices are at A, with the sum of the base Zs of the As whose vertices are at O. SUG. 6. Compare the sum of the vertical Zs at A with the sum of the vertical Zs at O. SUG. 7. Compare the sum of the Zs at A with four rt. Zs. Therefore 280 SOLID GEOMETRY. PROPOSITION XXXIII. THEOREM. 390. If two trihedral angles have the three face angles of one equal respectively to the three face angles of the other t the corresponding dihedral angles are equal. , Let A-BCD and A'-B'C'D' represent two trihedral angles whose face angles are equal ; viz., B A C = B'A ' ' , CAD=C'A'D and B A D = B' A D'. To prove that the corresponding dihedral angles CB A D and C' B' A' D ', etc., are equal. SUG. 1. Pass planes BCD and B' C D ', making the edges A B, A C, A D, A' B', A' C 1 and A' D 1 all equal. SUG. 2. In A B and A' B' take points M and M' so that A M = A' M 9 , and at M and M' pass planes J_ to A B and A' B' respectively. SUG. 3. The plane M O N must intersect B C and B D, or these lines extended, and the plane M' (7 N 1 must in- tersect B' C' and B' Z7, or these lines extended. Why ? SUG. 4. What relation do the lines MO and M N sus- tain to A B ? Why ? What relation do the lines M 1 a and M N' sustain to A B' ? Why ? LINES AND PLANES. 281 SUG. 5. Compare As B A C and B' A' C r . Compare Z.*A CandA' B' C'. SUG. 6. Compare As B MO and B' Af <7. (Art. 101.) Compare B O with B' a. SUG. 7. In a similar manner compare As B M N and ^ Af' A 77 . Compare ^ JV with B' N' . SUG. 8. Compare As C B D and C' B' D '. Compare Zs C B D and C' B' D' . SUG. 9. Compare As O B N and a B' W . Compare ON and N 1 . SUG. 10. Compare As O M N and <7 M' N'. Com- pare Zs O M N and (7 A/" N f . Therefore 391. SCHOLIUM. When the parts of A-B CD are ar- ranged in the same order as the respectively equal parts of A'- B' C' D' the two dihedral angles are equal, for they may then be superposed (i. e., placed one upon an- other so as to coincide throughout), but when the parts of one are arranged in the reverse order from the respect- ively equal parts of the other, the two trihedral angles are symmetrical. Note. The student should attempt to superpose A' - B' C' D' upon A-B CD. If A' B' be placed upon A B what must be consid- ered in order that A ' C' may fall upon A Cl When the face B' A' C 1 is thus made to coincide with the face B A C the edges A 1 D' and A D may fall upon the same or opposite side of the plane B A C. If the two edges A' D' and A D fall upon the same side of B A C the two trihedral angles are equal, but if they fall upon opposite sides of BAG the two trihedral angles are symmetrical. 392. An isosceles trihedral angle is a trihedral angle two of whose face angles are equal. 282 SOLID GEOMETRY. PROPOSITION XXXIV. THEOREM. 393. Two symmetrical isosceles trihedral angles are equal in all respects. Let A B CD and A' B' C' D' represent two symmet- rical isosceles trihedral angles in which the face angles B A C and CAD are equal also the face angles B' A' C' and C' A' D' are equal. To prove that the trihedral angles A-BCD andA'-B'C'H are equal in all respects. SUG. 1. Pass planes BCD and B' C D' , cutting the edges of the two trihedral Zs, so that A B, A C, A D, A' B', A' C and A 1 D> are all equal. SUG. 2. Compare the face Zs B A C and D' A' C'. SuG. 3. Compare the face Zs C A D and C' A' B' . SUG. 4. Superpose A'-B' C D 1 upon A-BCD, placing A' upon A, D' upon B and C' upon C. Why is this possible ? Where will the face C A' B' fall with respect to the face CA D! Why? SUG. 5. Where will the line A ' B' fall ? Why ? SUG. 6. Where will the face D' A ' B' lie ? Therefore PROPOSITIONS IN CHAPTER VI. PROPOSITION I. A plane may be passed through any straight line, but the line does not determine the plane. PROPOSITION II. A. plane is determined: I. By a straight line and a point without the line. II. By two interseciing straight lines. III. By two parallel straight lines. IV. By three points not in a straight line. PROPOSITION III. The intersection of two planes is a straight line. PROPOSITION IV. From a given point without a plane, one, and only one, perpen- dicular can be dropped to the plane, and the perpendicular is the shortest line from the point to the plane. PROPOSITION V. At a given point in a plane, one perpendicular, and only one, can be erected to the plane. PROPOSITION VI. If, from any point in the perpendicular to a plane, oblique lines be drawn, those which meet the plane at equal distances from the foot of the perpendicular are equal; and, of two unequal oblique lines, that which meets the plane at the greater distance from the foot of the perpendicular is greater. PROPOSITION VII. If a straight line is perpendicular to two lines of a plane at their point of intersection, it is perpendicular to the plane. SOLID GEOMETRY. PROPOSITION VIII. All the perpendiculars to a given line at the same point lie in the same plane, and that plane is perpendicular to the given line. PROPOSITION IX. If, from the foot of a perpendicular to a plane, a line be drawn perpendicular to any line of the plane, and, from the point of inter- section, a line be drawn to any point of the perpendicular, the las* line is perpendicular to the line of the plane. PROPOSITION X. If one of two parallel straight lines is perpendicular to a plane, the other is perpendicular to the same plane. PROPOSITION XI. If two lines are perpendicular to the same plane, they are parallel. PROPOSITION XII. If two straight lines are each parallel to a third straight line, they are parallel to each other. PROPOSITION XIII. Every plane containing one, and only one, of two parallel lines, is parallel to the other line. PROPOSITION XIV. If a line and a plane are parallel, the intersection of the plane with any plane containing the line is parallel to the line. PROPOSITION XV. Planes perpendicular to the same straight line are parallel. PROPOSITION XVI. The intersections of two parallel planes with a third plane are par- allel lines. PROPOSITION XVII. If two angles not in the same plane have their sides parallel and lying in the same direction from their vertices, they are equal and their planes are parallel. LINES AND PLANES. 285 PROPOSITION XVIII. A straight line perpendicular to one of two parallel planes is per- pendicular to the other also. PROPOSITION XIX. If three parallel planes are intersected by two lines, the segments of the lines are proportional. PROPOSITION XX. All plane angles of the same dihedral angle are equal. PROPOSITION XXI. If a plane be passed perpendicular to the edge of a dihedral angle, the lines of intersection wiih the faces form the plane angle of the dihedral angle. PROPOSITION XXII. Two dihedral angles are equal if their plane angles are equal. PROPOSITION XXIII. The ratio of two dihedral angles equals the ratio of their plane angles. PROPOSITION XXIV. If a straight line is perpendicular to a plane, every plane contain- ing that line is perpendicular to the plane. PROPOSITION XXV. If two planes are perpendicular to each other, a line drawn in one of them perpendicular to their intersection is perpendicular to the other. PROPOSITION XXVI. If two planes are perpendicular to a third plane, their intersection is perpendicular to that plane. PROPOSITION XXVII. Through a given straight line, one, and but one, plane can be passed perpendicular to a given plane. 286 SOLID GEOMETRY. PROPOSITION XXVIII. Every point in a plane which bisects a dihedral angle is equidis- tant from the faces of the dihedral angle. PROPOSITION XXIX. The projection of a straight line upon a plane is a straight line. PROPOSITION XXX. The angle which a straight line makes with its own projection upon a plane is the least angle the line makes with any line of the plane. PROPOSITION XXXI. The sum of any two face angles of a trihedral angle is greater than the third face angle. PROPOSITION XXXII. The sum of the face angles of any convex polyhedral angle is less than four right angles. PROPOSITION XXXIII. If two trihedral angles have the three face angles of one equal re- spectively to the three face angles of the other, the corresponding di- hedral angles are equal. PROPOSITION XXXIV. Two symmetrical isosceles trihedral angles are equal in all re- spects. CHAPTER VII. POLYHEDRONS. DEFINITIONS. 394. A polyhedron is a geometric solid bounded by planes. The bounding planes of a polyhedron are its faces, the lines in which the faces intersect are the edges, and the points in which the edges intersect are the vertices of the polyhedron. Any face designated may be considered the base of the polyhedron. 395. A straight line joining any two vertices not in the same face, is called a diagonal of the polyhedron. 396. The intersection of a polyhedron and a plane is called a plane section of the polyhedron. 397. Polyhedrons are classified according to the num- ber of their faces. A polyhedral angle requires at least three planes to meet at its vertex, and to completely inclose space re- quires at least one more plane; hence a polyhedron can- not have less than four faces. 398. A polyhedron of four faces is called a tetrahe- dron; one of five faces, a pentahedron; one of six faces, a hexahedron; one of eight faces, an octahedron; one of ten faces, a decahedron; one of twelve faces, a do- decahedron; one of twenty faces, an icosahedron, etc. 288 SOLID GEOMETRY. ICOSAHEDRON. DODECAHEDRON. OCTAHEDRON*. HEXAHEDRON. TETRAHEDRON. 399. A convex polyhedron is one in which every possible plane section is a convex polygon. PRISMS. 400. A prism is a polyhedron two of whose faces are polygons which are equal in all respects and are in par- allel planes, and whose remaining faces are parallelo- grams. 401. The bases of a prism are the equal faces in the parallel planes. The lateral faces are the remaining faces of the prism. 402. The basal edges of a prism are the intersections of the lateral faces with the bases. The lateral edges are the intersections of the lateral faces. 403. A right section of a prism is a section whose plane is perpendicular to the lateral edges of the prism. POLYHEDRONS. 289 404. The altitude of a prism is the perpendicular dis- tance between its bases. 405. Prisms are classified as triangular, quadrangu- lar, etc., according as their bases are triangles, quadri- laterals, etc. \D *\ RIGHT PRISM. OBLIQUE PRISM. 406. A right prism is one in which the lateral edges are perpendicular to the bases, and an oblique prism is one in which the edges are oblique to the bases. 407. A regular prism is a right prism whose bases are regular polygons. Ex. 280. Parallel lines in- tersecting the same plane make equal angles with it. Let A B and C D represent || lines, and B O and D S their respective projections on the plane M N. SUG. 1. Drop J_s A O and C S to the plane M N. SUG. 2. Compare /LBA O with ZDCS. (Art. 362.) SUG. 3. Compare Z B with Z D. 19 Geo. 290 SOLID GEOMETRY. PROPOSITION I. THEOREM. 408. The lateral edges of a prism are equal and parallel, and make equal angles with the plane of either 'base. Let A F t B H, etc., represent the lateral edges of a prism. To prove that A F, B H, etc. , are equal and parallel, and that they make equal angles with the plane of either base. SuG. 1. What kind of a figure is the lateral face A HI Then, what relation must A F and B H sustain to each other ? SuG. 2. What relation must any two lateral edges sus- tain to each other ? SUG. 3. Compare the Z. made by any two lateral edges with either base. Give auth. (See Ex. 280.) Therefore Ex. 281. If a plane intersects two parallel planes the interior dihedral angles on the same side of the cutting plane are supplements of each other. POLYHEDRONS. 291 PROPOSITION II. THEOREM. 409. Sections of a prism made by parallel planes are polygons which are equal in all respects. Let A C and ac be sections of the prism F K 9 made by parallel planes. To brove that the polygons A C and a c are equal in all respects. SUG. 1. What relation does A E sustain to a ? Why ? (Art. 361.) SUG. 2. What relation does E D sustain to e d, D C to d c, etc.? SUG. 3. Compare Zs A ED and a e d\ also Zs ED C and edc, etc. Give auth. Therefore Hx. 282. Bvery section of a prism made by a plane parallel to a lateral edge is a parallelogram. Ex. 283. If two planes are cut by a third plane so that the alternate interior dihedral angles are equal, and the edges of the dihedral angles thus formed are parallel, the two planes are parallel. 292 SOLID GEOMETRY. PROPOSITION III. THEOREM. 410. The lateral area of a prism equals the product of a lateral edge by the perimeter of a right section of the prism. Let C D be a lateral edge, and MP a right section of the prism A D. To prove that the area of the lateral faces of the prism equals C D 'multiplied by the perimeter of M P. SUG. 1. What relation does P O sustain to C D ? Why ? What relation does any side of the right sec- tion sustain to a lateral edge which meets that side ? Sue. 2. What is the area of the face C E, in terms of POsiud CD? SUG. 3. Express the area of any lateral face. SUG. 4. Add the areas of the several lateral faces, re- membering that the lateral edges are equal. Therefore Ex. 284. If two planes are cut by a third plane so that the corresponding dihedral angles are equal and the edges of the dihedral angles formed are parallel, the two planes are parallel. POLYHEDRONS. 293 Note. The algebraic expression ab+ac + ad, may be reduced, by factoring, to the form a (b-\-c-\-d). Compare with the former ex- pression the indicated sum of the areas of the lateral faces and re- duce it, by the algebraic process of factoring, to the product of a lateral edge by the perimeter of a right section of the prism. Ex. 285. In any trihedral angle the three planes passed through the edges and the bisectors of the respectively opposite face angles intersect in a straight line. Ex. 286. In any trihedral angle the three planes bi- secting the dihedral angles intersect in a straight line. Ex. 287. Determine a * > point in a plane the differ- M S ence of whose distances from p ..' \ two given points on oppo- site sides of the plane is the maximum. x ; ** SUG. Drop a _L to the SJ ^ plane from one of the points, as B, and extend it to M, an equal distance on the other side of the plane. Con- nect A and M, and extend the line to meet the plane at O. Prove that O is the required point. Ex. 288. Determine a point in a plane the sum of /J x / whose distances from two given points on the same side of the plane is the mini- mum. SUG. Drop a _L to the 'M plane from one of the points A, and extend to M, an equal distance beyond the plane. Connect M, the ex- tremity of the JL, with the other point B. Prove that O, the point in which MB intersects the plane, is the required point. 