mummmum X THE ELEMENTS OF GEOMETRY X BUSH AND CLARKE SILWR^BURDETT & COMPANY — — % > J MUWIHW W tlH IN MEMORIAM FLORIAN CAJORI e^ Digitized by the Internet Arciiive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofgeometOObushrich THE ELEMENTS OF GEOMETRY BY WALTER N. BUSH PRINCIPAL OF THE POLYTECHNIC HIGH SCHOOL SAN FRANCISCO AND JOHN B. CLARKE DEPARTMENT OF MATHEMATICS, POLYTECHNIC HIGH SCHOOL SAN FRANCISCO SILVER, BURDETT AND COMPANY NEW YORK BOSTON CHICAGO SAN FRANCISCO Copyright, 1905, 1909, Bt silver, burdett and company. CAJORI PREFACE The present text, the outgrowth of over twenty years' expe- rience with classes in geometry, carries the work from the simple elements necessary in the beginning class of the high school to the most advanced requirements of university prepa- ration. For many years we have followed in our own classes the plan here presented, with such modifications and improve- ments as experience has suggested. It is generally conceded that much of the pupil's difficulty in demonstration arises from his failure to grasp thoroughly and keep vividly in mind as separate and distinct statements, first, the exact data of the proposition, and, second, the precise fact to be established. To remove this stumbling-block, we have stated the hypothesis and conclusion separately for every theorem and corollary demonstrated. Through the " open " arrangement of the printed matter, we have sought to make each successive step stand out clearly ; and by so adjusting diagrams and text that in the course of any single demonstration it is unnecessary to turn the page, we have endeavored to avoid waste of effort on the part of the pupil. All the original exercises should be mastered with only such help as is given by the book itself. If a proposition has been found difficult for the average pupil, it has been broken up into a series of exercises in such sequence that the difficulties are presented one at a time and in natural order, the truth iV PREFACE of tht main propoiition being •stablished by means of these graded exercises. It has been our purpose to eliminate discouraging elements, to refresh the memory of the student before he begins inven- tive work, to arouse his interest and to inspire his confidence in his ability to discover hidden truths. We desire to express our acknowledgments for valuable counsel and suggestions to Professor F. N. Cole of Columbia University, to Professor Irving Stringham of the University of California, to Professor R. E. Gaines of Richmond College, Virginia, and to J. A. C. Chandler, LL.D., formerly Dean of Richmond Academy, Virginia. WALTER N. BUSH. JOHN B. CLARKE. San Francisco, Caxifoknia. PLAN AND SCOPE There are few students who fail to respond to the stimulus of original work in Geometry. The energy expended and enthusiasm displayed in the solution of exercises is in sharp contrast with their apathy toward the study of theorems in the text. That the student may arrive by the shortest path to the point where the real development of his mental power begins, where his interest in the independent solution of original exercises becomes an active part of his school life, it is of the first importance that the theorems of the text be so classified and demonstrated as to offer his attempts to master them the least resistance. It is of equal importance that these exer- cises should be so carefully graded as to stimulate and not discourage the pupil, and that some method of systematically attacking and solving • problems should be devised and pre- sented for his assistance. Nothing is presented in the pages that follow that has not stood the actual test of class-room experience for many years. In accordance with these essentials of a text for use in Geometry classes, we call special attention to the following features of this work: First. The classification of Definitions and Axioms. Second. The arrangement into groups of Theorems relating to the same topic. For example, theorems concerning isosceles triangles in the "Isosceles Triangle Group"; congruent tri- angles in the "Congruent Triangle Group"; comparison oi areas in the "Areal Katio Group." v vi PLAN AND SCOPE Third. The arrangement of original exercises, second in importance only to the grouping of theorems. The exercises are not only attached to the groups of theorems upon which their solution depends, but are graded according to their degree of difficulty. Fourth. The elimination of all theorems not essential to a clear understanding of the principles of Geometry; hence, a number of theorems found with their proofs in the usual text are given as exercises. Fifth. The compactness of each group and the simplicity and clearness of the demonstrations. With few exceptions, not more than three or four recitations are needed to master any one of the groups. Sixth. The helpful suggestions as to the method of solving original exercises. Seventh. A simple statement, with illustrations, of the close connection between Algebra and Geometry; classification of principles of the analogy between geometric and algebraic work into Indeterminate, Determinate, and Overde terminate groups. In the grouping of Theorems and in the arrangement and grading of Problems, there is constantly employed, as a valuable aid to the student, the principle of the Association of Similars. In the treatment of the Spherical Geometry, much space and time are saved by utilizing the common properties of the plane and sphere-surface ; and thus transferring, where pos- sible, the theorems and proofs of the plane to the sphere- surface. It is believed that this plan not only gives the pupil a clearer comprehension of the unity of the subject than he will otherwise obtain, but that it is also eminently suggestive of the unlimited possibilities of the extension of geometric truths to other surfaces. SUGGESTIONS TO TEACHERS Just as each brick in a building rests upon a brick below it, the whole superstructure standing upon a securely laid founda- tion, so the proof of each Theorem in Geometry rests upon the proof of the preceding Theorem, which in turn must finally depend upon the Definitions and Axioms. Definitions and Axioms, therefore, must be carefully studied and thoroughly understood. For example, to prove that the J3i sector of the Vertex Angle of an Isosceles Triangle is identical with the Altitude (Group IV, 1 6), we must know the definitions of the following words : Bisector, Vertex Angle, Isosceles Triangle, and Altitude. To understand the meaning of Altitude, we must know what a Perpendicular is, wdiich, in turn, requires that we know the meaning of a Bight Angle. The student, therefore, before attempting the proof of a Theorem, should be made to understand the meaning of each and every term found in the Hypothesis and Conclusion of the Theorem. To cultivate the habit of defining terms before using them, it affords the student valuable if not indispensable exercise to require of him frequent definitions of all the terms found in the Theorems and Corollaries, particularly of the first six groups. Beginners have great difficulty in keeping in mind the parts given in the Hypothesis distinct from those given in the Conclusion. During the process of demonstration they con- fuse what was to be proved with what was given. It will vii Tlii SUGGESTIONS TO TEACHERS relieve this confusion if they form the habit of marking the parts given in the Hypothesis by the usual symbols, i.e. : First. If lines are given parallel, by drawing the symbol Q across the lines. Second. If one line is greater than another, by drawing the symbol >. Third. If angles are given equal, by making the proper symbol at the vertex, just within the sides of the angle. Fourth. All parts mentioned in the Conclusion may be marked with an X • In drawing triangles, unless the triangle is Isosceles, Equi- lateral, or Right, it is well to adopt the following rule : First. Draw the base line. Second. Find approximately its mid-point. Third. Place the pencil or chalk to the left a convenient distance and erect an imaginary perpendicular. Fourth. At any suitable point on this perpendicular select the vertex of the vertex angle, from which draw the sides of the triangle. In this way the beginner may avoid the pitfalls of giving special proofs for the Isosceles, Equilateral, and Right Triangle that will not apply to the general Triangle. In drawing Parallelograms, unless a Rectangle, Square, or Rhombus be given in the Hypothesis, always construct a Rhomboid. At the beginning of the course, by written exercises and much blackboard work, familiarize the student with the use of symbols and abbreviations ; also with the freehand drawing of the altitudes of obtuse triangles. After the tirst month's work require frequent written exami- nations, insisting, as the course advances, that close attention be paid to the form in which written work is presented. Before the close of the course in Plane Geometry the student should be able to present his examination papers in the com- pact form found in the text. CONTENTS PAes Symbols : zi Abbreviations xii Definitions 1-9 Extension 1 Figures 2 Angles 4 The Circle and the Locus . ' 8 Axioms 9 Three Preliminary Theorems on Inequality .... 10 General Terms 11 Ten Easy Exercises in Geometrical Dramming .... 13 PLANE GEOMETRY I. The Group on Adjacent and Vertical Angles . . 19 11. The Parallel Group .23 III. The (2w-4) Right Angles Group .... 30 IV. The Group on Isosceles and Scalene Triangles . 35 V. The Group on Congruent Triangles .... 46 VI. The Group on Parallelograms 53 VII. The Group on Sum of Lines and Mid-joins . . 61 VIII. The Group on Points — Equidistant and Random . 69 Nine Illustrations of Elementary Principles of Loci . 74 IX. The Group on the Circle and its Related Lines . 78 X. The Group on Concurrent Lines of a Triangle . 91 Summary of Triangular Relations .... 99 XI. The Group on Measurement 100 Ratio and Proportion ....... 101 Method of Limits 101 ix X CONTENTS PAOK XII. The Group on Measurement of Anoles . . . Ill Hints to the Solution of Original Exercises . . . 120 Illustration of the Method of solving Original Problems 122 Theorems of Special Interest 127 Classification of Problems — Indeterminate, Determi- nate, and Overdeterminate 130 XIII. The Group on Areas of Rectangles and Other Polygons 138 XIV. The Pythagorean Group 149 XV. The Group on Similar Figures 160 XVI. The Group on Areal Ratios 176 XVII. The Group on Linear AppLiCATiiON of Proportion . 183 XVIII. The Group on Circumscribed and Inscribed Regular Polygons 210 XIX. The Group on the Area of the Circle . . . 220 XX. The Group on Concurrent Transversals and Normals 228 SOLID GEOMETRY XXL The Group on the Plane and its Related Lines . 233 XXII. The Group on Planal Angles 253 XXIII. The Group on the Prism and the Cylinder . . 266 XXIV. The Group on the Pyramid and the Cone . . . 287 XXV. The Group on the Sphere 311 XXVI. The Group on Geometry of the Sphere Surface; Briefly', Spherical Geometry .... 328 Correspondence between Plane and Spherical Geometry 331 Summary of Propositions Common to Plane and Spheri- cal Geometry 332 Notes and Biographical Sketches 347 Index . . 349 SYMBOLS 1. Letter points with capitals. 2. Letter lines with small letters. 3. Name angles, when there is italicized. no ambiguity, with small letters f^ (Greek word kentron) center in general. Ki Center of inscribed circle, i.e. in-center. Ke Center of escribed circle, i.e. ex-center. 4. Points Ka Center of circumscribed circle, i.e. circum-center. Ko Ortho-center (intersection of altitudes). ^9 Centroid or Center of Gravity of triangle (inter- section of medians). ■ Sides of a triangle : Use small letters corresponding to the capitals at the vertices opposite the respective sides ; a opposite A, etc. _L Perpendicular, or " is perpendicular to." 5. Lines Mid ± Mid-perpendicular. / Oblique. II Parallel. O Circumference or circle. r\ Arc. 6. Angle Rt. Z Angle in general. Right angle. 7. Triangle • A Rt. A Triangle in general. Right triangle. 4-side o 1 1 o Quadrilateral in general. 8. Quadri- lateral Parallelogram or rhomboid. Rectangle. Rhombus. ID Square. ■••• Therefore. Since or because. ,^ Similar. 9. Miscel- laneous ^ Congruent. Identical. > Is greater than. < Is less than. =: Approaches as a limit. For the plura il add the letter s. The cancella tion across a symbol means "not," e.g. : > means not greater than ; i - means not parallel. ABBREVIATIONS Alt. Alternate. 0pp. Opposite. Ax. Axiom. Prop. Proposition. Adj. Adjacent. Prob. Problem. Cone. Conclusion. Q.E.D. Quod erat demonstrandum Const. Construction. (which was to be proved). Cor. Corollary. Q.E.F. Quod erat faciendum Corr. Corresponding. (which was to be done). Def. Definition, Supp. Supplemental. Dem. Demonstration. Th. Theorem. Ex. Exercise. Vert. Vertical. Ext. Exterior. V. Vide (see). Hyp. Hypothesis. q.v. Quod vide (which see). Horn. Homologous. cf. Compare. Int. Interior. r. Radius. n-gon. Polygon. Groups, Theorems, and Corollaries are read as follows : II. 1. a means Group II, Theorem 1, Corollary a. zii THE ELEMENTS OF GEOMETRY DEFINITIONS EXTENSION The Definition of a mathematical term is its explanation in words familiar to the student. The test of a complete definition is that the subject and predicate may be interchanged without affecting the truth of the statement. Space extends about us on every side. Every material object occupies a portion of this space. This portion of space is called a geometrical solid or simply a Solid. The only properties of the solid with which geometry is concerned are its form and size, and its position with reference to other solids. A boundary of a solid is called its Surface. It is no part of the solid, and therefore has but two dimensions : length and breadth. A boundary of a surface is called a Line. It is no part of the surface, and has therefore but one dimension : length. A boundary of a line is called a Point. A point has no dimension, but position only. A point, line, or surface that divides any magnitude into two equal parts is called the Bisector of that magnitude. Any definite portion of a line is called a Line-segment. A Straight Line is a line that lies evenly between its extreme points; that is, if the ends of one segment may be placed upon the ends of a second segment, the segments must coincide throughout their whole extent. 1 2 THE ELEMENTS OF GEOMETRY A straight line conuecting any two points is called the Join of thos^ pojots. A Broken Line is a series of joins, any two consecutive joini having one point in common. A Curved Line is a line such that no segment of it is straight. Concurrent Lines are lines passing through the same point. A Transversal is any line intersecting a number of other lines. Note. — in the text the word line is used to mean a straight line. FIGURES A Figure or Complex is any collection of points, lines, or points and lines. Similar figures are figures having the same shape. Equal figures are figures having the same size. Congruent figures are figures having the same shape and the same size. The Test of Congruency is that one figure may be plaqed on the other so that every part of the first will coincide with the corresponding part of the second. Two figures thus placed are said to be in Coincident Super- position. A Plane is a surface such that if any two of its points be joined by a straight line, this line must lie wholly within the surface. A Plane Figure is one all the parts of which lie in the same plane. Note. — All figures hereafter defined are assumed to be plane figures. A Polygon is a portion of a plane bounded by a closed broken line called its Perimeter. When only the form is considered, the word polygon is fre- quently used to mean the perimeter of the polygon. DEFINITIONS 3 A three-sided polygon is called a Triangle. A four-sided polygon is called a Quadrilateral or 4-8ide. A five-sided polygon is called a Pentagon. A six-sided polygon is called a Hexagon. A seven-sided polygon is called a Heptagon. An eight-sided polygon is called an Octagon. A nine-sided polygon is called a Nonagon. A ten-sided polygon is called a Decagon. A twelve-sided polygon is called a Dodecagon. A fifteen-sided polygon is called a Pentadecagon. The Vertices of a polygon are the points in which its consecu- tive sides meet. A Diagonal of a polygon is the join of any two non-consecu- tive vertices. A Regular Polygon is a polygon whose angles are equal and whose sides are equal. Ex. 1. What is the test of a complete definition ? Ex. 2. Apply this test to the definition of a line-segment. Ex. 3. Define (a) Bisector. (&) Line, (c) JoiU' (d^ Concurrent lines, (e) Transversal. Ex. 4. Define a straight line. Can you place two equal portions of a barrel hoop in such a way that the ends of one will coincide with the ends of the other, but the portions (or segments) themselves will not coincide ? Can you place them so that they will coincide ? Since, then, equal segments of a curve may or may not be made to coincide, what is the word to emphasize in the definition of a straight line ? Ex. 5. Draw three concurrent lines. Ex. 6. Draw three non-concurrent lines. Ex. 7. Draw a transversal to two lines. ^x. 8. What are congruent figures ? Ex, 9. What is the test of congruency ? Ex. 10. Give illustrations of equal, similar, and congruent figured. Ex. 11. What are the two dimensions of a surface ? Ex. 12. Define a plane. Ex. 13. Using your ruler as a straightedge, show that the top of your desk is a plane. 4 THE ELEMENTS OF GEOMETRY ANQLBB An Angle is a figure formed bj two lines that meet ; the lints being called the Sides of the angle. The point of meeting of the sides of the angle is called the Vertex. The usual Method of Reading an angle is to read the three letters on the sides, placing the letter at the vertex between the other two. If there is no ambiguity, the angle may be read by the letter at the vertex. If, when the sides of an angle are produced, all the angles formed are equal, each angle is called a Right Angle, and the lines are said to be Perpendicular to each other. If a line bisects a second line and is also perpendicular to it, the first line is called the Mid-normal q or Mid-perpendicular to the second. ABC is an angle. Its vertex is B. If the side AB is produced to E and the side CB is produced to F, and if angles ABC^ ^ \.~ e CBE, EBF, and ABF are all equal, then they are right angles, and the line CF is perpendicular to the line AE. If the line CF also bisects the line AE, it is the Mid- normal to AE. j* By the distance of a point from a line is meant the perpen- dicular distance ; by the distance between two lines, the per- pendicular distance. Classes of Angles (a) As to their Algebraic Sign An angle may be considered as generated by the revolution of one line about a fixed point in a second line. The rotating line, when it comes to rest, is called the Termi- nal Line. The fixed line is called the Initial Line. DEFINITIONS If the rotating line moves anti-clock- wise, the angle generated we call Positive ; if clockwise, Negative. If the rotating line, moving from a position coincident with the initial line, complete a revolution, it generates four right angles, or a Perigon ; if half a revo- lution, it generates two right angles, or a Straight Angle. Note. —The size of an angle does not depend upon the length of its sides. Note. — In finding the sum, Z.a -f- Z&, of 2/:!, place the initial line and vertex of Z& on the terminal line and vertex of Aa. Then generate the Z & by the rotation indicated by its sign. Then the angle between the initial line of Za and the terminal line of Z & is Z a + Z 6. The Relative Direction of one line with respect to another is the angle that the first line makes with the second, both the size and sign of the angle being con- sidered. The direction of rotation of the line generating the angle will be considered positive unless the contrary is stated. Relative direction will be the only direc- tion considered in this book; absolute direction will not be discussed. (h) As to Size An Acute angle is an angle less than a right angle. An Obtuse angle is an angle greater than a right angle. An obtuse angle equal to two right angles is called a Straight angle. An Oblique angle is any angle that is not a right angle. 6 THE ELEMENTS OF GEOMETRY Wheu two angles are both acute or both obtuse, they are said to be of the Same Kind. If the sum of two angles equals one right angle, they are called Complemental angles. If the sum of two angles equals two right angles, they are called Supplemental angles. , (c) As to Location * Adjacent angles are angles that have a common side and a common vertex. Vertical angles are the alternate angles formed by two lines that cross each other. Angles formed by a Transversal with Two Other Lines Angles within the two lines crossed by a transversal are called Interior angles ; angles without, Exterior angles. Non-adjacent angles on the same side of the transversal are called Corresponding angles. There are three classes of corresponding angles : Corresponding Interior, Corresponding Exterior, and Corresponding Exterior- interior angles. Corresponding exterior-interior angles are Corresponding Non-adjacent angles, one of which is interior and the other exterior. Alternate angles are angles on opposite sides of the trans- versal that do not have the same vertex. There are three classes of alternate angles : Alternate Interior, Alternate Exterior, and Alternate Exterior-interior angles. Alternate exterior-interior angles are alternate angles, one exterior and the other interior, but not having a common vertex. Point out in the adjoining figure the following angles : acute, obtuse, oblique, supplemental, ad- jacent, opposite, of the same kind, corr. ext., corr. int., corr. ext.-int., alt. ext., alt. int., alt. ext. -int. DEFINITIONS 7 A Postulate is a construction admitted to be possible. Post. There may always be drawn a pair of lines that make equal corresponding exterior-interior angles with any trans- versal whatever cutting them. Two lines are Parallel when, if cut by any transversal what- ever, the corresponding exterior-interior angles are equal. Ex. 14. The ruler can be so placed as to lie wholly on the stovepipe. Why, then, is not the stovepipe a plane ? Ex. 15. Show that any two parallels have the same direction with re- spect to any transversal. Ex. 16. Define: (a) Angle. (6) Right angle, (c) Perpendiculars. (d) Perpendicular bisector (or mid-normal). (e) Oblique angles. (/) Angles of the same kind. Ex. 17. What is the complement of two fifths of a right angle ? Ex. 18. What is the supplement of two fifths of a right angle ? Ex. 19. Name the angles formed by the clock hands : (a) 10 minutes after three o'clock, (c) 20 minutes after three o'clock. (6) 15 minutes after three o'clock, (d) 30 minutes after three o'clock. Ex. 20. Draw two oblique angles ; draw two angles of the same kind. Ex. 21. Draw two angles that are : (a) Adjacent. (6) Adjacent and complemental. (c) Adjacent and supplemental, (d) Non-adjacent and supplemental. Ex. 22. An angle is one fifth its complement. What is the value of the angle ? Ex. 23. Two angles are equal and at the same time complemental. What is their value ? Ex. 24. Two angles are equal and at the same time supplemental. What is their value ? Ex. 25. Draw two angles that have a common side but not a common vertex. Ex. 26. Draw two angles that have a common vertex but not a common side. Ex. 27. Draw two lines so that a transversal crossing them makes : (a) Corresponding exterior-interior angles equal. (&) Corresponding interior angles equal, (c) All the angles equal. Ex. 28. In case (a) of the preceding question what other angles are equal ? 8 THE ELEMENTS OF GEOMETRV THE CIRCLE AND THE LOCUS A Circle is a portion of a plane bounded by a curved line, called the Circumference, every point of which is equidistant from a point called the Center. Any line from the center to the circumference is called a Radius. Ciiclefi having the same center are Concentric Circles. A Locus is a line or a complex, all points of which possess a common property that does not belong to any point without this line or complex. E.g. it is the common property of every point in the circumference of a circle that is a radial dis- tance from the center; the circumference is the locus of all points that are a radial distance from the center. Ex. 29. Do the two lines meet ? Ex. 30. In case (6) what other angles are equal ? Ex. 31. Do the two lines meet ? Ex. 32. In case (c) do the two lines meet ? The angles in case (c) are all of what kind ? Ex. 33. How may an angle be considered to be generated ? Ex. 34. Define a negative angle. Ex. 35. The hands of the clock are together at twelve o'clock. If the hour hand is stationary, what is the algebraic sign of the angle formed by the hands at ten minutes after twelve ? Ex. 36. If the hour hand is stationary and the minute hand is moved to the left, what is the algebraic sign of the angle formed at ten minutes before twelve ? Ex. 37. What is the sign of the angle generated by a line from the observer to the moon from moonrise to moonset ? Ex. 38. To mariners what is the fixed or known line ? What instru- ment on every sliip serves to determine this line ? Ex. 39. If a ship is sailing southeast, what angle does its course make with this known line ? Ex. 40. What, then, is the relative direction of the ship with respect to this fixed line ? Note. — Answer all questions concerning angles in terms of a right angle until the word degree has been defined. Ex. 41. What angle is one fifth its supplement ? DEFINITIONS 9 The truths of geometry are expressed by its definitions, axioms, and theorems. An Axiom is a statement, the truth of which is assumed. AXIOMS 1. Things equal to the same thing or equal things are equal to each other. 2. If equals be added to or subtracted from equals, the results will be equal. 3. If equals be multiplied or divided by equals, the results will be equal. 4. The whole equals the sum of its parts. Direct Inference : The whole is greater than any of its parts. 5. The intersection of two lines, straight or curved, fixes the position of a point. E.g. the intersection of the latitude and longitude of a ship at sea determines its position. 6. Two points fix the position of a line. E.g. if a railroad extending in a straight line passes two stations whose positions are known, the railroad is also determined or fixed in position. 7. A point and the direction of -a straight line determine its position. E.g. if a railroad extends northwest and passes a known station, the position of the railroad is known. a. From a point without a line but one perpendicular can be drawn to the line. b. At a point in a line but one perpendicular can be drawn to the line. 8. Through a point one line and only one can be drawn parallel to a given line. 9. Any figure may be transferred from one position in space to any other without change of size or shape. 10. If there be but one x and one y, then, from the fact that X is y, it necessarily follows that y is x. lU THE ELEMENTS OF GEOMETRY THREE PRELIMINARY THEOREMS ON INEQUALITY These propositions are given in many texts as axioms. They are proved, however, in the leading treatises on algebra. 1. If unequals are added to unequals in the same sense, the results will he unequal in the same sense, Dem. : Suppose a > 6, and c> e. To prove a-^c>h-\-e. a'>h. ,', a = h-\- some quantity, say x. c> c. .\ c — e-\- some quantity, say y. Thus a=-h + Xy c = e + y. .'. a + c = 5 + e + a; + 2^. (Ax. 2.) That is, a -f c > & + e. Q.E.D. 2. If equals are added to unequals, the results are unequal in the same sense, Dem. : Suppose a > h, and c = e To prove a -f c > 6 + e. a > 6. .*. a = & + some quantity, say x. Then a = & -}- », and c = e. .\ a + c = b •^e-\'X. (Ax. 2.) .*. a-f c> &4-e. Q.E.D. 3. If equals are subtracted from unequals, the results are unequal in the same sense. The proof of this proposition is precisely similar to that of Theorem. 2. Ex. 42. What is the supplement of the angle between the hands of a clock at five o'clock ? Ex. 43. What is the complement of 'the angle in the preceding ques- tion ? (v. Negative angles.) DEFINITIONS 11 GENERAL TERMS A Theorem is a statement to be proved. It consists of two parts : The Hypothesis (Hyp.), or supposition or premise. The Conclusion (Cone), or what is asserted to follow from the hypothesis. A Proof is a course of reasoning by which the truth of a theorem is established. The Converse of a theorem is obtained by interchanging the hypothesis and conclusion of the original theorem. Theorem : If ^ is B, then G is E. Converse : If (7 is E, then A is B. Note. — The converse is sometimes called the indirect theorem. The Contradictory of a theorem is true if the theorem is false, and vice versa. Theorem : If ^ is B, then C is E. Contradictory : If A is B, then C is not E. The Opposite of a theorem is obtained by making both the hypothesis and conclusion negative. Theorem : If A is B, then C is E. Opposite : If ^ is not B, then C is not E. A Reciprocal theorem is formed by replacing, when possible, in the original theorem, the words " point by line," " line by point," " angles of a triangle by the opposite sides of the tri- angle," " sides of a triangle by the opposite angles of the tri- angle," "opposite angles of a 4-side" by "the opposite sides of a 4-side," etc. Note. — For every statement in a proof a reason must be given. This reason must be : 1. A hypothesis. 3. A definition. 2. A construction. 4. An axiom. 5. A previously established theorem or corollary. A Corollary is a subordinate statement deduced from, or .suggested by, the main statement or its proof. 12 THE ELFMENTS OF GEOMETRY A iH-oblem requires the construction of a geometric figure that will satisfy given conditions. The Solution of a Problem consists of four parts : First: The Analysis, or course of reasoning by which the. method of constructing the required figure is discovered or rediscovered. Note. — The analysis of problems is explained in full under the article •' Helps to the Solution of Original Problems," Group XII. Second: The Construction of a figure after the method has been discovered. Third: The Proof that the figure satisfies the given condi- tions. Fourth: The Discussion of the changes in the number of figures that will satisfy the given conditions, made by a change in the size of the given magnitudes, in their relative position, or in both. A Proposition is a general terra applying to theorems or problems. A Scholium is a remark upon a particular feature of a propo- sition. A Lemma is a theorem introduced merely to be used in the proof of one immediately following. Plane Geometry is that branch of mathematics in which are considered the properties of magnitudes lying on the same plane. Suggestions for Class Work at Blackboard In lettering a fifjure avoid the use of letters that have the same sound ; as B, Z>, P, and T; JWand N, etc. Place uip of pointer within the figure you are to read. To move it from point to point is confusing. If reading a line-segment, place tip about the middle of the segment ; if an angle, place it near the vertex ; if a polygon of any kind, within the polygon. In drawing triangles make them, unless otherwise directed, scalene, with the angle at A greater than the angle at .B, and C the vertex angle. TEN EASY EXERCISES IN GEOMETRICAL DRAWING 13 I II J VI IX X. III VJI IV VIIl TEN EASY EXERCISES IN GEOMETRICAL DRAWING These problems are introduced at this point to familiarize the student with the use of the ruler and compasses and with geometrical terms. The constructions are simple, and serve to illustrate some of the practical applications of geometry. The student will observe that the constructions in Probs. II, V, VI, IX, and X are direct applications of the constructions in Prob. I, and that the others are also intimately connected with it. Thus Prob. I may be considered the string from which the other problems are sus- pended. This de- pendence is shown by the adjacent diagram. The proofs for these constructions are to be given as soon as the necessary theorems have been established. Eeferences to the fol- lowing problems are made in footnotes and exercises attached to these theorems. Summary Problem I. (a) Bisect a given line-segment, (b) Erect a mid -L to it. Problem II. Bisect a given arc. Problem III. Bisect a given angle. Problem IV. Ti'isect a given right angle. Problem V. Erect a A. to a given line at a given point in the line. Problem VI. Draw a ± to a given line from, a given point luithoiit the line. Problem VII. Inscribe a O in a given triangle. Problem VIII. Escribe a O to a given triangle. Problem IX. Circumscribe a O to a given triangle. Problem X. Find the center of a given O. 14 THE ELEMENTS OF GEOMETRY Problem I. (a) To bisect a given line-sec/vient. (b) To erect a mid Lto a given line-segment Given. The line-segment AB. j Required, (a) To bisect AB. (b) To ! erect a mid J. to AB. I A d B Const. With ^ as a center and a r = AB, ! describe arcs above and belpw AB. \ With 5 as a center and the same r, de- ^1^ scribe arcs above and below AB. E Let these arcs intersect above the line in C; below in E. Draw the join CE. Let it intersect AB in 0. Then (a) bisects AB. (b) CE is the mid ± to AB. Q.E.F. M\ Problem II. To bisect a given arc. Given. The arc AB, / Required. To bisect the arc AB. Const. Draw the chord AB. Construct the mid J_ to this chord by Prob. T. Let this mid _L intersect the arc AB in M. Then M bisects the arc AB. Q.E.F. Ex. 44. In what line do you find all the house* that are one mile from the county courthouse ? • • TEN EASY EXERCISES IN GEOMETRICAL DRAWING 15 Problem III. To hisect an angle. j,^ Given. Z MAL. .^rtlTrl™. \ Required. To bisect Z MAL. ^""""""^-^.i.,^^ Const. With ^ as a center and any r, describe an arc BC. Bisect the arc BC by Prob. II. The join of A and the mid-point of arc BC bisects /.MAL. Q.E.F Problem IV. To trisect a right angle. c Given. The rt. Z A. Required. To trisect rt. Z A. ^^ /^^ / V H Const. Lay off on AH any line-segment AB. With ^ as a center and AB as a r, describe an arc. With J5 as a center and the same r, describe an arc. Let these two arcs intersect at E. Draw EA and bisect Z BAE by Prob. III. ZBAT=Z. TAE = Z.EAC. Q.E.F. Problem V. To erect a perpendicular to a line at a point in the line. . Given. The line ^5 and P in ^5. ^j Required. To erect a X to AB j atP. A ^G P \i 'b Const. Lay off P(7=PQ. Construct the mid X PF to CQ by Prob. I. PP is the required X. Q.E.F. 10 THE ELEMENTS OF GEOMETRY Problem VI. To draw a perpendicular to a line from a point without the line. Given. The line ^JB and ^ the point P without AB. Required. To draw a ± from P to AB. a~^ fi X> b Const. With P as a center and any r > the distance from P to AB, describe an arc intersecting AB in C and Q. Construct the mid _L PH to the line-segment CQ by Prob. I. PH is the required ±. Q.E.F. Problem VII. To inscribe * a circle in a triangle. G Given. The A.4JBC. Required. To inscribe a O in A ABC. "^ A Const. Bisect Z. A and Z jB by Prob. III. Let the bisectors intersect at some point K. Construct a J. from ^ to ^B by Prob. VI. With ^ as a center and this ± as a r, describe a O. This will be inscribed in A ABC. Q.E.F. Ex. 45. What is the locus of a point one mile from a given point ? Ex. 46. What is the locus of a point that is h distant from a given point F7 Ex. 47. What is a proof or demonstration ? Ex. 48. Into what two parts may a theorem be separated ? A problem ? Ex. 49. Name the five classes of reasons, one of which must be given for every statement that is made in the proof. * A O is inscrioed in a A when the sides of the A are tangent to the O (i.e. touch the O in but one point). See p. 78. TEN EASY EXERCISES IN GEOMETlilCAL DRAWING 17 Problem VIII. To escribe ^ a circle to a ti'iangle. \P Given. The A ABQ. Required. To escribe a O to the b.ABC. Const. Produce the sides h and a. Bisect Z BAE and Z FBA by Prob. III. Let the bisectors intersect in some point K. Construct a _L from if to a (or to IS) produced. With ^ as a center and this _L as a ?', describe a O. This O will be escribed to A ABC. Q.E.F. Problem IX. To pass a circle through three points. Given. The three points, A, B, and C, / \ \ not coUinear. / -'^''V^ if:-""\ ! Required. To pass a O through A, B, \ ; \ / and a a\ r 7B I ,,' Const. Draw the joins AB and JBC. Construct the mid Js to AB and BC by Prob. I. Let these mid Js intersect at a point K. With KsiS a center and a r = KA, KB, or KC, describe a O. This O passes through A, B, and (7. Q.E.F. Note. — If A, B, and C are connected, the above G is said to be cir- cumscribed to A ABC. 1 To escribe a circle is to draw it tangent %o one side of a triangle and to the other two sides produced. 18 THE ELEMEIJTS OF GEOMETRY Problem X. To find the center of a given circle. Given. The O K, Required. To find its center. Const. Take any three points in the O, as A, B, and C. Draw the joins of any two of these, as AB and BC. Construct the mid Js to these joins by Prob. I. Let these mid J§ intersect at some point K. K is the center required. Q.E.F. Note. — Prob. X, after the three points have been selected, is evidently identical with Prob. IX. Note. — The proofs for the solution of these problems will be found as follows : — I.,V., VI.,p. 72. IL, p. 79. III., p. 112. IV., p. 49, with Def. p. 30. VII., p. 92, X., 1, a. VIII., p. 93, Scb. IX. and X., p. 94. Ex. 50. What is the opposite of a theorem ? Ex. 51. State the opposite of Group I, Theorem 2. PLANE GEOMETRY I. THE GROUP ON ADJACENT AND VERTICAL ANGLES PROPOSITIONS I. 1. If from the same point in a line any number of lines are drawn on the same side of the line, the sum of the successive angles formed equals two right angles. Hyp. If from a point in /q AB, PC and PE are drawn on the same side of AB, ^ forming successively ZBPC, Z CPE, and ZEPA, X P Cone. : then Z BPC + Z CPE + Z EPA = 2 rt. A. Dem. If PB rotates to PA, it generates a straight Z. (Def. of a straight Z.) In this process of rotation, PB generates successively Z BPC, ZCPE,?iXi^ZEPA. .'. Z BPC + Z CPE + Z EPA = 2 rt. A. (Ax. 4.) Q.E.D. I. 1. ScH. If the angles are on both sides of the line, their sum equals four right angles. Ex. 1. Z J. is the supplement of Z-B. They bear to each other the relation of 4 to 7. What part of a rt, Z is each ? Ex. 2. If Z. AMC, adjacent to Z CMB, is four thirds of a rt. Z, and Z CMB is four fifths of a rt. Z, are their exterior sides in the same straight line ? Ex. 3. Three successive angles about a point on one side of a straight line are in the ratio of the numbers 2, 3, and 5. What is the value of each angle ? 19 20 THE ELEMENTS OF GEOMETRY I. 2. If two angles are adjacent and have their ex- terior sides in the same straight line, they are supple- mental. Hyp. UZAPCandZCPB are adjacent, and if PB and PA are in the same straight line. > A P B Cone. : then Z CPA + Z CPB = 2 rt. A. Dem. If PB rotates to PAy it generates a straight Z. (Def. of a straight Z.) In this process of rotation, PB generates successively Z BPC and Z OPA .-. Z ^PC + Z CP^ = 2 rt. A. (Ax. 4.) Q.E.D. Note. — The above proposition, which is a slight modification of Theorem 1, is introduced to assist the pupil to a clear statement and understanding of its important converse, which directly succeeds. I. 3. If tivo angles are adjacent aiid supplemental, their exterior sides form, the same straight line. Hyp. If A ACL and x' ALCB are supplemental and adjacent, J" ^, n t Cone: then AG is in same straight line with CB. Dem. If CB is not in the same straight line with AC, draw CT that is in the same straight line. Then A LCT ■\-AACL = 2it A. (I. 1.) But ALCB-\-AACL = 2xtA. (Hyp.) .• ALCT=ALCB. (Ax. 1.) .-. CT falls on CB. (Ax. 7.) But CT was drawn in the same straight line with AC. .'. CB and CA are in the same straight line. Q.E.D I. ADJACENT AND VERTICAL ANGLES 21 I. 4. If two straight lines intersect, the vertical aiigles formed are equal. A Ryp. If AB intersects CE, Cone. : then Z APC = Z EPB. Dem. Z CPA + Z APE = 2 rt. A Z ^PS + Z ^P£; = 2 rt. A. (1.2.) (I. 2.) Z CP^ + Z ^P^ = Z ^P5 + Z APE. (Ax. 1.) .-. Z CPA = Z ^P^. (Ax. 2.) Q.E.D. The reference number only is given when the reason theorem belongs to the same group as the theorem in course of demonstration. Ex. 4. The hands of a clock at three o'clock form an angle equal to a rt. Z. This angle will fit the space about the pivot of the hands exactly four times. How many times is two thirds of a rt. Z contained in a peri- gon ? Four thirds of a rt. Z ? Four fifths of a rt. Z ? Ex. 5. The bisectors of two supplemental adjacent angles form a rt. Z. Ex. 6. If the bisectors of two adjacent A are ± to each other, the A are supplementary. Ex. 7. The bisector of an angle is, when produced, the bisector of its vertical angle. Ex. 8. The bisectors of a pair of vertical angles form the same straight line. Ex. 9. The bisectors of two pairs of vertical angles are perpendicular to each other. Ex. 10, If the sides of ZL CM are perpendicular to the sides of ZACB, prove that the angles are supplemental. Ex. 11. If through a point, A^ four straight lines, AB, AC, AE, AF, are drawn so that ZBAC= ZEAF, and ZBAF= ZCAE, then FAC and BAE are straight lines. Ex. 12. What relation does Ex. 11 bear to I, 4 ? 22 THE ELEMENTS OF QEOMETKY I. SUMMARY OF PROPOSITIONS IN GROUP ON ADJA- CENT AND VERTICAL ANGLES 1. If from the same point in a line any number of lines are draion on the same side of the line, the sum of the successive angles formed equals two right angles. ScH. If the angles are on both sides of the line, their sum equals four right angles. 2. If iivo angles are adjacent and have their exterior sides in the same straight line, they are supj^lemental. 3. If tivo angles are adjacent and supplemental, their exterior sides form the same straight line. 4. If two straight lines intersect, the vertical angles formed are, equal. 11. THE PARALLEL GROUP DEFINITIONS Two lines are said to be Parallel when they are so situated that if cut by any transversal the corresponding exterior- interior angles are equal. Two Lines Perpendicular to a Third. Direct inferences from the definition of parallels : (1) If two lines are perpendicular to a third, they are parallel. (2) A line perpendicular to one of two parallels is per- pendicular to the other. 11. 1. If two parallels are crossed hy a third line, the alternate interior angles are equal, F/ Hyp. If AB and CE are II, and are crossed by the transversal FK, c H/ Cone. : then AAGH^A GHE, Dem. Z FGB = Z GHE. (Def. of lis.) ZFGB=^ZAGH. [If two straight lines intersect, the vert. /4 formed are =.] (1. 4.) .-. Z AQH = Z GHE, (Ax. 1.) QE.D. 24 THE ELEMENTS OF GEOMETRY II. la. If two parallels are crossed by a third line, the alternate exterior angles are equal. Hyp. If AB and CE are II, and are crossed by the transversal KF K' Cone. ; then Z AGF = Z EHK Dem. ZAGF = ZBOH. [If two straight lines intersect, the vert. A formed are =.] (1. 4.) ZBGH^^ZGHC. (ILL) ZGHC = ZEHK. (L4.) .-. Z AGF = Z EHK (Ax. 1.) Q.E.D. II. 2. If two parallels are crossed hy a third line, the corresponding interior angles are supplermntaL Ff Hyp. If AB and CE are II, and are crossed by the transversal KFy K' Cone. : then Z BGII+ Z GHE = 2 rt. A. Dem. Z BGH + Z BGF = 2 rt. A. (1.) [If from the same point in a line any number of lines are drawn on the same side of the line, the sum of the successive A formed = 2 rt. A.~\ (I. 1.) Z BGF = Z GHE. (Def . of lU.) Substituting Z GHE for its equal Z BGFj we have, from (1), Z BGH+ Z GHE = 2 rt. A. Q.E.D. II. PARALLELS 25 II. 2 a. If two parallels are crossed hy a third line, the corresponding exterior angles are supplemental. Hyp. If AB and CE are II, and are crossed by the transversal KFy W Cone. : then AAGF-^Z. CHK = 2 rt. A Dam. ZAGF + ZAGH=2vt.A. [If from the same point in a line any number of lines, etc.] (I. 1.) Z AGH = Z CHK (Def . of lis.) .•. as in II. 2, Z AGF + ZCHK^ 2 it. A. Q.E.D. II. 3. If tivo lines are crossed hy a third so as to make the alternate interior angles equal, the lines are parallel Hyp. If TL crosses AB and CE so that the alt. int. AAVS and VSE are equal, Cone. : then AB II CE. Dem. ZAVS==ZTVB. [If two straight lines intersect, the vert. A, etc.] (I. 4.) ZAVS = ZVSE. (Hyp.) .-. Z TVB = Z VSE. (Ax. 1.) .-. AB II CE. (Def. of lis.) Q.E.D. 26 THE ELEMENTS OF GEOMETRY II. 3 a. If two lines are crossed hy a third so as to make the alternate exterior angles equal, the lines -are parallel, ^t Hyp. If AB and CE are crossed by the trans- versal TL, and if the alt. ext. A TVB and CSL are "o ^7" F equal, L/^ Cone. : then AB II CE. Dem. Z CSL = Z VSE. [If two straight lines intersect, the vert. Af etc.] (I. 4.) Z CSL = Z TVB. (Hyp.) .-. Z VSE = Z TFB. (Ax. 1.) .-. AB II Cii;. (Dei. of lis.) Q.E.D. II. 4. If two lines are crossed hy a third so as to make (1) the corresponding interior angles supplemental, or (2) the co7Tesponding exterior angles supp)lemental, the lines are parallel. t Hyp. (1) If the corr. int. ABVS and VSE are sup- plemental, Cone. : then AB 11 CE. Dem. Z.BVS + ZVSE = 2vt.A. (Hyp.) Z.BVS + ATVB = 2v%. A. [If from the same point in a line any number of lines, etc] (T. 1.) .-. Z VSE = Z TVB. (Ax. 1.) .-. AB II CE. (Def. of lis.) Q.E.D. sx JS n. PARALLELS 27 Hyp. (2) If the corr. ext. A TVB and LSE are supplemental, Cone. : then AB II CE. Dem. Similar to that of II. 4 (1). (Let the pupil supply the proof.) II. 5. (1) If two arigles have their sides respectively parallel, they are equal, if of the same kind. Fig. I. Fig. II. Hyp. (1) If Z.A and Z B are of the same kind, and if ACWBF A\ \ \ \ E E Ji\ \ \ \ K\ H and AE 11 BH, \ ''S H Cone. : then ZA = ZB in Fig. I and Fig. II. Dem. Fig. I. Extend AG to cross BH at K. ZA = Z HKL. (Def . of lis.) AB = A HKL. (Def. of lis.) ,\AA = Z.B. (Ax. 1.) Q.E.D. Note. — In Fig. I the sides of the angles extend in the same direction from the vertices. Dem. Fig. II. Extend AC to cross BH at K, AA = AAKH. (11. 1.) Z.B = Z.AKH, (Def. of lis.) .-. Z ^ = Z 5. (Ax. 1.) Q.E.D. Note. — In Fig. II the sides of the angles extend in opposite directions from the vertices. 28 THE ELEMENTS OF GEOMETRY II. 5. (2) If two angles have their sides respectively parallel, they are 8up>plemental, if of different kinds. Hyp. U ZAsLudZB are of different kinds, and if AF W BC and AE II HB, Cone. : then ZAis supplemental to Z B. Dem. Extend AE to intersect BC at T. ZAis supplemental to Z ATC. (II. 2.) ZATC = ZB. (Def. of lis.) .'. ZAis supplemental to Z B. (Ax. 1.) Q.E.D. Note. — Two sides ot ZA and Z.B extend in the same direction from the vertices A and B^ while the other two sides, BH and AE^ extend in opposite directions from A and B. The theorem may therefore be stated thus : If two angles have their sides respectively parallel, and two extend in the same, two in the opposite directions from their vertices, they are supplemental. Ex. 1. If two parallels be crossed by a transversal, and any angle is a right angle, what is the value of each of the others ? Ex. 2. Lines II to the same line are || to each other. Ex. 3. Show that Theorem .3 is the converse of Theorem 1. Ex. 4. Prove that a line parallel to the base of a triangle cuts off a A whose angles are respectively equal to the angles of the original A. Ex. 5. The bisectors of a pair of alt. int. A of parallels are paralleL . Ex. 6. The bisectors of a pair of corresponding interior angles are perpendicular to each other, Ex. 7. The parallel to the base of an isoangular triangle cuts off another isoangular triangle. n. PARALLELS 29 n. SUMMARY OP PROPOSITIONS IN PARALLEL GROUP 1. If two parallels are crossed by a third line, the alternate interior angles are equal, a If two parallels are crossed by a third line, the alternate exterior angles are equal. 2. If two parallels are crossed by a third line, the corresponding interior angles are supplemental. a If tivo parallels are crossed by a third line, the corresponding exterior angles are sup- plemental. 