LIBRARY OF THK University of California. ©IFT OF .WL'VL.''^.^ Class V \--' CHAUVENET'S TREATISE ON Elementary Geometry REVISED AND ABRIDGED BY W. E. BYERLY PROFESSOR OF MATHEMATICS IN HABVABD ONIVERSITY PHILADELPHIA J. B. LIPPII^COTT COMPANY 1892. 'aift^-^ as > Copyright, 1887, by J. B. LippIncott Compant. 3TER£OTYPERSAN0PRI!NT^s ] il l' PREFACE. In preparing this edition of Chauvenet's Geometry I have endeavored to compel the student to think and to reason for himself, and I have tried to emphasize the fact that he should not merely learn to understand and demonstrate a few set propositions, hut that he should acquire the power of grasping and proving any simple geometrical truth that may he set before him ; and this power, it must be remembered, can never be gained by memorizing demonstrations. Systematic practice in de- vising proofs of new propositions is indispensable. On this account the demonstrations of the main propo- sitions, which at first are full and complete, are gradually more and more condensed, until at last they are some- times reduced to mere hints, by the aid of which the full proof is to be developed; and numerous additional theorems and problems are constantly given as exercises for practice in original work. A syllabus, containing the axioms, the postulates, and the captions of the main theorems and corollaries, has been added to aid student and teacher in reviews and examinations, and to make the preparation of new proofs more easy and definite. 6 ELEMENTS OF GEOMETRY. In the order of the propositions I have departed con- siderably from the larger Chauvenet's Geometry, with the double object of simplifying the demonstrations and of giving the student, as soon as possible, the few theorems which are the tools with which he must most frequently work in geometrical investigation. Teachers are strongly advised to require as full and formal proofs of the corollaries and exercises as of the main propositions, and to lay much stress upon written demonstrations, which should be arranged as in the illustrations given at the end of Book I. In preparing a written exercise, or in passing a written examination, the student should have the syllabus before him, and may then conveniently refer to the propositions by number. In oral recitation, however, he should quote the full captions of the theorems on which he bases his proof. W. E. BYEELY. Cambridge, Mass., 1887. CONTENTS. PA OB INTKODUCTION 9 PLANE GEOMETEY. BOOK I. Rectilinear Figures 12 Exercises on Book 1 49 BOOK II. The Circle. Ratio. Incommensurables. Doctrine of Limits. Measure of Angles 55 Exercises on Book II 94 BOOK III. Theory of Proportion. Proportional Lines. Similar Poly- gons 101 Exercises on Book III 125 BOOK IV. Comparison and Measurement of the Surfaces of Recti- linear Figures 130 Exercises on Book IV 146 BOOK V. Regular Polygons. Measurement of the Circle 149 Exercises on Book V 169 Miscellaneous Exercises on Plane Geometry 173 Syllabus of Plane Geometry. Postulates, Axioms, and Theorems 182 7 8 CONTENTS. , GEOMETEY O^ SPACE. BOOK VI. PAGK The Plane. Diedral Angles. Polyedral Angles .... 195 Exercises on Book VI 220 BOOK VII. POLYEDRONS 224 Exercises on Book VII 254 BOOK VIII. The Three Bound Bodies. The Cylinder. The Cone. The Sphere. Spherical Triangles. Spherical Poly- gons 257 Exercises on Book VIII 287 BOOK IX. Measurement of the Three Bound Bodies 290 Exercises on Book IX 308 Miscellaneous Exercises on Solid Geometry 309 Syllabus of Propositions in Solid Geometry. Theorems . 311 ELEMENTS OF GEOMETRY. IKTEODUOTIOK 1. Every person possesses a conception of space indefi- nitely extended in all directions. Material bodies occupy finite, or limited, portions of space. The portion of space which a body occupies can be conceived as abstracted from the matter of which the body is composed, and is called a geometrical solid. The material body filling the space is called a physical solid. A geometrical solid is, therefore, merely the fornij or figure, of a physical solid. In this work, since only geometrical solids will be considered, we shall, for brevity, call them simply solids, and we shall define them formally, as follows : Definition. A solid is a limited or bounded portion of space, and has length, breadth, and thickness. 2, The boundaries of a solid are surface^. Definition. A surface is that which hae length and breadth, but no thickness. 1 If a surface is bounded, its boundaries are lir^es. If two surfaces intersect, their intersection is a line. Definition. A line is that which has length, but neither breadth nor thickness. 9 10 ELEMENTS OF GEOMETRY. If a line is terminated, it is terminated by points. If two lines intersect, they intersect in a point. Definition. A point has position, but neither length, breadth, nor thickness. 3. If we suppose a point to move in space, its path will be a line, and it is often convenient to regard a line as the path, or locus, of a moving point. If a point starts to move from a given position, it must move in some definite direction; if it continues to move in the same direction, its path is a straight line. If the direc- tion in which the point moves is continually changing, the path is a curved line. If a point moves along a line, it is said to describe the line. By the direction of a line at any point we mean the direc- tion in which a point describing the line is moving when it passes through the point in question. Definitions. A straight line is a line which -f f has everywhere the same direction. A curved line is one no portion of which, however short, is straight. A broken line is a line composed of different successive straight lines. 4. Definitions. A plane surface, or simply a plane, is a surface such that, if any two points in it are joined by a straight line, the line will lie wholly in the surface. A curved surface is a surface no portion of which, however small, is plane. 5. Definitions. A geometrical figure is any combination of points, lines, surfaces, or solids, formed under given condi- \ INTRODUCTION. 11 tions. Figures formed by points and lines in a plane are called plane figures. Those formed by straight lines alone are called rectilinear^ or right-lined^ figures; a straight line being often called a right line. 6. Definitions. Geometry may be defined as the science of extension and position. More specifically, it is the science which treats of the construction of figures under given con- ditions, of their measurement and of their properties. Plane geometry treats of plane figures. The consideration of all other figures belongs to the geom- etry of space, also called the geometry of three dimensions. 7. Some terms of frequent use in geometry are here de- fined. A theorem is a truth requiring demonstration. A lemma is an auxiliary theorem employed in the demonstration of another theorem. A problem is a question proposed for solu^ tion. An axiom is a truth assumed as self-evident. A postu- late (in geometry) assumes the possibility of the solution of some problem. Theorems, problems, axioms, and postulates are all called propositions. A corollary is an immediate consequence deduced from one or more propositions. A scholium is a remark upon one or more propositions, pointing out their use, their connection, their limitation, or their extension. An hypothesis is a sup- position, made either in the enunciation of a proposition or in the course of a demonstration. PLANE GEOMETRY. BOOK I RECTILINEAR FIGURES. ANGLES. 1. Definition. A jplane angle^ or simply an angle, is the amount of divergence of two lines which meet in a point or which would meet if produced (i.e., prolonged). The point is called the vertex of the angle, and the two lines the sides of the angle. From the definition it is clear that the mag- nitude of an angle is independent of the length of its sides. An isolated angle may be designated by the letter at its vertex, as " the angle O;" but when several angles are formed at the same point by different lines, as OA, OB, OC, we designate the angle intended by three 5 letters; namely, by one letter on each of its yC^'^^ sides, together with the one at its vertex, q— -i which must be written between the other two. Thus, with these lines there are formed three different angles, which are distinguished as AOB, BOC, and AOG. Two angles, such as AOB, BOO, which have the same vertex and a common side OB between them, are called adjacent. 12 BOOK I. 13 2. Definitions. Two angles are equal when one can be superposed upon the other, so that the vertices shall co- incide and the sides of the first shall fall along the sides of the second. Two angles are added by placing them in the same plane with their vertices together and a side in common, care being taken that neither of the angles is superposed upon the other. The angle formed by the exterior sides of the two angles is their sum. 3. A clear notion of the magnitude of an angle will be obtained by supposing that one of its sides, as OjB, was at first coincident with the other side OA^ and that it has revolved about the point (turning upon as the leg of a pair of dividers turns upon its hinge) until it has arrived at the posi- tion OB. During this revolution the movable side makes with the fixed side a varying angle, which increases by in- sensible degrees, that is, continuously ; and the revolving line is said to describe^ or to generate^ the angle AOB. By con-« tinning the revolution, an angle of any magnitude may be generated. 4. Definitions. When one straight line meets another, so as to make two adjacent angles equal, each of these angles is called a right angle; and the first line is said to be perpendicular to the second. Thus, if ^ 00 and BOC are equal angles, each is a right angle, and the line CO is perpendicular to AB. Intersecting lines not perpendicular are said to be oblique to each other. An acute angle is less than a right angle. An obtuse angle is greater than a right angle. 2 14 ELEMENTS OF GEOMETRY. 5. Definition. Two straight lines lying in the same plane and forming no angle with each other — that is, two straight lines in the same plane which will not meet, however far produced — are parallel. TKIANGLES. 6. Definitions. A plane triangle is a portion of a plane bounded by three intersecting straight lines; as A 50. The sides of the triangle are the portions of the bounding lines included between the points of intersection; viz., AB, BC, CA. The angles of the triangle are the angles formed by the sides with each other; viz., CAB, ABC, BCA. The three angular points. A, B, C, which are the vertices of the angles, are also called the vertices of the triangle. If a side of a triangle is produced, the angle which the prolongation makes with the adja- cent side is called an exterior angle ; as A CD. A triangle is called scalene (ABC) when no two of its sides are equal ; isosceles (DEF) when two of its sides are equal ; equilateral {GUI) when its three sides are equal. A right triangle is one which has a right angle ; as MNP, which is right-angled at ]V. The side MP, opposite to the right angle, is called the hypotenuse. The base of a triangle is the side upon which it is supposed BOOK I. 15 to stand. In general, any side may be assumed as the base ; but in an isosceles triangle BEF^ whose sides DE and DF are equal, the third side EF is always called the base. When any side ^(7 of a triangle has been adopted as the base, the angle BA G opposite to it is* called the vertical angle, and its angular point A the vertex of the triangle. The perpen- dicular AD let fall from the vertex upon the base is then called the altitude of the triangle. 7. Definition. Equal figures are figures which can be made to coincide throughout if one is properly superposed upon the other. Eoughly speaking, equal figures are figures of the same size and of the same shape ; equivalent figures are of the same size but not of the same shape; and similar figures are of the same shape but not of the same size. POSTULATES AND AXIOMS. 8. Postulate I. Through any two given points one straight line, and only one, can be drawn. Postulate II. Through a given point one straight line, and only one, can be drawn having any given direction. 9. Axiom I. A straight line is the shortest line that can be drawn between two points. Axiom II. Parallel lines have the same direction. 16 ELEMENTS OF GEOMETEY. PROPOSITION I.— THEOREM. 10. At a given point in a straight line one perpendicular to the line can be drawn, and but one. Let be the given point in the line AB. Suppose a line OD, constantly passing through 0, to revolve about 0, starting from the position OA and stopping at the position OB. The angle which 01) makes with OA will at first be less than the angle which it makes with OB, and will eventually become greater than the angle made with OB. ^ D Since the angle DOA increases continu- j / ously (3), the line OB must pass through ]/_ ... BOA one position m which the angles DOA and DOB are equal. Let OC he this position. Then 0(7 is perpendicular to ^jB by (4).* There can be no other perpendicular to AB at 0, for if OD is revolved from the position 0(7 by the slightest amount in either direction, one of the adjacent angles will be in- creased at the expense of the other, and they will cease to be equal. 11. Corollary. Through the vertex of any given angle one line can be drawn bisecting the angle, and but one. Suggestion. Suppose a line OD to revolve about O, as in the proof just given. * An Arabic numeral alone refers to an article in the same Book ; but in referring to articles in another Book, the number of the Book is also given. V, BOOK I. 17 PROPOSITION II.— THEOREM. 12. All right angles are equal. Let AOG and AIQfC be any two right angles. Superpose A'CfC upon AOC^ placing the point (J upon the point and making the line C/A' fall along the line OA; then will OC coincide with 00, for other- 5 — wise we should have two perpen- diculars to the line AB at the same point 0, which, by Propo- sition I., is impossible. The angles A'C/C and AOC are then equal by definition (2). B' (y PROPOSITION III.— THEOREM. 13. The two adjacent angles which one straight line makes with another are together equal to two right angles. If the two angles are equal, they are right angles by definition (4), and no proof is necessary. If they are not equal, as AOD and BOD, still the sum of AOD and BOB is equal to two right angles. Let OC be drawn at perpendicular to AB. The angle AOD is the sum of the two angles AOC and COD (2). Adding the angle BOD, the sum of the two angles AOD and BOD is the sum of the three angles AOC, COD, and BOD. The first of these three is a right angle, and the other two are together equal to the right angle BOC ; hence the sum of the angles AOD and BOD is equal to two right angles. 18 ELEMENTS OF GEOMETRY. 14. Corollary I. The sum of all angles having a common vertex^ and formed on one side of a straight line, ^ is two right angles. 15. Corollary II. The sum of all the angles that can be formed about a point in a plane is four right angles. EXERCISE. Theorem. — If a line is perpendicular to a second line, then reciprocally the second line is perpendicular to the first. 16. Definition. When the sum of two angles is equal to a right angle, each is called the complement of the other. Thus, J) 0(7 is the comple- ment of A ODj and A OD is the complement of DOC. When the sum of two angles is equal to b two right angles, each is called the supple- ment of the other. Thus, BOD is the supplement of AOD, and AOD is the supplement of BOD. It is evident that the complements of equal angles are equal to each other ; and also that the supplements of equal angles are equal to each other. BOOK I. 19 PROPOSITION IV.— THEOREM 17. If the sum of two adjacent angles is equal to two right angles^ their exterior sides are in the same straight line. Let the sum of the adjacent angles AOD^ BOD, be equal to two right angles ; then OA and OB are in the same straight line. For BOD is the supplement of A 02), and is therefore identical with the angle which OD makes with the prolongation of ^ (Proposition TIL). Therefore OB and the prolongation of AO are the same line. 18. Every proposition consists of an hypothesis and a con- clusion. The converse of a proposition is a second proposition of which the hypothesis and conclusion are respectively the conclusion and hypothesis of the first. For example, PropO' sition III. may be enunciated thus : Hypothesis — if two adjacent angles have their exterior sides in the same straight line, then — Conclusion — the sum of these adjacent angles is equal to two right angles. And Proposition TV. may be enunciated thus : Hypothesis — if the sum of two adjacent angles is equal to two right angles, then — Conclusion — these adjacent angles have their exterior sides in the same straight line. Each of these propositions is, therefore, the converse of the other. A proposition and its converse are, however, not always both true. 20 ELEMENTS OF GEOMETRY. PROPOSITION v.— THEOREM. 19. If two straight lines intersect each other, the opposite (or vertical) angles are equal. B Let AB and CD intersect in ; then will the opposite, or vertical, angles AOG and BOD be equal. For each of these angles is, by Proposi- tion III., the supplement of the same angle BOC, and hence, they are equal (16). In like manner it can be proved that the opposite angles AOD and BOC are equal. EXERCISES. 1. Theorem. — The line which bisects one of two vertical angles bisects the other. 2. Theorem. — The straight lines which bisect a pair of adjacent angles formed by two intersecting straight lines are per- pendicular to each other. Suggestion. Prove EOH = FOH. Q PROPOSITION VI.— THEOREM. 20. Two triangles are equal when two sides and the included angle of the one are respectively equal to two sides and the included angle of the other. In the triangles ABC, DEF, let AB be equal to DE, BC to EF, and the included ' AD D angle B equal to the included angle E; then the triangles are equal. For, superpose the .X* B C E F F E triangle ABC upon the triangle DEF, placing the point B BOOK I. 21 upon the point E, and making the side BC fall along the side EF. Then, since J5(7 is equal to EF by hypothesis, the point G will fall upon the point F. Since the angle B is equal to the angle E^ and the side BG has been made to coincide with the side EF^ BA must fall along ED^ by definition (2) ; and, as BA is equal to ED^ the point A will fall on the point D. Since the point G has been proved to coincide with the point F^ and the point A with the point D, the side GA must coincide with the side FD^ by Postulate I. (8). The two triangles have now been proved to coincide throughout, and are equal, by definition (7). 21. Scholium. When two triangles are equal, the equal angles are opposite to the equal sides. PROPOSITION VII.— THEOREM. 22. Two triangles are equal when a side and the two adjacent angles of the one are respectively equal to a side and the two adjacent angles of the other. In the triangles ABG, DEF, let BG be equal to EF, and let the angles B and G adjacent to BGhe respectively equal to the angles E and F adjacent to EF; then the triangles are equal. For, superpose the \ \ \ \ ..--''' / ..1.^^ , B C E F F E triangle ABG upon the triangle BEF, placing the point B upon the point E, and making the side BG fall along the side EF. Since J5(7 is equal to EF, the point G will fall upon the point F. Since the angle B is equal to the angle E, by hypothesis, 22 ELEMENTS OF GEOMETRY. and the side 5(7 has been made to coincide with the side EF^ BA must fall along ED^ and the point A will fall somewhere on the side ED, or on that side extended. Since the angle G is equal to the angle P, by hypothesis, and BC coincides with EF^ CA must fall along FD, and the point A will fall somewhere on the line FD, or on that line extended. Since A has been proved to lie upon ED, and also upon FD, it must coincide with the only point they have in com- mon, the point D. Hence the triangles coincide throughout, and are equal. PEOPOSITION VIII.— THEOREM. 23. In an isosceles triangle the angles opposite the equal sides are equal. Let ^5 and AC he the equal sides of the isosceles triangle ABC; then the angles B and C are equal. Through the vertex A draw a line AD, bisect- ing the angle BAC, and meeting the side BC t atD. / In the triangles ABD and ACD the side AB is equal to the side AChy hypothesis, the side AD is common, and the included angle BAD is equal to the included angle CAD by construction. The triangles are therefore equal, by Proposition YI., and the angle C of the one is equal to the angle B of the other, by (21). 24. Corollary. The straight line bisecting the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base. EXERCISE. Theorem. — An equilateral triangle is also equiangular. BOOK I. 23 PROPOSITION IX.— THEOREM. 25. Two triangles are equal when the three sides of the one are respectively equal to the three sides of the other. In the triangles ABC, DEF, let AB be equal to BE, AG to DF, and BG to EF ; then the triangles are equal. ^ For, suppose the triangle ^^ \ ^^ ABG to be placed so that its ^ ^ \^^ base ^(7 coincides with its equal EF, but so that the vertex A falls on the opposite side of EF from X), as at G, and join D and 6^ by a straight line. The triangle EBG is isosceles, since the side ED is equal to the side EG by hypothesis; therefore the angles EBG and EQD are equal, by Proposition YIII. The triangle FBG is isosceles, since the side FI) is equal to the side FG by hypothesis; therefore the angles FDG and FGD are equal, by Proposition YIII. If to the equal angles EDG and EGB we add the equal angles FBG and FGB, the sums will be equal, and we have the whole angle EBF equal to the whole angle EGF. The two triangles EBF and EGF have now the side EB equal to the side EG by hypothesis, the side BF equal to the side FG by hypothesis, and the included angle EBF proved equal to the included angle EGF. Hence the tri- angles are equal, by Proposition YI. EXERCISE. Theorem. — A line drawn from the vertex of an isosceles tri- angle to the middle point of the base is perpendicular to the base, and bisects the vertical angle. 24 ELEMENTS OF GEOMETRY. PROPOSITION X.— THEOREM. 26. Two right triangles are equal when they have the hypote- nuse and a side of the one respectively equal to the hypotenuse and a side of the other. In the right triangles ABC, A'B'G\ let the hypotenuse AB be equal to the hypotenuse A'B', and the side ^C be equal to the side BC^ ; then the triangles are equal. Extend the side BC to D, making CD equal to BC, and join A and D ; and extend B'C^ to iX, making C^jy equal to B'C, and join A' and IT. The triangle ADC and the triangle ABC having the side A in common ; the side CD equal to the side CB by con- struction ; and the included angle A CD equal to the included angle ACB, since they are adjacent angles and ACB is ^ right angle by hypothesis, are equal, by Proposition YI. In like manner the triangle A!D^C' may be proved equal to the triangle MB'C, The triangles BAD and B'A!D^ having the side AB equal to the side A!B! by hypothesis; the side BD equal to the side B'D'^ because they were constructed the doubles oi BG and B'C'^ which were equal by hypothesis; and the side AD equal to the side A!D\ since they have been proved to be equal respectively to the sides AB and A!B' ; are equal to each other, by Proposition IX., and B is equal to B'. The triangles ABC and A!B'C' have now been proved to have two sides and the included angle of the one respectively equal to two sides and the included angle of the other, and are equal, by Proposition VI. BOOK I. 25 PROPOSITION XI.— THEOREM. 27. If two angles of a triangle are equals the sides opposite to them are equals and the triangle is isosceles. Let the angles BAG and BCA of the triangle ABC be equal, then are the sides AB and BG equal. For, if AB and BG are not equal, one must be greater than the other. Suppose AB greater than BG. Then cut oif from AB a part AD equal to BG, and join D and G. Compare now the triangle ADG with the whole triangle ABG, of which it is a part. The two triangles have the side A C in common ; the side AD equal to the side BG hy construction; and the included angle A equal to the included angle BGA by hypothesis. Therefore, by Proposition YI., the triangles ADG and ABG are equal, which is impossible. Consequently, AB could not have been greater than BG. In like manner we can prove that BG cannot be greater than AB. Therefore, since neither can be greater than the other, AB and BG are equal. EXERCISE. Theorem. — An equiangular triangle is also equilateral. ^ r n (h 26 ELEMENTS OF GEOMETRY. PKOPOSITION XII.— THEOEEM. 28. If two angles of a triangle are unequal^ the side opposite the greater angle is greater than the side opposite the less angle. In the triangle ABC let the angle C be greater than the angle B ; then AB is greater than A C. For, suppose the line CD to be drawn, cutting off from the greater angle a part BCD = B. Then BDC is an isosceles triangle, by Propo- sition XI., and DC = DB. But in the triangle ADC we have ^D + DC > AC, by Axiom I. ; or, putting DB for its equal DC,AD + DB:>AC;ovAB:>Aa PROPOSITION XIII.— THEOREM. 29. If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. In the triangle ABC let the side AB be greater than the side BC; then will the angle C be greater than the angle A. For, if C is not greater than J., it must be equal to A or less than A. C cannot be equal to A, for in that case AB ^ ^ and BC would be equal, by Proposition XI. C cannot be less than J., for in that case AB would be less than BC, by Proposition XII. Therefore C is greater than A. BOOK I. 27 PROPOSITION XIV.— THEOREM. 30. If two triangles have two sides of the one respectively equal to two sides of the other, and the included angles unequal, the triangle which has the greater included angle has the greater third side. Let ABC and ABB be the two triangles in which the sides AB, AC are respectively equal to the sides AB, AD, but the included angle BAC is greater than the included angle BAD; then ^Ois greater than BD. For, suppose the line AE to be drawn, bisecting the angle CAD and meeting BC in E; join DE. The triangles AED and AEC are equal, by Proposition YI., and therefore ED = EC. But in the triangle BDE we have BE + ED:> BD, by Axiom I., and substituting EC for its equal ED, BE+ EC> BD, or BC > BD. PROPOSITION XV.— THEOREM. 31. If two triangles have two sides of the one respectively equal to two sides of the other, and the third sides unequal, the triangle which has the greater third side has the greater included angle. In the triangles ABC and DEF let AB = DE, AC = DF, and let 5(7 be greater than EF ; then will the angle A be greater than the angle D. For, if A were equal to D, BC would be equal to EF, by Proposi- tion YI. ; and if A were less than D, BC would be less than EF, by Proposition XIY. 28 ELEMENTS OF GEOMETRY. PEOPOSITION XVI.— THEOEEM. 32. From a given pointy without a straight line^ one perpen- dicular can be drawn to the line^ and but one. Let AB be the given line and P the given point. Take a second line DJS, and at some point F of DE let a perpendicular be erected (Proposition I.). Superpose this second figure upon the p ^ first, placing the line DF upon the line AB, and then move the fig- ure along, keeping DF ^ ^ J J J always in coincidence with AB, until the perpendicular FGr passes through P; we shall then have a perpendicular to AB drawn through P. Let PC in the figure below be this perpendicular. No other perpendicular from P can be drawn to the line AB. For, suppose that a second perpendicular FD could be drawn. Extend FG to P', making GF' equal to PC, and join J> andP'. The two triangles FGF and F'GF have the side FG equal to the side F'G by con- struction; the side GF common; and the included angle FGF equal to the included angle F'GD, by Proposition III. There- fore, by Proposition YI., the triangles are equal, and the angle FDG is equal to the angle F'FG. But FDG is a right angle by hypothesis; therefore F'DG must be a right angle, and FD and DF must lie in the same straight line, by Proposition lY. ; and we have two straight lines drawn between P and P', which, by Postulate I., is impossible. BOOK I. 29 Since this impossible result follows necessarily from the assumption that a second perpendicular can be drawn from P to AB, that assumption must be false. EXERCISES. 1. Theorem. — A perpendicular let fall from the vertex of an isosceles triangle upon the base bisects the base and bisects the vertical angle. 2. Theorem. — Two right triangles are equal when they have the hypotenuse and an adjacent angle of the one respectively equal to the hypotenuse and an adjacent angle of the other. Suggestion. Superpose the second triangle upon the first, making the given equal angles coincide. PROPOSITION XVII.— THEOREM. 33. The perpendicular is the shortest line that can be drawn from a point to a straight line. Let PC be the perpendicular and PD any oblique line from the point ^P to the line AB. Then PC is shorter than PD. Extend PC to P', making CP' equal to PC, and join D and P'. The triangles PCD and P'CD are equal, by Proposition YI. Therefore P'B = PD. PP' <^PP + DP', by Axiom I. Therefore PC, the half of PP', is less than PD, the half of PJ)P\ EXERCISES. 1. Theorem. — Two oblique lines drawn from a -point to a line, and meeting the line at equal distances from the foot of the per- pendicular from the given point, are equal. 2. Theorem. — Two equal oblique lines drawn from a point to a line meet it at equal distances from the foot of the perpen- dicular. 3* 30 ELEMENTS OF GEOMETEY. PROPOSITION XVIII.— THEOREM. 34. If a perpendicular is erected at the middle of a straight linCj then, 1st. Uvery point in the perpendicular is equally distant from the extremities of the line ; 2d. Every point without the perpendicular is unequally distant from the extremities of the line. Let AB be a finite straight line and CD a perpendicular at its middle point. 1st. Then is any point P on CD equi- distant from A and B. For, join P and A and P and J5. The triangles FCA and FOB are equal, by Proposition YI. ; therefore FA and FB are equal. 2d. Any point Q without the perpen- dicular is unequally distant from A and B. For, Q being on one side or the other of the perpendicular, one of the lines QA, QB must cut the perpendicular; let it be QA and let it cut in P; join FB. The straight line QB is less than the broken line QFB^ by Axiom I. ; that is, C^ < §P + FB. But FB = FA ; therefore QB <: QF + FA, or QB < QA. 35. Definition. A geometric locus is the geometric figure containing all the points which possess a common property, and no others. In this definition, points are understood to have a common property when they satisfy the same geometrical conditions. Thus, since all the points in the perpendicular erected at the middle of a line possess the common property of being equally distant from the extremities of the line (that is, BOOK I. 31 satisfy the condition that they shall be equally distant from those extremities), and no other points possess this property^ the perpendicular is the locus of these points; so that the preceding proposition is fully covered by the following brief statement : The perpendicular erected at the middle of a straight line is the locus of the points which are equally distant from the extrem- ities of that line, PROPOSITION XIX.— THEOREM. 36. Uvery point in the bisector of an angle is equally distant from the sides of the angle ; and every point not in the bisector, but within the angle, is unequally distant from the sides of the angle ; that is, the bisector of an angle is the locus of the points within the angle and equally distant from its sides. 1st. Let AD be the bisector of the angle JBAC, P any point in it, and PE, PF, the perpendicular distances of P from AB and AC ; then PE = PF. For, the right triangles APE, APF, having the angles PAE and PAF equal, and AP common, are equal (32, Exercise 2) ; therefore PE = PF. 2d. Let Q be any point not in the bisector, but within the angle; then the perpendicular distances QE and QH are unequal. For, suppose that one of these distances, as QE, cuts the bisector in some point P; from P let PF be drawn perpen- dicular to AC, and join QF. We have QJff and EF be cut by the line AB ; then will the angles CAB and ABF be equal. For, if they are not equal, draw through A sl line AG, making the angles GAB and ABF equal ; then, by Propo- sition XXIV., GA and EF are parallel, and we have two parallels to the same line EF drawn through the same point A, which is contrary to Proposition XXIII., and there- fore impossible. Hence the angles CAB and ABF are equal. 48. Corollary I. If two parallel lines are cut by a third straight line, any two corresponding angles are equal. 49. Corollary II. If two parallel lines are cut by a third straight line, the sum of the two interior angles on the same side of the secant line is equal to two right angles. EXERCISE. Theorem. — A line perpendicular to one of two parallel lines is perpendicular to the other. /f^^ Off rn^.'^^^ /^ . ay J^IFOUll^ 38 ELEMENTS OF GEOMETRY. PK0P08ITI0N XXVL— THEOREM. 50. The sum of the three angles of any triangle is equal to two right angles. Let ABC he any triangle ; then the sum of its three angles is equal to two right angles. Produce BGj and through G draw C£J parallel to BA» Since the line AG meets the parallel A lines AB and EG, the alternate-interior angles AGJ3 and BAG are equal, by Proposition XXV. Since the line BD cuts the parallel lines AB and EG, the corresponding angles EGD and ABG are equal, by Proposition XXY., Corollary I. Therefore the sum of the angles of the triangle is equal to BGA-\- AGE + EGD, which is two right angles, by Propo- sition III., Corollary I. 51. C0R01.LARY. If a side of a triangle is eoctended, the exterior angle is equal to the sum of the two interior opposite angles. EXERCISE. Theorem. — If the sides of an angle are respectively perpendicular to the sides of a second angle, the an- gles are equal, or supple- mentary. BOOK I. 39 POLYGONS. 52. Definitions. A polygon is a portion of a plane bounded by straight lines ; as ABODE. The bounding lines are the sides ; their sum is the perimeter of the polygon. The angles which the adja- cent sides make with each other are the angles of the polygon; and the vertices of these angles are called the vertices of the polygon. Any line joining two vertices not consecutive is called a diagonal; as AG. 53. Definitions. Polygons are classed according to the num- ber of their sides : A triangle is a polygon of three sides. A quadrilateral is a polygon of four sides. A pentagon has five sides ; a hexagon, six ; a heptagon, seven ; an octagon, eight ; an enneagon, nine j a decagon^ ten ; a dodec- agon, twelve ; etc. An equilateral polygon is one all of whose sides are equal ; an equiangular polygon, one all of whose angles are equal. 54. Definition. A convex polygon is one no side of which, when produced, can enter within the space enclosed by the perimeter; as ABODE in (52). Each of the angles of such a polygon is less than two right angles. It is also evident from the definition that the perimeter of a convex polygon cannot be intersected by a straight line in more than two points. A concave polygon is one of which two or more sides, when produced, will enter the space enclosed by the perimeter; as MNOPQ, of which OP and QP, when ^ produced, will enter within the polygon. The angle OPQ, formed by two adjacent re-entrant sides, 40 ELEMENTS OF GEOMETRY. is called a re-entrant angle ; and hence a concave polygon is ^OTCLQiimQ^ cdXlQ^ 2i re-entrant polygon. All the polygons hereafter considered will be understood to be convex. PROPOSITION XXVII.— THEOREM. 55, The sum of all the angles of any polygon is equal to twice as many right angles, less four, as the figure has sides. Join any point within the polygon to each of the ver- tices, thus dividing the polygon into as many triangles as it has sides. The sum of the angles of these tri- angles will, by Proposition XXYI., be twice as many right angles as the figure has sides. But the angles of the triangles form the angles of the polygon plus the angles at 0, which are equal to four right angles, by Proposition III., Corollary II. EXERCISE. 1. Theorem. — If each side of a polygon is extended, the sum of the exterior angles is four right angles. Suggestion. The sum of all the an- gles, exterior and interior, is obvi- ously twice as many right angles as the figure has sides. BOOK I. 41 QUADRILATERALS. 56. Definitions. Quadrilaterals are divided into classes, as follows : Ist. The trapezium (J.), which has no two of its sides parallel. 2d. The trapezoid (jB), which has two sides parallel. The parallel sides are called the hases^ and the perpendicular distance /^ ^ \ \ between them the altitude of the trapezoid. 3d. The parallelogram ( (7), which is bounded v r\ by two pairs of parallel sides. \ j \ The side upon which a parallelogram is supposed to stand and the opposite side are called its lower and upper bases. The perpendicular distance between the bases is the altitude. 57. Definitions. Parallelograms are divided into species, as follows : 1st. The rhomboid (a), whose adjacent 's* sides are not equal and whose angles are \ not right angles. 2d. The rhombus^ or lozenge (&), whose sides are all equal. 3d. The rectangle (c), whose angles are all equal, and therefore right angles. 4th. The square (d), whose sides are all equal and whose angles are all equal. The square is at once a rhombus and a rectangle. 42 ELEMENTS OF GEOMETRY. PROPOSITION XXVIII.— THEOREM. 58. Two parallelograms are equal when two adjacent sides and the included angle of the one are equal to two adjacent sides and the included angle of the other. AD == A'jy, and the angle BAD I j ^/ j = ^'J.'i/; then these parallelo- ^ ^ ^' ^ grams are equal. For they may evidently be applied the one to the other, so as to coincide throughout, (v. Proposition XXIII.) 59. Corollary. Two rectangles are equal when they have equal bases and equal altitudes. PROPOSITION XXIX.— THEOREM. 60. The opposite sides of a parallelogram are equal and the opposite angles are equal. Suggestion. Draw a diagonal AC. ACB J^ ^ and CAD are equal, by Proposition XXY. CAB and AGD are equal, by Proposition XXY. Hence the triangles ABC and ADC are equal, by Propo- sition YII. EXERCISES. 1. Theorem. — If one angle of a parallelogram is a right angle, all the angles are right angles, and the figure is a rectangle. 2. Theorem. — If two angles have the sides of one respectively parallel to the sides of the other, they are equal, or supplementary. 