294 SOLID GEOMETRY. PROPOSITION IV. THEOREM. 411. Two prism>s are equal in all respects if the three faces about a trihedral angle of one are respect- ively equal in all respects to the three faces about a trihedral angle of the other, and these faces are sim- ilarly placed in the two figures. Let A H and A' H' be two prisms in which the three faces A Z>, A G and A K are respectively equal in all re- spects to tfie three faces A'D' 9 A'G' and A K' , and are similarly placed. To prove that the prisms A H and A' H 1 are equal in all respects. SUG. 1. Compare the trihedral ^.s A and A '. (Arts. 390 and 391.) SUG. 2. Apply the prism A' H' to the prism A H, so that the face A' D r coincides with the face A D. Why can this be done? SUG. 3. In what plane will the face A' G' fall ? Why ? SUG. 4. In what plane will the face A' K 1 fall ? Why ? SUG. 5. Where will the line A' F' fall ? Why ? SUG. 6. Where will the point F' fall ? Where will the point G' fall ? Why ? Where will the point K' fall ? Why? POLYHEDRONS. 295 SUG. 7. Since F' t G and K' fall upon F, G and ^re- spectively, in what plane will the face F r I' fall ? Why ? (Art. 339.) SUG. 8. Since B' , G and C' fall upon B, G and C re- spectively, in what plane will the face B' ff fall ? Where will the line C' fT fall ? SUG. 9. Similarly, where will the line D' I' fall ? Where will the points f? and /' fall ? SUG. 10. Compare the prism A'ff with the prism A H. Therefore 412. COROLLARY. Two right prisms are equal in all respects if their altitudes are equal, and the bases of one are equal in all respects to the bases of the other. 413. A truncated prism is a portion of a prism in- cluded between a base and a section of the prism by a plane not parallel to the base. PROPOSITION V. THEOREM. 414. Two truncated prisms are equal in all respects if the three faces about a trihedral angle of one are respectively equal in all respects to the three faces about a trihedral angle of the other, and these faces are similarly placed in the two figures. SUG. Use the method of Prop. IV. Ex. 289. If a plane be passed through the edge of a dihedral angle in such a manner that every point in the plane is equidistant from the faces of the dihedral angle, the plane bisects the dihedral angle. SOLID GEOMETRY. PROPOSITION VI. THEOREM. 415. An oblique prism is equal in magnitude to a right prism whose base is a right section of the oblique prism and whose altitude is a lateral edge of the oblique prism. Let B K be an oblique prism and A'B C' D' E' a right section. To prove that the prism B K is equal in magnitude to a right prism having A' B 1 C' D' E' for a base and an alti- tude equal to E K. SUG. 1. Extend the lateral edge E K to K' ', making E K' = E K. Through K' pass a plane K' G' \\ to the plane E! B'. Extend the other lateral edges to meet the plane K' G in the points /', H' , G and F' . SUG. 2. Compare /' D r with K' E' and hence with K E. Compare / D with K E, Hence, compare /' D' with ID. Compare //' with D D' . SUG. 3. Compare K' I' with E'D'; also K' K with E'E. SUG. 4. Compare the trapezoid K' I with the trape- zoid E' D. SUG. 5. Similarly compare the trapezoid K 1 F with the trapezoid E' A. POLYHEDRONS. 297 SUG. 6. Compare Z I' K' F' with Z. D' E' A'. Com- pare the polygon K' G' with the polygon E' B' . SUG. 7. Compare the trihedral Z I' K' K F with the trihedral Z D'E' E A'. SUG. 8. Compare the truncated prism K' G with the truncated prism E' B. SUG. 9. Compare the right prism K' B' with the oblique prism K B in respect to magnitude. Therefore Ex. 290. Prove that the lateral area of a right prism is less than the lateral area of any oblique prism having the same base and an equal altitude. PARAU<>/'. To prove that the triangular prisms, EFH-A ana GFH-C, into which the pnsm is divided are equal in volume. SUG. 1. Pass a right section M N O P through the par- allelopiped. SUG. 2. Compare A M N O with A M P O. SUG. 3. Compare the triangular prism whose base is M N O and whose altitude is H D with the triangular prism whose base is M P O and whose altitude is H D, SUG. 4. How is the triangular prism whose base is M N O and whose altitude is H D related to the triangu- lar prism EP H-A ? SUG. 5. How is the triangular prism whose base is M P O and whose altitude is H D related to the triangular prism G F H-Ct SUG. 6. Compare triangular prism E H F-A with tri- angular prism G FH-C. Therefore 300 SOLID GEOMETRY. PROPOSITION IX. THEOREM. 423. Any parallelepiped is equal in volume to a rectangular parallelopiped having an equal altitude and a base equal in area. Let E Cite a given parallelopiped. To prove that E C is equal in volume to a rectangular parallelopiped having an equal altitude and a base equal in area. SUG. 1. Extend the edges A D, B C, F G and E H. Take E' If = Effand pass the right sections E' B' and If C JL to E' H'. SUG. 2. Compare the parallelepipeds h C and E' C' in volume. (Art. 415.) SUG. 3. Extend the edges A' B', D' C', E 1 F and If G'. Take S R = H' G' and pass the right sections 6" K and R L _L to S R. SUG. 4. Compare the parallelepipeds O M and E' C' in volume. (Art. 415.) SUG. 5 Compare the parallelepipeds E C and O M in volume. POLYHEDRONS. 301 SUG. 6. Since the plane R L was passed J_ to 5 R Y 2. e., _L to K L, what does Z. K L M equal ? Hence, what kind of a figure Js L N1 SUG. 7. Since the plane L R is JL to K L, what does Z A-ZP equal? SUG. 8. The plane E B' was passed _L to E IT. Hence, how is the plane O L related to L M"? Hence, what does Z. ML /'equal? SUG. 9. Since P L is J_ to L K and L M, how is P L related to the plane L Nt Hence what kind of a par- allelepiped is K Rt SUG. 10. Compare the area of K M with the. area of A' C', also the area of A' ' with the area of A C. Hence, compare the area of K M with the area of A C. SUG. 11. Compare the altitudes of these three par- allelepipeds. Therefore Ex. 293. A plane perpendicular to one edge of a trihe- dral angle intersects the faces in lines which form a right triangle. Hx. 294. The four diagonals of a rectangular parallel- epiped are equal to one another. Ex. 295. Any straight line drawn through the middle point of any diagonal of a parallelepiped terminating in two opposite faces is bisected at that point. Ex. 296. If from any point in space perpendiculars are drawn to the lateral faces of a prism, these perpendiculars are all in the same plane. Ex 297. If from any point in space perpendiculars are drawn to the lateral edges of a prism, these perpendicu- lars are all in the same plane. 302 SOLID GEOMETRY. PROPOSITION X. THEOREM. 424. Two rectangular parallelopipeds having equal bases have the same ratio as their altitudes. H Let A & and E H represent two rectangular parallelo pipeds whose bases A C and E G are equal, and whose al titudes are AB and E F respectively . 41 A D AB To prove that CASE I. When the altitudes A B and E F are com- mensurable. SUG. 1. Lay off a common measure upon the altitudes A B and E F, and let this common measure be contained m times in A B and n times in E F. A B SUG. 2. What does the ratio -=- equal? SUG. 3. Pass right sections of the parallelopipeds through the points of division of the altitudes A B and E F, thus dividing the given parallelopipeds into smaller parallelopipeds. Compare these smaller parallelopipeds in respect to volume. (Art. 412.) SUG. 4. Compare the number of small parallelopipeds into which the given parallelopipeds are divided with the number of segments into which the altitudes are divided. A D SUG. 5. What does the ratio - equal ? AD SUG. 6. Compare the ratios - and A B ^ . Jb Jrl L, r SUG. 7. What is the conclusion when the altitudes are commensurable ? POLYHEDRONS. 303 CASE II. When the altitudes are incommensurable. r> sS\ s\ B / F SUG. 1. If A B and E F are incommensurable, lay off any measure of R /''upon A B as many times as possible. There will be a remainder, as KB, less than the meas- ure. Why ? SUG. 2. Through K pass a right section K L ot the parallelepiped A D. A L A K SUG. 3. Compare the ratios -^r-^ and -~-~-. (Case L) , Jri L r SUG. 4. If a line shorter than KB be taken as a unit of measure of E F, the remainder will be some line MB, which is less than KB. Why ? SUG. 5. By making the unit of measure of E F smaller and smaller continually, the remainder MB is made to decrease indefinitely. Why? SUG. 6. What relation always exists between the ra- A N , A M TT7U _ tios -r; and --. Why? A N A M SUG. 7. Are -^-77 and -=,--- variables ? Why? What Jz, rl , r are their respective limits ? A N A M SUG. 8. Compare the limits of -frjj- and _ _ . Give LL ri h, r auth. SUG. 9. What is the conclusion when the altitudes are incommensurable ? Therefore - 304: SOLID GEOMETRY. 425. COROLLARY. If two rectangular parallelepipeds have two edges of one equal respectively to two edges of the other, the ratio of the parallelepipeds equals the ra- tio of the third edges. PROPOSITION XI. THEOREM. 426. The number of units of volume in a rectangu- lar parallelepiped is equal to the product of the num- ber of linear units in the edges meeting at any vertex. Let A represent any rectangular parallelopiped, and let a, b and c represent three edges meeting at a verteoo. Let V be the unit of measure for volume, the edge of U be- ing the linear unit u. Let ^ = m, = n, and r; m representing the number of linear units in a, n in b, and r in c. ^ To prove that -=j- = mxnxr. SUG. 1. Construct the parallelepiped B, two of whose edges meeting at a vertex are b and c and the third edge is u, the edge of the unit U. SUG. 2. Construct another parallelepiped C, whose edges meeting at a vertex are c (the remaining edge of A}, u and u. POLYHEDRONS. 305 Sue. 3. What does the ratio -^ equal? Give auth. n SUG. 4. What does the ratio -= equal? Give auth. SUG. 5. What does jj- equal ? Give auth. Therefore 427. SCHOLIUM I. In the applications of this theorem the three edges must be expressed in terms of the same unit, and the unit of volume must be a cube whose edge is the linear unit. 428. SCHOLIUM II. By comparison of the theorem with the definition of volume (Art. 420), it will be observed that the volume of a rectangular parallelepiped is equal to the product of the measures of three edges meeting at any vertex joined to the name of the unit of measure for volume. 429. SCHOLIUM III. The expression ''the product of the three dimensions," is a common abbreviation for the expression, "the product of the measures of three edges meeting at any vertex joined to the name of the unit of measure for volume." The product of three lines must not be interpreted in any other sense than that just stated. With this inter- pretation, Prop. XI is usually stated as follows: The volume of a rectangular parallelepiped is equal to the product ol its three dimensions. 430. SCHOLIUM IV. When each dimension of the rectangular parallelepiped is di- visible by the linear unit, which is the edge of the unit of volume, the truth of the theorem may be shown by dividing the parallele- piped into cubes, each equal to the unit of measure. This method is usually employed in arithmetic. 20 Geo. 306 SOLID GEOMETRY. 431. SCHOLIUM V. If the three edges of a rectangu- lar parallelepiped meeting at any vertex are equal, the volume is equal to the third power of the linear unit, and hence the third power of a number is called the cube of the number. PROPOSITION XII. THKOREM. 432. The volume of any parallelopiped is equal to the product of the area of its base by its altitude. SUG. See propositions IX and XI. 433. COROLLARY I. If two parallelepipeds have bases which are equal in area, the ratio of their volumes equals the ratio of their altitudes. SUG. 1. Let V, B and A represent respectively the volume, the area of the base and the altitude of one par- allelopiped, and F', B' and A' the volume, area of the base and altitude of the other parallelopiped. Then by the theorem V = By, A and V = B'xA'. SUG. 2. If now B = B', what is the ratio -^ equal to ? 434. COROLLARY II. If two parallelepipeds have equal altitudes, the ratio of their volumes equals the ra- tio of the areas of their bases. SUG. As in Cor. I, V= BxA, and V = B'xA'. If V A = A' what is the ratio -y^ equal to ? Ex. 298. In a given plane find a point which is equally distant from the vertices of a given triangle. POLYHEDRONS. 307 PROPOSITION XIII. THEOREM. 435. The volume of a triangular prism is equal to the product of the area of its base by its altitude. Let A B C-F represent a triangular prism. To prove that the volume of A B C-F is equal to the area of A B C multiplied by the altitude of the prism-. SUG. 1. Draw A D || to B C, and C D || to B A. Also draw E H || to F G, and G H \\ to F E. Connect D with H. SUG. 2. What kind of a figure is A B CD-F1 Why ? SUG. 3. What is the volume of A B C D-F equal to? Why? SUG. 4. Compare the prism A B C-F with the figure ABC D-F in respect to volume. Give auth. (See Prop. VIII.) SUG. 5. Compare the bases ABC and A B CD of the figures A B C-F and ABC D-F respectively . Also com- pare the altitudes of these two figures. SUG. 6. What is the volume of the prism A B C-F in terms of the area of its base and its altitude ? Therefore 308 SOLID GEOMETRY. PROPOSITION XIV. THEOREM. 436. The volume of any prism is equal to the prod- uct of the area of its base by its altitude. A B Let B K represent any prism, and A B CD E its base. To prove that the volume of B K is equal to the area of ABODE multiplied by the altitude of the prism. SUG. 1. Through any lateral edge, as A /% pass diag- onal planes A H, etc. SUG. 2. Into what kind of figures is^the given prism divided by these planes ? SUG. 3. What is the volume of A B C-G equal to? The volume of A CD- HI etc. Give auth. SUG. 4. Express the sum of these volumes in the sim- plest form. SUG. 5. What is the volume of B K in terms of its base and altitude ? Therefore 437. COROLLARY I. If two prisms have bases which are equal in area, their volumes have the same ratio as their altitudes. POLYHEDRONS. 309 438. COROLLARY II. If two prisms have equal alti- tudes their volumes have the same ratio as the areas of their bases. PYRAMIDS. 