3. If two lines are crossed by a third so as to make the alternate i?iterior angles equal, the lines are parallel. a If two lines are crossed by a third so as to make the alternate exterior angles equal, the lines are parallel. 4. If two lines are crossed by a third so as to make : (1) the corresponding interior angles supplemental, or (2) the corresponding exterior angles supplemental, the lines are parallel. 5. (1) If two angles have their sides respectively paral- lel, they are equal, if of the same kind. (2) If two angles have their sides respectively paral- lel, they are supplemental, if of different kinds. III. THE (2w-4) RIGHT ANGLES GROUP Briefly: The (2^ — 4) Group DEFINITIONS A triangle is a figure formed by the intersection of three lines not passing through the same point. If no two sides of a triangle be equal, the triangle is said to be Scalene (limping). If two sides of a triangle be equal, the triangle is said to be Isosceles. If two angles of a triangle be equal, the triangle is said to be Isoangular. If three sides be equal, the triangle is said to be Equilateral. If three angles be equal, the triangle is said to be Equiangular. If three angles be acute, the triangle is called an Acute Tri- angle. If one angle be obtuse, the triangle is called an Obtuse Tri- angle. If one angle be right, the triangle is called a Right Triangle. In a right triangle the side opposite the right angle is the Hypotenuse. An Altitude of a triangle is a perpendicular from a vertex to the opposite side. This side is called the Base. A Median of a triangle is a line from a vertex to the middle point of the opposite side. The Vertex Angle of a triangle is the angle opposite the base. The Exterior Angle of a triangle is the angle between one side and a second side produced. 30 III. THE (271-4) RIGHT ANGLES GROUP 31 III. 1. If a figure is a triangle, the sum of the inte" rior angles equals two right angles. Hyp. If /X /^ the figure / N. / ABC is a / \^ / triangle, / \. / Z Jii 1/ A £ £ Cone. : then ZA^/LC + Z ABC = 2 rt. A, Dam. Produce AB to any point E. Draw BF || AC, Then ZA = Z FBE. (Del of lie.) ZC^ZFBC. [If two lis are crossed by a third line, the alt. int. A, etc.] (II. 1.) But Z FBE + Z FBC + Z ABC = 2 vt A. [If from the same point in a line any number of lines are drawn on the same side of the line, the sum, etc.] (I. 1.) .-. ZA + ZC + ZABC = 2i±Z. Q.E.D. III. 1 a. One interior angle of a triangle is the supplement of the sum of the second and third angles. III. 1 h. In a right triangle, either acute angle is the complement of the other acute angle. Ex. 1. In a triangle Za = 2Zb, Zb = SZc. What is the value of Z rt, Z 6, and Zc? Ex. 2. In a triangle Za + Zb = ^rtZ, Za-Zb=i it. Z. What is the value of Z a, Z b, and Zc? Ex. 3. What is. the value of each acute Z in an isoangular right A ? Ex. 4. If 2 A of one A are equal respectively to 2 zl of another, the 3d A are equal. Ex. 5. The vertex Z of an isoangular triangle is f rt. Z. Find the value of the Z between the base and an altitude on one leg. THE ELEMENTS OF GEOMETRY III. 2. Any exterior angle of a triangle equals the sum of the two non-adjacent interior angles. Hyp. If the figure ABC is a triangle, ^ ^ Cone. : then Z BGE = ZA + Z B. Dam. Z ECB + Z BOA = 2 rt. A [If from the same point in a line any number of lines are drawn on the same side of the line, the sum, etc.] (I. 1.) ZA + ZB->rZACB = 2vt.A. (III. 1.) .-. ZBCE + ZBCA = ZA + ZB-\-ZACB. (Ax. 1.) .-. Z BCE = ZA + ZB. (Ax. 2.) Q.E.D. III. 2. SoH. The exterior angle of a triangle is greater than either of the non-adjacent interior angles. III. 2 a. The exterior vertex angle of an isoangular triangle equals twice either interior base angle. III. S. If a figure is a polygon of n sides, the sum of the interior angles, equals (2 ti — 4) right angles. "Ryp. If ABO ••• (? is a polygon of n sides, Cone: thenZA + ZB-^ZC, etc. = (2 w - 4) rt. A III. THE (2n-4) RIGHT ANGLES GROUP 33 Dem. From any point within the polygon draw the joins, OA, OB, OC, etc. We til us obtain as many triangles as there are sides, namely, n. The sum of the interior angles of each A = 2 rt. A. (III. 1.) .-. the sura of the int. A oi 7i A = n x2 rt. A, or 2n rt. A. But the base angles only of these triangles compose the angles of the polygon. .•. from the 2 nit. A we must subtract the sum of the angles about O, or 4 rt. A. (Sch. to I. 1.) .-. Z A-\-AB-\-A C, etc. =2 n rt. ^-4 rt. A, or (2 ri-4) rt. A. Q.E.D. III. 3 a. In a regular polygon each interior angle equals right angles, ^ ^ Hyp. If a polygon ABC "• iy/ of w sides is regular, Cone. : then any interior angle, as A, = — rt. A. n Dem. AA = AB = AC= etc. (Def. of regular polygon.) .-. AA + AB + AC+'" = nAA={2n-^)vt.A. (III. 3.) .-. AA = ^ of (2n-4)rt.zi = ?A=irt. A n n Q.E.D. Ex. 6. The sum of the exterior oblique angles of a right triangle equals 3 rt. A. Ex. 7. What is the sum of the interior angles __ in a 4-8ide ? a pentagon ? a hexagon? an octagon ? Ex. 8. What is the value of each interior angle in the above if each polygon is equiangular ? Ex. 9. How many sides has the regular polygon one of whose interior angles is f rt. Z ? f rt. Z ? | rt. Z ? | rt. Z ? 34 THE ELEMENTS OF GEOMETRY ni. SUMMARY OF PROPOSITIONS IN THE (2/} -4) QROUP 1. If a figure, is a triangle, the sum of the interior angles equals two right angles. a One interior angle of a triangle is the supple- ment of the sum of the second and third angles. h In a right triangle, either acute angle is the complement of the other acute angle. 2. Any exterior angle of a triangle equals the sum of the two non-adjacent interior angles. ScH. The exterior angle of a triangle is greater than either of the non-adjacent interior angles. a The exterior vertex angle of an isoangular tri- angle equals twice either interior hose angle. Z. If a figure is a polygon of n sides, the sum of the interior angles equals (2 ?i — 4) right angles. a In a regular polygon each interior angle equals right angles. Ex 10. Prove that the sum of the ext. zi of a polygon formed by pro- ducing the sides in order at all the vertices in succession equals 4 rt. A. Ex. 11. What polygon has the sum of the interior angles equal to three times the sum of the exterior angles ? One half the sum of the ext. A ? Ex. 12. If 2 A have their sides respectively ±, they are equal if of the same kind, and supplemental if of different kinds. Ex. 13. Can the plane space about a point be filled without overlapping by equiangular triangles ? By regular pentagons ? By regular octagons and squares ? By regular dodecagons and equilateral triangles ? Ex. 14. If in a triangle the altitudes Irom the extremities of the base be drawn to the two sides, prove that the angle formed by them is the supplement of the vertex angle. IV. GROUP ON ISOSCELES AND SCALENE TRIANGLES PROPOSITIONS The Isosceles Triangle IV. 1. If a triangle is isosceles, it is isoangular. Hyp. If A ABC is isosceles; that is if* a =6, Cone. : then ZA = ZB. Dam. Draw CT bisecting Z ACB. On CT as an axis fold over A CTB to plane of A CTA. Side a will fall on side h, '.• Z BCT= Z ACThy const. (Ax. 7.) B will fall on A, •/ side a = side b by hyp. TB coincides with TA. .-. ZA = ZB. (Ax. 6.) Q.E.D. Note. — It will assist the student to remember the sequence of the theorems in Group IV if he observes that in the hypothesis of both IV. 1. and IV. 4. are given the relations between the two sides of the tri- angle ; in the conclusion, the relations between the angles opposite those sides. 35 86 THE ELEMENTS OF GEOMETRY IV. 1 a. If the vertex angle of an isosceles triangle is bisected, the bisector is identical with (1) the altitude to the base, (2) the median to the base. Hyp. 1. If A ABC is isosceles, and if CT bisects ZACB, Cone. : then CT is identical with the altitude from C to AB. Hyp. 2. If A ABC is isosceles, and if CT bisects Z ACB, Cone. : then CT is identical with the median from C to AB. ScH. In an isosceles triangle, the altitude to the base is identical with the median to the base. (Ax. 10.) IV. 2. If a triangle is isoangular, it is isosceles. Hyp. If A ABC is isoangular ; that is, if ZA = AB, Cone.: then Dem. If CB does not equal CA, draw CB' that does. In other words, suppose AB'C is an isosceles triangle of which CA and CB' are the equal sides. Then Z CB'A = Z A. (IV. 1.) But ZB = ZA. (Hyp.) .-. Z CB'A = ZB. (Ax. 1.) .'. as CB and CB' have C in common, and as both make the same angle with AB, it follows that C^ must coincide with CB'. (Ax. 7.) But CB' was drawn equal to CA. .-. CB=CA. Q.E.D IV. ISOSCELES AND SCALExNE IRIANGLES 37 IV. 3. If the altitude of a triangle bisects the vertex angle J the triangle is isosceles. Hyp. If, in A ABG, CT±AB, and if Cr bisects ZACB, A Cone. : then A AGE is isosceles, Dem. On (72' as an axis revolve A II to plane of A I. CB will fall on CA, -.- ZBCT=ZACT by hypothesis. (Ax. 7.) B will fall on b, or its prolongation. TB will fall on TA, •.• Z GTB and ZGTA are right angles by hypothesis. (Ax. 7.) B falls on TA, or its prolongation. As B lies in b, and also in TA, it must fall on A, (Ax. 5.) .*. a = b, and A AGB is isosceles. Q.E.D. IV. 3 a. If the altitude of a triangle bisects the base, the triangle is isosceles. Hyp. If, in A ABG, GM± AB, and if AM= BM, Cone. : then A AGB is isosceles ; i.e. a = b. Dem. is similar to the preceding. Let the pupil supply it. Suggestion for Notation. — Letter foot of altitude, H] of median, M; of bisector of vertex angle, T. General Suggestions. Does not your greatest difficulty in proving a theorem lie in these two points : First, that you forget the hypothesis and conclusion ? Second, that you do not clearly remember the definition of the term you are using ? It is most important, therefore, that you should know what an altitude is ; what a scalene triangle is ; what a median is ; in short, the exact meaning of every term you use. 88 THE ELEMENTS OF GEOMETRY The Scalene Triangle IV. 4. In any triangle the greater angle lies opposite the greater side. Hyp. If, in A ABGf a>b, Cone. : then Z.A>AB. Dem. Lay off CL = b, and draw AL. AZ/^C is isosceles. (Def.) .-. ZCAL = ZCLA (IV. 1.) But ZCLA>ZB. [The exterior angle of a A is greater, etc.] (III. 1. Sch.) .'. its equal, Z CAL > ZB. .\ all the more is Z CAB > Z B, Q.E.D. Note. — It does not follow that Z A ia twice Z B, if side a is twice side b. The relative length of tiie sides of a triangle as compared with the size of the angles opposite which they lie, is treated in trigonometry. Ex. 1. If a triangle is equilateral, it is equiangular. Ex. 2. If a triangle is equiangular, it is equilateral. Ex. 3. If the vertex angle of an isosceles triangle is twice either base angle, what is the value of each angle of the triangle ? Ex. 4. Prove that the bisector of the exterior vertex angle of an isosceles triangle is parallel to the base. Ex. 5. State the converse of Ex. 4. Ex. 6. If the vertex angles of two isosceles triangles are supplemental, the base angles are complemental. IV. ISOSCELES AND SCALENE TRIANGLES 39 IV. 5. In any triangle the greater side lies opposite the greater angle. a Hyp. If, in A ABC, ZA>ZB, Cone: then a>b. Dem. a must be > 6, < b, or = b. If a<6, ZAb. Q.E.D. IV. 5 a. In any right triaiigle the hypotenuse is greater than either side. ScH. A perpendicular is the shortest distance from a point to a line. Note. — Distance from a point to a line is measured on the perpen- dicular through the. point. Ex. 7. if the sides of a regular hexagon be produced until they meet, prove that an "equilateral triangle is formed. Ex. 8. The angle between the base of an isosceles triangle and the altitude on one of the legs equals one half the vertex angle. Ex. 9. li AT bisects Z GAB. and we draw X^ = EA, prove that EL \\ CA. Ex. 10. If, through the vertex of the vertex angle S of the isosceles A ABC, LC is drawn ± CA and CM± CB^ prove Zi^Cilf is the double ofZA. (Fig. 1, p. 41.) Ex. 11.. If ^ACB and LCM (Fig. 1, p. 41) are isosceles, and the sides of the first are perpendicular to the sides of the second, prove that Z L is the complement of Z J.. 40 THE ELEMENTS OF GEOMETRY TV. 6. In any triangle^ if the altitude to the base is drawn, the side cutting off the greater distance from the foot of the altitude is the greater. ^ ^ ia) - - - ^' iij) Hyp. If, in any scalene A ABCf the altitude CL to the base AB is drawn, and LB > LA^ Cone. : then CB > CA. Dem. (a) If the altitude falls within the triangle, lay off LM^LA. A ACM is isosceles. (IV. 3 a.) .-. /.A==Z.LMC. A CLM is a right triangle, Z CLM being a right Z. (Hyp.) .-. Z.LMC is acute. [One interior angle of a triangle is the supplement of the sum of the second and third angles.] (III. 1 a.) .*. Z GMB is obtuse, being the supplement of Z LMQ, Z jB is acute, being < Z LilfO. [The exterior angle of a triangle is greater than either oppo- site interior angle.] (III. 2. Sch.) (IV. 5.) (IV. 3 a.) /. CB>CM. CM= CA /. CB>CA. Or thus : Z CMB > Z CLM== Z (7Z^ > Z5. .-. CB>CM. CM^ CA. .'. CB>CA. Q.E.D. Dem. (b) When the altitude falls without the triangle, the prbof is the same as in (ft), omitting the first step. IV. ISOSCELES AND SCALENE TRIANGLES 41 IV. 7. In any triangle, if the altitude to the base is draivn, the greater side cuts off the greater distance from the foot of the altitude. Hyp. If, in A ABC, CL ± AB, and if CB > CA, Cone. : then LB > LA. Dem. LB must be < LA, = LA, or > LA. If LB < LA, then CB < CA. (IV. 6.) If LB = LA, then CB = CA. (IV. 3 a.) Each of these conclusions is contrary to the hypothesis, viz., that CB > CA. .'. LB must be > LA. Q.E.D. Ex. 12. In Fig. 1, show that LM 11 AB. Ex. 13. In Fig. 2, show that if CH is the altitude of A ABC, CW the altitude of A CML, CH and CH' are in the same straight line. Ex. 14. If the bisectors of Z.A and ZB of tsABC (Fig. 3) intersect in M, and through M, EF is drawn || to AB, cutting AC in E and BC in JF', prove that the AAE3f and ilOF are isosceles. Ex. 15. Prove in the figure for Ex. 14 that EF = AE + BF. Ex. 16. Prove that if a leg of an isosceles triangle (Fig. 4) is extended its own length from the vertex, the join of its ex- tremity with the extremity of the base is perpendicular to the base. Fig. 4 42 . THE ELEMENTS OF GEOMETRY IV. SUMMARY OF PROPOSITIONS IN THE GROUP ON THE TRIANGLE The Isosceles Triangle 1. If a triangle is isosceles, it is isoangidar. a If the vertex angle of an isosceles triangle is bisected, the bisector is identical ivith (1) the altitude to the base, (2) the median to the base. ' ScH. In an isosceles triangle, the altitude to the base is identical with the median to the base. 2. If a triangle is isoangular, it is isosceles. 3. If the altitude of a triangle bisects the verteTi angle, the triangle is isosceles. a If the altitude of a triangle bisects the base, the triangle is isosceles. The Scalene Triangle 4. In any triangle the greater angle lies ojjposite the greater side. a In any right triangle the hypotenuse is greater than either side. ScH. A perpendicular is the shortest distance from a point to a line. 5. In any triangle the greater side lies opposite the greater angle. 6. In any triangle, if the altitude to the base is drawn, the side cutting off the greater distance from the foot of the altitude is the greater. 7. In any triangle, if the altitude to the base is drawn, the greater side cuts off the greater distance from the foot of the altitude. IV. ISOSCELES AND SCALENE TRIANGLES 43 Ex. 17. If one Z of a A = the sum of the other two, the A can be divided into isosceles A. Ex. 18. If, in the isosceles A ABC, Z C = f rt. Z, and BE ^/ is taken equal to AB, then the angles of AEAB equal the angles of the original triangle, and A ECB is isosceles. ^ Ex. 19. If, in an isosceles A ABC, from any point E in CB produced, EA is drawn, then ZBAC= ^ ^^^ + ZE^ Ex. 20. The angle formed by the bisectors of the interior base angles of a triangle equals a rt. Z + J the vertex angle. Ex. 21. The exterior base angle of an isosceles triangle equals the angle formed by the bisectors of the two interior jj; base angles : that is, Z CAE = ZAIB. A Ex. 22. In AABCnCT bisects Z C, and CH is an altitude, prove that ZHCT — ZA- ZB Ex. 23. Prove that in a 4-side the sum of three interior angles minus the exterior angle at the ^ fourth vertex equals 2 rt. A. H T Ex. 24. In any triangle the three new triangles formed by the bisectors of all the exterior angles of the triangle are mutually equiangular. Ex. 25. If, in an isosceles I\ABC, AE is drawn to any point, as E in BC, then Z CEA is greater than ZEAC. Ex. 26. If, in the isosceles A ^ BC, ^ is any point in BC, ^ prove that AE is greater than BE. Ex. 27. If, in A ABC, AE is perpendicular to BC, then AC+CB is greater than AE + EB. What is the greatest side of the triangle f .E 44 THE ELEMENTS OF GEOMETRY Ex. 28. If the vertex angle of an iso.scele8 triangle is twice the base angle, the bisector of the vertex angle divides the triangle into two isosceles triangles. Ex. 29. If, in an isosceles triangle, either base angle equals twice the vertex angle, the vertex angle is | rt. Z. Ex. 30. If, in an isosceles triangle, either base angle equals twice the vertex angle, the vertex angle is f rt. Z. Ex. 31. In a regular pentagon, what is the value of the interior Z. between two diagonals drawn from the same vertex ? Ex. 32. Prove that these two diagonals are equal. V. GROUP ON CONGRUENT TRIANGLES PROPOSITIONS V. 1. If two triangles have two sides aiid the included angle of the first equal to two sides a7id the included angle of the second, they are congruent Hyp. If, in A I and II, G AB = EF, CA = HE, and A A = ZE, /^\ A B Cone. : then A I ^ A II. Dem. Place A II on A I with EF in coincident superposition with AB, E falling on A. Then^i2'willfallon^(7,because,byhyp.,ZJ5;=ZA (Ax.7.) .-. H will fall on C, because EH= AC, (Hyp.) .-. HF coincides with CB. (Ax. 6.) .-.AI^AIi. Q.E.D. Ex. 1. If two triangles are congruent, the following homologous lines {i.e. lines having the same relative position) are equal : (a) the homologous medians ; (6) the homologous altitudes ; (c) the homologous bisectors of the interior angles. Ex. 2. If two altitudes of a triangle are equal, the triangle is isosceles. Ex. 3. Prove that the altitudes of an equilateral triangle are equal. Ex. 4. Two isosceles triangles are congruent if the vertex angle and its bisector in one are equal to the corresponding parts of the other. 45 46 THE ELEMENTS OF GEOMETRY V. 2. If two triangles have two angles and the included side of the first equal to two angles and the included side of the second, they are congruenti m Hyp. If, A I and II, ZA=ZE, AB=ZF, and AB=EF, Cone. : then A I ^ A II. Dem. Place A II on A I so that EF is in coincident super- position with AB, E falling on A. EH will fall on A C, since, by hyp., ZA = ZE. (Ax. 7.) .-. -fflies on AC, or its prolongation. Again, FH will fall on BC, since, by hyi^.,ZF=Z B. (Ax. 7.) .-. H lies on BC, or its prolongation. Since // lies on both AC and BC, it must lie at their inter- section. (Ax. 5.) .-. A I ^ A 11. Q.E.D. V. 3. If tioo triangles have three sides of the one equal to three sides of the other, they are congruent. Hyp. If, in A I and II, AO=EH, AB = EF, and BC^FH, E F Cone. : then A I = A II. Dem. Place A II in the position of A II' with EF in coinci- dent superposition with AB, E falling on A. Draw CH\ A CAH' is isosceles ; also A BCH'. (Hyp.) .-. Z1 = Z2 and Z3 = Z4. V, COHGRUENT .TRIANGLES 47 [If a triangle is isosceles, it is isoangular.] (lY. 1.) Zl + Z3 = Z2 + ^4; th2itis,ZACB = ZAH'B. (Ax. 2.) .-. A I ^ A II'. (V. 1.) .-. A I ^ A II. (Ax. 1.) Q.E.D. V. 4. If tioo right- triangles have the hypotenuse and a leg of the one equal to the hypotenuse and a leg of the other, they are congruent. 1' Hyp. If, in A I /' andll, C5 = J7Fand / CA = HE, and ZB / / and ZF are right / / / LI angles, / / A B E F Cone: then rt. AI ^ rt .All. Dem. Place A I in position of A I' so that CB is in coinci- dent superposition with HF, C falling on H. EFB is straight. (I. 2.) [If two supp. A are adj., their ext. sides, etc.] ZB^ = ZE. (An isosceles triangle is isoangular.) .-. ZEHF=ZFHB'. A I' ^ A II. [Two A and the included side of one equal, etc.] .-. A I ^ A II. (IV. 1.) (Why ?) (V. 2.) (Ax. 1.) Q.E.D. ScH. Since congruent triangles may be placed in coincident superposition, it follows that homologous altitudes, medians, angle bisectors, mid-joins, and all other crresponding parts are respectively equal. 48 THE ELEMENTS OF GEOMKTRV V. SUMMARY OF PROPOSITIONS IN THE GROUP ON CONGRUENT TRIANGLES 1. If two triangles have two sides and the included angle of the first equal to two sides and the included angle of the second, they are congruent. 2. If two triangles have two angles and the included side of the first equal to two angles and the included side of the second, they are congruent. 3. If two triangles have three sides of the one equal to three sides of the other, they are congruent. 4. If two right triangles have the hypotenuse and a leg of the one equal to the hypotenuse and a leg of the other, they are congruent. ScH. Since congruent triangles may be placed in coincident superposition, it follows that homologous altitudes, medians, angle bisectors, mid-joins, and all other corresponding parts are respectively equal. V. CONGRUENT TRIANGLES 49 PROBLEMS Prob. I. To construct a right triangle having one side and the hyj)otemcse given. Given. The sides 6 b and c. Required. To con- struct a triangle. JS \B OF Const. Take an indefinite line EF. At C, any point in EF, erect a ±, CA = b. With ^ as a center and a radius = c, describe an arc cutting EF, as in B. Rt. A ABO is the required right triangle. Q.E.F. Proof. If two right triangles, etc. (V. 4.) Pkob. II. To construct a triangle^ its three sides being given. Given. The three , sides, a, b, and c. a — Required, lo con- struct the triangle. ^ '. Const. Draw AB=:c. With ^ as a center and 5 as a radius, describe an arc. With JB as a center and a as a radius, describe an arc. Let these two arcs intersect at any point O. Draw AC and BC. A ABC is the required triangle. Q.E.F. Note. — No triangle can be constructed if c? + & < 0, 60 THE ELEMENTS OF GEOMETRY Prob. III. To construct an angle equal to a given angle. ^ Given. Z a. '/.. t. Required. To construct / \ / / •- \ an angle ^qual U^a. /_. \ A a' \ A F -B A^ i' Const. On AB lay off any distance, AF. On AC lay off any distance, AE. Draw the join EF. Construct a A whose sides are AF, FE, and AE, (Prob. II.) AA'C'B'^AACB, [Two A are ^ if three sides of one =, etc.] (V. 3.) . Then Z a' is the required angle. Q.E.r. Proof. Za'^Za. . (Horn. A oi = A.) Q.E.D. Prob. IV. TJirough a given point to draw a line parallel to a given line. Given. The point P and the line AB. Required. To construct a line 2. M through P parallel to AB. Const. Through P draw any line cutting AB, say at M. At P draw a line PL, making with PM an Z = to Z PMB. PL is the line required. Q.E.F. Proof. ^PMB=/.LPM. (Const.) .-. LPW AB. ' (11.3;) Q.ED. V. CONGRUENT TRIANGLES 61 Prob. V. Given tivo sides and an angle opposite one of them, to construct the triangle. Given. The sides a and &, and t^e angle A Required. To construct the triangle. Const. At one extremity of an indefinite line AL, construct an Z LAO = ZA. On AOlay oEAC=h With (7 as a center and radius equal to a, describe a circle. The point B, in which the circle cuts AL, will be the third vertex of the triangle. Q.E.F. Proof. To be supplied by the student. From the figures you will observe that several cases arise : 1. a 6 ; one triangle satisfies the conditions. 3. a = the perpendicular from C to AL ; triangle is a right triangle. 4. a < the ± from C to AL ; no A satisfies the conditions. Verify by drawing the figures for each case not shown. Ex. 5. If A ^50 is equilateral and ^£'=:BF = CH, show ilmt ^ECH, FAE, and FBUkvq con- gruent. Hence sfliow that A EFH is equilateral. Ex. 6. Show that in the A ABC there are three congruent 4-sides, Ex. 7. If HABC is equilateral and AE = CH = BF, show that ABE, BCF, and ^Cif are con- gruent. (See Fig., p. 520 52 THE ELEMENTS OF GEOMETRY Ex. 8. In the adjacent figure show that A ^^If, nCE, and ACF are congruent. (Data as in Ex. 7.) Ex. 9. Also show that ^AEL, FMB, and CHQ are congruent. Ex. 10. Hence prove that A LMQ is equi- lateral. Ex. 11. Show that ^ ALB, BMC, and CQA are congruent. Ex. 12. Prove that the 4-sides ^iilfJ?', ^MQJT, and CQLE are congruent. Ex. 13. Two triangles are congruent if two altitudes and a side to which one of them is drawn in one triangle equal the corresponding parts of the second triangle. Ex. 14. Two right triangles are congruent, if the altitude to the hypot- fnuse and the median to the hypotenuse in one triangle are equal to the corresponding parts of the second triangle. Ex. 15. All the theorems and corollaries concerning congruent triangles in the summary have a condition in common. "What is it ? What, then, may we infer must be one of the conditions in order to prove two triangles congruent ? Prove that two triangles are congruent if the following parts of one are equal, and are similarly situated, to the corresponding parts of the other : (Use the following notation: Z A, Z B, ZC; sides opposite these tri- angles, a, b, c; altitudes to a, 6, c, are ha, hb, he ; medians on a, b, c, are »ia. wift, mc ; angle bisectors are' ta, tj, tc; r is the radius of the inscribed circle ; Vc is the radius of the circumscribed circle.) Ex. 16. a, 6, and m^, Z A being obtuse in both triangles. Ex. 17. a, 6, and he Ex. 18. c, he, nic- Ex. 19. 6, Z A, he, ZA being acute in both triangles. Ex. 20. a, Z B, nie, Z B being obtuse. Ex. 21. Two angles and the side opposite one of them being given, to construct the triangle. Ex. 22: Two rt. L are ^ if an acute Z and a leg of one, or an acute Z and hypotenuse of one, equal the corresponding parts of the other. Ex. 23. Through a given point without a given line, to draw a line making a given Z with the given line. Ex. 24. Construct a triangle, given the two base angles and the bisec- tor of one of them. VI. GROUP ON PARALLELOGRAMS DEFINITIONS The Quadrilateral or Four-side A Quadrilateral, or 4-side, is a figure formed by the intersec- tion of four lines, no three of which pass through the same point. Its alternate angles are called opposite angles. A Trapezium is a 4-side upon which no conditions are im- posed. A Trapezoid is a 4-side having one pair of parallel sides, called the bases. A Parallelogram is a 4-side having two pairs of parallel sides. Note. — A parallelogram in the above general form is called a Rhomboid. A Rectangle is a parallelogram with two consecutive angles equal. A Rhombus is a parallelogram with two consecutive sides equal. A Square is a rhombus, one of whose angles is a right angle. The Mid-join of a Trapezoid is the line joining the mid-points of the non-parallel sides. The Median of a Trapezoid is the line joining the mid-points of the parallel sides. Note. — If the non-parallel sides of a trapezoid be produced until they meet, the median of the trapezoid becomes a part of the median of the triangle formed by one base of the trapezoid and the non-parallel sides produced. Some authorities give the name median to the mid-join. The definition above is more consistent w^ith the use of the term median in connection with triangles. An Isosceles Trapezoid is a trapezoid having its non-paralleJ sides equal. The Altitude of a Trapezoid or of a Rhomboid is the perpen- dicular distance between the bases. A Kite is a 4-side that has 2 pairs of adjacent sides equal. A Cyclic Four-side is one whose vertices lie in a circum- ference. 53 54 THE ELEMENTS OF GEOMETRY PROPOSITIONS VI. 1, If a 4:'side has two sets of opposite sides equal, it is a parallelogram. Hyp. If, in the /^^- '- 7^ 4.side A-Cy a = a' / i r and 6 = 6', A b B Cone. : then a II a' and 6 II 6' ; i.e., the 4-side A-E is a O. Dem. Draw the diagonal EB. AI^AII. [Two A ate ^ if three sides of the first, etc.] (V. 3.) .-. /. ABE = Z CEB. (Horn. Aoi^ A.) .-. 6 II 6'. [If two lines be crossed by a transversal, etc.] (II. 3.) Similarly, Z EBC = A AEB. .-. a II a'. .•. the 4-side A-0 is a, parallelogram. (Def. of a O.) Q.E.D. VI. la. If a ^'side is a parallelogram^ its opposite sides are equal. , Hyp. If the 4-side Z^^^-' / 7f A-C is a parallelo- / // ^^^^^-..^ A gram, A b Cone. : then a = a', 6 = 6'. Dem. Draw a diagonal. AI^AII. (V. 2.) .'. a = a', and h = 6'. (Hom. sides of ^ A.) Q.E.D. ScH. A diagonal divides a O into two congruent triangles. Note. — VI. 1. and VI. la. are "Direct" theorems; VI. 1'. and VI. I'a'., on page 65, are their "Reciprocals." See definition of reciprocal on p. 11. 7 VI. PARALLELOGRAMS 65 VI. 1'. If a Aside has tivo sets of opposite angles egiial, it is a parallelogravi. Hyp. If, in the 4-side a/ A-C, ZA = AC and / ZB = ZE, A b B Cone. : then a 11 a' and 6 ll 6'; i.e., the 4-side A-C is a parallelo gram. Dem. ZA==ZC, ZB = /.E. (Hyp.) ,-. ZA + Z.B=^Z.C + /.E. (Ax. 2.) But Z^ + Z.8 + Z(7 + Z^ = 4rt. A. [The sum of the interior angles of a 4-side = 4 rt. ^.] (III. 3.) .-. Z^ + Z5 = 2 rt. A (Ax. 2.) .-. a I! a'. [If two lines are crossed by a third, etc.] (II. 4 (1).) Similarly, & II &'. .'. the 4-side A-C is a parallelogram. (Del of a O.) Q.E.D. VI. 1' a', i/* a Aside is a parallelo gram, its opposite angles are equal Hyp. If the 4-side ^^ yO A-C is a parallelo- y A' gram, / / Cone: then Z^ = Z C7, and Z5 = ZjE?. Dem. a 11 a'. (Def. of a O.) .*. Z ^ is supplementary to Z jB. [If two parallels are crossed by a transversal, etc.] (II. 2.) Z C is supplementary to Z B, (Sam© reason.) A z^=za (Ax. 1.) Similarly. Z 7? =. Z i7. 9E.©. 66 THE ELEMENTS OF GEOMETRY VI. % If a ^'Side has one set of sides both equal and parallely it is a parallelogram, E h o Hyp. If a = a' and a i a', y /^^^\ A ' b' B Cone. : then 6 II 6'j i.e., the 4-side is a parallelogram. Dem. Draw the diagonal EB. AI^AII. [Two A are ^, if two sides and the included Z of the first = two sides and the included Z of the second.] (V. 1.) /. Z ABE = Z BEC. (Horn. ^ of ^ A are = .) .-. h II h\ [If two lines are crossed by a third so as to make the alter- nate interior angles equal, the lines are parallel.] (II. 3.) /. the 4-8ide ^-C is a parallelogram. (Def. of a O.) Q.E.D. VI. ^. If a ^-side is a parallelogram, the diagonals bisect each other, E & Hyp. If the 4-side /\i^ A-G is a parallel©- /^^-""t^^^^ gram, ■ A ■ h' B Cone. : then the diagonals AC and BE mutually bisect. Dem. h = h'. (VI. 1 a.) /.ACE = Z.CAB. [If two parallels are crossed by a third line, etc.] (II. 1.) Z BEC = Z ABE. (Same reason.) .-. AI^AII. (Why?) .-. AM= MG and BM= ME. (Horn, sides of ^ A.) Q.E.D, VI. PARALLELOGRAMS 67 VI. 4. If the diagonals of a parallelogram are equal, the parallelogram is a rectangle. Hyp. If, in the CJA-G, AC^BE, Cone: then AEAB^ZABG; i.e., the O A-G is a rectangle. Dem. AABE^AABG. [If two A have three sides of the first equal, etc.] (V.3.) .-. Z EAB = Z CBA. (Horn. ^ of ^ A.) But ZEAB + Z.CBA=:2vtA. [If two parallels are crossed by a third line, etc.] (II 2.) .-. Z^^jB = lrt. Z. (Ax. 2.) ZCT^ = lrt. Z. (Ax. 2.) Similarly, Z AEG = Z BGE, .*. O A-G is a rectangle. (Def. of 1 1.) Q.E.D. VI. 4: a. If ttvo rectangles have the base and altitude of the one equal to the base and altitude of the other, they are congruent. Suggestion. — It will be of much assistance to the student in the solution of original exercises, if he marks the parts given in the hypothesis so as to distinguish them from those mentioned in the conclusion. For example, a figure marked thus might be interpreted to mean that if in a parallelo- gram the diagonals are equal, the parallelo- gram is a rectangle. That is, significant symbols may be used to indicate the relations given in the hypoth- esis^ while the cross is used to refer to those of the conclusion. 58 THE ELEMENTS OF GEOMETRY Ex. 1. What are the different kinds of quadrilaterals, or 4-sides ? Ex. 2. What is the median and what the mid-join of a trapezoid? Ex. 3. Produce the non-parallel sides of a trapezoid until they meet What does the median (produced) of tlie trapezoid become? Ex. 4. What is the converse and what the reciprocal of a theorem? Find an illustration of each in Group VI. Find an illustration of each in Group V. Ex. 5. If through the vertices of any triangle lines are drawn parallel to the opposite sides, point out the three parallelograms formed. Ex. 6. Prove that the new triangle is four times the size of the original triangle. Ex. 7. Prove that if through the ends of each diagonal of a 4-8ide, parallels to the other diagonal are drawn, a parallelogram is formed which is twice as large as the 4-side. Ex. 8. Prove the converse of VI. 3. Ex. 9. The bisectors of the corresponding interior angled of two parallelg crossed by a transversal meet at right angles. Prove. Ex. 10. The bisectors of the interior angles of a parallelogram inclose a rectangle. Prove. Ex. 11. What is a square ? Ex. 12. The bisectors of the interior angles of a rectangle inclose a square. Prove. Ex. 13. The bisectors of the exterior angles of a rectangle inclose a square. Prove. Ex. 14. In what parallelograms do these bisectors of the interior angles coincide with the diagonals ? Ex. 15. In such parallelograms, what becomes of the inclosed rectangle ? See Ex. 10. Ex. 16. What is the sum of the interior angles of a triangle ? Ex. 17. What is the sum of the interior angles of a 4-side ? Ex. 18. Prove that if HE and LA bisect Z.E and Z A, Z HFL is supple- ment of ^^^ 2 2 (Fig. for Ex. 21.) Ex. 19. Prove that if HC and LB bisect ZC and Z.B, ^HJL is supple- int of :^ -h ^^. (Fig. for Ex. 21.) Ex. 20. Why is (Fig. for Ex. 21.) ^+^ the supple- 2 2 mentof ^+^? 2 2 Ex. 21. Prove that the four bisectors of the interior angles of a 4-side form a 4-side whose opposite angles are supplemental. VI. PARALLELOGRAMS 59 Ex. 22. Prove that the four bisectors of the exterior angles of a 4-side form a 4-side whose opposite angles are supplemental. Ex. 23. Prove that if two parallelograms have two sides and the in- cluded angle of one equal to two sides and the included angle of the other, they are congruent. Ex. 24. Prove that if the diagonals of a 4-side not only bisect but are also perpendicular to each other, tiie 4-side is a rhombus. Ex. 25. Prove that an isosceles trapezoid is isuangular. (Draw per- pendiculars to the longer base from the ends of the sliorier.) Ex 26. Prove that the diagonals of an iisosceles trapezoid are equal. Ex. 27. Why are the opp. A of an isosceles trapezoid supplemental ? E Ex. 28. If from the opposite vertices of a parallelogram perpendiculars are let fall on a diagonal, they are equal. A B Ex. 29. If a line is drawn parallel to the base of an isosceles triangle, it divides the triangle into an isosceles triangle and au isosceles trapezoid. Ex. 30. If, in a parallelogram ^-C, J5i^ ^ ^ ^ = EH, then the 4-side AFCH is a parallelo- gram. Ex. 31. If one of the diagonals of a 4-side bisects a pair of opposite angles, but the other diagonal does not, the 4-side is a kite. Ex. 32. Tlie 4-side formed by joining the ends of any two diameters of a circle is a rectangle. Ex. 33. The 4-side formed by joining the ends of two perpendicular diameters is a square. Ex. 34. The bisectors of the opposite angles of a rhomboid are parallel. Ex. 35. Any line through the mid-point of the diagonal of a parallelogram divides the parallelo- gram into two equal parts. Ex. 36. Express in terms of n the number of diagonals of an n-gon. 60 THE ELEMENTS OF GEOMETRY VI. SUMMARY OF PROPOSITIONS IN GROUP ON PARALLELOGRAMS \. If a ^-side has two sets of opposite sides equal, it is a parallelogram. a If a Arside is a parallelogram, its opposite sides are equal. ScH. A diagonal divides a O into two congruent triangles. 1'. If a 4rside has two sets of opposite angles equal, it is a parallelogram. a* If a 4t-side is a parallelogram, its opposite angles are equal. 2. If a Aside has one set of sides both equal and parallel, it is a parallelogram. S.Ifa 4rside is a parallelogram, the diagonals bisect each other. 4. If the diagonals of a parallelogram are equal, the parallelogram is a rectangle. a If two rectangles have the base and altitude of the one equal to the base and altitude of the other, they are congruent. VII. GROUP ON SUM OF LINES AND MID-JOINS PROPOSITIONS VII. 1. The sum of two sides of a triangle is greater than the third side. Hyp. If ABC is a triangle, Cone. : then h-\-a>c. Dem. If either a or 5 > c, no proof is required. If each side < c, draw CL _L c. Then AL and BL are each ± CL. Now AL < h, and BL < a. [A perpendicular is shortest distance from a point to a line.] (IV. 4 a, Sch.) .', AL-\- BL(= c) < 6 + a. (Preliminary Th. 1.) Q.E.D. VII. 1 a. The difference hetiveen any two sides of a tri- angle is less than the third side. (Use Preliminary Th. 3.) Ex. 1. Prove that a line from the vertex of the vertex angle of an isosceles triangle to any point in the base is less than either of the legs of the triangle. 61 62 THE ELEMENTS OF GEOMETRY VII. 2. Tlie sum of two lines drawn from any point within a triangle to the ends of one side is less than the sum of thetxoo other sides of the triangle, Px Hyp. If from Pi, Pi^ and Pjfi are drawn, and from P, FA and PB are drawn, but enveloped by PiA and PjB, A Cone: then P^A + P^B> PA + PB. Dem. Extend AP to intersect P^B at F. P,A + P,F>AF. (VII. 1.) BF+ PF > BP. (VII. 1.) .-. Pi^4-PiP+-BP+PP>-4P+PF+JBP. (Preliminary Th. 1.) Take away PF from both members of the inequality and we have p^A + P^B >PA + PB. (Preliminary Th. 3.) Q.E.D. VII. Z. If a series of parallels cut off equal segments on one transversal, (a) They will cut off equal segments on every trans- versal. \ \ \ \ X.^-^"^ Hyp. If G^TTand^O \ \ \ \^^^^^^\ are any transversals ; ' \ \ ^^•^'C^^-'-^'X \ AG^ BHy etc., a set of \^,^'^i\^'"jK \ \ parallels, and AB=BO ''o\""l\ \ \ \ = CE, etc., A b\ c\ F\ W~0 Cone. : then (a) GH= HJ= JK== KS, etc. Dem. Draw GL, HM, JQ, etc., all parallel to AO. The four-sides ABGL, BCMH, etc., are m. (Def. of O.) .-. GL = AB, HM= BC, JQ = CE, etc. (VI. 1 a.) Ex. 2. Show how theorem VII. 3 (a) may be practically utilized in solving the problem : To divide a given line-segment into any number of equal parts. VII, SUM OF LINES AND MID-JOINS 63 But AB = BC= CE, etc. (Hyp.) .-. GL = HM=JQ, etc. (Ax. 1.) Again, Z HGL = Z. JHM= Z KJQ, etc. (Def. of ils.) And Z HLG = Z JMH= Z A"Q,/, etc. (II. 5.) .-. A GHL ^ A HJM^A JKQ, etc. (V. 2.) .: GB=HJ=JK=KS,etG. Q.E.D. VII. oh. If the series consists of three consecutive parallels termiiiating in the transversals, the mid- parallel equals half the sum of the other tivo. That is, the 7nid-join of a trapezoid equals half the sum of the two parallel sides. . \ fiyp. If GA, HB, and \ _\-.r-rrr::^r JC are three successive - — qS^ "A \ parallels terminating in \ \ \ the transversals AO and \ \ \ GW, ^ ^V ^ Cone: then HB = i(GA-\-JC); that is, the mid-join of a trapezoid equals one half the sum of the bases. Dam. HJ^HG. (VII. 3 a.) .-. BH is the mid-join of the trapezoid AC JO, Draw GL and HQ W AO. GA = LB, CQ = LB + LH. (VI. 1 a.) Ce/= CQ + QJ, and LH= QJ. (Hom. sides of ^ A are =.) .-. GA + CQ + CJ= LB-\-LB + LH+CQ + QJ, (Ax. 2.) or GA+CJ=2LB + 2LH, QJ. %A±£i^LB + LH=HB. Q.E.D. 64 THE ELEMENTS OF GEOMETRY VII. 3 c. If in a triangle a parallel to the base in drawn through the 7nid-point of one side, it bisects the other side and equals half the base. Hyp. If GCJ is a A, and if B is the mid-point of GO, and BH || CJ, Cone. : then (1) and (2) QH^HJ HB=W' Dem. (1) The A GCJ" may be considered to be the trapezoid GACJoi VII. 3 (6) with the side GA reduced to a point. .-. the Dem. of VII. 3 (6) will apply to VII. 3 (c). Proof. Let the student give the proof in full. Ex. 3. If, from any point in the base of an isosceles triangle, lines be drawn parallel to the sides, the perimeter of the parallelogram formed equals the sum of the two equal sides of the isosceles triangle. G Ex. 4. If, from any point in the base of an isosceles triangle, perpendiculars to the equal sides are drawn, prove that their sum equals the altitude from either one of the base angles. (Draw FL ± BH. ) Ex. 5. Prove the foregoing proposition for the obtuse isosceles triangle. . Ex. 6. AB of the isosceles A is then the locus of what point ? Ex. 7. If, from any point within an equilateral tri- angle, perpendiculars be drawn to the sides, prove that the sum of these three perpendiculars equals the alti- tude of the triangle. (Draw LE through F \\ AB.) L, RQ Ex. 8. Prove that the mid-join of a triangle is parallel to the third •ide. VII. SUM OF LINES AND MID-JOINS ^5 VII. 4. Li a right triangle the median to the hypote- nuse equals half the hypotenuse. Hyp. If, in the rt. A ABC, M is the mid-point of the hypotenuse c, Cone: then OJf=?. Dem. Draw MJ II h. MJ II b and = \ .'. Z MJB is a rt. Z, and • A CMB is isosceles. [If the altitude of a A bisects the base, etc.] .-. MB = CM. But MB = MA. .'. CM=MA. . Qj^^MB±MA^c^ 2 2 (VII. 3 (c).) (Why?) (IV. 3 a.) (Hyp.) (Ax. 1.) Q.E.D. VII. 4 a. Conversely, if a median of a triangle equals half the side that it bisects, the triangle is right. Ex.9. The A ABC and AFC have AC common, CF=CB, and ZACB>ZACF. Show that AB > AF. Note that if Z FCB be bisected by CH, and HF be drawn, A FCH ^ A HCB (V. 1) and .-. FH= HB. But ylil+ FH>AF, whence ^^ >^i^; that is, If two triangles have two sides of the first equal respectively to two sides of the second and the included angles unequal, the third sides will be unequal ; the greater side will belong to the triangle having the greater included angle. Ex. 10. State and prove the converse of Ex. 9. 66 THE ELEMENTS OF GEOMETRY Ex. 11. Prove that if in a right triangle one acute angle is two thirds of a right angle, the hypotenuse equals twice the shorter side. (Draw the median to the hypotenuse.) Ex. 12. Prove that the joins of the mid-points of the adjacent sides of a 4-8ide form a parallelogram. (Draw the diagonals.) Ex. 13. Prove that the parallelogram formed in Ex. 12 is one half the orighial 4-8ide. Ex. 14. Prove that the mid-join of the diagonals of a trapezoid is parallel to the bases and equals one half their difference. (Draw the join of an end of the upper base and an end of the mid- join. Two congruent triangles are formed with the bases and diagonals.) Ex. 15. The mid-joins of the three sides of a triangle divide the triangle into three congruent triangles. Ex. 16. If we call ABEC a cross trapezoid, prove that AB-\-CE'\% less than AC ■\- BE. Ex. 17. In any 4-side the sura of the diagonals is less than the perimeter and greater than the semi- perimeter. Ex. 18. In the isosceles A -4 J5C, E is any point in AB. Through E to draw FL terminating in 6 and a produced so that FE = EL. Ex. 19. In the isosceles A ABO prove that if FL is bisected in E that AF = BL. Ex. 20. If, in a parallelogram (Fig. 1), Mis the mid- point of EC, and L is the mid-point of AB, show that EL !l BM. Ex. 21. If, in Fig. 1, EL intersects CA in E, and BM intersects CA in f, show that F is the mid-point of CB. Ex. 22. In PMg. 1, show that OF = CF 2 * Ex. 23. In Fig. 1, BM and CO are medians of A BCE. Prove by means of the three pre- ceding theorems that the medians of a triangle concur. Ex. 24. If two medians of a triangle intersect AO 2 BE 3 * in as in the adjoining figure, then 0M = and O^ = 4^ J that is, OM = ^, and ^0 =- 2 3 HnrT. — Draw CL y BO. Vn. SUM OF LINES AND MID^OINS 67 Ex. 25. If, in A ABO, AG and CH are two altitudes on a and c, respectively, and TK and MK are mid-perpendiculars to a and c, prove that Tir = AI 2 Hint. — Draw MF = ilf r and JPi? II CH. What triangles are con- gruent ? Ex. 26. The mid-joins of the four halves of the two diagonals of a rectangle form a second rectangle. Ex. 27. If, in A ABC, CT bisects the vertex angle and AE and BF are drawn perpendicular to CT, then, if M is the mid-point of AB, the join ME (or MF) = \{BC - AC). Ex. 28. If ^£' and jBjP are drawn perpendicular to the bisector of the exterior vertex angle, the join ME = \{BC-^ AC). Ex. 29. The Z QAB = lU CAB -AB). Ex. 30. The join of the mid-points of two opposite sides of a 4-side and the join of the mid- points of its diagonals are the diagonals of a parallelogram. Ex. 31. State at least three general methods of proving that : Lines are parallel. Lines are equal. Angles are equal. Angles are supplemental. 68 THE ELEMENTS OF GEOMETRY VII. SUMMARY OF PROPOSITIONS IN THE GROUP ON SUM OF LINES AND MID^OINS 1. The sum of two sides of a triangle is greater than the third side. (a) The difference between any two sides of a triangle is less than the third side. 2. The sum of two lines draivn from any point loithin a triangle to the ends of one side is less than the sum of the two other sides of the triangle. Z. If a series of parallels cut off equal segments on one transversal, (a) They will cut off equal segments on every transversal. (b) If the series consists of three consecutive paral- lels terminating in the transversals, the mid- parallel equals half the sum of the other two. That is, the mid-join of a trapezoid equals half the sum of the two parallel sides. (c) If in a triangle a parallel to the base is drawn through the mid-point of one side, it bisects the other side and equals half the base. 4. Ii a right triangle the median to the hypotenuse eqitals half the hypotenuse. (a) Conversely, if a median of a triangle equals half the side that it bisects, the triangle is right. VIII. GROUP ON POINTS — EQUIDISTANT AND RANDOM PROPOSITIONS VIII. 1. Every point on the mid-perpendicula?' of a Une-segment is equidistant fi^om the ends of the line- segment, r^' Hyp. If RM is a mid-perpen- dicular to AB, and P is any point in RM, Cone. : then Dem PM is the altitude to base of A ABP. (Def . of alt.) .*. A PAB is isosceles. (IV. 3 a.) PA = PB. (Def. of isos. A.) Q.E.D. What Ex. 1 Ex. 2 Ex.3 Ex. 4 Ex. 5 Ex. 6 Ex. 7 Ex. 8 is the locus of a point satisfying the following conditions ; At a distance a from a given point Q ? At a distance a from a given line AL ? At a distance a from a given circumference K? Equidistant from Q and B ? Equidistant from two intersecting lines AL and BM? Equidistant from two parallels ? Equidistant from two equal circumferences ? Equidistant from two concentric circumferences ? Find the points that satisfy the following conditions : Ex. 9. At a distance a from Q, and a distance b from a line AL. Ex. 10. At a distance a from ^, and a distance b from a O K. 69 70 THE ELEMENTS OF GEOMETRY VIII. 2. Every point loithout the mid-perpendicular of a line-segment is unequall]/ distant from the ends of the line-segment. p Hyp. If ML is a mid-perpen- dicular to AB, and if P is any point without ML, Cone. : then PA is not equal to PB. Dem. Let AP intersect the mid J_ ML in T, Draw TB, TB=TA. TB+TP>PB. TB.-{-TP=TA+ TP (Ax. 2) = PA, .'. PA does not equal PB. (VIII. 1.) (VII. 1.) Q.E.D. ScH. 1. The mid-perpendicular of a line-segment is the locus of points equidistant from the ends of the line-segment. ScH. 2. Two points, each of which is equidistant from the ends of a line-segment, determine the mid-perpendicular to the line. VIIT. 3. Every point in the bisector of an angle id equidistant from the sides of the angle. jj Hyp. If ^T is the bi- ^- sector of Z BAC, P any point in AT, and if PF and PE are perpendiculars to-4B and AC, respectively, Cone. : then P is equidistant from AB and AC. That is, PF=PE. VIII. POINTS — EQUIDISTANT AND RANDOM 71 Dem. ZEAP=ZFAP. (Hyp.) Z EPA = Z FPA. (Ax. 1.) AP=AP. .-. AEAP^AFAP. [Two Z and the included side of the first, etc.] (V. 2.) .-. PF= PE. (Horn, sides of ^ A.) Q.E.D. Vlli. 4. Every point ivitJiout the bisector of an angle IS unequally distant from the sides of the angle, G Hyp. If AT is the bi- sector of Z BAC, P any point ^-^""^^^X / without AT, and if PF and PE are perpendiculars to AB and AC, respectively, /^ B Cone. : then P is unequidistant from AB and AC. That is, PF does not = PE. • Dem. Draw MffX^C; also the join P5". PH>PE. [A ± is the shortest distance from a point to a line.] (IV. 4 a. Sch.) [The sum of two sides of a A is greater, etc.] (VII. 1.) MH=FM. (VIII. 3.) .-. FM+MP>PH. But FM+MP= PF. (Ax. 4.) /. PF> PII, which is greater than PE. .