3. Theorem. — Two parallel lines are everywhere equidistant. v. BOOK I. 43 PROPOSITION XXX.— THEOREM. 61. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Suggestion. Let AD be equal and parallel i^ ^ to BC. Draw a diagonal AC. / \ The triangles ABC and ABO are equal, £ by Proposition YI. Therefore the angles BAG and ACD are equal, and AB and CD are parallel, by Proposition XXIY. PROPOSITION XXXI.— THEOREM. 62. If the opposite sides of a quadrilateral are equals the figure is a parallelogram. Suggestion. Draw a diagonal, and prove the two triangles equal. PROPOSITION XXXII.— THEOREM. 63. The diagonals of a parallelogram bisect each other. A D Suggestion. The triangles AED and BEC are equal, by Proposition YII. EXERCISES. 1. Theorem. — The diagonals of a rectangle are equal. 2. Theorem. — The diagonals of a rhombus are perpendicular to each other. 3. Theorem. — If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. 4. Theorem. — If the diagonals of a parallelogram are equal, the figure is a rectangle. 5. Theorem. — If the diagonals of a parallelogram are perpen- dicular to each other, the figure is a rhombus. 44 ELEMENTS OF GEOMETRY. ARRANGEMENT OF WRITTEN EXERCISES. 64. In writing out a demonstration, brevity of statement and clearness of arrangement should be carefully studied, and symbols and abbreviations may bo used with profit. The following list is recommended : SYMBOLS AND ABBREVIATIONS. . • . therefore. Bef. definition. = equal to. Fost postulate. O equivalent to. Ax. axiom. > greater than. Prop. proposition. <; less than. Cor. corollary. II parallel to. Hyp. hypothesis. J_ perpendicular to. Cons. construction. /_ angle. Adj. adjacent. ^ angles. Inc. included. rt. /_ right angle. Alt.-int. alternate-interior. l\ triangle. Sup. supplementary. l^ triangles. Comp. complementary. rt. I\ right triangle. Q.E.B, , quod erat demonstran- / / parallelogram. dum (= which was / * / parallelograms. to be proved). O circle. circles. BOOK I. 45 65. In arranging a written demonstration, it is well to begin each statement on a separate line, giving the reason for the statement at the end of the line, if it can be written briefly, or in parenthesis immediately below the line, if it cannot be written briefly. The following examples of demon- strations prepared as written exercises, or for a written examination, will serve as illustrations. (1) PROPOSITION XII.— THEOREM. If two angles of a triangle are unequal^ the side opposite the greater angle is greater than the side opposite the less angle. In A ABC, let it be given that Z_ACBy /^B, we are to prove AB > J. (7. Draw CDj cutting off from /^ACB si part / BCD = /_B. Then in /\ BCD we have / BCD = l^B. Cons. BD = CD, Prop. XL and BD + DA = CD ^ DA. But AC<:CD-^DA. Ax. I. AC%^ 2d. Let CG, AB, be unequal chords ; OE, OH, their distances from the centre ; and let CG be less than ^5; then OlfyOE. For, since chord AB > chord CG, we have arc AB > arc CG ; so that if from C we draw the chord CD = AB, its subtended arc CD, being equal to the arc AB, will be greater than the arc CG. Therefore the perpendicular OS will intersect the chord CD in some point I. Drawing the per- pendicular OF to CD, we have, by the first part of the demonstration, OF = OE. But OH > 01, and 01 > OF (Proposition XYIL, Book I.) ; still more, then, is OS > OF, or OS^OE. If the chords be taken in two equal circles, the demonstra- tion is the same. 62 ELEMENTS OF GEOMETRY. 19. Corollary. Conversely^ in the same circle^ or in equal circles^ chords equally distant from the centre are equal ; and of two chords unequally distant from the centre, that is the greater whose distance from the centre is the less. EXERCISE. Theorem. — The least chord that can be drawn in a circle through a given point is the chord perpendicular to the diameter through the point Suggestion, v. Proposition XYII., Book I. TANGENTS AND SECANTS. 20. Definitions. A tangent is an indefinite straight line which has but one point in common with the circumference; as ACB. The common point, C, is called the point of contact, or the point of tan- gency. The circumference is also said to be tangent to the line AB at the point C, A secant is a straight line which meets the circumference in two points ; as JEF. Two circumferences are tangent to each other when they are both tangent to the same straight line at the same point. 21. Definition. A rectilinear figure is said to be circumscribed about a circle when all its sides are tangents to the circumference. In the same case, the circle is said to be inscribed in the figure. BOOK II. 63 PROPOSITION VIII.— THEOREM. 22. A straight line cannot intersect a circumference in more than two points. For, if the line could intersect the circumference in three points, the radii drawn to these points would meet the line at unequal distances from the perpendicular let fall from the centre of the circle upon the line, and would be unequal, by Proposition XXL, Book I. PROPOSITION IX.— THEOREM. 23. A straight line tangent to a circle is perpendicular to the radius drawn to the point of contact. For any other point of the tangent, 2)^/ as D, must lie outside of the circle, and ^y\\ X therefore the line OD^ joining it with a^^ \1 \ the centre, must be greater than the I q ] radius 0(7, drawn to the point of con- V / tact. ^ OCiSj then, the shortest line that can be drawn from to the tangent AB, and is therefore perpen- dicular to AB, by Proposition XYII., Book I. 24. Corollary I. A perpendicular to a tangent line drawn through the point of contact must pass through the centre of the circle. 25. Corollary II. If two circumferences are tangent to each other, their centres and their point of contact lie in the same straight line. Suggestion. Through their point of contact draw a line perpendicular to the tangent at that point, (v. Corollary I.) 64 ELEMENTS OF GEOMETRY. PROPOSITION X.— THEOREM. 26. When two tangents to the same circle intersect, the dis- tances from their j)oint of intersection to their points of contact are equal. Eor the right triangles OAP and OBP (Proposition IX.) are equal, by Proposition X., Book I. EXERCISES. 1. Theorem. — In any circumscribed quadrilateral, the sum of two opposite sides is equal to the sum of the other two opposite sides, 2. Theorem. — If two circumferences are tangent, and from any point, P, of the tangent at their point of contact, tangents are drawn to the two circles, the points of contact of these tangents are equally distant from P. PROPOSITION XI.— THEOREM. 27. Two parallels intercept equal arcs on a circumference. We may have three cases : 1st. When the parallels AB, CI), are both secants, then the intercepted arcs A G and BB are equal. For, let OM be the radius drawn perpendicular to the parallels. By Proposition YI., the point M is at once the middle of the arc AMB and of the arc CMD, and hence we have AM = BM and CM = DM, whence, by subtraction, AM— CM= BM — DM, AG = BD. that is, E M F G/^ <^ \ — V \d ( ^ \ A V \ / BOOK II. 65 2d. When one of the parallels is a secant, as AB, and the other is a tangent, as EF at Jf, then the intercepted arcs AM and BM are equal. For the radius OM drawn to the point of contact is perpendicular to the tangent (Proposition IX.), and consequently perpendicular also to its parallel AB ; therefore, by Proposition YI., AM = BM. 3d. When both the parallels are tangents, as JSF at Mj and GH at iV, then the intercepted arcs MAN" and MBN are equal. For, drawing any secant AB parallel to the tan- gents, we have, by the second case, AM = BM and AN = BM, whence, by addition, AM+A]Sr=BM+BW, that is, MAN=MBJSr; and each of the intercepted arcs in this case is a semi-circum- ference. MEASURE OF ANGLES. As the measurement of magnitude is one of the principal objects of geometry, it will be proper to premise here some principles in regard to the measurement of quantity in general. 28. Definition. To measure sl quantity of any kind is to find how many times it contains another quantity of the same kind, called the unit. Thus, to measure a line is to find the number expressing how many times it contains another line, called the unit of length, or the linear unit. The number which expresses how many times a quantity contains the unit is called the numerical measure of that quantity. 66 ELEMENTS OF GEOMETRY. 29. Definition. The ratio of two quantities is the quotient arising from dividing one by the oth-er : thus, the ratio of A to B is ~. To find the ratio of one quantity to another is, then, to find how many times the first contains the second; there- fore it is the same thing as to measure the first by the second taken as the unit (28). It is implied in the defini- tion of ratio that the quantities compared are of the same kind. Hence, also, instead of the definition (28), we may say that to measure a quantity is to find its ratio to the unit. The ratio of two quantities is the same as the ratio of their numerical measures. Thus, if P denotes the unit, and if P is contained m times in A and n times in B^ then A mP m B~~nP~ n 30. Definition. Two quantities are commensurable when there is some third quantity of the same kind which is contained a whole number of times in each. This third quantity is called the common measure of the proposed quantities. Thus, the two lines A and B are commensurable if there is some line, (7, which is contained a whole number of times in each, as, ^' ' ^ ' ' ' ' ' for example, 7 times in J., and 4 times £< — • — « — « — « in B. 0' — « The ratio of two commensurable quantities can, therefore, be exactly expressed by a number, whole or fractional (as in the preceding example by -), and is called a commensurable ratio. BOOK II. 67 31. Definition. Two quantities are incommensurable when they have no common measure. The ratio of two such quantities is called an incommensurable ratio. If A and B are two incommensurable quantities, their ratio is still expressed by - . 32. Problem. To find the greatest common measure of two quantities. The well-known arithmetical process may be ex- tended to quantities of all kinds. Thus, suppose AB and CD are two straight lines whose common measure is required. Their greatest common measure can- not be greater than the less line CD. ^' ' ' J"^ Therefore let CD be applied to AB as Cr—T-pD many times as possible, suppose three times, with a remainder EB less than CD. Any common measure of AB and CD must also be a common measure of CD and EB ; for it will be contained a whole number of times in CD, and in AE^ which is a multiple of OD, and therefore to measure AB it must also measure the part EB. Hence the greatest common measure of AB and CD must also be the greatest common measure of CD and EB, This greatest common measure of CD and EB cannot be greater than the less line EB ; therefore let EB be applied as many times as possible to CD, suppose twice, with a remainder FD. Then, by the same reasoning, the greatest common measure of CD and EB^ and consequently also that of AB and CD, is the greatest common measure of EB and FD. Therefore let FD be applied to EB as many times as possible: suppose it is contained exactly twice in EB without remainder; the process is then completed, and we have found FD as the required greatest common meas- ure. 68 ELEMENTS OF GEOMETRY. The measure of each lin-e, referred to FD as the unit, will then be as follows : we have EB = 2FD, CD = 2EB +FD = 4FD + FD = 6FD, AB = SCD + EB= WFD + 2FD = 17FD. The proposed lines are therefore numerically expressed, in terms of the unit FD, by the numbers 17 and 5 j and their .. . 17 ratio IS -— . 5 33. When the preceding process is applied to two quanti- ties and no remainder can be found which is exactly con- tained in a preceding remainder, however far the process bo contmued, the two quantities have no common measure ; that is, they are incommensurable, and their ratio cannot be exactly expressed by any number, whole or fractional. 34. As the student often has difficulty in realizing the possibility of an incommensurable ratio, and imagines that if two lines are given it must be possible to take a divisor so small that it will go exactly into each of them, it seems worth while to consider at some length an important exam- ple, — namely, the ratio of the diagonal of a square to one of the sides. Let the method of (32) be applied to finding the common measure of the diagonal and a side of the square ABCD. AO 18 clearly less than twice AB ; i.e., than AB -^ BC. Lay off on J. (7 AB\ equal to AB. Our problem is now reduced to finding a common measure of B'C and AB, or its equal, CB. BOOK II. 69 Erect at B' a perpendicular B'A' to AC. A'B\ B'C, and A'B are all equal (y. Exercise 23, Book I.). If, then, we lay off CB" equal to CB\ B'G goes into BC twice, with a remainder B"A'^ by which we must proceed to divide B'C. But A' B'C is half a square, precisely similar to ABC^ and in performing the division of B'C, or its equal, A'B'^ by A'B"^ we are merely repeating, on a smaller scale, the process just performed in dividing BC loy B'C This will lead us to another repetition, on a still smaller scale, and so on indefi- nitely, and we shall never reach an exact division. The diagonal and the side of a square have then no common divisor, and are absolutely incommensurable. 35. Although an incommensurable ratio cannot be exactly expressed by a number, a number can be found by the fol- lowing method that will approximately express it, and the approximation may be made as close as we choose. Suppose that - denotes the ratio of two incommensurable quantities, A and B. Let B be divided into n equal parts, n being some number taken at pleasure ; and then let A be divided by one of these parts. Suppose A is found to con- tain this divisor m times, with a remainder, which, of course, is less than the divisor; then - is an approximation to the value of - , and an approximation that may be made as close as we please by taking a sufficiently great value of n. For, if X is the magnitude of one of the parts into which B is divided, we have B = nXj while A^ mx and < (m -|- l)x. Hence ^>^and<(^^+^)^: B nx nx 70 ELEMENTS OF GEOMETEY. that is, ^ Hes between ^ and ^ + 1 ; B n n n and by increasing n we may make -, which is the difference A between two numbers, one less and one greater than -, as small as we choose, and may thus make the less number - n A as close an approximation to the value of - as we please. As a numerical example, take the ratio of the diagonal of a square to one of the sides (34). If the side is divided into three equal parts, the diagonal will contain one of these parts four times, with a remainder less than the divisor. - is then o an approximation, though a very rough one, to the value of 4 5 the ratio in question, which must lie between - and -. o o If the side is divided into five equal parts, the diagonal 7 will contain seven of them, and - is a closer approximation. 5 141 1414 and are still closer approximations. 100 1000 ^^ 36. Definition. A proportion is an equation between two itios. '] equation A A! ratios. Thus, if the ratio - is equal to the ratio — , the 4= ^ B B' is a proportion. It may be read, " Eatio of ^ to jB equals ratio of A' to 5'," or, "J. is to B as A' is to B'r BOOK II. 71 A proportion is often written as follows : where the notation A : B is equivalent to A -i- B. When thus written, A and B' are called the extremes^ B and A' the means, and B' is called a fourth proportional to A, B and A' ; the first terms, A and A\ of the ratios are called the ante- cedents ; the second terms, B and B\ the consequents. When the means are equal, as in the proportion A:B = B:C, the middle term B is called a mean proportional between A and Cj and C is called a 3f/^^r^ proportional to ^ and j5. 37. In cases where it is necessary to prove the equality of incommensurable ratios, it is usually best to employ what is called the method of limits. . 38. Definitions. A variable quantity, or simply a variable, is a quantity whose value is supposed to change. A constant quantity, or simply a constant, is a quantity whose value is fixed. The value of a variable may be changed at pleasure, in which case it is called an independent variable ; or it may be changed by changing at pleasure the value of some other variable or variables on which it depends, and in this case it is called a dependent variable. 39. Definition. If, by changing in some specified way the variable on which it depends, we can make a dependent vari- able approach as near as we please to some given constant, but can never make the values of the variable and the constant exactly coincide ; or, in other words, if we can make the dif- ference between the variable and the constant as small as we please, but cannot make it absolutely zero, the constant is called the limit of the variable under the circumstances specified. 72 ELEMENTS OF GEOMETRY. 40. For example, consider the fraction -, where n is sup- n posed to be an independent variable, — Le.^ one whose value may be changed arbitrarily and to any extent, — the fraction - is then a dependent variable. By increasing n at pleasure, _ may be made to approach as near as we please to the value zero, but can never be made exactly equal to zero. We say, therefore, that zero is the limit of -, as n is indefi- n nitely increased. Again, the numerical approximation to the value of an incommensurable ratio (v. 35) is a dependent variable^ depend- ing upon the arbitrarily chosen number, n, of equal parts into which the denominator of the ratio is divided, and it has been shown to differ from the actual value of the ratio by an amount less than -. By increasing n at pleasure we can make this difference as small as we please, but can never make it absolutely zero, for in that case we should have found a common measure of the incommensurable numerator and denominator of the given ratio. The actual value of an incommensurable ratio is, then, the limit approached by the approximation described in (35), as n is indefinitely increased. 41. The usefulness of the method of limits flows entirely from the following fundamental theorem, the truth of which is almost axiomatic. Theorem. — If two variables dependent upon the same variable are so related that they are always equals no matter what value is given to the variable on which they depend^ and if as the inde- pendent variable is changed in some specified way, each of them approaches a limit, the two limits must be absolutely equal. For, in considering two variables that are and that always BOOK II. 73 remain equal to each other, we are dealing with a single vary- ing value, — i.e., their common value, — and it is clear that a single variable cannot be made to approach as near as we please to two diiferent constant values at the same time, as if it is once brought between the two values in question, afterward, in approaching nearer to one, it must inevitably recede from the other. The student should study this demonstration in connection with that of Proposition XII., which follows. PROPOSITION XII.— THEOREM. 42. In the same circle, or in equal circles, two angles at the centre are in the same ratio as their intercepted arcs. Let AOB and AGO be two angles at the centre of the same circle, or at the centres of equal circles; AB and AC, their intercepted arcs ; then AOB ^AB AOC AC' o 1st. Suppose the arcs to have a common measure, x, which is contained m times in AB and n times in A C Then AB =t mx and AC = nx, and AB rnx m AC nx n Apply the measure x to the arcs AB and AC, and draw radii to the points of division. The angle AOB is thus divided into m parts, and the angle AOC into n parts, all of which are equal, by Proposition III., Corollary. Call any one of these smaller angles y ; then JLO^ = my and AOC == ny, and AOB rny m AOC ny n' Therefore A^ = ^, AOC AC' or (v. 36) AOB : AOC =z AB : AC. 74 ELEMENTS OF GEOMETRY. 2d. If the arcs are incommensurable, suppose the arc AG to be divided into any arbitrarily chosen number, w, of equal parts, and let one of the parts be applied as many times as possible to the arc AB ; let B' be the last point of division, and draw the radius OB'. By construction, the arcs AB' and AG are commensurable. Therefore, by the proof above, AOB' ^ A^ AOG AG' If, now, we change n the number of parts into which A G is divided, AB' and AOB' will change, and consequently AOB' , AB' .|| , AOB' . AB' .. AOG ""^ AG ^^^^ '^"^^'- AOG ^^^ AG "'' '^'^ ^""^- bles depending upon the same variable, n. By increasing n at pleasure we can make each of the equal parts into which AC is divided as small as we please, and consequently the remainder B'B, which is necessarily less than one of these parts, can be made as small as we please. It cannot, however, be made zero, for the arcs AB and AG are incommensurable, by hypothesis. It is clear, then, that if n is indefinitely increased, AB' will have AB for its limitj and A OB' will have A OB for its limit Hence -j^^ IS the limit of -j^, and 41 is the limit of 41?', AG AG as n is indefinitely increased. BOOK II. 75 A OB' AB' As the two variables and — — , both depending upon -4.C/0 Aiy n, are always equal, no matter what the value of w, and each approaches a limit as n is indefinitely increased, the two limits in question are absolutely equal (41). Hence AOB ^AB AOG AG' PROPOSITION XIII.— THEOREM. 43. The numerical measure of an angle at the centre of a circle is the same as the numerical measure of its intercepted arc^ if the adopted unit of angle is the angle at the centre which intercepts the adopted unit of arc. Let AOB be an angle at the centre 0, and AB its intercepted arc. Let -10(7 be the angle which is adopted as the unit of angle, and let its intercepted arc AC he the arc which is adopted as the unit of arc. By Proposition XII., we have AOB ^AB AOO AC' But the first of these ratios is the measure (28) of the angle AOB referred to the unit AOC; and the second ratio is the measure of the arc J.J5 referred to the unit AC. Therefore, with the adopted units, the numerical measure of the angle AOB is the same as that of the arc AB. 44. Scholium I. This theorem, being of frequent applica- tion, is usually more briefly, though less accurately, expressed t>y saying that an angle at the centre is measured by its inter- 76 ELEMENTS OF GEOMETRY. cepted arc. In this conventional statement of the theorem, the condition that the adopted units of angle and arc cor- respond to each other is understood ; and the expression " is measured by" is used for " has the same numerical measure as." 45. Scholium II. The right angle is, "by its nature, the most simple unit of angle ; nevertheless custom has sanctioned a different unit. The unit of angle generally adopted is an angle equal to ^ part of a right angle, called a degree, and denoted by the symbol °. The corresponding unit of arc is ^^^ part of a quadrant (10), and is also called a degree. A right angle and a quadrant are therefore both expressed by 90°. Two right angles and a semi-circumference are both expressed by 180°. Four right angles and a whole circum- ference are both expressed by 360°. The degree (either of angle or arc) is subdivided into minutes and seconds, denoted by the symbols ' and " : a minute being -^ part of a degree, and a second being -^ part of a minute. Fractional parts of a degree less than one second are expressed by decimal parts of a second. An angle, or an arc, of any magnitude is, then, numeri- cally expressed by the unit degree and its subdivisions. Thus, for example, an angle equal to ^ of a right angle, as well as its intercepted arc, will be expressed by 12° 51' 25''.714 46. Definition. When the sum of two arcs is a quadrant (that is, 90°), each is called the complement of the other. When the sum of two arcs is a semi-circumference (that is, 180°), each is called the supplement of the other. See (I., 16). BOOK II. 77 47. Definitions. An inscribed angle is one whose vertex is on the circumference and whose sides are chords ; as BA C. In general, any rectilinear figure, as ABCj is said to be inscribed in a circle when its angular points are on the circum* ference; and the circle is then said to be circumscribed about the figure. An angle is said to be inscribed in a segment when its vertex is in the arc of the segment, and its sides pass through the extremities of the subtending chord. Thus, the angle BAG is inscribed in the segment BAG. PROPOSITION XIV.— THEOREM. 48. An inscribed angle is measured by one-half its intercepted arc. There may be three cases : 1st. Let one of the sides AB of the inscribed angle BAG be a diameter; then the measure of the angle BAG is one-half the arc BG. For, draw the radius OG. Then, AOG being an isosceles triangle, the angles OAG and OGA are equal (I., Proposition YIII.). The angle BOG, an exterior angle of the triangle AOG, is equal to the sum of the interior angles OAG and OGA (I., Proposition XXYI., Cor- ollary), and therefore double either of them. But the an- gle BOG, at the centre, is measured by the arc BG (44) ; therefore the angle OAG is measured by one-half the arc BG. 7* 78 ELEMENTS OF GEOMETRY. 2d. Let the centre of the circle fall within the inscribed angle BAG ; then the measure of the angle BAC is one-half of the arc BC. For, draw the diameter AD. The meas- ure of the angle BAB is, by the first case, one-half the arc BD ; and the measure of the angle GAD is one-half the arc GD ; therefore the measure of the sum of the angles BAD and GAD is one-half the sum of the arcs BD and GD ; that is, the measure of the angle BAG is one-half the arc BG. 3d. Let the centre of the circle fall without the inscribed angle BAG ; then the measure of the angle BAG is one-half the arc BG. For, draw the diameter AD. The meas- ure of the angle BAD is, by the first case, one-half the arc BD ; and the measure of the angle GAD is one-half the arc GD ; therefore the measure of the difference of the angles BAD and GAD is one-half the difference of the arcs BD and GD ; that is, the measure of the angle BAG is one-half the arc BG. 49. Corollary. An angle inscribed in a semicircle is a right angle. EXERCISE. Theorem. — The opposite angles of an inscribed quadrilateral are supplements of each other. c D BOOK II. 79 PROPOSITION XV.— THEOREM. 50. An angle formed hy a tangent and a chord is measured by one-half the intercepted arc. Let the angle BAC be formed by the tangent AB and the chord AC; then it is measured by one-half the intercepted arc AMC. For, draw the diameter AD. The angle BAD, being a right angle (Prop- osition IX.), is measured by one-half the semi-circumference AMJD ; and the angle CAD is meas- ured by one-half the arc CD; therefore the angle BAC, which is the difference of the angles BAD and CAD, is measured by one-half the difference of AMD and CD ; that is, by one-half the arc AMC. Also, the angle B'AC is measured by one-half the inter- cepted arc ANC. For, it is the sum of the right angle B'AD and the angle CAD, and is measured by one-half the sum of the semi-circumference AND and the arc CD ; that is, by one-half the arc ANC. EXERCISE. Prove Proposition XY. by the aid of this figure, OE being a radius perpendic- ular to A C. Suggestion. Complements of the same angle are equal. 80 ELEMENTS OF GEOMETRY. PROPOSITION XVI.— THEOREM. 51. An angle formed by two chords^ intersecting within the circumference^ is measured by one-half the sum of the arcs inter- cepted between its sides and between the sides of its vertical angle. Let the angle AEC be formed by the chords AB, CD, in- tersecting within the circumference ; then will it be measured by one-half the sum of the arcs AG and BD^ intercepted be- tween the sides of AEG and the sides of its vertical angle BED, For, join AD. The angle AEG is equal to the sum of the angles EDA and EAD, and these angles are measured by one-half of AG and one-half of BD, re- spectively ; therefore the angle AEG is measured by one-half the sum of the arcs AG and BD. EXERCISE. Prove Proposition XYI. by the aid of this figure, DF being draT\jn parallel to AB. (v. Proposition XI.) PROPOSITION XVII.— THEOREM. 52. An angle formed by two secants, intersecting without the circumference, is measured by one-half the difference of the inter- cepted arcs. Let the angle BA G be formed by the se- cants AB and AG; then will it be measured by one-half the difference of the arcs BG and DE. For, join GD. The angle BDG is equal to the sum of the angles DAG and AGD ; therefore the angle A is equal to the differ- BOOK II. 81 ence of the angles BDC and ACD. But these angles are measured by, one-half of BG and one-half of JDE respec- tively; hence. the angle A is measured by one-half the differ- ence of BG aad DB, EXERCISE. Prove Proposition XYII. by the aid of Proposition XI., drawing a suitable figure. PROPOSITION XVIII.-THEOEEM. 53. An angle formed by a tangent and a secant is measured by half the difference of the intercepted arcs. For the angle A is equal to BDG minus ABD, by I., Proposition XXYI., Corollary. 54. Corollary. An angle formed by two tangents is measured by half the difference of the intercepted arcs. EXERCISE. 1. Prove Proposition XYIII. and its Corollary by the aid of Proposition XI. 2. Theorem. — If through the point of contact of two tangent circles, two secants are drawn, the chords joining the points where the secants cut the circles are parallel. Suggestion. FED = CEa, . • . DBE = GAE. Consider, also, the case where the given circles are internally tangent. 82 ELEMENTS OF GEOMETRY. PROBLEMS OF CONSTRUCTION. Heretofore our figures have been assumed to be constructed under certain conditions, although methods of constructing them have not been given. Indeed, the precise construction of the figures was not necessary, inasmuch as they were only required as aids in following the demonstration of principles. We now proceed, first, to apply these principles in the solu- tion of the simple problems necessary for the construction of the plane figures already treated of, and then to apply these simple problems in the solution of more complex ones. All the constructions of elementary geometry are effected solely by the straight line and the circumference, these being the only lines treated of in the elements ; and these lines are practically drawrij or described, by the aid of the ruler and compasses, with the use of which the student is supposed to be familiar. PROPOSITION XIX.— PROBLEM. 65. To bisect a given straight line. Let AB be the given straight Kne. "With the points A and B as centres, and with a radius greater than the half of AB, describe arcs D intersecting in the two points D and E. Through these points draw the straight line DJS, which bisects AB at the point G. For, J) and B being equally distant from A and B, the straight line DE is perpendicular to AB at its middle point (I., Proposition ^ XYIII.). -iB "jk" BOOK II. 83 PROPOSITION XX.— PROBLEM. 56. At a given point in a given straight line, to erect a perpen- dicular to that line. Let AB be the given line and G the given .^ ..^ point. Take two points, D and B, in the line and at equal distances from C. With D J"d o Fb and E as centres, and a radius greater than DCov GEj describe two arcs intersecting in F. Then CF is the required perpendicular (I., Proposition XYIII.). 57. Another solution. Take any point O, without the given line, as a centre. from O to C, describe a circumference and with a radius equal to the distance ^y intersecting J.jB in G and in a second **• -'' point D. Draw the diameter DOE^ and join EG. Then EG will be the required perpendicular ; for the angle EGD^ inscribed in a semicircle, is a right angle (Proposition XIY., Corollary). This construction is often preferable to the preceding, es- pecially when the given point G is at, or near, one extremity of the given line, and it is not convenient to produce the line through that extremity. The point O must evidently be so chosen as not to lie in the required perpendicular. ,t7l?I?B 84 ELEMENTS OF GEOMETRY. PROPOSITION XXI.— PROBLEM. 58. From a given point without a given straight line, to lei fall a perpendicular to that line. o Let AB be the given line and G the given point. With C as a centre, and with a radius a — "-^^ — I — r^rr- — b sufficiently great, describe an arc in- tersecting AB in D and E. With D and E as centres, and a radius greater than the half of DE^ describe two arcs intersecting in F. The line CF is the required perpendicular (I., Proposition XYIII.). 59. Another solution. With any point in the line AB as a centre, and with the radius 0(7, describe an arc GDE intersecting AB in D. With i) as a centre, and a radius equal to the distance DC, describe an arc in- ^ tersecting the arc CDE in E. The line /% CE is the required perpendicular. For, the point I) is the middle of the arc CJDE^ and the radius OD drawn to this point is perpendicular to the chord CE (Proposition YI.). li) PROPOSITION XXII.-PROBLEM. 60. To bisect a given arc or a given angle. Ist. Let AB be a given arc. Bisect its chord AB by a perpendicular, as in (55). This perpendicular also bisects the arc (Proposition YI.). BOOK II. 8) 2d. Let BAC be a given angle. With ^ as a centre, and with any radius, de- scribe an arc intersecting the sides of the angle in D and E, With D and E as centres, and with equal radii, describe arcs intersecting in F. The straight line AF bisects the arc DE^ and consequently also the angle BAG, 61. Scholium. By the same construction, each of the halves of an arc, or an angle, may be bisected ; and thus, by succes- sive bisections, an arc, or an angle, may be divided into 4, 8, 16, 32, etc., equal parts. PROPOSITION XXIII.— PROBLEM. 62. At a given point in a given straight line, to construct an angle equal to a given angle. Let J. be the given point in the straight line AB, and the given angle. With as a centre, and with any radius, describe an arc MJS^ terminated by the sides of the angle. With A as a centre, and with the same radius, OM, describe an indefi- nite arc BC. With ^ as a centre, and with a radius equal to the chord of MJ^, de- scribe an arc intersecting the indefinite arc BC in D. Join AI). Then the angle BAD is equal to the angle 0. For the chords of the arcs MN and BD are equal; therefore these arcs are equal, and consequently also the angles and A. 86 ELEMENTS OF GEOMETRY. PROPOSITION XXIV.— PROBLEM. 63. Through a given point, to draw a parallel to a given straight line. Let A be the given point, and JBC ^ the given line. ,\ From any point B in BO draw the a/. — \ jp straight line BAD through A. At the point A, by the preceding problem, b ' ^ construct the angle DAE equal to the angle ABC. Then AB is parallel to BC (I., Proposition XXIY., Corollary L). 64, Scholium. This problem is, in practice, more accurately solved by the aid of a triangle, con- structed of wood or metal. This triangle has one right angle, and its acute angles are usually made equal to 30° and 60°. Let A be the given point, and BG the given line. Place the tri- angle, BJFD, with one of its sides in coincidence with the given line BO. Then place the straight edge of a ruler, MN, against the side EF of the triangle. Now, keeping the ruler firmly fixed, slide the triangle along its edge until the side ED passes through the given point A. Trace the line EAD along the edge ED of the triangle ; then it is evident that this line will be parallel to BG. EXERCISE. Prohlem. Two angles of a triangle being given, to find the third, (v. I., Proposition XXYI., and I., Proposition III., Cor- ollary I.) BOOK II. ^7 PROPOSITION XXV.— PROBLEM. 65. Two sides of a triangle and their included angle being given, to construct the triangle. y & Let 6 and c be the given sides, and Xj^ A their included angle. Draw an indefinite line AE, and construct the angle EAF=^A. On AE take AC = b, and on AF take AB = c; join BC. Then ABC is the triangle required ; for it is formed with the data. With the data, two sides and the included angle, only one triangle can be constructed ; that is, all triangles constructed with these data are equal, and thus only repetitions of the same triangle (I., Proposition YL). 66. Scholium. It is evident that one triangle is always pos- sible, whatever may be the magnitude of the proposed sides and their included angle. PROPOSITION XXVL— PROBLEM. 67. One side and two angles of a triangle being given, to construct the triangle. Two angles of the triangle being given, ^^ ^^^^ the third angle can be found; and we c shall therefore always have given the two angles adjacent to the given side. Let, then, c be the given side, A and B the angles adjacent to it. Draw a line AB = c ; at J. make an angle BAD = A, and at B an angle ABE = B. The lines AD and BE intersecting in (7, we have ABC as the required triangle. With these data but one triangle can be constructed (I., Proposition YIL). 88 ELEMENTS OF GEOMETRY. 68. Scholium. If the two given angles are together equal to or greater than two right angles, the problem is impos- sible ; that is, no triangle can be constructed with the data ; for the lines AD and BG will not intersect on that side of AB on which the angles have been constructed. PROPOSITION XXVII.— PROBLEM. 69. The three sides of a triangle being given, to construct the triangle, a Let a, bj and c be the three given sides. 6. . Draw BC = a; with (7 as a centre and * a radius equal to b describe an arc ; with .B as a centre and a radius equal to c de- scribe a second arc intersecting the first in A. Then ABC is the required triangle. With these data but one triangle can be constructed (I., Proposition IX.). 70. Scholium. The problem is impossible when one of the given sides is equal to or greater than the sum of the other two (I., Axiom I.). PROPOSITION XXVIII.— PROBLEM. 7L Two sides of a triangle and the angle opposite to one of them being given, to construct the triangle. i a 1st. When the given angle A is acute, and the given side a, oppo- r/ site to it in the triangle, is less than VwiXo the other sjiven side c. X 7 I \ Construct an angle DAE = A. ^ ^,' '^z, ^ In one of its sides, as AD, take AB = c; with J5 as a centre and a radius equal to a, describe BOOK II. 89 an arc which (since a <^ c) will intersect AE in two points, C" and C", on the same side of A. Join BC and BG", Then either ABC or ABC" is the required triangle, since each is formed with the data ; and the problem has two solutions. There will, however, be but one solution, even with these data, when the side a is so much less than the side c as to be just equal to the perpendicular from B upon AE. For then the arc described from -S as a centre, and with the radius a, will touch AE m2i single point, (7, and the required triangle will be ABG^ right angled at C. 2d. When the given angle A is either acute, right, or obtuse, and the side a opposite to it is greater than the other given side c. The same construction being "^(""a' ^^^ made as in the first case, the arc described with j5 as a centre, and with a radius equal to a, will intersect AE in only one point, C, on the same side of A. Then ABC will be the triangle required, and will be the only- possible triangle with the data. The second point of intersection, C", will fall in EA pro- duced, and the triangle ABC thus formed will not contain the given angle. 