439. A pyramid is a polyhedron all but one of whose faces meet in the same point. A The point in which all the faces but one meet is called the vertex. The face which does not pass through the vertex is the base. Tiie faces which meet at the vertex are called lateral faces. The edges formed by the intersections of the lateral faces are called the lateral edges. The edges formed by the intersections of the base with the lateral faces are called the basal edges. 440. The altitude of a pyramid is the perpendicular distance from the vertex to the plane of the base. 441. A pyramid is called triangular, quadrangular, pentagonal, etc., according as its base is a triangle, quadrilateral, pentagon, etc. 442. A regular pyramid is one whose base is a regu- lar polygon and whose vertex is in a per- pendicular to the base erected at its mid- dle point. 443. The slant height of a regular pyramid is the perpendicular from the vertex of the pyramid to any basal edge. 310 SOLID GEOMETRY. 444. A truncated pyramid is the portion of a pyra- mid included between the base and a plane cutting all its lateral edges. 445. The frustum of a pyramid is a truncate pyra- mid in which the cutting plane is par- allel to the base. The section of the pyramid made by the cutting plane is the upper base of the frustum. 446. The altitude of the frustum of a pyramid is the perpendicular distance between its bases. PROPOSITION XV. THEOREM. 447. If a pyramid be cut by a vlane parallel to the base: I. The edges and altitude are divided proportion- ally. II. The section is a polygon similar to the base. Let A-G HKL.M represent a pyramid, BCDEF a section made by the plane S T par all el to the base, A O the altitude, and P the point in which A O intersects the plane S T. POLYHEDRONS. 311 T _ ,, ,AB AC AD 4 AP I. To prove that _ == _ = _, etc., - -^ . SUG. 1. Through A, pass a plane N R \\ to the plane GL. A B AC AP SUG. 2. Compare -^, ^^, -^Q> etc. (See Art. 364.) II. To prove that B CD E F is similar to G HKL M. SUG. 1. Compare As A B F and A G Min respect to form. Also As A F E and A ML, etc. BF FE ED SUG. 2. Compare the ratios -^, ^ z ^ ) etc. Give auth. SUG. 3. Compare Z. B F E with Z. G M L. Also /.FED with Z.MLK, etc. SUG. 4. Then, what relation does the polygon /? ^ sustain to the polygon G L ? SUG. 5. Notice the conclusions of parts I and II. Therefore - ADDITIONAL HELPS. In II, Sug. 1, ,#/MS U to GM, FE is || to ML. Why? (Art. 361.) DC* In II, Sug. 2. To compare the ratios - and y compare each with the ratio .. . 448. COROLLARY I. The areas of parallel sections of a pyramid are proportional to the squares of the distances of the cutting planes from the vertex. (See Art. 268.) 449. COROLLARY II. In pyramids whose bases are equal in area and whose altitudes are equal, sections at equal distances from the vertices are equal in area. 312 SOLID GEOMETRY. PROPOSITION XVI. THEOREM. 450. Two triangular pyramids having equal alti- tudes and bases equal in area f are equal in volume. B Let S-ABC and S'-A'B'C' represent two triangular pyramids having equal altitudes and bases ABC and A'B'C' equal in area. To prove that S-ABC and S'-A'B'C are equal in volume. SUG. 1. Divide the altitude A R of the pyramids into equal parts, and through the points of division, M, N and (9, pass planes || to the bases, thus making sections of the pyramids. Upon each section of S-A B C as up- per base construct a prism whose altitude is equal to the _L distance between the sections and whose lateral edges are || to the edge S A of the pyramid S-ABC. Simi- larly, upon each section of S'-A'B'C' as upper base, construct a prism whose altitude is equal to the _L dis- tance between the sections and whose lateral edges are || to the edge S' A ' of the pyramid S'-A'B'C. POLYHEDRONS. 313 Sue. 2. Compare the areas of the sections DBF and D'E'F'. Giveauth. SUG. 3. Compare the prisms D E F-A and D' E' F'-A ' ; also the two prisms adjacent to them; also the next two, etc. Give auth. SUG. 4, Compare the sum of the prisms in S-A B C with the sum of the prisms in S'-A'B'C'. SUG. 5. Is the answer to Sug. 4 true, whatever the number of parts into which the altitude is divided ? SUG. 6. If the number of parts into which the altitude is divided be increased continually the sum of the prisms in A- B C D will be a variable; also the sum of the prisms in A'-B' C D' will be a variable. Why? SUG. 7. How are these two variables related ? SUG. 8. What are their respective limits ? Therefore Ex. 299. The square of a diagonal of a rectangular parallelepiped equals the sum of the squares of the three dimensions. Ex. 300. Find the length of a diagonal of a rectangu- lar parallelepiped 1 ft. long, 4 in. wide and 3 ft. high. Ex. 301. Find the lateral area of a regular triangular prism whose basal edges are each 4 ft. and whose lateral edges are each 2 yds. Ex. 302. Find the total area of a regular triangular prism whose basal edges are each 4 ft., and whose lateral edges are each 8 ft. Ex. 303. Find the volume of a regular hexagonal prism whose basal edges are each 2 ft. , and whose lateral edges are each 2 yds. 314: SOLID GEOMETRY. PROPOSITION XVII. THEOREM. 451. The volume of a triangular pyramid is one third the volume of a triangular prism having the same base and altitude. Let A-B CD represent a triangular pyramid. To prove that the volume of A-B CD is one third the volume of a prism having BCD for a base and an altitude equal to the altitude of the pyramid. SUG. 1. Through B, draw BE || and equal to C A, and through D draw D F || and equal to C A. Con- nect E with A, E with F, and A with F, thus forming the A E A F. SUG. 2. What is the figure EA F-B CD thus formed ? Why? (Arts. 362 and 400.) SuG. 3. If, from the figure E A F-B CD, the pyramid A-B CD be removed there remains the figure A-B D FE. What is this figure which remains ? What kind of a quadrilateral is B D F E ? Why ? SUG. 4. Compare the two pyramids A-E B D and A-EFD. (Prop. XVI.) SUG. 5. Read the pyramid A-EFD as D-E A F, i. e., consider E A F as the base and D as the vertex, and then compare D-E A .Fwith A-B C D. (Prop. XVI.) POLYHEDRONS. 315 SUG. 6. What part of the whole figure is A-B CD ? Therefore 452. COROLLARY. The volume of a triangular pyra- mid is equal to one third of the product of the area of its base by its altitude. PROPOSITION XVIII. THEOREM. 453. The volume of any pyramid is equal to one third of the product of the area of its base by its alti- tude. O Let O-A BCDE represent a pyramid. To prove that the volume of O-A B CD E is equal to one third the area of the base ABODE multiplied by its alti- tude. SUG. 1. Through an edge, as O A, pass all possible diagonal planes. SUG. 2. What is the figure O-ABC1 the figure O-A CD ? etc. SUG. 3. What is the volume of O-A B C ? of O-A CD ? etc. Give auth. SUG. 4. Express, in its simplest form, the sum of the volumes of the figures O-A B C, O-A CD, etc. SUG. 5. What is the volume of the pyramid O-ABCDE ? Therefore 316 SOLID GEOMETRY. 454. COROLLARY I. If two pyramids have bases which are equal in area, their volumes have the same ratio as their altitudes. 455. COROLLARY II. If two pyramids have equal altitudes their volumes have the same ratio as the areas of their bases. PROPOSITION XIX. THEOREM. 456. The volume of the frustum of a triangular pyramid is equal to the sum of the volumes of three triangular pyramids whose common altitude is the altitude of the frustum and whose bases are respect- ively the upper base of the frustum, the lower base of the frustum, and a mean proportional between the bases of the frustum. B Let A B C-D E F represent the frustum of a triangular pyramid. To prove that the volume of A B C-D E F is equal to the sum of the volumes of three pyramids, each having the same altitude as the frustum, a?id whose bases are respectively ABC, DEF, and a mean proportional between ABC and D E F. POLYHEDRONS. 317 SUG. 1. Through the edge B C and the vertex D y pass a plane BCD. SUG. 2. If A B C be considered as the base of the pyra- mid D-A B C, how does the altitude of this pyramid com- pare with the altitude of the frustum ? Compare the pyr- amid D-ABC with one of the pyramids mentioned in the theorem. SUG. 3. Pass the plane C D E> cutting off the pyramid C-DE F. Compare this pyramid with one of the pyra- mids mentioned in the theorem. SUG. 4. The pyramid D-C B E still remains. Draw C M || to B E, and connect D with M. SUG. 5. Compare the pyramids D-C B E and D-C ME. (Art. 450.) SUG. 6. Consider C the vertex and D M E the base. How does the altitude of C-D M E compare with the altitude of the frustum ? SUG. 7. Draw MN\\ to C A. n &DME DE SUG. 9. -- ^FVF-ET = -TFF- Why? A N M E NE D F SUG. 10. Compare the ratios , . and ._ . Give ME NE auth. SUG. 11. Compare the ratios and A NME SUG. 12. What relation does D M E sustain to AsDFjE.ind N M El SUG. 13. Compare A NME with & A C B. SUG. 14. Then, what relation does D M E sustain to the bases D F E and A C B~t Compare the pyramid C-D M E with one of the pyramids mentioned in the theorem. SUG. 15. Review Sugs. 2, 3 and 14. Therefore - 318 SOLID GEOMETRY. PROPOSITION XX. THEOREM. 457. The volume of the frustum of any pyramid is equal to the sum of the volumes of three pyramids whose common altitude is the altitude of the frustum and whose bases are respectively the upper base of the frustum t the lower base of the frustum, and a mean proportional between the bases of the frustum. o Let L B represent tlie frustum of any pyramid. To prove that the volume of L B is equal to the sum of the volumes of three pyramids each having the same alti- tude as L, B, and whose bases are respectively the upper base L G of the frustum, the lower base E B of the frus- tum, and a mean proportional between the bases L, G and EB. SUG. 1. Let the lateral edges of the frustum L B be extended until they meet at the vertex O, thus forming the pyramid O-A B CDE. Let V-M N P be a triangu- lar pyramid whose altitude is equal to the altitude of O-A B CDE, and whose base MNP is equal in area to the base A B C D E. Let R S T be a section of the POLYHEDRONS. 319 triangular pyramid || to the base MNP, and the same distance from the vertex V that the plane L G is from the vertex O. SUG. 2. Compare the sections L G and R S T in re- spect to area. Give auth. (Art. 449.) Sue. 3. Compare the pyramids O-A B C D E and V-MNPm respect to volume. SUG. 4. Compare the pyramids O-FGHKL and V-S T R in respect to volume. SUG. 5. Compare the frustum L B with the frustum RS T-M N P in respect to volume. Compare the alti- tudes of these two frustums. SUG. 6. What is the volume of the frustum RST-MNP equal to? SUG. 7. What then is the volume of the frustum L B eqaul to ? Therefore - 458. COROLLARY. If the volume of the frustum of a pyramid be represented by F, the altitude by H, the area of the lower base by B, and the area of the upper base by b, the truth of the theorem may be expressed by the formula Similarly the volume of a prism is given by the formula V=HB, and the volume of a pyramid by the formula Ex. 304. Find the volume of a regular triangular prism each edge of which is 4 ft. Ex. 305. Find the lateral area of a regular pentagonal prism each edge of which is 3 in. 320 SOLID GEOMETRY. PROPOSITION XXI. THEOREM. 459. The lateral area of a regular pyramid is equal to one half the product of the perimeter of the base by the slant height. Let O-A BCDE represent a regular pyramid, and O M its slant height. To prove that the lateral area of the pyramid equals one half the perimeter of the base multiplied by O M. SUG. 1. What is the area of A O A E ? of A O E D ? of A O D C? etc. SUG. 2. What is the lateral area of the pyramid equal to ? Therefore Ex. 306. Theorem. The volume of a truncated trian- gular prism is equal to the sum of the volumes of three pyramids whose common base is the base of the prism and whose vertices are the vertices of the inclined section. I^et ABC-DEF represent a trun- J cated triangular prism. SUG. 1. Draw the lines A D, A F y B D, BE, C R and CF. POLYHEDRONS. 321 SUG. 2. Consider the truncated prism separated into the three figures B-D E F, B-A ED and B-A CD. SUG. 3. Compare -DFwiih one of the pyramids mentioned in the theorem. SUG. 4. Consider B-A E D as D-AEB. Show that D-A EB = D-A EF. Consider D-A EF as A-DEF and compare with one ot the pyramids mentioned in the theorem. SUG. 5. Compare B-A C D with B-ECD. Consider B-E CD as E-B CD. Compare E-B CD with E-FCD. Consider E-FCD as C-D E F, and compare with one of the pyramids mentioned in the theorem. PROPOSITION XXII. THEOREM. 460. The lateral area of the frustum of a regular pyramid equals the product of the slant height by one half the sum of the perimeters of the two bases. C. u Let G E represent the frustum of a regular pyramid and NM its slant height. To prove that the lateral area of the frustum G E equals M N multiplied by one half the sum of the perimeters of the two bases. SUG. 1. What is the area of the trapezoid G Ct of the trapezoid H D ? of K E ? etc. SUG. 2. What is the lateral area of the frustum? Therefore 21 Geo. 322 SOLID GEOMETRY. 461. A regular polyhedron is a polyhedron whose faces are all equal regular polygons and whose polyhe- dral angles are all equal. PROPOSITION XXIII. THEOREM . 462. Only five regular convex polyhedrons are possible. SUG. 1. The regular polygon which has the smallest number of sides is the equilateral A. SUG. 2. How many degrees are there in an Z. of an equilateral A ? SUG. 3. Is it possible for three equilateral As to meet so as to form a polyhedral Z. ? (Art. 389.) SUG. 4. What is the least number of equilateral As that can be used to enclose space ? There is a regular polyhedron having four equilateral As for faces. It is called a regular tetrahedron. SUG. 5. Is it possible for four equilateral As to meet at a vertex so as to form a polyhedral Z ? Why ? SUG. 6. How many equilateral As are required to completely enclose space if four As meet at each vertex ? There is a regular polyhedron formed by eight equal equilateral As. It is called a regular octahedron. SUG. 7. Is it possible for five equilateral As to meet at a vertex so as to form a polyhedral /L ? Why? SUG. 8. How many equilateral As are required to com- pletely enclose space if five As meet at each vertex ? POLYHEDRONS. 