-. PF>PE. Q.E.D. 72. THE ELEMENTS OF GEOMETRY 8cu. 1. Two points, each of which is equidistant from the sides of an angle, determine the bisector of the angle. ScH. 2. The bisector of an angle is the locus of all points equidistant from the sides of the angle. Gen. Sch. Two points in any straight line locus determine the locus. Note. — Sch. 1. of VIII. 2. affords the proof of the solutions of the following problems of the ten easy exercises in geometrical drawing (pp. 14-17): Prob. I. (a) Bisect a given line-sugment. (6) Erect a raid-perpendicular to a given line-segment. Prob. V. Erect a perpendicular to a given line at a given point' in the line. Prob. VI. Draw a perpendicular to a given line from a given point without the line. Find the points that satisfy the following conditions : Ex. 11. In a given line AL, and at a distance b from a given line Q. Ex. 12. In a given line AL^ and at a distance b from a OA'. Ex. 13. In a given circumference, and at a distance b from Q. Ex. 14. In a given circumference, and at a distance b from a line AL. Ex. 15. In a given circumference, and at a distance b from a second circumference. Ex. 16. In a given line, and equidistant from Q and B. Ex. 17. In a given circumference, and equidistant from Q and E. Ex. 18. At a distance a from a line AL, and equidistant from Q and B. Ex. 19. At a distance a from a given circumference, and equidistant from Q and B. Ex. 20. At a distance a from a given point Q, and equidistant from B and S. Ex. 21. At a distance a from a given point Q, and equidistant from two intersecting lines AL and AM. Ex. 22. At a distance a from a point Q, and equidistant from two parallels AL and BM. Ex. 23. At a distance a from a circumference, and equidistant from two intersecting lines. Ex. 24. Equidistant from Q and B, and also from two intersecting lines AL and BM. VIII. POINTS — EQUIDISTANT AND RANDOM 73 VIII. SUMMARY OP PROPOSITIONS IN THE GROUP ON POINTS — EQUIDISTANT AND RANDOM 1. Every point on the mid-jjerpendicular of a line- segment is equidistant from the ends of the line-segment. 2. Evey^y point loithout the mid-perpendicular of a line-segment is unequally distant from the ends of the line-segment. ScH. 1. The mid-perpendicular of a line-segment is the locus of points equidistant from the ends of the line-segment. ScH. 2. Two points, each of which is equidistant from the ends of a line-segment, determine the mid-perpendicular to the line. 3. Every point in the hisector of an angle is equidis- tant from the sides of the angle. 4. Every point ivithout the hisector of an angle is unequally distant from the sides of the angle. ScH. 1. Two points, each of which is equidistant from the sides of an angle, determine the bisector of the angle. ScH. 2. The bisector of an angle is the locus of all points equidistant from the sides of the angle. Gen. Sch. Two points in any straight line locus determine the locus. 74 THE ELEMENTS OF GEOMETRY NINE ILLUSTRATIONS OP ELEMENTARY PRINCIPLES OF LOCI 1. The locus of a point at a given distance from a given Point. 2. The locus of a point at a given distance from a given Straight Line. 3. The locus of a point at a given distance from a given Circle. 4. The locus of a point equidistant from two given Points. 5. The locus of a point equidistant from two given Straight Lines which intersect. 6. Tlie locus of a point equidistant from two Parallels. 7. The locus of a point equidistant from two Concentric Circles. 8. The locus of a point from which perpendiculars may be drawn to a given straight line, (a) to a given point in the line ; (b) through a given point without the line. 9. The locus of a point from which obliques may be drawn making a given angle with the line, (a) to a given point in the line ; (6) through a given point without the line. NINE EXERCISES IN LOCI Because of the frequent use of The questions on the left are the idea of the locus in the subse- given in familiar, everyday lan- quent demonstrations, it is of the guage. utmost importance that the pupil The statements below are an- become thoroughly familiar with swers to the questions opposite, these simple yet fundamental no- and are given in the language of tions of the locus. geometry. 1. What line contains all the 1. The locus of a point 1 mile houses that are 1 mile distant from distant from a given point is a cir- the city hall ? cle whose center is the given point Ans. The circle whose center is and whose radius is 1 mile, the city hall and whose radius is The locus of a point at a given 1 mile, distance from a given point is a cir- What is necessary to determine cle whose center is the given point the size and position of a circle ? and whose radius is the given dis- tance. Ex. Define a circle as a locus. 2. What line or lines contain all 2. The locus of a point 1 mile the houses 1 mile distant from a distant from a given line consists of main street in your city ? two parallels to the given line, one on each side, 1 mile from the line. VIII. POINTS — EQUIDISTANT AND RANDOM 75 Note. — By distance is meant, unless otherwise stated, the perpen- dicular distance. What is necessary to fix or de- termine the position of a line ? 3. What line contains all the flower pots that may be placed 10 feet from a circular path whose diameter is 100 feet ? The distance from a circle is always measured on a radius, or radius produced. What is the name of two or more circles that have the same center ? 4. What line contains all the hy- drants that may be placed equidis- tant from the ends of a straight street ? What determines the position of this line ? 5. What line contains all the hydrants that may be placed in a park so as to be equidistant from two intersecting straight paths ? . 6. What line contains all the points that are equidistant from the The locus of a point at a given distance from a given line consists of two parallels to the given line, one on each side, at a given distance from the line. Ex. By what axiom are the above lines determined in position ? Under which of the two ways of stating this axiom does the determi- nation directly fall ? 3. The locus of a point 10 feet from a given circle whose diameter is 100 feet, cormsis o/two concentric circles whose radii, respectively, are 60 feet and 40 feet. The locus of a point at a given distance a from a given circle whose radius is r, consists of two concentric circles whose radii, respectively, are r -\- a and r — a. Ex. What determines the posi- tion and size of a circle ? 4. The locus of a point equidis- tant .from two given points is the mid-perpendicular to the join of the two points. Ex. What is the direction of this locus with reference to the join of the two points ? (v. definition of direction.) Note. — The mid-perpendicular to the join of two points is also the locus of the centers of circles, any one of which passes through both points. 5. The locus of a point equidis- tant from two given intersecting straight lines consists of two lines, and bisecting the angles formed by the lines. 6. The locus of a point equidis- tant from two parallels is a line 76 THE ELEMENTS OF GEOMETRY two rails of a cable street railroad ? What determines the position of this line ? 7. What lino contains all the flower pots that may be placed equidistant from two concentric circular paths with radii of 50 feet and 76 feet, respectively ? 8. What line contains all the windows in a high building from which a boy may drop apples into a basket standing against the build- ing, on the level sidewalk ? 9. (1) Along what line should we find all the telegraph poles on which wires may be strung in north- east and southwest direction, to cross an east and west county road at the schoolhouse (a) on the county road, (6) 1 mile from the county road ? (2) Same question for a north- west and southeast telegraph line. parallel to them, midway between ihem. 7. The locus of a point equidis- tant from two concentric circles whose radii are, respectively, 50 feet and 75 feet, is a circle concentric with the given circles, whose radius is 62 J feet. The locus of a point equidistant from two concentric circles of radii a and 6 is a circle concentric with the given circles of radius ^-i— . 8. (a) The locus of a point from which perpendiculars may be drawn to a given line at a given point in the line is the perpendicular to the line at the given point. (6) The locus of a point from which perpendiculars may be drawn to a given line through a given point without the line is the perpendicular to the line from the point. 9. The locus of a point through which obliques may be drawn to a given line, making an angle equal to one half a right angle (a) at a given point in the line, (h) at a given point without the line, is the line through the given point (1) making half a right angle with the line ; (2)* making a negative angle equal to half a right angle with the line. Note. — In order to prove that a locus consists of a line, or a set of lines, it is necessary to show First, that every point on the line, or set of lines, fulfills the given conditions. Second, that no point without the line, or set of lines, does fulfill the given conditions. VIII. POINTS— EQUIDISTANT AND RANDOM 77 CHART PROBLEMS Note. — In order to answer the questions asked, students are at liberty to change dimensions and must sometimes for theoretical reasons deal with impractical conditions. Let us assume, for the purpose of illustration, that The Pirate'' s Chart gives the following description of the locations of his buried treasure : 1. The first is a half mile from an oak, and at the same time is three quarters of a mile from a chestnut. Locate the treasure. (v. Locus, Ex. 1.) When are there two possible locations ? When none ? [Draw a diagram for the above and for the following exercises. In the diagram make the given line or lines solid, the loci dotted.] 2. The second is a quarter mile from the shore of a shallow circular pond, whose radius is one mile, and simultaneously is a half mile from a neighboring straight beach. Locate the treasure. (v. Locus, Exs. 2 and 3.) When may there be eight such locations ? 3. The third is equidistant from the oak and the chestnut, and simul- taneously is one and a half miles distant from the shore of a neighboring circular pond, whose radius is one quarter of a mile. Locate the treasure. (v. Locus, Exs. 4 and 3.) When may there be four such locations ? 4. The fourth is equidistant from the turnpike and the valley road, and is simultaneously equidistant from the oak and the chestnut. Locate the treasure. (v. Locus, Exs. 5 and 4.) 5. The fifth lies on a line making with the turnpike a positive two thirds of a rt. Z, and passing through the oak ; also on a line making with- the turnpike a negative one half of a rt. Z, and passing through the chestnut. Locate the treasure. (v. Locus, Ex. 9.) Suppose the trees were (a) upon the turnpike, (6) remote from it. 6. The sixth lies on a perpendicular to the turnpike at the school- house ; also on a line passing through the oak and || to the valley road. Locate the treasure. (v. Locus, Ex. 8.) 7. The seventh lies on a perpendicular through the oak to the join of the oak and the chestnut, and is simultaneously one mile from the oak. Locate the treasure. (v. Locus, Exs. 8 and 1.) Note. — Descriptions 5. 6, and 7 are applications of Locus, Exs. 8 and 0. These two exercises should be given careful attention. IX. GROUP ON THE CIRCLE AND ITS RELATED LINES DEFINITIONS A Secant is a line cutting a circumference in two points. A Chord is the join of any two points on a circumference. The arcs that have the same extremities as a chord are said to be subtended by the chord. The greater of the two arcs is called the major, and the smaller the minor, arc. A Diameter of a circle or circumference is a chord that passes through the center. A Tangent is a line that touches a circle or circumference in but one point. Two circles are tangent to each other when they are tangent to the same line at the same point. An Arc is any part of a circumference. A Segment of a circle is that portion of the circle contained between an arc and the chord having the same extremities as the arc. This chord is said to subtend the arc. A Sector of a circle is that portion of the circle contained between two radii and the arc that they intercept. Corollaries op the Definitions 1. Circles with equal radii are congruent. (See Defini- tions, pp. 2, 8.) 2. A line tJiat intersects a circumference intersects it in two points and no more. 3. Any diameter of a circle bisects it. 4. A tangent may he considered as obtained by re- volving a secant about either point of secancy until the two points coincide. 78 IX. THE CIRCLE AND ITS RELATED LINES 79 PROPOSITIONS IX. 1. A radius perjyendicidar to a chord bisects the chord and its subtended arc, and convei^sely, F Hyp. If, in O K^ the fadius KF is perpen- dicular to the chord AB, Cone. : then (a) KF bisects the chord AB. (b) KF bisects the arc AFB. Bern, (a) Draw KA and KB. A AKB is isosceles. (Sides are radii.) .-. KF bisects the chord AB. [In an isosceles A the altitude to the base, etc.] (IV. 1 a, Sch.) Q.E.D. Dem. (b) On KF as an axis revolve the sector KBF to the plane of sector KFA. B will fall on A. (Why ?) Moreover, the arcs AF and FB will coincide throughout, as all radii are equal. .-. arc^i^= arc^F. Q.E.D. Hyp. Conversely, if, in a circle, a radius .KFbisects a chord J.5, Cone. : then /fjP ± chord AB. Dem. AK= BK. (Radii of same O.) .•. A ABK is isosceles. (Def. of isos. A.) .*. KF is the altitude to the base of isos. A ABK. [In an isosceles A the altitude to the base, etc.] (IV. 1 a, Sch.) .-. KF±choidAB. Q.E.D. THE ELEMKKTS OF GEOMETRY IX. 1 a. A radius 2}erpe7idicular to a chord is mid' ^erfendiadar to every chord parallel to the first - Hyp. If, in O 7i, the radius KF is perpendicular to the chord ^B, and if CJ57 is any chord parallel to the chord ABy Cone : then KF is the mid-JL to the chord CE. Dem. KFA. AB. (Hyp.) .-. KFA. CE. (i)ef. of lis, first inference.) .-. KF is the mid-X to CE, (IX. 1.) Q:ED. IX. 2. In the same circle, or in equal circles, equal chords are equidistant from the center, and conversely. Hyp. If, in O K, the chord AB = the chord CE, Cone. : then the ± KH = the ± KF. Dem. Draw the radii, KB, KA, KC, and KE, A KAB ^ A KCE. [If two A have three sides of one equal, etc.] (V. 3.) .-. KH = KF. (Sch. to Th.'s ^ A, Group V.) Q.E.D. Hyp. Conversely, if the ± KH= the l.KFm. the O K, Cone. : then the chord AB = the chord CE. IX. THE CIRCLE AND ITS RELATED LINES 81 Dem. It. A HKA ^ rt. A HKB ^ rt. A FKC ^ rt. A FKE. [Right A are ^ if a leg and hypotenuse of one, etc.] (V. 4.) " .-V AH= HB = CF= FE. (Horn, sides of ^ A.) .-. the chord ^5= the chord CE. (Ax. 2.) Q.E.D. IX. 3. Li the same circle, or in equal circles, equal angles at the center subtend equal arcs on the circum- ference, and conversely. Hyp. If, in the equal (D K and A^ Cone. : then arc AB = arc CE. Dem. Place Z K in coincident superposition with its equal Z Ki, A falling on (7; then B must fall on E. ' ' (Ax. 7.) KA=KiC and KB=KiE, being radii of circles equal by hyp. .-. arc AB will coincide with arc CE. (The CD are = by hyp.) Q.E.D. , Converse. Proof of the converse is left to the pupil. IX. 3 a. In the same circle, or in equal circles, equal chords subtend equal arcs, and conversely. Ex. 1. In any circle the greater of two arcs is subtended by the greater chord, and conversely. (Use IX. 3, and Ex. 9, p. 65.) Ex. 2. In any circle, of two unequal chords the one nearer the center is the greater, and conversely. Let AB and CE be the chords, KG, the distance of AB from the cen- ter iT, beina; greater than KF, the distance of CE from K. On KG lay off KM = KF, and draw through M, HL i. KM. HL = CE (iX. 2). But arc HABL > arc AB by tlie sum of arcs HA and B.L. Hence, chord HL > chord AB by Ex. 1. Ex. 3/' If two circles are concentric, tangents to the first circle that are chords of the second are equal. Ex. 4. If a radius can bedrawn bisecting the angle between two inter- secting chords, the chords are equal. 82 THE ELEMENTS OF GEOMETRY Tangency IX. 4. A radius to a point of tangency is perpendicu' E iV lar to the tangent Hyp. If TN is a tangent, and KT is a radius to the point of tangency, Cone. : then TKl. TN. Dem. Let KE be the join of A" and any point of T^ except T. Then E must be without the circle. (Def. of tangent.) .% KT^ Hyp. If PG i Kv j— -™^P and Pr are tan- V \ J^^^^ gents to O K, ^^ — *q Cone: then PG = PT. Dem. Draw radii to the points of tangency T and G ; also draw PK. py j_ j^y . _pQ _l j^q (ix. 4.) rt. A PKT ^ rt. A PGK. [Two right A are ^ if hypotenuse and leg, etc.] (V. 4.) .-. PG = PT. (Horn, sides of ^ A.) Q.E.D. IX. 5 a. The join of the center and the iiitersection of tioo tangents is the bisector of the angle made hy the tangents, and of the angle made hy the radii to the points of tangency. 84 THE ELEMENTS OF GEOMETRY IX. 6. If two circles intersect, the line of centers is the mid-perpendicular of the common chord. Hyp. If iT intersects O Ki in the points A and Bj Cone. : then KK^ is mid-_L to the common chord AB. Dem. Draw the radii KA, KB, and K^A and K^B. K is equidistant from A and B. {KA and KB being radii.) Ki is equidistant from A and B, {K^A and K^B being radii.) .'. KK^ is a mid-perpendicular to AB. [Two points equidistant from the ends of a line fix the mid- perpendicular to the line.] (VIII. 2, Sch. 2.) Q.E.D. IX. 7. i/* ^ic;o circles are tangent, their centers and the point of tangency are in the same straight line. Hyp. If Oir is tangent to O Ki at T, Cone. : then K, T, and K^ are in the same straight line. Dem. Draw NG^ a common tangent through the common point T. (Def. of tangent ©.) KTJlNG. [A radius to point of tangency is ± to tangent.] (IX. 4.) KiT±NG. (For the same reason.) /. -KTand KiT Sire in the same straight line. (Ax. 7.) Q.E.D IX. THE CIRCLE AND ITS RELATED LINES 85 IX. 8. If two circles are tangent, the distance hetiveen their centers is (a) the Slim of the radii, if the tangency is external ; (h) the difference of the radii, if the tangency is in- ternal. Fig. 2, Hyp. If ^1 and 7^2 are tangent circles at the point T, and their radii are r^ and ^2, Cone: then (Fig. 1) 7^1/^2 =ri+r2 and (Pig. 2) KJC2=ri—r2. Dem. Draw the common tangent LM through T. (Def. of tangent (D.) Then Ki, K^, and T are in the same straight line. (IX. 6.) .-. (a) K,T-\- TK^^KJ^:^, (Ax. 4.) and (6) K^ T- TK^ = K^K^ ; (Ax. 4.) that is, (a) K1K2 = ^i + »'2 and (&) K1K2 = i\ — r^. Q.E.D. IX. 8 a. Conversely. If the distance hetiveen the centers of tivo circles is equal to (a) the sum of the radii, the circles are tangeiit to each other externally ; (b) the difference of the radii, one circle is tangent to the other internally. Ex. 5. Show that if the distance between the centers of two circles is (a) greater than the sum of the radii, each circle is without the other ; (6) less than the sum of the radii, but greater than their difference, the circles intersect each other ; (c) less than the difference of the radii, one circle i& wholly within the other. 86 THE ELEMENTS OF GEOMETRY IX. SUMMARY OF PROPOSITIONS IN THE GROUP ON THE CIRCLE AND ITS RELATED LINES 1. A radius perpendicular to a chord bisects the chord and its subtended arc, and conversely, a A radius perpendicular to a chord is mid-per- pendicular to every chord parallel to the first 2. In the same circle, or in equal circles, equal chords are equidistant from the center, and conversely. 3. Li the same circle, or in equal circles, equal angles at the center subtend equal arcs on the circumference, and conversely. a In the same circle, or in equal circles, equal chords subtend equal arcs, and conversely. 4. A radius to a point of tangency is perpendicular to the tangent. a A line perpendicular to a radius at the outer extremity of the radius is the tangent to the circle at that point, h A perpendicular to a tangent at the point of tangency passes through the center of the circle. 5. Tangents from the same point to the same circle are equal. a The join of the center and the intersection of two tangents is the bisector of the angle made by the tangents, and of the angle made by the radii to the points of tangency. IX. THE CIRCLE AND ITS RELATED LINES 87 6. If huo circles intersect, the line of centers is the mid-perpendicular of the common chord. 7. If two circles are tangent, their cejiters and the point of tajigency are in the same straight line. 8. If two circles are tangent, the distance hetiveen their centers is (a) the sum of the radii, if the tangency is external ; (5) the difference of the radii, if the tangency is iiiternal, 8 a. Conversely. If the distance hetween the centers of two circles is equal to (a) the sum of the radii, the circles are tangent to each other externally ; (b) the difference of the radii, one circle is tangent to the other internally. The Isosceles Triangle as Part op the Sector of a Circle Note. — In an isosceles triangle, the altitude to the base is identical with the median to the base. (IV. 1 a, Sch.) Note. — Observe that if the vertex of the vertex angle of an isosceles triangle be taken as a center and a circle be described with either leg as a radius, the legs of the triangle are radii of the circle ; the base of the triangle is a chord, and the altitude to the base, the median, and the bisector of the vertex angle (which we have seen in (IV. 3) to be the same line) are a part of the radius perpendicular to the chord. 88 THE ELEMENTS OF GEOMETRY PROBLEMS Prob. I. To construct a tangent through a given point to a given circle. Given. A circle Ky and a point P. Required. To construct a tangent through P to the circle K. Case I. When P, the given point, is on the circle. What is the angle formed by a tangent and a radius drawn to the point of tangency ? What is the construction required ? Case II. When P, the given point, is without the circle. Const. Join P and K, and on PK as a diameter describe a circle intersecting the given circle in T and G. Then PT and PG are the required tangents. Proof. Let the pupil supply the proof. - Why is Z PTK a right angle ? Prob. II. To construct a common exterior tangent to two given circles ivhose radii are r a7id r. Given. Two circles, K and /fi, whose radii = r and r'. Required. To construct a com- mon exterior tangent. Const. With if as a center and a radius =r — r', draw the inner concentric circle. From Ki draw a tangent to this circle. (Prob. I.) Q.E.F. (VII. 4 rt.) IX. THE CIRCLE AND ITS RELATED LINES 89 Draw KT to the point of tangency, T, and produce it to meet the outer circle in N. Through iV^draw NL II TK,. (Prob. IV., p. 50.) Through K^ draw a line II /0»''and meeting NL, say at 6r. KG is the reqitired common exterior tangent. Q.E.F. Proof. The 4-side T— G is a parallelogram. (Def. of a O.) .-. K^G = TN. [Opposite sides of a parallelogram are equal.] (YI. 1 a.) But TN=r' by construction. .-. K,G = r' (Ax. 1.) But Z T is a right angle. (IX. 4.) .-. all the A of the 4-side T— G are right angles. (VI. 1' a'.) .*. iVGi^ is a common exterior tangent to© /f and ^1. (IX. 4 a.) Q.E.D. Note. — Another common tangent may be found, crossing KKi between the circles, and therefore called an interior tangent. In this case the first auxiliary circle has the radius = r + r' instead of r — r'. Let the student give the construction in full. Show that four com- mon tangents are possible. When may three only be drawn ? When two ? When one ? When none ? Ex. 6. (a) Draw a chord equal and parallel to a given chord. (6) Draw a chord equal and perpendicular to a given chord. Ex. 7. If, in a circle, two equal chords are drawn, and a radius is drawn to the end of each chord, the angles between the radii and the chords are equal to each other. c Ex. 8. In the same or equal circles the greater of two minor arcs is subtended by the greater chord. Ex. 9. If chord ^i?'= chord BC, then arc AB = arc CF. 90 THE ELEMENTS OF GEOMETRY Ex. 10. Show that two chords that are not diameters cannot bisect each other. Ex. 11. Prove by means of IX. 3 a that two triangles are congruent if tliree sides of the one equal three sides of the other. Ex. 12. If an inscribed polygon is equiangular, it is not necessarily equilateral. (y Ex. 13. If, in Fig. 1, the secant is drawn so that ^ AB = BKy show that Z CKF = 3 Z J. Ex. 14. Two parallel chords intercept equal arcs on the circumference. (Fig. 2.) Ex. 15. A chord and a parallel tangent inter- cept equal arcs on the circumference. Ex. 16. Draw : (1) Two common exterior tangents to two circles. (2) Two common interior tangents to two circles. Ex. 17. Prove that the above exterior tangents are equal. Ex. 18. In Fig. 3, O Xis tangent to O i^ at T, and is also tangent to LP at P. Why are K^ X, and T in the same straight line? Ex. 19. If O X is tangent to LF at P, why is Z LPA a right angle ? Ex. 20. If PA = KT, why is A KAX .isosceles ? Ex. 21. How, then, if PA and KA are given in position and length, may the points X and T be determined ? Ex. 22. Problem : Given the line LF, the point P in LF, and the O K, construct a circle that shall be tangent to LF a,t P and also tangent to O K. Ex. 23. Similarly, by laying off PA' (= PA) below LP, find the center X of a second circle that shall also be tangent to LF at P and likewise tangent to O K. Ex. 24. Show that the hypotenuse of a rt. A equals the sum of the two remaining sides, minus twice the radius of the in- scribed O. X. GROUP ON CONCURRENT LINES OF A TRIANGLE DEFINITION Lines are Concurrent when they intersect at the same point. PROPOSITIONS X. 1. The bisectors of the interior angles of a triangle concur. Hyp. If ABC is a triangle, Cone. : then the bisectors oi Z. A, Z. B, and Z. C concur. Dem. The bisector of Z ^. either meets, or is parallel to, the bisector of Z C. / a / n If the bisectors are II, the ^ + ^ = 2 rt. A (II. 2.) .'. ZA + ZC = 4: rt. A. (Ax. 3.) But this conclusion is impossible, (III. 1.) .'. the bisector of Z A must meet the bisector of Z C. Let the point of intersection be Ki. Ki, in the bisector of Z C, is equidistant from a and b, and Ki, in the bisector of Z A, is equidistant from b and c. [Every point in the bisector of an angle, etc.] (VIII. 3.) .*. Ki is equidistant from a and c. (Ax. 1.) .-. Ki is in the bisector of Z B. (VIII. 4, Sch.) .'. the bisectors of ZA,ZB, and Z C concur. Q.E.D. This point of concurrence is called the In-center. 91 92 THE ELEMENTS OF GEOMETRY X. 1 a. If the in-center he taken as a center, and the distance to any side as a radius, a circle may he drawn tange7it to the sides of the triangle. X. 2. .The hisectors of one interior angle of a triangle and the two exterior angles non-adjacent to it concur, Cy Hyp. If ABC is a triangle, Cone. : then the bisectors of ext. Z A, ext. A By and int. Z C concur. Dem. The bisector of Z (7 either meets, or is parallel to, the bisector of ext. Z A, ' If parallel, then ^^LAA = ^. [If two parallels be crossed by a third line, etc.] (II. 1.) .-. ext. Z^ = Za (Ax. 3.) But this conclusion is impossible, [The exterior angle of a A is greater, etc.] (III. 2, Sch.) .-. the bisectors of ext. A A and of Z C must meet, say at K^, K^, in the bisector of Z C, is equidistant from a and b (pro^ duced). [Every point in the bisector of an angle, etc.] (VIII. 3.) K^, in the bisector of ext. A A, is equidistant from c and b. (Same reason.) . , Kfis equidistant from c and a. (Ax. 1.) .*. K^ must lie in the bisector of ext. Z B. X. CONCURRENT LINES OF A TRIANGLE 93 [The bisector of an angle is the locus, etc.] (VIII. 4, Sch.) .*. the bisectors of ext. Z.A, ext. Z B, and int. Z C concur. Q.E.D. The point of concurrence is called an Ex-center of the tri- angle. There are two other ex-centers : viz., K^. on the bisector of Z A, and K^.. on the bisector oi Z.B. Sch. To three non-concurrent lines, three tangent circles, (besides the inscribed circle already indicated) may be drawn, with K^, K^', Ke" as centers, and the distances from these points to the corresponding lines as respective radii. Circles tangent to one side of a triangle and to the two other sides produced, are called Escribed Circles. Ex. 1. In A ABC, I is the center of the in- scribed circle. Show that Z / = rt. Z-\ ^• 2 Ex. 2. In A ABC, E is the center of the escribed circle. Show that Z ^ = rt. Z - — • 2 Prove that ZE oi Fig. 2 is the supplement of Z/of Fig. 1. Find the locus of the center of a circle that satisfies the following conditions : Ex. 3. That passes through Q, and has a radius a. Ex. 4. That touches a line AL, and has a radius a. Ex. 5. That touches a circumference, and has a radius a. Ex. 6. That passes through two given points. Ex. 7. That is tangent to two intersecting lines. Ex. 8. That is tangent to two parallel lines. Ex. 9. That is tangent to two equal circumferences. Ex. IQ. That is tangent to a line at a given point. Ex. 11. That is tangent to a circumference at a given point. 94 THE ELEMENTS OF GEOMETRY X. 3. The mid-perpendiculars to the three sides of a trimigle concur. Hyp. If ABC is a triangle, Cone. : then the mid-perpendiculars to a, 6, and c concur. Dem. The mid-perpendicular to a either intersects, or is parallel to, the mid-perpendicular to h. If they be parallel, AG and CB would have, (1) the same direction, •.• both would be perpendicular to these parallels ; and (2) the point G in common. .'. AG and GB would lie in the same straight line. (Ax. 7.) But this conclusion is contrary to the hypothesis. that ABG is a triangle. .-. the mid-perpendicular to a must intersect the mid-perpen- dicular to h in some point, say K,.. K^, in the mid-perpendicular to a, is equidistant from (7 and B. [Every point in the mid-perpendicular, etc.] (VIII. 1.) ifg, in the mid-perpendicular to 6, is equidistant from G and A. (Same reason.) .'. -H^c is equidistant from A and B. (Ax. 1.) .•. Kc is in the mid-perpendicular to c. [The mid-± to a line-segment, etc.] (VIII. 2, Sch.) .-. the mid-perpendiculars concur. Q.E.D. The point of concurrence is called the Circumcenter. Sch. Through three non-collinear points an unique circle may be drawn. X, CONCURRENT LINES OF A TRIANGLE 95 X. 4. The altitudes of a triangle concur. Hyp. If ABC is a triangle, ^7""- ^1 Cone. : then the altitudes AL, BT, and C/ concur. Dem. Through C, A, and B draw parallels to c, a, and 6, respectively. Produce these parallels to intersect in Cj, A^, and By. The 4-sides ABCB^ and ABA^C are ZI7. (Def. of O.) .-. ^iC= ^S; CA = AB. (VI. 1 a.) .'. B,C=CAi. (Ax. 1.) Again, (7/± BiA^. (Def. of lis, direct inference.) .*. CI is the mid-perpendicular to ByAy, Similarly, BT is mid-perpendicular to A^'ij -^-^j mid-per- pendicular to BiCi. .'. the altitudes of the original triangle are mid-perpendicu- lars to the sides of the larger triangle. But these mid-perpendiculars concur. (X. 3.) .*. the altitudes of the original triangle concur. Q.E.D. The point of concurrence is called the Orthocenter. Construct a circle that satisfies the following conditions : t Ex. 12. That has a given radius a, and passes through two given points. Ex. 13. That has a given radius a, and is tangent to two given inter- secting lines AL and BM. Ex. 14. That has a given radius a, and is tangent to two equal cir- cumferences. Ex. 15. That has a given radius a, and is tangent to a given line at a given point. 96 THE ELEMENTS OF GEOMETRY X. 6. The medians of a triangle concur. Hyp. If ABC is a triangle, Cone. : then the medians AH, BL, and CF concur. Dem. CF and BL intersect, or are parallel. If CF II BL, all points in CF must lie on the same side of BL. But, as a consequence of the definition of a median, BL must lie between c and a. ,'. C and F must lie on opposite sides of BL. .'. CFand BL must intersect, say at Kg. Draw AKg ; draw LJ and FO II ^^, ; also LF, and OJ". Then i J and i?^0 each II AK„ and = ^. [If in a A a parallel to the base, etc.] (VII. 3 c.) .'. the 4-side LJOF is a parallelogram. [A 4-side is a parallelogram if it has one set, etc.] (VI. 2.) .-. KgF=KgJ. [The diagonals of a parallelogram mutually bisect.] (VI. 3.) Now JKg^JC. (Const.) .-. KgF = ^. Similarly, K,L = ^ . .*. as any two medians cut off the same third of the third median, the three medians must concur. Q.E.D This point of concurrence is called the Center of Gravity, oi Centroid, of the triangle. Ex. 16. Construct a circle that has a given radius a, and is tangent to a given circumference at a given point. X. CONCURRENT LINES OF A TRIANGLE 97 Ex. 17. That has a given radius a, passes through a given point Q, and is tangent to a given line AL. Ex. 18. That has a given radius a, passes through a given point Q, and is tangent to a given circumference. Ex. 19. That has a given radius a, is tangent to a given line AL, and is also tangent to a given circumference. X. SUMMARY OF PROPOSITIONS IK THE GROUP ON CONCURRENT LINES OF A TFIANGLE 1. The bisectors of the interior angles of a triangle concur, a If the in-center he taken as a center, and the distance to any 07ie side as a radius, a circle may he drawn tangent to the sides of the triangle. 2. The hisectors of one interior angle of a triangle and the two exterior angles non-adjacent to it concur. ScH. To three non-concurrent lines three tangent circles may be drawn, with K^, K^., K^.. as centers and the distances from these points to the corresponding lines as respective radii. 3. The mid-perpendiculars to the three sides of a tri- angle concur, 4. The altitudes of a triangle concur. ScH. Through three non-collinear points an unique circle may be drawn. 5. The medians of a triangle concur. 98 THE ELEMENTS OF GEOMETRY Fig. 1. X. CONCURRENT LINES OF A TRIANGLE 99 SUMMARY OF TRIANGULAR RELATIONS Important Properties of the Angles of a Triangle (Fig. 1) 1. ZA + ZB-{- ZC = 2Tt. A. (in. 1.) 2. ZAis the supplement ot Z B -{- Z 0. (III. 1 a. ) 3. If Z C is a right angle, Z B is the complement of Z A. (III. 16.) 4. liZA = ZB, a = b. (IV. 2.) 5. ItZA = ZB=:ZC,a = b = c. 6. IiZA>ZB, a>b. (IV. 5.) 7. Ext. ZA = ZB-\-ZC. (in. 2.) 8. Ext. ZA> ZB or ZC. (IIL 2, Sch.) 9. It ZA = ZB, ext. ZC = 2ZAor2ZB. 10. The shape (not size) of a triangle is given by any two of its inde- pendent angles. Important Properties of the Lines op a Triangle (Fig. 2) 1. a + b>c. (VIL 1.) 2. lta = b,ZA=ZB. (IV. 1.) 3.lfa = b = c,ZA = ZB = ZC. 4. The bisector of Z J. is perpendicular to the bisector of ext. Z A. 5. The bisectors of ZA, Z B, and Z C concur at the in-center. (X.L) 6. The bisectors of ext. Z A, ext. Z B, and of Z C concur at an ex-center. (X. 2.) 7. The mid-Js to a-, &, and c concur at the circumcenter. (X. 3.) 8. The altitudes to a, b, and c concur at the orthocenter. (X. 4.) 9. The medians to a, 6, and e concur at the centroid. (X. 5. ) 10. The shape and size of a triangle are determined by any three independent parts. Important Properties of the Lines op a Right Triangle (Fig. 3) 1. The median to the hypotenuse equals J the hypotenuse. (VII. 4.) 2. If Z ^ = I rt. Z, the median to the hypotenuse equals b. 3. The altitude to a coincides with &, and vice versa. 4. The hypotenuse is the diameter of the circumcircle, (VII. 4.) 5. {a -{- b)— c is the diameter of the inscribed circle. (Ex. 24, p. 90.) 6. The hypotenuse, c, > a or b. (IV. 4 a.) XI. GROUP ON MEASUREMENT DEFINITIONS Measurement is (a) Direct, or (b) Indirect.* The Direct Measurement of a magnitude is the process of finding how many times it contains another magnitude of the same kind, which is called the unit of measure. E.g. length may be measured in feet, miles, meters, kilo- meters, etc. ; area in acres, square miles, hectares, etc., and weight in kilograms, pounds, tons, etc. Any line may be assumed as a Unit of Length. The Indirect Measurement of a quantity is the process of de- termining its size by comparing it with some other quantity, the changes in size of which correspond to changes in size of the first magnitude. E.g. the pressure of steam is measured by the changes in position of a hand on a dial plate. The height of a mountain is measured by the motion of the index on an aneroid barometer. The strength of an electric current is measured by the tem- perature to which it raises a wire of known dimensions. The pitch of an organ pipe is measured by the length of the pipe. The amount of acid in a solution is measured by the intensity of the color it produces in a piece of litmus^ paper. Angular Measure. Among geometrical magnitudes an angle is often measured by its intercepted arc, or by the quotient of the intercepted arc divided by the radius of the circle ' 1 Whether measurement is direct or indirect, the object attained is the same. The measure of a magnitude is a ratio ; therefore, always abstract. 100 XI. MEASUREMENT 101 whose center is the vertex of the angle. The-^tjber;!? cal!(?d; Radial Measure. ,>>.,., ^ ^ >', ^ Ratio and Proportion ; ' ' - , - • '' ' '' The Geometric Ratio of one magnitude to another is the quotient obtained by dividing the first by the second. Thus, the ratio of a to 6 is - = a h- & = a : 6. b The Antecedent of a ratio is the first, or dividend magnitude. The Consequent of a ratio is the second, or divisor magni- tude. The usual Sign of ratio is :, although any method of indi- cating division may be used. Both terms of any ratio may be multiplied or divided by the same quantity, without affecting the value of the ratio. Commensurable Ratios When the terms of a ratio can each be expressed as a multi- ple of a common unit, the terms are said to be commensurable with each other, and the ratio is said to Commensurable. In this case the ratio can always be expressed as a numerical fraction, both of whose terms are whole numbers. Problem I. To express the ratio of tivo line-segments. Given. AB and CE. , L_ Z GB Required. The ratio AB.CE. G Hi Apply CE to AB as often as possible, say twice, with a remainder GB, so that AB = 2CE-\-GB. Apply the remainder GB to CE as often as possible, say four times, with a remainder FE, so that CE=:4.GB + FE, and AB = SGB + 2FE+GB =:9GB + 2FE. 102 THE ELEMENTS OF GEOMETRY fVpply tht 'ffLSft remainder FE to OB as often as possible, say fo\:^r .ti|iif)9, withqut. remainder, so that '' '''•'>■' ■''• •''<'■''< '93= ^FE, and CE=16FE+ FE = 17 FE, and AB = 36 FE + 2 FE = SS FE. Thus the given lines have been expressed in terms of the common unit FE, and their ratio AB^ 3SFE ^3S CE 17 FE 17* The ratio ^ = ~ AB 38 Apply the same method to the solution of the following problems. Problem II. To express the ratio of two arcs of equal circles, arc AB and arc CE. Problem III. To express the ratio of two angles or sectors of equal circles, /. AKB and Z CK'E, ^ Incohmensubablb Ratios If, on the other hand, the terms of a ratio cannot be expressed as multiples of the same unit, the terms are incommensurable with each other, and the ratio is said to be Incommensurable. In this case no one of the remainder line-segments, arcs, angles, etc., of the process just indicated will be exactly con- tained in the preceding remainder line-segment, arc, angle, etc., no matter how long the process be continued. XI. MEASUREMENT 103 Hence, an incommensurable ratio cannot he expressed as a fraction the terms of which are whole numbers. Approximate Expression of Incommensurable Ratios Such a ratio can usually be expressed, however, in some form that will enable us to state the value of the ratio correctly to any required decimal place, or to any required degree of accuracy. Hence, all the ratios with which we shall deal may be ex- pressed, either exactly or approximately, as fractions. But the necessity for thus expressing them seldom arises. We shall be concerned chiefly with the relations between ratios, and one of the most important of these geometric relations is that of equality. This relation gives us the geometric proportion. A Geometric Proportion is an expression of equality between two or more geometric ratios. Thus, a:b = c :m, or, as oftener written, a:b :: c:m, and [jP:\JQ::h:h' are geometric proportions. If more than two ratios are compared, the proportion is said to be a Continued Proportion. Each ratio of a proportion is called a Couplet. The Extremes are the first and fourth terms of a proportion. The Means are the second and third terms of a proportion. It a:b::c: e, e is said to be a Fourth Proportional to a, b, and c. Similarly, 6 is a fourth proportional to a, c, and e, etc. li a:b ::b : c, cis said to be a Third Proportional to a and b. Similarly, a is a third proportional to b and c. li a: b::b: c, b is said to be a Mean Proportional between a and c. A Transformation of a proportion is a change in the propor- tion, either in the order of the terms or otherwise, that does not destroy the equality of the ratios. 104 THE ELEMENTS OF GEOMETRY A Derived Proportion is one obtained from a given proportion by transformation. Thus, a^:b^i:c^i€^ is derived from a :b ::c :e by cubing the terms of the latter. PROPOSITIONS XL 1. If four quantities are in j^roportmiy the prod- uct of the means equals the product of the extremes, and conversely. If a:b:ic:e, i.e. if - = ^, b e then ae = be. [If both members of the given equation be multiplied by be, the results will be equal.] (Ax. 3.) Q.E.D. Conversely, if the product of two quantities equals the product of two others, either set of factors may he made the extremes, and the other the means, of a proportion. Hyp. If ae = be, Cone. : then a:b::c:e, i.e. ^ = -. b e [If both members of the given equation be divided by be, the results will be equal.] (Ax. 3.) Q.E.D. ScH. 1. We have seen that in the original proportion the product of the means equals the product of the extremes. The test of the coiTectness of every derived proportion is that when the product of its extremes is placed equal to the product of its means, the resulting equation is the same as the equation similarly obtained from the origincd proportion, or may be reduced to the same. Hyp. If a : b : : c : e, Cone. : ; then (1) b : a : : e : c. Dem. If «=:^ then i = i. be a c b e b__e_ "a c XI. MEASUREMENT 105 Illustration. — If a:b: : c: e be the original proportion, then, by the test, a-{-b:b::c-\-e:e\s a correct derived pro- portion ; for, by the application of XI, 1 to each, we get in each case ae = be. (Let the student make the application.) ScH. 2. The most important transformations give the fol- lowing derived proportions : (Ax. 3.) Q.E.D. (This form is said to be derived from the given proportion by inversion.) Cone. (2) : a : e : : b : e. Dem. If ^ = ^, then ^x- = -x-. (Ax. 3.) be b c e c - — -, ' c e Q.E.D, (This form is said to be derived from the given proportion by alternation.) Cone. (3) : a + b : a : : c 4- e : c, and a-fb:b::c-He:e- Dem. If ^ = ^, then ^ + 1 = £ + 1. (Ax. 2.) be b e a + & _ c-f-e b ~ e ' Q.E.D. (This form is said to be derived from the given proportion by composition.) 106 THE ELEMENTS OF GEOMETRY Cone. (4) : a — b : a : : c — e : c, and a — b : b : : c — e : e. Dem. If ? = -, then, ?- 1=^-1. (Ax. 2.) be be " b " e ' Q.E.D. (This form is derived from the given proportion by division.) Cone. (5): a + b:a — b::c + e:e — e. Dem. Divide the last equation in conclusion (3) by the last equation in conclusion (4), member by member. ... ?L±:^ = ^±^. (Ax. 3.) a—b c—e Q.E.D. (This form is said to be derived from the given proportion by composition and division.) XI. 2. In any number of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. Hyp. If a: 6:: c: e::/:gr::etc., Cone.: then a + c+/+ '•• : & + « + g'-h •••: : a: 6:: c; e :: etc. Let ? = - = ^=^, beg whence a = br\ c = er\ f= gr, etc., and a + C+/+ ••• = 5r + er + grr + •••. Then a + c+f-i- ... = r(6 + e +^+ ...). b + e-\-g-i b e Q.E.D. XI. 3. If two proportions he multiplied together, term by term, the resulting products will he in propor- tion. Let the student supply the proof, by use of Ax. 3. XI. MEASUREMENT 107 a If any number of propor^tions he multiplied together, term by term, the resulting products will be in proportion. b Like powers of the terms of a proportion are in proportion. c Like roots of the terms of a proportion are in proportion. XI. 4. If the terms of a proportion be divided suc- cessively by the terms of a second, the residting quotients will be in proportion. Let the student supply the proof, using Ax. 3. Ex. 1. If 2/= c, what is the ratio of c to/? Ex. 2. If a = 3 e, what is the ratio of a to e ? Ex. 3. If a + e : a — e : : 7 : 5, what is the ratio of a to e ? What is the ratio of x and y in the following : Ex. 4. x + y:x::lS:i? Ex. 5. x-y:x::6:9? Ex. 6. x-hy:x-y::lZ:6? Give the name and value of x in each of the following proportions : Ex. 7. 5 : a; : : 10 : 12. Ex. 8. a; : 8 : : 10 : 16. Ex. 9. 5 : 11 : : 10 : a;. What is the name and what the value of x in the following : Ex. 10. 4 : X : : a : 36 ? Ex.11. rB:8::8:2? Give ten proportions that can be derived from the proportion : Ex. 12. 3 : 7 : : 9 : a:. Ex. 13. Test the correctness of your answer by showing that the product of the means and extremes in the derived proportions is identical with the product of the means and extremes in the original proportion. 108 THE ELEMENTS OF GEOMETRY Method op Limits DEFINITIONS A Variable is a quantity that in the course of a single dis- cussion is always changing its value. Thus, as the point X moves along the curve ABy its distance from the line AB, its distance from the point A^ and the projection AY, of this latter distance on the line AB, are all variables. A Constant is a quantity that does not change its value in the course of a single discussion. Thus, if the curve in the above figure be a semicircle on AB as a di- ameter, AB is a constant. The changes in the variables above mentioned produce no changes in AB. A Limit of a variable is a constant which the value of the variable may be made to approach as near as we please, but which the variable cannot be made to reach. That is, the dif- ference between the limit and the variable may be made less than any assignable quantity, but cannot be made zero. A G E F B To illustrate : Suppose a point moves from A toward jB so as to cover one half the distance in the first second, one half the remaining distance in the second, and so on. Will the moving point ever coincide with B ? In other words, will the variable distance covered by the moving point ever coincide with the constant line-segment AB? What, then, is the Limit of the variable distance gone over by the moving point ? If the distance passed over in the first second be called 1, that passed over in the second second will be \, that passed over in the third second will be i, and so on. The whole distance, therefore, say x, will be l + i + i + J+T^+^jH — • The greater the number of terms we take, the nearer x will approach the value 2. XL MEASUREMENT 109 1 + J + i = 1; 2- 1= i i+i+i+i =¥; 2-¥= i i + i + i + i + i^ = fi; 2 - fi = tV Thus, and so on. We can make x as near 2 as we please ; i.e. differ from 2 by as small a fraction as we choose by taking a sufficiently large number of terms. But, no matter how great the number of terms we take, their sum will never actually reach 2. /. 2 is said to be the limit of the sum of the terms. And AB is the limit of the sum of the segments J.C, CJS', EF^ etc. ; i.e, the Limit of the Variable Distance gone over by the moving point is AB. The symbol = is employed to denote the expression "ap- proaches as a limit," or any equivalent expression. Postulate of Limits. If, while approaching their respective limits, two variables are alivays equal, the limits are equal. For, since the two variables are equal at every stage of their progress, we have practically but one variable ; and it is impossible that one vari- able (increasing or decreasing) should be approaching two different limits at the same time. Direct inferences : (a) The limit of the pi*oduct of two variables is the product of their limits. (b) If two variables have a constant ratio, and each ap- proaches a limit, these limits, taken in the same order, have the constant limit of the ratios. Ex. 14. 25 and 49 are perfect squares. By what proposition does it follow that their product must be a perfect square ? Ex. 15. 27 and 125 are perfect cubes. By what proposition does it follow that 3375 is also a perfect cube? Ex. 16. Verify all the conclusions of XL 1, Sch. 2 by use of the test given in XL 1, Sch. 1. Ex. 17. Show that, if a : 6 : : 6 : c : : c : e, then a : e : : a* : 6*. 