72. Scholium. The problem is impossible when the given angle A is acute and the proposed side opposite to it is less than the perpendicular from B upon AE; for then the arc described from B will not intersect AE. The problem is also impossible when the given angle is right, or obtuse, if the given side opposite to the angle is less than the other given side ; for either the arc described from B would not intersect AEj or it would intersect it only when produced through A. 90 ELEMENTS OF GEOMETRY. EXERCISE. Problem. — The adjacent sides of a parallelogram and their included angle being given, to construct the parallelogram. PROPOSITION XXIX.— PROBLEM. 73. To find the centre of a given circumference, or of a given arc. Take any three points, A, B, and (7, in the given circumference or arc, and join them by chords AB, BC. The perpen- diculars erected at the middle points of these chords will intersect in the required centre (Proposition YI., Corollary I.). 74. Scholium I. Only one solution is possible ; for, since the centre is equidistant from B and (7, it must lie in the perpen- dicular erected at the middle point of BC (I., Proposition XYIII.), and since it is equidistant from A and B, it must lie in the perpendicular erected at the middle point of AB ; and these perpendiculars can have but one point in common. 75. Scholium IL The same construction serves to describe a circumference which shall pass through three given points, Aj B, C; or to circumscribe a circle about a given triangle, ABC; that is, to describe a circumference in which the given triangle shall be inscribed (47). 76. Scholium III. It follows from Scholium 1. that three points not in the same straight line will determine a circum- ference, — i.e, through three points not in the same straight line one circumference, and only one, can be drawn. Hence two circumferences cannot intersect in more than two points; for if they had three points in common they would coincide throughout. BOOK II. 91 PROPOSITION XXX.— PROBLEM. 77, At a given point in a given circumference^ to draw a tan- gent to the circumference. Let A be the given point in the given circumference. Draw the radius OA^ and at A draw BAC perpendicular to OA ; BG will be the required tangent (Propo- sition IX.). If the centre of the circumference is not given, it may first be found by the preceding problem, or we may proceed more directly as follows : take two points, D and E^ equidistant from A ; draw the chord DE^ and through A draw BAC parallel to BE. Since A is the middle point of the arc BE^ the radius drawn to A will be perpendicular to BE (Proposition YI.), and con- sequently also to BG ; therefore jBC is a tangent at A, PROPOSITION XXXI.— PROBLEM. 78. Through a given point without a given circle^ to draw a tangent to the circle. Let be the centre of the given circle and P the given point. Upon OP, as a diameter, describe a cir- cumference intersecting the circumference of the given circle in two points, A and A'. Draw PA and PA\ both of which will be tangent to the given circle. For, drawing the radii OA and OA', the angles OAP and OA'P are right 92 ELEMENTS OP GEOMETRY. angles (Proposition XIY., Corollary) ; therefore PA and PA^ are tangents (Proposition IX.). In practice, this problem is accurately- solved by placing the straight edge of a ruler through the given point and tangent to the given circumference, and then tracing the tangent by the straight edge. The pre- cise point of tangency is then determined by drawing a perpendicular to the tangent from the centre. 79. Scholium, This problem always admits of two solutions. EXERCISE. Problem. — To draw a common tangent to two given circles. Suggestion. For an ex- terior common tangent, in the larger circle draw a concentric circle whose radius is the difference of the radii of the given circles. For an interior common tangent, about one of the circles draw a concentric circle whose ra- dius is the sum of the radii of the given circles. BOOK II. 93 PROPOSITION XXXII.— PROBLEM. 80. To inscribe a circle in a given triangle. Let ABC be the given triangle. Bisect any two of its angles, as B and (7, by straight lines meet- ing in 0. From the point let fall per- pendiculars OX), OE^ OF, upon the three sides of the triangle ; these perpendiculars will be equal to each other (I., Proposi- tion XIX.). Hence the circumference of a circle, described with the centre 0, and a radius = OD, will pass through the three points D, E, F, will be tangent to the three sides of the triangle at these points (Proposition IX.), and will therefore be inscribed in the triangle. EXERCISE. Problem. — Upon a given straight line, to describe a segment which shall contain a given angle. Suggestion. Through one end of the given line AB draw a line BC, making with it the given angle. The two lines will be one a chord and the other a tangent. Hence the centre of the circle can be found. EXERCISES ON BOOK II. THEOREMS. 1. If two circumferences are tangent in- ternally, and the radius of the larger is the diameter of the smaller, then any chord of the larger drawn from the point of contact is bisected by the circumference of the smaller {v. Proposition XIV., Corollary, and Proposition VI.). 2. If two equal chords intersect within a circle, the segments of one are respectively equal to the segments of the other. What is the corresponding theorem for the case where the chords meet when produced? 3. A circumference described on the hypotenuse of a right tri- angle as a diameter passes through the vertex of the right angle. (v. Proposition XIV., Corollary.) 4. The circles described on two sides of a triangle as diameters intersect on the third side. Suggestion, Drop a perpendicular from the opposite vertex upon the third side. 5. The perpendiculars from the angles upon the opposite sides of a triangle are the bisectors of the angles of the triangle formed by joining the feet of the perpendiculars. Suggestion. On the three sides of the given triangle as diam- eters describe circumferences, {v. Exercise 3, Proposition XIV., and I., Proposition XXVI.). 6. If a circle is circumscribed about an equilateral triangle, the perpendicular from its centre upon a side of the triangle is equal to one-half of the radius. 94 BOOK II. 95 7. The portions of any straight line which are intercepted between the circumferences of two concentric circles are equal. 8. Two circles are tangent internally at P, and a chord AB of the larger circle touches the smaller at C; prove that PC bisects the angle APB. Suggestion. CPQ = BCP, BPQ = BAP, BCP—BAP = APC (I., Propo- sition XXVI., Corollary). 9. If a triangle ABC is formed by the intersection of three tangents to a circumference, two of which, A3f and AN, are fixed, while the third, BC, touches the circumfer- ence at a variable point P, prove that the perimeter of the triangle ^^C is constant, and equal to AM -\- AN, or 2 AN (Proposition X.). Also, prove that the angle P 0(7 is constant. 10. If through one of the points of intersection of two circum- ferences a diameter of each circle is drawn, the straight line which joins the extremities of these diameters passes through the other point of intersection, and is parallel to the line joining the centres. Suggestion. Draw the common chord and the line joining the centres, (v. Proposition VI., Corollary II., and Exercise 29, Book I.) 11. The difference between the hy- potenuse of a right triangle and the sum of the other two sides is equal to the diameter 'of the inscribed circle. I 96 ELEMENTS OF GEOMETRY. 12. A circle can be entirely sur- rounded by six circles having the same radius with it. 13. The bisectors of the vertical angles of all triangles having the same base and equal vertical angles have a point in common. Suggestion, The triangles may all be inscribed in the same circle. 14. If the hypotenuse of a right triangle is double one of the sides, the acute angles of the triangle are 30° and 60° respectively. 15. If, from a point whose distance from the centre of a given circle is equal to a diameter, tangents are drawn to the circle, they will make with each other an angle of 60°. LOCI. 16. Find the locus of the centre of a circumference which passes through two given points, (v. I., Proposition XVIII.) 17. Find the locus of the centre of a circumference which is tangent to two given straight lines, {v. I., Proposition XIX.) 18. Find the locus of the centre of a circumference which is tangent to a given straight line at a given point of that line, or to a given circumference at a given point of that circumference. 19. Find the locus of the centre of a circumference passing through a given point and having a given radius. 20. Find the locus of the centre of a circumference tangent to a given straight line and having a given radius. 21. Find the locus of the centre of a circumference of given radius, tangent externally or internally to a given circumference. BOOK II. 97 22. A straight line MN, of given length, is placed with its extremities on two given perpendicular lines ^^, CD; find the locus of its middle point P (Exercise 31, Book I.). 23. A straight line of given length is inscribed in a given circle; find the locus of its middle point, {v. Proposition VII.) 24. A straight line is drawn through a given point A^ intersecting a given circum- ference in B and C; find the locus of the middle point, P, of the intercepted chord BC. Note the special case in which the point A is on the given circumference. 25. From any point ^ in a given circumference, a straight line -4P of fixed length is drawn parallel to a given line MN ; find the locus of the extremity P, {v. I., Proposition XXX.) 26. From one extremity -4 of a fixed diameter ABy any chord AC is drawn, and at (7 a tangent CD. From P, a per- pendicular BD to the tangent is drawn, meeting AC in P, Find the locus of P. Suggestion. (Draw radius OC. v. I., Exercise 28.) 27. The base BC of a triangle is given, and the medial line BE, from jB, is of a given length. Find the locus of the vertex A. Suggestion. Draw AO parallel to EB. Since BO=BC, O is a fixed point; and since AO = 2BEy OA is a constant distance. 98 ELEMENTS OF GEOMETRY. PROBLEMS. The most useful general precept that can be given, to aid the student in his search for the solution of a problem, is the follow- ing : Suppose the problem solved, and construct a figure accord- ingly ; study the properties of this figure, drawing auxiliary lines when necessary, and endeavor to discover the dependence of the problem upon previously solved problems. This is an analysis of the problem. The reverse process, or synthesis^ then furnishes a construction of the problem. In the analysis, the student's in- genuity will be exercised especially in drawing useful auxiliary lines ; in the synthesis, he will often find room for invention in combining in the most simple form the several steps suggested by the analysis. The analysis frequently leads to the solution of a problem by the intersection of loci. The solution may turn upon the deter- mination of the position of a particular point. By one condition of the problem it may appear that this required point is neces- sarily one of the points of a certain line ; this line is a locus of the point satisfying that condition. A second condition of the problem may furnish a second locus of the point ; and the point is then fully determined, being the intersection of the two loci. Some of the following problems are accompanied by an analysis to illustrate the process. 28. To determine a point whose distances from two given inter- secting straight lines, AB^ A^B\ are given. •.... c Analysis. The locus of all the points which are at a given distance from AB consists of two parallels to ABy CE, and DF, each at the given distance from AB, The locus of all the points at a given distance from A^B^ consists of two parallels, C^E^ and D^F^^ each at the given distance from A^B\ The re- quired point must be in both loci, and therefore in their inter- section. There are in this case four intersections of the loci, and the problem has four solutions. Construction. At any point of AB, as A, erect a perpendicular CD, and make AC = AD = the given distance from AB ; through O and 2) draw parallels to AB. In the same manner, draw par- allels to A^B^ at the given distance A^C = A'D\ The intersec- E \ \ / A \»* ^^ \ /'o '\ \ y*. F BOOK II. 99 tion of the four parallels determines the four points P^, Pj? P^^ ^4» each of which satisfies the conditions. 29. Given two perpendiculars, AB and CD, intersecting in O, to construct a square, one of whose angles shall coincide with one of the right angles at O, and the vertex of the opposite angle of the square shall lie on a given straight line EF, (Two solutions.) 30. In a given straight line, to find a point equally distant from two given points without the line. 31. To construct a square, given its diagonal. 32. Through a given point P within a given angle, to draw a straight line, terminated by the sides of the anglfe, which shall be bisected at P. {v. Exercise 28, Book I.) 33. Given two straight lines which caimot be produced to their intersection, to draw a third which would pass through their intersection and bisect their contained angle. Suggestion. Find two points equidistant from the two lines. (v. I., Proposition XIX.) 34. Given the middle point of a chord in a given circle, to draw the chord. 36. To draw a tangent to a given circle which shall be parallel to a given straight line. 36. To draw a tangent to a given circle, such that its segment intercepted between the point of contact and a given straight line shall have a given length. Suggestion. The tangent, the radius drawn to the point of con- tact, and a line drawn from the centre to the end of the tangent form a right triangle, two of whose sides are known. A simple construction gives the hypotenuse. In general there are four solutions. Show when there will be but two ; also, when no solution is possible. 37. Through a given point within or without a given circle, to draw a straight line, intersecting the circumference, so that the intercepted chord shall have a given length. (Two solutions.) {v. Exercise 23 and Section 78.) 38. Construct an angle of 60°, one of 120°, one of 30°, one of 150°, one of 45°, and one of 135°. 100 ELEMENTS OF GEOMETRY. 39. Construct a triangle, given the base, the angle opposite to the base, and the altitude. Analysis. Suppose BAC to be the re- ^ s^^ quired triangle. The side BC being fixed ^ 7^^. ^\\ ^ in position and magnitude, the vertex A is / / ^X; 1 1 to be determined. One locus of A is an k.^^ '^-vV arc of a segment, described upon AB, con- ^ ^ taining the given angle. Another locus of J. is a straight line MN drawn parallel to ^(7, at a distance from it equal to the given altitude. Hence the position of A will be found by the intersection of these two loci, both of which are readily constructed. Limitation. If the given altitude were greater than the perpen- dicular distance from the middle of ^C to the arc BAC, the arc would not intersect the line MN, and there would be no solution possible. The limits of the data within which the solution of any prob- lem is possible should always be determined. 40. Construct a triangle, given the base, the medial line to the base, and the angle opposite to the base. 41. With a given radius, describe a circumference, 1st, tangent to two given straight lines ; 2d, tangent to a given straight line and to a given circumference ; 3d, tangent to two given circum- ferences ; 4th, passing through a given point and tangent to a given straight line ; 5th, passing through a given point and tan- gent to a given circumference ; 6th, having its centre on a given straight line, or a given circumference, and tangent to a given straight line, or to a given circumference. (Exercises 19, 20, 21.) 42. Describe a circumference, 1st, tangent to two given straight lines, and touching one of them at a given point (Exercises 17, 18) ; 2d, tangent to a given circumference at a given point and tangent to a given straight line ; 3d, tangent to a given straight line at a given point and tangent to a given circumference (Exer- cise 18) ; 4th, passing through a given point and tangent to a given straight line at a given point ; 5th, passing through a given point and tangent to a given circumference at a given point. 43. Draw a straight line equally distant from three given points. When will there be but three solutions, and when an indefinite number of solutions ? 44. Inscribe a straight line of given length between two given circumferences, and parallel to a given straight line. {v. Exer- cise 25.) BOOK III. PROPORTIONAL LINES. SIMILAR FIGURES. THEORY OF PROPORTION. 1. Definition. One quantity is said to be proportional to another when the ratio of any two values, A and JB, of the first, is equal to the ratio of the two corresponding values A' and 5', of the second ; so that the four values form the proportion (II., 36) A:B=A':B% or B B'' This definition presupposes two quantities, each of which can have various values, so related to each other that each value of one corresponds to a value of the other. An exam- ple occurs in the case of an angle at the centre of a circle and its intercepted arc. The angle may vary, and with it also the arc ; but to each value of the angle there corresponds a certain value of the arc. It has been proved (II., Proposi- tion XII.) that the ratio of any two values of the angle is equal to the ratio of the two corresponding values of the arc ; and, in accordance with the definition just given, this proposition would be briefly expressed as follows : " The angle at the centre of a circle is proportional to its intercepted arc." 2. Definition. One quantity is said to be reciprocally propor- tional to another when the ratio of two values, A and B^ of the first, is equal to the reciprocal of the ratio of the two «* 101 102 ELEMENTS OF GEOMETRY. corresponding values, A' and B\ of the second, so that the four values form the proportion A\B=B':A\ or A^B^ ^. _^ ^ B A' ' B'' For example, if the product p of two numbers, x and y, is given, so that we have xy=p, ^ then X and y may each have an indefinite number of values, but as X increases y diminishes. If, now, A and B are two values of rr, while A' and B' are the two corresponding values of ?/, we must have A XA' = p, BX B' = p, whence, by dividing one of these equations by the other, B^ B'~ ' and therefore A _ J_ ^ ^' . B A^ A" B' that is, two numbers whose product is constant are reciprocally proportional. 3. Let the quantities in each of the couplets of the pro- portion j = §,ovA:B = A':B', [1] be measured by a unit of their own kind, and thus expressed BOOK III. 103 by numbers (II., 28) ; let a and h denote the numerical meas- ures of A and B, a' and h' those of A' and B' ; then (II., 29) A^a A!. =9l B 6' B' b" and the proportion [1] may oe replaced by the numerical pro- portion 4. Conversely, if the numerical measures a, 6, a', h\ of four quantities, A^ J5, A\ B', are in proportion, these quantities themselves are in proportion, provided that A and B are quantities of the same kind, and J.' and B' are quantities of the same kind (though not necessarily of the same kind as A and B) ; that is, if we have a\h = a' :h\ we may, under these conditions, infer the proportion A:B = A' :B'. 5. Let us now consider the numerical proportion a\h = a' :h\ Writing it in the form and multiplying both members of this equality by hh\ we obtain ah' = a'b, whence the theorem : the product of the extremes of a (numer- icaT) proportion is equal to the product of the means. 4ii»tt. 104 ELEMENTS OP GEOMETRY. Corollary. If the means are equal, as in the proportion a : b = b : Cj we have b^ = ac, whence b = j/ac; that is, a mean proportional (II., 36) between two numbers is equal to the square root of their product. 6. Conversely, if the product of two numbers is equal to the product of two others, either two may be made the extremes, and the other two the means, of a proportion. For, if we have given ab' = a^bj then, dividing by bb', we obtain ^ = ^'ora:b = a':b\ b b Corollary. The terms of a proportion may be written in any order which will make the product of the extremes equal to the product of the means. Thus, any one of the following proportions may be inferred from the given equal- ity ab' = a'b : a :b =a' :b\ a :a'=b :6', b :a=b' :a'. b :b' = a :a', U '. a' =b '.a, etc. Also, any one of these proportions may be inferred from any other. 7. Definitions. When we have given the proportion a-.b = a' :b', and infer the proportion a: a' =b :¥, the second proportion is said to be deduced by alternation. When we infer the proportion b : a = b' '. a', this proportion is said to be deduced by inversion. BOOK III. 105 8. It is important to observe that when we speak of the products of the extremes and means of a proportion, it is implied that at least two of the terms are numbers. If, for example, the terms of the proportion A:B=A': B' are all lines^ no meaning can be directly attached to the products ^ X ^'j ^ X A\ since in a product the multiplier at least must be a number. But if we have a proportion such as A \ B = m : n^ in which m and n are numbers, while A and B are any two quantities of the same kind, then we may infer the equality nA = mB. Nevertheless, we shall, for the sake of brevity, often speak of the product of two lineSj meaning thereby the product of the numbers which represent those lines when they are measured hy a common unit 9. If A and B are any two quantities of the same kind, and m any number whole or fractional, we have, identically, mA A , mB~ B' that is, equimultiples of two quantities are in the same ratio as the quantities themselves. Similarly, if we have the proportion A:B = A':B% and if m and n are any two numbers, we can infer the pro- portions mA : mB = nA' : nB\ mA : nB = mA' : nB\ 106 ELEMENTS OF GEOMETRY. 10. Composition and Division. If we have given the pro- A A' portion - = — , we have, by alternation, B B' A' B'' Let r be the common value of these two ratios ; then 4, = r,and|=r, and A = rA', and B == rB\ Adding the second equation to the first, we have A + B^KA'+B'), or A_±B_ ^ A^B_ A' + B' A' B'' The proportions ^^±|-, = A, and ^,i|, = | are said to be formed from the given proportion A:B^=A':B\hj composition. If we subtract the equation B = rB' from A = rA!^ we have A—B = r(^Al — B'), whence, as above, A —B ^A A' — B' A" and A —B ^B A' — B' B" two proportions which are said to be formed from the given proportion A : B = A' : B^hj division. 11. Definition. A continued proportion is a series of equal ratios, as A:B=zA' :B'=:A" : B" = A'" : B"' = etc. i BOOK III. 107 12. Let r denote the common value of the ratio in the continued proportion of the preceding article ; that is, let B B' B" B"' then we have A=::Br, A'=B'r, A" = B'^r, A"' = B'^'r, etc., and, adding these equations, A-{.A' + A''+ A'" + etc. = (B-{-B' + B" + B'" + etc.) r, whence ^ + ^^+^"+^"^+etc. ^^^A^A^^^^^, BJ^B'^B^'^ B"' + etc. B B' that is, the sum of any number of the antecedents of a continued proportion is to the sum of the corresponding consequents as any antecedent is to its consequent. In this theorem the quantities J., 5, (7, etc., must all be quantities of the same kind. If, instead of a continued proportion, we have an ordinary proportion, the theorem just proved obviously holds good. 13. If we have any number of proportions, as a '. h =^ c '. dj a'-.h' = c' : d\ a" : 6" = c" : d'\ etc. ; then, writing them in the form a c a! c^ a" c" h~ d' h'~d" W~dr' ' and multiplying these equations together, we have aa' a" . . . c d d' . . . hh'h'' ...~ dd'd'\.: or aaW ...'.hh'h" ... = cdd'...:dd'd"...] that is, if the corresponding terms of two or more proportions are multiplied together, the products are in proportion. 108 ELEMENTS OF GEOMETRY. If the corresponding terms of the several proportions are equal, that is, ii a z= a' = a", b = b' = V'^ etc., then the mul- tiplication of two or more proportions gives that is, if four numbers are in jprojportion^ like powers of these numbers are in proportion. PEOPOETIONAL LI:N^ES. PROPOSITION I.— THEOREM. 14. A parallel to the base of a triangle divides the other two sides proportionally. Let DE be a parallel to the base, BC, of the triangle ABC; then AB:AD=AC :AE. ^ 1st. Suppose the lines J.J5, AD, to have # i^ a common measure which is contained m times in AB and n times in AD. Then / [S I ^^u' AB_m L A AD n ^ ^o Apply this measure to AB, and through the points of division draw lines parallel to the base BC o^ the triangle ; then through the points of intersection of these lines with A G draw lines parallel to AB. The small triangles thus formed are all equal, by Propositions XXIX. and YII., Book I. Hence the m parts into which ACiQ divided are all equal, and, as AE contains n of these parts, AC ^m AE n' Therefore AB ^ AC AD AE BOOK III. 109 2d. If AB and AD are incommensurable, let AD be divided into any arbitrarily chosen number n of equal parts, and let AB be divided by one ^ of these parts. Let B' be the last point of A division, 5 '5 beinff of course less than the di- / \ visor. Through B' draw B' C parallel to DE, / \ ~ Since AB' and AD are commensurable, / \ A W A C I \ = — — -, and this holds true no matter i//-— -\c" AD AE t \ what value may be given to n. By taking a sufficiently great value for /i, we can make B' come as near ''is we please to B ; but we cannot make B' and B coincide, since no divisor of AD can divide AB without remainder. AB' AC AB' and AC, and consequently and , are then ^ ^ AD AE' variables dependent upon the same variable, n; and, as we have seen above, they are equal, no matter what value is given to n. If n is indefinitely increased, and Therefore, by the fundamental theorem in the Doctrine of Limits (41, Book II.), these limits are equal, and therefore AB ^ AC AD~ AE' \ Compare this reasoning with that in II., 42. \ EXERCISE. \ Show that in Proposition I. AD : DB = AE : EC Q>. 10), and also that AB ^AD ^ DB AC AE EC' 10 AB' AD has the limit AB AD' AC AE has the limit AC AE 110 ELEMENTS OF GEOMETRY. PKOPOSITION II.— THEOREM. 15. Conversely, if a straight line divides two sides of a tri- angle proportionally, it is parallel to the third side. Let DE divide the sides AB, AC, of the tri- angle ABC, proportionally; then DE is par- allel to BC For, if DE is not parallel to BC, let some other line DE', drawn through D, be parallel to BC. Then, by Proposition I., AB:AD=AC:AE'. But, by hypothesis, we have AB:AD = AC:AE. Hence AC^ AE' AG AJEf whence it follows that AE' = AE, which is impossible unless DE' coincides with DE. Therefore DE is parallel to BC. EXERCISE. 1. Theorem. — The line bisecting the vertical angle of a triangle divides the base into segments proportional to the adjacent sides of the triangle. Suggestion. Through B draw a line parallel to the bisector and extend the side CA to meet it. The triangle EAB is isos- celes. .'.AE = AB. CE and CB are divided proportionally (Proposition I.). Hence CD : DB = CA : AB. 2. Prove the converse of Exercise I. (v. Proposition II.) BOOK III. Ill SIMILAR POLYGONS. 16. Definitions. Two polygons are similar when they are mutually equiangular and have their homologous sides pro- portional. In similar polygons, any points, angles, or lines, similarly situated in each, are called homologous. The ratio of a side of one polygon to its homologous side in the other is called the ratio of similitude of the polygons. PROPOSITION III.— THEOREM. 17. Two triangles are similar when they are mutually equi- angular. Let ABC, A'B'C, be mutually ^ equiangular triangles, in which y^i -4' A=A', B = B\ C= C; then / / ^f these triangles are similar. / / / / For, superpose the triangle b c b! c A'B'C upon the triangle ABC, making the angle A' coincide with its equal, the angle A, and let B' fall at h and C at e. Since the angle Abe is equal to B, he is parallel to BC (I., Proposition XXIY., Corollary I.), and we have (Proposition I.) AB:Ah=AC '.Ac, or AB\A'B'=^AC'.A'C'. If, now, we superpose A'B'C upon ABC, making B' co- incide with B, we may prove, in the same manner, that AB'.A'B'=BC'.B'C'; and, combining these proportions, AB ^ AC_ ^ BC_ PJ-, A'B' A'C B'C '- ^ Therefore the homologous sides are proportional, and the tri- angles are similar (16). / 112 ELEMENTS OF GEOMETRY. 18. Scholium I. The homologous sides lie opposite to equal angles. 19. Scholium II. The ratio of similitude (16) of the two similar triangles is any one of the equal ratios in the con- tinued proportion [1]. EXERCISE. Theorem. — The altitudes of two similar triangles are to each other in the ratio of similitude of the triangles. PROPOSITION IV.— THEOREM. 20. Two triangles are similar when an angle of the one is equal to an angle of the other, and the sides including these angles are proportional. In the triangles ABC, A'B'C, ^ ^' let A = A', and yf /j AB AC , .X / X / A'B' A'C' / Bf G' then these triangles are similar. ^ ^ For, place the angle A' upon its equal angle A ; let B' fall at h, and C at c. Then, by the hypothesis, ^ AB^AC Ah Ac Therefore he is parallel to BC (Proposition II.), and the tri- angle Ahc is similar to ABC (Proposition III.). But Ahc is equal to A'B'C; therefore A'B'C is also similar to ABC. BOOK III. 113 PROPOSITION v.— THEOREM. 21. Two triangles are similar when their homologous sides are propoj'tional. In the triangles ABC, A'B'C, let AB_^AG_^BG_, ry. A'B A'C B'G" •- ■' then these triangles are similar. ^ ^' For, take Ah == A!B', and Ac ' /I /I = ^'(7', and join .6 and c. /^ / / / Abe is similar to ABC, by Prop- y^' J n c osition lY. Therefore b o AB BO AB BC -, or Ab be ' A'B' be But, by hypothesis, AB ^ BC A'B' B'C' Hence The triangle A!B'G' is then equal to Abe, by I., Proposition IX., and is consequently similar to ABC. PROPOSITION VI.— THEOREM. 22. If two polygons are composed of the same number of tri- angles similar each to each and similarly placed, the polygons are similar. Let the polygon ABCD, /"^^^•^^^^ /t^^^ etc., be composed of the tri- ^/:;.'.. \^ A'k::y.. \n angles J.5 (7, A CD, etc.; and \ " '"^^^ V---^'^' let the polygon A'B'C'D\ V-^-^""''^ etc., be composed of the triangles A'B'C\ A'C'D\ etc., similar to ABC, ACD, etc., 2^ 10* -^ 114 ELEMENTS OF GEOMETEY. respectively, and similarly placed; then the polygons are similar. 1st. The polygons are mutually equiangular. For, the homologous angles of the similar triangles are equal ; and any two corre- sponding angles of the poly- gons are either homologous angles of two similar triangles, or sums of homologous angles of two or more similar trian- gles. Thus, B = B\; BCD = BCA -f ACD = B'C'A' + A'C'D'=B'C'D' ; etc. 2d. Their homologous sides are proportional. For, from the similar triangles, we have AB ^ BG ^ AC ^ CD ^ AD ^ DE ^ A'B' B'C A'C CD' A'D' D'E' ^ ^' Therefore the polygons fulfil the two conditions of similarity (16). \y— PEOPOSITION VII.— THEOREM. 23. Conversely, two similar polygons may be decomposed into the same number of triangles similar each to each and similarly placed. Let ABCD, etc., A'B'C'D', etc., be two similar polygons. From two homologous ver- tices, A and A', let diag- onals be drawn in each polygon; then the poly- gons will be decomposed as required. For, 1st. We have, by the definition of similar polygons, AB BC Angle B = B\ and A'B' B'C BOOK III. 115 therefore the triangles J.5(7and A^B'C are similar (Proposi- tion lY.). 2d. Since ABC and A!B'G' are similar, the angles BGA and B'C'A! are equal; subtracting these equals from the equals BCD and B'C'I)\ respectively, there remain the equals ACJ) and A' CD', Also, from the similarity of the triangles ABC and A'B'C, and from that of the polygons, we have AC BC CD A'C B'C CD'' therefore the triangles ACD and A' CD' are similar (Propo- sition lY.). Thus, successively, each triangle of one polygon may be shown to be similar to the triangle similarly situated in the other. PKOPOSITION VIII.— THEOREM. 24. The 'perimeters of two similar polygons are in the same ratio as any two homologous sides. For we have (see preceding figures) ^^ ^^ ^^ etc.; A'B' B'C CD' whence (12) AB -{- BC + CD -{- etc. ^ AB ^ BC A'B' + B'C+ CD' + etc. A'B' B'C etc. w .1^^ 116 ELEMENTS OF GEOMETRY. PROPOSITION IX.— THEOREM. 25. If a perpendicular is drawn from the vertex of the right angle to the hypotenuse of a right triangle : 1st. The two triangles thus formed are similar to each other and to the whole triangle ; 2d. The perpendicular is a mean proportional between the seg- ments of the hypotenuse ; 3d. Each side about the right angle is a mean proportional between the hypotenuse and the adjacent segment. Let G be the right angle of the triangle ABC, and CD the perpendicular to the hypotenuse; then, 1st. The triangles ACD and CBD are similar to each other and to ABC. For the triangles ACD and ABC have the angle A common, and the right angles ADCj ACBj equal; therefore they are mutually equiangular, and consequently similar (Proposition III.). For a like reason CBD is similar to ABCj and consequently also to ACD. 2d. The perpendicular CD is a mean proportional between the segments AD and DB. For the similar triangles ACD, CBD, give AD:CD = CD : BD. 3d. The side J. (7 is a mean proportional between the hy- potenuse AB and the adjacent segment AD. For the similar triangles, ACD, ABC, give AB:AC = AC:AD. In the same way the triangles CBD and ABC give AB:BC = BC :BD. BOOK III. 117 26. Scholium. If the lengths of the lines in the figure above are expressed in terms of the same unit, the results just obtained may be written (5, Corollary) Vlf =ABX I>B, TO' = AB X AD, 'mJ' = AB X BD. 27. Corollary. If from any point in the circumference of a circle a perpendicular is let fall upon a diameter, the perpen- dicular is a mean proportional between the segments of the diam- eter. Suggestion. Draw the chords A C and CB, (v. 11., Proposition XIY., Corollary.) '< ' PROPOSITION X.— THEOREM. 28. The square of the length of the hypotenuse of a right triangle is the sum of the squares of the lengths of the other two sides, the three lengths being expressed in terms of the same unit. Let ABC be right angled at C; then Xg' = TO' + BU'. For, by Proposition IX., we have TU' = AB X AD, and 277' = AB X BD, the sum of which is ■AJ]' + 'mj'=.AB X {AD ^ BD) = AB X AB = A^. 29. Scholium I. By this theorem, if the numerical measures of two sides of a right triangle are given, that of the third is found. For example, if ^(7= 3, BO = 4; then AB = 1/ [3^ + 4^ = 5. If the hypotenuse, AB, and one side, AC, are given, we have BC^ = TB'' — X0\' thus, if there are given AB = b, AC = 3, then we find BO = ^^[5^ — 3'] = 4. 118 ELEMENTS OF GEOMETRY. 30. Scholium II. If J. is the diagonal of a square ABCD, we have, by the preceding theorem, whence and, extracting the square root, AC -j£ = \/^= 1.41421 + ad inf. Since the square root of 2 is an incommensurable number, it follows that the diagonal of a square is incommensurable with its side. (v. II., 34.) 31. Definition. The projection of a point A upon an indefinite straight line XY is the foot P of the perpendicular let fall from the point upon the line. The projection of a finite straight line AB upon the line XY is the distance FQ between the projections of the extremities oi AB. If one extremity B of the line AB is in A the line- XY^ the distance from B to P (the projection of A) is the projection of AB on XY; for the point B is in this case its own projectioik PM, EXERCISES. 1. Theorem. — In any triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of these sides ^dnd the projection of the other upon that side. y^ BOOK III. 119 PCy, Fig. 1, or YiQ. 2. Suggestion. Let (7 be the acute angle in question in Fig. 1 or Fig 2. A Fig. 1. == XF' + (5(7 -^'^ XF' + (PC == 2T^ + TO' + :BT' - 2BC X Pc; = Xa' + IB^C — 2BC X PC, 2. Theorem. — J/i an obtuse angled triangle, the square of the side opposite to the obtuse angle is equal to the sum of the squares of the other two sides, increased by twice the product of one of these sides and the projection of the other upon that side. PROPOSITION XI.— THEOREM. 32. If two chords intersect within a circle, their segments are reciprocally proportional. For the triangles APB' and A'PB are mu- tually equiangular (II., Proposition XIY.), and therefore similar (Proposition III.). Hence AP : A'P = PB' : PB. 33. Scholium. If the lengths of the lines in question are expressed in terms of the same unit, the result above can be written APXPB= A'P X PB', and the proposition may be stated : if through a fixed point within a circle any chord is drawn, the product of the lengths of its segments is the same whatever its direction. 120 ELEMENTS OF GEOMETRY. EXERCISE. Theorem. — Either segment of the least chord that can he drawn through a fixed point is a mean proportional between the segments of any other chord drawn through that point, (v. II., 19, Exercise.) PROPOSITION XII.— THEOREM. 34. If two secants intersect without a circle, the whole secants and their external segments are reciprocally proportional. For the triangles PAB^ and PA'B are mu- tually equiangular (II., Proposition XIY.), and therefore similar. Consequently PB : PB' = PA' : PA. 35. Corollary. If a tangent and a secant intersect, the tangent is a mean proportional between the whole secant and its external seg- ment. Suggestion. Show that the triangles PA T and PTB are similar. 36. Scholium. If the lengths of the lines are expressed in terms of the same unit, the result of (34) can be written PB X PA = PB' X PA\ and Proposition XII. can be stated : if through a fixed point without a circle a secant is drawn, the product of the lengths of the whole secant and its external segment has the same value in whatever direction the secant is drawn. BOOK III. 121 EXERCISE. Theorem. — If from any point on the common chord of two intersecting circles^ produced^ tangents are drawn to the two circles^ the lengths of these tangents are equal. PEOBLEMS OF CONSTEUCTIOK PROPOSITION XIII.— PROBLEM. 