323 Perhaps the question in Sug. 8 cannot be answered without the aid of a figure, but by means of a figure con- structed of cardboard, as suggested below, it is easily seen that there is a regular polyhedron formed by twenty equal equilateral As. It is called a regular icosahedron. SUG. 9. Is it possible for six equi- lateral As to meet at a vertex so as to form a polyhedral Z. ? Why ? SUG. 10. Are any other regular con- vex polyhedrons possible whose faces are equilateral As ? SUG. 11. How many degrees in each Z. of a square? SUG. 12. Is it possible for three squares to meet at a vertex so as to form a polyhedral 4L ? Why ? SUG. 13. How many squares are required to com- pletely enclose space if three squares meet at each vertex ? There is a regular polyhedron formed by six equal squares. It is called a regular hexahedron or cube. SUG. 14. Is it possible for more than three squares to meet at a vertex so as to form a polyhedral Z ? Why ? Are any other regular convex polyhedrons possible whose faces are squares ? SUG. 15. How many degrees in each Z of a regular pentagon ? SUG. 16. Is it possible for three regular pentagons to meet at a vertex so as to form a polyhedral Z ? V/hy ? SUG. 17. With three regular pentagons meeting at each vertex, how many will be required to completely enclose space ? SOLID GEOMETRY. The question in Sug. 17 may be too difficult without the aid of a figure, but by means of a fig- ure constructed of cardboard as suggested below, it is easily seen that there is a regular convex polyhedron formed by twelve regular pentagons. It is called a dodecahedron. SUG. 18. Is it possible for more than three regular pentagons to meet at a vertex so as to form a polyhedral ^ ? Why ? Are any other regular convex polyhedrons possible whose faces are regular pentagons ? SUG. 19. How many degrees in each Z. of a regular hexagon ? Is it possible for three or more regular hexa- gons to meet at a vertex so as to form a polyhedral ^ ? Why ? Are any regular convex polyhedrons possible whose faces are regular hexagons ? SUG. 20. Are any regular convex polyhedrons possible whose faces are regular polygons of more than six sides ? SUG. 21. Five regular convex polyhedrons have been enumerated; are any others possible ? Therefore 463. SCHOLIUM. The five regular polyhedrons may be constructed of cardboard, as follows: Cut the material in the shape of the following patterns, then cut it half through along the lines separating the polygons. Fold it over and join the edges. Paste a strip of paper neatly over the joined edges. TETRAHEDRON. HEXAHEDRON. OCTAHEDRON. POLYHEDRONS. 325 AAAAA vvvvv DODECAHEDRON. ICOSAHEDRON. Ex. 307. To construct a regular tetrahedron. SUG. Construct an equilateral A for the base. At the center erect a _L . With a radius equal to a side of the A and with a center at one vertex of the A, cut off the perpendicular. Ex. 308. To construct a regular hexahedron. Ex. 309. To construct a regular octahedron. SUG. Construct a square, and at its center erect a _L . With a vertex of the square as a center, and a radius equal to one side of the square, cut off the _L on each side of the square. Ex. 310. To construct a regular icosahedron. SUG. Construct a regular pentagon, and at its center erect a _L . With a vertex of the pentagon as a center, and a radius equal to one side of the pentagon, cut off the _L. Join the point on the _L to each vertex of the pentagon thus forming a pentagonal pyramid. The poly- hedral ^. at the vertex is one of the polyhedral ^s of the icosahedron required. In a similar manner, construct polyhedral ^s at the vertices of the pentagon first drawn. PROPOSITIONS IN CHAPTER VII. PROPOSITION I. The lateral edges of a prism are equal and parallel, and make equal angles with the plane of either base. PROPOSITION II. Sections of a prism made by parallel planes are polygons which are equal in all respects. PROPOSITION III. The lateral area of a prism equals the product of a lateral edge by the perimeter of a right section of the prism. PROPOSITION IV. Two prisms are equal in all respects if the three faces about a tri- hedral angle of one are respectively equal in all respects to the three faces about a trihedral angle of the other, and these faces are simi- larly placed in the two figures. PROPOSITION V. Two truncated prisms are equal in all respects if the three faces about a trihedral angle of one are respectively equal in all respects to the three faces about a trihedral angle of the other, and these faces are similarly placed in the two figures. PROPOSITION VI. An oblique prism is equal in magnitude to a right prism whose base is a right section of the oblique prism and whose altitude is a lateral edge of the oblique prism. PROPOSITION VII. Opposite faces of a parallelepiped are parallelograms which are equal in all respects. POLYHEDRONS. 327 PROPOSITION VIII. A plane passed through the diagonally opposite edges of a paral- lelopiped divides it into two triangular prisms which are equal in magnitude. PROPOSITION IX. Any parallelepiped is equal in volume to a rectangular parallele- piped having an equal altitude and a base equal in area. PROPOSITION X. Two rectangular parallelepipeds having equal bases have the same ratio as their altitudes. PROPOSITION XI. The number of units of volume in a rectangular parallelepiped is equal to the product of the number of linear units in the edges meet- ing at any vertex. PROPOSITION XII. The volume of any parallelepiped is equal to the product of the area of its base by its altitude. PROPOSITION XIII. The volume of a triangular prism is equal to the product of the area of its base by its altitude. PROPOSITION XIV. The volume of any prism is equal to the product of the area of its base by its altitude. PROPOSITION XV. If a pyramid be cut by a plane parallel to the base: I. The edges and altitude are divided proportionally. II. The seetion is a polygon similar to the base PROPOSITION XVI. Two triangular pyramids having equal altitudes and bases equal in area, are equal in volume. 328 SOLID GEOMETRY. PROPOSITION XVII. The volume of a triangular pyramid is one third the volume of a triangular prism having the same base and altitude. PROPOSITION XVIII. The volume of any pyramid is equal to one third of the product of the area of its base by its altitude. PROPOSITION XIX. The volume of the frustum of a triangular pyramid is equal to the sum of the volumes of three triangular pyramids whose common alti- tude is the altitude of the frustum and whose bases are respectively the upper base of the frustum, the lower base of the frustum, and a mean proportional between the bases of the frustum. PROPOSITION XX. The volume of the frustum of any pyramid is equal to the sum of the volumes of three pyramids whose common altitude is the altitude of the frustum and whose bases are respectively the upper base of the frustum, the lower base of the frustum, and a mean proportional between the bases of the frustum. PROPOSITION XXI. The lateral area of a regular pyramid is equal to one half the prod- uct of the perimeter of the base by the slant height. PROPOSITION XXII. The lateral area of the frustum of a regular pyramid equals the product of the slant height by one half the sum of the perimeters of the two bases. PROPOSITION XXIII. Only five regular convex polyhedrons are possible. CHAPTER VIII. THE THREE ROUND BODIES. THE CYLINDER. 464. A cylindrical surface is a surface formed by a moving straight line which al- ways remains parallel to itself and continually touches a given curved line. The moving straight line is called the generatrix, and the given curved line is called the directrix. 465. A straight line in a cylindrical surface which occupies one of the positions of the generatrix is called an element of the surface. 466. A cylinder is a figure bounded by a cylindrical surface and two parallel planes. The plane surfaces are called the bases of the cylinder, and the cylindrical sur- face is called the lateral surface of the cylinder. The perpendicular dis- tance between the bases is the altitude of the cylinder. 467. A right section of a cylinder is the intersection of the cylinder and a plane perpendicular to an element of the surface. 330 SOLID GEOMETRY. 468. A circular cylinder is a cylinder whose bases are circles. 469. A right cylinder is a cylinder whose elements are perpendicular to its bases. 470. A cylinder of revolution is a right circular cylinder. It may be generated by the revolution of a rectangle about one of its sides. 471. Similar cylinders of revolution are cylinders generated by the revolution of similar rectangles about homologous sides. 472. A tangent line to a cylinder is a line which touches the cylindrical surface in a point but does not intersect the surface at that point. 473. A tangent plane to a cylinder is a plane which touches the cylindrical surface along one of the elements but does not intersect the surface along that element. PROPOSITION I. THEOREM. 474. Every section of a cylinder made by a plane containing an element is a parallelogram. Let the plane A D contain the element A B, THE CYLINDER. 331 To prove that the section A B C D is a parallelogram. SUG. 1. D is one point common to the plane and the surface of the cylinder. If a line be drawn through D || to B A will it lie in the plane A D ? Why ? Will it lie in the surface of the cylinder ? Why ? Will it coin- cide with the intersection D C? How, then, is D C re- lated to B A ? . SUG. 2. How is A C related to B D ? Give auth. SUG. 3. What kind of quadrilateral is A B D C1 Therefore 475. COROU,ARY. Every section of a right cylinder containing an element is a rectangle. Ex. 311. Find the lateral area of a regular pentagonal pyramid whose basal edge is 2 ft. and whose slant height is 1 yd. Ex. 312. Find the total area of a regular quadrangular pyramid whose basal edge is 8 ft. and whose lateral edge is 5 ft. Ex. 313. Find the volume of a regular quadrangular pyramid whose basal edge is 8 ft. and whose slant height is 5 ft. Ex. 314. Find the volume of a regular triangular pyr- amid whose basal edge is 4 ft. and whose altitude is 1/3 ft. Ex. 315. A pyramid whose altitude is ^16" ft. is cut into two parts of equal volume by a plane parallel to the base. Find the distance of the cutting plane from the vertex. 332 SOLID GEOMETRY. PROPOSITION II. THEOREM. 476. The bases of a cylinder are equal. Let AEbea cylinder ivhose bases are ABC and I>EF. To prove that the bases ABC and D E F are equal. SUG. 1. I/et B and Cbe any two points in the perime- ter of the upper base, and let B F be a plane containing the line B C and the element B E. Compare B C and E F. SUG. 2. Let A represent any other point in the perim- eter of the upper base, and A D the element through A. Pass a plane through the || lines A D and B E ; also a plane through the || lines A D and C F. Compare B A and E D ; also C A and F D. SUG. 3. Compare As A B C and D E F. Give auth. SUG. 4. Consider the upper base placed upon the lower base with B C upon E F. Where will the point A fall ? Since A is any point in the perimeter of the upper base, where will trie perimeter of the upper base lie with re- spect to the perimeter of the lower base ? Therefore 477. COROU,ARY I. Any two parallel sections cut- ting all the elements of a cylinder are equal. THE CYLINDER. 333 478. COROLLARY II. All sections of a circular cyl- inder- parallel to the bases are equal to the bases, and the straight line which connects the centers of the bases passes through the centers of the sections. SUG. Draw two diameters of one of the bases, and through these diameters and elements of the cylinder pass planes. These planes will intersect the sections in diam- eters. Why ? 479. A cylinder is inscribed in a prism when each lateral face of the prism is tan- gent to the cylinder and the bases of the prism circumscribe the bases of the cylinder. When a cylinder is inscribed in a prism the prism is said to be circumscribed about the cyl- inder. 480. A cylinder is circumscribed about a prism when each edge of the prism is an element of the cylinder, and the bases of the prism are inscribed in the bases of the cylinder. When a cylinder is circumscribed about a prism the prism is said to be inscribed in the cylinder. Ex. 316. The base of a pyramid is a square whose side is 6 ft. , and the lateral area of the pyramid is f of its total area; find the altitude and the slant height of the pyramid. 334: SOLID GEOMETRY. PROPOSITION III. THEOREM. 481. The area of the lateral surface of a cylinder is equal to the perimeter of a right section multiplied by an element of the surface. A Let AGbea cylinder, KM a right section, and AEan element. To prove that the area of the lateral surface equals the perimeter of the section KM multiplied by A B. SUG. 1. Inscribe a prism within the cylinder. SUG. 2. What is the lateral area of the prism equal to ? (Art. 410.) SUG. 3. Let the number of sides of the bases of the prism be increased by bisecting the arcs subtended by the sides and joining the points of bisection to the adjacent vertices. SUG. 4. As the number of faces of the prism increases the lateral area varies. Why ? SUG. 5. What is the limit of the lateral area of the prism ? SUG. 6. The perimeter of the polygon inscribed in the right section approaches what limit ? THE CYLINDER. 335 SUG. 7. As the lateral area of the prism increases what does it remain equal to ? (Art. 410.) SUG. 8. What, then, is the area of the lateral surface of the cylinder equal to ? Therefore 482. COROLLARY I. The lateral area of a cylinder of revolution equals the perimeter of the base multiplied by the altitude. This fact may be expressed by the formula wherein A represents the area of the lateral surface, H the altitude, and R the radius of the base of the cylinder of revolution. 483. COROLLARY II. The lateral areas of similar cyl- inders of revolution are to each other as the squares of their altitudes or as the squares of the radii of their bases. A, H and R repre~ sent respectively the area, altitude, and radius of the base of one cylinder and a, h and r the area, altitude, and radius of the base of the other cylinder. Then A= and a= A C) ^ J~) T_T JO T y S & 71- J\. JT2 J\, ft a ~~ %nrh ' r ~h~ Hence = But 7- X why ? Hence = -^, or ^-. a r 2 h* 336 SOLID GEOMETRY. MODEL. PROPOSITION III. THEOREM. 