110 THE ELEMENTS OF GEOMETRY XI. SUMMART OF PROPOSITIONS IN THE GROUP ON MEASUREMENT 1. If four quantities are in 'proportion^ the product of the means equals the product of the extremes, and conversely, 2. In any number of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. 3. If two proportions he midtiplied together, term hy term, the resulting products will be in proportion. a If any number of proportions he midtiplied together, term by term, the resulting products will be in proportion, b Like powers of the terms of a proportion are in proportion. e Like roots of the terms of a proportion are in proportion, 4. If the terms of a proportion he divided successively hy the terms of a secojid, the resulting quotients will he in proportion. Postulate of Limits. If, while approaching their respective limits, two variables are always equal, the limits are equal. XII. GROUP ON MEASUREMENT OF ANGLES DEFINITIONS A Central Angle is an angle whose vertex is at the center of a circle. An Inscribed Angle is an angle formed by two chords inter- secting on the circumference. An Escribed Angle is an angle formed by one chord with another chord produced. An angle is inscribed in a segment when its vertex is in the arc of the segment and its sides pass through the extremities of this arc. The angle between two curves at any point of intersection is the angle formed by the tangents to the curves at this point. If the angle between two curves is right, the curves are said to cut each other orthogonally. Thus, a AT, BT (Fig. 1, Fig. 2) be tangent to the circles K and i, respectively, Z.ATB is the angle between the circles at T. If, as in Fig. 2, Z A TB is right, the circles are said to cut each other orthogonally. The angle KTL, between the radii to T, is supplemental to Z.ATB (IX, 1 and II, 5). Hence, \i /.ATB is right, Z KTL is right, and con- versely, that is, Two circles cut each other orthogonally when the radii to either point of intersection are perpendicular to each other. Ill 112 THE ELEMENTS OF GEOMETRY PROPOSITIONS XII. 1. In the same circUy or equal circles, central angles are to each other as their intercepted arcs. Hyp. Case I. If O A^ = O /li, and arc AB is to arc CE as any two whole numbers, say 6 to 4 (commensurable), Cone. : then Z AKB : /. CK^E : : arc AB : arc CE. Dem. Arcs AB and CE have a common measure. (Hyp.) Apply it to each of the arcs, and suppose it is contained six times in AB and four times in CE. * Join the points of division with the centers of the circles. All the central angles thus formed will be equal. (IX. 3.) .-. Z^7fJ3:ZC/A^::6:4. But arc^l^rarcO^ : : 6 : 4. .-. Z AKB. A CK^E'.'. arc ^5: arc CE. Q.E.D. Hyp. Case II. If Oir=0^i, and arc AB and arc CE are incommensura- ble, Cone. : then Z AKB : Z CK^E : : arc AB : arc CE. Dem. Divide arc AB into any number of equal parts, say 4, each less than arc CE. Let AM be one of these parts, and be contained in CE twice, with a remainder LE. As the remainder is always less than the divisor, it follows that if we increase the number of equal parts into which AB is divided, we diminish both the divisor and the remainder. XII. MEASUREMENT OF ANGLES 113 But, as the arcs AB and CE are incommensurable (Hyp.), the remainder can never be 0. ,\ arc CL = arc CE, and Z CK^L = Z CK,E. (XI. Def. of Limit.) ^"* 44^ always = ^^^> (XII. 1, Case I.) Z. (JKiL arc iJL .-. Z ^^5 : Z CK^E : : arc ^5 : arc CE. (XI. Post. Limits.) Q.E.D. ScH. Heretofore; we have measured angles directly (see p. 100), using the right angle as the unit of measure. This unit is inconvenient, however, as its use requires us to employ fractions too frequently. The foregoing theorem, due to Thales of Miletus (640 b.c), introduced a very simple method of indirect measurement. Any change in the magnitude of the central angle produces a proportional change in the intercepted arc. Thus, if the cen- tral angle be doubled, the intercepted arc is doubled; if the angle be trebled, the arc is trebled ; and so on. Hence, the intercepted arc may be taken as the measure of the central angle. This very important theorem may be stated thus : A central angle is measured by its intercepted arc. The circumference is divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes, and each minute into 60 equal parts, called seconds. For brevity, an angle measured by an arc of 1° is called an angle of 1° ; etc. Note. — This division of the circumference, known as the sexagesimal division (or division by sixties) , is due to the Babylonians. The Babylonian year consisted of 12 months of 30 days each, or 360 days. Accordingly the zodiac was divided into 12 signs of 30 degrees each, making 360 degrees for a complete circle. These people were also familiar with the fact that the radius of a circle, used as a chord, divides the circle into 6 equal parts of 60 degrees * each. Hence, arose the custom of subdividing the degree by sixties, into minutes, etc., as indicated above. 114 THE ELEMENTS OF GEOMETRY XII. 2. An inscribed angle is measured by half the intercepted arc. Case (1) B Case (2) li Hyp. If Z 5 is inscribed in OKy and if AC is the inter- cepted arc, Cone: then Z^ is measured by E^jd^. ^ 2 Case (1). When one arm of the angle is a diameter through B. Case (2). When the two arms are on opposite sides of the diameter through B. Case (3). When both arms are on the same side of the diam- eter through B. Dem. (1) Draw the radius CK, forming the isosceles A CKB. •Then ZB = Z.a (IV. L) .-. ZB = :^/.AKa (III. 2 a.) But Z AKC is measured by arc AG, (XII. 1. Sch.) .*. Z jB is measured by \ arc AC. Dem. (2) Draw the diameter BM. Then Z MBC is measured by | arc MC. (XII. 2, Case (1).) Z ABM IB measured by \ ^.tgAM. (XII. 2, Case (1).) .-. Z MBC + Z ABM is measured by ^ arc MC + ^ arc AM. .'. Z ABC is measured by 4 arc^C. Q.E.D. Dem. (3) Both arms on the same side of the diameter through B. Proof to be supplied by the pupil. Q.E.D. XII. MEASUREMENT OF ANGLES 115 XII. 2 a. An angle formed by a tangent and a chord is measured by half the intercepted arc. Hyp. If AO and EF inter- sect at C, and if ^Ois a chord and EF is a tan- gent, Cone. : then Z AOF is measured by 2 Dem. Draw the diameter CL. Z LCF is a right angle. [A radius to point of tangency A. the tangent.] (IX. 4.) ZACF=Tt.Z.LCF+ZLCA. (Ax. 4.) Kt. Z LCF is measured by ESJ^, [A right angle is measured by one half a semicircumference.] (XII. 1, Sch.) /LLC A is measured by ?^^^. (XII. 2.) .-. Z^OT is measured by arc XmC arc ^i^ ^ arc ^mC •^2^2 2 Q.E.D. ScH. Angles inscribed in the same segment are equal. If the segment is greater than a semicircle, the angles are acute. If the segment is a semicircle, the angles are right. If the segment is less than a semicircle, the angles are obtuse. Ex. 1. The bisectors of the vertex angles of all triangles on the same base and inscribed in the same segment are concurrent. 116 THE ELEMENTS OF GEOMETRY XII. 3. An angle ivhose vertex lies between the center and the circumference is measured by half the sum of its intercepted arcs. Hyp. If AC and BE intersect at V lying between the center and the cir- cumference, Cone. : then ZAVB is measured by arc ^g + arc C^ ^ Dem. Draw BC. ZAVB = ZC+ZB. [The ext. Z of a A equals the sum, etc.] (III. 2.) But Z C is measured by ^:E£AM. (XII. 2.) And Z 7? is measured by ^^^/^^. (XII. 2.) 2 arc AB + arc CE Z A VB is measured by Q.E.D. Ex. 2. The opposite angles of an inscribed 4-side are supplemental. Ex. 3. The bisector of any interior angle of an inscribed 4-side and the bisector of the opposite exterior angle intersect on the circum- ference. Ex. 4. How many degrees are there in the arc that subtends an in- scribed angle of 25° ? Of 25° 40'? Of 25° 40' 84"? Ex. 5. What arc measures the supplemental adjacent angles of the preceding inscribed angles ? Ex. 6. What are these angles called ? Ex. 7. An angle between the center and the circumference is 40° 30'. What is the sum of the arcs that measure it ? Ex. 8. An angle of 70° and its supplement are formed by a tangent and a chord. What is the value of each arc that subtends these angles ? Ex. 9. The arc of a segment is 140°. What is the value of each angle inscribed in this segment ? _<^ XII. MEASUREMENT OF ANGLES 117 XII. 4. An angle formed hy two secants intersecting without the circle is measured hy half the difference of the intercepted arcs. Hyp. If AQ and BE inter- sect at V lying without the cir- cumference, and AC and BE are both secants, Cone. : then ZAVBi^ measured by arc ^^ -- arc (7^ ^ Dem. Draw BC. AACB = ZAVB + ZB. (1.) [The ext. Z of a A equals the sum, etc.] (III. 2.) .-. AAVB = ZACB~ZB. ((1.) by transposition.) But Z ACS is measured by H2^. (XII. 2.) ZBis measured by ^^5-^. (XII. 2.) .'. ZAVB is measured by . arc AB arc CE _ arc AB — arc CE 2 2 ~ 2 ■ * Q.E.D. Prove by means of XII. 2 : Ex. 10. That an isosceles triangle is isoangular. Ex. 11. That an isoangular triangle is isosceles. Ex. 12. That the sum of the interior angles of a tri- *igle equals two right triangles. Ex. 13. That the sum of the exterior angles of a triangle equals four right angles. Ex. 14. That two mutually equiangular triangles inscribed in the same circle are congruent. 118 THE ELEMENTS OF GEOMETRY XII. 4 a. An angle formed hy a tangent and a secant is measured hy half the difference of the intercepted arcs. Hyp. liAQ is a secant and BE is a tangent (XII. 4 a), or if both are tan- gents (XII. 46), Fig. 2. Cone. : then Z AVB is measured by Fig. 3. arc AB — arc CE Dem. Similar to Demonstration of XII. 4. (Let the pupil give it in full. Note that when the secant VB of Fig. 1 is turned on F as a pivot until it becomes a tangent, the points B and E become coincident, as shown in Fig. 2, and may be denoted by a single letter. Observe also that if the secant VA of Fig. 1 be likewise turned on F as a pivot until it becomes a tangent, the points A and C of Fig. 1 become coincident, as shown in Fig. 3, and may be denoted by a single letter.'i XIT. 4 h. An angle formed hy tivo tangents is meas- ured hy half the difference of the intercepted arcs. Ex. 15. The 4-side ABCE is inscribed in a circle. Zi^ is 28°; ZBOC is 82^ How many degrees in each of the arcs CB and AE ? How many degrees : Ex. 16. In Z OBF &ndZOCF? Ex. 17. In ZBAE and ZBAC? Ex. 18. In Z GEO ? Ex. 19. What kind of angle with reference to the circle is Z GEO ? Ex. 20. What is the value of all the angles that are inscribed in the major segment that stands on the chord AE ? XII. MEASUREMENT OF ANGLES 119 XII. SUMMARY OF PROPOSITIONS IN THE GROUP ON MEASUREMENT OF ANGLES 1. In the same, or equal circles, central angles vary {are to each other) as their intercepted arcs. ScH. A central angle is measured by its intercepted arc. 2. An inscribed angle is measured by half the inter- cepted arc. a An angle formed by a tangent and a chord is measured by half the intercepted arc. 3. An angle ivhose vertex lies between the ceiiter and the circumference is measured by half the sum of its intercepted arcs. 4. All angle formed by two secants intersecting ivith- out the circle is measured by half the difference of the intercepted arcs. a An angle formed by a tangent and a secant is measured by half the difference of the intercepted arcs. b An angle formed by two tangents is measured by half the difference of the intercepted arcs. 120 THE ELEMENTS OF GEOMETRY Hints to the Solution of Original Exercises In solving a problem in algebra we proceed as follows : 1. We assume that we have the quantity required and call it x. 2. We form an equation in which x may be surrounded by a number of modifiers — coefficients, exponents, etc. The value of x is found when by a transformation or by a series of transformations, x stands alone on one side of the equation^ while the modifiers in some form appear on the other. In solving a problem in geometry we proceed in a similar way : 1. We assume that we have the figure that satisfies the conditions given in the problem. This assumed figure corresponds to the x of algebra. 2. We ask ourselves what follows from this assumption ; that is, what definitions, axioms, or previously established theorems, corollaries, or problems are suggested by the assumed figure. .3. We ask what one of these theorems or combination of them may be applied in the actual construction of the required figure. These applied propositions correspond to the modifiers ofx in algebra. The drawing of such auxiliary line or lines as will make it possible to apply a suggested theorem, or a combination of suggested theorems, as well as the discovery of these theorems, is the test of the inventional power of the student. No rule can be made that will tell him what theorem to select or what line to draw, but the systematic attack, persistently made, familiarizes him with the principles of geometry. The solution of a problem consists of : 1. The analysis as outlined above. • 2. The construction. 3. The proof. 4. The discussion. Many problems in the beginning of the course in algebra may be solved without the use of x. That is, they may be considered problems in arithmetic. So, in geometry, it is by no means always necessary to give the analysis of the problem ; that is, to assume we have the required figure, etc. We pass, however, from the simple to the complex. We learn best how to use the method in problems where it is indispensable by applying it to the solution of simpler problems first. XII. MEASUREMENT OF ANGLES 121 PROBLEMS, EXERCISES, AND SPECIAL THEOREMS Problems Prob. I. O71 a given line to construct a circular segment which shall contain a given angle. Given. A line-segment AB and an angle E. Required. On AB as a chord to construct a circular segment capable of containing an angle equal to Z E. Const. Construct an Z FRH= Z E. (V. Prob. III.) With AB as a radius, and any point J in BF as a cen- ter so taken that the arc described will cut RH, describe an arc. Let it intersect BH in Q, Draw QJ, and draw a circle through B, Q, and J. (Prob. IX., Ex. in drawing, p. 17.) The segment QRJ is the segment required. Q.E.D. Dem. Any angle inscribed in this segment is measured by one half the same arc that subtends Z FRH. (XII. 2.) all such angles equal Z.FBH—Z.E. Q.E.D. A second construction in common use is the following : Erect a mid ± to AB. At B make Z ABC = Z E, and erect BMA.BC. The O with K, the intersection of these Js, as center and KB as radius gives the required segment. 122 THE ELEMENTS OF GEOMETRY Illustration of the Method of Solving Original Problems Prob. II. Gfiven the base, vertex angle, and sum of the legs of a triangle, construct it. Given. The afb c, the vertex Z C, and the sum a + b. Required. A triangle whose base equals c, vertex angle equals Z C, and the sum of the legs equals a -\-b. Analysis. Assume A ABC has c' = c, Z.C = Z.C, and a' H- 6' = a -f- 6. It follows that if AC be produced, making CB' = a', and if BB' be drawn, A CB'B is isosceles. Suggested theorems are : III. 2 a and IV. 1 ; also Axioms 7 and 5. Applicable theorems, etc. : All the above; for the ZJB' and (IV. 1 and III. Ex. 4.) Z. C .'. if we start with a + 6, since ZJ5' = -— -, the position of B'B is known. ^ (Ax. 7.) And, since 5 is c distant from A and also somewhere in BB' or its extension, B is known. (Ax. 5.) Similarly the position of BC is known. (Ax. 7.) Hence, C is known. (Ax. 5.) .*. we have discovered how to construct the required triangle. the Z CBB' each equal — • Note. — An analysis when complete shows us clearly the method of construction. In short, the construction then becomes merely an exercise in geometrical drawing. XII. MEASUREMENT OF ANGLES 123 Const. Let AB' = a + 6. :^ At B' construct an angle with AB' == — —• / \ With ^ as a center and a radius equal to c, /^n\^ describe arc BB". G't J/^^ Draw BO making Z B'BG^ Z. B\ /^^^\ Draw AB. ^r^.___>4 A ^5(7 is the required triangle. Q.E.F. Proof. ZB'BG=ZB' = ^' (Construction.) /. A B'BC is isoangular. .-. ZACB = 2ZB'. [The ext. vert. Z of an isoangular A equals, etc.] (III. 2 a.) .-. Z^(75 = givenZa (1) Again, B'G=Ba [An isoangular triangle is isosceles.] (IV. 2.) .-. AC+BC=AG+B'G=a + b. (2) AB = c. (Construction.) (3) Q.E.D. Discussion. A second triangle fulfilling the given conditions may be formed ; for the circle of line (3) of the above construc- tion cuts B'B in a second point, B". Therefore, draw B"G' just as -BC was drawn and draw the join B"A. A AB"0' is the second triangle meeting the given conditions. If the circle with ^ as a center and c as a radius is tangent to BB', the two triangles coincide and become identical. If this circle does not cut BB', no such triangle can be con- structed. Exercises The following exercises are on the loci of vertices of the equal vertex angles of triangles standing on the same base; of the incenters of such triangles ; of their excenters, circum- centers, orthocenters, and centroids. 124 THE ELEMENTS OF GEOMETRY Ex. a. Hyp. If any number of angles equal Z C, and their sides pass through the ends of the given line- segment ABy Cone. : then the locus of their vertices is a circular segment on AB as a chord, and capable of containing an angle equal to Z C. Demi. On AB, equal to the given line, as a chord, construct a circular segment capable of containing an angle equal to the given Z C. (Prob. I. Group XII.) The arc of this segment is the locus required. For, (1) Any angle inscribed in this segment equals Z C, -: it is measured by H^A^. (XII. 3.) (2) No angle whose vertex is without the circle, subtended by AB and to the left of it, can equal Z C. For, Draw the auxiliary HB. Z AHB {= ZC. Why ?) is greater than Z V. [The ext. Z of a A > either non-adj. int. Z.] (III. 2, Sch.) Similarly it may be shown that an angle whose vertex lies within the circle cannot equal Z C. .'. the arc of the segment to the left of AB, on AB as a chord, is the locus of the vertices of all angles to the left oi AB subtended by AB. The locus of the vertices of such angles on the right of AB and equal to ZC will be the arc of a segment on AB as a chord on the right of AB, and equal to that on the left. Q.E.D. Note 1. — If the given angle be acute, both arcs will be major. If the given angle be right, both arcs will be semicircles. If the given angle be obtuse, both arcs will be minor. Note 2. — The pupil will remember that in order to prove that a line is the locus of a point, two things must be established. What are they ? Ex. b. The locus of the incentere of triangles whose base is AB and whose vertex angles always = Z C, is : The arc of a segment on AB as a chord that will ZC^ 2 * contain an Z = rt. Z + XII. MEASUREMENT OF ANGLES 125 Proof. Z AKiB = 1 rt. Z + ^. (IV. Ex. 20.) .'. by XII. Ex. (a) the locus of Kt is the arc of a segment on AB as a Q.E.D. chord that will contain an angle = rt. Z ^ Ex. c. The locus of the excenters of triangles whose base is AB and whose vertex angles always equal Z (7, is : The arc of a segment on AB as a chord that will contain an Z = rt. Z Proof. KeA and KgB bisect ext. Z A and ext. Z B, respectively. Draw KiA and KiB bisecting int. Z A and int. Z J5, respectively. Z iTi = rt. Z + — . (v. preceding Ex. 6.) Z KiAKe and Z KiBK^ are right angles. Why ? .-. Z iTi + Z /Q = 2 rt. A. 2 .'. the locus of Ke is the arc of a segment on AB as a chord that will ZC _ 2 contain an Z = rt. Z Q.E.D. Ex. d. The locus of the circumcenters of triangles whose base is AB and whose vertex angles always equal Z (7, is : The arc of a segment on AB as a chord that will contain an Z = 2 Z C. Proof. Kc is by hypothesis the center of the cir- cumcircle of triangle ABC. (v. X. 3.) .'. Z AKcB is a central angle and is measured by the arc AB. (XII. 1, Sch.) But Z C is inscribed in this circle. .". Z C is measured by half the arc AB. :. ZAKcB = 2ZC. :. the locus of Kc is the arc of a segment on AB as a chord that will contain an angle equal 2ZC. Q.E.D. 126 THE ELEMENTS OF GEOMETRY Ex. «. The locus of the orthocentera of triangles whose base is AB and whose vertex angles always equal Z C, is : The arc of a segment on AB as a chord that will contain an angle equal to the supplement of Z. C. Proof. • Z AKoB = /. LKo T. ^ •/ Z CLKo is a right angle and Z CTKo is a right angle, Z C is the supplement of Z LKoT. :. Z C is the supplement of Z AKoB. :. the locus of Ko is the arc of a segment on AB as a chord that will contain an angle equal to the supplement of Z C. Ex. /. If the opposite angles of 4-side ABCE are supplemental, show that a circle passing through A, B, and C, will also pass through E ; that is, if the oppo- site angles of a 4-side are supplemental, it is a cyclic. Ex. g. Let the student state the locus of each center given. Ex. 21. Describe a circle. Draw a 4-side so that one of its angles shall be formed by two tangents, one Jsy a tangent and a chord, one by a tangent and a secant, and the fourth by two secants. Point out the measure of each of the above four angles. Ex. 22. If two chords are perpendicular to each other in a circle, the sum of either set of opposite intercepted arcs is a semicircle. Ex. 23. If at the vertex of an inscribed equilateral triangle a tangent is drawn, find the angle between the tangent and each of the sides meeting at the vertex. Ex. 24. Tangents are drawn at the vertices of an inscribed triangle. Two angles of the inscribed triangle are respectively 70° and 80°. Find the angles of the circumscribed triangle. XII. MEASUREMENT OF ANGLES 127 Theorems of Special Interest Pedal Triangle Th. 1. If, in A ABC, P, E, and D are the feet of the altitudes on c, a, and h, respectively, A F J5;P, and DB bisects Z EDP. Dem. CP, BD, and AE concur at K„. (X. 4.) The 4-side PKoDA is cyclic. [If the opposite Z of a 4-side be supp., it is cyclic] (Ex./, p. 126.) .-. Z.DPIC = Z.DAK., (XII. 3.) Both are measured by ^^^-- — -- The 4-side APEC is cyclic. [Rt. A CEA and CPA stand on the same hypotenuse AC.^ .-. Z.DAK.= Z.CPE. Both are measured by arc 0^ n ^ J A CPE. .'. ZDPK, .\ CP bisects Z DPE. Similarly, AE bisects ZD^P and DB bisects Z.PDE Note. — APJED is known as the pedal triangle of A ABC. (XII. 3.) (Ax. 1.) Q.E.D. Ex. 25. If, through the point of con- tact of two circles tangent externally, a straight line is drawn terminating in the circles, the tangents at the extremities of this line are parallel. 128 THE ELEMENTS OF GEOMETRY Nine-Point Circle Theorem Th. 2. If, in a A ABC, ^ H, L, and T be feet of altitudes ; M, E, and D feet of medians, and S, O, and F mid-points of BK„j CK^, and AK„, respectively, Cone. : then H, L, T, M, Ey D, S, Gy F are cyclic. Dem. 1. Pass a circle through M, E, and D. First, prove that this circle passes through H] second, through!^. ME W b and MDW a. [The mid-join of 2 sides of a A is II to the 3d side.] (VII. Ex. 8.) .-. 4-side MEGD is a parallelogram. (Def. of O.) .-. Z DME = Z ACB. (VI. 1' a'.) A DHC is isosceles. (VII. 4.) .-. Z DHC = Z DCH, and similarly, Z CHE = Z HCE. .-. ZDHCi-^CHE=Z.DCH+ZHCE{=Z.ACB). (Ax. 2.) .-. z z>ir^ = z i)io;. (Ax. i.) .-. a circle through M, E, and D passes through H. [The arc DME is the locus of vertices of all angles = ZDME, etc.] Similarly, this identical circle passes through L and T. Q.E.D. 2. Draw the joins TF and ZD. TF divides rt.A TAK„ into two isosceles triangles. (Why ?) LD divides rt. A LCA into two isosceles triangles. These two triangles have Z CAL in common, and are there- fore mutually equiangular. .-. Z TEA = Z ADL, whose supplements, A TFL and TZ)!., are therefore equal. .-. a circle passing through T, D, and L must pass through F. Similarly, it may be shown that this circle also passes through S and O, Q.E.D. XII. MEASUREMENT OF ANGLES 129 Theorem of Orthogonal Circles Th. 3. If circles of a given radius are drawn so as to cut a given O Z orthogonally, Cone. : then the locus of their centers is a circle concentric with Z, whose radius is the hypotenuse of a right triangle whose legs are the radius of O Z and the given radius. Dem. Draw any radius of Z, as KP, and at P erect a per- pendicular to KP and equal to the given radius. With K as a center and KL as a radius, describe the O M. This circle is the locus of the centers described in the hy- pothesis, for XPand KP are of constant lengths, and the Z LPK is always a right angle. .-. LK, the hypotenuse, must always be of the same length. .*. the O Jf, with the center K and a radius equal to LK, is the locus required. ^ Q.E.D. Ex. 26. If the angle formed by two tangents is 50°, how many degrees in each of the intercepted arcs ? Ex. 27. A circle is circumscribed about a triangle. Prove that the radii drawn to the extremities of the base form an angle equal to twice the angle at the vertex of the triangle. 130 THE ELEMENTS OF GEOMETRY Classification op Problems — Indeterminate, Determinate, and Overdeterminate Kinds of Equations In algebra the student has learned that there are three classes of simultaneous equations, to wit : Indeterminate, having an indeterminate number of roots. Determinate, having a determinate number of roots. Overdeterminate, having, in general, no roots. So in geometry problems may be similarly classified, to wit : Indeterminate, in which too few conditions are imposed to give a determinate or definite number of figures that will satisfy the given conditions. E.g. draw a circle tangent to a given line at a given point in the line. An indeterminate number of such circles may be drawn. Determinate, in which enough conditions are imposed to give a determinate number of figures that will satisfy the given conditions. E.g. draw a circle, with a given radius, that shall be tangent to a given line at a given point in the line. Two such circles may be drawn. Overdeterminate, in which too many conditions are imposed. E.g. draw a circle, with a given radius, that shall be tangent to a given line at a given point in the line, and at the same time (simultaneously) pass through a given point. In general, no such circle can be drawn. EXERCISES IN INDETERMINATE, DETERMINATE, AND OVERDETERMINATE PROBLEMS 1. Add to the following indeterminate problems a condition that will make each determinate : (a) Draw a line through a given point. (b) Draw a perpendicular to a given line. (c) Draw a circle tangent to two intersecting lines. (d) Construct a A, having given one side and an angle adjacent to it. (e) Construct a triangle, having given 2 A. 2. By what axiom is the following problem determinate ? Draw the bisector of the vertex angle of a given triangle. XII. MEASUREMENT OF ANGLES 131 3. Why is the following problem overdeterminate ? Draw a bisector of the vertex angle of a triangle that shall be perpendicular to the base. 4. Arrange a summaiy of quadrilateral relations, similar to that of tri- angular relations on page 99, giving ten properties of the angles and lines of a 4-side. Problems — Their Classification Illustrated Indeterminate Determinate Overdeterminate Construct : Construct : Construct : 1. A point a distant A point a distant A point a distant from P. from P, and b from P, b distant distant from P^. from Pi, and c distant from P. 2. A point a distant A point a distant A point a distant from a given from I, and b dis- from I, b distant line I. tant from P. from F, and c dis- tant from Pi. 3. A point a distant A point a distant A point a distant from a given from O K, and b from Q K,b dis- OK. distant from a tant from I, and c given line I. distant from l^. 4. A point equidis- A point equidis- A point equidistant tant from two tant from two from two paral- parallels. parallels, and a lels, a distant . distant from from OA" and b OK distant from P. 5. A point equidis- A point equidis- A point equidistant tant from two tant from two from two concen- concentric cir- concentric cir- tric circles whose cles whose radii cles whose radii radii are a and 6, are a and b, re- are a and b, and and at the same spectively. at the same time time c distant • c distant from from OK and e QK distant from I. 182 THE ELEMENTS OF GEOMETRY Ex. 28 (a). Parallel chords intercept equal arcs. Ex. 28 (6). If the opposite ends of two parallel chords are joined, two isosceles triangles are formed. Ex. 29. If a 4-side is circumscribed about a circle, prove that the sum of two opposite sides equals the sum of the other two sides. Ex. 30. If A is any point in a diameter, B the (j extremity of a radius perpendicular to the diameter, E the point in which AB meets the circumference, C the point in which the tangent through E meets the diameter produced, then AC = EC. Ex. 31. If two circles are internally tangent, and the diameter of the less equals the radius of the larger, the circumference of the less bisects every chord of the larger which can be drawn through the point of contact. Ex. 32. Two circles are internally tangent in the point E, and AB is a chord of the larger circle tangent to the less in the point C. Prove that EC bisects Z^i?5. Ex. 33. Show that in a circumscribed hexagon the sum of one set of alternate sides (first, third, fifth) equals the sum of the other set (second, fourth, sixth). Show also that the sum of one set of alternate sides of a circumscribed octagon equals the sum of the other set. Ex. 34. Show that any circumscribed polygon with an even number of sides has the sum of one set of alternate sides equal to the sum of the other set. Ex. 35. Inscribe a square in a given circle. Show how to obtain from this square, by bisection of sides, etc., a regular inscribed octagon and a regular inscribed hexa-decagon. Ex. 36. An equilateral inscribed polygon is equiangular. XII. MEASUREMENT OF ANGLES 183 Ex. 37. What is the converse of Ex. 36 ? Is it true ? Illustrate your answer by a figure. Ex. 38. If through one of the points of intersec- tion of two circles tlie diameters of the circles be drawn, the join of the other extremities of these diameters passes through the other point of intersec- tion of the circles. Ex. 39. The join spoken of in Ex. 38 is parallel to the line of centers of the circles. Ex. 40. This join is longer than any other line through a point of intersection of the circumferences and terminated by them. Ex. 41. What is a cyclic 4-side ? ■ Ex. 42. What kind of angle with reference to the circle is AECB'i ZECF2 Ex. 43. What is the measure of Z ECF ? Why ? Ex. 44. What is the measure of Z EFC ? Ex. 45. What is the measure of Z 00^ ? Oi ZEQO? Ex. 46. If FH bisects ZF and ML 'bisects Z M, show that : (a) Arc LE — arc CJ= arc LA — arc JB. (6) Arc LE + arc JB = arc LA + arc CJ. (c) Arc LEC + arc JB = arc LA + arc EC J. Ex. 47. The first member of (c) is the measure of what angle ? Ex. 48. The second member of (c) is the measure of what angle ? Ex. 49. Therefore, what kind of triangle is FOQ ? Ex. 50. Why, then, is FH ± LM? Ex. 51. Similarly, prove that A MGK is isosceles. Ex. 52. Combine 41, 46, and 50 into one theorem. Ex. 53. If two straight lines are drawn through the point of contact of two tangent circles, the chords of the arcs intercepted by these lines are parallel. Ex. 54. What parallelograms may be inscribed in a circle ? Ex. 55. The apparent size of a circular object is determined by the angle between two tangents drawn from the eye to the object. What is the locus of the point from which a given circle always appears to have the same size ? 134 THE ELEMENTS OF GEOMETRY B Ex. 56. Given the rt. A ABC. At any point // of the hypotenuse AB erect a ±HE. Let HE intersect BC (produced) in F. Draw AF and GB. Let OB (produced) meet AF in Q. As the ±HE moves along AB, what is the locus oiQ? AM Ex. 67. What part of the hypotenuse of a right triangle is the median to the hypotenuse ? Ex. 58. Into what kind of triangles does the median to the hypotenuse divide the right triangle ? Ex. 59. Draw a right triangle and its altitude to the hypotenuse. Name the three sets of complemental angles in your figure. Ex. 60. What is the measure of an inscribed angle ? If, in the adjacent figure, CE± BA, OF A. BE, and the 4-side is inscribed, point out three angles equal to Z BOF^ and give reasons. Ex. 61. Give two reasons why ^FEO = Z CAO. Ex. 62. Prove that if the diagonals of a cyclic 4-side be perpendicular to each other, and from their intersection a perpendicular be let fall on one side of the 4-side, this per- pendicular will bisect the opposite side. In the following exercises the angles, sides, and principal lines of a triangle are represented thus : Angles: A, B, C. Sides : a, b, c opposite ZA, ZB, and Z C, respectively. Altitudes : ha, hb, and he, altitudes to sides a, b, and c, respectively. Medians : nia, W5, and »ic, medians to sides a, b, and c, respectively. Bisectors of A : t^, ta, and ta^ bisectors of A, B, and C, respectively. (Note. — When one angle of an isosceles triangle is given, all the angles are given.) Construct an isosceles triangle, given : Ex. 63. c and ZA. Ex. 64. c and Z C. Ex. 65. c and the radius of the inscribed circle. Ex. 66. c-\- a and Z B. (Anal. : Extend c to C, making BC = c -\- a. Draw CC. Use IIL 2 o») Ex. 67. a and he. Ex. 69. Z B and he. Ex. 68. c and he. Ex. 70. Z C and a 4- 6. ',^^^^ XII. MEASUREMENT OF ANGLES 135 Ex. 71. c and hi. Ex. 73. Z C and a + c. Ex. 72. he and Z C Ex. 74. Z C and wij. Ex. 75. 6 + /ic and Z C. (Anal. : Let /ic with its extension to ^=^c+&- Draw WA. Use III. 2 a.) Ex. 76. 6 + Ac- and c. /Anal. : Let Ac with its extension to JBT = Ac + 6. Take -• Use III. 2. j Ex. 77. a 4- 6 4- c and Z A. (Note. — If one acute angle of a right triangle is given, the other is also given.) Construct a right triangle (right angle at C), given: Ex. 78. c and Z A. Ex. 79. c and Ac. Ex. 80. c and the radius of the inscribed circle. Ex. 81. Z A and the radius of the inscribed circle. Ex. 82. a and the radius of the inscribed circle. Ex. 83. rtic and Ac. (Anal. : Use VIL 4.) Ex. 84. Z^and Ac. Ex. 85. The two segments of c mad6 by Ac. Ex. 86. The two segments of c made by tc. (Anal. : Extend tc to meet O on c. Use XII. 2.) Ex. 87. c and the distance from the vertex of Z C to a given line. Ex. 88. a + h and Z A. Ex. 89. The radius of the inscribed and the radius of the circumscribed circle. Ex. 90. a — h and c. Ex. 91. c — a and Z A. • (Anal. : Z 5 is known. Const, a A, given c — a and its A.) Ex. 92. a + 6 + c and AA. Construct an equilateral triangle, given : Ex. 93. The perimeter. Ex. 94." The altitude. Ex. 95. a + A. Ex. 96. The radius of the inscribed circle. Ex. 97. The radius of the circumscribed circle. Construct a triangle, given : Ex. 98. The perimeter and Z A and Z B. Ex. 99. c, Ac, and Z C. Ex. 100. c, nic, and Z C. Ex. 101. c, Ac, and ha. (Use IX. 4.) 136 THE ELEMENTS OF «^;E0]VIETRY Ex. 102. ./ie, ^Ay and Z B. Ex. 104. a, c, and mo. Ex. 103. a, hei and c. Ex. 105. c, Ac» and wip. Ex. 106. a, 6, and Wc. (Double the median. Use Conv. of VI. 3.) Ex. 107. »/io /Ja» and /ij. (Double the median. Use IX. 4.) Construct a square, given : Ex. 108. Its apothem. Ex. 109. The difference between its diagonal and its side. Construct a rhombus, given : Ex, 110. The two diagonals. Ex. 111. One diagonal and a side. Construct a trapezoid, given : -ft/^5___£ — --::^np .^^^- Notation : ^^^^""""'^''^^^^^^ Ex. 112. a, c, ZA, and ZB. A '^ B Ex.113, a, c, A, and Z A Ex.115, a, 6, c, and ^^. Ex. 114. c, e, AC, and BE. Ex. 116. c, a, &, e, and AC. Construct a rhomboid, given : Ex. 117. c and ^C and BE. Ex. 119. AC, h, and ZA. Ex. 118. c, AC, and /i. Ex. 120. AC, h, and b. Ex. 121. If a 4-side is circumscribed about a circle, the central angles subtended by the opposite sides are supplemental. Ex. 122. In rt. A ABC, inscribe OK. Ex. 123. If the points of tangency of this O be N, G, and T, why does BG = BN? Ex. 124. Show that 4-side KGATia a square. Ex. 125. Show CB = CA + BA minus the diameter of the inscribed circle. Ex. 126. Show that the diameter of the circumcircle plus the diameter of the incircle equals the sum of the legs of the right triangle. Ex. 127. If AB is produced to the left, mak- ing AC' = AC, what is the value of Z AC C? Ex. 128. If BC is the sum of the legs in the Tt.AABC, and BC is the diameter of the cir- cumcircle, what are the two loci of C ? Ex. 129. Construct a right triangle, given the sum of the legs and the radius of the incircle. XII. MEASUREMENT OF ANGLES 137 Ex. 130. If the radius of the circum-O is given, and also that of the in-O, how from these do you obtain the hypotenuse and the sum of the legs ? Ex. 131. Construct a right triangle, given the radius of the incircle and radius of the circumcircle. Ex. 132. If, in A ABC, CT bisects ZACB and AL is drawn perpen- dicular to or, by what theorem is A ^C^ isosceles ? Ex. 133. If M is the mid-point of AB, why is LM parallel to QB ? Ex. 134. Prove that LM = ^(a-b). Ex. 135. If BL' is drawn perpendicular to CT, prove that the join L'M is also = J(a — b). Ex. 136. If CT bisect ext. Z BCA', and from J5 a ± to CT' is drawn, then the join of M and the foot of this perpendicular = l(a -\-h). Ex.137. ZLAM=ZCAB-ZCAQ. then, that z LAM = \{Z CAB - ZB). Ex. 138. If CH±AB, prove that ZHCT=^X^C^S-^S)- Ex. 139. State the preceding theorem in general terms. Ex. 140. Draw FQ, and prove that as AAFQ is isosceles (Why ?), ZQFB = ZCAB-ZB. Ex. 141. Show that Z AQB = Z ACB -{- ^ the supplement of ZACB. Ex. 142. Construct a triangle, having given the base, the difference of the two sides, and the vertex angle. Ex. 143. If from Jf, the middle point of the arc AMB, any two chords MF, MQ are drawn, cutting the chord AB in E and 0, then CEFG is cyclic. XIII. GROUP ON AREAS OP RECTANGLES AND OTHER POLYGONS (Briefly, the Areal Group) DBTINITIONS The Area of a plane figure is the ratio of its surface to some other surface taken as a unit of area. This Unit of Area is usually a square whose base and altitude are each a unit of length. A Polygon is circumscribed to a circle when the sides of the polygon are tangents to the circle. A Polygon is inscribed in a circle when the sides of the polygon are chords of the circle, that is, when its vertices are in the circumference. PROPOSITIONS XIII. 1. Two rectangles (a) having equal bases, vary as their altitudes ; (6) having equal altitudes, vary as their bases. Hyp.L (a) If, in the a's A-Q and E-L, the bases AB and EF are equal, M IT A BE M Cone. : then □ A-G : czi E-L : : altitude AM: altitude EH. 138 ^^>^ XIII. AREAS OF POLYGONS 139 Case I. Commensurable case. Dem. If AM and EH have common measure (v. XI. Prob. I.), say m, apply it as many times as possible to AM and to EH. Suppose it is contained in AM seven times, in EH five. r. AM:EH::7:5. Through the points of division draw parallels to the bases. The cjA-C will be divided into seven ^ n's, and the nuE-L will be divided into five ^ a's. (VI. 4 a.) Furthermore, all of these smaller a's are ^. (VI. 4 a.) .-. rDA-C:u3E-L::7:5. .• □ A-C:\I3E-L : ; altitude AM: altitude EH. (Ax. 1.) Q.E.D. K ■ E G A Case II. Incommensurable case ; that is, h and hi have no common measure. Dem. Divide AB into any number of equal parts, say n. Apply one of these as a divisor to CE until there is a re- mainder LE less than the divisor. Dmw LMWHC. □ G-B : n 0-M: : AB : CL. (XIII. 1. (a) Case I.) Now, if we decrease the divisor, we decrease the remainder without affecting the equality of the quotients. That is, as LE = 0. and CL = CE, so cnCM=u3 CK. (XI. Def . of Limit.) 140 THE ELEMENTS OF GEOMETRY The ratios □ G-B : □ C-M and AB : GL, however, remain equal as they approach their limits, viz. : □ G-B : □ C-Zrand AB : CE. '. UDG-B:n2C-K::AB:CE. (Case I.) [If, while approaching their respective limits, etc.] (XI. Post. Limits.) Q.E.D. Hyp. 1. (b) If, I H in a's I and II, the altitudes h, h^ are equal, and the bases are b and 61, /( Cone. : then □ I:nII::6:6i. Dem. The proof of 1. (b) is exactly similar to that of 1. (a). Q.E.D. XIII. 1 a. Any tivo rectangles are to each other as the products of their bases and altitudes. Hyp. If the two a's I and II have bases b and 61, and altitudes h and ^1, Cone: then / /', // //, III b b, unl'.nnlli'.b 'h:,bi'\. Dem. Construct a □ III with base equal to b and altitude equal to ^1. □ I : □ III : : 7i : ^1. (1) (XIII. 1 (a).) □ III : □ II : : 6 : 6,. (2) (XIII. 1 (6).) Multiply proportion (1) by (2), and we have □ I : □ II : : 6 • 7i : 61 • ^1. [If two proportions be multiplied together, etc.] (XI. 3.) Q.E.D. XIII. AREAS OF POLYGONS 141 XIII. 1 h. Any two parallelograms are to each other as the products of their bases and altitudes. Hyp. If two UJA-B and C-E have b and b^, h and Ji^ as bases and altitudes, respectively, Cone. : then O A-B : EJ C-E -.'.b-h-.b^- hy. Dem. Draw the perpendiculars to meet sides (produced if necessary) as in figure. Et. A I ^ rt. A I' and rt. A II ^ rt. A II'. [If two rt. A have the leg and hypotenuse of one, etc.] (V. 4.) .-. r]A-B=^\ziA-R) CJC-E = C3C-H. b and 7i are identical in O A-B and □ A-R. bi and h^ are identical in O C-E and □ C-H. Now □ A-R : □ C-H ::b'Ji:b,' h,. (XIII. 1 a.) .'. CJA-B : O C-E ::b'h:b,' K Q.E.D. ScH. A parallelogram is equal to a rectangle of the same base and altitude. Ex. 1. Two rectangles are equal. The bases are 27 and 18, respec- tively, and the altitude of the first is 8. What is the altitude of the second ? Ex. 2. Construct three equal triangles on the same base, the first of which shall be acute, the second right, and the third obtuse. Ex. 3. A number of equal triangles stand on the same base. How do their altitudes compare ? Show, then, that the locus of their vertices consists of two lines parallel to the base. Ex. 4. The base of a triangle is divided into five equal parts, and the points of division are joined to the vertex. How do the altitudes of the resulting triangles compare ? How, then, do the areas of these triangles compare ? 142 THE ELEMENTS OF GEOMETRY XIII. 1 c. Any two triangles are to each other as the products of their bases and altitudes. Hyp. lithe A ABO q .^ andEFHhdiveABsiud /^V / ^^(T^^T- n^ EF for bases, and CQ / X^ / y'^^'^^^X^ and HA for altitudes, a Q B A E F Cone: then A ABG: A EHF: : □ oiAB • CQ : aof EF - HA. Dem. Complete the parallelograms as in the figures. OA-MiOE-L:: a oi AB - CQ:C2 of EF: HA. (XIII. 1 5.) But A ABC is i O A-M and A EFH is ^ O E-L. [The diagonal of a O divides it into two ^ A.] (VI. 1 a, Sch.) .-. AABO.AEHF: .nnoiAB- CQ: □ of EF - HA. Q.E.D. Sch. 1. Parallelograms (or triangles) with equal bases are to each other as their altitudes. Sch. 2. Parallelograms (or triangles) with equal altitudes are to each other as their bases. Sch. 3. If parallelograms (or triangles) have equal altitudes and equal bases, they are equal. Ex. 5. How would you divide a triangle into n equal parts by lines passing through the vertex ? Ex. 6. If the altitude, 26 ft., of a rectangle is to be reduced to 20 ft, how much must be added to the base, 30 ft., to keep the area un- changed ? Ex. 7. If one angle of a right triangle be 30°, how does the hypotenuse compare with the shortest side ? Find the area of a right triangle, one of whose sides is 12 ft. and one of whose angles is 30°. Ex. 8. Show, by drawing a figure, that the square on one half a line is one fourth the square on the line. Ex. 9. Show also that the square on one third a line is one ninth the square on the line. XIII. AREAS OF POLYGONS 143 XIII. 2. Tlie area of a parallelogram equals the product of its base and altitude. Hyp. If O A-0 has ^ 6 as a base and h as an altitude, r 1 \ Cone. : then the area of O A-G —h'Ji. Dem. On any line as UN assumed as a unit of length (v. Def. in XI.) construct 2,UU-T. This square may be taken as the unit of area (v. Def. in XIII.). .-. the n U-T=l. nJA-CiU U-T::b-h:l x 1. [Any two UJ are to each other, etc.] (XIII. 1 b.) .'. O A-0 : 1 : : 6 • 7i : 1 ; that is, the area of O A-C = b -h. (Def. of area (XIII.).) Q.E.D. XIII. 2 a. The area of a triangle equals one half the product of its base and altitude. Hyp. If b and h are the base and altitude, respectively, of A ABC, Cone. : then the area of A ABO =^b -h. Dem. Complete UJ A-E as in figure. AABO = ^IZIA-E. [The diagonal of a O divides it into 2 ^ A.] (VI. 1, Sch.) The area of O A-E = b'h. (XIII. 2.) .-. the area of A ABO= \b'K Q.E.D. Ex. 10. The legs of one right triangle are 8 ft. and 6 ft. ; of another 5 ft. and 12 ft. What is the ratio of their areas ? Ex. 11. A parallelogram has a base of 9 ft. and an altitude of 16 ft. What is the side of a square equivalent to the parallelogram ? / 144 THE ELEMENTS OF GEOMETRY Areas of Irregular Figures ScH. To obtain the area of an irregular polygon, join any point within the polygon with the vertices of the figure. Find the area of each triangle thus formed. The sum of these areas equals the area of the polygon. XIII. 2 h. Tlie area of a trapezoid equals the product of the altitude and the mid-join of the non-parallel sides. Hyp. If the 4-side A-C is a -9 _-**— trapezoid, MJ its mid-join, and Ci^ its alti- tude. Cone. : then the area of trapezoid A-C = □ of CF • MJ, Dem. Draw AC. The area of A ACB =CF'^' (XIII. 2 a.) The area of A AEQ =.CF'^' (XIII. 2 a.) .-. A^OB + AACE =CF- ^^'^ ^^ . (Ax. 2.) That is, the area of trapezoid A-C = OF- ^^ + C^ But 4B±CE^^j [The mid-join of a trapezoid = \ the sum of II sides.] (VII. 3 &.) .-. the area of the trapezoid A-G equals the rectangle of CF'MJ. Q.E.D. Ex. 12. In the last problem substitute "triangle" for "parallelo- gram," and solve. Ex. 13. If the area of a trapezoid be 65 sq. ft., and the parallel sides respectively 10 ft. and 16 ft., vs^hat is the altitude ? Ex. 14. A square and a rectangle have the same perimeter. Which hais the greater area ? Why ? XIII. AREAS OF POLYGONS 145 XIII. 3. The area of a circumscribed polygon equals one half the ijroduct of its perimeter and the radius of the inscribed circle. Hyp. If an n-gon ABC-F is circum- scribed to a circle of center 7ij and of radius r^, Cone. : then the area of n-gon ABC-F = i(AB ■^BC+CE+ "')r,. Dem. Draw K^A, KiB, etc. ; also K^L, KtM, etc. K^L, KiM, etc., are altitudes of AAK^B, BK^C, etc. (IX. 4.) That is, the radius of the inscribed circle equals the altitude of the triangles that make up the n-gon. But the area of AAKiB = —'^\. Similarly, for the remaining triangles. .-. the sum of the areas of the triangles, or the area of the n-gon ABC-F = i {AB + BC + CE + "-) - r^. Q.E.D. Ex. 15. What is the area of a triangle if the base is 1384 ft., and the altitude is 256 ft. ? Ex. 16. What is the area of a rt. isosceles A one leg of which is 1414 ft. ? Ex. 17. What is the area of a right triangle whose perimeter is 840 ft., and whose sides are to each other as 3 : 4 : 5 ? Ex. 18. What is the area of a right triangle whose perimeter is 300 ft., if its sides are as 5 : 12 : 13 ? Ex. 19. The base of a triangle is to its altitude as 11 : : 60 ; the area of the triangle is 1320 sq. ft. What is the length of the base ? Ex. 20. The altitude of a trapezoid is 16 ft. ; the mid-join is 32 ft. What is the area of the trapezoid ? Ex. 21. A A whose altitude is 10 ft. and base 24 ft. is transformed into a rhombus. Its longer diagonal is 16 ft. What is the length of the shorter? What are the dimensions of a rectangle if the Ex. 22. Area is 3822 sq. ft. and the sides are as 6 : 13 ? Ex. 23. Area is 59100 sq. ft. and perimeter 994 ft. ? 