37. To divide a given straight line into any given number of equal parts. Let AB be the given line. Through A draw an indefinite line AX^ upon which lay off the given number of equal distances, each distance being of any convenient length ; through M the last point of division on AX draw MB^ and through the other points of division of AX draw parallels to MB^ which will divide AB into the required number of equal parts. This follows from the first part of the proof of Proposition I. PROPOSITION XIV.— PROBLEM. 38. To divide a given straight line into parts proportional to two given straight lines. Let it be required to divide AB into parts proportional to M and N, From A draw the indefinite line AX^ upon which lay off AG = M and CD = N. Join BB^ and through G draw GE parallel to DB. Then we shall have AE :JEB=AC:GD, by Propo- sition I. I 122 ELEMENTS OF GEOMETRY. EXERCISE. Problem. — To divide a given straight line into parts proportional to given straight lines. ...•'5 M N PROPOSITION XV.— PROBLEM. 39. To find a fourth proportional to three given straight lines. Let it be required to find a fourth pro- portional to M, JV, and P. Draw the in- definite lines AJT, AY, making an angle with each other. Upon AJC lay off AB = M, AD = N; and upon ^ Ylay off ^(7 = P; join BG, and draw BE parallel to BC ; then AEi^ the required fourth pro- portional. For we have (Proposition I.) AB : AD=AG: AE, or M : W= P : AE. EXERCISE. Prohlem.- lines. To find a third proportional to tvx) given straight PROPOSITION XVI.— PROBLEM. 40. To find a mean proportional between two given straight lines. n Let it be required to find a mean pro- portional between M and N. Upon an indefinite line lay ofl AB=M,BG = JSF; upon AG describe a semi-circumference, and at B erect a perpendicular, BD, to AG. Then BD is the required mean proportional (Proposi- tion IX., Corollary). BOOK III. 123 41. Definition. When a given straight line is divided into two segments such that one of the segments is a mean pro- portional between the given line and the other segment, it is said to be divided in extreme and mean ratio. Thus, AB is divided in extreme and mean ratio ^ H — I at C,ii AB\AC=:AG : CB. PROPOSITION XVII.— PROBLEM. 42. To divide a given straight line in extreme and mean ratio. Let AB be the given straight line. At B erect the per- pendicular BO equal to one-half of AB. With the centre and radius OB^ describe a circumference, and through A and draw AO cutting the circumference first in D and a second time in D'. Upon AB lay ofl AG == AD. Then AB is divided at C in extreme and mean ratio. For we have (Proposition XIL, Corollary) ^ ^J • ^^^ AD' :AB=AB:AD or AC, [1] whence, by division (10), AD' — AB:AB = AB — AC:Aa, or, since DD' =20B = AB, and therefore AD' — AB = AD^ — DD' = AD=AC, AO:AB = CB:AC, and, by inversion (7), AB:AC = AC:CB; that is, AB is divided at C in extreme and mean ratia 124 ELEMENTS OF GEOMETRY. PROPOSITION XVIII.— PROBLEM. 43. On a given straight line^ to construct a polygon similar to a given polygon. Let it be required to construct upon A'B' a polygon similar to ABCDEF, Divide AB CDEF mio tri- angles by diagonals drawn from A. Make the angles B'A'C and A'B'C equal to BAG and ABC respec- tively; then the triangle A'B'C will be similar to ABC (Proposition III.). same manner construct the triangle A'D'C similar to ADC, A' WD' similar to AED, and A'E'F' similar to AEF. Then AIB'C'D'E'F' is the required polygon (Proposition YL). In the EXERCISES ON BOOK III. (1 THEOREMS. 1. If two straight lines are intersected by any number of par- allel lines, the corresponding segments of the two lines are pro- portional, {v. Proposition I.) 2. The diagonals of a trapezoid divide each other into segments which are proportional. 3. In a triangle any two sides are reciprocally proportional to the perpendiculars let fall upon them from the opposite vertices. 4. The perpendiculars from two vertices of a triangle upon the opposite sides divide each other into segments which are recipro* cally proportional. 6. If the three sides of a triangle are respectively perpendicular to the three sides of a second triangle, the triangles are similar. 6. If ABC and A^BO are two triangles Laving a common base BC and their vertices in a line AA^ parallel to the base, and if any parallel to the base cuts the sides AB and ^ C in Z> and jE7, and the sides A^B and A^C in D^ and E^^ then DE=^J)'E' (Proposition III.). ""7. If two sides of a triangle are divided propor- tionally, the straight lines drawn from correspond- ing points of section to the opposite angles intersect on the line joining the vertex of the third angle and the middle of the third side. Suggestion. Draw the line ADE through the in- tersection of B^C and BC^. B'E'D and CED are similar ; similar BC ^ B'C Hence EC B'E' DC B'D DC B'D BC B'DC^ and BDC are ^ AB^C^ and ABC are similar AB AB'' EC B'C AB'E' and ABE are similar ; BE AB AB' BE B'E' B'E' B'E' and BE = EC, 11* 125 126 ELEMENTS OF GEOMETRY. 8. The difference of the squares of two sides of any triangle is equal to the difference of the squares of the projections of these j--^ sides on the third side (Proposition X.). ^ ^ 9. J[f from any point in the plane df a polj[gon\peitoendiculars ^"^^ -^alfe^rawn to all the sides, the two siiml of\the^qu\res of the alternate segments of the sides are equkl. ^ ^ iW \J 10. If through a point P in the circum- ference of a circle two chords are drawn, the chords and the segments cut from them by a line parallel to the tangent at P are re- ciprocally proportional. Suggestion, Prove PAB and Pba similar. -^ -41 11. If three circles int^fSect, their three common chords pass through the same point, {v. Proposition XI.) 12. If two tangents are drawn to a circle at the extremities of a diameter, the por- tion of any third tangent intercepted be- tween them is divided at its point of con- tact into segments w^hose product is equal to the square of the radius. Suggestion. Prove AOB a right triangle. 13. The perpendicular from any point of a cir- cumference upon a chord is a mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord. Suggestion. PBD and PAE are similar ; . • . PB^PD PA PE PCE and PAD are similar PA PC PD PE' Hence ^ = PA PA PO' BOOK III. 127 14. If two circles touch each other, secants drawn through their point of contact and terminating in the two circumferences are divided proportionally at the point of contact, {v. II., 54, Exer- cise 2.) ^ - ' ^ 15. If two circles are tangent exter- nally, the portion of their common tangent included between the points of contact is a mean proportional be- tween the diameters of the circles. Suggestion. Show that OBO^ is a right triangle. 16. If a fixed circumference is cut by any circumference which passes through two fixed points, the com- mon chord passes through a fixed point. Suggestion. PA . PB = PC. PD = PT^ by Proposition XII. and Corollary. Join P with (7^, and show that PC^ will cut both circles at the same distance from P, and will be their common chord. "^ ''-::.-.-.-...-=''£ ^ ^f :z X LOCI. 17. From a fixed point O, a straight line OA is drawn to any point in a given straight line JJfJV, and divided at P in a given ratio m : n (ie, so that OP:PA='m:n); find the locus of P. (v. Proposition II.) A' 18. From a fixed point O, a straight line OA is drawn to any i:)oint in a given cir- cumference, and divided at P in a given ratio ; find the locus of P. Suggestion. PC is a fixed length. 19. Find the locus of a point whose distances from two given straight lines are in a given ratio. (The locus consists of two straight lines.) ^^r f. l^i^r-*^ ELEMENTS OF GEOMETRY. / yy- 20. Find the locus of the points which divide the various chords of a given circle into segments whose product is equal to a given constant, k'^ (33, Exercise). 21. Find the locus of a point the sum of whose distances from two given straight lines is equal to a given constant, k. {v, I., Exercise 10.) 22. Find the locus of a point the difference of whose distances from two given straight lines is equal to a given constant, k. Suggestion, Reduce it to I., Proposition XIX., by drawing a third line parallel to the more distant of the given lines at a distance from it equal to k, ^^ PROBLEMS. 23. To divide a given straight line into three segments. A, B, and C, such that A and B shall be in the ratio of two given straight lines M and iV, and B and C shall be in the ratio of two other given straight lines P and Q. ■24. Through a given point, to draw a straight line so that the portion of it intercepted between two given straight lines shall be divided at the point in a given ratio. Suggestion. Through the point draw a line parallel to one of the given lines, {v. II., Exercise 32.) 25. Through a given point, to draw a straight line so that the distances from two other given points to this line shall be in a given ratio. Suggestion. Divide the line joining the two other given pointe 26. To determine a point whose distances from three given in- deiinite straight lines shall be proportional to three given straight lines. (Exercise 19.) 4LJ' A 'h^sq 27. In a given triangle ABC, to inscribe a square DEFQ. (Exercises 6 and II., 29.) FG i:r.~ BOOK m. -Hr 129 ^ 28. Griven two circumferences inter- secmng in -4, to draw through A a i l^ant, JSACj such that AB shall be TO AO in Si given ratio. Suggestion. Divide 00^ in the given ratio, (v. Exercise 1.) 29. To aescribe a circumference passing through two given points A and B and tangent to a given circumference O. Analysis. Suppose ATB is the re- guired circumference tangent to the given circumference at T, and ACDB any circumference passing through A and B and cutting the given circum- ference in C and D. The common chords AB and CD^ and the common tangent at T^ all pass through a com- mon point P (Exercise 16) ; from which a simple construction may be inferred. There are two solutions given by the two tangents that can be drawn from P. <| 30. To describe a circumference passing through two given ^ points and tangent to a given straight line. (Two solutions.) {v. Proposition XII., Corollary.) 31. To describe a circumference passing through a given point and tangent to two given straight lines, {v. Exercise 13.) \y -'lY,'//. <^>^xjeJ?i >^ir BOO COMPARISON AND MEASUREMENT OP THE SURFACES OP RECTILINEAR PIGURES. 1. Definition. The area of a surface is its numerical meas- ure, referred to some other surface as the unit ; in other words, it is the ratio of the surface to the unit of surface (II., 29). The unit of surface is called the superficial unit. The most convenient superficial unit is the square whose side is the linear unit. 2. Definition. Equivalent figures are those whose areas are equal. PROPOSITION I.— THEOREM. 3. Parallelograms having equal bases and equal altitudes are equivalent. Let ABCD and AJECF be two parallelograms having equal bases and equal altitudes. ^ ^ Superpose the second upon the first, making the equal bases coincide. Since the altitudes are equal, the upper bases will lie in the same straight line. The triangles ABE and CDF are equal (I., Proposition YL). If the triangle CDF is taken from the whole figure, ABFC, the first parallelogram ABCD is left; if the equal triangle ABE is taken from the same figure, the second parallelogram AECF is left. The magnitudes of the two parallelograms are therefore equal, and the parallelograms are equivalent. 4. Corollary. Any parallelogram is equivalent to a rectangle having the same base and the same altitude. 130 BOOK IV. 131 PROPOSITION II.— THEOREM. 5. Two rectangles having equal altitudes are to each other as their bases. D O D F I. Let ABCD and AEFD be two rectangles having equal altitudes ; then are they to each other as AB : AE. 1. Suppose the bases have a common measure which is contained m times in AB and n times in AK Then we have AB AE Apply this measure to the two bases, and through the points of division draw perpendiculars to the bases. The two rec- tangles are thus divided into smaller rectangles, all of which are equal, by I., Proposition XXYIIL, Corollary, and of which ABCD contains m and AEFD contains n. Then ABOD m and consequently AEFD ABCD ^ AEFD~ n AB AE' c c 2. If the bases are incommensurable, divide AE in any arbitrarily chosen number n of equal parts, and apply one of these parts to AB. Let B' be the last point of division, B'B being of course less than the divisor. Since AB' and AE are commensurable, we have = AEFD AB' £f B AE , and this holds, no matter what the value of n. If, now, 132 ELEMENTS OF GEOMETRY. C O n is increased at pleasure, we can make J5'J5, and consequently B'BGG'j as small as we please, but cannot make them absolutely zero. Hence, as n is indefinitely in- creased, AB' has AB for its limit, AB'C'D has ABGD for its Hmit, Bf B has . for its limit. and AEFD AB' AEFD , „ has — — for its limit. AE AE Therefore, by II., Theorem^ Doctrine of Limits, ABCD AB AEFD ^. (^v. IL, 42, and III., 14.) 6. Corollary. Two rectangles having equal bases are to each other as their altitudes. Note. In these propositions, by " rectangle" is to be under- stood " surface of the rectangle." v^Vfiv^jfi^^-^ V PBOPOSITION III.— THEOKEM. 7. Any two rectangles are to each other as the products of their bases by their altitudes. Let B and i2' be two rec- tangles, k and k their bases, h and h' their altitudes ; then B kXh JS^ be a third rectan- gle, having the same base k BOOK IV. 133 as the rectangle -B, and the same altitude h! as the rectangle R' ; then we have, by Proposition II., Corollary, and Propo- sition II., S h" R' k" and multiplying these ratios, we find R' Ji'X h'' 8. Scholium. It must be remembered that by the product of two lines is to be understood the product of the numbers which represent them when they are measured by the linear unit (III., 8). PEOPOSITION IV.— THEOREM. 9. The area of a rectangle is equal to the product of its base and altitude. Let R be any rectangle, k its base, and h its altitude numerically ex- pressed in terms of the linear unit ; and let Q be the square whose side is the linear unit ; then, by Proposition III., R _k X h Q IX 1 = k X h. R But since Q is the unit of surface, - = the numerical meas- ure, or area, of the rectangle, R, (1) ; therefore Area of R = k X h. 12 134 ELEMENTS OF GEOMETRY. 10. Scholium I. When the base and altitude are exactly divisible by the linear unit, this proposition is rendered evi- dent by dividing the rectangle into squares each equal to the superficial unit. Thus, if the base contains 7 linear units and the alti- tude 5, the rectangle can obviously be divided into 35 squares each equal to the superficial unit ; that is, its area = 5x7. The propo- sition, as above demonstrated, is, however, more general, and includes also the cases in which either the base or the altitude, or both, are incommensurable with the unit of length. 11. Scholium II. The area of a square, being the product of two equal sides, is the second power of a side. Hence it is that in arithmetic and algebra the expression " square of a number" has been adopted to signify "second power of a number." We may also here observe that many writers employ the expression "rectangle of two lines" in the sense of "product of two lines," because the rectangle constructed upon two lines is measured by the product of the numerical measures of the lines. PROPOSITION v.— THEOREM. 12. The area of a parallelogram is equal to the product of its base and altitude. For, by Proposition I., the parallelogram is equivalent to a rectangle having the same base and the same altitude. BOOK IV. { I '^ < / ' 135 PROPOSITION VI.— THEOREM. 13. The area of a triangle is equal to half the product of its base and altitude. Let ABC be a triangle, k the numerical f- --^ measure of its base BC^h that of its alti- •'' ^^^^"^^^ / •* tude AB^ and 8 its area \ then B^~h o""d S=ikXh. For, through A draw AE parallel to CB, and through B draw B£J parallel to CA. The triangle ABC is one-half the paral- lelogram AEBC (I., Proposition IX.); but the area of the parallelogram = k y^ h ; therefore, for the triangle, we have S=lk X h. / 14. Corollary I. A triangle is equivalent to one-half of any i^parallelogram having the same base and the same altitude. J 15, Corollary II. Triangles having equal bases and equal \laltitudes are equivalent. 16. CoROLLA'feY III. Triangles having equal altitudes are to each other as their bases ; and triangles having equal bases are to each other as their altitudes. X PROPOSITION VII.— THEOREM. / 17. The area of a trapezoid is equal to the product of its alti^ tude by half the sum of its parallel bases. Let ABCD be a trapezoid ; MN= h, its altitude ; AD = a, BC = b, its parallel bases ; and let S denote its area ; then S=i(^a + b)X h. 136 ELEMENTS OF GEOMETRY. A M For, draw the diagonal AC. The altitude of each of the triangles ABC and ABC is equal to h, and their bases are respectively a and b ; the area of the first is ^ a y^ h^ that of the second is ^ 6 X ^ ; and the trapezoid being the sum of the two triangles, we have b n o S=iaXh-]-ibXh = i(a + b)Xh. 18. Scholium. The area of any polygon may be found by finding the areas of the several triangles into which it may be decomposed by drawing diagonals from any vertex. The following method, however, is usually preferred, espe- cially in surveying. Draw the longest diagonal AD of the proposed polygon ABCDEF; and upon AD let fall the perpendiculars BM^ CW, UP, FQ. The polygon is thus decomposed into right triangles and right trapezoids, and by measuring the lengths of the perpendiculars and also of the distances AM, MN^ ND, AQ, QP, PD, the bases and altitudes of these triangles and trape- zoids are known. Hence their areas can be computed by the preceding theorems, and the sum of these areas will be the area of the polygon. i PROPOSITION VIII.— THEOREM. ■' 19. Similar triangles are to each other as the squares of their homologous sides. Let ABC, A'B'C, be similar tri- angles; then ABC ^ ZB^ BOOK IV. Let AD and A'D' be the altitudes ; then 137 ABC _ ^AD X BC _ 4^ BC A'B'C ^A'D' X B'C A'D' '^ B'C But the triangles ABB and A'D'B' are similar (III., Propo sition III.) ] therefore AB AB A'D' A'B" and from the similarity of ABC and A'B'G\ BC AB hence and we have B'C A'B" AB BC ^ TB'' A'B' ^ B'G' A^'' ABC ^ X5' EXERCISE. \ I ^ Theorem. — Ti^o triangles having an angle of the one equal to an angle of the other are to each other as the products of the ^ sides including the equal angles. L^ Suggestion. Let ABE and ABC be the two triangles. Draw BE^ and compare the two triangles with AEB. (v. Proposition YI., Corollary III.) 12» 138 ELEMENTS OF GEOMETRY. PROPOSITION IX.— THEOREM. 20. Similar polygons are to each other as the squares of their homologous sides. Let ABCDEF, A' B' C D' E' F' , be two similar polygons, and denote their surfaces by /Sand >S"; then b ^ ^ c' • For, let the polygons be p decomposed into homolo- gous similar triangles (III., Proposition YII.). The ratio of the surfaces of any pair of homologous triangles, as ABC and A'B'C, ACD and A' CD', etc., will be the square of the ratio of two homologous sides of the polygons (Proposition YIII.) ; therefore we shall have ABO ACD ADE AEF Tff A'B'C A' CD' A'D'E' A'E'F' Z^' Therefore, by addition of antecedents and consequents (III., 12). ABC + ACD + ADE + AEF ^S^ Z^ A'B'C + A' CD' + A'D'E' + A'E'F' S' ATB"^' PROPOSITION X.— THEOREM. 21. The square described upon the hypotenuse of a right tri- angle is equivalent to the sum of the squares described on the other two sides. BOOK IV. 139 Let the triangle ABC be right angled at C; then the square AIT, described upon the hypotenuse, is equal in area to the sum of the squares AF and £D, described on the other two sides. For, from C draw CP perpen- dicular to AB and produce it to meet KH in Z. Join CK, BG. Since ACF and ACB are right angles, CF and CB are in the same straight line (I., Proposition lY.) ; and for a similar reason AG and CD are in the same straight line. In the triangles OAK, GAB, we have AK equal to AB, being sides of the same square; AC equal to AG, for the same reason; and the angles CAK, GAB, equal, being each equal to the sum of the angle GAB and a right angle; there- fore these triangles are equal (L, Proposition VI.). The triangle CAK and the rectangle AL have the same base AK; and since the vertex C is upon LP produced, they also have the same altitude; therefore the triangle CAK is equivalent to one-half the rectangle AL (Proposition VI Corollary I.). ' The triangle GAB and the square AF have the same base AG; and smce the vertex B is upon FG produced, they also have the same altitude ; therefore the triangle GAB is equiva- lent to one-half the square AF (Proposition VI., Corollary I ) But the triangles CAK, GAB, have been shown to be equal • therefore the rectangle AL is equivalent to the square AF In the same way it is proved that the rectangle BL is equivalent to the square BD. 140 ELEMENTS OF GEOMETRY. Therefore the square AH^ which is the sum of the rec- tangles AL and BL, is equivalent to the sum of the squares AF and BD. 22. Scholium. This theorem is ascribed to Pythagoras (born about 600 B.C.), and is commonly called the Pythagorean Theorem. The preceding demonstration of it is that which was given by Euclid, in his Elements (about 300 B.C.). It is important to observe that we may deduce the same result from the numerical relation XS^ ==:AC^ -\- ^FU^ already established in III., Proposition X. For, since the measure of the area of a square is the second power of the number which represents its side, it follows directly from this numer- ical relation that the area of which AB^ is the measure is equal to the sum of the areas of which AC and BTf are the measures. EXERCISES. 1. Theorem. — If the three sides of a right triangle be taken as the homologous sides of three similar polygons. constructed upon them, then the polygon constructed upon the hypotenuse is equiva- lent to the sum of the polygons constructed upon the other two sides, (v. Proposition IX.) 2. Theorem. — The squares on the sides of a right triangle are proportional to the segments into which the hypotenuse is divided by a perpendicular let fall from the vertex of the right angle, (v. Figure, Proposition X.) BOOK IV. 141 PEOBLEMS OF C0:N^STEUCTI0K. PROPOSITION XI.— PROBLEM. 23. To construct a triangle equivalent to 4: given polygon. Let ABCDEF be the given polygon. ' Take any three consecutive vertices, as J., 5, (7, and draw the diagonal AC' Through B draw BP parallel to AO Wk meeting DC produced in P; join AP. The triangles APG, ABC, have the same base AG ; and since their vertices, P and 5, lie on the same straight line BP parallel to AG^ they also have the same altitude ; therefore they are equiva- lent. Therefore the pentagon APDEF is equivalent to the hexagon ABGDEF. Now, taking any three consecutive ver- tices of this pentagon, we shall, by a precisely similar con- struction, find a quadrilateral of the same area ; and, finally, by a similar operation upon the quadrilateral we shall find a triangle of the same area. Thus, whatever the number of the sides of the given poly- gon, a series of successive steps, each step reducing the num- ber of sides by one, will give a series of polygons of equal areas, terminating in a triangle. PROPOSITION XII.— PROBLEM. 24. To construct a square equivalent to a given parallelogram or to a given triangle. ^ ^ 1st. Let J. (7 be a given parallelogram, k its \ ^i \ base, and h its altitude. Find a mean proportional x between h and j y k, by III., 40. The square constructed upon X will be equivalent to the parallelogram, since x^ = h y^ k. \J 142 ELEMENTS OF GEOMETRY. n 2d. Let ABC be a given triangle, a its base, and h its altitude. A Find a mean proportional x between a and ih; the square constructed upon x will be equivalent to the triangle, since ^ a u = a X i h = i ah. I 1 25. Scholium. By means of this problem and the preceding, a square can be found equivalent to any- given polygon. Pi- PEOPOSITION XIII.-PROBLEM. 26. To construct a square equivalent to the sum of two or more given squares, or to the difference of two given squares. 1st. Let m, w, ^, q, be the sides of given squares. Draw AB = m, and BC = n, perpendic- ular to each other at B; join AC. Then (Proposition X.) ZU^ ^ m^ -\- n\ Draw CD = p perpendicular to A (7, and join AD. Then AJf = TU' + p' = m' + n^ + _p^ Draw DB = q perpendicular to AD, and join AK Then TE^ = Jlf -\- q' = m,' + n^ _|_ p^ _|_ ^2 Therefore the square con- structed upon AE will be equivalent to the sum of the squares constructed upon m, 7i, p, q. In this manner may the areas of any number of given squares be added. 2d. Construct a right angle ABC, and lay off BA = n. With the centre A and a radius = m, describe an arc cutting BCinC. Then u BOOK IV. 143 :ro' = zu^ - zs' m' therefore the square con- M structed upon BC will be equivalent to the difference of the squares constructed upon m and n. EXERCISE. Prohlem.^Upon a given straight line, to construct a rectangle equivalent to a given rectangle. \ PBOPOSITION XIV.-PROBLEM. 27. To construct a rectangle, having given its area and the sum of two adjacent sides. Let MN be equal to the given sum of the adjacent sides of the required rec- ^ .^ tangle ; and let the given area be that of i,,..^.^::C:r\9 the square whose side is AB. '' ^ — ^^;--— Upon MJSr as a diameter describe a semicircle. At M erect MP = AB per- pendicular to MN, and draw PQ parallel to MN, intersecting he circumference in Q. From Q let fall QR perpendicular to MN; then MR and RN are the base and altitude of the PROPOSITION XV.-PEOBLEM. 28. To find two straight lines in the ratio of the areas of two given polygons, • Let squares be found equal in area to o the given polygons respectively (23 and /T^v. 24). Upon the sides of the right angle '^ [ ^_j, ACB, take CA and CB equal to the sides of these squares, join AB, and let fall CD perpendicular to B N 144 ELEMENTS OF GEOMETRY. AB. Then, by (III., 26), we have lU' = AD X AB, VF^ = JDB X AS- Hence IV ^ AD . VB' DB' therefore AD and DB are in the ratio of the areas of the given polygons. EXERCISE. Problem. — To find a square which shall he to a given square in the ratio of two given straight lines, (y. 28.) PROPOSITION XVli— PROBLEM.X)^ Q/n 29. To construct a polygon similar to a given polygon P and equivalent to a given polygon Q. Find M and iV, the sides of squares respectively equal in area to P and (23 and 24). I ^ 1 I ^ 1 Let a be any side of P, and find a fourth proportional a' to Mj i^T, \ p' | and a; upon a', as a homologous ^ — -f — ' side to a, construct the polygon P' similar to P; this will be the required polygon. For, by construction, M_a jsr a" therefore, taking the letters P, Q, and P', to den0te the areas of the polygons, P^3P^a^ Q W' a'' ' BOOK IV. 145 But, the polygons P and P' being similar, we have, by (Proposition IX,), P' a'^ ' and comparing these equations, we have P' = Q. Therefore the polygon P' is similar to the polygon P and equivalent to the polygon Q, as required. EXERCISE. Problem. — To construct a polygon similar to a given polygon^ and whose area shall he in a given rhtio to that of the given polygon, (v. 28, Exercise, an^III., 43.) ^ ■J # G k ^^^ i.^ IS D f EXERCISES ON BOOK IV. THEOREMS. ^r ^ 1. Two' triangles are equivalent if they have two sides of the (^r'^^iiAM) one respectively equal to two sides of the other, and the included At angle of the one equal to the supplement of the included angle of the other. -— ^^Vz. The two opposite triangles formed by joining any point in the interior of a parallelogram 4o its four vertices are together equivalent to one-half the parallelogram. 3. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezoid to the extremities of the opposite side is equivalent to one-half the trapezoid, {v, I., Ex- ercise 24.) 4. The figure formed by joining consecutively the four middle points of the sides of any quadrilateral is equivalent to one-half the quadrilateral, (v. I., Exercise 32.) „ V. 6. If in a rectangle ABCD we draw the diagonal AC^ inscribe a circle in the triangle ABC^ and from its centre draw OE and Oi^ perpendicular to AD and DC respectively, the rectangle OD will be equivalent to one-half the rectangle ABCD. C 6. The area of a triangle is equal to one-half the product of its perimeter by the radius of the inscribed circle. ^ 7. The area of a rhonabus is one-half the product of the diag- onals. .\,w^j'«' BOOK IV. 147 - '" 8. The straight line joining the middle points of the parallel sides of a trapezoid divides it into two equivalent figures. (4:'j ^. Any line drawn through the point of intersection of the diag- onals of a parallelogram divides it into two equal quadrilaterals. 10. In an isosceles right triangle either leg is a mean propor- tional between the hypotenuse and the perpendicular upon it from the vertex of the right angle. 11. If two triangles have an angle in common, and have equal areas, the sides about the equal angles are reciprocally propor- V tional. . .Ivt \'^' The perimeter of a triangle is to a side as the perpendicular fronythe opposite vertex is to the radius of the inscribed circle. ! : (v.yExercise 6.) f yi3. Two quadrilaterals are equivalent when the diagonals of , V one are respectively equal and parallel to the diagonals of the I ^o^her. ^ 14. The sum of the perpendiculars from any point within an ■ ♦ equilateral convex polygon upon the sides is constant. y^ Suggestion. Join the point with the vertices of the polygon. ^ 15. The lines joining two opposite vertices of a parallelogram ^ with the middle points of the sides form a parallelogram whose ^^^area is one-third the a^ea of the given parallelogram. A ^.^ 16. The sum of the squares on the segments of two j)erpendic- ular chords in a circle is equivalent to the square on the diameter. 17. Let ABC be any triangle, and "^ upon the sides AB, AC, construct parallelograms AD, AF, of any mag- nitude or form. Let their exterior *^ sides DE, FG, meet in M; join MA, ^ and upon BC construct a parallelo- gram BK, whose side BH is equal and parallel to MA, Then the par- allelogram BK is equivalent to the sum of the parallelograms AD and AF, {v. Proposition I.) From this deduce the Pythagorean Theorem. «y^^ V 18. Prove, geometrically, that th^ square described upon the sum of two straight lines is equivalent to the sum of the squares described on the two lines plits twice their rectangle. Note. By the "rectangle of two lines" is here meant the rec- tangle of which the two lines are the adjacent sides. QV' 19. Prove, geometrically, that the square described upon the difference of two straight lines is equivalent to the sum of the squares described on the two lines minus twice their rectangle. ELEMENTS OP GEOMETRY. V 20. Prove, geometrically, that the rectangle of the sum and the difference of two straight lines is equivalent to the difference of the squares of those lines. PROBLEMS. .4 V u 21. To construct a triangle, given its angles and its area (eq to that of a given square). Suggestion. Construct any triangle having the given angles. The problem then reduces to (29). 22. Given any triangle, to construct an isosceles triangle of the same area, whose vertical angle is an angle of the given triangle. {v. 19, Exercise.) 23. Given any triangle, to construct an equilateral triangle of he same area. {v. Exercise 21.) 24. Bisect a given triangle by a parallel to one of its sides, (v. Proposition VIII. and 28.) 25. Bisect a triangle by a straight line drawn through a given point in one of its sides, (v. 19, Exercise.) 26. Inscribe a rectangle of a given area in a given circle. Suggestion. Draw a diagonal of the rectangle. The problem can then be reduced to inscribing in the given circle a right ;riangle of given area. 27. Given three points, A, B^ and (7, to find a fourth point P, such that the areas of the triangles APB, APC, BPG, shall be equal. (Four solutions.) (v. III., Exercise 19.) h \ ;^*(..«' ' BOOK T. REGULAR POLYGONS. MEASUREMENT ®P THE CIRCLE. 1. Definition. A regular polygon is a polygon which is at once equilateral and equiangular. The equilateral triangle and the square are simple exam- ples of regular polygons. The following theorem establishes the possibility of regular polygons of any number of sides. PROPOSITION I.— THEOREM. 2. If the circumference of a circle be divided into any number of equal parts, the chords joining the successive points of division form a regular polygon inscribed in the circle ; and the tangents drawn at the points of division form a regular polygon circum^ scribed about the circle. Let the circumference be divided into the equal arcs AB, BC, GD^ etc. ; then, 1st, drawing the chords AB, BC, CD, etc., ABGD, etc., is a regular inscribed polygon. For its sides are equal, being chords of equal arcs ; and its angles are equal, being inscribed in equal segments. 2d. Drawing tangents at A, B, C, etc., the polygon GHK^ etc., is a regular circumscribed polygon. For, in the triangles AGB, BHC, CKD, etc., we have AB = BC = CD, etc., and the angles GAB, GBA, HBC, HOB, etc., are equal, since each 13* 149 150 ELEMENTS OF GEOMETRY. is formed by a tangent and chord and is measured by half of one of the equal parts of the circum- ference (II., Proposition XY.) ; therefore these triangles are all isosceles and equal to each other. Hence we have the an- gles a = H= K, etc., and AG = GB =.BH = HG = CK, etc., from which, by the addition of equals, it follows that GH = HK, etc. 3. Corollary I. If the vertices of a regular inscribed poly- gon are joined with the middle points of the arcs subtended by the sides of the polygon, the joining lines will form a regular inscribed polygon of double the number of sides. 4. Corollary II. If at the middle points of the arcs joining adjacent points of contact of the sides of a regular circumscribed polygon tangents are drawn, a regular circumscribed polygon of double the number of sides will be formed. 5. Scholium. It is evident that the area of an inscribed polygon is less than that of the inscribed polygon of double the number of sides ; and the area of a circumscribed polygon is greater than that of the circumscribed polygon of double the number of sides. EXERCISE. Theorem. — If a regular polygon is inscribed in a circle, the tangents drawn at the middle points of the arcs subtended by the sides of the inscribed polygon form a circumscribed regular polygon, whose sides are par- allel to the sides of the inscribed poly- gon, and whose vertices lie on the radii drawn to the vertices of the inscribed polygon. c A'{ ""~--^A e/7 \ /\ V \f V"^'-^ ^^\/ u BOOK V. 151 PROPOSITION II.— THEOREM. 6. A circle may be circumscribed about any regular polygon ; and a circle may also be inscribed in it. Let ABCD ... be a regular polygon. Through A' and B\ the middle points of AB and BG^ draw perpendiculars, and connect 0, their point of intersec- tion, with all the vertices of the poly- gon and with the middle points of all the sides. The triangles OA'B and OB'B are equal, by I., Proposition X. OB'B and OB'G are equal, by I., Proposition YI. The angle OBB' is one-half of ABG ; . • . 005' is one-half of the equal angle BGD. Hence the triangles OB'G and OGG' are equal, by I., Proposition YI. By continuing this process we may prove all the small triangles equal. 0, then, is equidis- tant from all the vertices, and therefore with as a centre a circle may be circumscribed about the polygon. is also equidistant from all the sides, and therefore with as a centre a circle may be inscribed in the polygon. 7. Definitions. The centre of a regular polygon is the common centre, 0, of the circumscribed and in- scribed circles. The radius of a regular polygon is the radius, OJ., of the circumscribed circle. The apothem is the radius, O-ff, of the inscribed circle. The angle at the centre is the angle, AOB, formed by radii drawn to the extremities of any side. ^o A vv ^E 152 ELEMENTS OF GEOMETRY. 8. The angle at the centre is equal to four right angles divided by the number of sides of the polygon. 9. Since the angle ABC is equal to twice ABO, or to ABO + BAO, it fol- lows that the angle ABC of the poly- gon is the supplement of the angle at the centre. PROPOSITION in.— THEOREM. 10. Begular 'polygons of the same number of sides are similar, Ijet ABCI)JE,A'B'C'iyE\ be regular polygons of the same number of sides ; then they are similar. For, 1st, they ape mutu- ally equiangular, since the magnitude of an angle of either polygon depends only on the number of the sides (8 and 9), which is the same in both. 2d. The homologous sides are proportional, since the ratio AB : A'B' is the same as the ratio BG : B'C, or CD : CD', etc. Therefore the polygons fulfil the two conditions of simi- larity. 11. Corollary. The perimeters of regular polygons of the same number of sides are to each other as the radii of the cir- cumscribed circles, or as the radii of the inscribed circles ; and their areas are to each other as the squares of these radii, (y, III., Proposition YIII., and lY., Proposition IX.) BOOK V. 153 PROPOSITION IV.— THEOREM. 12. The area of a regular polygon is equal to half the product of its perimeter and apothem. For straight lines drawn from the centre to the vertices of the polygon divide it into equal triangles whose bases are the sides of the polygon and whose common altitude is the apothem. The area of one of these triangles is equal to half the product of its base and altitude ; therefore the sum of their areas, or the area of the polygon, is half the product of the sum of the bases by the common altitude; that is, half the product of the perimeter and apothem. EXERCISE. Theorem. — The area of any polygon circumscribed about a circle is half the product of its perimeter by the radius of the circle. PROPOSITION v.— THEOREM. 13. An arc of a circle is less than any line which envelops it and has the same extremities. Let AKB be an arc of a circle, AB its chord ; and let ALB, AMB, etc., be any lines enveloping it and terminating at A and B. Of all the lines AKB, ALB, AMB, etc., which can be drawn (each including between itself and the chord AB the segment, or area, AKB), there must be at least one minimum or shortest line, since all the lines are obviously not equal. Now, no one of the lines ALB, AMB, etc., envel- oping AKB, can be such a minimum ; for, drawing a tangent CKL to the arc AKB, the ImeAGJCLB is less than ACLDB ; 154 ELEMENTS OF GEOMETRY. therefore ALB is not the minimum ; and in the same way it is shown that no other enveloping line can be the minimum. Therefore the arc AKB is the minimum. 14. Corollary. The circumference of a circle is less than the perimeter of any poly- gon circumscribed about it. 15. Scholium. The demonstration is applicable when AKB is any convex curve whatever. PKOPOSITION VI.