484. The area of the lateral surface of a cylinder is equal to the perimeter of a right section multiplied by an element of the surface. B Let A G^e a cylinder, KM a right section, and AEan element. To prove that the area of the lateral surface equals the perimeter of the section K M multiplied by A E. Inscribe a prism within the cylinder. The lateral area of this prism is equal to the perimeter of the right section K L M N multiplied by an edge of the prism, as A E. (Art. 410.) Now, let the number of sides of the bases of the prism be increased by bisecting the arcs subtended by the sides and joining the points of bisection to the adjacent ver- tices. As the number of sides of the bases of the prism increases the number of faces of the prism increases. As the number of faces of the prism increases the lat- eral area of the prism also increases; for the perimeter of a right section increases, and hence this perimeter multi- plied by an edge of the prism increases. THE CYLINDER. 837 The lateral area of the prism approaches the lateral area of the cylinder as a limit. The perimeter of a right section of the prism ap- proaches the perimeter of a right section of the cylinder as a limit. Hence, the lateral area of a prism is one variable, and the perimeter of a right section of the prism multiplied by an edge is another variable, and these two variables are always equal. (Art. 410.) Hence the limits of these variables are also equal. (Art. 170.) Therefore the area of the lateral surface of a cylinder is equal to the perimeter of a right section multiplied by an element of the surface. Ex. 317. If from any point within an equilateral tri- angle perpendiculars be drawn to the three sides, the sum of those perpendiculars is equal to the altitude of the tri- angle. SUG. Divide the given A into three As whose common vertex is at the point from which the _Ls are drawn. Ex. 318. If from any point within a regular tetrahe- dron perpendiculars be drawn to the four faces, the sum of those perpendiculars is equal to the altitude of the tetrahedro i. SUG. Divide the tetrahedron into four triangular pyra- mids whose common vertex is at the point from which the _Ls are drawn. Ex. 319. The lateral areas of similar pyramids have the same ratio as the squares of their altitudes, or as the squares of any two homologous edges. 22 Geo. 338 SOLID GEOMETRY. PROPOSITION IV. THEOREM. 485, The volume of a cylinder is equal to the area of its base multiplied by its altitude. Let A G be a cylinder whose base is EG. To prove that the volume of the cylinder A G equals the area of the base E G multiplied by the altitude. SUG. 1. Inscribe a prism within the cylinder and then let the number of the faces of the prism increase indef- initely. What is the volume of the prism always equal to? (Art. 436.) SUG. 2. What two variables are always equal ? SUG. 3. What is the limit of the volume of the prism ? SUG. 4. What is the limit of the area of the base of the prism multiplied by its altitude? Therefore - 486. COROLLARY I. The volume of a cylinder of rev- volution may be expressed by the formula wherein V represents the volume, //'the altitude, and R the radius of the base of the cylinder of revolution. CONES. 339 486. COROLLARY II. The volumes of similar cylin- ders of revolution are to each other as the cubes of their altitudes or the cubes of the radii of their bases. Let F, H and R represent respectively the volume, altitude and radius of the base of one cylinder, and v, h and r the volume, altitude and radius of the base of the other cyl- inder. Then and Hence But Hence - = , or -T^-. h* CONES. 487. A conical surface is a curved surface formed by a moving straight line which passes through a given fixed point and contin- ually touches a given curve. The mov- ing straight line is called the generatrix, and the given curve is called the di- rectrix. 488. A straight line in a conical sur- face which occupies one of the positions of the generatrix is called an element of the surface. 489. A cone is a figure bounded by a conical surface and a plane. The plane is called the base of the cone, the conical surface is called the lateral surface or the convex surface of the cone, and the point in which the ele- ments meet is called the vertex of the 34:0 SOLID GEOMETRY. cone. The perpendicular distance from the vertex to the base is the altitude of the cone. 490. A circular cone is a cone whose base is a circle. A straight line joining the vertex with the center of the base is called the axis of the cone. 491. A right circular cone is a cone whose axis is perpendicular to its base. A right cir- cular cone is a cone of revolution. It may be generated by revolving a right triangle about one of its legs. The hypotenuse in any one of its positions is an element of the cone and is called the slant height of the cone. 492. Similar cones of revolution are cones that may be generated by the revolution of similar right triangles about homologous sides. 493. A tangent line to a cone is a line which touches the conical surface in a point but does not intersect the surface at that point. 494. A tangent plane to a cone is a plane which touches the conical surface along one of the elements but does not intersect the surface along that element. 495. A frustum of a cone is that part of a cone included be- tween the base and a section paral- lel to the base. The base of the cone is the lower base of the frus. turn, the section parallel to the base of the cone is the upper base of CONES. 341 the frustum. The perpendicular distance between the bases is the altitude of the frustum. The slant height of the frustum of a cone of revolution is that part of the slant height of the cone which is included between the bases of the frustum. PROPOSITION V. THEOREM. 496. Every section of a cone made by a plane pass- ing through the vertex is a triangle. Let AS C be a section of a cone by a plane passing through the vertex A. To prove that A B C is a triangle. SuG. 1. What kind of a line is A C? Why? (Art. 488.) What kind of a line is A B ? B C? Why ? SUG. 2. What is the figure A B C? Therefore Hx. 320. The volumes of two similar pyramids have the same ratio as the cubes of their altitudes, or as the cubes of any two homologous edges. 342 SOLID GEOMETRY. PROPOSITION VI. THKORKM. 497, Every section of a circular cone made by a plane parallel to the base is a circle. Let BOCbe the base of a circular cone, and ESFa sec- tion parallel to the base. To prove that the section E S Fis a circle. SUG. 1. ]>t A O be the axis of the cone, and 5 the point in which the axis intersects the section E S F. I^et E and F be any two points whatever in the perimeter of the section E S F. SUG. 2. Through the axis A O and the point E pass a plane intersecting the base in the line O B and the sec- tion in the line S E. Also through the axis A O and the point F pass a plane intersecting the base in the line O C, ancl the section in the line 6* F. SUG. 3. 5 E is || to O B, and 5 F is || to O C. Give auth. (Art. 361.) AS S E SUG. 4. Compare r-^ with -==. Give auth. Also Si U U > AS ^SF compare _ with ^ CONES. 343 9* f? *? /** SUG. 5. Compare - with -. Give auth. SUG. 6. The base of the cone is a circle whose center is at O. Why? (Art. 490.) SUG. 7. Since O B and O C are equal, S E and 5 .F are how related ? SUG. 8. Since E and ^ are any two points in the per- imeter of the section, what kind of a figure is the sec- tion E S F? Therefore -- 498. COROLLARY. *The axis of a circular cone passes through the centers of all sections parallel to the base. 499. A cone is inscribed in a pyr- amid when the vertex of the cone co- incides with the vertex of the pyramid and the base of the cone is inscribed in the base of the pyramid. When a cone is inscribed in a pyra- mid the pyramid is said to be circumscribed about the cone. 500. A cone is circumscribed about a pyramid when the vertex of the cone coincides with the vertex of the pyramid and the base of the cone is circumscribed about the base of the pyramid. When a cone is circumscribed about a pyramid the pyramid is *aid to be inscribed in the cone. Ex. 321. If a cone is inscribed in a pyramid prove that each lateral face of the pyramid is tangent to the cone. 344 SOLID GEOMETRY. PROPOSITION VII. THEOREM. 501. The area of the convex surface of a cone of revolution is equal to one half the product of the per- imeter of its base by its slant height. Let A-B CDEbea cone of revolution. To prove that the area of the convex surface of the cone is equal to one half the product of the perimeter of its base by its slant height. SUG. 1. Inscribe a regular pyramid within the cone. SUG. 2. What is the lateral area of the pyramid equal to? Complete the demonstration by a method similar to that of Prop. III. Therefore - 502. COROLLARY I. The truth of the theorem may be expressed by the formula in which A represents the area of the convex surface, R the radius of the base, and 6" the slant height of the cone of revolution. CONES. 345 503. COROLLARY II. The lateral areas of two similar cones of revolution are to each other as the squares of their slant heights, or the squares of their altitudes, or the squares of the radii of their bases. See method of Art. 483. Ex. 322. Prove the above proposition by using a cir- cumscribed instead of an inscribed pyramid. Ex. 323. If a cone is circumscribed about a pyramid prove that each lateral edge of the pyramid is an element of the cone. PROPOSITION VIII. THEOREM. 504. The area of the convex surface of a frustum of a cone of revolution is equal to one half the product of its slant height by the sum of the perimeters of its "bases. SUG. Express the lateral area of the circumscribed frustum of a pyramid. Let the number of lateral faces of the frustum of the pyramid increase, and employ the method of limits in a manner similar to that of Prop. III. Therefore 34:6 SOLID GEOMETRY. PROPOSITION IX. THEOREM. 505. The volume of a cone equals one third of the product of the area of its base by its altitude. Let A-B CDE represent a cone. To prove that the volume of the cone equals one third the product of the area of its base by its altitude. SuG. 1. Inscribe a pyramid within the cone. SuG. 2. What is the volume of the pyramid equal to ? SUG. 3. Complete the demonstration by a method sim- ilar to that of Prop. IV. Therefore - 506. COROI^ARY I. The volume of a circular cone may be expressed by the formula wherein V represents the volume, H the altitude, and R the radius of the base of the cone. 507. COROU,ARY II. The volumes of similar cones of revolution are to each other as the cubes of the radii of their bases, or the cubes of the altitudes, or the cubes of the slant heights. SUG. Prove this by a method similar to that used in Art. 486. THE SPHERE. 347 THE SPHERE. 508. A sphere is a solid bounded by a surface all the points of which are equally distant from a fixed point within it. The fixed point is called the center of the sphere, and the bounding surface is called the surface of the sphere. A sphere may be generated by the revolution of a semi- circle about its diameter. In higher branches of mathematics the word sphere is used to denote what is here spoken of as the surface of the sphere, but in this book a sphere is considered as a solid instead of the surface which bounds the solid. 509. The radius of a sphere is any straight line drawn from the center to the surface of the sphere. The diameter of a sphere is any straight line drawn through the center, terminated both ways by the surface of the sphere. From the definition of a sphere, all radii of the same sphere are equal, and since a diameter is twice the radius, it follows that all diameters of the same sphere are equal. 510. A line is tangent to a sphere when it has only one point in common with the sphere. A plane is tangent to a sphere when it has only one point in common with the sphere. Two spheres are tangent to each other when they have only one point in common. Ex. 324. The slant height of a cone of revolution is 13 ft., and its altitude is 12 ft. Find the lateral surface and the volume of the cone. 348 SOLID GEOMETRY. PROPOSITION X. THEOREM. 511. Every section of a sphere made by a plane is a circle. Let the plane MN intersect the sphere O, in the section A SB. To prove that the section A S B is a circle. SUG. 1. From the center of the'sphere O drop a J_ O S to the section, and from 5 draw SA and SB to any two points in the perimeter of the section. Connect A and O ; also B and O. SUG. 2. Compare A O and O B. Compare As A O S and B O S. SUG. 3. Compare 5 A and 5 B. SUG. 4. What is the section ? Therefore QUERY. Suppose a plane passes through the center of a sphere, is the section a circle ? Why ? The section of a sphere made by a plane is called a circle of the sphere. 512. COROU,ARY I. The line drawn from the center of a sphere to the center of a circle of a sphere is perpen- dicular to the circle of the sphere. THE SPHERE. 34:$ 513. COROLLARY II. If two circles of a sphere are equally distant from the center, they are equal. 514. COROLLARY III. Of two circles of a sphere un- equally distant from the center, that one is greater which is nearer the center. 515. COROLLARY IV. All circles of a sphere which contain the center are equal. 516. A great circle of a sphere is one which contains the center of the sphere. A small circle of a sphere is one which does not con- tain the center of the sphere. 517. The poles of a circle of a sphere are the extrem- ities of the diameter which is perpendicular to the plane of the circle. This diameter is often called the axis of the circle. Ex. 325. Prove Prop. VIII by considering the convex surface of the frustum of a cone as the difference between the convex surfaces of two cones. SUG. Let A denote the area of the convex surface, R the radius of the base, and 6" the slant height of the larger cone, and a, r and s the area of the convex sur- face, the radius of the base, and the slant height of the smaller cone. Then A itRS, and a = 7trs. Find A a to agree with the theorem. Ex. 326. Prove that the convex surface of the frustum of a cone of revolution is equal to the circumference of a section equidistant from the bases multiplied by the slant height. 350 SOLID GEOMETRY. PROPOSITION XI. THEOREM. 518. Two intersecting great circles of a sphere bi- sect each other. B Let A&CE and B D FE be two great circles of a sphere intersecting in the line J> E. To prove that the circles A DCE and B DFE bisect each other. SUG. 1. Is the center of the sphere in the plane of the QADCEl Why? SUG. 2. Are all points in the circumference of the O A D CE equally distant from the center of the sphere ? Why? SUG. 3. Locate the center of the O A D CE with re- spect to the center of the sphere. SUG. 4. Locate the center of the O B DFE with re- spect to the center of the sphere. SUG. 5. Are the centers of the Os^Z>C^ and ^>/^E jn the line of intersection D E? SUG. 6. What is a line through the center of a O called? SUG. 7. Does the line DE> bisect each of the QsADCE DFE? Therefore THE SPHERE. 