146 THE ELEMENTS OF GEOMETRY b c c 1) h h Ex. 24. In a parallelogram a line is drawn that cuts off one fourth of one side and three fifths of tlie opposite side. What part of the parallelo- gram is each trapezoid thus formed ? Ex. 25. Show geometrically that if h and c rep- c resent lines, (6 + c)a = 62 +2 6c + c\ State the proposition in words, i.e. without the use of symbols. Ex. 26. Show geometrically that if 6 and c represent lines, (6 - c)2 = 62 + c2 - 2 6c. J = 62 _ 2 6c + c2. State the proposition in words. b be c b-c c (b-c)^ bo c Ex. 27. Show, from the figure, that if 6 and c be any two lines, (6 + c)(6-c) = 62~c2. State the proposition in words. b< c ^ I c (T I b-c b+c Ex, 28. If, in a trapezoid ABCE^ AC and BE^ the two diagonals, are drawn, prove AABC = AABE. Ex. 29. If, in the same trapezoid, the diagonals intersect in M, prove thsii /\A3IE= A BMC. Ex. 30. If two equal triangles have the same base, and lie on opposite sides of this base, the join of the vertices of the triangles is bisected by the common base. Ex. 31. If, in a O ABCE, perpendiculars from B and E are let fall on the diagonal AC, prove that they are equal. Ex. 32. In the above parallelogram, if P is any point in the diagonal AC, prove that A APB = A APE. Ex. 33. Prove that if from a vertex of a parallelogram a line is drawn to the mid-point of one of the opposite sides, it cuts off one third of the diagonal it intersects. Ex. 34. Prove that the line referred to in the preceding theorem cutg off a triangle equal to one fourth of the parallelogram. XIII. AREAS OF POLYGONS 147 XIII. SUMMARY OF PROPOSITIONS IN THE AREAL GROUP 1. Tloo rectangles (a) having equal bases, vary as their altitudes, (b) having equal altitudes, vary as their bases. a. Any two rectangles are to each other as the products of their bases and altitudes. b. Any tivo parallelograms are to each other as the products of their bases and altitudes. ScH. A parallelogram is equal to a rectangle of the same base and altitude. c. Any two triangles are to each other as the products of their bases and altitudes. ScH. 1. Parallelograms (or triangles) with equal bases are to each other as their* altitudes. 2. Parallelograms (or triangles) with equal altitudes are to each other as their bases. 3. Parallelograms (or triangles) with equal bases and equal altitudes are equal. 2. The area of a parallelogram equals the product of its base and altitude. a The area of triangle equals one half the product of its base and altitude. ScH. Areas of irregular figures. b The area of a trapezoid equals the product of the altitude and the mid-join of the non- parallel sides. 3. The area of a circumscribed polygon equals one half the product of its perimeter and the radius of the inscribed circle. X A 148 THE ELEMENTS OF GEOMETRY PROBLEM Prob. I. To reduce a polygon to an equivalent triangle. Given. The poly- gon ABCEF. Required. To re- duce it to an equiva- lent triangle. Const. Extend AF indefinitely. Draw FC, a diagonal. Draw EQ II CFj meeting AF produced in Q, Draw CQ. ^ q^^ ^ ^ q^q [A having = bases and = altitudes are = .] (XIII. 1. c, Sch. 3.) .-. ABCF + A CEF = ABCF + A CFQ. (Ax. 2.) .-. n-gon ABCQ = n-gon ABCEF. Similarly, draw OA, BK, and CK. The number of vertices of n-gon is thus reduced to three. .-. A KCQ = n-gon ABCEF. Q.E.P. Ex. 35. What is the centroid of a triangle ? (X. , 6. ) If G be the centroid of ABC, what is the ratio of the areas of ACM and AGM? Of the areas of BCM and BGM? Of the areas of ^ G^^ and A CB ? Ex. 36. How, then, do the AAGB, AGC, and BGO compare in area ? State as a theorem the property of the centroid thus established. Ex. 37. The centroid of a material triangle uniform in thickness and of the same material throughout is called the "center of gravity" of the tri- angle. How might this name be suggested by the property just established ? Ex, 38. On a side of a given triangle as a base, to construct a triangle equal to the first and having its vertex on a given line. Ex. 39. Show that the greatest number of solutions possible in Ex. 38, is two. Under what conditions is the solution impossible ? Under what conditions is the problem indeterminate ? Draw figures to illustrate your answers. XIV. PYTHAGOREAN GROUP DEFINITIONS Projection on a Line The Projection of a Point on a straight line is the foot of the perpendicular from the point to the line. The line on which the perpendicular is dropped is called the Base of Projection. The Projection of a Line-Segment is that portion of the base of projection which lies between the projections of the extremities of the given line-segment. _^ 2? Projection of AB Note. — This group is named after Pythagoras, an eminent Greek mathematician of the sixth century b.c, who was the first to publish a proof of the first theorem of the group. The truth of this proposition was known before his time, but a proof had long been sought in vain. The theorem, the 47th of Euclid, is often called the pons asinorum of geometry. Ex. 1. In the right, acute, and obtuse A I, II, III, draw the projec- tions of a on b, and of & on a ; also of a on c, and of c on a. C 140 150 THE ELEMENTS OF GEOMETBY A k ^^' in \vr l^hv:^.- / / / \ / \ / / \ / \ 1 \ PROPOSITIONS XIV. 1. If a triangle is right, the square on the hy- Xiotenuse equals the sum of the squares on the legs. Hyp. If, in a rt AABC, ZO is the right angle, Cone. : then the D on c = the D on a 4- the D on h. Dem. Draw the altitude CJ and extend it to L on the D on c. Draw CHy CQ, BE, and AF. AB = AH and AE = AC. (Def . of D.) Z.EAB=:Z. CAH. (Each = a rt. Z + Z CAB.) .'. A ABE ^ A CAH. (V. 1.) A ABE=:i the D on b} (Having the same base EA and = altitudes.) [The area of a A = ^ the product of b • h.] (XIII. 2 a.) A CAH =^^ the nnA-L} (Having the same base AH and = altitudes.)' .-. i D on 6 = ^ the □ A-L. (Ax. 1.) .-. the D on 6 = the cz] A-L. (Ax. 2.) Similarly, we may show that the D on a = □ L-B. .'. the D on a + the D on 6 = □ A-L -\- □ L-B = D on c. Q.E.D. XIV. 1 a. The square on a leg of a right triangle equals the difference between the square on the hypote- nuse and the square on the other leg. 1 Draw altitude in A ABE from BtoEA;iJiA CAH from O to AH. XIV. PYTHAGOREAN GROUP 151 XIV. 2. In any triangle, the square on a side opposite an acute angle equals the sum of the squares on the other tico sides, diminished by twice the rectangle of either of these sides and the projection of the other upon it f K c Kyp. If, in the A ABC, Z.C is acute ; CT, the pro- jection of b on a, and CM, the projection of a on b. H K Ck)nc. : then D onc=D ona+D on6 — 2.nofa« GT, or 2'niotb'CM. Dem. Draw the altitudes and extend them to meet sides of the D's as shown in the figure. These altitudes must pass through Jf and T. (Def. of projection.) If, as in figure for XIV. 1, CH and BG were drawn, A CAH would be ^ A BAG. ... □ A-K=n2 G-M. (Ax. 3.) Similarly, C=i 5- JT = □ jE7-J5. . Again, if BQ and AF were drawn, ABCq^h.AGF. .'. □ M-Q = Z3T-F. (Ax. 3.) cz] A-K+ □ B-K = □ G-M + □ E-B, (Ax. 2.) = D on 6 - □iW-Q + D on a - □ T-F, or (since □ T-F = □ M-Q) = non6 + nona-2a M-Q, = non6 + nona — 2aof6 and CM, = Don6 + nona — 2nofa and CT. Q.E.D. 152 THE ELEMENTS OF GEOMETRY XIV. 3. In any obtuse triangle, the square on the side opposite the obtuse angle equals the sum of the squares on the other two sides, increased by twice the rectangle of either of these sides and the projection of the other upon it Hyp. If, in the A ABC, ZC is obtuse; CT the pro- jection of b on a, and CM the projection of a on b, Cone. : then nonc = nona + non6 + 2«aof6 and CM, = D on a + D on & + 2 . □ of a and or. Dem. Draw the altitudes of A ABC and produce them as in the figure ; also draw AF and BG. If EB and CH be drawn, □ E-M may be proved = □ A-K. [Each □ being twice one of the ^ A ABEsiRd ACH (why ^ ?).] Similarly, - □ B-S = □ B-IC (Draw CL and AQ and give remainder of proof.) .-. cDA-K-h\Z}B-K=c3E-M-\-C3B-S. That is, the D on c = □ E-M + □ B-S. But nDE-M= the D on 6 + □ O-M, and □ B-S = the D on a + □ C-S. .-. the D on c = the D on 6 + the D on a + □ G-M+ □ C-S. But □ G-M = □ C-S, -.' A ACF ^ A GGB. (Why ?) But ^G-S^CT'CFov CT'a,B.xidin^G-M:=b'CM. . . G on c = D on 6 4- D on a -f 2 • □ of a and CT, = Don6 + Dona-|-2-aof6 and CM. Q.E.D. XIV. PYTHAGOREAN GROUP 153 Ex. 2. The radius of a circle is r. A tangent of length 1 is drawn to this circle. Draw the hypotenuse and find an expression for its length. What is the locus of the extremity of this tangent ? Ex. 3. A ladder 50 ft. long just reaches the top of a wall 40 ft. high. If the ground be level, how far is the foot of the ladder from the wall ? Ex. 4. (a) The length of a chord is 12 ft. What is its distance from the center of a circle whose radius is 20 ft. ? (6) If the length of a chord be 2 a, how far is this chord from the center of a circle of radius r ? Ex. 5. What is the length of a chord in a circle of radius r, and at a distance d from the center ? Ex. 6. Find the area of the cross section of a ditch, the section being an isosceles trapezoid 20 ft. wide at the bottom, 30 ft. wide at the top, and each slope 25 ft. from top to bottom. Ex. 7. If a public square is 200 yds. on a side, how much is gained by crossing the square from corner to corner on a diagonal walk instead of using the sidewalk around the square ? Ex. 8. Find the side and area of the □ inscribed in a O of radius a. Ex. 9. The sides of a triangle are 6, 8, and 10, respectively. What kind of triangle is it ? Why ? Ex. 10. The sides of a A are 5, 7, and 10, respectively. What kind of A is it ? Ex. 11. If the three sides of a triangle are given, we may find : (1) The lengths of the projections of two of the sides on the third side; (2) The length of the altitude to the third side ; (3) The area of the triangle, in the following manner : (1) Let the sides of the triangle be 8, 11, and 14. Let X and y be the projections on 14 of 11 and 8, respectively. x + y = 14. x^ — y2 = 57 (XIV. 1 and •.• the altitude is common to both A.) 14(ic -y)z=67,orx-y = f|. Knowing x ■{■ y and x — y, we may find x and y. (2) Knowing either x or y, we may find the altitude by XIV. 1 a. (3) Knowing the base and altitude, we have the area. 154 THE ELEMENTS OF GEOMETRY XIV. SUMMARY OF PROPOSITIONS IN PYTHAGOREAN GROUP 1. If a triangle is right, the square on the hypotenuse equals the sum of the squares on the other two sides. a The square on the side of a right triangle equals the difference between the square on the hy- potenuse and the square on the other side. 2. In any triangle, the square on a side opposite an acute angle equals the sum of the squares on the other tivo sides, diminished by tivice the rectangle of either of these sides and the projection of the other upon it. 3. In any obtuse triangle, the square on the side oppo- site the obtuse angle equals the sum of the squares on the other two sides, increased by twice the rectangle of either of these sides and the projection of the other upon it. XIV. PYTHAGOREAN GROUP 155 Ex. 12. Having the length of two sides of a A and of the altitude to the third side, how do you find the length of the third side ? Ex. 13. If each side of an equilateral triangle equals 10, show that the altitude is V75 =by/S, and the area 25 V3. Ex. 14. In an equilateral triangle, if one of the sides is a and the altitude is h, show that h=- V3. 2 Ex. 15. What is the area of an equilateral triangle whose side is 20 ft. ? Ex. 16. Prove that in any triangle, ABC, if p and q are the segments of c made by an altitude on c, and a and b are the remaining sides, then a + b : p + q : : p — q : a — b. Ex. 17. Prove that if a line-segment makes an Z of 60° with the base of projection, the length of the projection is one half the original line-segment. Ex. 18. If a line-segment of length a makes an angle of 30° with the base of projection, how long is the projection of a ? Ex. 19. The sides of a trapezoid are 12, 32, 12, and 40, respectively. Find its area. Ex. 20. An extension ladder, 75 ft. long, just reaches a window 60 ft. from the ground. How far is the foot of the ladder from the side of the building ? Ex. 21. The sides of a right triangle are three consecutive integral numbers. What is the length of the hypotenuse ? Ex. 22. What is the length of the diagonal of a rectangular floor 24 ft. by 22 ft. ? Ex. 23. The diagonal of a rectangle is 2.9 ft. ; the perimeter is 8.2 ft. What is the length of the rectangle ? Ex. 24. The hypotenuse of a right triangle is 58 ft. ; one leg is 42 ft. What is the length of the altitude to the hypotenuse ? Ex. 25. The sides of a triangle are 5, 6, and 7. Find the segments of each side made by the altitude upon it. Ex. 26. Draw the projection of the line-segment on the base of pro- jection in each of the following cases : 1. The line-segment above the base of projection. 2. The line-segment meeting the base of projection. 3. The line-segment intersecting the base of projection. Ex. 27. If, in a scalene triangle, a median is drawn, prove that one of the angles it forms with the side is acute and the other is obtuse. 156 THE ELEMENTS OF GEOMETRY Ex. 28. Hence, prove that in any triangle the sum of the squares on two sides equals twice the square on half the third side plus twice the square on the median to that side. (XIV. 2 and 3, and combine by addition.) Ex. 29. Prove that in any triangle the difference of the squares on any two sides equals twice the rectangle of the third side and the projection of the median on that side. (Use XIV. 2 and 3 ; combine by subtraction.) Ex. 30. Verify by means of Ex. 28 the theorem that the square on the hypotenuse equals the sum of the squares on the two sides. Ex. 31. Prove that the sum of the squares on the four sides of a parallelogram equals the sum of ---'''^''/2^^^^^-''e, AE C £( EB /"VV 1 \ 'FH~~GH~'m' ^ ' '^ XVI. GROUP ON AREAL RATIOS PROPOSITIONS XVI. 1. If two triangles have an angle of one equal to an angle of the other, they are to each other as the rectangles of the sides respectively including the equal angles, A A c L B E g F Hyp. If, in the A ABC and EFG, Z.A = Z £7, Cone. : then A^5(7: A EFQ : : □ 6 • c : □/• g. Dem. Place A EFQ on A ABG so that it takes the position AJL. BmwJB. AAJL lAAJB .'.AL'.AB. (1) [A with equal altitudes are to each other, etc.] (XIII. 1 c, Sch. 2.) A AJB : A ABC : : AJ: AC. (Same reason.) (2) .-. AAJL : A ABC: : AL • AJ: AB - AC (Multiplying (1) by (2).) /. AEFG:AABC:icnf'g:C3b-c. Or, by inversion, A ABC : A EFG ::c3b • c:c3f' g. Q.E.D 176 XVI. AREAL RATIOS 177 XVr. 2. If two triangles are similar, tJmj are to each other as the squares on any two homologous sides. Q // Xe A c B E g" F Hyp. If the triangles I and TI are similar, Cone. : then Al : All :: a' : e^:: h'' : f :: c^ : g\ Dem. fjl=l^- (XVI.1.)(1) Now - = -. (Horn, sides of ^^ A.) / e Substituting in (1) for - , its equal, -, we have Al ' All ' ' ct^ ■ e^ and similarly as b^ : f^. Q.E.D. Ex. 1. Two equilateral A are as 5 : 4. Compare their altitudes. Ex. 2. The base of one equilateral triangle equals the altitude of another. What is the ratio of their areas 1 Ex. 3. The homologous sides of two similar polygons are in the ratio of 3 : 7. What is the ratio of the areas of the polygons ? Ex. 4. What are homologous lines of similar figures ? Ex. 5. Why are the pefrimeters of similar polygons homologous lines ? Ex. 6. If two similar polygons have equal perimeters, what do you know of their areas ? Ex. 7. Under what conditions are two dissimilar triangles equal ? Ex. 8. What is the relation between the areas of two triangles that have an angle of one equal to an angle of the other ? Ex. 9. Draw a triangle. Draw a second triangle whose vertex angle is the supplemental adjacent angle of the other. Prove that the areas of these triangles vary (or are to each other) as the rectangles of the sides including the supplemental angles. 178 THE ELExMENTS OF GEOMETRY XVI. 3. If tioo polygons are sirnilar, they are to each other as the squares of any tivo homologous sides. A B A B' Ryp. If polygon ABC-" is similar to polygon ^'^'(7 ••• , and if their areas be Q and Q', respectively, Cone. : then Q:Q'::A^: A^'\ Dem. Draw ^ FO, FB, and FC, FB'. [Then the triangles of the first polygon are similar to the similarly placed triangles of the second polygon.] (XV. 5.) . ^^ = ^ ^^^ ^ FS ^ A BFC ^ FC' ^ A FEG " JiB^ AFA'B' fW AB'FO WG^ AFE'C' (XVI. 2.) . AjP.^^ ^ A BFC ^ AFEC f ^ aS \ ,. .s '' AFA^B' ABF'O AFE'C\ JTb'')' ^ ^' '^ A FAB + BFO + FEG A FAB AB" (XL 2.) AF'A'B' + BF'C' + FE'G' AFA'B' jr^ That is, Q:Q':: AB" : jVW. XVI. 3 a. TTie areas of tioo similar polygons are to each other as the squares of any tioo homologous lines. 0em. AB : A'B' r.FBt F'B'. (Hom. sides ^ A.) .-. Aff : aW i'.fS: FB'-. (XI. 3, b.) But Q : Q'.-.A'b': AB'\ (XVI. 3.) .. QiQ'iiFB'iFB'', XVI. AREAL RATIOS 179 XVI. SUMMARY OP PROPOSITIONS IN THE GROUP ON AREAL RATIOS 1. If two triangles have an angle of one equal to an angle of the other, they are to each other as the rectan- gles of the sides respectively including the equal angles. 2. If two triangles are similar, they are to each other as the sguares on any two homologous sides. 3. If two polygons are similar, they are to each other as the squares of any tivo homologous sides. a TJie areas of any ttvo similar polygons are to each other as the squares of any two homolo- gous lines. 180 THE ELEMENTS OF GEOMETRY PROBLEMS pROB. I. On a given line that is to he homologous to a given side of a given polygon, to construct a poly- gon similar to the given polygon. Given. Any line I, and any polygon P. ^ Required. A polygon on I, ~ P and of which the side equal to I shall be homologous to AF. Const. Lay off A'F = I, at A' construct ZA' = ZA; at F' construct Z F' = Z F. Find a fourth proportional to AF, A'F', and AB. Lay off this fourth proportional on the terminal line of Z A', as A'B'. [Completion of construction and proof left to student as an exercise.] pROB. 11. To construct a polygon similar to each of tivo given similar j^olygons and equal to their difference. HixT. — Construct a right triangle as in Prob. I, and use XVI. 3. pROB. III. To construct a polygon similar to each of two given polygons and equal to their sum. Hint. — Construct a right triangle as in XIV. Prob. I, and use XVI. 3. Ex. 10. Draw a triangle. Draw a second triangle within it, two of whose sides are perpendicular to two sides of the first. Prove that the areas of these triangles vary as the rectangles of the sides that are perpen- dicular to each other. XVI. AREAL RATIOS 181 The areas of two similar triangles are respectively 196 sq. ft. and 256 sq. ft. Ex. 11. What is the ratio of any pair of their homologous sides ? Ex. 12, What is the ratio of the rectangles of the sides including a pair of homologous angles ? To construct a triangle similar to ABO and satisfying the following conditions : Ex. 13. Having a perimeter three times as long as the perimeter of ABC. ,Ex. 14. Having an area equal to four ninths the area of ABO. Ex. 15. Having an area twice as great as that of ABO. Ex. 16. The areas of two triangles, I and II, having one angle in com- mon, are 20 sq. ft. and 60 sq. ft., respectively. The sides in triangle I about the common angle are 5 ft. and 6 ft. One of the corresponding sides of triangle II is 12 ft. What is the length of the other side ? Ex. 17. The sides of a triangle are 4, 9, 10. Divide the triangle into two equal parts in three ways by drawing, in succession, parallels to the three sides. G /\ Ex. 18. Prove Theorem 2 by using the adjoining / j \ figure, OH being the altitude of the l^ABO. e/—^ -\^ A H B Ex. 19. Each side of a regular pentagon is 3. Construct a similar pentagon twice as large as the first. Ex. 20. To construct a hexagon similar to a given hexagon and having one third the area of the given hexagon. Ex. 21. The areas of two similar polygons are 324 sq. ft. and 576 sq. ft. They are divided into three sets of similar triangles by diagonals drawn from each of two homologous vertices. What is the ratio of the areas of corresponding pairs of the similar triangles ? What is the ratio of the homologous diagonals ? Ex. 22. Given two similar hexagons Q and B. To construct a hexagon similar to Q and B and equal to their sum. Ex. 23. Given two similar pentagons Q and R. To construct a penta- gon similar to Q and B and equal to their difference. 182 TITK ELEMENTS OF GEOMETKY Ex. 24. Ciwn any number of similar polygons. Construct a polygon similar to each and equal to their sum. (Use XVI. a and XIV. Prob. III.) Ex. 25. To divide a triangle into two equal parts by a line parallel to a given line. Ex. 26. To divide a triangle into two equal parts by a line through a given point on one side of the triangle. Ex. 27. To divide a triangle into three equal parts by parallels to one side. Ex. 28. What does the area of a triangle equal in terms of the base and altitude ? Ex. 29. The sides of a triangle are a, 6, and c. The altitudes to these sides are h„, hb, and he. Show that a- ha = b-hi = C' he. Ex. 30. Divide each member of the preceding equation by ha • h and show that a _b _ c h ha hnhh he Ex. 31. Place ikl^ = x and construct x (a fourth he proportional to fto, ^j, and he). Ex. 32. If X equals a line m, then — = ^- = -£. . hh ha m Ex. 33. Why, then, would a triangle whose sides are Aj, ha, and m be similar to a triangle whose sides are a, ft, and c ? Ex. 34. Construct such a triangle, A'B'C, and draw the altitude corresponding to he of A ABC. Produce this altitude, if necessary, to equal he. Through its foot, draw the parallel to c'. Produce the sides, if necessary, to meet this parallel. Prove this new triangle congruent with A ABC. Ex. 35. Construct a triangle having given the three altitudes. XVII. GROUP ON LINEAR APPLICATION OF PROPORTION PROPOSITIONS XVII. 1. If the bisectors of the interior and exterior angles at the vertex of a triangle are draion, these bisectors will divide the base into segments proportional to the other two sides. Hyp. If CT bisects AACB (Fig. 1), or /.AGH (Fig. 2), Cone. : then, in either figure, AT :TB::AC: BC. Dem. Case I. Extend BC, making E0=: AC. Draw EA. A CAE is isoangular. (IV. 1.) .-. AACB = 2Z.E. (in. 2 a.) .-. Z BCTf= ^4^) = ^ -27. (Ax. 2.) .-.CTWAE. (Def. oflls.) .-. AT:TB::EC: BC (XV. 1.) .Substituting AC for EC, we have AT:TB::AO:Ba Q.B.P. 188 184 THE ELEMENTS OF GEOMETRY Dem Case II. Make EC = AG. Draw EA. A CAE is isoangular. (IV. 1.) .-. ZACH = 2 Z CEA (III. 2 a.) .• CTII ^^. (Det of II..) .-. AT' TB'.'.AC'BC. Q.E.D. Det. When a line-segment is divided internally and exter- nally in the same absolute geometric ratio, the line-segment is said to be divided harmonically. XVII. 3 a The bisectors of the interior and exterior vertex angles of a triangle divide the base harmonically in the ratio of the sides, jj^ flyip. If CT and CT bisect mt Z.AOB and ext. Z. ACH, respectively, and cut AB in T and T', Cone. : then AT - TB Dem i^=^'y also ^^=^~' (XVII 1.) TB CB TB CB AT'.TB: AT : T B. (Ax. 1.) Q.E.D. JCVIl. 1 If T and T divide AB harmonically, then the points a and B divide the line TT harmonicalty For, if AT: TB::Ar: T'B, then, by alternation, AT : AT' ii TB: T'B. That is, the ratio of the distances of A from T and T' equals the ratio of the distances of B from T and T', A^ B and T, T' are called two pairs of conjugate hctrmorvk paints* jT XVII. LINEAR APPLICATION OF PROPORTION 185 XVII. 2 The perimeters of similar polygons are to each other as atiy two homologous lines Hyp. If p and p' be the perimeters of two similar polygons, AB and A'B two homologous sides, and AE and A'E' any two homologous lines (e.g. ho- mologous diagonals), E f> Cone. : then pip'i.AB. A'B> . : AE A'E Pern. AB : A'B' : : BC - B'C : : CE : C'E', etc. (Horn, sides of ' AB ^BC "A'B' AB-tBC-hCE-t A'B' -\-B'C' -\- C'E' -I- • • A'B' B'C [In a series of equal ratios the sum, etc. ;] i.e. p:p'::AB: A'B', etc. But AE : A'E' ..AB: A'B' .'. p:p':'.AB: A'B' ::AE: A'E' w polygons.) etc. (XL 2.) (XV, 5.) (Ax. 1.) Q.E.D. Ex. 1. The sides of a triangle are 7, 9, and 12. Find the segments of the side 12 made by the interior and exterior bisectors of the opposite angle. Ex. 2. To divide a line-segment harmonically in a given ratio ; i.e. so that the ratio of the segments of internal and external division shall equal the ratio of two given lines a and b. Ex. 3. If, in the last exercise, a = b, what becomes of the external point of division of the given line-segment ? Ex. 4. If the diagonal of one pentagon is twice as long as the corre- sponding diagonal of a similar pentagon, How does the perimeter of the first figure compare with that of the second ? Ex. 5. If the diagonal of a pentagon is equal to the sum of the corre- sponding diagonals of two pentagons similar to the first, What relation exists between the perimeter of the first figure and the perimeters of the other two ? 186 THE ELEMENTS OF GEOMPrrRY XVII. 3. If two angles have their vertices at the cen- ters of two unequal circles, the angles are to each other as the arc of the first divided hy its radius is to the arc of the second divided hy its radius. E o T Hyp. If Z BCE and Z K have their vertices C and K at the centers of circles of radii r and r', respectively, cone. : then Z JSC£ : Z /r :: 5]:«^ : 51^. r r Dem. With C as a center and a radius = r\ describe arci^G^. Produce arc FG, making arc FH = arc LM. Draw CH. Then ZFCH=ZK. [In equal CD equal A at the centers intercept, etc.] (IX. 3.) .-. ZFCG:ZFCH::2iTcFG:a,vcFH. [In equal ® central A vary as their intercepted arcs.] (XII. 1.) But AFCG = Z.BCE. ZBOEiZFCH:: arc FG SiTcFH r' r' But the sectors BCE and FCG have = central A. . arci^G^ arc^^ (XV. 7.) ZBCE-.ZFCH:: .. ZBCE:ZK:: arc BE . arc FH r r' BicBE SiTcLM Q.E.D. XVII. LINEAR APPLICATION OF PROPORTION 187 ScH. Upon this theorem is based what is called the Radial method of measuring angles. The unit angle of this system is of course the angle whose meas- ure is 1 ; i.e. the angle for which — — 1 ; or arc = radius. This unit is called a Radian, and will hereafter be shown to denote an angle of about 57°. 3. The advantage of radial measure over the ordinary measure- ment in degrees is that the former does not depend at all upon the size of the circle. The radian is practically the only unit employed in advanced work. Application to the Sides of a Right Triangle and THE Line-segments dependent upon the Altitude XVII. 4. Lemma. If the altitude of a right triangle is drawn to the hypotenuse, the three triangles of the resulting figure are similar. Hyp. If, in rt. A ABC, the altitude CH is drawn to the hypotenuse, Cone. : then rt. A I ~ rt. A ABC^ rt. A II. Dem. Et. A .45(7 ~ rt. A L •.• Z^ is common to both of them. [Two rt. A are ~ if an acute Z of one, etc.] (XV. 2 a.) Similarly, rt. A ABC ^ rt. A IL /. the z^ of rt. A I = respectively the A of rt. A II. (Ax. 1.) .-. rt.AI'-rt.AIL (XV. 2.) .-. rt. AI~rt.A^BO'-rt. AIL OE.D 188 THE ELEMENTS OF GEOMETRY XVII. 5. Jf the altitude of a right triangle to the hypotenuse is drawn: (a) The altitude is a mean proportional between the segments of the hypotenuse. (p) Either leg is a mean proportional {or geometric mean) between the hypotenuse and the segment adjacent to that leg. (c) The segments of the hypotenuse are to each other as the squares on the legs respectively adjacent to them. (d) The square of the length of the hypotenuse equals the sum of the squares of the lengths of the legs. Hyp. If CL is ^ the altitude to AB^ the hypotenuse of the rt. A ABC, Cone. : then (a) ALiCLi.CL: LB, or CZ' = AL - LB. (b) ALiCAi.CA: AB, or CA" = AL - AB. (c) AC'iBC'-.'.ALiLB. (d) ac' + bc'^ab'. Dem. (a) Tit A ALC <^ vt A CLB. (XVII. 4.) .-. AL:CL::CL: LB. (Horn, sides of ^ A, etc.) .. CL' = AL'LB. (XL 1.) Q.E.D. Dem. (6) m. A 4LC '^ rt. A ACB. (XVII. 4.) .-. AL:CA::CA:AB, or CA' = AL'AB', (1) also, LB :BC::BC: AB, ot BC" = LB - AB. (2) Q.E.D. XVII. LINEAR APPLICATION OF PROPORTION 189 Dem. (c) Divide (1) by (2), member by member, and we have AG" : BC^ iiAL: LB. Q.E.D. Dem. (d) Add (1) to (2), member to member, and we have AG" -hB(f= (AL + LB) • AB = AB'AB = aS. Q.E.D. Def. If be any point on the straight line AB (whether between the points A and B, or on AB produced), the distances OA and OB are called the segments of AB made by the point 0. Exe 6. What are homologous sides of similar triangles ? Name the sets of similar triangles in the figure of XVII, 4. Read the homologous angles of each set ; also the homologous sides. Ex. 7. Find a mean proportional to a and h by using XVII. 6 a. Ex. 8. Find a third proportional to a and 6 by using XVII. 6 a. Ex. 9. Given a = 8 ft., c = 13 ft., and s=z\\lt. Find/. c Ex. 10. Given : perimeter of small triangle is 46 ft. ; of large, 138 ft. What is the ratio of any two homologous sides ? Ex. 11. In two similar triangles the sides of the first are 4 ft., 9 ft., and 11 ft. The shortest side of the second is 12 ft. What is the length of the other two sides •? What is the ratio of their perimeters ? The sides of a triangle are 7 ft., 10 ft., and 12 ft. Find : Ex. 12. The segments determined on each side by the bisector of the interior angle at the opposite vertex. Ex. 13. The segments determined on each side by the bisector of the exterior angle at the opposite vertex. Ex. 14. The projections of each side on the other two. Ex. 15. The projection of the bisector of each interior angle on the opposite side. 190 THE ELEMENTS OF GEOMETRY XVII. 6. If through a fixed point any line is drawn cutting a circle in two points, the rectangle of the seg- ments of this line is constant, in whatever direction the line is drawn. Three Cases. (1) T'he fixed point within the circle. (2) The fixed point on the circumference, (3) The fixed point loithout the circle. Casb (1). Case (3). Hyp. If is a fixed point, and AB is any line through 0, arid EG is any other line through 0, each cutting the circle, Cone. : then the □ of J50 • 0^ = the □ of ^0 • OC. Dem. Draw AE and BC in each of the three cases. AAOE-^ABOO. '.' ZA = ZC. ('Each is measured by ^^^^^ .\ ZAOE = ZBOa (Why?) .-. A AOE ~ A BOC, and OA:OE::00: OB. (XV. 2.) .% the rectangle of BO - 0A = the rectangle of EO • 00. Q.E.D ScH. The proportion OA : OE : : OC: OB may be written OA^qc O^^i . OB OE OB' OE ' OC XVII. LINEAR APPLICATION OF PROPORTION 191 That is : The ratio of two segments equals the reciprocal of the ratio of the two corresponding segments ; in other words, the segments are reciprocally proportional. XVII. & a. If from the same point to the same circle a tangent and a secant he drawn, the tangent ivill he a mean proportional hetween the secant and the part without .the circle. ^ . Hyp. If from 0,0T f \}^^ a tangent, and OB a I ^^ ^.^^-^n/ secant are drawn to \ ^...^^^^ jl^ Cone. : then OB.OT.iOT: OA, or 0T' = 0B • OA. Dem. Draw AT and BT. A OAT r^ A 0TB. (XV. 2.) .-. OB:OT::OT: OA, or OT' = OB - OA (XL 1.) Q.E.D. ScH. OA is a tliird proportional to OB and OT. OB is a third proportional to OA and OT. The sides of a triangle are 7 ft., 10 ft., and 12 ft. Find : Ex. 16. The projection of the bisector of one of the exterior angles on the opposite side. Ex. 17. The lengths of any of the angle bisectors. Ex. 18. The area of the triangle. Ex. 19. The acute angles of a right triangle are 60° and 30°. What is the ratio of the segments into which the shortest side is divided by the bisector of the opposite angle ? Ex. 20. Given a circle of radius a ; through a point at a distance 2 a from the center, tangents are drawn. Find : (a) The length of each tangent. (6) The length of the chord of contact. (c) The angle between the tangents. (d) The an^le at the center subtended by the chord of contact. (e) The distance of the chord of contact from the center. 192 THE ELEMENTS OF GEOMETRY XVII. SUMMARY OF PROPOSITIONS IN THE GROUP ON THE LINEAR APPLICATION OF PROPORTION Application to Angle Bisectors, Perimeters, AND TO Angular Measurement 1. If the bisector of the interior and exterior angles at the vertex of a triangle are drawn, these bisectors will divide the base into segments proportional to the other two sides, a The bisectors of the interior and exterior vertex angle of a triangle divide the base harmoni- cally in the ratio of the sides. 2. The perimeters of similar polygons are to each other as any two homologous lines. 3. If two angles have their vertices at the centers of two unequal ciixles, the angles are to each other as the arc of the first divided by its radius is to the arc of the second divided by its radius. ScH. Unit of angular magnitude : radian. , Application to the Sides of a Eight Triangle and the Line-segments dependent upon the Altitude 4. Lemma. If an altitude of a light triangle is drawn to the hypotenuse, the thtee triangles of the resulting figure will be similar. 5. If the altitude of a right triangle to the hypotenuse is drawn, XVIL LINEAR APPLICATION OF PROPORTION 198 (a) The altitude is a mean proportional between the segments of the hypotenuse. (b) Either leg is a mean proportional (or geometric mean) hetiveen the hypotenuse and the segment adjacent to that leg. (c) The segments of the hypotenuse are to each other as the squares on the legs respectively adjacent to them. (d) The square of the length of the hypotenuse equals the sum of the squares of the lengths of the legs. Application to Chords, Tangents, and Secants 6. If through a fixed point any line is drawn cutting a circle in two points, the rectangle of the segments of this line is constant, m whatever direction the line is drawn. Three Gases. (1) The fixed point within the circle. (2) The fixed point on the circumference. (3) The fixed point loithout the circle. ScH. The ratio of two corresponding segments equals the reciprocal of the ratio of the other two corresponding segments. That is, the segments are reciprocally proportional. a. If from the same point to the same circle a tan- gent and a secant are drawn, the tangent ivUl he a mean proportional hetiveen the secant and the part without the circle, ScH. On third proportionals. 194 THE ELEMENTS OF GEOMETRY PROBLEMS Prob. I. To find a fourth proportional to three given l^nes. a b q X Given. The three lines, a, b, and c. Required. A fourth proportional to a, 6, and c. Analysis. Suppose x to be the required fourth proportional. Then a:b::c:x. Suggested Theorem, XVII. 6, Cases (1) and (3). Fio. 1. The " dash and dot " Hue in figures is the line required. Const. Case (1). Draw a circle whose diameter > 6 -f c. (Fig. 1.) In this circle draw a chord BC = b + c. Take OA = a and produce it to intersect circle at X. Then OX is the required fourth proportional. (XVII. 6 (1).) Q.E.F. Const. Case (2). Draw a circle whose diameter > 6 — c. (Big. 2.) In this circle draw a chord B0= b — c, . Produce it so that CO = c. Take OA = a. Then OX is the required fourth proportional. (XVII. 6 (3).) ...Q.E.F. XVII. LINEAR APPLICATION OF PROPORTION 195 Note. — The student should suggest still other ways of laying off the given segments in Problem I. a; is a fourth proportional to a, 6, and c in any one of the following arrangements : (1) ff : 6 : : c : X. (3) a : a; : : 6 : c. (2) a : & : : X : c. (4) x : 6 : : a : C. Also in their transformations. ScH. If two of the given line-segments are equal, e.g. if 6 = c, each of the constructions given above will determine a third proportional to a and h, a and c, or h and c, provided that the line-segments that become equal are not in the same couplet of the proportion. Prob. II. To construct a mean proportional to two given lines. Given. The lines a and h. Required. To con- struct a mean propor- ^ ^ ^ t.'^ o f ^ ^ Fig. 1. Fig. 2. tional to a and o. Const. Case (1). Draw EF= h and lay off EQ = a. Describe a semicircle on EF as a diameter. At Q erect a perpendicular to EF intersecting semicircle in T. ET is the required mean proportional. Proof. EF: ET: : ET: EQ ox h : ET: : ET: a. [Either leg of a rt. A is a mean proportional, etc.] (XVII. 5 h.) .-. ET of Fig. 1 is a mean proportional to a and b. Const. Case (2). Draw EQ = a and EF= b. Then QF=a-{-b. On QF as a diameter describe a semicircle. At E erect a ± to QF intersecting the semicircle at T ET is the required mean proportional. Proof. QE:ET::ET: EF or a : ET: : ET: b. For QTF is 2i right A. (Why ?) .-. ET of Fig. 2 is a mean proportional to a and b. 196 THE ELEMENTS OF GEOMETRY Prob. III. To draw a tangent of given length to a circle so that the chord of the secant drawn from the outer extremity of the tangent to the same circle shall equal a given line. Given. The leogth q, the line d, and the circle K. Required. To draw a tangent = q, so that the chord of the secant from the outer extremity of the tangent = d. Analysis. Suppose tangent AT= q and that AO — AB = d. It follows that if /Of be drawn _L BC, M is the mid-point oiBC. (Why?) Propositions, etc., suggested are : Find the locus of the mid-points of all chords in the given circle equal to (7, and Draw to the given circle a tangent equal to q, and From the outer extremity of this tangent draw a tangent to the locus circle. We have now discovered a method of constructing the line required. Discussion. As two tangents may be drawn from A to the locus circle, there are evidently two solutions. These solutions become coincident ' when d equals the diameter of the given circle. There is no solution when d is greater than the diameter of the given circle. Dei A line is divided in extreme and mean ratio (or in golden section) when one segment of the line is a mean proportional between the whole line and the other segment. Ex. 21. The line of centers of two circles is divided externally in the ratio of their radii by its point of intersection with an external common tangent. XVII. LINEAR APPLICATION OF PROPORTION 197 Prob. IV. To divide a given line in extreme and mean yxtio. AB Q' Given. The line AB. Required. To divide A'B in extreme and mean ratio. Const. Draw KB ± AB. Let it equal or be greater than From A draw a secant AM to this circle so that the chord LM=AB. (Prob. III.) With ^ as a center, and a radius equal to AL, describe an arc cutting AB, as at Q. Then AB is divided at Q in extreme and mean ratio. That is AB '.AQ'.:AQ: QB. Q.E.F. Proof. AB is tangent to large circle K. [If a line is ± to a radius at its extremity, etc.] (IX. 4 a.) .-. AMiAB:: AB.AL. (1) .\ AM-AB'.AB::AB-AL:AL. (From (1) by division.) But AB = LM, and AL^AQ. (Const.) .-. AM- AB = AM- LM = AL, and AB-AL = AB-AQ=QB. .-. AQ.AB:: QB:AQ. (2) .-. AB:AQ::AQ: QB. (From (2) by inversion.) Q.E.D. Discussion. With ^ as a center, and a radius equal to AM, describe an arc, cutting AB produced to the left, as at Q'. Then AB:AQ'::AQ':Q'B. Prove by taking (1) by composition. .-. Q' is a second point of golden section. 198 THE ELEMENTS OF GEOMETRY Prob. V. To construct a square that shall he three times a given square. ^"-----J Given. The square A-C. / j ^\ Required. To construct a A[ B \ \ J r ----i----f--- - square equal to three times \ ^**a the square A-C. J /7---J — J Const. Produce ABy making AO = 3 AB. Produce AO, making a' = a. On AH as a diameter, describe a semicircle. At O erect a perpendicular to AH, meeting the semicircle in J. JG is the side of the required square. Proof is left to the pupil. Def. To transform a figure means to change it to another figure that is eqiial to the first. Prob. VI. To transform a rectangle into a square, ^F Given. Then ^-5. ^Q Required. To trans- form it into an equal square. Const. Produce EC, making a' = a. . On EG as a diameter describe a semicircle. At C erect a perpendicular to EG, meeting the semicircle inF. CF is the side of the required square. Proof is left to the pupil. Ex. 22. The line of centers of two circles is divided internally in the ratio of their radii by its point of intersection with an internal common tangent. Ex. 23. The points of intersection of the pairs of internal and external common tangents to two circles divide the line of centers harmonically. XVII. LINEAR APPLICATION OF PROPORTION 199 Pkob. VII. To transform a triangle into a square. Construct a mean proi^ortional between the base and half the altitude. Ex. 24. Two sides of a right triangle are 20 and 21. Determine : (a) The projections of these sides on the hypotenuse. (6) The altitude on the hypotenuse. Ex. 25. If the segments of a diameter be 3 and 7, what is the length of the perpendicular to the diameter at the point of division and extending to the circumference ? Ex. 26. Show how to construct \/l4 geometrically. Ex. 27. What is the expression for the rectangle of the segments of a diameter determined by any point in the diameter ? From this expression show that the rectangle of the segments of a diameter is greatest when the segments are equal. Ex. 28. If two chords be drawn from any point of a semicircle to the extremities of a diameter, the squares of the chords will be to each other as the projections of the chords on the diameter. Ex. 29. The area of a triangle equals the rectangle of its base by half the altitude or 6 . -• 2 Ex. 30. Construct a square equal to twice a given triangle. Ex. 31. Construct a square equal to the sum of two given triangles. Ex. 32. Construct a square equal to the sum of a given triangle and a given parallelogram. Ex. 33. Construct a square equal to the sum of a given pentagon and a given rectangle. Ex. 34. What is the length of the side of a square equal to an equi- lateral triangle each side of which is 10 ft. ? What is the geometric meaning of the following expressions? Ex. 35. c h Ex. 36. x=^AB^', x=^IAB\ Ex. 37. a; = V2 AB^ - □ of CE - HJ. Ex. 38. X = \/2 J of a • 6. Ex. 39. ^Mq^-'iAw' 200 THE ELEMENTS OF GEOMETRY E Prob. VIII. To construct a rectangle whose perimeter is equal to twice the length of a given line and whose area equals that of a given square^ Given. The line /, and the square IS Required. To con struct a rectangle whose perimeter equals 2 1 and whose i^j-ea equals «S Analysis. Suppose CD R has perimeter = 2 ^ and area = C S. That-is, that rectangle of XY- XZ = AB^, (1) and XY-{'XZ=l. (2) It follows from (1) and (2) that AB is a mean proportional between two line-segments whose sum = I. Several previously established theorems are suggested by the last statement, viz. : XVII. 5 (a) and 5 (5) ; XVII. 6 and 6 a ; Problems I, II, III. Of these, let us try XVII. 5 a and XVII. 6. Case 1. C4.se 2. Const Case 1. Describe a seipicircle on EF— I as diameter. Let m be the locus of a point that is AB distance from EF. From (7, the intersection of this locus with the circle, draw CQ±EF. The rectangle of EQ • QF is the rectangle required. Q.E.P XVII. LINEAR APPLICATION OF PROPORTION 201 Proof. CQ = AB, •■■ m is the locus of all points AB distant from EF ^^^, ^^.^^^^ ^ EQCQ:^ CQ . QF or CQ' = rectangle of EQ - QF. (.XVI1.5a) QE.D Const. (Jase 2 Draw any circle whose diameter > I Draw a chord CJ= 2 AB. Draw the circle that is the locus of mid-points of chords —I. From Q, the mid-point of CJ, draw a chord of the larger circle that is tangent to the locus circle. The rectangle of EQ • QF is the rectangle required QE.P Proof to be given by the pupil from XVII. 6. Ex 40. If a tangent to a circle be terminated by two parallel tangents, the rectangle of the segments of this tangent determined by the point of tangency equals the square on the radius. Ex.. 41, Given a circle of radius 20 ft. ; through a point 16 ft. from the center a chord is drawn. Find the rectangle of the segments into which the chord is divided at the point. • Ex. 42. Given a circle of radius 24 ft. ; through a point 40 ft. from the center a tangent is drawn, and also a secant 58 ft. long. Find : (a) The length of the tangent. (6) The length of the chord cut from the secant, (c) The distance of the secant from the center. Ex. 43. The radius of a circle is 12 ft. ; tangents are drawn to this circle through a point 20 ft. from the center. What is the length of the chord joining the points of tangency ? Ex. 44. Given a line m and a parallelogram of base b and altitude h. Use Theorem 6 to construct on m as a base a rectangle equal in area to the given parallelogram. Ex. 45. To pass a circle through two given points and tangent to a given line. (Use XVII. 6 a.) Ex. 46. To transform a parallelogram into an equal square. Ex. 47. To transform a scalene triangle into an equilateral triangle of the same area. 202 THE ELEMENTS OF GEOMETRY Prob. IX. To construct a square that shall he to a given square in a given ratio. ,^i^ ^^N k a / \ \ '/i^ p ^^^1 jj Am B k € Given. The square on a and the ratio k : m. Required. To construct a square that shall be to a* as A; is to m. Const. On indefinite line AL, lay off AB = m and BC=k. On AC as a diameter describe a semicircle. At B erect a perpendicular to AC, cutting semicircle in 0- Draw OA and OC and lay off OQ = a. Draw GE II AC. OE is the side of the required square. Q.E.P. Proof. A FOQ ^ A BOA and A FOE '^ A BOO. (XV. 2.) and But or .-. m : GF k'.FE ,'. m : GF .: m:k OF'.FE m: k :BO:FO (Hom. sides of ^A.) : BO : FO. (Hom. sides of ^A.) :k:FE (1). (Ax. 1.) : GF: FE. (Taking (1) by alt.) : OG' : O^, (XVII. 5 (c).) : a^ : OE' QJSJ>. Prob. X. To construct a polygon similar to a given polygon and having a gioen ratio to it. Given. The polygon P, and the ratio k : m. Required. To con- struct a polygon similar to P and having with P the ratio k : m. XVII. LINEAR APPLICATION OF PROPORTION 203 Analysis. Suppose X ^ P and X: P::k:m. If X^P, then X:P::YZ':Aff; (XVI. 3.) and, if X: P::k:m, then k:m::YZ': Iff. (Ax. 1.) This proportion suggests XVII. 5 (c) and the preceding problem. Hence, the construction. Const. Construct a line A'B' such that the square upon it shall be to the square on AB as A; : m. (Prob. IX.) On this line AB', homologous to AB, construct a polygon Q similar to P. Q is the polygon required. Q.E.F. Proof. (Let pupil give proof.) Ex. 48. To transform a scalene triangle into an equal isosceles triangle with a given base angle. Ex. 49. To construct a square that shall be to a given pentagon as 4 is to 9. Ex. 50. How does an angle inscribed in a segment compare with the angle between the base of the segment and a tangent at either extremity of the base ? State the converse of the proposition. Ex. 51. A circle is described on a given line AH as a diameter. A£ is a fixed chord. BF is drawn perpendicular to AH. AJ is any chord cutting BF in 3. Why is Z AJB = Z ABF ? Ex. 52. Why, then, is AB tangent to a circle through E, B, and J? Prove, then, that AJ? = rectangle of AE • A J. Ex. 53. (a) Show that the 4-sideC^^J is cyclic. (&) Noting that ZABH is a right angle, and that BC is the altitude on the hypotenuse, give another proof that AB^ = rectangle of AE • AJ. Ex. 54. What is the locus of a point E that divides all chords through A so that EA is a third proportional to AB and the whole chord, say AJ? 204 THE ELEMENTS OF GEOMETRY Pros. XI. To construct a polygon whose sides shall be in a given ratio to the sides of a given polygon, E Given. Any rj-gon _1'_ u4BC'--- and two lines ^''^^S'-^^J^^-.y^^^o /^ \ k and w, representing ""~""*"^\-7r y^\ ^ any given ratio. m i ''••■r'^.*"~-\ ^ / Required. Ann-gon -i -^ ~ LMO"- whose sides are to the sides of n-gonABC"' in the ratio of k to m. That is, the ratio of similitude of the required to the given n-gon is A: : m. Const. Find a fourth proportional to k, m, and any side AB of n-gon ABC '". (v. Prob. I.) Let this line be LM, and draw it parallel to AB at any con- venient distance from AB. Draw AL and BM, and let them intersect at K. Draw KC, KE, and KF. Through L draw LQ II AF and terminating on KF, Similarly, draw QP and PO. Draw OM, The w-gon LMO '"is the n-gon required. Q.E.P. Proof. LM: AB ::k:m (Const.) KM:KB::KL:KA::LQ:AF, etc. KOiKO. .-. OM II BC. [If a line divides two sides of a A proportionally, etc. ] (XV. 1 a.) .-. n-gon LMO n-gon ABC — (XV. 6 a.) and its ratio of similitude to n-gon ABC ••• is A; : m. Q.E.D. Ex. 55. What is locus of J (Fig. for Ex. 53) so taken that the rectangle of AE • AJ = AB^, where ^ is a fixed point in the mid-perpendicular to the given line-segment FB ? -fflCki. XVII. LINEAR APPLICATION OF PROPORTION 205 pROB. XII. To construct a rectangle ivhose area shall equal a given square, and the difference of ivhose base and altitude shall equal a given line. Given. The square Q and the line d. Required. To con- struct a rectangle whose area shall equal Q, and the difference of whose base and altitude shall equal d. Analysis. Suppose [Z}A-L= the rectangle required. That is, that □ A-L = D Q and that x — y = d. If □ A-L =DQ, then X'y = q\ (1) (XIII. 2.) Equation (1) suggests any one of the theorems concerning mean proportionals, the most convenient of which for practical purposes is that on the tangent and secant. Again, equation (1) is indeterminate ; but, if we examine it and remember that x — y = d, we see that the problem may now be stated: Draw a tangent of a given length (q) to a circle so that the chord of the secant drawn from the extremity of the tangent to the same circle shall equal a given line (d). (XVII. Prob. III.) We have, therefore, discovered or rediscovered the method of constructing the required rectangle. Por the tangent equals the side of the square, the whole secant equals the base, and its ex- ternal segment equals the altitude, of the required rectangle. Ex. 56. If AH is the diameter of a circle, and CE is perpendicular to AH produced, prove that the locus of a point J which divides any line from A to BF so that the rectangle of AE • AJ= rectangle of AH' AC, is the circle on AH SiS a diameter. (Note that 4-side HCJE is cyclic.) E B 206 THE ELEMENTS OF GEOMETRY pROB. XIII. To construct a polygon similar to one given polygon and equal in area to another given polygon. Given. The / ^^^^^ polygons P an (IQ. ^^\ / ^-^^ ^/ Required. lo \ "^ / \ ^ \ X I construct a poly- \ I \ ^^ \ / • '^ \ r» AH \ L gon similar to F ^' b and equal in area to Q. Analysis. Suppose polygon X ~ P and = Q. If X - P, then P.XiiAff: A^\ (XVI. 3.) In this proportion substitute for X its equal Q, and we have PiQiiAffiA^. This proportion contains but one unknown quantity, namely, The question arises. How shall we construct, geometrically. Now P and Q are dissimilar polygons by hypothesis ; but, if we change each to an equal square whose sides are respectivelv m and k, proportion (1) becomes .