— THEOREM. 16. If the number of sides of a regular polygon inscribed in a circle be increased indefinitely, the apothem of the polygon will approach the radius of the circle as its limit. Let AB be a side of a regular polygon in- scribed in the circle whose radius is OA; and let OD be its apothem. Whatever the number of sides of the poly- gon OB < OA, by I., Proposition XYII. OA < AD + OD (I., Axiom I.) • .. OA — OD <, AD, and consequently OA — OD <^ AB. The perimeter of the polygon is manifestly less than the circumference of the circle. If n is the number of sides of the polygon, AB is less than one-nth of the circumference. Therefore, by taking a sufficiently great value of n, we can make AB, and consequently OA — OD, as small as we please. Since OA — OD can be made as small as we please by increasing the number of sides of the polygon, but cannot be made absolutely zero, OA is the limit of OD, as the number of sides of the polygon is indefinitely increased (II., 39). BOOK V. 155 PROPOSITION VII.— THEOREM. 17. The circumference of a circle is the limit which the perim- eters of regular inscribed and circumscribed polygons approach when the number of their sides is increased indefinitely ; and the area of the circle is the limit of the areas of these polygons. Let AB and CD be sides of a regular in- scribed and a similar circumscribed polygon (v. Proposition I., Exercise) ; let r denote the apothem OE, R the radius OF^ p the perim- eter of the inscribed polygon, P the perim- eter of the circumscribed polygon. Then we have (Proposition III., Corollary) whence, by division (III., 10), P — p __ R — r R or P Now, we have seen in Proposition YI. that by increasing the number of sides of the polygons the diiference R — r may P be decreased at pleasure ; consequently, since — - does not in- R P crease, — - X (-^ — r), or P — p, may be decreased at pleasure. R But P being always greater, and p always less, than the cir- cumference of the circle, the difference between this circum- ference and either P or ^ is less than the difference P -~ p^ and consequently may be made as small as we please by increasing the number of sides of the polygons, and sinco it obviously cannot be made absolutely zero, the circumference is the common limit of P and p, as the number of sides of the polygons is indefinitely increased. IjiisaL. 156 ELEMENTS OF GEOMETRY. Again, let s and S denote the areas of two similar inscribed and circumscribed polygons. The difference between the triangles COD and A OB is the trapezoid CABD, the measure of which is ^ (CD + AB) X ^^; therefore the difference between the areas of the polygons is C F D ±\^ ^\/« N ^ ^ consequently, S-sSf=^(7xP. [1] 24. Corollary. The area of a circle is equal to the square of its radius multiplied by the constant number tz. Suggestion. If we substitute for C in [1] its value 27zR (20), we have EXERCISE. Theorem. — The area of a sector is equal to half the product of its arc by the radius. Suggestion. Compare the sector with the whole circle. 160 ELEMENTS OF GEOMETRY. PEOBLEMS OF COIS^STEUCTIOK AND COMPUTA- TIOK PKOPOSITION X.— PROBLEM. 25. To inscribe a square in a given circle. B Draw any two diameters A C, BJ), per- pendicular to each other, and join their extremities by the chords ABj BC, CD, DA; then ABCD is an inscribed square. 26. Corollary. To circumscribe a square about a circle, draw tangents at the extremities of two perpendicular diam- eters AC, BD. 27. Scholium. In the right triangle ABO we have AB' = 'UA' + 'UB'=2'UA\ whence AB =OA.y^, by which the side of the inscribed square can be computed, the radius being given. PROPOSITION XI.— PROBLEM. 28. To inscribe a regular hexagon in a given circle. Suppose the problem solved, and let ABCDEF be a regular inscribed hexa- gon. Draw the radii OA and OB. The angle A OB is measured by \ of the cir- cumference, and therefore contains 60°. OAB and OB A are therefore together equal to 180° — 60°, or 120° ; and, since they are equal, each is 60°, and the tri- angle OAB is equilateral, and th^gfor^the side of the in- scribed regular hexagon is equal to thfe radius of the circle. BOOK V. 161 Consequently, to inscribe a regular hexagon, apply the radius to the circle six times as a chord. 29. Corollary. To inscribe an equilateral triangle^ join the alternate vertices of the regular hexagon. 30. Scholium. Since OB bisects the arc ABC, it bisects the chord AG Bit right angles ; and since in the isosceles triangle A OB, AH is perpendicular to the base, it bisects the base, and OII=WB = WA; that is, the apothem of an inscribed regular triangle is equal to one-half the radius. In the right triangle AHO, AH^ = WC — OTT' = TJA^ — {lOAy=^lUA\2^n^ whence AC =^ OA-^/S, by which the side of the inscribed tri- angle can be computed from the radius. The apothem of the regular inscribed hexagon is equal to OA 2 AS 1/5- PROPOSITION XII.— PROBLEM. 31. To inscribe a regular decagon in a given circle. Suppose the problem solved, and let ABC .,. . . Jv be a regular inscribed deca- gon. Join AFj BGr ; since each of these lines bisects the circumference, they are diameters and intersect in the centre 0. Draw BK intersecting OA in M. The angle AMB is measured by half the sum of the arcs KF and AB (II., Proposition XYI.), — I 14* t'X^ ELEMENTS OF GEOMETRY. that is, )by two divi^ns of the circumference ; the inscribed angle^ikHJ^ is measured by half the arc BF^ — that is, also, by two divisions; therefore AMB is an isosceles triangle, and MB = AB. Again, the inscribed angle MBO is measured by half the arc KG, — ^that is, by one division ; and the angle MOB at the centre has the safne measure ; there- fore 0MB is an isosceles triangle, and OM = MB = AB. The inscribed angle MBA, being measured by half the arc AK, — that is, by one division, — is equal to the angle AOB. Therefore the isosceles triangles AMB and A OB are mutually equiangular and similar, and give the proportion whence OA'.AB^AB: AM; OA XAM = A^=(m\- that is, the radius OA is divided in extreme and mean ratio at Jf (III., 41) ; and the greater segment OJf is equal to the side AB of the inscribed regular decagon. Consequently, to inscribe a regular decagon, divide the radius in extreme and mean ratio (III., 42), and apply the greater segment ten times as a chord. o 32. Corollary. To inscribe a regular pentagon J join the alternate vertices of the regular inscribed decagon. BOOK V. 163 PROPOSITION XIII.— PROBLEM. 33. To inscribe a regular pentedecagon in a given circle. Suppose AB is the side of a ■ f regular inscribed pentedecagon, ^n*-- - -y^ or that the arc AB is ^ of the ^'^^===*==>.-^i^i;:i^^^'''^ circumference. JSTow, the fraction J^. == i. _ -i^ ; therefore the arc AB is the difference between \ and -^ of the circumference. Hence, if we inscribe the chord AG equal to the side of the regular inscribed hexagon, and then CB equal to that of the regular inscribed decagon, the chord AB will be the side of the regular inscribed pentedecagon required. 34. Scholium. Any regular inscribed polygon being given, a regular inscribed polygon of double the number of sides can be formed by bisecting the arcs subtended by its sides and drawing the chords of the semi-arcs (Proposition I., Cor- ollary I.). Also, any regular inscribed polygon being given^ a regular circumscribed polygon of the same number of sides can be formed (Proposition I.). Therefore, by means of the inscribed square, we can inscribe and circumscribe, succes- sively, regular polygons of 8, 16, 32, etc., sides ; by means of the hexagon, those of 12, 24, 48, etc., sides ; by means of the decagon, those of 20, 40, 80, etc., sides ; and, finally, by means of the pentedecagon, those of 30, 60, 120, etc., sides. Until the beginning of the present century, it was sup- posed that these were the only polygons that could be con- structed by elementary geometry ; that is, by the use of the straight line and circle only. Gaxjss, however, in his Disqui- sitiones Arithmeticce^ Lipsise, 1801, proved that it is possible, by the use of the straight line and circle only, to construct regular polygons of 17 sides, of 257 sides, and in general of >)'' nxXcj__ ^M3 -u^^JM " F E G D \ ^^^^P "-^4 \ \ 1 \\i , / 164 - any number oF'sT^Fg which can be^ expressed by 2" + !> ^ being an integer, provided that 2" + 1 is a prime number. I^ROPOSITION XIV.— PROBLEM. 35. Given the perimeters of a regular inscribed and a similar • circumscribed polygon, to compute the perimeters of the regular inscribed and circumscribed polygons of double the number of sides. Let AB he a side of the given inscribed polygon, and CD a side of the similar circumscribed poly- gon, tangent to the arc AB Sit its middle point B. Join AE, and at A and B draw the tangents AF and BG ; then AE is a side of the regular inscribed polygon of double the number of sides, and FG is a side of the circumscribed polygon of double the number of sides. Denote the perimeters of the given inscribed and circum- scribed polygons by p and P, respectively ; and the perimeters of the required inscribed and circumscribed polygons of double the number of sides by y and P', respectively. Since 0(7 is the radius of the circle circumscribed about the polygon whose perimeter is P, we have (Proposition III., Corollary) ?=oc^^qc p OA OE' and since OF bisects the angle COE, we have (III., 15, Ex- ercise) OC^CF, OE FE' therefore P^CF p FE' BOOK V. 165 whence, by composition, P + p ^ CF-\-FE ^ CE • 2p 2FE Fa' I^ow, FG is a side of the polygon whose perimeter is P', and is contained as many times in F' as GE is contained in P; hence (III., 9) CE P FG F" and therefore P+P P^ whence 2p F' F + p m Again, the right triangles AES and EFW are similar, since their acute angles EAS and FEJ}^ are equal, and give AH^EN AE EF Since AH and AE are contained the same number of times in p and p\ respectively, we have AH ^p , AE ^' and since EN and EF are contained the same number of times in »' and P', respectively, we have ^^ EW^p^, EF P" therefore we have p'=y^^lxrT\ ^2^ Therefore, from the given perimeters p and P we compute P' by the equation [1], and then with p and P' we compute p' by the equation r2"|. whence 166 ELEMENTS OF GEOMETRY. PROPOSITION XV.— PROBLEM. 36. To compute the ratio of the circumference of a circle to its diameter^ approximately. Method op Perimeters. — In this method, we take the diameter of the circle as given and compute the perimeters of some inscribed and a similar circumscribed regular poly- gon. We then compute the perimeters of inscribed and cir- cumscribed regular polygons of double the number of sides, by Proposition XIY. Taking the last-found perimeters as given, we compute the perimeters of polygons of double the number of sides by the same method ; and so on. Each com- putation gives us, of course, a pair of values between which the value of the circumference must lie. As we continue the process, these values will come nearer and nearer to the actual value of the circumference (Proposition YII.), and we may thus obtain as close an approximation to that value as we please. Taking, then, the diameter of the circle as given = 1, let us begin by inscribing and circumscribing a square. The" perimeter of the inscribed square = 4 X ? X i/^= 2y^ (27) ; J that of the circumscribed square = 4 ; therefore, putting p = 2^2 =t 2.8284271, S r I '^ ' } we find, by Proposition X., for the perimeters of the circum- scribed and inscribed regular octagons. P' = ^P^^ = 3.3187085, p' = yp xP' = 3.0614675. BOOK V. 167 Then, taking these as given quantities, we put P = 3.3137085, p = 3.0614675, and find by the same formulae for the polygons of 16 sides P' == 3.1825979, / = 3.1214452. Continuing this process, the results will be found as in the following TABLE.* Number Perimeter of Perimeter of of sides. circumscribed polygon. inscribed polygon. 4 4.0000000 2.8284271 8 3.3137085 3.0614675 16 3.182.5979 3.1214452 32 3.1517249 3.1365485 64 3.1441184 3.1403312 128 3.1422236 3.1412773 256 3.1417504 3.1415138 512 3.1416321 3.1415729 1024 3.1416025 3.1415877 2048 3.1415951 3.1415914 4096 3.1415933 3.1415923 8192 3.1415928 3.1415926 From the last two numbers of this table we learn that the circumference of the circle whose diameter is unity is less than 3.1415928 and greater than 3.1415926 ; and since, when the diameter = 1, we have (7 == tt (20), it follows that TT = 3.1415927 within a unit of the seventh decimal place. * The computations have been carried out with ten decimal places in order to insure the accuracy of the seventh place, as given in the table. 168 ELEMENTS OF GEOMETRY. 37. Scholium. Archimedes (bom 287 b.c.) was the first to assign an approximate value of ;:. By a method similar to that above, he proved that its value is between 3^ and 3^J, or, in decimals, between 3.1428 and 3.1408; he therefore as- signed its value correctly within a unit of the third decimal place. The number 3^, or %^^ usually cited as Archimedes' value of Tz (although it is but one of the two limits assigned by him), is often used as a sufficient approximation in rough computations. Metius (a.d. 1640) found the much more accurate value 1^, which correctly represents even the sixth decimal place. It is easily remembered by observing that the denominator and numerator written consecutively, thus, 113 1 355, present the first three odd numbers each written twice. More recently, the value has been found to a very great number of decimals, by the aid of series demonstrated by the Differential Calculus. Clausen and Dase, of Germany (about A.D. 1846), computing independently of each other, carried out the value to two hundred decimal places, and their results agree to the last figure. The mutual verifica- tion thus obtained stamps their results as thus far the best established value to the two-hundredth place. (See Schu- macher's Astronomische Nachrichten, No. 589.) Other com- puters have carried the value to over five hundred places, but it does not appear that their results have been verified. The value to fifteen decimal places is TT = 3.141592653589793. For the greater number of practical applications, the value t: = 3.1416 is sufficiently accurate. *~— - (^ IP EXERCISES ON BOOK V. THEOREMS. ^1. An equilateral polygon inscribed in a circle is regular. ^ 2. An equilateral polygon circumscribed about a circle is reg- ular if the number of its sides is odd. ^ 3. An equiangular polygon inscribed in a circle is regular if the number of its sides is odd. \ 4. An equiangular polygon circumscribed about a circle is reg- ular. \i 6. The area of the regular inscribed triangle is one-half the area of the regular inscribed hexagon. / 6. The area of the regular inscribed hexagon is three-fourths of that of the regular circumscribed hexagon. I 7. The area of the regular inscribed hexagon is a mean propor- tional between the areas of the inscribed and circumscribed equi- lateral triangles. (L \ 8. A plane surface may be entirely covered (as in the construc- tion of a pavement) by equal regular polygons of either three, four, or six sides. 9. A plane surface may be entirely covered by a combination of squares and regular octagons having the same side, or by dodecagons and equilateral triangles having the same side. 10. If squares be described on the sides of a regular hexagon, and their adjacent external vertices be joined, a regular dodeca- gon will be formed. 11. The diagonals of a regular pentagon form a regular pentagon. 12. The diagonals joining alternate vertices of a regular hexagon enclose a regular hexagon one-third as large as the original hex- agon. . \ 15 169 k 170 ELEMENTS OF GEOMETRY. 13. The area of the regular inscribed octagon is equal to the product of the side of the inscribed square by the diameter. Suggestion. A quarter of the octagon is the sum of two triangles having as a common base the side of the inscribed square, and having the radius as the sum of their altitudes. 14. The area of a regular inscribed dodecagon is equal to three times the square of the radius. 15. Prove the correctness of the following construction : If AB and CD are two perpendicular diameters in a circle, and E the middle point of the radius 0(7, and if EFis taken equal to EA^ then OF is equal to the side of the regular inscribed decagon, and AF is equal to the side of the regular inscribed pentagon, {v. III., 42.) 16. From any point within a regular polygon of n sides, per- pendiculars are drawn to the several sides ; prove that the sum of thea^ perpendiculars is equal to n times the apothem. Suggestion. Join the point with the vertices of the polygon, and obtain an expression for the area in terms of the perpendiculars : then see Proposition IV. 17. The side of the regular inscribed triangle is equal to the hypotenuse of a right triangle of which the sides of the inscribed square and of the regular inscribed hexagon are the sides, {v. IV., Proposition X.) 18. If a is the side of a regular decagon inscribed in a circle whose radius is i?, u a = | (1/5-1). B Suggestion. By (31), - = a H a BOOK V. 171- 19. If a = the side of a regular polygon inscribed in a circle whose radius is i?, and a' = the side of the regular inscribed polygon of double the number of sides, then a^2 =r{2R — VU^ a 0, Suggestion. ABC and ADO are similar. Hence ^ = ^ and ^UfVbut 37)^ = {2RY — a'\ c 7 a" \ ''' 20. If a = the side of a regular pentagon inscribed in a circle whose radius is i?, then R. 21. If a = the side of a regular octagon inscribed in a circle whose radius is B^ then a = BV2 — y± 22. If a = the side of a regular dodecagon inscribed in a circle whose radius is i2, then =ri2l/2-l/5. 23. The side of the regular inscribed pentagon is equal to the hypotenuse of a right triangle whose sides are the radius and the side of the regular inscribed decagon. 24. The area of a ring bounded by two concentric circumfer- ences is equal to the area of a circle having for its diameter a chord of the outer circumference tangent to the inner circumfer- ence. 25. If on the legs of a right triangle, as diameters, semicircles are described external to the triangle, and from the whole figure a semicircle on the hypot- enuse is subtracted, the remainder is equivalent to the given triangle. I^*r%.r 172 ELEMENTS OF GEOMETRY. 26. If on the two segments into which a diameter of a given circle is divided by any point, as diameters, semi-circumferences are described lying on op- posite sides of the given diameter, the sum of their lengths is equal to the length of a semi- circumference of the given circle, and a line which they form divides the circle into two parts whose areas are to each other as the seg- ments of the given diameter. 27. If a diameter of a given circle is divided into n equal parts, and through each point of division a curved line of the sort described in the last problem is drawn, these lines will divide the circle into n equivar lent parts. 28. If a circle rolls around the circumference of a circle of twice its radius, the two circles being always tangent internally, the locus of a fixed point on the cir- cumference of the rolling circle is a diameter of the fixed circle. ^^vsIfUA-^ygasguareis subdivided into n^ equal squares, nbSngany^^i^vm^number, and in each of these smaller squares a circle is in- scribed, the sum of their areas is equal to the area of the circle inscribed in the original square. MISCELLANEOUS EXERCISES PLANE GEOMETRY. THEOREMS. 1. The sum of the three straight lines drawn from any point within a triangle to the three vertices is less than the sum and greater than the half sum of the three sides of the triangle. 2. If one of the acute angles of a right triangle is double the other, the hypotenuse is double the shortest side. 3. If from any point within an equilateral triangle perpendiculars to the three sides are drawn, the sum of these lines is constant, and equal to the perpendicular from any vertex upon the opposite side. 4. Lines drawn from one vertex of a parallelogram to the middle points of the opposite sides trisect a diagonal. 6. The bisectors of the angles con- tained by the opposite sides (produced) of an inscribed quadrilateral intersect at right angles. 15* 174 ELEMENTS OF GEOMETRY. 6. If AOB is any given angle at the centre of a circle, and if BC can be drawn meeting AO produced in C, and the circumference in Z>, so that CD shall be equal to the radius of the circle, then the angle C will be equal to one-third the angle AOB. Note, There is no method known of drawing J5(7, under these conditions, and with the use of straight lines and circles only, AOB being any given angle ; so that the trisection of an angle^ in general, is a problem that cannot be solved by elementary geom- ♦*try. 7. If through P, one of the points of intersection of two circumfer- ences, any two secants, APB^ CPD^ are drawn, the straight lines, AC, DB, joining the extremities of the secants, make a constant angle PJ, equal to the angle 3fPN formed by the tangents at P. 8. If a figure is moved in a plane, it may be brought from one position to any other by revolving it about a certain fixed point ; that is, by causing each point of the figure to move in the circum- ference of a circle whose centre is the fixed point. 9. If a square DEFO is inscribed in a right triangle ABC, so that a side DE coincides with the hypotenuse BC (the vertices F and O being in the sides AC and AB), then the side DE is a mean proportional between the segments BD and EC of the hypotenuse. 10. If the middle points of the sides of a triangle are joined by straight lines, the medial lines of the triangle thus formed are the medial line^ of the original triangle, and the perpendiculars from, the vertices upon the opposite sides are the perpendiculars at the middle poihts of the sides of the original triangle. 11. If O is the centre of the circle circumscribed about a triangle ABC, and P is the intersection of the perpendiculars from the angles upon the opposite sides, the perpendicular from O upon the side JSC is equal to one-half the distance AP. MISCELLANEOUS EXERCISES. 175 12. In any triangle, the centre of the circumscribed circle, the intersection of the medial lines, and the intersection of the per- pendiculars from the angles upon the opposite sides, are in the same straight line ; and the distance of the first point from the second is one-half the distance of the second from the third. 13. If two circles intersect in the points A and B, and through A any secant CAD is drawn terminated by the circumferences at C and i), the straight lines BC and BD are to each other as the diameters of the circles. 14. If through the middle point of each diagonal of any quad- rilateral a parallel is drawn to the other diagonal, and from the intersection of these parallels straight lines are drawn to the middle points of the four sides, these straight lines divide the quadrilateral into four equivalent parts. 15. If three straight lines Aa, Bb, Ce^ drawn from the vertices of a triangle ABC to the opposite sides, pass through a com- mon point O within the triangle, then Oa , Ob ,0g ^. Aa"^ Bb~^ Cc 16. If from any point O within a tri- angle ABC any three straight lines, Oa, Ob, Oc, are drawn to the three sides, and through the vertices of the triangle three straight lines, Aa^, Bb\ Co', are drawn parallel respectively to Oa, Ob, Oc, then Oa ^^\\ Q^ —I Aa^ T Bb' "^ Cc' 17. The area of a circle is a mean proportional between the areas of any two similar polygons, one of which is circumscribed about the circle and the other isoperimetrical with the circle. {Galileo^ 8 Theorem.) 18. Two diagonals of a regular pentagon, not drawn from a common vertex, divide each other in extreme and mean ratio. 176 ELEMENTS OF GEOMETRY. LOCL 19. The angle ACB is any inscribed an- gle in a given segment of a circle ; ^C is produced to P, making CP equal to GB ; find the locus of P. 20. The hypotenuse of a right triangle is given in magnitude and position ; find the locus of the centre of the inscribed circle. 21. The base BC of a triangle ABC is given in position and magnitude, and the vertical angle J. is of a given magnitude ; find the locus of the centre of the inscribed circle. 22. From a given point O, any straight line OA is drawn to a given straight line MN^ and OP is drawn making a given angle with OA^ and such that OP is to OA in a given ratio ; find the locus of P. With the same construction, if OP is so taken that the product OP. OA is equal to a given con- stant ; find the locus of P. 23. From a given point O, any straight line OA is drawn to a given circumfer- ence, and OP is drawn making a given angle with OA, and such that OP is to OA in a given ratio ; find the locus of P. With the same construction, if OP is so taken that the product OP. OA is equal to a given constant ; find the locus of P. 24. One vertex of a triangle whose angles are given is fixed, while the second vertex moves on the circumference of a given circle ; what is the locus of the third vertex ? 25. Through A, one of the points of intersection of two given circles, any secant is drawn cutting the two circumferences in the points B and C; find the locus of the middle point of BG. A MISCELLANEOUS EXERCISES. 177 PROBLEMS. 26. Describe a circle through two given points which lie outside a given line, the centre of the circle to be in that line. Show when no solution is possible. 27. In a given circle, inscribe a chord of a given length which produced shall be tangent to another given circle. 28. Through P, one of the points of intersection of two circum- ferences, draw a straight line, terminated by the circumferences, which shall be bisected in P. 29. Through one of the points of intersection of two circum- ferences draw a straight line, terminated by the circumferences, which shall have a given length. 30. In a given triangle ABC, to inscribe a parallelogram DEFO, such that the adja- cent sides DE and DO shall be in a given ratio and contain a given angle. 31. Construct a triangle, given its base, the ratio of the other two sides, and one angle. 32. To determine a point in a given arc of a circle, such that the chords drawn from it to the extremities of the arc shall have a given ratio. 33. To find a point within a given triangle, such that the three straight lines drawn from it to the vertices of the triangle shall make three equal angles with each other. 34. Inscribe a trapezoid in a given circle, knowing its area and the common length of its inclined sides. 35. To construct a triangle, given one angle, the side opposite to that angle, and the area (equal to that of a given square). 36. Divide a given circle into a given number of equivalent parts, by concentric circumferences. Also, divide it into a given number of parts proportional to given lines, by concentric circumferences. 178 ELEMENTS OF GEOMETRY. 37. A circle being given, to find a given number of circles whose radii shall be proportional to given lines, and the sum of whose areas shall be equal to the area of the given circle. 38. In a given equilateral triangle, inscribe three equal circles tangent to each other and to the sides of the triangle. Determine the radius of these circles in terms of the side of the triangle. 39. In a given circle, inscribe three equal circles tangent to each other and to the given circle. Determine the radius of these circles in terms of the radius of the given circle. NUMERICAL EXAMPLES. Note.— The following approximate values are close enough for ordinary purposes : tt = 2^2^ ^2 = ||, |/8 = |f, /S = ff. Radius of earth = 3960 miles. \^' 40. The vertical angle of an isosceles triangle is 36°, and the length of the base is 2 feet ; find the base angles, the length of the bisector of a base angle, and the length of a side of the given triangle. Ans. TZ°, 2 feet, (1 + V5) feet. I 41. One angle of a triangle is 60°, the including sides are 8 feet and 8 feet ; find the area and the third side. Ans. 61/3 square feet, 7 feet. .^ ' V 42. The three sides of a triangle are 9 inches, 10 inches, and 17 Lo inches, its area is 36 square inches ; find the area of the inscribed ^^ circle. Ans. 47r. "^ (/ 43. The adjacent sides of a parallelogram are 12 feet and 14 feet, the area is 120 square feet ; find the long diagonal. Ans. 24 feet. 44. The area of a right triangle is 6 square feet, the length of the hypotenuse is 5 feet ; find the other sides. Ans. 3 feet, 4 feet. 45. ©btain a formula connecting the length of a chord ^, its distance from the centre c?, and the radius r. Ans. - = r^ — cP, 4 46. Obtain a formula for the length ^ of a common tangent to two circles, given tlie radii r, r'', and the distance between the centres d. Ans. {r — r^y 4-^ = ^2 for external tangent. [r -f r^f + ^ = c^ for internal tangent. 47. Through what angle does the hour-hand of a clock move in 1 hour? in 1 minute? Through what angle does the minute- hand move in 1 minute ? What angle do the hands of a clock make with each other at ten minutes past three ? at quarter of six ? Ans. 35°, 97° 30^. / 179 180 ELEMENTS OF GEOMETRY. 48. Two secants cut each other without a circle, the intercepted arcs are 12° and 48° ; what is the angle between the secants ? Two chords intersect within the circle, a pair of opposite inter- cepted arcs are 12° and 48° ; what is the angle between the chords ? 49. Tw^o tangents make with each other an angle of 60° ; re- quired the lengths of the arcs into which their points of contact divide the circle, given radius equals 7 inches. Ans. 14| inches, 29^ inches. 60. A swimmer whose eye is at the surface of the water can just see the top of a stake a mile distant ; the stake proves to be 8 inches out of water ; required the radius of the earth. Am. 3960 miles. 61. A passenger standing on the deck of a steamer about to start observes that his eye is on a level with the top of the wharf, which he knows to be 12 feet high ; when they have steamed 8^ miles the wharf disappears below the horizon ; required the radius of the earth. Ans. 3974 miles. 62. How many miles is the light of a light-house 150 feet high visible at sea ? A7is. 15. 63. On approaching Portland from the sea, Mount Washington is first visible 12 miles from shore ; Portland is 85 miles from Mount Washington ; required the height of the mountain. Ans. 6270 feet. 64. The latitude of Leipsic is 51° 21^ that of Venice 45° 26^ and Venice is due south of Leipsic ; how many miles are they apart? Use 4000 miles as the earth's radius. Ans. 413 miles. 66. The latitude of the Peak of Teneriffe is about 30° N. ; the rising sun shines on its summit on the 21st of March 9 minutes before it shines on its base ; required the height of the mountain. Ans. About 12,000 feet. 66. A quarter-mile running-track 10 feet wide, with straight parallel sides and semicircular ends, is to be laid out in a rectan- gular field 220 feet wide. How long must the field be in order that a runner, keeping in the middle of the track, may have one- quarter of a mile to cover? how much can he gain by keeping close to the inner edge of the track ? what is the area of the field ? of the portion encircled by the track? of the track itself? Ans. 550 feet ; 31f feet ; 121,000 square feet ; 97,428^ square feet ; 13,200 square feet. 67. The fly-wheel of an engine is connected by a belt with a smaller wheel driving the machinery of a mill. The radius of the NUMERICAL EXAMPLES. 181 fly-wheel is 7 feet ; of the small wheel, 21 inches. How many revolutions does the small wheel make to one of the fly-wheel ? The distance between the centres of the two wheels is 10^ feet. What is the length of the connecting band ? Ans. 51 feet 2 inches. 58. If from each vertex of a regular polygon as a centre, with a radius equal to one-half the side, an arc is described outward from side to side of the polygon, an ornamental figure much used in architecture is formed. Such a figure formed on a polygon of numerous sides is often used as a rose-window. The figure bounded by three arcs is called a trefoil ; by four arcs, a quatre-foil ; by five arcs, a cinque-foil. Find the area of a tre- foil, given the distance be- tween the centres of adja- cent arcs equal to 21 inches. Ans. 7.338 square feet. 69. A rose-window of six lobes is to be placed in a circular space 42 feet in diam- eter. How many square feet of glass will it contain ? Ans. 1123.8 square feet. SYLLABUS OF PLANE GEOMETRY. POSTULATES, AXIOMS, AISTD THEOKEMS. BOOK I. Postulate I. Through any two points one straight line, and only one, can be drawn. Postulate II. Through a given point one straight line, and only one, can be drawn having any given direction. Axiom I. A straight line is the shortest line that can be drawn between two points. Axiom II. Parallel lines have the same direction. Proposition I. At a given point in a straight line one perpendicular to the line can be drawn, and but one. Corollary. Tiirough tlie vertex of any given angle one straight line can be drawn bisecting the angle, and but one. Proposition II. All right angles are equal. ' -^ Proposition III. The two adjacent angles which one straight line makes with another are together equal to two right angles. Corollary I. The sum of all the angles having a common ver- tex, and formed on one side of a straight line, is two right angles. Corollary II. The sum of all the angles that can be formed about a point in a plane is four right angles. 182 SYLLABUS OF PLANE GEOMETRY. 183 Proposition IV. If the sum of two adjacent angles is two right angles, their exterior sides are in the same straight line. Proposition V. If two straight lines intersect each other, the opposite (or ver- tical) angles are eq^ual. Proposition VI. Two triangles are equal when two sides and the included angle of the one are respectively equal to- two sides and the included angle of the other. Proposition VII. Two triangles are equal when a side and the two adjacent angles of the one are respectively equal to a side and the two adjacent angles of the other. Proposition VIII. In an isosceles triangle the angles opposite the equal sides are equal. Corollary. The straight line bisecting the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base. Proposition IX. Two triangles are equal when the three sides of the one are respectively equal to the three sides of the other. Proposition X. Two right triangles are equal when they have the hypotenuse and a side of the one respectively equal to the hypotenuse and a side of the other. Proposition XI. If two angles of a triangle are equal, the sides opposite to them are equal, and the triangle is isosceles. Proposition XII. If two angles of a triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. 184 ELEMENTS OF GEOMETRY. Proposition XIII. If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. Proposition XIV. If two triangles have two sides of the one respectively equal to two sides of the other, and the included angles unequal, the tri- angle which has the greater included angle has the greater third side. Proposition XV. If two triangles have two sides of the one respectively equal to two sides of the other, and the third sides unequal, the triangle which has the greater third side has the greater included angle. Proposition XVI. From a given point without a straight line one perpendicular can be drawn to the line, and but one. Proposition XVII. The perpendicular is the shortest hne that can be drawn from a point to a straight line. Proposition XVIII. If a perpendicular is erected at the middle of a straight line, then every point on the perpendicular is equally distant from the extremities of the line, and every point not on the perpendicular is unequally distant from the extremities of the line. Proposition XIX. Every point in the bisector of an angle is equally distant from the sides of the angle ; and every point not in the bisector is un- equally distant from the sides of the angle ; that is, the bisector of an angle is the locus of the points within the angle and equally distant from its sides. Proposition XX, A convex broken line is less than any ether line which envelops it and has the same extremities. SYLLABUS OF PLANE GEOMETRY. 185 Proposition XXI. If two oblique lines drawn from a point to a line meet the line at unequal distances from the foot of the perpendicular, the more remote is the greater. Proposition XXII. Two straight lines perpendicular to the same straight line are parallel. Proposition XXIII. Through a given point one line, and only one, can be .drawn parallel to a given line. Proposition XXIV. When two straight lines are cut by a third, if the alternate- interior angles are equal, the two straight lines are parallel. Corollary I. When two straight lines are cut by a third, if a pair of corresponding angles are equal, the lines are parallel. Corollary II. When two straight lines are cut by a third, if the sum of two interior angles on the same side of the secant line is equal to two right angles, the two lines are parallel. Proposition XXV. If two parallel lines are cut by a third straight line, the alter- nate-interior angles are equal. Corollary I. If two parallel lines are cut by a third straight line, any two corresponding angles are equal. Corollary II. If two parallel lines are cut by a third straight line, the sum of the two interior angles on the same side of the secant line is equal to two right angles. Proposition XXVI. The sum of the three angles of any triangle is equal to two right angles. Corollary. If one side of a triangle is extended, the exterior angle is equal to the sum of the two interior opposite angles. Proposition XXVII. The sum of all the angles of any convex polygon is equal to twice as many right angles, less four, as the figure has sides. 16* 186 ELEMENTS OF GEOMETRY. Proposition XXVIII. Two parallelograms are equal when two adjacent sides and the included angle of the one are equal to two adjacent sides and the included angle of the other. Corollary. Two rectangles are equal when they have equal bases and equal altitudes. Proposition XXIX. The opposite sides of a parallelogram are equal, and the oppo- site angles are equal. Proposition XXX. ' If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Proposition XXXI. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. Proposition XXXII. The diagonals of a parallelogram bisect each other. BOOK II. PROPOSITIONS. Postulate. A circumference may be described with any point as centre and any distance as radius. Proposition I. Two circles are equal when the radius of the one is equal to the radius of the other. Proposition II. Every diameter bisects the circle and its circumference. Proposition III. In equal circles, or in the same circle, equal angles at the centre intercept equal arcs on the circumference. Corollary. Conversely, in the same circle, or in equal circles, equal arcs subtend equal angles at the centre. SYLLABUS OF PLANE GEOMETRY. 