351 PROPOSITION XII. THEOREM. 519. Three points on the surface of a sphere deter- mine a circle of the sphere. SUG. 1. How many points determine a plane? SUG. 2. A plane intersects a sphere in what kind of a figure ? Therefore 520. COROLLARY. Two points on the surface of a sphere determine a great circle of a sphere if the points are not the extremities of a diameter of the sphere. SUG. What third point is known in addition to the two given points ? Ex. 327. Prove Prop. IX by circumscribing a pyramid about the cone. Ex. 328. The volume of the frustum of any cone is equal to one third the altitude multiplied by the sum of the upper base, the lower base, and a mean proportional between the two bases. (See Art. 457.) Ex. 329. The total area of a right circular cylinder is 80 n square ft. and the radius of the base is 5 ft. Find the altitude of the cylinder. Ex. 330. The circumference of the base of a cone of revolution is 6 ft. , and the altitude of the cone is n ft. Find the volume of the cone. 'Ex. 331. The altitude of the frustum of a cone is \ the altitude of the entire cone. Find the relation between the volumes of the frustum and the entire cone. Ex. 332. Two circular cylinders have the same altitude but the volume of one is four times the volume of the other. Find the relation between the radii of the bases of the two cylinders. 352 SOLID GEOMETRY. PROPOSITION XIII. THEOREM. 521. The shortest distance on the surface of a sphere between any two points on that surface is the arc, not greater than a semi-circumference, of a great circle which joins them. Let A and B be two points on the surface of a sphere, and let A B be the arc, not greater than a semi-circle, which joins A and B, and let A CMDB be any other line on the surface of the sphere joining A and B. To prove that A B is less than A CMDB. SuG. 1. Connect any point of A CMDB, as M, with A and B, by arcs of great Os, as MA and- MB. Con- nect A, Maud B with O, the center of the sphere. SuG. 2. In the trihedral Z. O-A MB compare the face Z A 0j5 with the sum of the face Zs A OM&n&MOB. SUG. 3. Compare the arc A B with the sum of the arcs A M and MB. (Art. 174.) SUG. 4. In the same way connect C, any point in the line ACM, with A and M by arcs of great Os. Also connect D in the line MD B with M and B by arcs of great Os. SUG. 5. Compare the sum of the great O arcs A C, CM, M D and D B with the sum of the great O arcs A M and MB, and also with A B. THE SPHERE. 353 SUG. 6. Continue to take points in the line A C M B D and proceed as before. The sum of the arcs of great Os is a variable. Why? What is its limit? Why? SUG. 7. How does each succeeding value of the vari- able compare with the one preceding it ? SUG. 8. Then, how must A C M D B compare with A SI Therefore 522. By the distance between two points on the surface of a sphere is meant the distance measured on the arc of a great circle. Ex. 333. The portion of a tetrahedron cut off by a plane parallel to any face is a tetrahedron similar to the given tetrahedron. Ex. 334. Find the volume of the solid generated by revolving an equilateral triangle whose side is 6 ft. about one of its sides. SUG. Notice that two cones are generated whose bases coincide, and whose vertices are on opposite sides of the common base. Ex. 335. Compare the volumes of the solids generated* by revolving a rectangle whose dimensions are a and b successively about the adjacent sides. Ex. 336. The volume of a cylinder of revolution is equal to one half of the product of its lateral area by its radius. Ex. 337. In each of two right circular cylinders the altitude is equal to the diameter, and the volume of one is Y V that of the other. Find the relation of their alti- tudes. 23 Geo. 354 SOLID GEOMETRY. PROPOSITION XIV. THEOREM. 523. All points in the circumference of a circle are equally distant from either of its poles. Let A DC B be the circumference of a circle, and E and G its poles. To prove that all points in the circumference A D C B are equally distant from E, and also equally distant from G. SUG. 1. ]>t M be the center of the O A D CB. The straight line passing through O and M will determine E and G, the poles of iheQADCfi. Why ? (Arts. 512 and 517.) SUG. 2. The distances of all points in the circumfer- ence A D CB from any point in E G are how related ? (Art. 345.) SUG. 3. Then the distances of all points in the circum- ference A D C B from E are how related ? The distances from G are how related ? SUG. 4. But the distances considered in Sug. 3 are chords of great Os. Why ? SUG. 5. How, then, must the distances measured on the surface of the sphere compare ? Give auth. Therefore THE SPHERE. 355 524. The distance of any point in the circumference of a circle from the nearer pole of that circle is called the polar distance of the circle. 525. COROLLARY I. The polar distance of a great circle is a quadrant, i. e. t an arc of ninety degrees. SUG. What is the Z. at the center of a sphere that subtends the polar distance of a great circle ? 526. COROLLARY II. A point which is at the dis- tance of a quadrant from each of two points on the sur- face of a sphere is a pole of a great circle passing through those points. SUG. L F, A C = D E 9 B C = F E, Z. < = Z. A Z B = ^_ F and ^ C = ^ E. Also, let ABC and DEF be isosceles triangles, i. e., let A B = A C and DE = 1) F. To prove that the triangles ABC and DEF are equal. SUG. 1. Compare D E with A B. Give the reason for your statement. SUG. 2. Compare DF with A C. SUG. 3. Compare Z D with Z. A. SUG. 4. Place DEF upon A B C so that D E shall fall upon AB and D F upon A C. Why can this be done? SUG. 5. Where will E fall ? Where will F fall ? SUG. 6. Compare each part of DEF with a corre- sponding part of A B C. Therefore 362 SOLID GEOMETRY. 543. If from the vertices of a spherical triangle as poles, arcs of great circles be drawn, a spherical triangle is formed which is called the polar of the first. If from A, B and C as poles, arcs of great circles be drawn, a triangle D E F\& formed which is the polar of the triangle ABC. If entire circles be drawn they will intersect to form eight spherical triangles, but the polar of the given tri- angle A B C is that one of the eight triangles whose vertices lie on the same sides of the arcs of the given tri- angle as the respectively homologous vertices of the given triangle, and no side of which is greater than 180 degrees. PROPOSITION XIX. THEOREM. 544. If a spherical triangle D E F is the polar of another spherical triangle ABC, then the triangle A B C is the polar of D E F. Liet the spherical triangle D E F be the polar ofABC. To prove that A B C is the polar of D E F. THE SPHERE. 363 SUG. 1 . A is the pole of what arc ? SUG. 2. How many degrees from A to SUG. 3. C is the pole of what arc ? SUG. 4. How many degrees from C to El SUG. 5. E is the pole of the arc A C. Why ? SUG. 6. F is the pole of the arc A B. Why ? SUG. 7. D is the pole of the arc B C. Why ? Therefore Ex. 340. A sphere may be inscribed in any tetrahe- dron. Ex. 341. One, and only one, surface of a sphere may be described through any four points not in .the same plane. SUG. 1. Let A, B, C and D be the four given points. SUG. 2. What is the locus of points equally distant from A and #? SUG. 3. What is the locus of points equally distant from B and C? SUG. 4. What is the intersection of these two loci ? SUG. 5. What is the locus of points equally distant from CandZ>? SUG. 6. Does this last locus intersect the one referred to in Sug. 4 ? Ex. 342. A straight line tangent to a circle of a sphere lies in a plane which is tangent to the sphere at the point of contact. Ex. 343. A plane tangent to a sphere is perpendicular to the radius through the point of contact. SOLID GEOMETRY. PROPOSITION XX. THEOREM. 545. In two polar triangles each angle of one is measured by the supplement of the side opposite it in tine other. Let ABC and D E F be two polar triangles. To prove that the angle D is measured by the supple- ment of B C ? SUG. 1. Extend D E and D F until they intersect B C in M and N respectively. SUG. 2. B is the pole of what arc ? SUG. 3. How many degrees in B Nt SUG. 4. C is the pole of what arc? SUG. 5. How many degrees in C Ml SUG. 6. How many degrees in B N + M C? How many in BC+ M Nt SUG. 7. What is the supplement of B C? SUG. 8. What is the measure of Z. D ? Why ? Therefore THE SPHERE. 365 PROPOSITION XXI. THEOREM. 546. Two triangles on the same sphere, or equal spheres, having two sides and the included, angle of one equal respectively to two sides and the included angle of the other, are either equal or symmetrical* AD A C F c Let ABC and D E F be two spherical triangles in which the angle D equals tlie angle A, and the sides A B and A C are equal respectively to the sides D E and D F. To prove that the triangles ABC and D E F are either equal or symmetrical. CASE I. When the parts are arranged in the same order. SUG. 1. Place the A D E F upon the A A B C, so that D shall fall upon A, D F upon A C, and D E upon A B. Why can this be done ? SUG. 2. Where will F fall ? Where will E fall ? SuG. 3. Then how are the As A B C and DEF related? SUG. 4. What is the conclusion in Case I ? CASE II. When the parts are arranged in reverse order. SuG. 1. Construct A A Cthe symmetrical of A A B C (Fig. at right). Then, since two sides and the included Z. of DEF are = to two sides and the included Z! of ABC, each to each, but arranged in the reverse order, how do the parts of D E F compare with those of A E C, each to each ? SUG. 2. Compare A ^^Cwith A DEF. (Case I.) SuG. 3. Since A A E C is the symmetrical of A B C t how do As ABC and DEF compare ? SUG. 4. What is the conclusion in Case II ? Therefore 366 SOLID GEOMETRY. PROPOSITION XXII. THEOREM. 547. Two triangles on the same sphere, or equal spheres, having two angles and the included side of one equal respectively to two angles and the included side of the other, are equal or symmetrical. SUG. See figure and suggestions of Prop. XXI. Therefore PROPOSITION XXIII. THEOREM. 548. Two triangles on the same sphere, or equal spheres, having the three sides of one respectively equal to the three sides of the other, are either equal or symmetrical. A D Let ABC and D E F be two spherical triangles on the same spJiere, or equal spJieres, having the three sides of one equal respectively to the three sides of the other. To prove that ABC and DBF are either equal or sym- metrical. SUG. 1. Connect the vertices of each A with the cen- ter of the sphere on which the A is situated. SUG. 2. Compare the face Z. of the trihedral ^.s O and 5. (Art. 144.) THE SPHERE. 367 SUG. 3. Compare the dihedral Zs of the two trihedral Zs. (Art. 390.) SUG. 4. Compare the Zs of one A with the corre- sponding Zs of the other A. (Art. 534.) SUG. 5. Compare the A A B C with the A DBF, first, when the parts are arranged in the same order, and second, when they are arranged in reverse order. Therefore PROPOSITION XXIV. THEOREM. 549, Two triangles on the same sphere, or equal spheres, having the three angles of one respectively equal to the three angles of the other, are equal or sym- metrical. Let A and B represent two spherical triangles on the same sphere, or equal splieres, having the three angles of one equal respectively to tJie three angles of the other. To prove that A and B are either equal or symmetrical. SUG. 1. lyet C and D be the polar As of A and B re- spectively. SUG. 2. How do the sides of C and D compare ? (Art. 545.) SUG 3. How do the Zs of C and D compare ? (Art. 548.) SUG. 4. Then how do the sides of A and B compare ? SUG. 5. How are the As A and B related ? Therefore 308 SOLID GEOMETRY. 550. A convex spherical polygon is a spherical polygon, no side of which, if extended, would pass within the polygon. PROPOSITION XXV. THEOREM. 551. The sum of the sides of a convex spherical polygon is less than the circumference of a great circle. SUG. 1. Connect each vertex of the polygon with the center of the sphere, thus forming a polyhedral Z. . SUG. 2. Compare the sum of the face <>Z.B=Z.F and /_ C = Z E. To prove that the triangles ABC and DBF are equal in area. SUG. 1. Let O be the pole of a small O through the points A, B, and C. Draw the arcs of great Os O A, OB, and OC. SUG. 2. Compare the arcs O A, OB, and O C . SUG. 3. At D, draw D S, making Z. FD S = Z. B A O, and take DS = A O. Draw the arcs of .great Os SF and SE. SUG. 4. In the spherical As F D S and B A O, corn- pare D F with A B, DS with A O, and Z FD S with Z B A O. Compare the As FD S and B A O. (Art. 546.) Compare SF with O B. SUG. 5. DSF and A OB are symmetrical isosceles spherical As. Why ? Compare these As in respect to area. (Art. 542.) SUG. 6. In a similar manner, compare the spherical As D S E and A O C in respect to area. THE SPHERE. 371 SUG. 7. Compare the spherical As ESF and COB in respect to area. SuG. 8. Compare the spherical As DEF and ABC in respect to area. Therefore 556. A lune is a portion of the surface of sphere a in- cluded between two semi-circumferences ot A great circles, as A B D C . 557. The angle of a lune is the angle between the two semi-circumferences which bound the lune. Ex. 348. Prove Prop. XXVII when the pole of the small circle through A, B, and Cis without the triangle. Ex. 349. If the area of the convex surface of a right circular cone is twice the area of its base, prove that the slant height of the cone is equal to the diameter of its base. Ex. 350. The radius of the base of a right circular cone is 5 in., and the number of square inches in the area of the convex surface of the cone is equal to the number of cubic inches in the volume of the cone. Find the altitude and the slant height of the cone. Ex. 351. Which generates the greater volume, the revolution of a rectangle about its shorter or its longer side? Ex. 352. What is the locus of a point in space the sum of the squares of whose distances from two fixed points is equal to the square of the distance between the two points. 372 SOLID GEOMETRY. PROPOSITION XXVIII. THEOREM. 