-. m:k::AB:A'B'. That is, A'B' is a fourth proportional to m, k, and AB. Hence, the construction. Const. Change P to an equal triangle ; also Q. (XIII. Prob. I.) Find a square equal to each of these triangles. (XVII. Prob. VII.) Find a fourth proportional to the sides of these squares and AB. (XVII. Prob. I.) On this fourth proportional construct a polygon similar to P. (XVII. Prob. XII.) It will be the polygon required. Q.E.P. Proof is left to the pupil. XVII. LINEAR APPLICATION OF PKOPORTION 207 Ex.57. A ^CB is inscribed. CT bisects ZACB pendicular to AB. Why does CT pass througli F ? Wliy is FJa. diameter of tlie circumcircle ? Why is F the mid-point of arc AB ? Why is TF a fourtli proportional to JF, LFy and CF? Ex. 58. Let x= TF, t=zCT, s = LF, and d = JF. Then x td: is :x + t, or x^ + tx = ds. Solve Construct x. FJ is a mid-per- GJ ^(^ ± V4 L_i of d • s + D on 0- Ex. 59. By the aid of the foregoing, construct a triangle, having given the base, the vertex angle, and the bisector of the vertex angle. n Ex. 60. Given, A ABC inscribed in a circle. CE and CF are so drawn that ^ACE=AFCB. Prove that A ACE -^ A FCB. Hence, show that CA'CB=CE' CF. Ex. 61. In the figure, CF bisects ZACB. Circumscribe a circle to A ABC, and produce Ci^to L. Show that AACF-^ LCB. Why, then, is AC . CF. : CL: CB? Ex. 62. Why does the rectangle of CF- FL - rectangle oi AF- FB ? Prove, then, that the rectangle of ^C- CB= Ci^^ + rectangleof J.i''- FB. Ex. 63. State the above equation in general terms. Ex. 64. If LJ is a diameter of the circumcircle of A ABC, K the center of the incircle, and if KB is drawn, and KE±CB, show that Z ABL = ZLCB. Show that ZKBA = ZKBC, and therefore that ZLBK= ZLKB, and therefore LK= BL. Ex. 65. Why is A LJB a right triangle ? Why is rt. A LJB ~ KEC ? Why, then, does the □ of ZJ" • KE = [:3ot BL - KC = n3oi LK - KC? 208 THE ELEMENTS OF GEOMETRY Ex. 66. Prove from Exercise 66 that the rectangle of the diameter of the circunieircle of a triangle, and the radius of the incircle, equals the rec- tangle of the segments of any chord of the circumcircle that passes through the center of liie incircle. Ex. 67. If ^f is the mid-point of /TA'i, and RA is a perpendicular to A'A'i at any point R, then LK^^ - UP = 2 KK\ • RM. ( Why 1) Prove that if L be taken anywhere in tlie ±RA, say at J, JTi^ - 7a^ = 2 KKi . RM. Ex. 68. Of what point, then, is the ± RA the locus ? Ex. 69. In the expression lA"i^ - lK- = 2 im AT/Ti • RM, construct RM. (Factor the first member. XVII. Prob. 1.) Ex. 70. Find, then, the locus of a point the difference of the squares of whose distances from two given points equals a given square. Ex. 71. If two circles are drawn with any radii, and K and Ki as centers, and trom any point L in a perpendicular to KKi, as AR, tan- gents are drawn to these circles, show that LG^- LT^ = 2 AVi'i . RM-(r'^ - r^). Ex. 72. In the second member of the last equation, is there any term whose value changes as L moves up or down the ±AR? If, then, the value of L^ — LT^ is the same, or constant, no matter where, in the ±AR, L may lie, construct the locus of a point the differ- ence of the squares of the tangents from which to two given circles equals a given square. Def. The Radical Axis of two circles is the locus of a point from which the tangents to the two circles are equal. A In such case RA is called the radical axis j^ of the two circles. Ex. 73.^ If OTT intersect QK^ show that the radical axis of the two circles is their ■ common chord produced. Note, — The radical axis of two circles is the locus of the centers of circles which cut the two circles orthogonally. XVII. LINEAR APPLICATION OF PROPORTION 209 Ex.74. In a system of three circles the radical axes concur. For, if tan- gents be drawn to the three circles from the point where two of the radical axes intersect, these tangents will be equal. Therefore, the third radical axis must pass through this point. Def. This point of concurrence is called the Radical Center of the three circles. - Thus, if RA is so taken that LG^ - LT^ = 0, then LG'^ = LT^, or LG = LT. ....... Ex. 75. If ^B is divided internally in theratio of a : b at the point M, and also divided externally at L in the same ratio, and if on ML as a diameter a circle is described, and the point F is any point in the circle, prove the following relations : (1) 3IFL is a right angle. (2) The rectangle otAM- LB = the rectangle of AL . MB. (3) Prove that the circle on ML as a diameter is the locus of a point the ratio of whose distances from A and B is that of a : Z). (4) If AM = MB, that is, if the ratio a : 6 = 1, what becomes of the point L ? (5) If the ratio a : 6 is less than one, where does the point L reappear ? Ex. 76. The apparent size of an object is determined by the angle that it subtends at the eye of the observer. Thus, if at any point Q, the tangents to two unequal © K and L make equal angles, i.e. if /.BQS= ZHQM, the circles will appear equal, as seen from Q. Draw the radii LB and KM, and show that, if Q be such a poin t, A LQB is similar to A KQM. Hence, show that if a and b be the radii of the circles, QL : QK::a:b. Hence, find the locus of the point from which two unequal circles seem to be equal. ( Ex, 68. ) Def. This locus is called the Circle of Similitude of tlie given circles. XVIII. GROUP ON CIRCUMSCRIBED AND INSCRIBED REGULAR POLYGONS DEFINITIONS A Regular Polygon is a polygon that is both equiangular and equilateral. The Apothem of a regular polygon is the radius of its inscribed circle. The Radius of a regular polygon is the radius of the circum- scribed circle. 'The Center of a regular polygon is the common center of the inscribed and circumscribed circles. PROPOSITIONS XVIII. 1. If a polygon is regular, (1) A circle may he circumscribed about the polygon. (2) A concentric circle may be inscribed in the polygon. B.YP- If a polygon ABC-E is regular, Cone. : then (1) a circle may be passed through A, B, O^ etc., (2) a concentric circle may be described tangent to AB, BO, etc. Dem. (1) Pass a circle through A, B, and C; let its center be K. Join K to the vertices and to the mid-points of the sides of the polygon (by KA, KB, etc. ; KH, KL, etc.). Kii±Ba 210 XV III. CIRCUMSCRIBED AND INSCRIBED POLYGONS 211 [A radius ± to a chord bisects the chord, etc.] (TX. 1.) Kevolve KHCE on KH as an axis, until it falls on KHBA. The angles at H are right angles, as just shown. .-. HC takes the direction of HB. (Ax. 7.) HC=^ HB (Const.). .-. C falls on B. A HCE = Z HBxi and CE = BA. (Def. of regular polygon.) .-. first, CE takes the direction BA, and second, E falls on A ; i.e, KA^KE) and the circle through A, B, and C passes through E. Similarly, this circle may be shown to pass through any other vertex of the polygon. (2) The sides AB, BC, etc., are equal chords of the same circle. .*. these sides are equidistant from the center. [In the same O equal chords are equidistant, etc.] (TX. 2.) .-. A circle with K as center and a radius = KL = KH, etc., is tangent to every side of the polygon. Q.E.D. Ex, 1. In a regular n-gon, the central angle is the supplement of any one of the interior angles. Ex. 2. Divide a regular dodecagon into triangles by drawing radii, Join any two alternate vertices. Prove, by finding the area of the triangles crossed by the join, that : The area of a regular dodecagon equals three times the square on the radius. Ex. 3. Draw a figure showing a circumscribed equilateral polygon that is not regular. Ex. 4. If a circumscribed polygon is equilateral, the polygon will be regular, provided the number of sides be odd. (Use IX. 5 and V. 1.) Ex. 5. Explain why the polygon of Ex. 4 may not be regular if the number of sides be even. That is, show that while there will be two sets of equal angles when the number of sides is even, the angles of one set will not necessarily be equal to those of the other. 312 THE ELEMENTS OF GEOMETRY XVIII. 2. Conversely. If a polygon is inscribed in a circle and circumscribed to a concentric circle, the polygon is regular, 2? Hjrp. If a poly- gon ABC-0 is in- scribed in a Oe and also circumscribed to a concentric O i. Cone. : then the polygon is regular. Dem. AB, EC, etc., tangents to the O i, are perpendicular to the radii KL, KH, etc., of the Or. [A tangent is A. to the radius through the point, etc.] (IX. 4.) But these tangents to the O i are chords of the O e. (Hyp.) Again the distances of these chords from the common center of the © i and e are equal. .•. the chords themselves are equal. [Equal chords are equidistant from center, etc.] (IX. 2.) .*. the polygon is equilateral. Again the arcs AB, BC, etc., are all equal. [In the same O, or in equal (D, equal chords, etc.] (TX. 3 a.) Each angle of the polygon intercepts {n — 2) of these equal arcs, if n be the number of sides in the polygon. .*. each angle of the polygon is measured by ^{n — 2) of these equal arcs. [An inscribed Z is measured by one half, etc.] (XII. 3.) .*. each angle has the same measure as any other, n being the same for all. .*. the polygon is equiangular. •.• the polygon is both equilateral and equiangular, it is regular. ^^^ ° Q.E.D. XVIII. 2 a. The area of a regular polygon equals one half of the product of its perimeter and apothem, (See XIII. 3.) XVIII. CIRCUMSCRIBED AND INSCRIBED POLYGONS 213 XVIII. 3. If an inscribed polygon is equilateral^ the polygon is regular. Prove as above that the polygon is equiangular. XVIII. L If a regular hexagon is inscribed in a circle, the side of the polygon equals the radius of the circle, A Hyp. If a hexagon AB-G is regular and inscribed in a OK, Cone. : then the side of the hexagon equals the radius of the circle. Dem. Join K to the vertices of the hexagon, and draw the apothem, KQ. The A GKF is isosceles. (KG = KF= radius of given O.) Z GKF= I rt. Z. (Central Z of a hexagon.) .:. each of the two other angles = f rt. Z, and GKF is an equilateral triangle. .-. G^i^= G^= the radius. Q.E.D. Ex. 6. Draw figures showing inscribed equiangular polygons that are not regular. Ex. 7. If an inscribed polygon is equiangular, the polygon is regular, provided the number of sides be odd. (Use IX. 3 a and XII. 3.) Ex. 8. Explain why the polygon may not be regular if the number of sides be even. Ex. 9. Construct an angle of 4° 30'. Ex. 10. Construct an angle of 72°. Ex. 11. Construct an angle of 24°. 214 THE ELEMENTS OF GEOMETRY XVIII. 5. The radius is the limit to which the apo- them of the inscribed reguJwr polygon approaches, as the number of sides is increased indefinitely. Hyp. If AB be a side of a regular w-gon of apothem. KLy Cone. : then KL = KA as n is indefinitely increased. Dem. AK- KL < AL. (VII. 2 a. ) AL=^\AB. As n increases, AB diminishes, and may be made as small as ^® P^^^s®- .-.. AL = 0. (Def. of limit.) .*. AK— KL = ; (Same reason.) i.e. KL = AK. (Same reason.) Q.E.D. XVIII. 5 a. The radius of a regular circumscribed n-gon approaches as a limit the radius of the inscribed circle as n is indefinitely increased. x — ^r. i ^^ — yM Let the student supply the proof, which is exactly similar to the above, using the adjoin- ing figure. JC XVIII. 5 b. The square on the apothem approaches as a limit the square on the radius. (Fig. of Theorem.) XVIII. 6. The circumference is the common limit to which the perimeters of similar inscribed and circum- scribed regular polygons approach as the number of sides is increased indefinitely. Hyp. If C is the circumference of a circle, of radius r, and P and F are the. perimeters of the regular circumscribed and similar inscribed n-gons, XVIII. CIRCUMSCRIBED AND INSCRIBED POLYGONS 216 Cone. : then, as n is indefinitely increased, P=C and P' = (7. Dem. Let AB and EH be the sides of the n-gons ; KM and KL, the apothems. Then P.P.: KM : KL. (XVII. 2.) .-. P-P'.P.'. KM- KL : KM. (XI. 1. Sch. Cone. (4).) .-. (P- P') ' KM= P' (KM- KL) = P • {r- KL). (XL 1.) .-. P-P'=P' ^'~^^^ . (Ax. 3.) /Of ^ ^ But r - KL = (XVIII. 5.), and. P decreases as n increases. (Why ?) ... p.rzi^ = 0; i.e. P-P' = 0. (Def. Limit.) . Now C>P', and (7^^, then E C EC E'C EC, and the first fraction is less than the second. .-. E' must coincide with E, and AE, BF, CG concur. Q.E.D. 230 THE ELEMENTS OF GEOMETRY XX. 2. Three concurrent perpendiculars divide the sides of a triangle so that the sum of the squares of one set of alternate segments equals the sum of the squares of the other set, and conversely. Hyp. If / ^^, OE, OF, OG are concurrent F/ J§ on the sides of the A ABC, Cone: then AS" + BE" + CF^ = G^ + EG^ + Fj". Dem. . Join with A, B, and G, A0>--AG'=GG' = B0'-- GS, (XIV. 1 a, and Ax. 1.) Similarlj on the other sides. .-. ag^-bg'+be^-ec' + cf^-Fa'= AO'-Bd'-\-BO'-C&+CO'-AO' = 0. Transposing, ag'+be''-\-gf' = gb^+ec^+f3'. Conversely. If AG^ + BE'-{-CF''=GB' + EC^fFA', Cone. : then perpendiculars to the sides of the triangle at E, F, and G concur. Dem. Suppose the perpendiculars do not concur ; that those at F and E meet at 0, while the perpendicular from to AB meets AB at G'. Then AG'' + BE'+CF'=G^ + EC' + FA'. (1) (By the direct Th.) AG' + BE'-^CF^=GB'-\-EC'-hFA\ (2) (By Hyp.) Q.E.D. XX. CONCURRENT TRANSVERSALS AND NORMALS 231 Subtracting (2) from (1), member from member, . which is impossible unless G= G'-, f or if (r ^ G', the first member is + and the second is — , or vice versa, and a + quantity cannot equal a — quantity. .-. the perpendicular from on AB must fall at G; i.e. the three perpendiculars to the sides at E, F, and G must concur. Q.E.D. Ex. 1. Show, by using XX. 1, that the following sets of angle-trans- versals of a triangle are concurrent : (a) The medians. (6) The altitudes. (c) The joins of the vertices to the points of contact of the inscribed circle. (d) The bisectors of. the interior angles. (e) The bisectors of two exterior angles and an interior angle not adjacent to either of these exterior angles. Ex. 2. Show that the perpendiculars erected to the sides of a triangle at the points of contact of the escribed circles are concurrent. Ex. 3. Show from XX. 2 that if each of three circles intersect both the others, the three common chords are concurrent. 282 THE ELEMENTS OF GEOMETRY XX. SUMMARY OP PROPOSITIONS IN THE GROUP ON CONCURRENT TRANSVERSALS AND NORMALS 1. If three transversals through the vertices of a tri- angle are concurrent, the product of one set of three alternate segments determined by the transversals on the sides of the triangle equals the product of the other set, and conversely. 2. Three concurrent perpendiculars divide the sides of a triangle so that the sum of the squares of one set of alternate segments equals the sum. of the squares of the other set, and conversely. SOLID GEOMETRY XXI. GROUP ON THE PLANE AND ITS RELATED LINES DEFINITIONS A Plane has already been defined to be a surface such that if any two of its points be joined by a straight line, this line will lie wholly within the surface. A plane is said to be determined when it fulfills such condi- tions that its position is fixed. No two planes can fulfill the same set of determining condi- tions without coinciding throughout their whole extent. Corollaries of the Definition (a) A straight line and a point without the line deter- mine a plane. Dem. Let AB be the given line j^ and C the given point. 4^::r::^r2\ Let MQ be any plane through AB. yS^"^^ \ ,^t^^ Revolve MQ on AB as an axis, ^'^^^^^b' "^(i until it contains C. Let this position of the plane be Bh. The plane BL is fixed. For if it be revolved in either direction about AB^ it will no longer contain C Q.E.D. 284 THE ELEMENTS OF GEOMETRY (6) Three points nob in the same straight line deter- mine a plane. Dem. Join any two of the points; apply Cor. (a). (c) Two intersecting lines determine a plane. {d) Two parallels determine a plane. (e) The plane determined hy a line and a point is identical with tJie plane determined hy this line and the parallel to it thai contains tJie given point. (/) A straight line cannot intersect a plane in more than one point. (g) If four points^ A, B, C, and E, are not in the same plane (i.e. are not coplanar), no three of the points can he collinear. Dem. If, for example, AyB, and C lie in the same straight line, then this line determines with E a plane, i.e. all four points are coplanar, which contradicts the hypothesis. .-. no three of the points can be collinear. ^ Q.E.D. The point in which a line meets a plane is called the Foot of the Line. A Perpendicular to a Plane is a line perpendicular to every line of the plane that passes through its foot. The Projection of a Point on a plane is the foot of the perpen- dicular from the point to the plane. The Projection of a Line (straight or curved) on a plane is the locus of the projections of the points of the line. The Angle that a straight line makes with a plane is the angle formed by the line and its projection on the plane. A line is parallel to a plane when any plane through the line intersects the given plane in a line parallel to the given line. ,^-^^1^ XXI. THE PLANE AND ITS RELATED LINES 235 PROPOSITIONS XXI. 1. The line of intersection of two planes is straight. ^^^^^^^^^^^^^^^^^^^^ Hyp. If plane CF intersects plane EG in the line AB, Cone. : then AB is a straight line. Dem. If AB is not a straight line, it must contain at least a third point that is not in the same straight line with A and B. But three points not in the same straight line determine the position of a plane. (Def. of Plane, Cor. (&).) That is, if A, B, and the third point are not in the same straight line, planes FC and EG coincide. But this conclusion is contrary to the hypothesis. .'. AB is a straight line. ^ Q.E.D. Ex. 1. To make sure that a surface is perfectly "flat," a mechanic applies his " straightedge " to the surface in various directions and sees that the "straightedge" touches the surface along its whole length in every position. On what definition is his action based ? Ex. 2. How may three points be so situated that more than one plane may be passed through them ? Ex. 3. Show that four different planes may be passed so as to contain three out of four given points, if no three of the points be collinear. In how many planes would each of the four given points lie ? THE ELEMENTS OF GEOMETRY XXI. 2. If through a given point a perpendicular is drawn to a plane, it is the only perpendicular that can be drawn through the point to the plane. p R/ ^^^Fm /-^ \^^m / Z^ ■^■^1 Hyp. Case I. If is a given point in plane JfQ, and PO is perpendicular to plane 3fQ, Cone. : then PO is the only perpendicular that can be drawn to plane MQ, at 0. Dem. If possible, let TtO be a second ± to MQ, at 0. Let the plane of PO and RO intersect the plane MQ, in OA. Then RO must be ± to OA. (Def. of JL to a plane.) But PO is perpendicular to OA. (Def. of ± to a plane.) .'. we have OA in plane of PO and RO perpendicular to PO and RO. But this conclusion is impossible. (Ax, 7. Direct Inf.) .-. PO is the only perpendicular that can bo drawn to plane Mq at 0. Q.E.D. Ex. 4. Hold two pencils so as to show that if two lines do not intersect, and are not parallel, a plane cannot contain both of them. XXI. THE PLANE AND ITS RELATED LINES 237 O Hyp. Case II. If \ is a given point jH- without the plane MQ, and PO is perpen- dicular to plane MQ, Cone. : then PO is the only perpendicular that can be drawn to plane MQ through 0. Dem. If possible, let OL be a second perpendicular to MQ through 0. Draw LP. Then, in the plane OPLj we have OP and OL each perpen- dicular to PC. But this conclusion is impossible. (Ax. 7. Direct Inf.) .-. PO is the only perpendicular that can be drawn to MQ through 0. ^^^ Ex. 5. Why are the projections of straight lines on a plane always straight ? Show that if the projection of a line on a plane is straight, the line need not be straight. Ex. 6. Show how a circle may be so situated with respect to a plane that its projection on the plane will be a straight line. Ex. 7. Show that if the projection of the line AB on each of two inter- secting planes be straight, the line itself must be straight. Ex. 8. Why does a three-legged stool always stand firmly on a level floor, while a table or chair with four legs may be unsteady ? Ex. 9. Show that if two lines lie in the same plane, they must either intersect or be parallel. , Ex. 10. Show that if four lines concur, the greatest number of planes that can be determined by the lines two and two is six. 288 THE ELEMENTS OF GEOMETRY XXI. 3. A line perpendicular to two lines at their intersection is perpendicular to the plane of these lines, A ^^ yii' -~— ..NjV ^y^ B M^ / / \ \ / \ V / \. / 1 A' Hyp. If AB is perpendicular to BM and BL at B, and if OK is the plane determined by BM and BL, and if BR is any other line of OK through B, Cone. : then AB ± BE, or AB is perpendicular to plane OK. Dem. Draw EC, any line of OK Suppose EC cuts BL in C, BR in H, and BM in E. Produce AB to A\ making BA' = AB. Draw AC, AH, AE, and A'C, A'H, A'E. EB and CB are mid-perpendiculars to AA'. (Const.) .-. EA = EA', CA = CA'. EG is common to the A AEC, A' EC. .-. AAEC^AA'EC. (V. 3.) .-. ZACH=ZHCA'. (Rom. A of ^ A.) .'. AACH^AHCA'. (V. 1.) .-. HA = HA', (Hom. sides of ^ A.) .'. two points (B and H) of BR are equidistant from the ends of AA'. .: EB is a mid-perpendicular to AA'. .', AB is perpendicular to any line of OK passing through B. .-. AB ± plane OK. ^ Q.E.D. XXI. THE PLANE AND ITS RELATED LINES XXI. 3 a. If three lines are perpendicular to a given line at a given point, these perpendiculars lie in a plane perpendicular to the given line at the given point. XXI. 3 6. The plane mid-normal to the join of two points is the locus of all points equidistant from the given points. XXI. 3 c. The plane through a given point perpen- dicular to a given line, is unique, whether the given point he on the given line or ivithout the given line. XXI. 4. Of all straight lines drawn from a given point to a given plane, (1) The perpendicular is the shortest line, and con- versely. Hyp. If PO is perpendicular to plane MQ, and PA is any other line from P to plane MQ, Cone: then POAR. Dem. Lay off -80= BR, and draw AC AC=AR. (Why?) But AF>AC. .-. AF>AR. Proof of converse is left as an exercise for the pupil. Q.E.D. XXI. THE PLANE AND ITS liELATED LINES Theorem of the Three Perpendiculars 241 XXI. 5. If from the foot of a perpendicular to a plane a line is drawn at right angles to a second liiie of the plane, and if the point of intersection of these two lines is joined to any point in the perpendicidar, this third line is perpendicular to the second line of the plane. Hyp. If AB is perpendicular to plane MQ, and if BE in MQ is drawn from the foot of AB perpendicular to LF, any line of MQ, and if from E, the point of intersection of BE and LF, AE is drawn to A, any point in AB, Cone: then AE ± FL. Dem. Take EU = EC. Draw BR, BC, AE, and AC. ABEC^ABEB. (V. 1.) .-. BC==BR. (Horn, sides of ^ A.) .-. AC=AB. ■ (XXI. 4. (2).) ,\AE±FL. (VIII. 2. Sch. 2.) Q.E.D. XXI. 5 a. The second line FL in the above figure is perpendicular to the plane of the first and third lines, namely, plane AEB. ^ . ^ 242 THE ELEMENTS OF GEOMETRY XXI. 5 b. If two lines intersect at right angles, and through their point of intersection a perpendicular to the second is drawn without the plane of the lines, the first is the projection of the third perpendicular on the plane of the given lines. XXI. 6. If one of two parallels is perpendicular to a plane, the other is also perpendicular to the plane, and conversely. » M /E G' Hyp. If AB II CE^ and AB is perpendicular to plane MQ, Gone. : then CE is perpendicular to plane MQ. Dem. Draw BE\ also, in plane MQ, GF ± BE at E; draw AE. GF is perpendicular to plane of AE and EB. (XXI. 5 a.) But CE lies in plane EAB. (XXI. Def. of plane, (e).) .-. GE± CE. (Def. J to a plane.) .-. CE±GE. But CE± BE. (Def. lis, 2d Dir. Inf.) .-. CE± plane MQ. (XXI. 3.) Q.E.D. Conversely. If AB and CE are perpendicular to plane MQ, Cone. : then AB II CE. XXI. THE PLANE AND ITS RELATED LINES 243 Dem. If EC is not parallel to BA, draw ET that is. Then ^T is perpendicular to plane MQ. (XXI. 6.) But EC is perpendicular to plane MQ, (Hyp.) And EC is the only perpendicular that can be drawn to plane MQ at E. (XXI. 2.) .*. ET and EG must coincide. .-. AB II GE. Q.E.D. XXI. 6 a. If two lines are parallel to a third, they are pai^allel to each other. Hyp. If AB II GR and EF II CRy Cone: then A E G M ■ \ ■ ^ A AB II EF. Dem. Pass a plane MQ perpendicular to GR. Then AB and ^i^ are each perpendicular to MQ. (XXI. 6.) .-. ^5 II EF. (XXI. 6. Conv.) Q.E.D. Ex. 11. A ruled surface is one that may be generated by the motion of a straight line. Show that a plane is a ruled surface. Ex. 12. To get a "straightedge" we sometimes fold a sheet of paper and use the edge of the fold. What proposition are we illustrating ? Ex. 13. In how many positions can the pendulum of a clock be per- pendicular to the floor on which the clock stands ? Why ? Ex. 14. To find whether or not a square post is perpendicular to a floor, a carpenter applies a square to the floor and post on two sides of the post. On what theorem is his action based? Does it make any difference what sides of the post he selects ? Why ? Ex. 15. To keep a vertical sign in an upright position, it is fastened at the foot to two horizontal crosspieces nailed together in the shape of an X. What proposition is illustrated by this device ? 244 THE ELEMENTS OF GEOMETRY XXI. 7. If two angles not in the same plane have their sides respectively parallel and lying on the same side of the join of their vertices, the angles are equal. Hyp. If A'V II AL and A'R II AR, and these lines lie on same side of the join AA^ and also lie in different planes, Cone: then ZL'A'E' = ZLAR. Dem. Lay off AB=A'B', and AO=A'a'; and join CO, BB\ BO, and B' v. The 4-sides A-B' and A-C are parallelograms. (VI. 2.) .-. BB' = AA' and CO = AA'. (VI. 1 a.) .-. BB'=Ca. (Ax. 1.) BB' II CO, (XXI. 6 a.) ••. the 4-side B-C is a parallelogram. (VI. 2.) .-. BC=B'a. .-. A JBC^ ^ A B'OA'. (V. 3.) .-. Z L'A'R' = Z X^iJ. (Hom. Aof ^ A.) Q.E.D. XXI. THE PLANE AND ITS RELATED LINES 245 XXI. %. If a line is parallel to a plane, any parallel to the given line through a point of the plane lies wholly in the plane. Hyp. If AB is parallel to plane MQ, and CE II AB through G, a point of MQ, Cone. : then CE lies wholly in the plane MQ. Dem. But one parallel to AB can pass through C. (Ax. 7.) The line of intersection of the planes ABC and MQ is paral- lel to AB and passes through C. (Def. of II to a plane.) .-. this line of intersection is the parallel CE, that is, CE lies wholly in the plane MQ. Q.E.D. Ex. 16. How many lines may be drawn perpendicular to a given line at a given point ? How are all these lines situated ? Ex. 17. How many lines can be drawn perpendicular to a given line through a point without this line ? Prove. Ex. 18. Prove that one plane, and only one, may be passed perpen- dicular to a given line at a given point. Ex. 19. Show how to pass a plane through a given point perpendicular to a given line passing through the point. Ex. 20. Show how to pass a plane through a given point perpendicular to a line that does not pass through the point. Ex. 21. What surface is generated by the hand of a clock as it passes around the dial ? Why ? Ex. 22. A vertical flagstaff 75 ft. high stands in the center of a grass- plot 40 ft. in diameter. How far is the top of the pole from any pohit in the circumference of the grassplot ? Ex. 23. The blades of a windmill are 15 ft. long, and are fastened to the axle at an angle of 60°. How far does the tip of a blade travel in 10 minutes, when the wheel is making 80 revolutions a minute ? 246 THE ELEMENTS OF GEOMETRY XXI. 8 a. Conversely. If a line is parallel to a line of a plane f it is parallel to the plane. A B Hyp. UABWCEy M I a line of the plane / 2>L= 'i?^ MQ, / c Cone. : then AB is parallel to plane MQ. Dem. Through AB pass any plane AR. Let this plane intersect MQ in DR. If DR -^ AB, DR ^ CE. (XXI. 6 a.) If DR^ CE, draw DS II CE. DS lies in MQ. (Def. of plane, (d).) DSWAB. (XXI. 6 a.) *. DS is coplanar with AB. (Def. of plane, (d).) But DR is coplanar with AB. (Const.) .-. the two planes ABDR and ABDS have three points {A, B, D) common. But this is absurd. (Def. of plane, (6).) .-. to suppose DR -^ AB is absurd. .-. DR II AB. That is, AB is parallel to plane MQ. (Def. of II to plane.) Q.E.D. Ex. 24. The arm of a derrick is 50 ft. long ; it is so fastened to the mast as to revolve at a constant angle of 30° with the vertical upright. How far does the end of the arm travel in a quarter revolution ? Ex. 25. The ceiling of a room is 10 ft. high. How would you deter- mine, by means of a 12-ft. pole, a point in the floor directly under a gas drop in the ceiling ? Ex. 26. If two columns are perpendicular to the same floor, how are they situated with respect to each other ? XXI. THE PLANE AND ITS RELATED LINES 247 XXI. S b. If a line is parallel to each of two planes, it is parallel to their line of intersection. Hyp. If AB is II to plane EF, and also to plane EG, G Cone. : then AB II EC. Dem, A parallel to AB through C must lie in the plane EF. (XXI. 8.) This parallel must also lie in the plane EG. (XXI. 8.) .-. the parallel must be CE. ^ Q.E.D. Ex. 27. A column is perpendicular to a level floor. The capital and base of a second column in the same wall are respectively 20 ft, from the capital and base of the first. By what proposition do you know the second column to be vertical ? Ex. 28. How would you make use of XXL 5 6 to let fall a perpen- dicular to a plane from a point without the plane ? Ex. 29. Show how to draw, through a given point in a plane, a per- pendicular to the plane by using XXI. 5 6. Ex. 30. Show how to pass a plane through a given point parallel to a given line. Ex. 31. How many planes may be passed through a given point parallel to a given line ? Ex. 32. Show how to draw a line through a given point parallel to a given plane. Ex. 33. How many lines may be drawn through a given point parallel to a given plane ? Ex. 34. If a number of lines be drawn through a given point parallel to a given plane, how will these lines be situated ? Why ? Ex. 35. Show that if a number of lines be parallel to the same plane, and one of the lines intersect all the others, then all the lines must lie in the same plane. Ex. 36. If a number of planes be passed through a given point parallel to a given line, how will these planes b^ situated With respect to each other ? Why ? 248 THE ELEMENTS OF GEOMETRY XXI. SUMMARY OF PROPOSITIONS IN THK GROUP ON THE PLANE AND ITS RELATED LINES 1. The line of intersection of two planes is straight 2. If through a given point a perpendicular is draivn to a given plane, it is the only one that caii he drawn through the point to the plane. Case I. Point in plane. Case II. Point without plane, 3. A line perpendicular to each of two other lines at their intersection is perpendicular to the plane of these lines, a. If three lines are perpendicular to a given line at a given point, these perpendiculars lie in a plane perpendicular to the given line at the given point. h. The plane mid-normal to the join of two points is the locus of all points equidistant from the given points. c. The plane through a given point perpendicular to a given line, is unique, ivhether the given point he on the given line or without the given line. 4. Of all straight lines drawn from a given point to a given plane, (1) The perpendicular is the shortest line, and con- versely. XXI. THE PLANE AND ITS RELATED LINES 251 PROBLEMS XXI. Prob. 1. Through a given point to draw a perpendicular to a given plane. A \\ I \ M ! \ n M. F B Q B Case L Case II. Given. The plane MQ and the point A. Required. A perpendicular to MQ through A. Case I. A is without MQ. Const. Draw any line, BG, in MQ. Draw^E±£a On MQ draw EF A. BO. Dt^w AG±EF. AG is the perpendicular required. Dem. The demonstration is supplied by XXI. 5 b. Case II. J. is in MQ. Const. Draw any line, AE, in MQ. In MQ dvsiw BG±AE. Draw EF, without MQ, and perpendicular to BG. In the plane of FE and AE draw AG ± AE. AG is the perpendicular required. » Dem. The demonstration is supplied by XXI. 5 a. Q.E.P. Q.E.F. 252 THE ELEMENTS OF GEOMETRY XXI. Prob. 2. To draw a common perpendicular to two lines not in the same plane. Q Given. The lines AB and CE, not in the same plane. Required. A common perpendicular to AB and CE. Const. Through any point of CE, as F, draw FO II AB. CE and FO determine a plane MQ. (Def. of plane, (a).) Project BA on this plane, by the perpendiculars HL, AS. Let CE intersect SL in 0. Through draw 0J± to the plane MQ. (XXI. Prob. 1.) OJ is the perpendicular required. Dem. AB II MQ. (XXI. 8 a.) AB li SL. (Def. of II to a plane.) OJJL SL. (Def. of ± to plane.) OJ II AS. (XXI. 6, converse.) .'. OJ lies in the plane through 0, xS", and ^ ; i.e. in the plane of the parallels AB and SL. .'. OJ intersects AB, and OJAS is a rectangle. (Def. of □.) i.e. 0J± AB. But 0J± CE. (Def. of ± to a plane.) .*. OJ is a common perpendicular to ^IB and CE. ^ ^ Q.E.D. ScH. But one common perpendicular can be drawn to AB and CE ; for a line perpendicular to AB must be perpendicu- lar to SLf which is parallel to ^B. A common perpendicular to AB and CE must therefore be perpendicular to CE and SL, and hence perpendicular to MQ at 0. But the line perpendic- ular to MQ at is unique j that is, the common perpendicular is unique. XXII. GROUP ON PLANAL ANGLES DEFINITIONS (a) Dihedrals A Dihedral angle, or simply a dihedral, is the figure formed by two planes that intersect. The Faces of a dihedral are the planes by which it is formed (AE, FG). f G The Edge of a dihedral is the line of intersection of the faces {CE). When two planes intersect so that the four dihedrals formed are equal, each of the dihedrals is called a Right Dihedral, and the planes are said to be perpendicular to each other. The terms acute, obtuse, oblique, angles of the same kind, com- plemental, supplemental, adjacent, opposite, exterior, interior, corre- sponding, alternate, etc., have the same meaning in solid geometry as in the plane, the face of the dihedral replacing the side of the plane angle. Method of reading Dihedrals. A dihedral may be read, when there is no ambiguity, by merely reading the edge; as, dihedral CE. Otherwise, the angle is indicated by reading one letter from each face, with the edge between these letters ; thus, A-CE-G. The dihedral may be assumed to be generated by the revolu- tion of a plane, upon the edge as an axis, from an initial plane (as FG) to a terminal plane (as AE). As in plane geometry, so in solid geometry, rotation is Posi tive when anti-clockwise, and Negative when clockwise. 253 254 THE ELEMENTS OF GEOMETRY The Rectilineal Angle of a dihedral is the plane angle formed by two lines, one drawn in each face, and perpendicular to the edge at the same point. Note. — It is easily shown that (a) If two dihedrals are equal, their rectilinear angles are equal, and conversely. (6) Any two dihedrals are to each other a.s their rectilinear angles. Accordingly, the rectilineal angle is usually called the Measure of the dihedral. A Transversal Plane is a plane intersecting a number of other planes. Parallel Planes. Two planes are said to be parallel when they are so situated that if any transversal plane cuts them, the corresponding exterior-interior dihedrals are equal and have their edges parallel. The plane through a given point and parallel to a given plane is unique. (6) POLYHEDRALS A Polyhedral (n-dral) is the figure formed by the intersection of three or more planes at a single point. Dihedral and polyhedral angles are called, collectively, Planal Angles. The Vertex of the polyhedral is the point in which the planes intersect (0). The Faces of a polyhedral are the intersecting planes (AOB, BOC, etc.). The Edges of a polyhedral are the lines of intersection of the faces (0-4, OB, etc.). XXII. PLANAL ANGLES 255 The Face Angles of a polyhedral are the plane angles formed at the vertex by consecutive edges {Z.AOC, Z BOC, etc.). Symmetric Polyhedrals. As the faces and edges of a poly- hedral are infinite (unlimited) in extent, the polyhedral will have two parts, lying on opposite sides of the vertex, the angles of which, dihedral and plane, are equal in pairs. Thus, B-AO-C = B'-A'O-C. (Their faces are the same planes.) Similarly, A-BO-C = A'-B'O-C, etc. Again, Z AOC = Z A'OC (vertical angles), etc. But the equal angles of the two parts of the polyhedral occur in reverse order, as indicated by the arrowheads. For this reason the two parts of the polyhedral are called Symmetrical Polyhedrals. This name is due to Legendre. Unless the contrary is stated, or evidently implied, we shall, in using the word polyhedral, refer to the part on one side of the vertex only. The other part will be called the symmetrical. Polyhedrals are classified according to the number of faces as tri(3)hedrals, tetra(4)hedrals, penta(5)hedrals, etc. The Trihedral is analogous to the triangle, the faces of the trihedral corresponding to the sides of the triangle, and the dihedrals of the trihedral to the angles of the triangle. Hence, the following terms will be self-explanatory : Scalene, Isoangular, Isosceles, Equiangular. Equilateral, A Rectangular trihedral is one that has a single right dihedral. A Bi-rectangular trihedral is one that has two right dihedrals, and no more. A Tri-rectangular trihedral is one that has three right dihe- drals. Method of reading Polyhedrals. A polyhedral is read by taking first the letter at the vertex, and then, in succession, the letters at the other extremities of the edges, as 0-ABCE. 256 THE ELEMENTS OF GEOMETRY If no misunderstanding is likely to result, the letter at the vertex may be used alone, as in the case of plane angles. A Convex Polyhedral is one whose intersection with any plane not passing througli its vertex is a convex polygon. PROPOSITIONS (a) Dihedrals XXII. 1. In a right dihedral a line drawn in one face, perpendicular to the edge, is perpendicular to the other face. Hyp. If P-SO-Q is a right dihedral, and if AB of plane PO is perpendicular to SO, the edge of the dihedral, Cone. : then ^B is perpendicular to plane MQ. Dem. Through A draw CR in plane MQ, perpendicular to SO. Z CAB is the measure of the dihedral P-SO-Q. (Def. of measure of a dihedral.) .*. Z OAB is a right angle. But AB±SO. (Hyp.) .-. AB is perpendicular to plane MQ. (XXI. 3.) Q.E.D. XXII. PLANAL ANGLES 257 XXII. 1 a. A perpendicular to either face of a right dihedral, at any point of the edge, lies in the other face. Hyp. If AB is perpendicular to plane MQ, one face of the dihedral F-SO-Q, Cone. : then AB lies in plane OP. Dem. A line in plane OP perpendicular to SO is perpendicu- lar to plane MQ. (XXII. 1.) Only one _L to plane MQ can be drawn at J.. (XXI. 2.) But plane OP is perpendicular to plane MQ. (Hyp.) .-. AB must lie in plane OP. Q.E.D. Ex. 1. If a line and a plane are each perpen- dicular to a second plane, the line and plane are parallel. Ex. 2. If a line a be perpendicular to a plane MQ, and the lines b and c be both perpendicular to a, then 6 and c will both be parallel to plane J M <7 Ex. 3. If two intersecting lines be parallel to a given plane, the plane of the lines will be parallel to the given plane. Ex. 4. Hence, show how to pass, through a given point, a plane parallel to a given plane. 258 THE ELEMENTS OF GEOMETRY XXII. 2. If each of two intefi^secting i^lanes is per- pendicular to a third, their line of intersection is per- pendicular to the third. Hyp. If plane CE is perpendicular to plane MQ, and plane FG is perpendicular to plane MQ, and AB is their line of intersection, Cone. : then AB is perpendicular to plane MQ, Dem. Dihedral AG is a. right dihedral. (Hyp.) .-. a ± to plane MQ at A lies in plane FG. (XXII: 1 a.) Also a X to plane MQ at A lies in plane CE. (XXII. la.) .-. AB is perpendicular to plane MQ. Q.E.D. Ex. 5. If one of two parallels is parallel to a given plane MQ, the other is also parallel to MQ. M Ex. 6. If each of two planes is parallel to a third, these two planes are parallel to each other. Ex. 7. Through a given point but one plane can be passed parallel to two lines that are not parallel to each other. ZZ7 XXII. PLANAL ANGLES 259 XXII. 2 a. If a line is per^Jendicular to a plane, every plane through the line is also perpendicular to the plane. gc Hyp. If AB is perpendicular to plane MQ, and jf^ plane GF is drawn / through ABj / ^ F Q Cone. : then plane GF is perpendicular to plane MQ. Dem. Draw AJS± CF and in plane MQ. AB ± AE. (Def . of ± to a plane.) AE1.CF. (Const.) .•. AE is perpendicular to plane GF. (XXI. 3.) But rt. Z EAB is the measure of dihedral Q-FC-G. ,\ plane GF is perpendicular to plane MQ. Q.E.D. Ex. 8. What is the smallest number of plane angles that can be brought together at a point to form a convex polyhedral ? What must be true of the sum of these angles if they form a polyhedral ? Ex. 9. If three equilateral triangles be brought to- gether so as to have a common vertex, what will be the sum of the plane angles at this vertex ? Why, then, must a polyhedral be formed ? Ex. 10. Show that a polyhedral may be formed with four equilateral triangles ; also with five. Ex. 11. Show that no polyhedral can be formed by using any larger number of equilateral triangles than five. Ex. 12. Show that if all the faces of a polyhedral be regular w-gons, the only polyhedrals possible are those in which w = 3, 4, or 5. 260 THE ELEMENTS OF GEOMETRY (h) TOLYHEDRALS XXIT. 3. ITie sum of the two face angles of a tri- hedral is greater than the third angle. Note. — No proof is necessary unless the third angle is greater thap each of the others. Hyp. If F-ABC is a trihedral angle, and Z AFC is greater than either Z AFB or Z BFC, Cone. : then Z AFB + Z BFC > Z AFC Dem. Draw FR iii plane AFC so that Z AFB = Z AFB. Lay off FE = FB and pass a plane through B and B, cutting the edges in A, B, and O. Draw AB, BC, and AC. A AFB ^ A AFB. (V. 1.) .-, AB = AB. But AB + BC> AC. (VIT. 1.) .-. BC > BC. (Preliminary Th. 3.) .-. ZBFC>ZBFC [If two A have two sidee of one equal, etc.] (VII. Ex. 9.) .-. ZAFB-{-ZBFC>ZAFB-hZBFC( = ZAFC). Q.E.D. XXII. PLANAL ANGLES 261 XXII. 4. The sum of the face angles of any convex polyhedral is less than four right aiigles. Hyp. If i^ is a convex polyhedral, Cone. : then Z AFB + Z BFG + Z CFE, etc., < 4 rt. A. Dem. Pass a plane cutting the edges of polyhedral F in A, B, C, .... From 0, any point in this plane, draw OA, OB, 00, •••. In the base there are n triangles ; there are also n triangle faces in solid angle F. .'. the sum of the int. A of the base A = 2n rt. A, (III. 1.) and the sum of the int. A of the face A = 2 n rt. A (III. 1.) Now Z GAB of the base ', R on R', and Q on Q'. (These faxies are ^ by hyp.) .*. the plane of D, R, and Q must fall on the plane of D', R\ and Q'. (Def. of plane (b).) And as the truncated prisms are right, Jf must fall on M' and L on L'. (XXI. 2.) .-. the right truncated prism AM is congruent with right truncated prism A'M'. '^ Q.E.D. XXIII. THE PRISM AND THE CYLINDER 271 XXIII. 2 a. 2\vo right prisms having congruent bases and equal altitudes are congruent. XXIII. 3. Any oblique prism is equal to a right prism of ivhich the altitude equals a lateral edge of the oblique prism and the bases are right sections of the oblique prism. Hyp. If the right prism GJ' has its bases right sections of the oblique prism AE' and its altitude J J' equal to the edge EE', Cone. : then right prism GJ^ equals oblique prism AE', Dem. yle/and A' J' are right truncated prisms. (Const.) Their bases A-E and A-E' are congruent. (XXIII. 1 a.) The lateral faces BG and B'G' are congruent. Likewise, the lateral faces BI and BT are congruent. .-. Et. Tr. Prism AJ^ Rt. Tr. Prism A' J'. (XXIII. 2.) To each of the right truncated prisms add the right truncated prism GE', and we have right prism GJ^ equals oblique prism AE'. (Ax. 2.) Q.E.D. 272 THE ELEMENTS OF GEOMETRY XXIII. 4. Two rectangular parallelepipeds that have equal bases are to each other as their altitudes. ^ Q' IP A' Hyp. If the two rectangular parallelepipeds Q and Q' have equal bases and their altitudes are AB and A'B', Cone. : then rectangular parallelepiped Q : rectangular paral- lelepiped Q'l'.AB: A'B'. Case I. AB and A'B' commensurable. Case II. AB and A'B' incommensurable. Dem. Case I. Find a common measure of AB and A'B'. Let it be contained in AB five, and in A'B' three times. Then AB : A'B' : : 5 : 3. Through the points of division draw planes parallel to the bases. The small rectangular parallelepipeds thus obtained are all congruent. (XXIII. 2 a.) In Q there are five, in Q' three of these equal parallelepipeds. .-. Q: Q'l'.AB: A'B'. (Ax. 1.) Q.E.D. Ex. 13. The volume of a regular octagonal prism of altitude 8 is equal to the volume of a regular hexagonal prism of altitude 12. The radius of the base of the octagonal prism is 6. Find the lateral surface of each prism. XXm. THE PRISM AND THE CYLINDER 273 Q Sir" Q F llllii^^Hiiiiii Dem. Case II. Divide AB into any number of equal parts. Suppose one of these equal parts is contained in AB^ three times, with a remainder MB\ Through M pass a plane parallel to the base. Then ^^ .Q.: AM : AB. (Case I.) If the number of equal parts in AB be indefinitely increased, the remainder JfJ3' will be indefinitely decreased, but can never equal zero, because AB and A'B^ are incommensurable. .-. A'M approaches AB' as a limit, and Q" approaches Q' as a limit. limit, and -^ approaches ^ as a limit. But -^ is always equal to AB^ : AB, (Case I.) .*. the limits of the variables being equal, Et. parallelepiped Q' : rt. parallelepiped Q-.-.AM'. AB. Q.E.D. ScH. Two rectangular parallelepipeds which have two dimen- sions in common are to each other as their third dimensions. AM , AB' — — approaches — - — as a AB ^^ AB Ex. 14. The volume of each of two prisms is 1386 cu. ft. The base of the first is an equilateral triangle whose altitude is 20 ft. The base of the second is a square, each side of which is 20 ft. Find the ratio of the altitudes of the prisms. 274 THE ELEMENTS OF GEOMETRY XXIII. 4 a. Tloo rectangular parallelepipeds which have equal altitudes are to each other as their bases. A / /r / /I;/ / / f l>' A— / ^• / / / ^ Hyp. If the rectangular parallelepipeds Q and Q' have their altitudes equal, and the base of Q, a • 6 and of Q', c • e, Cone. : then Q : Q' : : a . & : c . e. Dem. Construct a rectangular parallelepiped Q" whose alti- tude is ^, and whose base is a • e. Then Q : Q" : : 6 : e. (XXIII. 4. Sch.) But Q" : Q' : : a : c. (XXIII. 4. Sch.) .-. Q : Q' : : a . 6 : c . e. (By mult.) Q.E.D. XXIII. 5. Two rectangular parallelepipeds are to each other as the products of their three dimensions, Q Hyp. If Q and Q' are two rectangular paral- lelepipeds whose bases are a • h and a' • h\ re- spectively, and whose altitudes are c and c', respectively. Cone : then Q : Q' : : a • 6 • c : a' • 6' • c'. Dem. Construct a rectangular parallelepiped Jf, whose base is a • 6 and whose altitude is c'. XXIII. THE PRISM AND THE CYLINDER 275 Then Q And M .