187 Proposition IV. In equal circles, or in the same circle, equal arcs are subtended by equal chords. Corollary. Conversely, in equal circles, or in the same circle, equal chords subtend equal arcs. Proposition V. In equal circles, or in the same circle, the greater of two un- equal arcs is subtended by the greater chord, the arcs being each less than a semi-circumference. Corollary. Conversely, in equal circles, or in the same circle the greater of two unequal chords subtends the greater arc. Proposition VI. The diameter perpendicular to a chord bisects the chord and the arcs subtended by it. Corollary I. The perpendicular erected at the middle point of a chord passes through the centre of the circle. Corollary II. When two circumferences intersect, the straight line joining their centres bisects their common chord at right angles. Proposition VII. In the same circle, or in equal circles, equal chords are equally distant from the centre ; and of two unequal chords the less is at the greater distance from the centre. Corollary. Conversely, in the same circle, or in equal circles, chords equally distant from the centre are equal ; and of two chords unequally distant from the centre, that is the greater whose distance from the centre is the less. Proposition VIII. A straight line cannot intersect a circle in more than two points. Proposition IX. A straight line tangent to a circle is perpendicular to the radius drawn to the point of contact. Corollary I. A perpendicular to a tangent line drawn through the point of c(mtact must pass through the centre of the circle. Corollary II. If two circumferences are tangent to each other, their centres and their point of contact lie in the same straight line. V^ 188 ELEMENTS OF GEOMETRY. Proposition X. When two tangents to the same circle intersect, the distances from their point of intersection to their points of contact are equal. Proposition XI. Two parallels intercept equal arcs on a circumference. Doctrine of Limits. — Theorem. If two variables dependent upon the same variable are so re- lated that they are always equal, no matter what value is given to the variable on which they depend, and if, as the independent variable is changed in some specified way, each of them ap- proaches a limit, the two Umits must be absolutely equal. Proposition XII. In the same circle, or in equal circles, two angles at the centre are in the same ratio as their intercepted arcs. Proposition XIII. The numerical measure of an angle at the centre of a circle is the same as the numerical measure of its intercepted arc, if the unit of angle is the angle at the centre which intercepts the adopted unit of arc. Proposition XIV. An inscribed angle is measured by one-half its intercepted arc. Corollary. An angle inscribed in a semicircle is a right angle. Proposition XV. An angle formed by a tangent and a chord is measured by one- half the intercepted arc. Proposition XVI. An angle formed by two chords intersecting within the circum- ference is measured by one-half the sum of the arcs intercepted between its sides and between the sides of its vertical angle. Proposition XVII. An angle formed by two secants intersecting without the cir- cumference is measured by one-half the difference of the inter- cepted arcs. SYLLABUS OF PLANE GEOMETRY. 189 Proposition XVIII. An angle formed by a tangent and a secant is measured by one- half the difference of the intercepted arcs. Corollary. An angle formed by two tangents is measured by one-half the difference of the intercepted arcs. BOOK III. THEOREMS. Proposition I. A parallel to the base of a triangle divides the other two sides proportionally. Proposition II. If a straight line divides two sides of a triangle proportionally, it is parallel to the third side. Proposition III. Two triangles are similar when they are mutually equiangular. Proposition IV. Two triangles are similar when an angle in the one is equal to an angle in the other, and the sides including these angles are proportional. Proposition V. Two triangles are similar when their homologous sides are proportional. Proposition VI. If two polygons are composed of the same number of triangles, similar each to each and similarly placed, the polygons are similar. Proposition VII. Two similar polygons may be decomposed into the same num- ber of triangles, similar each to each and. similarly placed. Proposition VIII. The perimeters of two similar polygons are in the same ratio as any two homologous sides. 190 ELEMENTS OF GEOMETRY. Proposition IX. If a perpendicular is drawn from the vertex of the right angle to the hypotenuse of a right triangle : 1st. The two triangles thus formed are similar to each othei and to the whole triangle ; 2d. The perpendicular is a mean proportional between the seg^ ments of tlie hypotenuse ; 3d. Each side about the right angle is a mean proportional be- tween the whole hypotenuse and the adjacent segment. Corollary. If from any point in the circumference of a circle a perpendicular is let fall upon a diameter, the perpendicular is a mean proportional between the segments of the diameter. Proposition X. The square of the length of the hypotenuse of a right triangle is the sum of the squares of the lengths of the other two sides, the three lengths being expressed in terms of the same unit. Proposition XI. If two chords intersect within a circle, their segments are re- ciprocally proportional. Proposition XII. If two secants intersect without a circle, the whole secants and their external segments are reciprocally proportional. Corollary. If a tangent and a secant intersect, the tangent is a mean proportional between the whole secant and its external segment. BOOK lY. THEOREMS. Proposition I. Parallelograms having equal bases and equal altitudes are equivalent. Corollary. Any parallelogram is equivalent to a rectangle having the same base and the same altitude. SYLLABUS OF PLANE GEOMETRY. 191 Proposition II. Two rectangles having equal altitudes are to each other as their bases. Corollary. Two rectangles having equal bases are to each other as their altitudes. Proposition III. Any two rectangles are to each other as the products of their bases by their altitudes. Proposition IV. The area of a rectangle is equal to the product of its base and altitude. Proposition V. The area of a parallelogram is equal to the product of its base and altitude. Proposition VI. The area of a triangle is equal to half the product of its base and altitude. Corollary I. A triangle is equivalent to one-half of any paral- lelogram having the same base and the same altitude. Corollary II. Triangles having equal bases and equal altitudes are equivalent. Corollary III. Triangles having equal altitudes are to each other as their bases, and triangles having equal bases are to each other as their altitudes. Proposition VII. The area of a trapezoid is equal to the product of its altitude by half the sum of its parallel bases. Proposition VIII. Similar triangles are to each other as the squares of their homol- ogous sides. Proposition IX. Similar polygons are to each other as the squares of their homol- ogous sides. Proposition X. The square described upon the hypotenuse of a right triangle is equivalent to the sum of the squares described on the other two sides. 192 ELEMENTS OF GEOMETRY. BOOK V. THEOREMS. Proposition I. If the circumference of a circle be divided into any number of equal parts, the chords joining the successive points of division form a regular polygon inscribed in the circle ; and the tangents drawn at the points of division form a regular polygon circum- scribed about the circle. Corollary I. If the vertices of a regular inscribed polygon are joined with the middle points of the arcs subtended by the sides of the polygon, the joining Unes will form a regular inscribed polygon of double the number of sides. Corollary II. If at the middle points of the arcs joining ad- jacent points of contact of the sides of a regular circumscribed polygon tangents are drawn, a regular circumscribed polygon of double the number of sides will be formed. Proposition II. A circle may be circumscribed about any regular polygon, and a circle may also be inscribed in it. Proposition III. Regular polygons of the same number of sides are similar. Corollary. The perimeters of regular polygons of the same number of sides are to each other as the radii of the circumscribed circles, or as the radii of the inscribed circles ; and their areas are to each other as the squares of these radii. Proposition IV. The area of a regular polygon is equal to half the product of Its perimeter and apothem. Proposition V. An arc of a circle is less than any line which envelops it and has the same extremities. Corollary, The circumference of a circle is less than the perim- eter of any polygon circumscribed about it. SYLLABUS OF PLANE GEOMETRY. 193 Proposition VI. If the number of sides of a regular polygon inscribed in a circle be increased indefinitely, the apothem of the polygon will ap- proach the radius of the circle as its limit. Proposition VII. The circumference of a circle is the Umit which the perimeters of regular inscribed and circumscribed polygons approach when the number of their sides is increased indefinitely ; and the area of the circle is the limit of the areas of these polygons. Proposition VIII. The circumferences of two circles are to each other as their radii, and their areas are to each other as the squares of their radii. Corollary I. The circumferences of circles are to each other as their diameters, and their areas are to each other as the squares of their diameters. Corollary II. The ratio of the circumference of a circle to its diameter is constant. Proposition IX. The area of a circle is equal to half the product of its circum- ference by its radius. Corollary. The area of a circle is equal to the square of its radius multiplied by the constant number tt. GEOMETRY OE SPACE. In Plane Geometry we have considered merely figures com- posed of lines and points, all of which are supposed to lie in the same plane (v. Introduction, 5 and 6), and in the proposi- tions and definitions of the preceding five books it has been tacitly assumed that the figures in question are plane figures. In many of the propositions and definitions this limitation is essential to the truth of the proposition ; for example. Propo- sitions I. and XXII., Book I., and Definition 20, Book II. In others the demonstration given is inconclusive without the limitation in question, although the proposition is true even when the limitation is removed; for example. Exercise 1, Proposition XXIII., Book I. While in propositions concern- ing equal polygons, which depend for their proof directly or indirectly upon a superposition of one polygon upon the other, the limitation is obviously of no importance; for ex- ample. Propositions YI., YII., and IX., Book I. It is, then, important, when we use the theorems of Plane Geometry in proving theorems of the Geometry of Space, to satisfy our- selves that they are still true in the figures with which we are concerned. 194 BOOK YI THE PLANE. POLYBDRAL ANGLES. 1. Definition. A plane has already been defined as a surface such that the straight line joining any two points in it lies wholly in the surface. JIT Thus, the surface ikfiV" is a plane, if, A / and B being any two points in it, the /—4- ?- straight line AB lies wholly in the sur- face. The plane is understood to be indefinite in extent, so that, however far the straight line is produced, all its points lie in the plane. But to represent a plane in a diagram, we are obliged to take a limited portion of it, and we usually represent it by a parallelogram supposed to lie in the plane. 2. Definition. A plane is said to be determined by given lines or points when one plane, and only one, can be drawn con- taining the given lines or points. PKOPOSITION I.— THEOREM. K^^Through any given straight line a plane may be passed ; hut the line will not determine the plane. Let AB be a given straight line. A straight line may be drawn in any plane, and the position of that plane may be changed until the line drawn in it is brought into coincidence with AB. We shall then have a plane passed through AB ; and this plane 195 r ~ J 7^ !.-'' '/ aJ. /I B L- 1 r|o ^ ELEMENTS OF GEOMETRY. may be turned upon AB as an axis, and made to occupy- as many different positions as we choose, and as in each of these positions it is a plane through AB, we may have as many planes as lL we choose through AB ; conse- quently AB does not determine (2) a plane. PBOPOSITION II.— THEOKEM. / 4. ^ plane is determined^ 1st, by a straight line and a point / without that line ; 2d, by two intersecting straight lines ; 3d, by 7 three points not in the same straight line ; 4th, by two parallel ! straight lines. Ist. Through a given line AB a plane may be passed, and may then / -^^ must therefore coincide with CD, by I., Postulate 11. Hence,- ^ AB and CD are parallel, (f "^^ T''- ff • ^ ^•■' f '^-^. tT^l^ 15. Corollary. Two lines paraltet to the seme lirie are par- allel to each other. For they have the same direction. ^..^^ PROPOSITION VI.— THEOREM. v^5r-v X^ <^ ^ BOOK VI. 207 25. Corollary. Two straight lines perpendicular to the same plane are parallel to each other, EXERCISE. Theorem. — Two parallel planes are everywhere equally distant, BIEDRAL ANGLES.— ANGLE OF A LINE AND PLANE, ETC. 26. Definition. When two planes meet and are terminated by their common intersection, they form a diedral angle. Thus, the planes AE, AF^ meeting in AB^ and terminated by AB^ form a diedral angle. The planes AE^ AF, are called the faceSy and the line AB the edge, of the diedral angle. A diedral angle may be named by four letters, one in each face and two on its edge, the two on the edge being written between the other two ; thus, the angle in the figure may be named DABC. When there is but one diedral angle formed at the same? edge, it may be nariied by two letters on its edge ,- thus, in the preceding figure, the diedral angle DABC may be named the diedral angle AB. 27. Definition. The angle CAD formed hj two straight lines AC, AD, drawn, one in each face of the diedral angle, per- pendicular to its edge AB at the same point, is called the plane angle of the diedral angle. EXERCISES. 1. Theorem. — All plane angles of the same diedral angle are equal, (v. Proposition X.) 2. Theorem. — If a plane is drawn perpendicidar to the edge uj a diedral angle, its intersections with the faces of the diedral angle form the plane angle of the diedral angle. 208 ELEMENTS OF GEOMETRY. Br c D' Ff 28. A diedral angle DABG may be conceived to be gener- ated by a plane, at first coincident with a fixed plane AE, revolving upon the line A ^ as an axis until it comes into the position AF. In this revolution a straight line CA, perpen- dicular to AB, generates the plane angle CAD. 29. Definition. Two diedral angles are equal when they can be placed so that their faces shall coincide. Thus, the diedral angles CABD, G'A'B'D', are equal, if, when the edge A'B' is applied to the edge AB and the face A'F' to the face AF^ the face A'E' also coincides with the face AE. Since the faces continue to co- incide when produced indefinitely, it is apparent that the magnitude of the diedral angle does not depend upon the extent of its faces, but only upon their relative position. 30. Definition. Two diedral angles GABD, DABEj which have a common edge AB and a common plane BD between them, are called adjacent. Two diedral angles are added together by placing them adjacent to each other. Thus, the diedral angle GABE is the sum of the two diedral angles GABD and DABE. E' ,,^ 31. Definition. Two planes are perpen- dicular when the plane angle of the di- edral angle which they form is a right angle. The diedral angle is then called a right diedral angle. M / cA VI N BOOK VI. 209 1/ B' C F' PROPOSITION XII.— THEOREM. (^ Two diedral angles are equal if their plane angles are equal. Let the plane angles CAD and C'A'D' of the diedral angles CABD, C'A'B'D', be equal; then are the diedral angles equal. For, superpose C'A'B'iy upon CABD, making the plane angle C'A'D' coincide with its equal CAD ; then the planes of these angles will coincide (Proposition II.). A'B' and AB^ being now perpendicular to the same plane at the same point, must coincide (Proposition III., Corollary) ; and, finally, the planes B'C and 50 will coincide, and B'D' and BD (Propo- sition 11.) . Therefore the diedral angles are equal. PROPOSITION XIII.— THEOREM. 331 Two diedral angles are in the same ratio as. their angles. Let CABD and GEFH be two ' diedral angles; and let CAD and GEH be their plane angles. Suppose the plane angles have a common measure, contained m times in CAD and n times in GEH; we have, then, CAD ^m GEH n plane H V- ~ 1 1 II 1 III 1 ill 1 1 1 1 1 1 II F a - -1 V ^^^^ -ij Apply this measure to CAD and GEH^ and through the lines of division and the edges of the given diedral angles o 18* 210 ELEMENTS OF GEOMETRY. pass planes, thus dividing CABD into m and GEFII into n smaller diedral angles. Each of these small diedral angles has one of the parts into which CAD is divided, or one of the parts into which GEH is divided, as its plane angle, because AB is per- pendicular to the plane of CAD, and EF to the plane of GEH, by Proposition lY. These small diedral angles are, then, all equal, by Proposition XII., and we have ?Fr::rl " ^^-\ ■\i Therefore CABD ^m GEFH n' CABD ^ CAD GEFH GEH' The proof is extended to the case where the given planej angles are incommensurable, by the method exemplified in the proof of II., Proposition XII., of III., Proposition I., and of IV., Proposition II. 34. Scholium. Since the diedral angle is proportional to its plane angle (that is, varies proportionally with it), the plane angle is taken as the measure of the diedral angle, just as an arc is taken as the measure of a plane angle. Thus, a diedral angle will be expressed by 45° if its plane angle is expressed by 45°, etc. PROPOSITION XIV.— THEOREM. V^Sa If a straight line is perpendicular to a plane, every plane passed through the line is also perpendicular to that plane. BOOK VV 211 Let AB be perpendicular to the plane MJSf; then any plane PQ, passed through AB, is also perpendic- ular to JfiV. For, at B draw BC^ in the plane MWj perpendicular to the intersection BQ. Since AB is perpendicular to the plane 3IW, it is perpendicular to BQ and BG; therefore the angle ABC is the plane angle of the diedral angle formed by the planes PQ and MN; and since the angle ABC is a right angle, the planes are perpendicular to each other. PROPOSITION XV.-THEOBEM. (^g) If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their intersection, is perpendicular to the other. Let the planes PQ and JfZV be perpendic- ular to each other ; and at any point B of their intersection BQ let BA be drawn, in the plane PQ, perpendicular to BQ; then BA is perpendicular to the plane 3fN. For, drawing BC, in the plane MW, per- pendicular to BQ, the angle ABC is a right angle, since it is the plane angle? of the right diedral angle formed by the two planes ; therefore AB, perpendicular to the two straight lines BQ, BC, is perpendicular to their plane MW (Proposition IV.). 37. Corollary I. If two planes are perpendicular to each other, a straight line drawn through any point of their intersec- tion perpendicular to one of the planes will lie in the other, (v. Proposition III., Corollary.) 212 ELEMENTS OF GEOMETRY. 38. Corollary II. If two planes are perpendicular^ a straight line let fall from any point of one plane perpendicular to the other will lie in the first plane, (v. Proposition III.) PROPOSITION XVI.— THEOREM. 39. If two intersecting planes are each perpendicular to a third plane, their intersection is also perpendicular to that plane. Let the planes PQ, BS, intersecting in the line AB, be perpendicular to the plane MN ; then AB is perpendicular to the plane M]Sf. For, if from any point A of A B a per- pendicular be drawn to MNj this perpen- dicular will lie in each of the planes PQ and BS (Proposition XY., Corollary II.), and must therefore be their intersection AB. EE3 PROPOSITION XVII.— THEOREM. fwi Through any given straight line a plane can be passed perpendicular to any given plane. Let AB be the given straight line, and MW the given plane. From any point A of AB let AC he drawn perpendicular to MN, and through AB and A C pass a plane AD. This plane is perpendicular to MN (Proposition XIY.). Moreover, since, by Proposition XY., Corollary IL, any plane passed through AB perpendicular to MN must contain the perpendicular A C, the plane AD is the only plane per- pendicular to MN that can be passed through AB, unless AB is itself perpendicular to MN, in which case every plai^ through AB is perpendicular to MN. ^ a BOOK VI. 213 EXERCISE. Theorem. — The locus of the points equally distant from two given -planes is ttie plane bisecting the diedral angle be- tween the given planes, (v. I., Proposition XIX.) 41. Definitions. The projection of a point A upon a plane MN is the foot a of the perpendicular let fall from A upon the plane. The projection of a line ABODE . . . upon a plane MN is the line abcde . . . containing the projections of all the points of the line ABODE . . . upon the plane. PROPOSITION XVIII.— THEOREM. (42^ The projection of a straight line upon a plane is a straight line. Let AB be the given straight line, and MI^ the given plane. The plane Ab, passed through AB perpendicular to' the plane MN, contains all the per- pendiculars let fall from points of AB upon MN (Proposition XY., Corollary /V ^ i^u\t i *^ II.); therefore these perpendiculars all t^t^^ . ^^^'^a-y.^AJ meet the plane MN in the intersection ab of the perpeiidic- i , V^ ular plane with MN. The projection of AB upon the plane MN is, consequently, the straight line ab. 214 ELEMENTS OF GEOMETRY. 43. Scholium. The plane Ah is called the projecting plane of the straight line AB upon the plane MN. PKOPOSITION XIX.— THEOREM. 44. The acute angle which a straight line makes with its own projection upon a plane is the least angle which it makes with any line of that plane. Let Ba be the projection of the straight line BA upon the plane MN, the point B being the point of intersection of the line BA with the plane ; let BC be any other straight line drawn through B in the plane; then the angle ABa is less than the angle ABC. For, take BG = Ba, and join AC. In the triangles ABa, ABC, we have AB common, and Ba = BC; but Aa ASC. For in the face ASC draw SD making the angle ASD equal to ASB, and through any point D of SB draw any straight line ADC cutting SA and SG; take SB = SD, and join AByBG. The triangles ASD and ASB are equal, by the construction (I., Proposition YI.), whence AD = AB. Now, in the tri- angle ABC, we have AB + BC>AG, and, subtracting the equals AB and AD, BC > DC; therefore, in the triangles BSC and DSC, we have the angle BSC > DSC (I., Proposition XY.), and adding the equal angles ASB and ASD, we have ASB + BSC >-ASC. BOOK VT. 217 PROPOSITION XXI.— THEOREM. [5^ The sum of the face angles of any convex polyedral angle is less than four right angles. Let the polyedral angle S be cut by a plane, making the section ABCDE^ by hypothesis, a convex polygon. From any point within this polygon draw OA, OB, OC, OB, OK The sum of the angles of the triangles ASB, BSC, etc., which have the common vertex S, is equal to the sum of the an- gles of the same number of triangles A OB, BOO, etc., which have the common vertex 0. But in the triedral angles formed at A, B, 0, etc., by the faces of the polyedral angle and the plane of the polygon, we have (Proposition XX.) SAB + SAB > EAB, SBA + SBC> ABC, etc. j hence, taking the sum of all these inequalities, it follows that the sum of the angles at the bases of the triangles whose vertex is S is greater than the sum of the angles at the bases of the triangles whose vertex is 0; therefore the sum of the angles at >S^ is less than the sum of the angles at 0; that is, less than four right angles. r> 218 ELEMENTS OF GEOMETRY. *>C? PROPOSITION XXII.-THEOREM. 55. If two triedral angles have the three face angles of the one respectively equal to the three face angles of the other^ the cor- responding diedral angles are equal. In the triedral angles S and s, let ASB = ash, ASC = asc, and BSC = bsc ; then the diedral angle SA is equal to the diedral angle sa. \ .^ On the edges of these angles take the six equal distances SA, SB, SC, sa, sb, sc, and draw AB, BC, AC, ah, he, ac. The isosceles triangles SAB and sab are equal, having an equal angle included by equal sides, hence AB = ah ; and for the same reason, BG = hc, AC = ac ; therefore the triangles ABC and abc are equal. At any point D in SA, draw DE in the face ASB and DF in the face ASC, perpendicular to SA; these lines meet AB and AC, respectively, for, the triangles ASB and ASC being isosceles, the angles SAB and SAC are acute; let E and F be the points of meeting, and join EF. Now on sa take sd = SB, and repeat the same construction in the triedral angle s. The triangles ADE and ade are equal, since AB = ad, and the angles at A and B are equal to the angles at a and d; hence AE = ae and BE = de. In the same manner we have BOOK VI. 219 AF = of and DF = df. Therefore the triangles AEF and aef are equal (I., Proposition YI.)j and we have FF = ef. Finally, the triangles FDF and edf, being mutually equilat- eral, are equal; therefore the angle FDF, which measures the diedral angle SA, is equal to the angle edf, which meas- ures the diedral angle sa, and the diedral angles SA and sa are equal (Proposition XII.). In the same manner it may be proved that the diedral angles SB and SO are equal to the diedral angles sb and sc, respectively. Scholium. It follows that the polyedral angles >S^ and s are either equal or symmetrical. Both cases are represented in the figure. EXERCISES ON BOOK YI. THEOREMS. 1. If a straight line AB is parallel to a plane JfJV, any plane perpendicular to the line AB is perpendicular to the plane 3lJS\ (v. Proposition VI., Exercise.) 2. If a plane is passed through one of the diagonals of a paral- lelogram, the perpendiculars to this plane from the extremities of the other diagonal are equal. 3. If the intersections' of a number of planes are parallel, all the perpendiculars to these planes, drawn from a common point in space, lie in one plane. Suggestion. Through the common point pass a plane perpen- dicular to one of the intersections, {v. Proposition XV., Corol- lary II.) 4. If the projections of a number of points on a plane are in a straight line, these points are in one plane. 5. If each of the projections of a line AB upon two intersecting planes is a straight line, the line AB is a straight line. 6. Two straight lines not in the same plane being given : 1st, a common perpendicular to the two lines can be drawn ; 2d, the common perpendicular is the short- . est distance between the two lines. o e d Suggestion. Let AB and CD be the two given lines. Pass through AB a plane MN parallel to CX>, and through AB and CD pass planes perpendicular to MN. Their inter- section Co is the required common perpendicular. CD and cd are par- allel, by 18, Exercise. N 2d. Any other line BF joining AB and CD is greater than JEIT, the perpendicular from B to cd (Proposition XV.), and therefore greater than Cc. 220 M ,1 1 l'\ /i i / i V " ' J \ 7 BOOK VI. 7. If two straight lines are intersected by three parallel planes, their corresponding segments are proportional, {v. Proposition VIII.) 8. A plane passed through the middle point of the common perpendicular to two straight lines in space, and parallel to both these lines, bisects every straight line joining a point of one of these Hues to a point of the other, {v. Exercise 7.) 9. In any triedral angle, the three planes bisecting the three diedral angles intersect in the same straight line. {v. 40, Exer- cise.) 10. In any triedral angle, the three planes passed through the edges and the bisectors of the opposite face angles respectively in- tersect in the same straight line. Suggestion. Lay off equal distances SAj SB, SC, on the three edges, and pass a plane through Aj B, C. The intersections of the three planes in question with ABCsive the medial lines of ABC, and have a common intersection, and the line joining this common intersection with S lies in the three planes. 11. In any triedral angle, the three planes passed through the bisectors of the face angles, and perpendicular to these faces respectively, intersect in the same straight line. Suggestion. Use the same construction as in Exercise 10. Then the intersections of the three planes with ABC are perpendicular to the sides of ABC at their middle points, and have a common intersection. 12. In any triedral angle, the three planes passed through the edges, perpendicular to the opposite faces respectively, intersect in the same straight line. Suggestion. At any point A of one of the edges, draw a plane ABC perpendicular to the edge SA. The intersections of the three planes with ABC are the perpendiculars from the vertices of ABC^ upon the oppo- site sides, and have a common intersection, {v. Proposition XVI.) 19* 222 ELEMENTS OF GEOMETRY. LOCI. 13. Find the locus of the points in space which are equally distant from two given points. 14. Locus of the points which are equally distant from two given straight lines in the same plane. 15. Locus of the points which are equally distant from three given points. 16. Locus of the points which are equally distant from three given planes, {v. 40, Exercise.) 17. Locus of the points which are equally distant from three given straight lines in the same plane. \ 18. Locus of the points which are equally distant from the three edges of a tried ral angle (Exercise 11). 19. Locus of the points in a given plane which are equally dis- tant from two given points out of the plane. 20. Locus of the points which are equally distant from two given planes, and at the same time equally distant from two given points. PROBLEMS. In the solution of problems in space, we assume, — 1st, that a plane can be drawn passing through three given points (or two in- tersecting straight lines) and its intersections with given straight lines or planes determined; and, 2d, that a perpendicular to a given plane can be drawn at a given point in the plane, or from a given point without it. The actual graphic construction of the solutions belongs to Descriptive Geometry. 21. Through a given straight line, to pass a plane perpendicular to a given plane, (v. Proposition XVII.) 22. Through a given point, to pass a plane perpendicular to a given straight line. Suggestion. If the given point is in the given line, pass two planes through the given line, and draw in each of them, through the given point, a line perpendicular to the given line. The plane determined by these lines is the perpendicular plane required. {v. Proposition IV.) If the given point is not in the given line, pass a plane through it and the given line, and in this plane, through the given point, draw a line parallel to the given line. A plane through the given point, perpendicular to this second line, is the plane required, (v. Proposition XI.) BOOK VI. 223 23. Through a given point, to pass a plane parallel to a given plane, (v. Proposition IX., Corollary.) 24. To determine that point in a given straight line which is equidistant from two given points not in the same plane with the given line. (v. Exercise 13.) 25. To find a point in a plane which shall be equidistant from three given points in space. 26. Through a given point in space, to draw a straight line which shall cut two given straight lines not in the same plane. Suggestion. Pass a plane through the given point and through one of the given lines ; the line through the given point and the point where the plane cuts the second given line is the solution required. 27. Through a given point, to draw a straight line which shall meet a given straight line and the circumference of a given circle not in the same plane. (Two solutions in general.) 28. In a given plane and through a given point of the plane, to draw a straight line which shall be perpendicular to a given line in space. Suggestion. Draw a plane through the given point and perpen- dicular to the given line. Its intersection with the given plane is the solution required. 29. Through a given point ^ in a plane, to draw a straight line AT in that plane, which shall be at a given distance PT from a given point P without the plane. Suggestion. Drop a perpendicular from P to the plane, and with the foot of this perpendicular as a centre, and with a radius equal to a side of a right triangle whose hypotenuse is PT, and whose other side is the length of the perpendicular, describe a circum- ference in the plane. A tangent from A to this circumference is the solution required, {v. 12, Exercise 2.) 30. Through a given point A, to draw to a given plane M a straight line which shall be parallel to a given plane N and of a given length. BOOK YIL POLYBDRONS. 1. Definition. A polyedron is a geometrical solid bounded by planes. The bounding planes, by their mutual intersections, limit each other, and determine the faces (which are polygons), the edgeSj and the vertices of the polyedron. A diagonal of a polyedron is a straight line joining^ any two of its vertices not in the same face. The least number of planes that can form a polyedral angle is three ; but the space within the angle is indefinite in extent, and it requires a fourth plane to enclose a finite por- tion of space, or to form a solid ; hence the least number of planes that can form a polyedron is four. 2. Definition. A polyedron of four faces is called a tetra- edron ; one of six faces, a hexaedron ; one of eight faces, an octaedron ; one of twelve faces, a dodecaedron ; one of twenty faces, an icosaedron. 3. Definition. A polyedron is convex when the section formed by any plane intersecting it is a convex polygon. All the polyedrons treated of in this work will be under- stood to be convex. 4. Definition. The volume of any polyedron is the numer- ical measure of its magnitude, referred to some other poly- edron as the unit. The polyedron adopted as the unit is called the unit of volume. To measure the volume of a polyedron is, then, to find its ratio to the unit of volume. 224 BOOK VII. 225 The most convenient unit of volume is the cube whose edge is the linear unit. 5. Definition. Equivalent solids are those which have equal volumes. PEISMS AND PARALLEL0PIPED8. 6. Definitions. A prism is a polyedron two of whose oppo- site faces, called hases^ are in parallel planes, and whose lateral edges (that is, the edges intersecting the bases) are all parallel to the same line. From this definition werreadily deduce the following consequences : 1st. Any two lateral edges of a prism are parallel (VI., Proposition Y., Corollary). 2d. All the lateral faces of a prism are parallelograms (YI., Proposition YIII.). Hence all the lateral edges are equal. 3d. The bases of a prism are equal polygons (YI., Proposi- tion X.). The lateral faces of' a prism constitute its lateral or convex surface. The altitude of a prism is the perpendicular distance be- tween the planes of its bases (YI., Proposition IX.) A triangular prism is one whose base is a triangle ; a quad- rangular prism, one whose base is a quadrilateral ; etc. 7. Definitions. A right prism is one whose lateral edges are perpendicular to the planes of its faces (YI., Proposition XI.) In a right prism, any lateral edge is equal to the altitude. The lateral faces of a right prism are perpen- dicular to the bases (YI., Proposition XIY.) 226 ELEMENTS OF GEOMETRY. An oblique prism is one whose lateral edges are oblique to the planes of its bases. In an oblique prism, a lateral edge is greater than the altitude. 8. Definition. A regular prism is a right prism whose bases are regular polygons. 8. Definition. Ifaprism,A5CZ)jE^-JF; is intersected by a plane GK, not pj^rallel to its base, the portion of the prism included between the base and this plane, namely, \A 5 (7D.&- QHIKLy is called a truncated prism. 10. Definition. A right section of a prism is the section made by a plane passed through the prism perpendicular to one of its lateral edges. A right section is perpendicular to all the lateral edges (YI., Proposition XI.) and to all the lateral faces (YI., Propo- sition XIY.) of the prism. 11. Definition. A parallelopiped is a prism whose bases are parallelograms. It is there- fore a polyedron all of whose faces are par- allelograms. From this definition and YI., Proposi- tion X., it is evident that any two oppo- site faces of a parallelopiped are equal parallelograms. 12. Definition. A right parallelopiped is a par- allelopiped whose lateral edges are perpendic- ular to the planes of its bases. Hence, by YI., 6, its lateral faces are rectangles ; but its bases may be either rhomboids or rectangles. BOOK VII. 227 A rectangular parallelopiped is a right parallelepiped whose bases are rectangles. Hence it is a parallelopiped all of whose faces are rectangles. 13. Definition. A cube is a rectangular parallelo- piped whose edges are all equal. Hence its faces are all squares. PROPOSITION I.— THEOREM. 14. The sections of a prism made by parallel planes are equal polygons. For the portion of the prism in- cluded between the two sections is a new prism (6). Therefore its bases, which are the sections in question, are equal. 15. Corollary. Any section of a prism made by a plane parallel to the base is equal to the base. EXERCISE. Theorem. — In a rectangular parallelopiped, the four diagonals are equal to each other ; and the square of a diagonal is equal to the sum of the squares of the three edges which meet at a common vertex. 228 ELEMENTS OF GEOMETRY. PROPOSITION II.— THEOREM. 16. The lateral area of a prism is equal to the product of the perimeter of a right section of the prism by a lateral edge. Let AD' be a prism, and GHIKL a right section of it; then the area of the convex surface of the prism is equal to the perimeter GHIKL multiplied by a lateral edge AA'. For, the sides of the section GHIKL^ being perpendicular to the lateral edges AA', BB' (YI., 6), etc., are the altitudes of the parallelograms which form the convex surface of the prism, if we take as the bases of these parallelograms the lateral edges AA\ BB\ etc., which are all equal (6). Hence the area of the sum of these parallelo- grams is GH X AA' -\-HIX BB' + etc. = {GH-\-HI+ etc.) X AA'. 