558. If two arcs of great circles intersect on the sur- face of a hemisphere, the sum of the areas of the oppo- site spherical triangles thus formed is equal to the area of a lune whose angle is the angle of the intersect- ing arcs. A ^ ^ B Let the arcs of great circles ACE and BCD in- tersect at the point C on the hemisphere ABED, thus forming the opposite spherical triangles ACB and D CE. To prove that the sum of the areas of the triangles ACB and D C E is equal to the area of a lune whose angle is equal to the angle ACB. SUG. 1. Extend the arcs ACE and B CD to complete the great Os A CEF and B CDF. SUG. 2. CDFE is a lune whose A \$DCE. Why ? SUG. 3. Compare the arcs C B and D F\ also the arcs A C and EF\ also the arcs A B and DE. SuG. 4. Compare the As A CB and EFD in respect to area. (Arts. 548 and 555.) SUG. 5. Compare the sum of the areas of the As ABC and CD E with the area of the lune CDFE. Therefore THE SPHERE. 373 EXERCISES. 353. What is the locus of a point in space which is at a given distance from a fixed straight line ? 354. To cut a cylinder of revolution by a plane paral- lel to an element in such a manner that the section shall be a rectangle equal in all respects to "the rectangle which generates the cylinder. 355. A right circular cylinder whose altitude is 5 ft. and the radius of whose base is 3 ft. is cut by a plane parallel to the base at such a distance from the base that the area of the section is a mean proportional between the areas of the convex surfaces of the two parts into which the cylinder is divided. Determine the lengths of the segments of the altitude. 356. Prove that the total surface of a right circular cylinder is equal to the circumference of its base multi- plied by the sum of its altitude and the radius of its base. 357. If four similar cylinders of revolution have their altitudes proportional to the numbers 3, 4, 5 and 6, prove that the volume of the largest cylinder is equal to the sum of the volumes of the other three. 358. If two regular tetrahedrons of different volumes have the edges of one parallel respectively to' those of the other, prove that the lines joining corresponding vertices meet in the same point. 359. The dimensions of a rectangular parallelepiped are 3, 4 and 12 ft. Find the circumference of a great circle of the circumscribing sphere. 360. What is the volume of a piece of timber 15 ft. long, whose bases are squares, each side of one base be- ing 14 in., and each side of the> other base 12 in ? 374 SOLID GEOMETRY. SPHERICAL, AREAS. 559. A spherical degree, or a degree of spherical surface, is one three hundred and six- tieth of the surface of a hemisphere. 560. If three great circles are per- pendicular to one another, eight spher- ical triangles are formed, each having three right angles. Bach of these tri- angles is called a tri-rectangular triangle. A tri-rectangular triangle has ninety spherical degrees. PROPOSITION XXIX. THEOREM. 561. The area of a lune is to the area of the surface of a sphere as the angle of the lune is to four right angles. A Let A J5 D C represent a lune tvhose angle is B A C. To prove that the area of A B D C is to the area of the surface of the sphere as the angle BAG is to four right angles. Let B C be an arc of a great O whose poles are A and D, and let O be the center of the sphere. THE SPHERE. 375 CASK I. When Z B O C and four rt. Z.s are commen- surable, SUG. 1. Draw lines dividing the Z. B O C and 4 rt. Z.s into equal parts, each of which is a common unit of measure of Z. B O C and 4 rt. Z^s. Let this unit be contained m times in Z. B O C, and n times in 4 rt. Z!s. SUG. 2. What is the ratio - Z B 4 *T equal to? 4. rt. Z_s SUG. 3. Pass a plane through the lines of division of the Z.s and the edge of the lune A D. These planes divide the surface of the sphere into small lunes. SUG. 4. Compare these small lunes in respect to area. SUG. 5. How does the number of lunes compare with the number of angles ? surface of lune SUG. 6. What is the ratio - equal to ? surface of sphere SUG. 7. Compare answers to Sugs. 2 and 6. What is the conclusion when Z. B O C and 4 rt. ^s are commen- surable ? CASK II. When Z. BOC and four rt. Z.s are incommen- surable. SUG. Employ the method of limits similar to Arts. 171, 254, 424, etc. Therefore 562. COROLLARY. A lune contains twice as many spherical degrees as its angle contains angular degrees. Let 5 represent the number of spherical degrees in the surface of the lune, and A the number of angular degrees in the Z. of the lune. Then, as there are 720 spherical degrees in the surface of a sphere, it is evident from the proposition that 5 A 720 = : 360 * Hence S= < 2A. 376 SOLID GEOMETRY. 563. The spherical excess of a spherical triangle is the excess of the sum of its angles over two right angles. PROPOSITION XXX. THEOREM. 564. The number of spherical degrees in a spheri- cal triangle is equal to the n.uiriber of angular degrees in its spherical excess. Let A It C be a spherical triangle. To prove that the number of spherical degrees in ABC is equal to the mimber of angular degrees in ^_ A -f ^_ B + Z.C- 180. SUG. 1. Let one side of the A, as B C, be extended to form a complete great B C D E. Let B A and C A be extended to meet the great B C D E in D and E respectively. SUG. 2. BCD,BAD t m&BED are semi- circum- ferences of great Os. Why ? SUG. 3. &ABC+&A ED = lune whose Z is A Why? SUG. 4. &A B C+ &A CD= lune whose Z is B Why? SUG. 5. &ABC+&ABE = lune whose Z is C Why? THE SPHERE. 377 SUG. 6. How many spherical degrees in a lune whose ^ is A ? in a lune whose ^ is B ? in a lune whose /_ is C? Give auth. SuG. 7. How many spherical degrees in a hemisphere ? SUG. 8. If the number of spherical degrees in A A B C be represented by m, then, from the equations in Sugs. 3, 4 and 5, 2 m + 360 equals what ? SUG. 8. What does w equal ? Therefore 565. SCHOLIUM. To say that a spherical triangle contains a certain number, say n, of the spherical de- grees, is simply to say that the surface of the triangle is equal to -~ of the surface of a hemisphere, or to of the surface of a sphere. The whole surface of a sphere, expressed in square measure, will be found in proposition XXXIII, and then the area of the surface of a spherical triangle can be expressed in square measure, by means of the theorem just given, when the angles of the tri- angle and the radius of the sphere are known. Ex. 361. How many spherical degrees are there in a spherical triangle whose angles are 200, 140 and 100 ? Ex. 362. What part of the surface of a sphere is a spherical triangle whose angles are 120, 140, 160 ? Ex. 363. What part of the surface of a sphere is a spherical triangle each angle of which is 90 ? How many spherical degrees in the same triangle ? Ex. 364. Find the angles of an equiangular spherical triangle whose surface is y 1 ^ of the surface of a sphere. 378 SOLID GEOMETRY. 566. A zone is that portion of the surface of a sphere which is included between two parallel planes. The circumferences of the sections of the sphere made by the planes are called the bases of the zone, and the perpendicular distance between the parallel planes is called the altitude of the zone. 567. A spherical segment is a portion of a sphere included between two parallel planes. The sections of the sphere by the planes are called the bases of the segment, and the perpendicular distance be- tween the planes is called the altitude of the segment. A zone is the spherical surface of a segment. The portion of a sphere cut off by any plane is considered as a spherical segment for it is included between the cutting plane and a plane tangent to the sphere parallel to the cutting plane. For the same reason, the portion of the surface of a sphere cut off by any plane is considered as a zone. In this case the segment and zone. are are said to be of one base. 568. If a semi-circle be revolved about its diameter as an axis, a sphere will be generated; any arc of the semi-circum- ference will generate a zone. 569. A spherical sector is the por- tion of a sphere generated by the revolu- B tion of a circular sector about a diameter. The pupil should form a mental picture of the different varieties of spherical sectors and describe them. For example, if the semi- circle A D B be revolved about the diameter, the circular sector, AOC, will generate a spherical sector whose surface is a zone of one base generated by the arc A C, and a convex conical surface generated by the radius O C. The circular sector COB generates a spherical sector whose surface is a zone of one base generated by the arc C D B and a concave conical surface generated by the radius O C. THE SPHERE. 37$ EXERCISES. 365. Construct a semi-circle, and in it a circular sector which, if revolved, will generate a spherical sector hav- ing two concave conical surfaces. Describe the zone. 366. Construct a circular sector which will generate a spherical sector whose surface will be a concave conical surface, a plane surface, and a zone. 367. Is a hemisphere a spherical sector? 368. What is the locus of a point in space which is at a given distance, a, from a given plane, and at a given distance greater than a from a given point in the given plane ? Ex. 369. Compute the lateral area, the total area, and the volume of the frustum of a cone of revolution, given the altitude of the cone, 20 ft.; the diameter of the lower base, 16 ft.; the diameter of the upper base, 12 ft. 370. A cylinder of revolution, and the frustum of a cone of revolution, have the same lower base and the same altitude; what must be the ratio of the radii of the two bases of the frustum in order that the volume of the frustum may be one half the volume of the cylinder ? 371. If a rectangle revolves about one of its sides the volume generated is 288 ft cubic ft., but if it revolves about the adjacent side the volume generated is 384 n cubit ft; find the diagonal of the rectangle. 372. A regular quadrangular pyramid has each basal edge equal to 12 ft., and each lateral edge equal to 10 ft. ; find the altitude of the pyramid. , 373. Find the radius of the base of a cylinder of revo- lution whose altitude is 1 yd., and whose volume is 48 n cubic ft. 380 SOLID GEOMETRY. PROPOSITION XXXI. THEOREM. 570. The area of the surf ace generated by the revo- lution of a straight line about an axis in its plane, but not crossing it, is equal to the projection of the line upon the ajcis multiplied by the circumference of a circle whose radius is a perpendicular erected at the middle point of the line and limited by the axis. Let H S be the axis of revolution, A B the line which revolves to generate the surface, CD the projection of AB upon K S 9 M the middle point of A B, and M O the per- pendicular to the line A B, at M. To prove that the surface generated by A B is equal to C D multiplied by the circumference of a circle whose radius is M O. CASE I. When A B does not meet the axis. SUG. 1. A B generates the convex surface of the frus- tum of a cone of revolution. What is the area of this surface equal to? (Art. 504.) SUG. 2. Draw A C, MN, and B D _L to R 5, and A P II to R S. THE SPHERE. 381 SUG. 3. Prove that MN equals one half otAC + BD, and hence the circumference of a generated by M equals one half the sum of the circumferences of the up- per and lower bases of the frustum. SUG. 4. The As MNO and A PB are similar. Why ? Compare the ratio -TT with -TFTT and hence with -rrrf- SUG. 5. I^et a be the measure of A B, c the measure of C D, r the measure of M O, and n the measure of M N. SUG. 6. Compare the ratios and r n SUG. 7. Compare a n and c r. SUG. 8. The measure of the surface generated by A B equals 2 n a n. Why ? It also equals 2 n cr. Why ? SUG. 9. What is the circumference of a O whose ra- dius is r equal to ? Hence, what is the area of the sur- face generated by A B ? CASE II. When one point, as A, is on the axis. SUG. This case is obtained from Case I, by making A C equal to nothing. Examine the reasoning in Case I, and see that it holds when A C = 0. CASK III. When A B is parallel to the axis. SUG. This case is obtained from Case I, by making A C = B D. Examine the reasoning in Case I and see that it holds when A C = B D. Therefore - QUERY. What is the surface generated by A B called when A is on the axis ? What when A B is parallel to the axis ? 382 SOLID GEOMETRY. PROPOSITION XXXII. THEOREM. 571. The area of a zone equals the product of its altitude by the circumference of a great circle. luet XCT represent a semi-circle, and A D an arc which, revolved around X Y, generates a zone. To prove that the area of the zo?ie is equal to its altitude multiplied by the circumference of a great circle. SUG. 1. Divide the arc A D into any number of equal parts, as A B> B C, etc., and draw the chords of these arcs. SuG. 2. How do these chords compare in length ? SUG. 3. At the middle points of these chords erect J_s terminating in the axis X Y. SUG. 4. Where will these J_s intersect ? Why ? SUG. 5. How do these J_s compare in length ? Why ? SUG. 6. What is the area generated by the chord A B ? by the chord B C? etc. SUG. 7. Express, in its simplest form, the area gener- ated by all these chords taken together. SUG. 8. Bisect the arcs, draw chords to the points of bisection, and proceed as before. THE SPHERE. 383 SUG. 8. Let the number of equal parts into which the given arc is divided be continually increased, then the surface generated by the chords of these arcs taken to- gether is a variable. But during its variation it is al- ways equal to what ? SUG. 9. What two variables are here always equal ? SUG. 10. What are the respective limits of these va- riables ? SUG. 11. What is the area of the zone? Therefore - 572. COROLLARY. The above theorem may be ex- pressed by the formula in which 5 represents the area of the surface of the zone, H its altitude, and R the radius of the sphere on which the zone lies. Ex. 374. Prove that the sum of the angles of a convex spherical polygon of n sides is greater than 2 n 4 and less than 6 n 12 right angles. SUG. Divide the spherical polygon of n side into n 2 spherical triangles by arcs of great circles drawn from one vertex to each non-adjacent vertex. Ex. 375. Prove that the number of spherical degrees in a convex spherical polygon of n sides is equal to the number which expresses the excess of the sum of its angles over 2 n 4 right angles. See Sug. in Ex. 374. Ex. 376. The volume of a rectangular parallelepiped is 6,720 cubic inches, and its edges are in the ratio of the numbers 3, 5 and 7; find the three ed;es. 384 SOLID GEOMETRY. PROPOSITION XXXIII. THEOREM. 573. The area of the surface of a sphere equals the product of its diameter by the circumference of a great circle. SUG. Employ the method of Prop. XXXII, or use that proposition as authority, taking a semi-circumfer- ence as the given arc. 574. COROW,ARY I. The trutn. of the theorem may be expressed by the formula S=%7tRD, or, 5=477- . 