-. Q M::c:c'. (XXIII. 4.) Q'::a'b:a' 'b\ (XXIII. 4 a.) Q'::a'b'c:a''b' 'C', (By mult.) Q.E.D. XXIII. 5 a. Tlie volume of a rectangular parallele- piped equals the product of its three dimensions. -Q- Hyp. If Q is a rec- tangular parallelepiped whose dimensions are a, ft, and c ■"" ilMIHi Cone. : then volume of Q = a • 6 • c. Dem. Construct a cube U, whose edge is the linear unit. Then Q:U::a-b'C:l -1-1. (XXIII. 5.) But Q:U is the volume of Q. (Del of vol.) AT a ' b • c J And — = a -b ' c. .\ the volume of Q = a • 6 • c. Q.E.D. ScH. The volume of a rectangular parallelepiped equals the product of its base by its altitude. Ex. 15. The interior dimensions of a water tank are 8 ft., 4 ft., and 5 ft., respectively. How many gallons will the tank hold ? Ex. 16. How much will it cost to line the tank with zinc at S5^ a square yard, allowing a waste of y\ of the material for seams ? Ex. 17. The convex surface of a right circular cylinder is equal to the total surface of a cube ; the diameter of the cylinder and its altitude each equals 10. Find the edge of the cube. Ex. 18. The altitude of a right circular cylinder is a, the radius of its base is b. To find the radius of a circle equal in area to the convex sur- face of the cylinder. 276 THE ELEMENTS OF GEOMETRY XXIII. 6. The volume of any parallelepiped equals the product of its base and its altitude. Hyp. If P'H is any parallelepiped whose base is the paral- lelogram P-H' and whose altitude is the perpendicular between the bases, Cone. : then the volume of PH equals the area of its base times its altitude. Dem. Produce PO, making KL = PO, and through K and L pass planes J'JIif and Q'Nl. KL. Extend the faces HP, H'P', PO', and GH' to intersect the planes J'3/ and Q'N, forming the right parallelepiped MQ'. Oblique parallelepiped P'H = right parallelepiped MQ'. (XXIII. 3.) Again, produce N'Q' making Q'F' = N'Q' and through Q' and F' pass planes Q'Jand F'R± Q'F'. Extend the faces Q'M' and LM, Q'N and M'K to intersect the planes Q'e/and F'R, forming the rectangular parallelepiped Q'R. (Def. of rt. parallelepiped.) Rt. parallelepiped MQ'=TQQ,t. parallelepiped Q'R. (XXIII. 3.) .•. rect. parallelepiped Q'J?=obl. parallelepiped P'H. (Ax. 1.) Now the volume of a rectangular parallelepiped equals the product of the base and an altitude. (XXIII. 5 a.) And as the base and an altitude of Q'R equal, respectively, the base and altitude of PH, the volume of PH equals the area of its base times the altitude. Q.E.O. XXIII. THE PRISM AND THE CYLINDER 277 XXIII. 7. The plane through two diagonally opposite edges of a parallelepiped divides the figure into two equal triangular prisms. yj Hyp. If AG is a parallelepiped A^'/^/^-^--^J / and a plane ACOR is passed through 'yT^-^^lZ^^rf/f AR and CQy / eI _jt Cone. • then triangular prism ACB-F equals triangular prism ACE-H. Dam. Through S, any point of ATI, pass a plane SMQK perpendicular to RA and intersecting ACOR in SQ. Plane BR is parallel to plane CH. (Def. of parallelepiped.) .-. SKW MQ, (Def. of II planes.) Similarly, " SM II KQ. .*. the 4-side S-Q is a parallelogram. (Def. of O.) .-. A SKQ ^ A SQM. (VI. 1 a. Sch.) Prism ABC-F equals a right prism whose base is A SKQ and whose altitude is FB. (XXIII. 3.) Prism ACE-H equals a right prism whose base is A SRQ and whose altitude is FB (or EH). (XXIII. 3.) But these right prisms are equal. (XXIII. 2 a.) .'. triangular prism ACB-F equals triangular prism ACE-H. (Ax. 1.) Q.E.D. Ex. 19. The total surface of a right circular cylinder whose height is twice the radius of its base is equal to the surface of a cube. If the edge of the cube is 12 in., what is the lieight of the cylinder ? Ex. 20. The altitude of a cylinder is 12 ft. Its base is a circle of radius 8 in. Find the volume and the total surface of the cylinder. Ex. 21. Find the diameter of a cylindrical tank 10 ft. deep, whose capacity is 8000 gal. 278 THE ELEMENTS OF GEOMETRY XXIII. 8. The volume of a triangular prism equals the product of its base by its altitude. Hyp. If ABG-F is a triangular prism whose base is ABG and whose altitude is ET^ Cone. : then the volume of AGB-F=A ABG x ET. Dem. Complete the parallelograms ABGO and EFQH. Draw OS" completing the parallelepiped ABGO-F. The volume of parallelepiped ABGO-F =^n ABGO x ET. (XXIII. 6.) But the volume of prism AGB-F equals one half that of parallelepiped AGBO-F. (XXIII. 7.) And A ABG== \ n ABGO. (VI. 1 a. Sch.) .-. volume of prism AGB-F = A ABG x ET. (Ax. 3.) Q.E.D. Ex. 22. How much sheet iron would it take 'to make such a tank (Ex. 21), allowing y^ for waste and seams ? Ex. 23. A cylindrical pipe of diameter 20 in. discharges 800 gal. a second. What is the velocity of the water in the pipe ? Ex. 24, The total surface of a right circular cylinder whose height is three times the diameter of its base is 2513.28 sq. ft. Find the volume of the cylinder. XXIII. THE PRISM AND THE CYLINDER 279 XXIII. 8 a. The volume of any prism equals the product of its base by its altitude. Hyp. If ABCO-F is any prism, and ABCOJ is its base, and ET is its altitude, Cone. : then its volume equals the product of its base times its altitude. Dem. Through EA and OH, EA and CQ, pass planes. IHiese planes divide the prism into triangular prisms. The volume of each triangular prism equals its base times its altitude. (XXIII. 8.) But the sum of the bases of the triangular prisms equals the base of the given prism, and the altitude of the triangular prisms is the altitude of the given prism. .*. the volume of prism J.S(70-jP equals the product of its base times its altitude. Q.E.D. Ex. 25. The diameter and the altitude of a cylinder are each equal to the edge of a cube. What is the ratio of the volumes of the two figures ? Ex. 26. A cubic foot of cast iron weighs 445 lb. What is the weight : a cast-iron pipe nal measurement ? 280 th:e elements of geometry XXIII. 8 b. Any two prisms are to each other as the products of the bases by the altitudes. If the bases are equal, the prisms are to each other as the altitudes. If the altitudes are equal, the prisma are to each other as the bases. If the bases are equal and also the altitudes, the prisms are equal. XXIII. SUMMARY OF PROPOSITIONS IN THE GROUP ON (a) THE PRISM 1. Parallel sections of a prism are congruent a. The bases of a prism are congruent 2. Tioo right truncated prisms are congruent, if three faces including a trihedral of the one are equal respeo- tively to three faces including a trihedral of the other and are similarly placed. a. Two right prisms having equal bases and equal altitudes are congruent. 3. Any oblique prism is equal to a right prism of ichich the altitude equals a lateral edge of the oblique prism, and the bases are right sections of the oblique prism. 4- Tioo rectangidar parallelepipeds that have equal bases are to each other as their altitudes. a. Two rectangidar parallelepipeds that have eqwA altitudes are to each other «5 their bases^ XXIII. THE PRISM AND THE CYLINDER 281 5. Two rectangular parallelepipeds are to each other as the products of their three dimensions. a. The volume of a rectangular parallelepiped equals the product of its three dimensions. 6. The volume of any parallelepiped equals the prod- uct of its base by its altitude. 7. The plane through two diagonally opposite edges of a parallelejnped divides the figure into ttvo equal triangular prisms. 8. The volume of a triangular prism equals the prod- uct of its base by its altitude. a. The volume of any prism equals the product of its base by its altitude. b. Any two prisms are to each other as the products of the bases by the altitudes. If the bases are equals the prisms are to each other as the altitudes. If the altitudes are equals the prisms are to each other as the bases. Jf the bases are equal, and also the altitudes, the prisms are equal. 282 THE ELEMENTS OF GEOMETRY DEFINITIONS (b) The Cylinder A Cylindrical Surface is a surface generated by a straight line that moves parallel to its first position along a curve not co- planar with the moving line. The moving line is called the Generatrix. The curve that directs the motion is called the Directrix. The successive positions of the generatrix are called the Elements of the Surface. A Cylinder is a solid inclosed by a cylindrical surface and two parallel planes. The Bases of a cylinder are the parallel plane sections. The Elements of the Cylinder are the portions of the elements of the cylindrical surface determined by the bases. The Altitude of a cylinder is the perpendicular distance between the bases. A Right Section of a cylinder is a section made by a plane perpendicular to an element. Cylindbbs A Right cylinder is a cylinder the elements of which are perpendicular to the bases. A Circular cylinder is a cylinder the bases of which are circles. A point is said to revolve around a fixed line when it gen- erates a circle whose plane is perpendicular to the fixed line and having its center on the fixed line. XXIII. THE PRISM AND THE CYLINDER 283 The fixed line is called the Axis of Revolution, or simply the axis. A line or surface is said to revolve about the axis, when every point in the moving line or surface revolves about the axis. The surface generated by the revolution of a line (straight or curved) about an axis is called a Surface of Revolution. The volume (or solid) generated by the revolution of a sur- face about an axis is called a Volume (or Solid) of Revolution. The axis of revolution is often called the Axis of the Surface or Volume generated by the revolution. Corollaries of the Definitions (a) Any section of a cylinder through an eleinent is a parallelogrmiv. (h) A right circular cylinder is a cylinder of revolution. As the circle has been shown (XVIII) to be the limit of the regular polygon as the number of sides is increased indefi- nitely, so the circular cylinder is the limit in surface and vol- ume of the prism with regular bases, as the number of sides of the bases is increased beyond any assignable number. (It will be a good exercise for the student to give the detailed proof of the statement.) Accordingly, every proposition that is true of every prism with a regular base, whatever may he the number of lateral faces, is true of the circular cylinder. We therefore obtain from the corresponding propositions of XXIII (a), the summary on the following page. Ex. 27. A horse power is the force necessary to raise 33,000 lb. 1 ft. in 1 min. The cylinder of an engine is 4 ft. in diameter and 6 ft. high ; the piston is 6 in. thick and the piston-rod 8 in. in diameter. Find the horse power of the engine when it is making 200 revolutions a minute with a steam pressure of 60 lb. to the square inch. Ex. 28. Show that no polyhedron can have less than four faces nor lesB than six edges. 284 THE ELEMENTS OF GEOMETRY ZXni. SUMMART OF PROPOSITIONS IN THE GROUP ON (6) THE CIRCULAR CYLINDER 1. Parallel sections of a circular cylinder are con- gruent a. The bases of a circular cylinder are congruent. 2. The volume of a circular cylinder equals the prod- uct of the area of its base by its altitude. a. Jf B. be the altitude of any circular cylinder, and R the radius of either base, the volume of the cylinder equals irR H. b. If B. be the altitude of any rigid circular cylin- der, and R the radius of either base, the area of the convex surface of the cylinder equals 2irRH. c. The volumes of any tioo circular cylinders are to each other as the products of the areas of the bases by the altitudes. If the bases are equal, the cylinders are to each other as the altitudes. If the altitudes are equal, the cylinders are to each other as the areas of the bases. If the bases are equal and also the altitudes, the cylinders are equal. XXin. THE PRISM AND THE CYLINDER 285 Ex. 29. Every plane section of a parallelepiped is a parallelogram if the plane of tiie secLion intersects 4 parallel edges. Ex. 30. What kind of quadrilateral is cut from a parallelepiped by a diagonal plane ? How do the diagonals of the section cut each other ? Ex. 31. Show that the diagonals of a parallelepiped concur in a point at which each is bi.sected. Def. The point in which the diagonals of a parallelepiped concur is called the center of the parallelepiped. Ex, 32, Show that the converse of this proposition is true. Ex, 33. Show that any line that passes through the center {K) of a parallelepiped and terminates in opposite faces of the parallelepiped is bisected at K. Ex, 34, Show that the sum of the squares of the diagonals of a paral- lelepiped equals the sum of the squares of the edges. Ex. 35, Show that the diagonals of a rectangular parallelepiped are equal. Ex, 36. Prove that the converse of the proposition is also true. Ex, 37. The edges of a rectangular parallelepiped are 12, 15, and 20 ft., respectively. What is the length of the diagonal ? Ex. 38. Show how to construct a parallelepiped that shall have its edges on three given straight lines. Ex, 39. A plane through any edge of the upper base and the diagonally opposite edge of the lower base of a prism cuts from the figure a rectangle. What kind of prism is the original figure ? Ex. 40, Every plane, through an edge of the upper base, that cuts the lower base of a prism, cuts from the prism a parallelogram. What kind of prism is the original figure ? Ex, 41, The volume of any regular prism is equal to the product of the lateral area by the apothem of either base. Ex. 42. Show how to cut from a cube a regular hexagon. Ex. 43. Show that if a prism be cut by two ^ planes and the cor- responding sides of the sections be produced, the points of intersection of these sides will be collinear. Ex. 44. Show that the volume of a prism equals the product of the area of a right section by the length of a lateral edge. Ex. 45. The total surface of a circular cylinder is equal to the convex surface of a cylinder having the same base as the given cylinder, and having an altitude equal to the altitude of the given cylinder plus the radius of the base. 286 THE ELEMENTS OF GEOMETRY Ex. 46. The volume of a cylinder equals the product of the area of a right section by an element of the cylinder. Ex. 47. Show that the volume of a cylinder is equal to its convex sur- face multiplied by ^ the radius of its base. Ex. 48. What is the locus of a point whose distance from a given straight line is equal to 10 in. ? Ex. 49. What is the locus of a point whose distance from a given plane is a and whose distance from a given line parallel to the plane is 6 ? Ex. 50. What is the locus of a point in a plane at a distance d from a line that intersects the plane ? Ex. 51. What surface is generated by the axis of a circular cylinder of radius a that rolls on the inner surface of a circular cylinder of radius b ? Ex. 52. Find a point equidistant from two given points, A and B, and also at a distance d from a given straight line. Ex. 53. Find a point equidistant from three given points, A, B, and C, and also at a given distance from a given straight line. XXIV. GROUP ON THE PYRAMID AND THE CONE DEFINITIONS (a) The Pyramid A Pyramid is a polyhedron, one face of which is a polygon, while the other faces are triangles that have a common vertex. This common vertex is called the Vertex of the pyramid. The Base of a pyramid is the polygon on which the pyramid is supposed to rest. If all the faces of a pyramid are triangles, any one may be taken as the base. The faces of a pyramid other than the base are called the Lateral Faces. The sum of the areas of the lateral faces is called the Lateral (or Convex) Surface of the pyramid. The Lateral Edges are the edges that meet in the vertex. The Altitude of a pyramid is the perpendicular dropped from the vertex to the base. 287 288 THE ELEMENTS OF GEOMETRY Pyramids are said to be Triangular, Quadrangular, Pentagonal, etc., according to the number of sides of the bases. A Regular pyramid is one whose base is a regular polygon that has for its center the foot of the altitude of the pyramid. Corollary op the Definition (a) The lateral faces of a regular pyramid are congruent isosceles triangles. The Slant Height of a regular pyramid is the altitude of any of its lateral faces. A Truncated pyramid is that portion of a /^""^^ pyramid which is comprised between the base / ! \ and a plane section not parallel to the base. / ^^--^-.^ \ A Frustum of a pyramid is that portion of a pyramid which is comprised between the base and a plane section parallel to the base. A Prismoid is a polyhedron, two of whose fares, called Bases, are parallel, while the other faces (lateral faces) are triangles or trapezoids that have their vertices at the vertices of the The Altitude of a frustum or of a prismoid is the perpendicu- lar distance between the bases. The Slant Height of a reornlar frustum (i.e. a frustum cut from a regular pyramid) is the altitude of any of its lateral faces. XXIV. THE PYRAMID AND THE CONE 289 PROPOSITIONS (a) Pyramids XXIV. 1. If a set of lines he cut hy three parallel 'planes, the lines are cut proportionally. Hyp. liAB and CF are cnt by the parallel planes SR, FO, and QM'in B, E, A, and F, H, C, Cone. : then AE : EB:: CH: HF. Dem. Draw the ioin ^F cutting plane PO in G. Draw the joins BF, EG, GH, and AC. Then But Note. — So also, AB:AE::CF: CH, and AB:BE::CF: HF. ScH. The same course of reasoning]: may be extended to any number of lines and any number of planes* AC II GH and EG II BF. (Def. of 11 planes.) \ AE.EB.'.AG: GF. (XV. 1.) AG: GF:: CH: HF. (XV. 1.) •. AE:EB::CH:HF. (Ax. 1.) Q.E.D. 290 THE ELEMENTS OF GEOMETRY XXIV. 2. Any section of a pyramid parallel to the base is similar to the base. a Hyp. If, in the pyramid S-ABCER, the section FOHKM is piirallel to ABCER, Cone. : then ?i-gon ABC n-gon FGH •". Dem. AB II FG and RA II MF. (Def. of II planes.) .-. Z RAB = Z3IFG. (XXI. 7.) Similarly, Z ABC = Z FGH; Z BCE =Z GHK, etc. j i.e. ABCE and FGHKare mutually equiangular. Again, A SIIG -^ASCB; A SGF ~ A SB A, etc. ;;XV. 2.) And •. BC'.GH AB.FG : AB : FG : ^>S' : GS. (Hom. sides of ~ A.) : jB/S' : (rxS'. (Same reason.) : BC'.GH. . (Ax. 1.) Similarly, the other pairs of homologous sides are propor- tional .-. w-gon ABCE w-gon FGHK - • -. (Def. of '^ figs.) Q.E.D. Ex. 1. From a point A without a plane obliques are drawn terminating on the plane. On each oblique a point P is so taken as to divide the line in the ratio of 2 : 3. Show that the locus of P is a plane. (XXIV. 1.) XXIV. THE PYRAMID AND THE CONE 291 XXIV. 2 a. The perimeters of parallel sections of a pyramid are to each other as the distances of the sec- tions from the vertex. / I J / ^/vM,'^ i \/ 1/ Hyp. If 8M\^ perpendicular to plane FQ in 3f and perpen- dicular to plane AE in i, and if section FORK-" is parallel to section ABCE • • •, Cone. : then perim. FQHK: perim. ABCE"-: : SM: SL. Dem. Through S pass a plane JT parallel to plane VQ parallel to plane AE. Perim. FGHK: perim. ABCE -..FG: AB. (XXIV. 2.) But FG: AB ::SG: SB. (Hom. sides of ~ A.) And SG:SB::SM:SL'.:SF: SA, etc. (XXIV. 1.) .-. perim. FGHK: perim. ABCE ::SM: SL. (Ax. 1.) Q.E.D. Ex. 2. AB and CE are two lines not in the same plane. Any number of lines QB terminating in AB and CE are bisected at D. Show that the locus of Z) is a plane through the mid- point M of the common ± FG. Pass a plane through M parallel to AB and CE. Through AB and CE pass planes parallel to the first plane. Use XXIV. 1. 292 THE ELEMENTS OF GEOMETRY XXIV. 2 b. The areas of two parallel sections of a pyramid are to each other as the squares of their dis- tances from the vertex. 7 Hyp. If SM is perpendicular to plane FQ in M&nd perpen- dicular to plane AE in L, and if section FGHK •••is parallel to section ABC •••, Cone : then area of w-gon FORK: area of w-gon ABCE : : SM^ : SI?. Dem. w-gon FGHK : rj-gon ABCE:: F(f : AIT. (XVI. 3.) But FG:AB::SG:SB::SM:SL. (XXIV. 1.) .-. FW : A& ::SG^:SB':: SM"" : SL\ (XI. (C).) .-. area of n-gon FGHK: area of n-gon ABCE : : SM^ : SL\ Q.E.D. Ex. 3. The base of a pyramid is a regular octagon whose radius is 10 ft. The altitude of the pyramid is 24 ft. Find tlie perimeter and the area of a section parallel to the base and 6 ft. from the vertex. Ex. 4. Two pyramids have the same volume. The area of the base of the first is 120 sq. yd. and its altitude 60 ft. The altitude of the second is 35 ft. What is the area of its base ? Ex. 5. The altitude of a regular pyramid is 15 ft. Its base is a square, each side of which is 4 ft. Find the volume of the pyramid. Find also the lateral surface and the total surface of the pyramid. XXIV. THE PYRAMID AND THE CONE 293 XXIV. '^ c. If tivo pyramids have equal altitudes and equal bases, sections parallel to their bases and equally distant from their vertices are equal. V Hyp. If A ACE = A FOH, altitude SL = altitude VM, and SV = VM', Cone: then n-gon A'C'E' = n-gon F'G'H'. Dem. Area ACE : area A'C'E' ::STJ: SL'\ (XXIV. 2 6.) Area FGH : area F'G'H' : : VM' : VM^'. But ST : SIJ' : : VM' : VM^. (Hyp.) .-. area ACE : area FGH : : area A'C'E' : area FG'H'. But area ACE = area FGH. .«. area A'C'E' = area i^'G^'^. Q.E.D. Note. — From the fourth proportion of the demonstration, it follows that if two pyramids have equal altitudes, the areas of sections parallel to the bases and equidistant from them have the same ratio as the bases. Ex. 6. The base of a regular pyramid is a dodecagon whose radius is 20 in. The volume of the pyramid is 4000 cu. in. Find the altitude. Ex. 7. Find the total surface and the volume of a regular triangular pyramid, each edge of wliich is a. Ex. 8. Each lateral edge of a regular triangular pyramid is 29 ft. ; the altitude is 21 ft. Find the total surface and the volume of the pyramid. 294 THE ELEMENTS OF GEOMETRY XXIV. 3. Two triangular pyramids that have equal bases and equal altitudes are equal. Hyp. If, in the triangular pyramids E and E', base ABC = base A'B'C'j and if the altitudes of the two pyramids are equal, Cone. : then Pyr. E = Pyr. E' Dem. If ^ :,fc E', let E-E' = r. Divide altitude AT into any number, say four, equal parts. Through each point of division pass parallel sections. On the base ABC construct a prism whose lateral edges are parallel to AE and whose altitude equals \ AT. Similarly, construct a prism upon each section as a base. This set of prisms is circumscribed about Pyr. E. With the topmost section of Pyr. E* as an upper base, con- struct a prism whose lateral edges are parallel to A'E' and whose altitude equals \ AT. Similarly, construct prisms with the remaining sections as upper bases. This set of prisms is inscribed in Pyr. E'. Each prism in Pyr. E' equals the prism next above it in Pyr. E. (XXIII. 8 b.) .'. the difference between the sums of prisms of Pyr. E and Pyr. E' is the lowest prism of Pyr. E. XXIV. THE PYRAMID AND THE CONE 295 Let sum of prisms in Pyr. E = S, and sum of prisms in Pyr. E' = S', and let volume of lowest prism in E = v. Then S-S' = v. But E to the vertices of the prismoid. Draw diagonals BI, BR, etc. in each trapezoid face. Pyramid K-ABG has h^ for base and ~ for altitude. z (Def. of mid-section.) 61. (XXIV. 4.) vol. K-ABG = Similarly, vol. K-ILJ The bases of the other set of pyramids are A, cut in mid- joins by the plane MQ of the mid-section. Let K-IBL be any one of these pyramids. Pyramid K-IBL ^ Area IBL Pyramid K-FBG ~ Area FBG (XXIV. 1.) Draw KG, KF. ^BL' BG' = 4. (■.'BL = 2BG.) (XXIV. 4 6. Sell.) (XVI. 2.) XXIV. THE PYRAMID AND THE CONE 303 But K-FBG = B-KFG = - • area KFO. (Why ?) .-. K-IBL = 4 (K-FBG) = ^ • area KFG. The sum of the areas of the A, KFG, etc. = m. (Ax. 4.) .*. the sum of the pyramids whose bases are the lateral faces of the prismoid = — - m. .-. F=|(6i + &2 + 4m). ' Q.E.D. ScH. The formula just established is often called the Pris- moidal Formula. Ex. 30. The elements of a right circular cone make an angle of 45° with the plane of the base. Find the ratio of the area of the base to the convex surface of the cone. Ex. 31. The elements of a right circular cone make an angle of 60° with the plane of the base. Find the ratio of the area of the base to the total surface of the cone. Ex. 32. Find the volume of a right circular cone of which the altitude is 6 ft. and the base has a radius of 30 in. Ex. 33. Find the volume of an oblique cone of which the altitude is 12 yd. and the base has a diameter of 3 yd. Ex. 34. Find the volume of a cone whose base has a radius of 10 ft. and whose slant height is 27 ft. Ex. 35. The convex surface of a right circular cone is 523.6 sq. ft. ; the slant height is 20 ft. Find the total surface and the volume of the cone. Ex. 36. The slant height of a right circular cone is 60 ft. ; each element makes an angle of 60° with the plane of the base. Find the total surface and the volume of the cone. Ex. 37. The volume of a right circular cone the radius of whose base is 12 in. is a cubic foot. Find the altitude and the convex surface of the cone. Ex. 38. One angle of a right triangle is 30°. Find the ratio between the volumes of the cones generated by revolving the triangle first about the shorter leg as an axis and then about the longer. 804 THE ELEMENTS OF GEOMETRY XXIV. SUMMARY OF PROPOSITIONS IN THE GROUP ON (a) THE PYRAMID 1. If a set of lines he cut hy three parallel planes, the lines are cut proportionately. 2. Any section of a pyramid parallel to the hose is similar to the base. a. The p)erimeters of parallel sections of a pyramid are to each other {vary) as the distances of the sections from the vertex. 6. The areas of two parallel sections of a pyramid vary as the squares of the distances of the sections from the vertex. c. If two pyramids have equal bases and equal altitudes, any sections of the tivo pyramids equidistant from the vertices are equal. 3. Tioo triangular pyramids that have equal bases and equal altitudes are equal 4. The volume of a triangular pyramid is one third the product of its base and altitude. a. The volume of any pyramid equals one third the product of its base and altitude. b. The volumes of any two pyramids are to each other as the products of their bases and altitudes. ScH. 1. If the bases are equal, the pyramids are to each other as their altitudes. If the altitudes are equal, the pyramids are to each other as their bases. XXIV. THE PYRAMID AND THE CONE 305 ScH; 2. The volume of any polyhedron may be found by dividing the figure into pyramids and adding the volumes of these pyramids. 5. The volume of a triangular frustum equals one third the product of the altitude into the sum of the tivo bases and a mean proportional between them. 6. The volume of any frustum equals one third the product of the altitude into the sum of the two bases and a mean proportional between them. 7. The lateral surface of a regular pyramid equals one half the rectangle of the slant height and the perime- ter of the base. a. The lateral surface of a regular pyramid equals the rectajigle of the slant height and the perimeter of a section parallel to the base and midway betiveen the vertex and the base (called the mid-section). 6. The lateral surface of a regular frustum equals the rectangle of the slant height and the perimeter of the mid-section. c. The lateral surface of a regular frustum equals the rectangle of the slant height and half the sum of the bases. 8. The volume of a prismoid equals one sixth the product of its altitude by the sum of the areas of the bases and four times the area of the mid-section. ScH. The formula just established is often called the Pri«- moidal Formula. 806 THE ELEMENTS OF GEOMETRY DEFINITIONS (b) The Conb A Conical Surface is a surface generated by a straight line that moves along a fixed c\irve and always passes through a fixed point not coplanar with the curve. The moving line is called the Generatrix. The fixed curve is called the Directrix. It is not necessary that the directrix be a plane curve. . Tlie successive positions of the generatrix are called the Elements of the Surface. The fixed point through which the generatrix passes is called the Vertex. The Nappes of the surface are the two parts into which it is divided at the vertex. A Cone is a portion of space bounded by one nappe of a coni- cal surface and a plane not passing through the vertex. The Base of the cone is the plane section that forms part of its bounding surface. The Elements of the Cone are the segments of the elements of the conical surface determined by the vertex and the base. The Altitude of a cone is the perpendicular from the vertex to the base. A Circular cone is one that has a circle for its base. A Right cone is one in which the altitude falls at the center of the base. XXIV. THE PYRAMID AND THE CONE 307 The terms Truncated and Frustum have the same meaning in the ease of the cone as in that of the pyramid. Note. — A frustum of a right circular cone is called a right circular frustum. COEOLLARIES FKOM THE DEFINITIONS (a) Every section of a cone that passes through the vertex is a triangle. (b) The right circular cone is a cone of revolution, of which the axis is the altitude. General Sch. As the Circle is the limit of the Regular Polygon when the number of sides of the latter is increased beyond any assignable value, so the Circular Cone is the limit of the Pyramid with a Regular Base when the number of lateral faces is increased beyond any assignable value. Hence, whatever propositions ai^e true of such a Pyramid, irrespective of the number of its lateral faces, are true of the Circular Cone. Accordingly, from the corresponding propositions on the pyramid, we have the following summary : XXIV. SUMMARY OF PROPOSITIONS IN THE GROUP ON (6) THE CIRCULAR CONE 9. Sections of a cone parallel to the base are similar to the base. 10. If sections of a cone are parallel to the base, (a) their circumferences are to each other as their distances from the vertex, and (h) their areas are to each other as the squares of their distances from the vertex. 11. The convex surface of a right cii^cular cone equals one half the product of the circumference of its base by its slant height. 808 THE ELEMENTS OF GEOMETRY CL If R is the radius of the base of a right circular cone, and H' the slant height of the cone, the convex surface of the cone equals ttRH', and the total surface equals 7rR(H' + R). b. The convex surface of a right circidar frustum equals the product of the slant height by the circumference of the mid-section. c. If Ri is the radius of the upper base of a right circular frustum, Rg the radius of the lower base, and h' the slant height, the convex sur- face of the frustum equals 7rli'(Ri4-R2)- 12. The volume of a circular cone equals one third the product of the area of its base by its altitude. a. If R is the radius of the base of a circular cone and H the altitude, the volume of the cone equals ^ ttR^H. h. Any two cones are to each other as the products of the areas of the bases by the altitudes. If the altitudes are equal, the cones are to each other as the areas of the bases. If the bases are equal, the cones are to each other as the altitudes. If the bases are equal and also the altitudes, the cones are equal, 13. Tlie volume of the frustum of a cone equals one third the product of the altitude by the sum of the areas of the bases and a mean proportional between them. XXIV. THE PYRAMID AND THE CONE 309 Ex. 39. The radius of the upper base of a right circular frustum is 25 in. ; the radius of the lower base is 36 in. ; the slant height is 15 in. Find the convex surface and also the total surface of the frustum. Ex. 40. Show that the convex surface of a right circular frustum equals the convex surface of a cylinder whose altitude is half the sum of the diameters of the bases of the frustum and whose diameter is the slant height of the frustum. Ex. 41. The radius of the upper base of a right circular frustum equals 12 ft. ; the slant height is 14 ft. ; the convex surface is 4928 sq. ft. What is the radius of the lower base ? (Take ir = ^^^.) Ex. 42. The radii of the bases of a right circular frustum are 16 in. and 24 in. respectively. Its convex surface is one half its total surface. What is the slant height of the frustum ? What is the altitude of the frustum ? Ex. 43. The convex surface of a right circular frustum is three fourths of the convex surface of the cone from which it has been cut. Find the ratio of the altitude of the frustum to the altitude of the cone. Ex. 44. The radius of the upper base of a frustum is 5 ft. ; the radius of the lower base is 8 ft. ; the altitude is 12 ft. What is the volume of the frustum ? Ex. 45. The radii of the bases of a right circular frustum are 20 ft. and 24 ft. respectively. Its volume is 4928 cu. ft. What is its altitude ? Ex. 46. The volume of a right circular frustum is 2200 cu. ft. ; the slant height is 13 ft. and the altitude is 12 ft. Find the radii of the bases. Ex. 47. How much sheet tin will be required to construct a water pail 18 in. in diameter at the top, 14 in. in diameter at the bottom, and having a capacity of 8 gal., allowing a waste of 10 per cent -in the material in seams and cuttings ? Ex. 48. The volume of a right circular frustum is three fourths of the volume of the cone from which it has been cut. What is the ratio of the altitude of the frustum to the altitude of the cone ? Ex. 49. The area of the lower base of a right circular frustum is four times the area of the upper base. The volume of the frustum equals the volume of a right circular cylinder whose base is the upper base of the frustum. Find the ratio of the altitude of the cylinder to the altitude of the frustum. Ex. 50. From a right circular frustum whose upper diameter is 2 a and lower diameter 4 a is bored out, as shown in the figure, a right circular cone whose base has a diameter 2 a. Find the ratio of the volume of the original frustum to the volume of the hollow frustum. 810 THE ELEMENTS OF GEOMETRY A •" G is a prismoid ; MIDS is a mid-section parallel to the bases. /l, any point whatever of MIBS\ is joined to the vertices of the prismoid. Denote the altitude by h, the upper base by 6, the lower by 6', and the mid-section by m. Ex. 51. Show that the volume of pyramid K-ABCT=h, 6 and the volume of pyramid K-EFGIi=-b'. Draw KH ± DI and OJJLhoWi CB And KH. .Draw OQ through H and KL±OQ. , (XXI. Prob. 2.) ^ • ^ OQ ± DI and KL ± BCGF. (XXI. 5, 5 a, 5 6, 3, 3 a.) Ex. 52. Show that OJ- KH=^ KL- OH, whence, 2 OJ • KH - DI=KL'20H. DI, and the volume of pyramid K-CBFG -\KL' OQ-DI 2l\DKI ^ADKI. Ex. 53. Hence show, by considering all the pyramids, that the volume of the prismoid = - (6 + &' + 4 w). 6 Ex. 54. Show that if the prismoid be the frustum of a pyramid, the above expression may be reduced to that given in XXIV. 6. Ex. 55. Show that any plane through the center of a parallelepiped divides the figure into two equal parts. Ex. 56. Divide a parallelepiped into two equal parts by a right section. Ex. 57. Divide a parallelepiped into two equal parts by a plane parallel to a given plane. Ex. 58. A parallelepiped is given and also two lines in space. Show how to pass a plane parallel to the given lines that shall divide the paral- lelepiped into two equal parts. m XXV. GROUP ON THE SPHERE DEFINITIONS A Spherical Surface is a surface generated by the revolution of a semicircumference (the generatrix) about its diameter as an axis. A Sphere is a portion of space inclosed by a spherical surface. Note. — As in the case of circle and circumference, the terms sphere and spherical surface are used interchangeably where no confusion is likely to result. The Radius of the generatrix is the radius of the sphere. A Great Circle of a sphere is a circle on the sphere whose plane passes through the center of the sphere. A Small Circle of a sphere is a circle on the sphere whose plane does not pass through the center of the sphere. The Axis of a Circle of a sphere is the perpendicular to the plane of the circle at its center. The Poles of a circle are the points in which the axis of the circle intersects the surface of the sphere. The Polar Distance of a point on a circumference on the sphere is the length of the great circle arc joining the point to the nearer pole of that circumference. A Plane is Tangent to a sphere when it has one point, and only one, in common with the surface of the sphere. A sphere is Circumscribed to a polyhedron when all the ver- tices of the polyhedron lie on the surface of the sphere. A sphere is Inscribed in a polyhedron when the faces of the polyhedron are all tangent to the sphere. A Zone is a portion of the surface of the sphere comprised between two parallel planes. 311 812 THE ELEMENTS OF GEOMETRY A Spherical Segment is a portion of the volume of the sphere comprised between two parallel planes. A Spherical Sector is the portion of the sphere generated by the revolution of a sector of the generatrix. PROPOSITIONS XXV. 1. Every plane section of a sphere is a circle. . F Hjrp. If the plane MQ cuts the sphere S in the line ABC •••, Cone. : then ABC ••• is a circle. Dem. Through S draw PP perpendicular to plane MQ, and intersecting the plane MQ in K. Suppose T to be any point whatever in ABC •••. Draw /ST and /i'r. Let ST=Ry KT=r, and SK=d. Then, in the right triangle SKT, r^VR'-d\ (XIV. la.) But R is constant for the given sphere, andd is constant for the given plane. .*. r is constant for all points in their line of intersection. "That is, the distance of T from K' is the same for all posi- tions of T. - .-. ABC ••• is a 6if6le" whose center is h": - . . . ; OLD ♦'..,-v4l XXV. THE SPHERE 813 XXV. 1 a. The join of the center of the sphere and the center of any circle on the sphere is the axis of the circle, and conversely, XXV. 1 K The locus of the centers of all spheres that pass through three given points is the axis of the circle that passes through the points, XXV. 1 c. Circles cut out by planes equidistant from the center of the sphere are equal, and conversely. Dem. T = Vi2' - d\ (XIV. 1 a.) .-. if d remains the same, r must remain the same, and if r remains the same, d must remain the same. Q.E.D. XXV. Id. Of two circles cut by planes unequally distant from the center, the one nearer the center is the greater, and conversely. Dem. r = Vi2' - d\ (XIV. 1 a.) When d = R,r=^B'-R^ = 0. As d diminishes, R^ — d^ increases. .-. r, which = -vR^ — d^, increases. .*. the nearer a circle is to the sphere center, the greater is the radius of the circle ; i.e. the greater is the circle. Q.E.D. XXV. 1 e. The polar distances of all points in the circumference of a circle of the sphere are equal. ScH. From this property the polar distance of the points of a circle are called arc-radii of the circle. The arc-radius of a great circle is a great-circle quadrant, called simply a quadrant. The arc-radius of a small circle is less than a quadrant. 814 THE ELEMENTS OF GEOMETRY XXV. 1/ Three points on the sphere surface {not on the same great circle, no two of which are the extremities of a diameter) are necessary and sufficient to determine a small circle of the sphere. XXV. 1 g. Two points on the sphere surface (not the extremities of a diameter) are necessary and sufficient to determine a great circle of the sphere, XXV. 2. A plane perpendicular to a radius at its extremity is tangent to the sphere, and conversely. Hyp. If the plane MQ is perpendicular to /Sr at T, Cone: then plane MQ is tangent to the sphere; and con- versely. Dem. Let DL be any plane perpendicular to ST^ and cutting the sphere. All points common to DL and the sphere lie in the circle ABC, of radius r = -^B?-^. (XIV. 1 a.) Move the plane DL away from S, keeping it ± to ST. As d increases, r decreases ; until, when %,e. the circle ABC which contains all points common to the plane and the sphere becomes itself a point, T. .'. this point is a unique point common to the sphere and the plane DL, i.e. DL is tangent to the sphere. (Def. of tangent plane.) XXV. THE SPHERE 315 But when d = R, the planes MQ and DL are both perpen- dicular to /ST at T. .'. plane MQ is identical with plane DL. (XXI. 3 a.) .-. plane MQ is tangent to the sphere at T, and is unique. Q.E.D. XXV. 3. Through any four poi7its, not in the same plane, a unique sphere may he passed. TLyp. If A, B, C, E, are not in the same plane, Cone. : then a unique sphere may be passed through A, B, C, and E. Dem. No three of the points can be collinear. (Def. of plane, g, XXI.) The locus of the centers of spheres through A, B, and G is KL, perpendicular to plane ABC at K, the center of the circle through A, B, and O. (XXV. 1 b.) All points equidistant from A and E lie on the plane MQ perpendicular to AE. (XXI. 3 b.) This plane must cut KL, as at S. (Why ?) .-. S is equidistant from A, B, C, and E. (Why ?) .*. a sphere with S as center and SA as radius passes through A, B, C, and E. Again, the point S is unique. (Why ?) .*. the sphere through A, B, C, and E is unique. Q.E.D. S16 THE ELEMENTS OF GEOMETRY XXV. 3 a. The perpendiculars to the faces of a tetra- hedron at their circumcenters are concurrent. Outline Dem. Draw AB, BC, etc., forming the tetrahedron ABCE. Then K is the circum center of triangle ABC, and KL the axis of the circle through A, B, and C. Similarly with each of the other faces of ABCE. .'. the perpendiculars concur at the center of the circum- sphere. (XXV. 1 a.) Q.E.D. XXV. 3 b. The six planes mid-normal to the edges of a tetrahedron have a unique point in common. XXV. ^. If R is the radius of a sphere, the area of the surface of the sphere equals 4 ttRK ■ p A' ^ ^R^ V ^w M B' 1 \ > C i D ^ -4 L ,J Hyp. If R is the radius of the sphere generated by the semicircle HBL, Cone. : then the area of the surface is 4 vR^. Dem. Let AB be a side of a regular n-gon inscribed in the circle, A'B' its projection on the axis HL, and M' the projection of its middle point M. XXV. THE SPHERE 317 Draw AG parallel to HL, and draw MS'. The surface generated by AB when HBL generates the sphere, or Surf. AB=2 ttMM' • AB. [The convex surface of the frustum, etc. (XXIV. 11 h.).] But rt. A MM'S' ^ rt. A ABG. (XV. Exs. 50, 51.) .-. AB : S'M=: AG{= A'B') : MM'. .-. AB'MM'=S'M'A'B'. .-. Surf. AB = 27rS'M • A'B'. Similarly for the surface generated by HA, BC, etc. .-. the surface generated by the semi-polygon HAB ... Z = 2 7rS'M(HA' + A'B' + B'C + •••) = 2 ttS'M- HL = 2 7rS'M'2 R =4:7rR- S'M. But if n be indefinitely increased, the semi-polygon = the generatrix, and the surface generated by the semi-polygon = the sphere surface and S'M^R. .*. the sphere surface = 4 ttR . i? = 4 irR^. Q.E.D. XXV. ^a. If h is the altitude of a zone on a sphere of radius E, the area of the zone equals 2 irRli. Outline Dem. The surface generated by any portion of the semi-polygon, e.g. ABC, is 2 ttR • A'C. At the limit, this sur- face becomes a zone of altitude A'C = li. Ex. 1. Find the surface and the vohime of a sphere whose radius is 10 ft. Ex. 2. Find the total surface of a hemisphere whose radius is 12 ft. Ex. 3. Tlie surface of a sphere is 1256.64 sq. ft. What is the radius ? Ex. 4. Show that the area of the surface of a sphere is the same as that of a circle wliose radius is the diameter of the sphere. Ex. 5. Find the area of a zone whose altitude is 10 ft. and which is situated on a sphere of 20 ft. radius. 818 THE ELEMENTS OF GEOMETRY XXV. 5. If E is the radius of a sphere^ the volume of the sphere is | nli^. Hyp. If B is the length of the radius of the sphere s, Cone. : then the volume of the sphere S is ^ irB?. Dam. Circumscribe the cube P^ about S. Draw SA, SB, etc., thus dividing P^ into pyramids. Volume of pyramid S-ABCE = ABCE - \ R. (XXIV. 4 a.) Similarly, for the other pyramids of Pj. .-. volume of P^ — (surface of Pj) • \ R. At Q, •••, where SA, etc., cut the surface of S, pass planes tangent to S and truncating (cutting off the corners) symmet- rically the cube Pi. Denote the polyhedron thus obtained by Pg. Draw SL, SO, etc., dividing Pa into pyramids. Volume of pyramid S-LMO = LMO - i R, (XXIV. 4.) Similarly, for the other pyramids of Pg. .% volume of P^, = (surface of P^ • J -B. By passing a new set of tangent planes truncating Pg, we obtain a new circumscribed polyhedron, Pg, such that volume of Pg = (surface of Pg) • \R. This process, may be continued indefinitely. But each successive polyhedron is nearer in volume to the sphere than the preceding. Again, the volume of any polyhedron thus obtained will be greater than the volume of S, since its vertices are without S, and its faces are tangent planes. XXV. THE SPHERE 319 But the excess of volume may be made as small as we please by continuing the truncation far enough. .% if P be any of the polyhedra, volume of P = volume of S. But volume of P= (surface of P) • | R. .', volume oi S = (surface oi S) • \E = 4:7rR''iE (XXV. 4.) Q.E.D. XXV. 5 a. The volume of a spherical sector equals one third the product of the zone that forms its base and the radius of the sphere. Ex. 6. The surface of a sphere is equal to the surface of a cube. Find the ratio of the diameter of the sphere to the edge of the cube. Assuming the radius of the earth to be 3967.2 miles, and the chord of an arc of 23 1° to be .407 of the radius, find : Ex. 7. The area of each Frigid Zone. Ex. 8. The area of each Temperate Zone. Ex. 9. The area of the Torrid Zone. Ex. 10. The total surface of a cylinder of revolution is equal to the total surface of a hemisphere. Find the ratio of the two volumes, if the diameter of the cylinder equals its altitude. Ex. 11. The total surface of a cone of which the altitude equals the radius of the base is equal to the surface of a sphere. Find the ratio of the altitude of the cone to the diameter of the sphere. Ex. 12. Show that if spheres be described on the sides of a right triangle as diameters the surface of the sphere on the hypotenuse will be equal to the sum of the surfaces of the other two spheres. Ex. 13. The radii of two spheres are a and b. Show how to construct the sphere whose area is equal to the sum of the areas of the given spheres. Ex. 14. Show that if spheres be described on three concurrent edges of a rectangular parallelepiped as diameters the sum of the surfaces of the spheres will equal the surface of the sphere described on the diagonal of the parallelepiped as diameter. 820 THE ELEMENTS OF GEOMETRY XXV. SUMMART OP PROPOSITIONS IN THE GROUP ON THE SPHERE 1 . Every plane section of a sphere is a circle. a. Tliejoin of the center of a sphere and the center of any circle on the sphere is the axis of this circle, and conversely. b. The locus of the centers of all spheres that pass through three given points is the axis of the circle that 2^cisses through the points. c. Circles cut ojf by planes equidistant from the center of the sphere are equal, and conversely. d. Of tioo circles cut by planes unequally distant from the center of the sphere, the one nearer the center is the greater, and conversely. e. The polar distances of all points in the circum- ference of a circle of the sphere are equal. ScH. From this property the polar distances of the points of a cirple are called arc-radii of the circle. The arc-radius of a great circle is a great-circle quadrant, called simply a quadrant. The arc-radius of a small circle is less than a quadrant. / Three points on the sphere surface {not on the same great circle, and no two of which are the extremities of a diameter) are necessary and siifficient to determiiie a small circle of the sphere. g. Tivo points on the sphere surface {not the extremi- ties of a diameter) are necessary and sufficient to determine a great circle of the sphere. XXV. THE SPHERE 321 2. A plane perpendicular to a radius at its extremity is tangent to the sphere, and conversely. 