17. Corollary. The lateral area of a right prism is equal to the product of the perimeter of its base by its altitude. PROPOSITION III.— THEOREM. 18. Two prisma are equal, if three faces including a triedral angle of the one are respectively equal to three faces similarly 'placed including a triedral angle of the other,— Let the triedral angles A and a of the prisms ABCDE- A\ abcde-a', be contained by- equal faces similarly placed, namely, ABCDE equal to abcde, AB' equal to ab', and AE' equal to aef ; then the prisms are equal. BOOK VII. 229 For, superpose the second prism upon the first, making the base ahcde coincide with the equal base ABCDE. Since the diedral angles ah and AB are equal and ae and AE are equal (YI., Proposition XXII.), the plane ah' will coincide with the plane AB'^ and the plane ae' with the plane AE', Hence the intersection aa' will fall along the intersection AA', As the faces ah' and AB' are equal, and have now been suitably superposed, they must coincide throughout^ and a'h' will coincide with A'B'. For the same reason,^ a'e' will coincide with A'E'. Consequently, the plane determined by a'h' and aV, namely, the plane of the upper base of the second prism, will coincide with the plane of the upper base of the first prism. Any lateral edge, as eef^ will fall along the corre- sponding lateral edge EE'^ for they are now parallel to the same line AA'^ and have a point e of one coinciding with a point E of the other. They have thus the same direction ^ and a point in common, and must coincide throughout^ by I., Postulate II. Since all the lateral edges of the second prism coincide with the corresponding lateral edges of the first,, the planes of all the corresponding lateral faces must coincide. There- fore, as all the corresponding faces of the two prisms coincide (the bases included), the prisms are equal. 19. Corollary I. Two truncated prisms are equal, if three faces including a triedral angle of the one are respectively equal to three faces similarly placed including a triedral angle of the other. For the preceding demonstration applies whether the planes A'B'C'D'E' and a'h'c'dle are parallel or inclined to the lower bases. 20. Corollary II. Two right prisms are equals if they have equal bases and equal altitudes. 20 230 ELEMENTS OF GEOMETRY. / In the case of right prisms, it is not necessary to add the condition that the faces shall be similarly placed ; for if the two right prisms ABC-A! ^ dbc-a\ can- not be made to coincide by placing the base ABC upon the equal base ahc ; yetj by inverting one of the prisms and applying the base ABC to the base a'6V, they will coincide. PROPOSITION IV.— THEOREM. V 21. Any oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. JjQtABCDE-A^hG the oblique prism. At any point F in the edge AA\ pass a plane perpendicular to AA' and forming the right section FGHIK. Produce AA' to F\ making FF' z= AA\ and through F' pass a second plane perpendicular to the edge AA', intersecting all the faces of the prism produced, and forming another right section F'G'H'I'K' parallel and equal to the first. The prism FGHIK-F' is a right prism whose base is the right section and whose altitude FF' is equal to the lateral edge of the oblique prism. The solid ABCDE-F is a truncated prism which is easily shown to be equal to the truncated prism A'B'C'D'E'-F' (Proposition III., Corollary I.). Taking the first away from the whole solid ABCDE-F', there remains the right prism ; BOOK VII. 231 taking the second away from the same solid, there remains the oblique prism ; therefore the right prism and the oblique prism have the same volume ; that is, they are equivalent. ^V^..^^ PROPOSITION v.— THEOREM. 26. Any parallelopiped is equivalent to a rectangular parallel- opiped of the same altitude and an equivalent base. Let ABGD^A' be any oblique parallel- opiped whose base is A BCD J and altitude B'O. Produce the edges AB,A'B',I)C,D'C'; in AB produced take FG = AB, and through F and G pass planes FF'FI, GG'H'H, perpendicular to the produced edges ; then the given parallelopiped and the right parallelo- piped FF'FI-H are equivalent, by Proposition lY. Produce, now, the edges of this second parallelopiped ZF, FF\HG, H'G'; in IF produced take NK^IF, and through .N and K pass planes KLL'K' smd NMM'N' perpendicular to the produced edges. Then the second parallelopiped and the parallelopiped NMM'W-K are equivalent, by Proposi- tion lY. Consequently, the given parallelopiped and the parallelopiped NMM'N'-K are equivalent. The last-named parallelopiped is a right parallelopiped, by construction, since the face KLL'K' was drawn perpendicular to the lateral edges. Moreover, as the planes KL' and KN' are perpendic- ular, the first to KI, the second to AG, they are perpendicular to the plane AHK, by YI., Proposition XIY., and their inter- 232 ELEMENTS OF GEOMETRY. section KK' is perpendicular to AHK(Y1.^ Proposition XYI.)> and therefore to KL (YL, 6). Hence the ^ ^' '' ^' base KLL'K' is a rectangle, and the par allelopiped Xii'X'-iV^is a rec- tangular parallelopi- ped. If, now, we take KIjMN as its base, its altitude is equal to that of the given parallelopiped, since the planes AUK and A'H'K' are parallel; and its base is equivalent to ABGD^ since each of them is equivalent to FGHI (lY., Proposition I.). ^ PROPOSITION VI.— THEOREM. 23. The plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms. Let ABCD-A' be any parallelopi- ped ; the plane A CCA', passed through its opposite edges AA' and CC, divides it into two equivalent triangular prisms ABC- A' and ADC-A\ Let FGHI be any right section of the parallelopiped, made by a plane perpendicular to the edge A A'. The intersection, FH, of this plane with the plane AC is the diagonal of the parallelogram FGHI, and divides that par- allelogram into two equal triangles, FGH and FIH. The oblique prism ABG-A' is equivalent to a right prism whose BOOK VII. 233 base is the triangle FGH and whose altitude is AA' (Propo- sition lY.); and the oblique prism ADC- A' is equivalent to a right prism whose base is the triangle FIH and whose alti- tude is AA\ The two right prisms are equal (Proposition III., Corollary II.) ; therefore the oblique prisms, which are respectively equivalent to them, are equivalent to each other. i- PKOPOSITION VII.— THEOREM. 24. Tioo rectangular parallelopipeds having equal bases are to each other as their altitudes. Let P and Q be two rectangular parallelopipeds having equal bases, and let AB and CD be their alti- tudes. Ist. Suppose the altitudes have a common measure, which is con- tained m times in AB and n times in CD. Then we have AB ^m CD~ n' Apply this measure to AB and CJ), and through the poinds of division draw planes perpendicular to AB and CD. P will thus be divided into m, and Q into n, smaller parallelo- pipeds, all of which will be equal, by Proposition I., Corol- lary, and Proposition III., Corollary II. Hence p \ N \ !\ \ K \ I\ \ \ Therefore P Q m n' P^AB i Q CD' The proof may be extended to the case where the altitudes are incommensurable, by the method exemplified in the proof 20* ^/ 234 ELEMENTS OF GEOMETRY. J of II., Proposition XII., of III., Proposition I., and of lY., Proposition II. ^'25. Scholium, The three edges of a rectangular parallelo- $ piped which meet at a common vertex are called its dimen- sions j and the preceding theorem may be expressed as follows : Two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions. f PROPOSITION VIII.— THEOREM. ^6. T.0 rectangular parallelopi,e,s Kaun, e.ual aUitudes are to each other as their bases. Let a, b, and c be the three dimen- sions of the rectangular parallelo- piped P; m, w, and c those of the rectangular parallelopiped Q; the dimension c, or the altitude, being common. Construct B, sl third rectangular parallelopiped, having the dimensions m, b, and c. If a and m are taken as the alti- tudes of P and JR, their bases are equal, and, by Proposition YII., P^ a 1^ m* If b and n are taken as the altitudes of R and §, theii bases are equal, and, by Proposition VII., R^b, . Q n' and, multiplying these ratios together, P^ a X b ' Q m X n p F^ Q rA 1 J A \ n' BOOK vir. awRft 235 But a y^ b is the area of the base of P, awFm X w is the area of the base of Q; therefore P and Q are in the ratio of their bases. 27. Scholium. This proposition may also be expressed as follows : Two rectangular parallelopipeds which have one dimension in common, are to each other as the products of the other two dimensions. PROPOSITION IX.— THEOREM. 28. Any two rectangular parallelopipeds are to each other as the products of their three dimensions. Let a, by and c be the three dimen- sions of the rectangular parallelo- piped P; m, n, and p those of the rectangular parallelopiped Q. Construct P, a third rectangular parallelopiped, having the dimen- sions a, 6, and p. By Proposition YII. we have p \ K C 1 L._P_ ! P P c and by Proposition YIII., P _ axb , Q m Xn' and, multiplying these ratios together, P axb X c Q mXnXp \ K V 4 236 ELEMENTS OF GEOMETRY. PROPOSITION X.— THEOREM. /4^ The volume of a rectangular parallelopiped is equal to the product of its three dimensions, the unit of volume being the cube whose edge is the linear unit. Let a, b, c, be the three dimen- sions of the rectangular parallelo- piped P; and let Q be the cube whose edge is the linear unit. The three dimensions of Q are each equal to unity, and we have, by the preceding proposition, p \ K i p Q ax b X c 1 X 1 X 1 = aXbXc. P., Now, Q being takeA as the unit of volume, - is the numer- ical measure, or volume of P, in terms of this unit (4); therefore the volume of P is equal to the product ^ X ^ X <^- 30. Scholium I. Since the product a X b represents the base, when c is called the altitude, of the parallelopiped, this proposition may also be expressed as follows : The volume of a rectangular parallelopiped is equal to the product of its base by its altitude. 31. Scholium II. When the three dimen- sions of the parallelopiped are each ex- actly divisible by the linear unit, the truth of the proposition is rendered evident by dividing the solid into cubes, each of which is equal to the unit of volume. Thus, if the three edges which meet at a common vertex A are, respectively, equal to 3, 4, and 5 times the linear unit, these edges may be divided respectively into 3, 4, and 5 t\ ^ "^^^ V. ^ v. ^ BOOK VII. 237 equal parts, and then planes passed through the several points of division at right angles to these edges will divide the solid into cubes, each equal to the unit cube, the number of which is evidently 3x4x5- But the more general demonstration, above given, includes also the cases in which one of the dimensions, or two of them, or all three, are incommensurable with the linear unit. 32. Scholium III. If the three dimensions of a rectangular parallelopiped are each equal to <2, the solid is a cube whose edge is a, and its volume is a X <^ X <^ = <^' ; or, the volume of a cube is the third power of its edge. Hence it is that in arithmetic and algebra the expression " cube of a number" has been adopted to signify the " third power of a number." PROPOSITION XI.— THEOREM. 33. The volume of any parallelopiped is equal to the product of the area of its base by its altitude. For, by Proposition Y., the volume of any parallelopiped is equal to that of a rectangular parallelopiped having an equiv- alent base and the same altitude (30). PROPOSITION XII.— THEOREM. ' 34. The volume of a triangular prism is equal to the product of its base by its altitude. Let ABC-A' be a triangular prism. In the plane of the base complete the parallelogram ABGD, and then through D draw a line DD' parallel to AA\ and through DD' and CO', and DD' and AA\ pass planes, thus constructing the n/ 238 ELEMENTS OF GEOMETRY. parallelopiped ABCD-D'. The given prism is half of the parallelepiped, by Proposition YI., and it has the same altitude. The volume of the parallelopiped is equal to its base BD multiplied by its altitude (Proposition XI.) ; therefore the volume of the triangular prism is equal to its base ABC, the half of BD, multiplied by its altitude. . 35. Corollary. The volume of any prism is equal to the product of its base by its altitude. Let ABCDE-A' be any prism. It may be divided into triangular prisms by planes passed through a lateral edge AA' and the several diagonals of its base. The volume of the given prism is the sum of the vol- umes of the triangular prisms, or the sum of their bases multiplied by their common ''- altitude, which is the base ABODE of the given prism multi- plied by its altitude. PYKAMIDS. 36. Definitions. A pyramid is a polyedron bounded by a polygon and triangular faces formed by the intersections of planes passed through b the sides of the polygon and a common point out of its plane ; as S-ABGDE. The polygon, ABODE, is the base of the pyramid; the point, S, in which the triangular faces meet, is its vertex; the triangular faces taken together constitute its lateral, or convex, surface; the area of BOOK VII. 239 this surface is the lateral area; the lines SA^ SB^ etc., in which the lateral faces intersect, are its lateral edges. The altitude of the pyramid is the perpendicular distance SO from the vertex to the base. A triangular pyramid is one whose base is a triangle; a quadrangular pyramid, one whose base is a quadrilateral ; etc. A triangular pyramid, having but four faces (all of which are triangles), is a tetraedron ; and any one of its faces may be taken as its base. 37. Definitions. A regular pyramid is one whose base is a regular polygon, and whose vertex is in the perpendicular to the base erected at the centre A of the polygon. This perpendicular is called the axis of the regular pyramid. From this definition it can be readily shown that the lateral edges of a regular pyramid are / all equal, and hence that the lateral faces are /'' equal isosceles triangles, ^"^L^^ The slant height of a regular pyramid is the perpendicular from the vertex to the base of any one of its lateral faces. It is the common altitude of all the lateral faces. 38. Definitions. A truncated pyramid is the portion of a pyramid included between its base and a plane cutting all its lateral edges. When the cutting plane is parallel to the base, the trun- cated pyramid is called a frustum of a pyramid. The altitude of a frustum is the perpendicular distance between its bases. In a frustum of a regular pyramid, the lateral faces are equal trapezoids; and the perpendicular distance between the parallel sides of any one of these trapezoids is the slant height of the frustum. 240 ELEMENTS OF GEOMETRY. V PEOPOSITION XIII.— THEOREM. 39. If a pyramid is cut by a plane parallel to its base : 1st, the edges and the altitude are divided proportionally ; 2d, the section is a polygon similar to the base. Let the pyramid S-ABCDE, whose alti- tude is SOj be cut by the plane abode par- allel to the base, intersecting the lateral edges in the points a, 6, c, d^ e, and the alti- tude in ; then 1st. The edges and the altitude are dir voided proportionally. Pass a plane through the altitude SO and any lateral edge SA, cutting the base in AO and the section in ao. By YI., Proposition YIII., _,ab, be, cd, . . . ao are parallel respectively to AB, BC, CD, ...AO. Therefore, by III., Proposition I., 8a Sb Sc Sd So SA SB SG SD" ' SO' 2d. The section abode and the base are similar. For they are mutually equiangular, by YI., Proposition X., and by similar triangles we have ab Sa be Sb cd Sc AB ~'SA' BC~ 'SB' CD ~ SC whence ab AB bc_ BC cd CD and the homologous sides of the polygons are proportional. Therefore the section and the base are similar. BOOK VII. 241 40. Corollary 1. If a pyramid is cut by a plane parallel to its base, the area of the section is to the area of the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. For abcde ^ a^ ^^ jy^ Proposition IX. ; ABODE Z3^' -^ ' "^ but ab , Sa So AB~~ SA~ SO Therefore abcde So ABCDE SO' 41. Corollary II. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to their bases and at equal distances from their vertices are equivalent. PROPOSITION XIV.— THEOREM. 42. The lateral area of a regular pyramid is equal to the product of the perimeter of its base by one-half its slant height. For, let S- ABCDE be a regular pyr- amid ; the lateral faces SAB, SBC, etc., being equal isosceles triangles, whose bases are the sides of the regular poly- gon ABCDE and whose common alti- tude is the slant height SH, the sum of their areas, or the lateral area of the pyramid, is equal to the sum of AB, BC, etc., multiplied by \SH. L g 21 V 242 ELEMENTS OF GEOMETRY. 43. Corollary. The lateral area of the frustum of a regular pyramid is equal to the half sum of the perimeters of its bases multiplied by the slant height of the frustum. PROPOSITION XV.— THEOREM. 44. If the altitude of any given triangular pyramid is divided into equal parts, and through the points of division planes are passed parallel to the base of the pyramid, and on the sections made by these planes as upper bases prisms are described having their edges parallel to an edge of the pyramid and their altitudes equal to one of the equal parts into which the altitude of the pyramid is divided, the total volume of these prisms will approach the volume of the pyramid as its limit as the number of parts into which the altitude of the pyramid is divided is indefinitely increased. Let S-ABC be the given trian- gular pyramid, whose altitude is A T. Divide the altitude A T into any number of equal parts Ax, xy, etc., and denote one of these parts by h. Through the points of division x, y, etc., pass planes parallel to the base, cutting from the pyramid the sections DBF, GUI, etc. Upon the triangles DEF, GHI, etc., as upper bases, construct prisms whose lateral edges are parallel to SA, and whose altitudes are each equal to h. This is effected by passing planes through EF, HI, etc., parallel to SA. There will thus be formed a series of prisms DEF- A, GSI-Dj^tc, inscribed in the pyramid. /(■^ rror^ tW IvI^A ' BOOK VII. 243 Again, upon the triangles ABC^ DEF, GHI^ etc., as lower bases, construct prisms whose lateral edges are parallel to SAj and whose altitudes are each equal to h. This also is effected by passing planes through BG, JSF, HI^ etc., parallel to SA. There will thus be formed a series of prisms ABC-D, DEF-G, etc., which may be said to be circumscribed about the pyramid. The total volume of the inscribed prisms is obviously less and the total volume of the circumscribed prisms is obviously greater than the volume of the pyramid. Each inscribed prism is equivalent to the circumscribed prism immediately above it, since they have the same base and equal altitudes. Consequently, the difference between the total volume of the inscribed prisms and the total volume of the circumscribed prisms is the volume of the lowest cir- cumscribed prism ABC-D, and therefore the difference be- tween the total volume of the inscribed prisms and the volume of thp pyramid is less than the volume of ABC-D. "By increasing at pleasure the number of parts into which the altitude AT is divided, we can make the volume of ABC-D as small as we please, since we diminish its altitude at pleasure without changing its base. Therefore we can make the difference between the total volume of the in- scribed prisms and the volume of the pyramid as small as we please ; but, as we have seen above, we cannot make it absolutely zero. Hence the volume of the pyramid is the limit of the total volume of the inscribed prisms, as the number of parts into which the altitude AT ia divided is indefinitely increased. 244 ELEMENTS OF GEOMETRY. PROPOSITION XVI.— THEOREM. 45. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. Let S-ABC and 8'-A'B^C' be two triangular pyramids G A! having equivalent bases ABC^ A'B'C'^ in the same plane, and a common altitude A T. Divide the altitude A T into any arbitrarily chosen number n of equal parts, Ax^ xy^ yz^ etc., and through the points of division pass planes parallel to the plane of the bases, inter- secting the two pyramids. In the pyramid S-ABG inscribe a series of prisms whose upper bases are the sections BEF, GHIj etc., and in the pyramid S'-A'B'C inscribe a series of prisms whose upper bases are the sections D'B'F'., G'HT, etc. Since the corresponding sections are equivalent (Propo- sition XIII., Corollary II.), the corresponding prisms, having equivalent bases and equal altitudes, are equivalent; there- fore the sum of the prisms inscribed in the pyramid S-ABG is equivalent to the sum of the prisms inscribed in the pyramid S'-A'B'C; that is, if we denote the total volumes of the two series of prisms by V and F', we have BOOK VII. 245 no matter what the value of n. If we vary w, V and F' obviously vary. If n is indefinitely increased, V has the volume of the pyr- amid S-ABG, and F' the volume of the pyramid S'-A'B'C, as its limit (Proposition XY.). Therefore, by III., Theorem of Limits, these volumes are equal. PROPOSITION XVII.— THEOREM. i|fp A triangular pyramid is one-third of a triangular prism of the same base and altitude. Let S-ABC be a triangular pyramid. Through A and C draw the lines AE and CD parallel to BS, Through AE and CD, which are parallel, by VI., Proposition V., Corollary, pass a plane, and through S pass a second plane parallel to ABC. The prism ABC-E has the same base and alti- tude as the given pyramid, and we are to prove that the pyramid is one-third of the prism. Taking away the pyramid S-ABC from the prism, there remains a quadrangular pyramid whose base is the parallelogram ACDE and vertex S. The plane SEC^ passed through SE and SC^ divides this pjrramid into two triangular pyramids, S-AEC and S-ECD, which are equiva- lent to each other, since their triangular bases AEC and ECD are the halves of the parallelogram A CDEj and their common altitude is the perpendicular from S upon the plane ACDE (Proposition XVI.). The pyramid S-ECD may be regarded as having ESD as its base and its vertex at C; therefore it is equivalent to the pyramid S-ABC, which has an equivalent base and the same altitude. Therefore the three pyramids into which the prism is divided are equivalent to each other, and the given pyramid is one-third of the prism. 21* ; V t 246 ELEMENTS OF GEOMETRY. 47. Corollary. The volume of a triangular pyramid is equal to one-third of the product of its base by its altitude. ^^^ PEOPOSITION XVIII.— THEOREM. 48. The volume of any pyramid is equal to one-third of the product of its base by its altitude. For any pyramid, S-ABCDB, may be di- vided into triangular pyramids by passing planes through an edge SA and the diagonals ABj ACj etc., of its base. The bases of these pyramids are the triangles which compose the base of the given pyramid, and their common altitude is the altitude SO of the given pyramid. The volume of the given ^ pyramid is equal to the sum of the volumes of the triangular pyramids, which is one-third of the sum of their bases multi- plied by their common altitude, or one-third of the product of the base ABODE by tte altitude SO. 49. Scholium. The volume of any polyedron may be found by dividing it into pyramids, and computing the volumes of these pyramids separately. The division may be effected by taking a point within the polyedron and joining it with all the vertices. The polyedron will then be decomposed into pyramids whose bases will be the faces of the polyedron, and whose common vertex will be the point taken within it. PROPOSITION XIX.— THEOREM. 60. A frustum of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases, of the frustum. 1 '^ ' 1V BOOK vir. ;,'; 247 Let ABC-D be a frustum of a triangular pyramid, the plane DEF being parallel to the base ABC. Through the vertices A, B, and ^ J' C pass a plane ^-E^C, and through // y^\/ \ the vertices E, D, and pass a /y^ ' \y^\ \ plane ^DC, dividing the frustum /^ I / /' a\ into three pyramids. ^C^^/i'J!jf---j53^^'^ The first of these, ABG-E, has ^^^l^""'^'^ for its base the lower base of the (^^^ frustum, and for its altitude the altitude of the frustum ; the second, DEF-C, has for its base the upper base of the frus- tum, and for its altitude the altitude of the frustum. It remains to show that the third, ACD-E, is equivalent to a pyramid having for its altitude the altitude of the frustum, and for its base a mean proportional between the bases of the frustum. Through E in the plane ABED draw a line EEf parallel to AD, and through E', D, and C pass a plane. EE' is parallel to the plane ACFD, by YI., Proposition YI. Therefore the pyramids ACD-E and ACD-E' are equivalent, since they have the same base and equal altitudes. If we take D as the vertex and AE'C as the base of ACD-E' , it has for its altitude the altitude of the frustum. Through F in the plane ACFD draw FF' parallel to AD, AE'F'-D is a prism, and consequently its bases DEF and AE'F' are equal^ and E'F' is parallel to EFy and therefore to 5a AE'F' ^ AF' AE'C AC' since the triangles AE'F' and AE'C have the same altitude. AE'C AE' n .. = -j^-, for the same reason. AJjC ab 248 ELEMENTS OF GEOMETRY. AF' AE' AG AB by III., Proposition I. Therefore DEF ^ AE'C AE'C ABC' and the base of the pyramid AE'C-D is a mean proportional between the bases of the frustum. ^1. Corollary. A frustum of any pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases, of the frustum. Suggestion, Let ABCDE-F be a frustum of any pyramid S-ABCDE. Construct a triangular pyramid, S'-A'B'C, having the same altitude as S-ABCDE, and a base A'B'C equivalent to ABODE and in the same plane with it. Let the plane of the upper base of the given frustum be pro- duced to cut the triangular pyramid in F'G'T. The upper bases of the frustums are equivalent, by Proposition XIII.) BOOK VII. 249 Corollary II., and the frustums themselves are equivalent, since the pyramids are equivalent and the pyramids above the frustums are equivalent. "^ PROPOSITION XX.— THEOREM. 52. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism^ and whose vertices are the three vertices of the inclined section. F Let ABC-BEF be a truncated tri- angular prism whose base is ABC and inclined section BEF. Pass the planes AEC and DEC, dividing the truncated prism into the three pyramids E-ABC, E-ACI), and E-CBF. The first of these pyramids, E-ABC^ has the base ABC and the vertex E. The second pyramid, E-A CD, is equivalent to the pyramid B-ACB ; for they have the same base ACB, and the same altitude, since their vertices E and B are in the line EB parallel to this base. But the pyramid B-ACB is the same as B-ABC ; that is, it has the base ABC and the vertex B. The third pyramid, E-CBF, is equivalent to the pyramid B-ACF ; for they have equivalent bases CBF and ACF in the same plane, and also the same altitude, since their ver- tices E and B are in the line EB parallel to that plane. But the pyramid B-ACF is the same as F-ABC; that is, it has the base ABC and the vertex F. Therefore the truncated prism is equivalent to three pyra- mids whose common base is ABC and whose vertices are E^ D, and F, 250 ELEMENTS OF GEOMETRY. THE KEGULAR POLYEDRONS. 53. Definition. A regular polyedron is one whose faces are all equal regular polygons and whose polyedral angles are all equal to each other. PROPOSITION XXI.— THEOREM. 64.. Only five regular (convex^ polyedrons are possible. (^The faces of a regular polyedron must be regular polygons, \ and at least three faces are necessary to form a polyedral \ angle ; moreover, the sum of the face angles of a polyedral / angle must be less than four right angles )(YI., Proposition XXI.). 1st. The simplest regular polygon is the equilateral tri- angle, and, since each angle of an equilat- eral triangle is an angle of 60°, three equi- lateral triangles «an be combined to form a polyedral angle. It is probable, then, that a regular polyedron can be formed bounded by equilateral triangles and hav- ing three at each vertex. There is such a regular polyedron. It has four faces, and is called the regular tetraedron. Since four angles of 60° are less than four right angles, four equilateral triangles can be combined to form a polyedral BOOK VII. 251 angle. It is probable, then, that a regu- lar polyedron can be formed bounded by- equilateral triangles and having four at each vertex. There is such a regular polyedron. It has eight faces, and is called the regu- lar octaedron. Since five angles of 60° are less than four right angles, five equilateral triangles can be combined to form a polyedral angle. It is probable, then, that a reg- ular polyedron can be formed bounded by equilateral triangles and having five at each vertex. There is such a regular polyedron. It has twenty faces, and is called the regular icosaedron. No regular polyedrons bounded by equi- lateral triangles and having more than five at a vertex are possible. For six or more angles of 60° cannot form a poly- edral angle. 2d. The next regular polygon to the equilateral triangle, in order of simplicity, is the square, each of whose angles is a right angle. Three right angles can be combined to form a polyedral angle. It is probable, then, that a regular polyedron can be formed bounded by squares and having three at each vertex. There is such a regular polyedron. It has six faces, and is called the cuhe^ or the regular hexaedron. No regular polyedrons bounded by squares and having more than three at a vertex are possible. For four or more right angles cannot form a polyedral angle. 252 ELEMENT8 OP GEOMETRY. 3d. The next regular polygon is the regular pentagon, each of whose angles contains 108° (I., Proposition XXYII.). Three angles of 108° each can be combined to form a poly- edral angle. It is probable, then, that a regular polyedron can be formed bounded by regular pentagons and having three at each vertex. There is such a regular polyedron. It has twelve faces, and is called the regular dodecaedron. !No regular polyedrons bounded by regular pentagons and having more than three at a vertex are possible. For four or more angles of 108° cannot be combined to form a polyedral angle. 4th. Each angle of the regular hexagon contains 120°. No regular polyedron can be formed bounded by hexagons. For three or more angles of 120° cannot be combined to form a polyedral angle. No regular polyedron can be formed bounded by reg- ular polygons of more than six sides. For it follows, from I., 55, Exercise, that the greater the number of sides in a regular polygon the greater the magnitude of its angles, and since, as we have seen, the angles of a hexagon are too great to allow the existence of a polyedral angle whose plane faces are regular hexagons, those of any regular polygon of more than six sides will be too great. Therefore the only possible regular polyedrons are the five we have figured. 55. Scholium I. It must be observed that we have not attempted to prove that the five regular polyedrons are pos- sible. This can be done by showing how to construct them ; but the investigation is difficult and tedious. BOOK VII. 253 56. Scholium II. The student may derive some aid in com- prehending the preceding discussion of the regular polyedrons by constructing models of them, which he can do in a very simple manner, and at the same time with great accuracy, as follows. Draw on card-board the following diagrams ; cut them out entire, and at the lines separating adjacent polygons cut the card-board half through ; the figures will then readily bend into the form of the respective surfaces, and can be retained in that form by gluing the edges. V Tetraedron. Hexaedron, Octaedron. V Dodccaedron. Tcosaedron. EXERCISES ON BOOK VIL THEOREMS. 1. The volume of a triangular prism is equal to the product of the area of a lateral face by one-half the perpendicular distance of that face from the opposite edge. 2. The lateral surface of a pyramid is greater than the base. Suggestion. Join the projection of the vertex on the base with the corners of the base. 3. At any point in the base of a regular pyramid a perpendic- ular to the base is erected which intersects the several lateral faces of the pyramid, or these faces produced. Prove that the sum of the distances of the points of intersection from the base is con- stant. Suggestion. The distances in question are proportional to the distances of the foot of the perpendicular from the sides of the base, and these distances have a constant sum. {v. V., Exer- cise 16.) 4. Two tetraedrons which have a triedral angle of the one equal to a triedral angle of the other, are to each other as the products of the three edges of the equal triedral angles, {v. IV., 19, Ex- ercise.) 6. In a tetraedron, the planes passed through the three lateral edges and the middle points of the edges of the base intersect in a straight line. 254 BOOK VII. 255 Suggestion. The intersections of the planes with the base are medial lines of the base. Therefore they intersect in the line joining the vertex with the point of intersection of the medial lines of the base. 6. The lines joining each vertex of a tetraedron with the point of intersection of the medial lines of the opposite face all meet in a point, which divides each line in the ratio 1 : 4. Note. This point is the centre of gravity of the tetraedron. Suggestion. If AF and DG are two of / the lines in question,, they must intersect, / since they both lie in the plane passed / g through AD and the middle point E of ^V/ the opposite edge. Moreover, since EF M = lED and EG = lEA (I., Exercise 38), GFis, parallel to AD and is equal to ^AD. C Whence HF = iHA and GH = iHD. The lines through C and B will also each cut off \ of AF, Hence the four lines have a common intersection. 7. The straight lines joining the middle points of the opposite edges of a tetraedron all pass through the centre of gravity of the tetraedron, and are bisected by the centre of gravity, {v. III., Exercise 7.) ^ 8. The plane which bisects a diedral angle of a ie^raedron divides the opposite edge into segments which are proportional to the areas of the adjacent faces. Suggestion. Consider the volumes of the two parts into which the tetraedron is divided. 9. If a, 6, e, d, are the perpendiculars from the vertices of a tetraedron upon the opposite faces, and a\ 6'', g\ d^, the perpen- diculars from any point within the tetraedron upon the same faces respectively, then abed Suggestion. Join the point in question with the vertices of the tetraedron, and compare the volumes of the four tetraedrons thus obtained with the volume of the given tetraedron. 256 ELEMENTS OF GEOMETRY. 10. The altitude of a regular tetraedron is equal to the sum of the four perpendiculars let fall from any point within it upon the four faces. 11. Any lateral face of a prism is less than the sum of the other lateral faces, {v. Proposition II.) PROBLEMS. 12. Given three indefinite straight lines in space which do not intersect, to construct a parallelopiped which shall have three of its edges on these lines, {v. VI., Exercise 8.) 13. Within a given tetraedron, to find a point such that planes passed through this point and the edges of the tetraedron shall divido^lie tetraedron into four equivalent tetraedrons. {v. Exer- cise 6.) ' BOOK YIIL THE THREE ROUND BODIES. 1. Op the various solids bounded by curved surfaces, but three are treated of in Elementary Geometry, — namely, the cylinder^ the cone^ and the sphere^ which are called the three ROUND BODIES. THE CYLINDER. 2. Definitions. A cylindrical surface is a curved surface gen- erated by a moving straight line which continually touches a given curve, and in all of its positions is parallel to a given fixed straight line not in the plane of the curve. Thus, if the straight line Aa moves so as continually to touch the given curve ABCD, and so that in any of its positions, as ^6, Cc, Dd, etc., it is parallel to a given fixed straight line 3im, the surface ABCDdcba is a cylindrical surface. If the moving line is of indefinite length, a surface of indefinite extent is generated. The moving line is -called the generatrix ; the curve which it touches is called the directrix. Any straight line in the surface, as Bb^ which represents one of the positions of the generatrix, is called an element of the surface. To draw an element through any given point of a cylin- drical surface, it is sufficient to draw a line through the point parallel to the given fixed straight line, or parallel to ap element (I., Postulate II.). r . 22* 257 258 ELEMENTS OF GEOMETRY. In this general definition of a cylindrical surface, the direc- trix may be any curve whatever. Hereafter we shall assume it to be a closed curve, and usually a circle, as this is the only curve whose properties are treated of in element- ary geometry. 3. Definition. The solid Ad bounded by a cylindrical surface and two par- . allel planes, ABD and ahd^ is called a cylinder ; its plane surfaces ABD, abd, are called its bases; the curved surface is sometimes called its lateral surface; and the perpendicular distance between its bases is its altitude. The elements of a cylinder are all equal. A cylinder whose base is a circle is called a circular cylinder. 4. Definition. A right cylinder is one whose elements are perpendicular to its base. 5. Definition. A right cylinder with a circular base, as ABCa, is called a cylinder of revolution, because it may be generated by the revolution of a rectangle A Ooa about one of its sides, Oo, as an axis ; the side Aa generating the curved ^ surface, and the sides OA and oa generating the bases. The fixed side Oo is the axis of the cylinder. The radius of the base is called the radius of the cylinder. m\- PROPOSITION I.— THEOREM. 6. Every section of a cylinder made by a plane passing through an element is a parallelogram. Let Bb be an element of the cylinder Ac ; then the section BbdD, made by a plane passed through Bb, is a parallelogram. BOOK VIII. 259 The line Dd in which the cutting plane intersects the curved surface a second time is an ele- ment. For, if through any point D of this intersection a straight line is drawn parallel to Bh^ this line, by the definition of a cylindrical surface, is an element of the surface, and it must also lie in the plane Bd ; therefore this element, being common to both surfaces, is their intersection. The lines BD and hd are parallel (VI., Proposition YIII.), and the elements Bh and Dd are parallel ; therefore Bd is a parallelogram. 7. Corollary. Every section of a right cylinder made by a plane perpendicular to its base is a rectangle. PROPOSITION II.— THEOREM. 