2 , in which represents the surface of the sphere, R its radius, and D its diameter. 575. CORONARY II. The area of the surface of a sphere equals the area of four of its great circles. (See second formula in Cor. I.) 576. CORONARY III. The area of the surface of a sphere equals the area of a circle whose radius is equal to the diameter of the sphere. (See second formula in Cor. I.) 577. CORONARY IV. The areas of the surfaces of two spheres have the same ratio as the squares of their radii or the squares of their diameters. 578. COROU,ARY V. The area of a spherical degree may be expressed as nor . - . (Art. 559.) 1 2i\) Ex. 377. Find, in square feet, the area of a spherical triangle whose angles are 95, 140 and 216, the radius of the sphere being 15 in. THE SPHERE. 385- PROPOSITION XXXIV. THEOREM. 579. The volume of a sphere is equal to the area of its surface multiplied by one third of its radius. SUG. 1 . Circumscribe a polyhedron about the sphere. SUG. 2. Join each vertex of the polyhedron with the center of the sphere, and pass planes through these lines and the edges of the polyhedron. Pyramids are thus formed. Why ? SUG. 3. Compare the altitude of these pyramids with the radius of the sphere. SUG. 4. What is the volume of each pyramid equal to ? SUG. 5. What is the volume of the polyhedron ? SUG. 6. Circumscribe a polyhedron of a greater num- ber of sides about the sphere. What is its volume in terms of its surface ? SUG. 7. Finish the demonstration. Therefore - 580. COROLLARY I. The volume of a sphere may be expressed by the formula in which V represents the volume and R the radius of the sphere. 25 Geo. 386 SOLID GEOMETRY. 581. COROLLARY II. The volumes of two spheres have the same ratio as the cubes of their radii, or the cubes of their diameters. 582. A spherical pyramid is a solid bounded by a spherical polygon, and the planes of the sides of the polygon, as O-ABCDE. The vertex of the pyramid is at the center of the sphere. The spherical polygon is the base of the spherical pyramid. Ex. 378. The volume of a spherical pyramid is equal to the area of its base multiplied by one third of the ra- dius of the sphere. SUG. Employ the method of Prop. XXXIV. Ex. 379. The volume of a spherical sector is equal to the area of its zone multiplied by one third of the radius of the sphere. SUG. Employ the method of Prop. XXXIV. Ex. 380. Prove that the volume of a sphere is twice the volume of a cone whose altitude is equal to the diam- eter of the sphere, and the radius of whose base is equal to the radius of the sphere. Ex. 381. The radius of a sphere is 3 in., and the area of a spherical triangle A B C on this sphere is 11.7 it. The angles A and B are 72 and 115 respectively. Find the angle C of the spherical triangle. Ex. 382. On the same sphere, or equal spheres, zones of equal altitudes are equal in area. THE SPHERE. 387 PROPOSITION XXXV. PROBLEM. 583. To find the volume of a spherical segment. Let A B D C generate a spJierical segment by revolving about A C. To find the volume of the segment. SUG. 1. Join B and D to the center of the semi- circle O. Sue. 2. Find the volume of the spherical sector gen- erated by revolving the circular sector BOD. (Ex. 379.) SUG. 3. Add the volume of the cone generated by re- volving the A B A O. SUG. 4. Subtract the volume generated by revolving the A D C O. Ex. 383. Find the volume of a spherical sector having a zone of 10 in. altitude on a sphere of 20 in. radius. Ex. 384. Find the volume of a spherical segment of 8 in. altitude, one base of which passes through the cen- ter of the sphere, the radius of the sphere being 20 in. 388 SOLID GEOMETRY. EXERCISES. 385. A sphere is cut by parallel planes so that the diameter is divided into ten equal parts: compare the area of the zones; also the volumes of the spherical sectors whose spherical surfaces are the respective zones. 386. If the diameter of the sphere is 30 ft., find the volumes of the first, fifth, and seventh segments re- ferred to in exercise 385. 387. Find the volume and area of the surface oi the sphere in exercise 386. 388. The sides of a triangle, on a sphere whose ra- dius is 10 ft., are, respectively, 95, 117, and 37. Find the area in square feet of its polar triangle. 389. Assuming the earth to be a sphere whose di- ameter is 7912 miles, how many square miles upon its surface. 390. If the average specific gravity of the earth is 7, what is its weight expressed in tons. 391. Assuming the diameter of the earth to be 8,000 miles, and that of the moon 2,000; how do the amounts of light reflected from them to a point in space equally distant from each compare ? 392. With the same assumption as that of exercise 391, what is the ratio oi the volumes of the earth and moon? 393. A triangle on a 12 in. globe has for its angles 140, 119, and 196 respectively; compute its area. 394. The diameter of a sphere is equal to the altitude of a cone of revolution and of a cylinder of revolution, and the radius of the sphere is equal to the radius of the cone and of the cylinder. Prove that the volumes of the cone, sphere, and cylinder are proportional to the num- bers 1, 2, and 3. PROPOSITIONS IN CHAPTER VIII. PROPOSITION I. Every section of a cylinder made by a plane containing an element is a parallelogram. PROPOSITION II. The bases of a cylinder are equal. . PROPOSITION III. The area of the lateral surface of a cylinder is equal to the perim- eter of a right section multiplied by an element of the surface. PROPOSITION IV. The volume of a cylinder equals the area of a right section multi- plied by a lateral edge. PROPOSITION V. Every section of a cone made by a plane passing through the ver- tex is a triangle. PROPOSITION VI. Every section of a circular cone made by a plane parallel to the base is a circle. PROPOSITION VII. The area of the convex surface of a cone of revolution is equal to one half the product of the perimeter of its base by its slant height. PROPOSITION VIII. The area of the convex surface of a frustum of a cone of revolution is equal to one half the product of its slant height by the sum of the perimeters of its bases. PROPOSITION IX. The volume of a cone equals one third of the product of the area of its base by its altitude. 390 SOLID GEOMETRY. PROPOSITION X. Every section of a sphere made by a plane is a circle. PROPOSITION XI. Two intersecting great circles of a sphere bisect each other. PROPOSITION XII. Three points on the surface of a sphere determine a circle of the sphere. PROPOSITION XIII. The shortest distance on the surface of a sphere between any two points on that surface is the arc, not greater than a semi-circumfer- ence, of a great circle which joins them. PROPOSITION XIV. All points in the circumference of a circle are equally distant from either of its poles. PROPOSITION XV. Given a material sphere, to find its radius. PROPOSITION XVI. A plane perpendicular to a radius of a sphere at its extremity, is tangent to the sphere. PROPOSITION XVII. A spherical angle is equal to the dihedral angle formed by the planes of the two arcs, and is measured by the arc of a great circle described from the intersection of the arcs as a pole, and intercepted between them. PROPOSITION XVIII. Two symmetrical isosceles spherical triangles can be made to co- incide, and are equal. PROPOSITION XIX. If a spherical triangle D E F is the polar of another spherical tri- angle ABC, then the triangle A B C is the polar of D E 2\ THE SPHERE. 391 PROPOSITION XX. In two polar triangles each angle of one is measured by the sup- plement of the side opposite it in the other. PROPOSITION XXI. Two triangles, on the same sphere, or equal spheres, having two sides and the included angle of one equal respectively to two sides and the included angle of the other, are either equal or symmetrical. PROPOSITION XXII. Two triangles, on the same sphere, or equal spheres, having two angles and the included side of one equal respectively to two angles and the included side of the other, are either equal or symmetrical. PROPOSITION XXIII. Two triangles, on the same sphere or equal spheres, having the three sides of one respectively equal to the three sides of the other, are either equal or symmetrical. PROPOSITION XXIV. Two triangles on the same sphere, or equal spheres, having the three angles of one respectively equal to the three angles of the other, are either equal or symmetrical. PROPOSITION XXV. The sum of the sides of a convex spherical polygon is less than the circumference of a great circle. PROPOSITION XXVI. The sum of the angles of a spherical triangle is greater than two and less than six right angles. PROPOSITION XXVII. Two symmetrical spherical triangle are equal in area. PROPOSITION XXVIII. If two arcs of great circles intersect on the surface of a hemisphere, the sum of the areas of the opposite spherical triangles thus formed is equal to the area of a lune whose angle is the angle of the intersect- ing arcs. 392 SOLID GEOMETRY. PROPOSITION XXIX. The area of a lune is to the area of the surface of a sphere as the angle of the lune is to four right angles. PROPOSITION XXX. The number of spherical degrees in a spherical triangle is equal to the number of angular degrees in its spherical excess. PROPOSITION XXXI. The area of the surface generated by the revolution of a straight line about an axis in its plane, but not crossing it, is equal to the pro- jection of the line upon the axis multiplied by the circumference of a circle whose radius is a perpendicular erected at the middle point of the line and limited by the axis. PROPOSITION XXXII. The area of a zone equals the product of its altitude by the cir- cumference of a great circle. PROPOSITION XXXIII. The area of the surface of a sphere equals the product of its diam- eter by the circumference of a great circle. PROPOSITION XXXIV. The volume of a sphere is equal to the area of its surface multi- plied by one third its radius. PROPOSITION XXXV. To find the volume of a spherical segment. INDEX. Abbreviations, 0. Acute triangle, IS. Acute angle, (5. Adjacent angles, 3. Alternate exterior, Jl. Alternate interior, 41. Alternation. 144. Altitude, <>!>, -js-.j. rsn'.i. :;io. :;; Angles, I. :;i Angle at the center, 21 1 Antecedent, KM. Apothem, '211. Arc. 83 Area, 1 7i. isc,, 374. Axiom, s, K>. -j|2. Axis of a circle, :! !'.. Basal edges, 288, :!<>!>. is, <;.. Broken line, 3. Center, S2. '.Ml. Chor'J I. Element, 339. Enunciation, 7. Equiangular polygon, 08. Equilateral polygon, (>S. Equilateral triangle, 1 s. Exterior angles, 41, 48. (17. Extremes, 137. Fourth proportional, 107. Frustum, 310, 340. Gauss, 238. General enunciation, 7. Generatrix, 329. Geometrical figure, 2. Geometry, 3. Great circle of sphere, 349. Greatest common unit, 101. Hexagon, 08. Homologous, 152. Hypotenuse, 38. Hypothesis, 7. 394: INDEX. Icosahedron, 325. Incommensurable number, 103, 104. Incommensurable quantity, 102. Incommensurable ratio, 103. Indirect method, 54, 55. Inscribed, 112, 110, 117, 333, 343. Interior angles, 41, 67. Inversion, 145. Isosceles triangle, 18. Isosceles trihedral, 28i Lateral edges, 288, 309. Lateral faces, 288. Leg, 09. Limits, 104. Line, 2. Locus of a point, 27, 28, 261, 262. Logical terms, 7. Lune, 371. Magnitudes, 3, 4, 33. Means, 137. Mean and extreme ratio, 204. Mean proportional, 140. Measurement, 99. Measure of a quantity, 99. Models, 12, 15, 16, 20, 23, 32, 36, 46, 109, 180. Notes: Area, 182. Base, 69. Circle, 82. How to study, 11. Indirect demonstration, 54. Intersection of loci, 179. Limits, 104. Parallelograms, 70. Property and definition, 72. Propositions, 139, 140, 142, 148. Ratio, 100, 179. Suggestions, 122. Number, 99, 100. Oblique angle, 6. Obtuse angle, 6. Obtuse triangle, 18. Octahedron, 322. Opposite angle, 6. Opposite interior angles, 49. Parallelogram, 68. Parallel, 40, 254. Parallelepiped, 297. Pentagon, 68. Pentahedron, 287. Perimeter, 67. Perpendicular, 6, 246. Pi (if), 225, 2->8, 2:5:1. Plane angle of a dihedral, 264. Plane, or plane surface, 3. Plane figure, .'5. Plane geometry, 3. Plane triangle, 18. Point, 2. Point of tangency, 83. Polar distance, 35(5. Polar triangle, 362. Poles of a circle, 349. Polygons, 67. Polyhedral angles, 277. Polyhedron, 287. Postulate, 8, 118. Premises, 7, 30. Prism, 3SS. Problem, 8, 118-132, 166-171, 198-201, 230-238. Product of base and altitude, 181 Projection of a point or a line, 19-1 , 275. Proof, 7. Property, 72. Proportion, 137. Proposition, 8. Pure number, 100. Pyramids, 5I09. Pythagorean proposition, 190. Quadrant, 355. Quadrilateral, 68. Quantity, 99. Quotient, 100. Radius. 82, 211, 347. Ratio, 99, 100. 103, 137. Ratio of similitude, 153. Kr:ctangle, 69, 70. INDEX. 395 Rectangular parallelepiped, 297. R- ctilinear figui Red net ii> ad absitrdiun, 5~>. Re-entrant angl< Regular polygons, 2(M Rt-gular polyhedrons, 322. Regular prisms, '.'>'.' R. -pillar pyramids, 309. Rhomb... (1, Ml). Rhombus, 70. Right angle. 5. Right circular cone, 310. Right cylinder, :;:',(). Right parallelopiped, 297. Right section ot prism, 288. Right triangle, 18. Scalene triangle, 18. Scholium, 8, 13!). Secant, S3. Secant line, 41. Sector, S3. Segment, 83. Sides of an angle, 5. Sides of a polygon, (57. Sides of a triangle, 18. Similar arcs, etc., 2MD Similar polygons, l.VJ Similitude, 1."):; Slant height, 3(>D Small circle of sphere, .'! ID Solid, 1. Solid geometry, 4, 311. Special enunciation, 7. Sphere, 347. Spherical area, 371. Spherical polygon, 3.~>D. Spherical pyramid, 380. Spherical sector, 378. Spherical segment, 378. Spherical triangle, 359. Squar- Square of a line, 181, 190. Straight angle, .">. Straight edge, 118. Straight line, 3. Subtend, 83. Supplements, 7. Supplementary adjacent angles, 7. Surface, 1, 2, 347. Symbols and abbreviations, 9. Symmetrical spherical triangle. 360. Symmetrical trihedral angles. 277. Tangent, 83, 340, 347. Terms of a proportion, 137. Tetrahedron. 287. Theorem, 7. Third proportional, 168. Transversal, 11 , Trapezium, C.S. Trapezoid, OS. Triangle, IS, <>s Trihedral angle, 2? 7. Trirectangular triangle, 37 1. Truncated pyramid, 310. Turn through an angle, 4. Unit of measure, 90, 111, I 7<>, 181. Variable, 104, 105. Vertex of an angle, 5. Vertical angles, G. Vertical angle of a triangle, 18. Vertices of a triangle, IS. Vertices of a polygon, 07. Zone, 378. YC 22379 M306168 V3 THE UNIVERSITY OF CALIFORNIA LIBRARY