3. Through any four points not in the same plane a unique sjjhere may he passed. a. The p)erpendiculars to the faces of a tetrahedron erected at their circumcentevs are concurrent. b. TJie six planes mid-normal to the edges of a tetrahedron have a unique point in common. A. If R is the radius of a sphere, the area of the sur- face of the sphere is 4 ttR^. a. If h. is the altitude of a zone on a sphere of radius R, the area of the zone equals 2'n-Rli. 5. If R is the radius of a sphere, the volume of the sphere equals | irR^. cu The volume of a spherical sector equals one third the product of the zone that forms its base and the radius of the sphere. Ex. 15. Find the volume of a sphere that will just fit into a cubical box each edge of the inside measurement of which is 10 in. Ex. 16. Find, to within .01 ft., the edge of a cube whose volume equals the volume of a sphere of radius 12 ft. Ex. 17. The volume of a sphere is 7241} cu. ft. What is the radius ? Ex. 18. The number that expresses the volume of a certain sphere is the same as the number that expresses the surface of the sphere. Find the radius of the sphere. Ex. 19. A cube and a sphere have the same surface. Which tas the greater volume ? Prove. 822 THE ELEMENTS OF GEOMETRY PROBLEMS XXV. Prob. 1. To find the diameter of a material sphere. G' Given. The material sphere AB ••• i?. Required. The diameter of this sphere. Const. With any point, as Aj as a pole, and any convenient opening of the dividers, describe a circle, B--- C. In this circle select any 3 points, F, E, 6r, and with the dividers set off E'F' = EF, F'G' = FG, and G'E' = GE so as to form the triangle E'F'G' congruent to triangle EFG. Find the circum-radius of this triangle, K'E', which will also be the circum-radius of EFG, i.e. the radius of circle B >" C. Construct rt. A^i^J^f with hypotenuse HM = chord AE and \egLM=K'E'. Erect a perpendicular to HM 3it M, and extend this perpen- dicular to meet HL, say at Q. HQ is the diameter required. Proof. Suppose the diameter AR drawn ; also ACy CR. rt. A HLM ^ rt. A ARE. (Const.) .-. In vtAACR, HMQ, ^A = ZH. (Hom.Zs^A.) Z AdR = Z. HMQ, (Both rt. A) XXV. THE SPHERE 323 AC=HM. ' (Const.) .-. rt. A ACli ^ rt. A HMQ. (V. 2.) .-. AB = HQ. (Horn. s's. -^ A.) Q.E.D. Ex. 20. An iron cannon ball 12 in. in diameter weighs 225 lbs. What is the diameter of a ball of the same material weighing 1800 lbs. ? Ex. 21. The volume of a sphere equals the volume of a cube. Find the ratio of the diameter of the sphere to the edge of the cube. Find also the ratio of the diameter of the sphere to the diagonal of the cube. Ex. 22. The volume of a sphere is equal to the volume of a cone whose slant height is double the radius of its base. Find the ratio of the total surfaces of the two figures. Ex. 23. The sura of the surfaces of three spheres is equal to a circle of which the radius is twice the diagonal of an oblong block whose edges are a, 6, and c. The volumes of the three spheres are in the ratio of a^ :b^ : c^ Find the radii of the spheres. Ex. 24. A sphere just fits into a regular triangular prism each base edge of which is a. Find the volume of the sphere. Ex. 25. The surface of a cube equals the surface of a sphere. Find the ratio of the volume of the cube to the volume of the sphere. Ex. 26. A sphere and a regular tetrahedron have the same surface. Find the ratio of their volumes. Find the ratio : Ex. 27. Of the surface of a sphere to the surface of the circumscribed cube. Ex. 28. Of the surface of a sphere to the surface of the inscribed cube. Ex. 29. Of the volume of a sphere to the volume of the circumscribed cube. Ex. 30. Of the volume of a sphere to the volume of the inscribed cube. Def. A Principal Section of a Surface of Revolution is a section that passes through the axis. The principal sections of a certain right circular cylinder are squares each side of which is 2 a. Find the ratio ; ♦ Ex. 31. Of the convex surface of the cylinder to the surface of the inscribed sphere. 824 THE ELEMENTS OF GEOMETRY Ex. 32. Of the total surface of the cylinder to the surface of the inscribed sphere. Ex. 33. Of the volume of the cylinder to the volume of the inscribed sphere. Note. — The three preceding problems were first solved by Archimedes. Ex. 34. Of the convex surface of the cylinder to the surface of the circumscribed sphere. Ex. 35. Of the total surface of the cylinder fb the surface of the cir- cumscribed sphere. Ex. 36. Of the volume of the cylinder to the volume of the circum- scribed sphere. Ex. 37. Of the volume of the circumscribed sphere to the volume of the inscribed sphere. The vertex angle of a right circular cone is 60° ; its slant height is 2 a. Find the ratio : Ex. 38. Of the convex surface of the cone to the surface of a sphere of radius a. Ex. 39. Of the total surface of the cone to the surface of the sphere of radius 2 a. Ex. 40. Of the total surface of the cone to the surface of a sphere whose radius is the altitude of the cone. Ex. 41. Of the volume of the cone to the volume of the sphere of radius a. Ex. 42. Of the volume of the cone to the volume of the sphere whose radius is the altitude of the cone. Ex. 43. Of the volume of the cone to the volume of the sphere whose surface equals the convex surface of the cone. Ex. 44. Of the volume of the cone to the volume of the sphere whose surface equals the total surface of the cone. The surface of a sphere equals the total surface of a right circular cone whose principal sections are equilateral triangles ; and also equals the total surface of a right circular cylinder whose principal sections are squares. Find the ratio : Ex. 45. Of the volum^of the sphere to the volume of the cylinder. Ex. 46. Of the volume of the sphere to the volume of the cone. Ex. 47. Of the volume of the cylinder to the volume of the cone. XXV. THE SPHERE 325 Ex. 48. Of the convex surface of the cylinder to the convex surface ot the cone. Ex. 49. An iron sphere whose radius is 8 in. is melted and recast in the form of a hollow right circular cylinder whose altitude and interior diameter are each in. Eind the thickness of the cylinder. Ex. 50. Find the weight of a hollow spherical cast-iron shell whose exterior diameter is 20 in. and whose interior diameter is 17 in. Ex. 51, A hollow spherical shell has a capacity of 2 gal. ; its exterior diameter is 12 in. Find the thickness of the shell. Ex. 52. Find the locus of the center of a sphere 10 in. in diameter, the surface of which passes throutih a given point A. What is the locus of a point at a constant distance (d) from a given point ? What is the locus : Ex. 53. Of the center of a sphere passing through two given points, A and B ? Ex. 54. Of the center of a sphere of radius r whose surface passes through two given points, A and B ? Ex. 55. Of the center of a sphere whose surface passes through three given points, A, B, and C ? Ex. 56. Of the centers of spheres whose surfaces all contain a given circle ? What is the locus of the center of a sphere : Ex. 57. Tangent to three given planes ? Ex. 58. Tangent to three intersecting lines ? Ex. 59. Tangent to a given plane at a given point ? Ex. 60. Tangent to a given cylindrical surface at a given point ? Ex. 61. Tangent to a given sphere at a given point ? Ex. 62. Tangent to a cone of revolution at a given point ? Ex. 63. Tangent to two concentric spheres ? Find the locus of the center of a sphere of given radius r that satisfies the conditions that follow : Ex. 64. Tangent to a given plane. Ex. 65. Tangent to two intersecting planes. Ex. 66. Tangent to a cylindrical surface of revolution. Ex. 67. Tangent to a conical surface of revolution. 826 THE ELEMENTS OF GEOMETRY Ex. 68. Tangent to a sphere of radius m. Ex. 69. Tangent to two cylindrical surfaces of revolution, of diameters a and b. Ex. 70. Tangent to two spheres of equal radius a. Ex. 71. To find a point equidistant from three given points and also equidistant from two other points not in the plane of the first three. To construct a sphere of radius r : Ex. 72. That shall pass through a given point and be tangent to two given planes. Ex. 73. That shall pass through two given points and shall also be tangent to a given sphere. Ex. 74. That shall pass through a given point and shall also be tangent to a cylinder of revolution along a given element. Ex. 75. That shall pass through two given points and shall also be tangent to a cylinder of revolution along a given element. Ex. 76. That shall pass through a given point and be tangent to a given sphere and a given plane. Ex. 77. Tangent to three given planes. Ex. 78. Tangent to three given concurrent lines. Ex. 79. That shall pa.ss through three given points. Ex. 80. Tangent to two intersecting given lines and passing through a given point. Ex. 81. Tangent to two non-intersecting lines and passing through a given point. Ex. 82. Tangent to two given spheres of radii & and c, respectively. and passing through a given point. Show that, given a point and a sphere, Ex. 83. In general, an infinite number of tangent lines can be drawn through the given point tangent to the given sphere. Ex. 84. These tangents either lie in the same plane or form th# elements of a cone. Ex. 85. The cone of tangents is a cone of revolution. Ex. 86. In general, an infinite number of planes may be passed through the point tangent to the sphere. Ex. 87. These planes will each be tangent to the cone of tangent lines through the point. Ex. 88. When will it be impossible to pass more than one tangent plane through the point ? XXVo THE SPHERE 327 Ex. 89. When will it be impossible to pass either a tangent line or a tangent plane through the point ? To pass a plane tangent to a given sphere, Ex. 90. At a point on the sphere. Ex. 91. Through a point without the sphere. Ex. 92. If AB and AEare tangent to the sphere S, show that ABS is a right triangle. Show further that AB^ = A0- AS. Hence, show how to find FC when AF and the radius SB are given. Ex. 93. An electric light is placed at ^, 3 ft. from the nearest point of a sphere 20 ft. in diam- eter. Find the area of that portion of the sphere which is illuminated by the light at A. Ex. 94. How far is the light A from the surface when the illuminated portion of the sphere is ^ of the whole surface ? When the illuminated portion is ^ of the whole ? Ex. 95. Why is the illuminated portion of the surface always less than | the whole surface ? Ex. 96. A-BGE is a zone of altitude AC. If AB, BF be drawn, what is the value of Z ^ ? In the A ABF^ what relation connects AB, AC, AF? (XVII. 5 (a).) Prove that the area of the zone A-BGE = the area of a circle of which AB is the radius. Ex. 97. The altitude CA of the zone A-BFE equals H) the radius of the sphere equals B. What is the area of the zone ? What is the vol- ume of the spherical sector S-ABFF? Show that, from the rt. A BCS, BC^ = 2E-H-H\ Find the expression for volume of cone S-BFE. Ex. 98. Show that the volume of the spherical segment A~BFE = 2 ttB'-H t(B H)(2B' H-m) 3 ~1LKs2B^ 3 ^ (iB-II)(2B ^)} = ^'(3i?-^). XXVI. GROUP ON GEOMETRY OF THE SPHERE SURFACE; BRIEFLY, SPHERICAL GEOMETRY DEFINITIONS The Distance between two points on the surface is the shorter arc of a great circle that passes through the points. The Arc-Radius of a small circle on the sphere is the shorter polar distance of any one of its points. A Spherical Angle is the figure formed by two great-circle arcs that intersect in one point. The Angle between any two circles that have a common point is the plane angle formed by tangents to the circles at this point. A spherical angb, therefore, is the same as the angle between the tangents to its sides drawn at the point of inter- section. Thus, the spherical angle ABC is the same as the plane angle FBH formed by the tangents FB and HB, to the sides AB and BC, respectively, at B. ' Corollaries to the Definitiox (a) Any spherical angle is the measure of the dihedral formed by the -planes of its sides. (b) The measure of a spherical angle (ABC), plane angle FBH, is the are AC intercepted hy its sides on the great circle of whi 880 THE ELEMENTS OF GEOMETRY Two spherical polygons are Symmetric, when the parts of the one are respec- tively equal to the parts of the other, but are arranged in opposite order. If two symmetric polyhedrals have their vertex at the center of the sphere, they cut from the surface two symmetric spherical polygons. Thus Z AOB = A A' OB'. (Vertical angles.) .*. arc AB = arc A'B'. (Measures of equal angles.) Again, dihedral A-BO-C = dihedral A'-B'0-C\ (YeTtical dihedrals.) .-. AB— /LB\ (Measures of = dihedrals.) Similarly, for the other parts of the two figures. The opposite order of parts in the two figures is shown by the arrow heads. Symmetric polygons so situated that the joins of correspond- ing vertices concur at the center of the sphere are said to be in Perspective with regard to this center. Two symmetric polygons cannot, in general, be placed in coincident superposition. For, on account of the curvature of the sphere surface, it will be impossible to make the parts of the one coincide with the corresponding parts of the other without breaking or tearing the surface. For the same reason, it is impossible to revolve one of two symmetric polygons having homologous sides in common about this common side, so as to make the polygons coincide. That is, revolution, as a step in a proof, cannot occur in the geometry of the sphere-surface ; an arc cannot be taken as an axis. It will be shown, however (XXVI. 3 a), that if polygons are symmetric, they are equal in area. A spherical polygon is the Polar of a second, when the ver- tices of the first are the poles of the sides of the second. A spherical polygon is Supplemental to a second, when the XXVI. SPHERICAL GEOMETRY 331 angles of the first are the supplements of the corresponding sides of the second. The Spherical Excess of a polygon of % sides is the excess of the sum of its angles over (2 n — 4) right angles. A Lune is a portion of a sphere-surface bounded by two great semicircles. An Ungula or Spherical Wedge is the portion of the volume of the sphere comprised between two great Semicircles. Corollaries to the Definitions. (a) A lune is the same fraction of the total surface of a sphere that the angle of the lune is of four right angles. (b) An ungula is the same fraction of the volujne of the sphere that the angle of the ungula is of four right angles. Preliminary Scholium Correspondence between Plane and Spherical Geometry The Line. Since any two points (not the extremities of a diameter) determine a great circle of the sphere (XXV. 1 g), the great-circle of the sphere corresponds to the straight line of the plane. A great-circle arc may therefore be called simply a line of the sphere-surface. But two great-circle arcs perpendicular to a third meet in the poles of this third, and from either of these poles an infinite number of perpendiculars can be drawn to the great-circle arc. Parallels. Hence, there are no lines on the sphere corre- sponding to the parallels of the plane. Axioms. It follows that : (1) There is no parallel axiom on the sphere surface. (2) All the axioms of the plane, except Axioms 7 and 8, are true on the surface of the sphere. 882 THE ELEMENTS OF GEOMETRY The Circle. Three non-coUinear points in a plane determine a circle. Three non-collinear points on a sphere-surface determine a small circle of this surface. The small circle of the sphere- surface therefore corresponds to the circle of the plane, the word Pule (meaning the nearer pole) replacing the word Center of the plane geometry. Propositions. The propositions of spherical geometry are therefore derived from those of plane geometry by replacing the " straight line " of the plane by the " great-circle arc," or simply the "line" of the sphere-surface, and dropping all propositions that cannot he proved without using Axiom 7 or Axiom 8. The proofs for the sphere-surface are identical with those already given for the plane where Axioms 7 and 8 are not involved. Summary of Propositions Common to Plane and Spherical Geometry The following summary presents the geometry of the sphere as it corresponds to the geometry of the plane. Plane Sphekb Group I. All propositions true on the sphere. Group II. No propositions true on the sphere. Group III. No propositions true on the sphere. Group IV. All propositions true on the sphere. Group V. All propositions true on the sphere, if "congruent" be replaced by " congruent or symmetric." Group VI. No propositions true on the sphere. Group VII. Only Theorems 1 and 2 true on the sphere. Groups VIII., IX., X., XI. All propositions true on the sphere. Group XII. Only Theorem 1 true on the sphere, but not used as a basis of meas- urement. XXVI. SPHERICAL GEOMETRY Areal Measurement in the Plane. The plane is boundless (i.e. without boundaries of any sort), and is infinite in extent. The ratio of the surface of a plane figure to the total surface of the plane on which it lies cannot be expressed in numbers. Hence, a purely arbitrary unit, the square on the linear unit, is selected, and the areal unit changes when the linear unit changes. All the propositions of plane geometry that deal with area involve the use of the square, Neither the square nor any analogous figure can be drawn on the sphere-surface. Areal Measurement on the Sphere. The sphere-surface, though boundless, is not infinite (XXV. 4). The ratio of the surface of a spherical polygon to the whole sphere-surface on which it lies can be expressed in numbers. Hence, the sphere-surface itself (or some definite fraction of it) is the natural unit of area for the figures that lie upon it, a unit that never changes for a given surface. Accordingly, the treatment of surface measurement on the sphere is essentially different from the treatment of this subject in the plane, and no propositions of plane geometry that deal with areas or areal relations can have any correspondents on the sphere-surface. Again, the Doctrine of Similarity, with all its varied applica^ tions, is based on Axioms 7 and 8. There is, therefore, no Theory of Similarity in spherical geometry. For these reasons, no use can be made of the plane geometry beyond Group XII. Detailed proofs will be required only for those propositions which are peculiar to the sphere-surface, or which have been proved in plane geometry by use of Axioms 7 and 8, when a proof not involving these axioms might have been used. 884 THE ELEMENTS OF GEOMETRY PROPOSITIONS XXVI. 1. The shortest line that can he drawn on the surface of the sphere, between any two points c^i the surface, is the great-circle arc {not greater than a semi- circumference) that joins them. Hyp. If AB is a great-circle arc less than a semicircle, and ALFB any other line whatever from Ato B, Cone: then AB < ALFB. Dem. With ^ as a pole and arc-radius less than ABy de- scribe O My cutting AB in G and ALFB in L. With 5 as a pole and an arc-radius BC, describe O Q, cutting ALFB in F, This circle will be tangent to O iWat G. (IX. 8 a, converse (a).) Whatever the form of the line AL may be, a line equal to AL may be drawn from A to G. For, if the zone whose base is O Jf be revolved on the axis of O Jf, Ly moving along the circumference, will reach the posi- tion C, and the line AL will itself extend from A to G. Similarly, a line equal to BF can be drawn from B to G. But L is without O Q (O Q is externally tangent to O Jf at (7). .-. BF+ FL > BG. (Preliminary Th. 1.) .-. BF + FL + LA> BG + GA] (Preliminary Th. 1.) t.e. BFLA > BA. Q.E.D. XXVI. SPHERICAL GEOMETRY 336 ScH. If A and B be extremities of a diameter, an infinite number of equal great-circle arcs may be drawn from A to B, but each will be less than any line not a great-circle arc that can be drawn from A to B. Note. — In speaking of two points on the sphere, we shall hereafter assume, unless the contrary be stated or evidently implied, that these points are not the extremities of a diameter of the sphere. The proof of the theorem emphasizes the analogy between the great circle on the sphere and the straight line of the plane. Each is deter- mined by two points of the surface ; each measures the shortest distance on the surface between any two of its points. Ex. 1. Show that a spherical triangle may have 3 obtuse angles. Ex. 2. Two spherical polygons of the same number of sides are equal, if the sum of the angles of the one equals the sum of the angles of the other. Ex. 3. Two spherical right triangles are congruent or symmetric, if the oblique angles of the one equal respectively the oblique angles of the other. Ex. 4. Two birectangular triangles are congruent or symmetric, if the oblique angle of the one equals the oblique angle of the other. Ex. 5. If the diagonals of a spherical polygon bisect each other, the opposite sides are equal. Ex. 6. If two spheres intersect, the tangent lines drawn to the spheres from any point in the plane of the circle of intersection are equal. Def. If two spheres be generated by the revolution of two circles about their line of centers as an axis, the radical axis of the circles (XVII. Ex. 72, Def. ) will generate a plane perpendicular to the line of centers of the spheres. (XXI. 3.) This plane is called the Radical Plane of the Spheres. Ex. 7. The radical plane of two spheres is the locus of the point from which pairs of equal tangent lines may be drawn to the spheres. Ex. 8. The radical planes of three spheres taken two and two inter- sect in a line perpendicular to the plane of the centers of the spheres. The radius of a sphere is 20 ft. Find the area : Ex. 9. Of a triangle on the sphere whose angles are 82°, 104°, and 84°, respectively. Ex. 10. Of a birectangular triangle whose vertex angle is 72°. Ex. 11. Of a trirectangular triangle. 886 THE ELEMENTS OF GEOMETRY XXVI. 2. Isosceles symmetric triangles are congruent. Hyp. If the A ABC, A'B'C are symmetric, and if AB =:AC,A'B' = A'a, Cone: then A ABC ^ A A'B'C. Dem. AB = AC=A'B'=A'C'. (Hyp.) Place A ABC on A A'B'C, so that A shall fall at A' and B shall fall at C. The two triangles fall on the same side of A'C, since the parts occur in reverse order. ZA = Z A'. (Def. Sph. n-gon. d.) .'. AC takes the direction oi A'B'. AC=A'B'. (Hyp.) /. C falls at B'. .'. BC coincides with B'C [Two points, etc., determine a great circle.] (XXV. 1 g.) .-. A^^C^A^'^'C". Q.E.D. XXVI. SPHERICAL GEOMETRY 837 XXVI. 3. Symmetric triangles are equal. Hyp. If A ABC is symmetric to A A'B'C, Cone: then AABC = AA'B'C'. Dem. Place A ABC in perspective with AA'B'C, with respect to S. Let Q be the pole of the circle through A, B, and C. [Three points, etc., determine a small circle.] (XXV. 1 /.) Draw QSQ' and the joins QA', Q'B', and Q'C. Trihedral S-ABQ is symmetric with trihedral S-A'B'Q'. (Def. of symmetric trihedrals.) .-. AABQis symmetric with AA'B'Q'. (Def. of symmetric w-gons, etc.) .-. QA= Q'A' and QB= Q'B'. But QA =QB= Q'A' = Q'B' ; (XXV. 1 e.) i.e. A AQB and A' Q'B' are isosceles. .: AAQB^A A'Q'B\ (XXVI. 2.) Similarly, A BQC ^ A B'Q'O, and AAQG^AA'Q'Q'. .\AAQB+ABQG-{-AAQC=AA'Q'B'-^AB'Q'C'-hAA'Q'a; K& A ABC = AA'B'C. Q.E.D. 838 THE ELEMENTS OF GEOMETRY XXVI. 3 a. Symmetric polygons are equal Outline Dem. Divide the polygons into triangles by diagonals from homologous vertices. These triangles will be symmetric in pairs and therefore equal in pairs, therefore the polygons are equal. Q.E.D. XXVI. 4. If one triangle is polar to another, the second is polar to the first. Hyp. If A, B, C, are Llie poles of B'C, A'C, A'B\ respectively, Cone. : then A\ B\ C", are the poles of BC, AC, AB, respec- tively. Dem. A is the pole of B'C, (Hyp.) .-. ^(7' = 90°. [The arc-radius of a great O is a quadrant (XXV. 1 e.) Sch.] Similarl}^, BC = 90°. .*. A and B lie in a great circle of which C is the pole. But only one great O can pass through A and B, (XXV. Ig.) and the side AB is a great-circle arc. .-. C is the pole of the side AB. XXVI. SPHERICAL GEOMETRY 339 Similarly, B' is the pole of the side AC and A is the pole of the side BG. ^ „ ^ Q.E.D. ScH. If the sides of AA'B'C be ex- tended to meet in A", B", C", three new triangles are formed each of which is B\ polar to A, B, O. A triangle therefore has four polars. But in speaking of the polar of a triangle, we always mean the central figure, for which each of the distances Aa\ Bb', Co', is less than a quadrant. XXVI. 4 a. Jf one polygon is polar to another, the second is polar to the first. XXVI. 4:b. A trirectangular triangle is its own polar (i.e. is self-polar). The radius of a sphere is 20 feet. Find the area : Ex. 12. Of a quadrilateral whose angles are 112°, 85°, 92°, and 126°, respectively. Ex. 13. Of a pentagon, whose angles expressed in radians, are 2.25, 2.83, 2.7, 3.72, and 1.7, respectively. Ex. 14. Of a hexagon each of whose angles is 130°. The radius of a sphere is 20 ft. A plane passes through the sphere at a distance 12 ft. from the center. Find : Ex. 15. The total surface of the minor spherical segment cut off by the plane. Ex. 16. The angles of an equiangular triangle on the sphere whose area is equal to the zone surface of the segment. Ex. 17. The angles of an equiangular triangle equal in area to the total surface of the segment. Ex. 18. The third angle of a triangle, two of whose angles are 100° and 75°, and which equals in area one face of the inscribed cube. 840 THE ELEMENTS OF GEOMETRY XXVI. 6. Polar triangles are supplemental, 0' Hyp. If A ABC is the polar of Cone: thenZ^=180"'-a'; ZB=180°-6'; Za=180°-c', and Z^' = 180°-a; Z5'=180°-6; ZC' = 180°-c. Dem. Extend AB, AC to meet B'O as at E and F. Z.A\3 measured by FE. (Def. of measure of spherical Z.) C is the pole of ABE. (Def. of polar n-gons.) .-. C'E = 90°', i.e. C'F+FE = 90°. Similarly, • FB' = 90°. .-. C'F-i-FB' -{-FE = 1S0°', i.e, C'B' + FE = 180°. .-. FE = 180° - C'B', or Z^ = 180°-a'. Similarly, for Z B and Z (7. In like manner, we may show that Z.A' = 180° — a (by extending BC as shown), etc. Q.E.D. XXVI. 5 a. Polar polygons are supplemental. XXVI. 5 h. Two triangles are congruent or symmetinc, if the angles of the one are equal respectively to the angles of the other. Dem. Let P and P be the polars of the given triangles. The sides of P equal the sides of P'. (XXVI. 5.) .-. P is congruent (or symmetric) to P. (V. 3.) XXVI. SPHERICAL GEOMETRY 341 .*. the angles of P equal the angles of P. • (Def. of symmetric n-gons.) .*. the sides of the given triangles are equal, respectively. (XXVI. 5.) .*. the given triangles are congruent or symmetric. (V. 3.) Q.E.D. XXVI. 6. The sum of the interior angles of a tri- angle lies hetiveen two right angles and six right angles. Hyp. If ABC is a spherical triangle, Cone: thenZ^ + ZS + ^C>2 rt.z^and<6 rt. A Dem. Let A ^'^'C be the polar of A ABC. Then Z J = 2 rt A - a'. (XXVI. 5.) Z 7i = 2 rt. A - b'. Z(7=2rt.Zs-c'. .-. Zyl + Z5 + Z<7=6rt.Zs-(a'-f 6'4-c'). But a' + ft' 4- c' > and < 4 rt. A. (Cor. to the def. XXVI. c.) .-. Z ^ + Z5 -}- Z (7> 6 rt. Z - 4 rt. Zs, or 2 rt. A, and < 6 rt. Z — 0, or 6 rt. Z. Q.E.D. XXVI. 6 a. The sum of the interior angles of a jjohj- gon lies hetiveen (2 n — 4) right angles and 2 n right angles (n being the numher of sides of the polygon). 842 THE ELEMENTS OF GEOMETRY XXVI. 7. If two great semicircles intersect on the surface of a hemisphere, the sum of either set of opposite triangles equals the lune whose angle is the corresponding angle of the semicircles. Hjrp. If the lines ACA', BCB' meet the base of the hemi- sphere S in Af A\ By B', Cone. : then A ABO + A A'B'C = lune OA'C'B'. Dem. CA' + A'C' = CA'C = 180°. (Def . of great circle.) CA' + AC = ACA' = 180°. (Same reason.) .-. CA' + AC= CA' + A'C. (Ax. 1.) .-. AC=A'a. (Ax. 2.) Similarly, BC=B'C', and AB = A'B'. .'. A ABC is symmetric and equal to A A'B'C. (XXVI. 3.) .-. A ABC + A A'B'C = A A'B'C + A A'B'C (Ax. 2.) = lune CA'CB' (or lune C). Q.ED. XXVI. SPHERICAL GEOMETRY 348 XXVI. 7 a. A trirectangular triangle is one eighth of the surface of the sphere on which it lies. ScH. Unit of Area. Angular Unit. The trirectangular triangle, or one eighth of the sphere-surface, will be taken as the unit of areal measure on the sphere, and the right angle as the unit of angle measure. Angles must be expressed in terms of the right angle in all formulae for area. Thus, a lune whose angle is a right angle is twice the tri- lectangular triangle; i.e. its area is 2; a lune whose angle is 50°, or I rt. Z, is f of 2, or -LQ ; a lune whose angle is A has an area of 2 A, etc. The area of the sphere is 8; that of the hemisphere is 4. XXVI. 8. The area of a spherical triangle equals its spherical excess. ^ Hyp, If the angles of the A ABC be expressed in rt. A, and the trirectangular A be taken as the unit of area, Cone. : then the area of A ABC =^ + ^ + (7—2. Dem. On the hemisphere ABB'A\ A ABC -{-AAB'C = lune B = 2B, (XXVI. 7. Sch.) A ABC + A A'BC = lune A = 2A, (Same Eeason.) A ABC -\-AA'B'0 = lune (7=2 0. (Same Eeason.) But ^ A ABC+A AB'G+A A'BC+A A'B'C^the hemisphere = 4. (XXVI. 7. Sch.) .-. (by addition) 2 A ABC + 4 = 2 (^ + J5 + C), or AABC + 2 = A-\-B+C. .: AABC = A-\-B+C-2. Q.E.D. 344 THE ELEMENTS OF GEOMKTRY XXVI. 8 a. The area of a spherical polygon equals its spherical excess. Let the student supply the proof by dividing the polygon into triangles and using the theorem. Ex. 19. The angles of an equiangular triangle equal in area to the con- vex surface of the cone of tangents extending to the sphere from a point whose distance from the center is 35 ft. Note. — The few exercises appended to Group XXVI are intended merely as illustrations of the method of utilizing the theorems and exer^ cises of the plane geometry as exercises in spherical geometry. As stated in the General Scholium, every proposition of the plane geometry that does not involve Ax. 7 or Ax. 8 as a necessary element in its proof, is true on the sphere-surface as in the plane. In the following problems, the word line (unqualified) means arc of a great circle; the word circle and the symbol O (unqualified) mean small circle. All data are supposed to be given on the sphere-surface. What is the locus of a point : Ex. 20. At a given distance from a given point ? Ex. 21. At a given distance from a given line ? Ex. 22. At a given distance from a given circle of radius r ? Ex. 23. Equidistant from two given points ? Ex. 24. Equidistant from two given lines ? What is the locus of the centers of circles : Ex. 25. Of given radius, passing through a given point ? Ex. 26. Of given radius tangent to a given line ? Ex. 27. Of given radius tangent to a given circle ? Ex. 28. Passing through two given points ? Ex. 29. Tangent to two given lines ? Ex. 30. Tangent to two concentric circles ? Ex. 31. Tangent to two equal circles? Ex. 32. Bisect a given line-segment. Ex. 33. Circumscribe a O about a A. Ex. 34. Bisect a given Z. Ex. 35. Inscribe a O in a given A. Ex. 36. Escribe a O to a given ^ XXVI. SPHERICAL GEOMETRY 345 XXVI. SUMMARY OF PROPOSITIONS IN THE GROUP ON SPHERICAL GEOMETRY 1. The shortest line that can he draivn on the sj^here, between any two points on the surface, is the great-circle arc (not greater than a semicircumference) that joins them. ScH. There is no " shortest line " on the sphere-surface be- tween the extremities of a diameter of the sphere. 2. Isosceles symmetric triangles are congruent. 3. Symmetric triangles are equal. a. Symmetric polygons are equal. 4. If one triangle is polar to another, the second is polar to the first. ScH. A triangle has four polars. a. If one polygon is polar to another, the second is polar to the first, h. A trirectangular triangle is its own polar (i.e. is self-polar), 5. Polar triangles are supplemental. a. Polar polygons are supplemeiital. b. Tivo triangles are congruent or symmetric, if the angles of the one are equal respectively to the angles of the other. 6. The sum of the interior angles of a triangle lies hetioeen tivo right angles and six right angles. 346 THE ELEMENTS OF GEOMETRY a. The sum of the intenor angles of a polygon lies heiiveen (2ii-4) right angles and 2n right angles y n being the number of sides of the polygon. 7. If two great semicircles intersect on the surface of a hemisphere, the sum of the opposite triangles formed equals the lune whose angle is the angle of the two semicircles. a. A trirectangular triangle is one eighth of the surface of the sphere on which it lies. ScH. Unit of Area and Angular Unit. 8. The area of a spherical triangle equals its spheri- cal excess. a. The area of any spherical polygon equals its spherical excess. NOTES AND BIOGRAPHICAL SKETCHES Alexandrian School (The First). From earliest times to about 100 n.c. To tliis school belonged Euclid. The Second Alexandrian School began with the Christian Era. Menelaus, Theon, and Hypatia belonged to this school. ApoUonius, Greek. Lived in third century b.c. He ranked with Euclid and Archimedes. Known as the "Great Geometer." Archimedes, Greek. Third century b.c. Born in Syracuse. Greatest mathematician of antiquity. Ceulen, Ludolp van, Netherlands. Carried the value of ir to 36 places, known as Ludolp's Numbers. Ceva, G. (1048-1737), Italian. Discovered the theorem that bears his name. Chasles, M. (1793-1880), French. He did much to develop modern projective geomet^-y. Cissoid. A curve by means of which may be found two mean propor- tionals between two given straight lines. Cube, Duplication of. One of the three ancient, unsolved problems of geometry, the other being the quadrature of a circle and the trisection of an angle, to which the ancients devoted much time. It is now generally admitted that they cannot be solved by the geometry of the straight line and circle. Descartes, RenI (1596-1650), French. He introduced into geometry the use of the algebraic equation for the purpose of analysis. Euclid (about 300 B.C.), Greek. Lived in Alexandria. Probably went there from Athens. Not much is reliably known of him. His "Elements," containing 13 books, has been the basis of all elementary geometrical instruction for over 2000 years. His geometry does not touch upon the subject of mensuration. Eudoxus (408 b.c). Pupil of Plato. Astronomer and legislator, as well as mathematician. Euler, L. (1707-1783), Swiss. He wrote a great number of works on mathematics. He introduced much of current notation into trigononaetry ; also use of tt. Golden Section. It cuts a line in extreme and mean ratio. It was much studied by Eudoxus. 347 348 THE ELEMENTS OF GEOMETRY Harmonics. Its fundamental theorem was discovered by Serenus of tlie Second Alexandrian School. Hippocrates of Chios (4:50 n.c), Greek. Developed the subject of similar figures ; also discovered important relations between areas of circles. He was the first author of an elementary text-book on geometry. Legendre, A. M. (1762-18;i3), French. Wrote an " Elements of Geome- try " in 1794, which was largely adopted in continental Europe and the United States as a substitute for the more difficult Euclid. Menelaus (98 a.d.), Greek. Of the Second Alexandrian School. The theorem of Menelaus is the foundation of the modern theory of trans- versals. He contributed largely to our knowledge of trigonometry. IT. For its history and values see Cajori's " History of Mathematics." Pascal (1632-1662), French. A great mathematician and metaphy- sician. Plato (429-348 b.c), Greek. A pupil of Socrates and at head of a school in Athens. Many of Euclid's definitions are ascribed to him. He was the first to use the method of analysis to discover proof of theorem. He added much to geometrical knowledge. Poncelet, J. V. (1788-1867), French. He investigated and developed modern projective geoinetry. Ptolemaeus, Claudius (about 139 a.d.), Egyptian. A celebrated a.s- tronomer. The fundamental idea of his system as opposed to that of Copernicus was that the earth is the center of the universe, and that the sun and planets revolve about it. Pythagoras (580 b.c), Greek, but lived many years in Egypt. In lower Italy he founded a school for the teaching of mathematics, philoso- phy and the natural sciences, chiefly the first. Steiner (1796-1863), Swiss-German. " The greatest geometrician since Euclid." He laid the foundation for modern synthetic geometry. Thales (640-546 b.c), Greek. Lived long in Egypt. He measured the height of the Pyramids from their shadows. He was one of the Seven Wise Men of Greece. INDEX Algebra and geometry, connected, 120. Alternation, proportion by, 105. Altitude of cone, 306. cylinder, 282. frustum of pyramid, 288. prism, 267. prismoid, 288. pyramid, 287. rhomboid, 53. trapezoid, 53. triangle, 30. Analysis of problems, 12, 120. Angle, 4. acute, 5. between any two circles, 328. between two curves, 111. central, 111. dihedral, 253. escribed. 111. exterior, of triangle, 30. initial line of, 4. inscribed, 111. negative, 5. oblique, 5. obtuse, 5. polyhedral, 254. positive, 5, rectilineal, 254. right, 4. same kind of, 6. spherical, 328. straight, 5. terminal line of, 4. tetrahedral, 255. trihedral, 255. vertex of, 4. Angles, adjacent, definition of, 6. adjacent and vertical, Group on, 19-22. alternate exterior, 6. alternate interior, 6. classes of, as to their algebraic sign, 4. as to their size, 5. as to their location, 6. complemental, 6. corresponding, 6. exterior, 6. interior, 6. measurement of, 100-119. right, Group on, 32-34. supplemental, 6. vertical, 6, 19. Antecedent, 101. Apothem, 210. Arc, 78. Arc-radius, 313. Area, definition of, 138. of the circle, Group on, 220-224. unit of, 138. Areal measurement in the plane, 333. on the sphere, 333. ratios. Group on, 176-179. Areas of rectangles and other poly- gons. Group on, 138-148. Axioms, 9, 331. Axis of circle of a sphere, 311. revolution, 283. the surface, 283, radical, 208. Babylonians, 113, Note.; J49 350 INDKX Base of cylinder, 282. prism, 266. projection, 149. pyramid, 287. triangle, 30. Biographical sketches, 847, 348. Bisector, definition of, 1. Center, circum-, of triangle, 94. ex-, of triangle, 03. in-, of triangle, 91. of circle, 8. of gravity, 96. of parallelepiped, 286. of regular polygon, 210. of similitude, 160. ortho-, of triangle, 95. radical, of three circles, 209. Centroid of triangle, 96. Chart problems, 77. Chord, definition of, 78. Circle, area of. Group on, 220- 227. axis of , 311. center of, 8. circumference of, 8. great, 311, 331. small, 311. Circle and its related lines, Group on, 79-90. Circles, circumscribed, 138. concentric, 8. escribed, 93. inscribed, 93. Circumference, 8. "Circumscribed figures, 138, 210-216, 311. Coincident superposition, 2, 47. Commensurable ratios, 101 , 102. Complemental angles, 6. * Complex figure, definition of, 2. Composition, proportion by, 105. Composition and division, propor- tion by, 106. Conclusion, definition of, 11. Concurrent lines of triangle, Group on, 91-99. transversals and normals, Group on, 228-232. Cone, altitude of, 306. base of, 306. circular, 306-308. definition of, 306. elements of, 306. frustum of, 807. Group on, 306-:^08. of revolution, 307. right, 307. truncated, 307. Congruent figures, definition of, 2. triangles, Group on, 45-48. Conical surface, definition of, 306. directrix of, 300. elements of, 300. generatrix of, 306. nappes of, 306. vertex of, 300. Conjugate harmonic points, 184. Consequent, definition of, 101. Constant, definition of, 108. Construction of a figure, 12. Continued proportion, 103. Contradiction, definition of, 11. Converse of a theorem, 11. Convex spherical polygon, 329. surface of a pyramid, 287. Corollary, definition of, 11. Corresponding points, 160. Couplet, definition of, 160, Cube, definition of. 267. Cyclic four-side, 53. Cylinder, altitude of, 282. bases of, 282. circular, 282-284. definition of, 282. elements of, 282. Group on, 266-282. of revolution, 283. right, 282. right section of, 282. INDEX 351 Cylindrical surface, definition of, 282. directrix of, 282. elements of, 282. generatrix of, 282. Decagon, 3. Definition, 1. Degree, 113. Determinate problems, 130, 131. Determination of circle, 331. locus, 73. plane, 233, 234. Diagonal of polygon, 63. Diameter of circle, 78. Dihedral angle, definition of, 253. edge of, 253. faces of, 253. measure of, 254. method of reading, 253. rectilineal angle of, 254. right, 253. Discussion of problem, 12. Distance from point to line, 38. on surface of sphere, 328. Division, harmonic, 184. proportion by, 106. Dodecagon, 3. Dodecahedron, 266. Edge of dihedral angle, 253. Edges of polyhedral angle, 266. polyhedron, 254. Elements of cone, 306. conical surface, 306. cylinder, 282. cylindrical surface, 282. Equal figures, 2. Escribed circles, 93 ; angles, 111. Ex-center of triangle, 93. Exterior angle of triangle, 30. Extreme and mean ratio, 196, 197. Extremes, definition of, 103. Faces of dihedral angle, 253. polyhedral angle, 254. polyhedron, 266. Figures, areas of irregular, 144. definitions of^ 2. similar. Group on, 160-175. Foot of a line, 234. Fourth proportional, 172, 194. Frustum of cone, 307, 308. pyramid, 288, 298, 299, 301, 305. Golden Section, 190. Harmonic division, 184, Heptagon, 3. Hexagon, 3. Homologous lines, 160. Hypotenuse, 30, 38. Hypothesis, definition of, 11. Icosahedron, 266. In-center of triangle, 91. Incommensurable ratios, 102. Indeterminate problems, 130, 131. Inscribed polygon, 138. Inversion, proportion by, 105. Isoangular triangle, 30, 32. Isosceles triangle, definition of, 30. part of sector of circle, 78. Isosceles and scalene triangles, Group on, 35-44. Join, definition of, 2. Lateral area of prism, 267. edges of pyramid, 287. faces of pyramid, 287. surface of pyramid, 287. Limit, definition of, 108. Limits, method of, 108, 109. postulate of, 109. Line, broken, 2 ; curved, 2. definition of, 1. ' . 352 INDEX Line, on sphere, 331. straight, 1. Linear application of proportion, Group on, 18;{-209. Lines, concurrent, 2. homologous, 100. parallel, 7, 23, 234. perpendicular, 4, 23, 234. Lines and mid-joins. Group on, 01, 68. Line-segments, 1, 189. projection of, 149. Locus, definition of, 8. determination of, 73. exercises on, 123-126. illustrations of elementary princi- ples of, 74-77. Lune, angle of, 331. area of, 331. definition of, 331. Mean proportional, 103. Means, definition of, 103. Measure, angular, 100, radical, 101. Measurement, Group on, 100-110. of angles, Group on, 111-137. Median of trapezoid, 53. triangle, 80. Method of limits, 108, 109. Mid-joins and lines. Group on, 01-08. Mid-normal, or mid-perpendicular, 4, 09, 70. Nappes of conical surface, 306. Nine-point circle theorem, 128. Normals, concurrent, Group on, 228-232. Numerical measure, 100. Octagon, 3. Octahedron, 266. Opposite of theorem, 11. Orthocenter of triangle, 95. Orthogonal circles, theorem of, 129. Orthogonally, defined. 111. Parallel group, 23-20. lines, 7, 23, 234. lines to a plane, 245, 247. planes, 254. Parallelepiped, definition of, 207. division of, 277. rectangular, 207, 272-276. volume of, 270. Parallelogram, definition of, 63. Parallelograms, areas of, 141-143. Group on, 53-t5(). Pedal triangle, 127. Pentagon, definition of, 3. Perigon, definition of, 5. Perimeter, computation of, 222, 223. definition of, 221. Perpendicular lines, 4, 23. to a plane, 234. Perpendiculars, theorems relating to, 230-239, 241-243. Perspective, 160. PI anal angles, Group on, 253-205. Plane, definition of, 2, 233. figure, 2. geometry, definition of, 12. tangent to sphere, 311. Plane and its related lines. Group on, 233-252. Point, definition of, 1. Points, equidistant and random, Group on, 09-77. Polar distance of circle, 311. of spherical polygon, 330. Poles of a circle, 311. Polygon, angles of, 32, 33. circumscribed, 138. definition of, 2, 3. diagonal of, 53. inscribed, 138. regular, 3, 170, 210. similar, 100, 166, 167, 170, 178, 186, 206. spherical, 329, 330. Polygons, area of, Group on, 138- 148. INDEX 353 Polygons, circumscribed and in- scribed regular, Group on, 210-219. Polyhedra, classified, 266. similar, 268. Polyhedral, convex, 250. definition of, 254. edges of, 254. face angles of, 255. faces of, 254. planal angles of, 254. vertex of, 254. Polyhedrals, method of reading, 255. symmetric, 255. Polyhedron, convex, 266. definition of, 266. edges of, 266. faces of, 266. sections of, 266. vertices of, 266. volume of, 268. Postulate, definition of, 7. of limits, 109. Principal section, definition of, 323. Prism, altitude of, 267. bases of, 266. definition of, 266. Group on, 282-284. lateral area of, 267. lateral edges of, 267. lateral faces of, 266. oblique, 267. pentagonal, 267. (quadrangular, 267. regular, 267. right, 267. right section of, 267. right truncated, 267. triangular, 267. truncated, 267. volume of, 279. Prismoid, definition of, 288. Prismoidal formula, 303. Problem, definition of, 12. solution of, 12, 120. Problems, classification of, 130, 131. Projection, base of, 149. of line on plane, 234. of line-segment, 149. of point on line, 149. of point on plane, 234. Proof, definition of, 11. Proportion, 103-107, 191. linear application of. Group on, 183-209. Proportional, construction of, 194, 195. Proposition, definition of, 12. Pyramid, altitude of, 287. base of, 287. definition of, 287. frustum of, 288. lateral faces of, 287. lateral (or convex) surface of, 287. quadrangular, 288. regular, 288. slant, 288. triangular, 288. truncated, 288. vertex of, 287. volume of, 297. Pythagorean group, 149-159. Quadrilateral, or four-side, 53. cyclic, 53. Kadian, definition of, 187. Kadical axes, 208. center, 209. plane, 335. Radius of circle, 8. regular polygon, 210. sphere, 311. Ratio, extreme and mean, 196, 197 of similitude, 160. Ratio and proportion, 101-104. Ratios, areal, Group on, 176-179. commensurable, 101, 102. incommensurable, 103, 104. 354 INDEX Reciprocal proportion, 191. theorem, 11. Rectangle, definition of, 53. Rectangles, areas of, Group on, 138- 140. Relative direction, 5. Rhomboid, 53. Right angle, definition of, 4. Right angles group, 30-34. Ruled surface, 243, Ex. 11. Scalene triangle, definition of, 30. Group on, 38-41. Scholium, 12. Seciant, 78. Section, principal, 323. Sector of circle, 78. spherical, 312. Segment of circle, 78. line, 1, 189. spherical, 312. Sides of an angle, 4. Similar figures, definition of, 2. Group on, 160-175. Similarity, doctrine of, 333. Similitude, center of, 160. ratio of, 160. Solid, definition of, 1. geometry, 266. Solution of problem, 12, 120. Sphere, area of surface of, 316. areal measurement on, 333. axis of circle of, 311. circumscribed about, 311. definition of, 311. diameter of, 311. great circle of, 311. Group on, 311-327. inscribed in polygon, 311. plane tangent to, 311. polar distance of, 311. poles of, 311. radius of, 311. small circle of, 311. Sphere surface. Group on geom- etry of, 328-346. volume of, 318, 319. Spherical angle, 328. excess, 331. geometry. Group on, 328-346. polygon, 329. sector, 312. surface, 311. triangle, 329. wedge, 331. S(iuare, definition of, 53. Square roots of numbers, 219. Straight line, 1, 2 ; angle, 5. Sum of angles, 5, 19. Superposition, coincident, 2. Supplemental angles, 6. Surface, conical, 306. cylindrical, 282. definition of, 1. spherical, 311. Tangent to a circle, 78. plane to sphere, 311. Tetrahedrals, 255. Tetrahedron, 266. Theorem, definition of, 11. Theorems of special interest, 127-130. on inequality, 10. Third proportional, 103, 195. Transformation, definition of, 103. of figures, 199, 203. Transversal, definition of, 2. plane, 254. Transversals, concurrent, Group on, 228, 229. Trapezium, definition of, 53. Trapezoid, altitude of, 53. area of, 144. definition of, 53. isosceles, 53. median of, 63. mid-join of, 53. Triangle, acute, 30. altitude of, 30. INDEX 355 Triangle, area of, 143. base of, 30. concurrent lines of, Grpup on, 91-99. definition of, 3. equiangular, 30. equilateral, 30. exterior angle of, 30. hypotenuse of, 30. isoangular, 30. isosceles. 30, 35-37. median of, 30, m. obtuse, 30. right, 30. scalene, 30, 38-41. spherical, 329. vertex angle of, 30. Triangles, congruent. Group on, 45- 52. similar, 160, 103-165. Triangular relations, summaiy of, 98, 99. Trihedral, birectangular, 265. definition of, 255. rectangular, 255. trirectangular, 256. Ungula, 331. Unit of area, 138. length, 100. surface (sphere), 333. volume, 268. Variable, 108. Vertex of angle, 4. conical surface, 306. polyhedral, 254. pyramid, 287. angle of triangle, 30. Vertical angles, Group on, 19-22. Volume of polyhedron, 268. unit of, 268. Wedge (spherical), 331. Zon*, 311. UNIVERSITY OF CALIFORNIA LIBRARY This book is DUE on the last date stamped below OCT 17 1947 250ct5'^'? 20Feb5 2. mOrfSSBW *S^^ 14Nov'54DS ^95^\-^ FEB 7 1955 14Jun'55lW 4Jun'59BB ' JUN4 M " lA ?ec'60HJ DEC ! " 1560 B 21-100w-12,'46(A2012sl6)4120 911228 3^ THE UNIVERSITY OF CA^FORNIA LIBRARY '■l