8. The bases of a cylinder are equal. Let BD be the straight line joining two points of the perimeter of the lower base, and let a plane passing through BD and the ele- ment Bb cut the upper base in the line bd ; then BD = bd (Proposition I.). Let A be any third point in the perim- eter of the lower base, and Aa the corre- sponding element. Through the parallels Aa and Bb pass a plane, and through Aa *^ and Dd pass a plane. Then AB = ab and AD = ad (Propo- sition I.) ; and the triangles ABD^ abdj are equal. Therefore, if the upper base be applied to the lower base with the line bd in coincidence with its equal BD^ the triangles will co- incide and the point a will fall upon A ; that is, any point a of the upper base will fall on the perimeter of the lower base, 260 ELEMENTS OF GEOMETRY. and consequently the perimeters will coincide throughout. Therefore the bases are equal. 9. Corollary I. Any two parallel sec- »w/.^^ / tions of a cylindrical surface are equal. /-^^^^^i^V For these sections are the bases of a A>-^^- / cylinder. / .../^/'^ 10. Corollary II. All the sections of a circular cylinder parallel to its bases are equal circles; and the straight line joining the centres of the bases passes through the centres of all the parallel sections. This line is called the axis of the cylinder. Suggestion. In the base draw two diameters, and through these diameters and elements of the cylinder pass planes. They will cut all the sections in diameters, and their line of intersections will pass through all the centres. 11. Definition. A tangent plane to a cylinder is a plane which passes through an element of the curved surface without cut- ting the surface. The element through which it passes is called the element of contact THE CONE. 12. Definition. A conical surface is a curved surface gener ated by a moving straight line which continually touches a given curve, and passes through a given fixed point not in the plane of the curve. Thus, if the straight line SA moves so as continually to touch the given curve ABGD, and in all its positions, SB, SC, SBj etc., passes through the given fixed point aS, the surface S-ABCD is a conical surface. BOOK VIII. 261 The moving line is called the generatrix ; the curve which it touches is- called the directrix. Any straight line in the surface, as SBj which represents one of the posi- tions of the generatrix, is called an element of the surface. The point S is called the vertex. The straight line joining any point of a conical surface with the vertex is obviously an element. If the generatrix is of indefinite length, as ASa^ the whole surface generated consists of two symmetrical portions, each of indefinite extent, lying on opposite sides of the vertex, as S-ABGD and S-abcd, which are called nappes ; one the upper, the other the lower, nappe. 13. Definition. The solid S-ABCD, bounded by a conical surface and a plane ABD cutting the surface, is called a cone ; its plane surface ABD is its base, the point ;S^ is its vertex, and the perpendicular distance SO from the vertex to the base is its altitude. A cone whose base is a circle is called a circular cone. The straight line drawn from the vertex of a circular cone to the centre of its base is the axis of the cone. 14. Definition. A right circular cone is a cir- cular cone whose axis is perpendicular to its base, as S-ABCD. The right circular cone is also called a cone of revolution, because it may be generated by the revolution of a right triangle, SA 0, about one of its perpendicular sides, SO, as an axis ; the hypotenuse SA generating the curved surface, and the remaining perpendicular side OA generating the base. 262 ELEMENTS OF GEOMETRY. PROPOSITION III.— THEOREM. 15. Every section of a cone made by a plane passing through its vertex is a triangle. Let the cone S-ABCD be cut by a plane SBC^ which passes through the vertex S and cuts the base in the straight line BC ; then the section SBC is a tri- angle; that is, the intersections SB and SG with the curved surface are straight lines. For the straight lines joining S with B and C are elements of the surface, by the definition of a cone, and they also lie in the cutting plane ; therefore they coincide with the inter- sections of that plane with the curved surface ; and BG^ being the intersection of two planes, is a straight line. PROPOSITION IV.— THEOREM. 16. If the base of a cone is a circle^ every section made ty a plane parallel to the base is a circle. Let the section abc, of the circular cone S-ABGj be parallel to the base. Let be the centre of the base, and let be the point in which the axis SO cuts the plane of the parallel section. Through SO and any number of ele- ments SAj SBj etc., pass planes cutting the base in the radii OA^ OB, etc., and the parallel section in the straight lines oa, ob, etc. is parallel to OA, and ob to OB, we have Since oa ofi So „„j ob So „i^ ^ oa ob BOOK VIII. 263 But OA = OB, therefore oa = oh ; hence all the straight lines drawn from o to the perimeter of the section are equal, and the section is a circle. 17. Corollary. The axis of a circular cone passes through the centres of all the sections parallel to the base, 18. Definition. A tangent plane to a cone is a plane which passes through an element of the curved surface without cutting this surface. The element through which it passes is called the element of contact. Cv//^^^'^^^ THE SPHERE. 19. Definition. A sphere is a solid bounded by a surface all the points of which are equally distant from a point within, called the centre. A sphere may be generated by the revolu- tion of a semicircle ABC about its diameter J. (7 as an axis ; for the surface generated by the curve ABC will have all its points equally distant from the centre 0. A radius of the sphere is any straight line drawn from the centre to the surface. A diameter is any straight line drawn through the centre and terminated both ways by the surface. Since all the radii are equal and every diameter is double the radius, all the diameters are equal. 20. Definition. It will be shown that every section of a sphere made by a plane is a circle ; and, as the greatest pos-- sible section is one made by a plane passing through the centre, such a section is called a great circle. Any section made by a plane which does not pass through the centre is called a small circle. 264 ELEMENTS OF GEOMETRY. 21. Definition. The poles of a circle of the sphere are the extremities of the diameter of the sphere which is perpen- dicular to the plane of the circle j and this diameter is called the axis of the circle. PROPOSITION v.— THEOREM. 22. Every section of a sphere made by a plane is a circle. 1st. If the -plane passes through the centre of the sphere, the lines .^-^^---^ joining points on the perimeter of aJifs//.V}liyss^\c the section with the centre of the / ^^^vHI]^?'^ \ sphere are radii of the sphere, and ^F;---— --^------^ are therefore all equal. Consequently \ j J it is a circle with its centre at 0. X,^^ I ^^/ 2d. If the plane does not pass ^ through the centre of the sphere, as abcj draw a diameter EOD pei*pendicular to the section and meeting it at o. If points a, b, c, of the perimeter are joined with 0, and also with 0, the triangles aeO, boO, coO, are all equal (I., Proposition X.). Therefore ao, bo, co, etc., are all equal, and the section is a circle with its centre at o. 23. Corollary I. The axis of a circle on a sphere passes through the centre of a .circle. 24. Corollary II. All great circles of the same sphere are equal. 25. Corollary III. Every great circle divides the sphere into two equal parts. Suggestion. Superpose one part upon the other, (v. II., Proposition II.) BOOK VIII. 265 26. Corollary IY. Any two great circles on the same sphere bisect each other ; for the common inter- section AB of their planes passes through the centre of the sphere and is a diame- ter of each circle. 27. Corollary Y. An arc of a great circle may he drawn through any two given points of the surface of the sphere^ and, unless the points are the opposite eoctremities of a diameter, only one such arc can he drawn; for the two points, together with the centre 0, determine the plane of a great circle whose circumference passes through the points. If, however, the two given points are the extremities A and 5 of a diameter of the sphere, the position of the circle is not determined, for the points J., 0, and B, being in the same straight line, will not determine a plane (YI., Proposi- tion I.). 28. Corollary YI. An arc of a circle may he drawn through any three given points on the surface of the sphere; for the three points determine a plane which cuts the sphere in a circle. EXEBCISE. Theorem. — The greater the distance of the plane of a small hsircle from the centre of the sphere, the less the circle. li 23 266 ELEMENTS OF GEOMETRY. PROPOSITION VI.— THEOREM. 29. All the points in the circumference of a circle of the sphere J,- are equally distant from either of its poles. Let abed be any circle'of the sphere, ..-s^^^--^ and PP' the diameter of the sphere „/^'Wt'^' — ^\j perpendicular to its plane; then, by I %. the definition (21), P and P' are the ^^^ poles of the circle ahcd^ and, by Prop- \ \^ osition Y., Corollary I., FT' passes \ through 0, the centre of ahcd. Join ^1^ P with any points, a, 6, c, on the cir- cumference of the circle. Then Pa, P6, Pc, are equal, since the triangles Poa, Tob^ Poc^ are equal, by I., Proposition VI. Hence all the points of the circumference abed are equally distant from the pole P. For the same reason, they are equally distant from the pole P'. 30. Corollary I. All the arcs of great circles drawn from a pole of a circle to points in its circumference are equal, since their chords are equal chords in equal circles. By the distance of two points on the surface of a sphere is usually understood the arc of a great circle joining the two points. The arc of a great circle drawn from any point of a given circle abc, to one of its poles, as the arc Pa, is called the polar distance of the given circle, and the distance from the nearest pole, is usually understood. 31. Corollary II. The polar distance of a great circle is a quadrant ; thus, PA, PB, etc., P'A, P'B, etc., polar distances of the great circle ABCD, are quadrants ; for they are the measures of the right angles A OP, BOP, AOP', BOP', etc., whose vertices are at the centre of the great circles PAP'j PBP', etc. BOOK VIII. 267 In connection with the sphere, by a quadrant is usually to be understood a quadrant of a great circle. 32. Corollary III. If a point on the surface of a sphere is at a quadrant's distance from each of two given points of the surface which are not opposite extremities of a diameter, it is the pole of the great circle passing through them. Suggestion. Let P be at a quadrant's distance from B and C; then FOB and POC are right angles, and FO is perpen- dicular to the plane ABCD. 33. Scholium. By means of poles, arcs of circles may be drawn upon the surface of a sphere with the same ease as upon a plane surface. Thus, by revolving the arc Fa about the pole P, its extremity a will describe the small circle abd ; and by revolving the quadrant FA about the pole P, the extremity A will describe the great circle ABD. If two points, B and (7, are given on the surface, and it is required to draw the arc ^C, of a great circle, between them, it will be necessary first to find the pole P of this circle ; for which purpose, take B and G as poles, and at a quadrant's distance describe two arcs on the surface intersecting in P. The arc BC can then be described with a pair of compasses, placing one foot of the compasses on P and tracing the arc with the other foot. The opening of the compasses (distance between their feet) must in this case be equal to the chord of a quadrant ; and to obtain this it is necessary to know the radius of the sphere. 34. Definition. A plane is tangent to a sphere when it has but one point in common with the surface of the sphere, 35. Definition. Two spheres are tangent to each other when their surfaces have but one point in common. 268 ELEMENTS OF GEOMETRY. PROPOSITION VII.— THEOREM. 36. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. For any other line drawn from the centre of the sphere to the plane must reach beyond the sur- face of the sphere, and therefore must be greater than the radius. The radius is, then, the shortest line that can be drawn from the centre of the sphere to the plane, and is consequently perpendicular to the plane (YI., Proposi- tion III.). 37. Corollary. Conversely, a plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere. 38. Scholium. Any straight line A T, drawn in the tangent plane through the point of contact, is tangent to the sphere. Any two straight lines, AT, A T\ tangent to the sphere at the same point A, determine the tangent plane at that point. PROPOSITION VIII.— THEOREM. 39. The intersection of two spheres is a circle whose plane is perpendicular to the straight line joining the centres of the spheres, and whose centre is in that line. Through the centres and (7 of the two spheres let any plane be passed, cutting the spheres in great circles which intersect each other in the points A and B ; the chord AB is bisected at G by the line 0(7 at right angles (II., Proposition YI., Corollary II.). If we now Tv^ BOOK VIII. 269 revolve the plane of these two circles about the line 0(7, the circles will generate the two spheres, and the point A will describe the line of intersection of their surfaces. Moreover, since the line AC will, during this revolution, remain perpen- dicular to 0(7, it will generate a circle whose plane is per- pendicular to 0(y, and whose centre is C. SPHERICAL ANGLES. 40. Definition. The angle of two curves passing through the same point is the angle formed by the two tangents to the curves at that point. This definition is applicable to any two intersecting curves in space, whether drawn in the same plane or upon a surface of any kind. , PROPOSITION IX.— THEOREM. M. The angle of two arcs of great circles is equal to the angle of their planes^ and is measured by the arc of a great circle described from its vertex as a pole and included between its sides (produced if necessary). Let AB and AB' be two arcs of >^ j, great circles, A T and A T' the tan- / i \^V~-^ gents to these arcs at A^ and the / j W \b centre of the sphere. TA and T'A ^L^ 4*r~"f3)^ lie in the planes of their arcs, and \^^~~"^ \ I / are perpendicular to the radius OA X^^ ! J/ drawn to their point of contact. ^ They form, then, the plane angle measuring the diedral angle formed by the planes of the arcs ; but, by (40), the angle which they form is equal to the angle of the two arcs. 270 ELEMENTS OF GEOMETRY. Now let CO' be the arc of a great circle described from A as a pole aiid intersecting the arcs AB^ AB' (produced if necessary), in G and C. The radii OG and OG' are perpendicular to J.0, since the arcs AG, AG', are quadrants (Proposition YI., Corollary II.); therefore the angle GOG' is a plane angle of the diedral angle A 0, and is equal to TAT', or to BAB', and it is obviously measured by the arc GG'. 42. Corollary. All arcs of great circles drawn through the pole of a given great circle are perpendicular to its circumfer- ence ; for their planes are perpendicular to its plane (YI., Proposition XIY.). SPHERICAL POLYGONS. 43. Definition. A spherical polygon is a portion of the sur- face of a sphere bounded by three or more arcs of great circles, as ABGD. Since the planes of all great circles pass through the centre of the sphere, the planes of the sides of a spherical polygon form, at the centre 0, a polyedral angle of which the edges are the radii drawn to the ver- tices of the polygon, the face angles are angles at the centre measured by the sides of the polygon, and the diedral angles are equal to the angles of the polygon (Proposition IX.). Since in a polyedral angle each face angle is assumed to be less than two right angles, each side of a spherical polygon will be assumed to be less than a semi-circumference. BOOK viir. 271 A spherical polygon is convex when its corresponding poly- edral angle at the centre is convex (VI., 52). A diagonal of a spherical polygon is an arc of a great circle joining any two vertices not consecutive. 44. Definition. A spherical triangle is a spherical polygon of three sides. It is called right angled^ isosceles^ or equilat- eral^ in the same cases as a plane triangle. 45. In consequence of the relation established between polyedral angles and spherical polygons (43), it follows that from any property of polyedral angles we may infer an anal- ogous property of spherical polygons. Eeciprocally, from any property of spherical polygons we may infer an analogous property of polyedral angles. The latter is in almost all cases the more simple mode of procedure, inasmuch as the comparison of figures drawn on the surface of a sphere is nearly if not quite as simple as the comparison of plane figures. 46. Arcs of great circles on the same sphere can be super- posed and made to coincide just as straight lines are super- posed and made to coincide. We have merely to place one point of the first arc on some given point of the second, and, keeping this point fixed, to turn the first arc about it as a pivot, until some second point in the arc not diametrically opposite the fixed point falls on the second arc. The two arcs must then coincide throughout, by Proposition Y., Corol- lary Y. Equal angles formed by arcs of great circles on the surface of the same sphere can be superposed and made to coincide just as equal plane angles are superposed and made to co- incide ; that is, if the vertex of the first angle is placed upon the vertex of the second, and one side of the first placed upon the corresponding side of the second, the other side of 272 ELEMENTS OF GEOMETRY. the first will coincide with the other side of the second. For, if the two given angles are equal, their diedral angles are equal (Proposition IX.). If the vertices of the angles co- incide, the edges of the diedral angles coincide ; if a side of the first angle is placed on a side of the second, one face of the first diedral angle coincides with one face of the second. The remaining faces of the diedral angles must then coincide, and consequently the remaining sides of the given angles coincide. 47. Definition. Two spherical triangles are symmetrical if all the parts of one are respectively equal to the parts of the other, but the corresponding parts are arranged in opposite orders in the two triangles. Two symmetrical triangles, as ABC^ ABC'j in the figure cannot be made to coincide; for, to bring the vertex C upon the corre- sponding vertex C, the second triangle would have to be turned over, and the two convex surfaces would thus be brought together. 48;^here is, however, one exception to the last remark, — namely, the case of symmetrical isosceles trian- gles, Por, if ABC is an isosceles spherical triangle and AB = AC, then, in its symmetrical triangle, we have A'B' = A!C\ and con- sequently AB = A'C, AG = A'B\ and, since the angles A and A' are equal, if AB be placed on A'G\ AG will fall on its equal A'B\ and the two triangles will coincide through- out. 49. Definition. If from the vertices of a spherical triangle as poles, arcs of great circles are described, these arcs form /}l\ / 4 BOOK VIII. 273 by their intersection a second triangle, which is called the polar triangle of the first. Thus, if Aj B, and C are the poles of the arcs of great circles, B'C'j A'C\ and A'B\ respectively, A'B'C is the polar triangle of ABC. Since all great circles, when completed, intersect each other in two points, the arcs jB'C", A'C, A'B\ if produced, will form three other triangles ; but the triangle which is taken as the polar triangle is that whose vertex A\ homologous to A, lies on the same side of the arc BC as the vertex A; and so of the other vertices. PROPOSITION X.— THEOREM. 50y^f the first of two spherical triangles is the polar triangle of the second, then, reciprocally, the second is the polar triangle of the first. Let A'B'C be the polar triangle of ABC; then is ABC the polar triangle of A'B'C, For, since A is the pole of the arc B'C, the point 5' is at a quadrant's distance from A ; and, since C is the pole of the arc A'B', the point ^' is at a quadrant's distance from C ; therefore B' is the pole of the arc AC (Proposition YI., Corollary III.). In the same manner it is shown that A' is the pole of the arc BC, and C" the pole of the arc AB. Moreover, A and A' are on the same side of B'C\ B and B' on the same side of A'C, C and C on the same side of A'B' ; therefore ABC is the polar triangle of A'B'C. 274 ELEMENTS OF GEOMETRY. PROPOSITION XI.— THEOREM. 5lX/« t.o polar triangles, ea^U anyle of one is measure, l>y the supplement of the side lying opposite to it in the other. Let ABC and A'B'C be two polar tri- angles. Let the sides AB and AC, produced if necessary, meet the side B'C in the points h and c. The vertex A being the pole of the arc he, the angle A is measured by the arc he (Proposition IX.). Now, B ' being the pole of the arc Ac, and C the pole of the arc Ah^ the arcs B'c and G'h are quadrants ; hence we have B'C -\- he =^B'c -\- C'h = Si semi-circumference. Therefore he, which measures the angle A, is the supplement of the side B'G\ In the same manner it can be shown that each angle of either triangle is measured by the supplement of the side lying opposite to it in the other triangle. .52. Scholium. Let the angles of the trian- gle ABC be denoted by A, B, and C, and let the sides opposite to them, namely, BC, AC, and AB, be denoted by a, h, and c, respec- tively. Let the corresponding angles and sides of the polar triangle be denoted by A', B', C, a', h', and c\ Also let both angles and sides be expressed in degrees. Then the preceding theorem gives the following relations : A +a'=B -\-h':=^C +c'^ 180°. A' +a :=:^B' -\rh =C' + c = 180°. BOOK VIII. 275 PROPOSITION XII.— THEOREM. 53. Two triangles on the same sphere are either equal or sym- metrical, when two sides and the included angle of one are re- spectively equal to txoo sides and the included angle of the other. In the triangles ^50 and DEF, let the angle A be equal to the angle D, the side AB equal to the side DE, and the side A G equal to the side DF. 1st. When the parts of the two trian- gles are in the same order, ABC can be applied to DEF^ as in the corresponding case of plane triangles (I., Proposition VI.), and the two triangles will coincide ; therefore they are equal. 2d. When the parts of the two triangles are in inverse order, let DE'F be the symmetrical triangle of BEF, and therefore having its angles and sides equal, respectively, to those of DEF. Then, in the triangles ABC and DE'F, we shall have the angle BAC equal to the angle E'DF, the side AB to the side DE', and the side AG to the side DF, and these parts arranged in the same order in the two triangles ; therefore the triangle ABG is equal to the triangle DE'F, and consequently symmetrical with DEF, 54. Scholium. In this proposition, and in the propositions which follow, the two triangles may be supposed on the same sphere, or on two equal spheres. 276 ELEMENTS OF GEOMETRY. PKOPOSITION XIII.— THEOREM. 55. Two triangles on the same sphere are either equal or sym- metrical^ when a side and two adjacent angles of one are equal respectively to a side and two adjacent angles of the other. Let the triangles ABC and BEF have the side a equal to the side d^ and the c' / "^ X6' angles B^ C, equal respec- tively to the angles E, F ; then are the triangles equal. Construct the polar tri- angles of ABC and DEF. We have h' = e', c' = /', and A' == D\ by Proposition XI. Then A'B'C and D'E'F' are equal or symmetrical, by Proposition XII. Therefore their polar triangles ABC, DEF, are equal or symmetrical. 58. Scholium. The proposition might be proved by direct superposition, as in I., Proposition YII. PROPOSITION XIV.— THEOREM. 57. Two triangles on the same sphere are either equal or sym- metrical, when the three sides of one are respectively equal to the three sides of the other. For if their vertices are joined with the centre of the sphere, the triedral angles thus formed have the three face angles of the one respectively equal to the three face angles of the other, and consequently, by VI., Proposition XXII., their corresponding diedral angles are equal. The given triangles are, then, mutually equilateral and mutually equiangular, and are equal or symmetrical. i BOOK vni. 277 58. Scholium. The proposition can be proved as in I., Prop- osition IX. PROPOSITION XV.— THEOREM. 59. If two triangles on the same sphere are mutually equi- angular, they are also mutually equilateral, and are either equal or symmetrical. . Let the spherical trian- '^'"'^ ^^^^^ gles M and JV be mutually equiangular. Let M' be the polar tri- angle of M, and iV' the polar triangle of iV. Since M and N are mutually equiangular, their polar triangles M' and iV' are mutually equilateral (Proposition XL); therefore, by Proposition XIY., the triangles M^ and N' are mutually equi- angular. But M' and iV' being mutually equiangular, their polar triangles M and iV are mutually equilateral. Conse- quently, M and N are either equal or symmetrical. 60. Scholium. It may seem to the student that the pre- ceding property destroys the analogy which subsists be^^een plane and spherical triangles, since two mutually equiangular plane triangles are not necessarily mutually equilateral. But in the case of spherical triangles the equality of the sides follows from that of the angles only upon the condition that the triangles are constructed upon the same sphere or on equal spheres ; if they are constructed on spheres of diiferent radii, the homologous sides of two mutually equiangular tri- angles will no longer be equal, but will be proportional to the radii of the sphere ; the two triangles will then be similar, as in thje case of plane triangles. 24 278 ELEMENTS OF GEOMETRY. EXERCISES. 1. Theorem. — In an isosceles spherical triangle the angles opposite the equal sides are equal. 2. Theorem. — The arc drawn from the vertex of an isosceles spherical triayigle to the middle point of the base is perpendicular to the base, and bisects the vertical angle. 3. Theorem. — If two angles of a spherical triangle are equal, the triangle is isosceles. PROPOSITION XVI.-^THEOREM. 61. Any side of a spherical triangle is less than the sum of the other two. Let ABC be a spherical triangle; then any. side, as AC, is less than the sum of the other two, AB and BC. / For, in the corresponding triedral / angle formed at the centre of the /^'l-"'''' sphere, we have the angle AOC less o**' than the sum of the angles AOB and BOC (VI., Proposition XX.) ; and since the sides of the tri- angle measure these angles, respectively, we have AC <^ AB + BC. EXERCISES. 1. Theorem. — If two angles of a spherical triangle are un- equal, the side opposite the greater angle is greater than the side opposite the less angle, (v. I., Proposition XII.) 2. Theorem. — If two sides of a spherical triangle are unequal, the angle opposite the greater side is greater than the angle oppo- site the less side. (v. L, Proposition XIII.) BOOK VIIT. 279 PEOPOSITION XVII.— THEOEEM. 62. The sum of the sides of a convex spherical polygon is less than the circumference of a great circle. For the sum of the face angles of the corresponding poly- edral angle at the centre of the sphere is less than four right angles (YI., Proposition XXI.). PEOPOSITION XVIII.— THEOEEM. ^ 63. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles. For, denoting the angles of a spherical tri- angle by A, B, C, and the sides respectively opposite to them in its polar triangle by a', ^< h', d, we have (Proposition XI.) A = 180° — o!, J?=: 180° — 6', 0= 180° —c', the sum of which is A^ B -^C ^ 540° — {a! + 6' + c'). But a' ^y -\- c' AB, and subtracting from the two members of this mequality the equal arcs AM and AC, wq have BM > BC; therefore ikf lies without the circumference whose pole is B. ^ 286 ELEMENTS OF GEOMETRY. Now let AFGB be any line from A to 5, on the surface of the sphere, which does not pass through the point C, and which therefore cuts the two circumfer- ences in different points, one in F^ the other in G. Then a shorter line can be drawn from A to B^ passing through G. For, whatever may be the nature of the line AF^ an equal line can be drawn from Ato C; since, if AC and AF be conceived to be drawn on two equal spheres having a common diameter passing through A, and therefor" having their surfaces in coincidence, and if one of these spfh^res be turned upon the common diameter as an axis, the point A will be fixed and the point F will come into coincidence with C; the surfaces of the two spheres continuing to coincide, the line AF will then lie on the common surface between A and C. For the same reason, a line can be drawn from B to 0, equal to BG. Therefore a line can be drawn from A to B, through G, equal to the sum of AF and BG, and consequently less than AFGB. The shortest line from A to B therefore passes through C; that is, through any, or every, point in AB. Consequently it must be the arc AB itself. EXERCISES ON BOOK VIII. THEOREMS. 1. A SPHERE can be circumscribed about any tetraedron. Suggestion. The locus of the points equally distant from A, B^ and C is the perpendicular EM erected at the centre of the circle circumscribed about ABC (VI., Exercise 15.) The locus of the points equally distant from B^ C, and D is tne perpendicular FN^ and both EM and FN lie in the plane perpendicular to BC Sit its middle point, since that plane contains all the points equally distant from B and C. EM and FN therefore intersect, and O, their point of intersec- tion, is equally distant from the four ver- tices of the tetraedron. There is only one such point. Therefore only one sphere can be circumscribed about a tetraedron. 2. The perpendiculars erected at the centres of the four faces of a tetraedron meet in a point. 3. A sphere can be inscribed in any tetra- edron. Suggestion, The locus of the points equally distant from two faces of the tetraedron is the plane bisecting the diedral angle be- tween them. 4. The planes bisecting the six diedral angles of a tetraedron intersect in a point. 287 288 ELEMENTS OF GEOMETRY. LOCI. 6. Locus of the points in space which are at a given distance from a given straight line. 6. Locus of the points which are at the distance a from a point Ay and at the distance b from a point B. 7. Locus of the centres of the spheres which are tangent to three given planes. 8. Locus of the centres of the sections of a given sphere made by planes passing through a given straight line. Suggestion. Pass a plane through the centre of the sphere per- pendicular to the given straight line. Then see II., Exercise 24. 9. Locus of the centres of the sections of a given sphere made by planes passing through a given point. PROBLEMS. 10. Through a given point on the surface of a sphere, to pass a plane tangent to the sphere, (v. Proposition VII., Corollary.) 11. Through a given straight line without a sphere, to pass a plane tangent to the sphere. Suggestion. Through the centre of the sphere pass a plane per- pendicular to the given line. In this plane, from its point of intersection with the line, draw a line tangent to the circle in which the plane cuts the sphere. A plane through the tangent line and the given line is the tangent plane required. (Two solu- tions.) 12. Through a given point without a sphere, to pass a plane tangent to the sphere. 13. To cut a given sphere by a plane passing through a given straight line so that the section shall have a given radius. SuggeMion. Pass a plane through the centre of the sphere per- pendicular to the given line. Then v. II., Exercise 37. BOOK VIII. 289 14. To construct a spherical surface with a given radius— Ist, passing through three given points ; 2d, passing through two given points and tangent to a given plane, or to a given sphere ; 3d, passing through a given point and tangent to two given planes, or to two given spheres, or to a given plane and a given sphere ; 4th, tangent to three given planes, or to three given spheres, or to two given planes and a given sphere, or to a given plane and two given spheres. 15. Through a given point on the surface of a sphere, to draw a great circle tangent to a given small circle. Suggestion, With the pole of the small circle as a pole, and with a polar distance equal to the polar distance of the small circle plus a quadrant, describe a second small circle. With the given point as a pole describe a great circle. A point of intersection of this great circle with the second small circle will be the pole of the great circle required. 16. To draw a great circle tangent to two given small circles. 17. At a given point in a great circle, to driaw an arc of a great circle which shall make a given angle with the first. b BOOK IX. MEASUREMENT OP THE THREE ROUND BODIES. THE CYLINDER. 1. Definition. The area of the convex, or lateral, surface of a cylinder is called its lateral area. 2. Definition. A prism is inscribed in a cylinder when its base is inscribed in the base of the cylinder and its lateral edges are elements of the cyl- inder. It follows that the upper base of the prism is inscribed in the upper base of the cylinder. To inscribe, then, a prism of any given number of lateral faces in a cylinder, we have merely to inscribe in the base a polygon of the given number of sides, and through the vertices of the polygon to draw elements of the cylinder. Planes passed through adjacent elements will form the lateral faces of the prism which is obviously wholly contained in the cylinder. 3. Definition. A prism is circum- scribed about a cylinder when its base is circumscribed about the base of the cylinder and its lateral edges are parallel to elements of the cylinder. It follows that its lateral faces are tangent to the lateral faces of the cylinder (YIII., 11) ; for any face, as AB'y contains the element bb\ since it contains the 290 BOOK IX. 291 parallel line AA^ and the point b (YI., Proposition II.), and, by YIII., Proposition I., it cannot cut the surface of the cyl- inder again jinless AB cuts the base again ; and that its upper base is circumscribed about the upper base of the cylinder. The cylinder is obviously wholly contained in the prism. 4. Definition. A right section of a cylinder is a section made by a plane perpendicular to its elements ; asabcdef. The intersection of the same plane with an inscribed or circumscribed prism is a right section of the prism. 5. Definition. Similar cylinders of revolution are those which are generated by similar rectangles revolving about homolo- gous sides. PKOPOSITION I.— THEOREM. 6. If a prism whose base is a regular polygon be inscribed in or circumscribed about a given cylinder^ its volume will ap- proach the volume of the cylinder as its limit, and its lateral sur- face will approach the lateral surface of the cylinder as its limit as the number of sides of its base is indefinitely increased. For, if we could make the base of the prism exactly coincide with the base of the cylinder, the prism and the cylinder would coincide throughout, and their volumes would be equal and their lateral surfaces equal. But, by increasing the number of sides of the base of the prism. 292 ELEMENTS OF GEOMETRY. we can make it come as near as we please to coinciding with the base of the cylinder (Y., Propo- sition YII.) ; we can then make the prism and cylinder fail of co- incidence by as small an amount as we choose. Consequently, by increasing at pleasure the number of sides of the base of the circum- scribed or inscribed prism, we can make the difference between the volumes of prism and cylinder, and between the lateral surfaces of prism and cylinder, as small as we choose, but cannot make it absolutely zero. 7. Scholium. The proposition just proved is true when the base of the prism is not a regular polygon ; but it is only for the case of the regular polygon that a rigorous proof has been given in Book V. PEOPOSITION II.— THEOREM. 8. The lateral area of a cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of the surface. Let ABCBEF be the base and AA' any element of a cylinder, and let the curve abcdef be any right section of the surface. Denote the perimeter of the right section by P, the element AA' by E^ and the lateral area of the cylinder by S. Inscribe in the cylinder a prism ABCDEF A' of any arbitrarily chosen BOOK rx. 293 number n of faces. The right section, abcdef, of this prism will be a polygon inscribed in the right section of the cylin- der formed by the same plane (4). Denote the lateral area of the prism by s, and the perimeter of its right section by p ; then, the lateral edge of the prism being equal to E, we have (VII., Proposition II.) s = pX E, no matter what the value of n. If n is indefinitely increased, s approaches the limit JS (Proposition I.), and p X E, the limit P X E. Therefore, by III., Theorem of Limits, S = P X E. 9. Corollary I. The lateral area of a cylinder of revolution is equal to the product of the circumference of its base by its altitude. This may be formulated, U if R is the radius of the base and H the altitude 10. Corollary II. The lateral ^ ^ areas of similar cylinders of revo- lution are to each other as the squares of their altitudes, or as the squares of the radii of their bases. S _ 2t.R.H s Suggestion. R H r ' h by (5). ^ r* jff» = -- = -—, smce - = 27:r.h R H 25* h<^ 294 ELEMENTS OP GEOMETRY. PBOPOSITION III.— THEOREM. , 11. The volume of a cylinder is equal to the product of its base by its altitude. Let the volume of the cylinder be denoted by F, its base by B, and its altitude by S. Let the volume of an inscribed prism be denoted by F', and its base by J5'; its altitude will also be -H, and we shall have (VII., Proposi- tion XIL, Corollary) no matter what the number of faces of the prism. If the number of faces of the prism is indefinitely in- creased, F' has the limit F, and ^' X -S" the limit B X S, Therefore v = bxb:. 12. Corollary I. For a cylinder of revolution this proposition may be formulated^ V = izB^.H. (Y., Proposition IX., Corol- lary.) 13. Corollary II. The volumes of similar cylinders of revo- lution are to each other as the cubes of their altitudes^ or as the cubes of their radii. THE CONE. 14. Definition. The area of the convex, or lateral, surface of a cone is called its lateral area. BOOK IX. 15. Definition. A pyramid is inscribed in a cone when its base is inscribed in the base of the cone and its vertex coincides with the vertex of the cone. It follows that the lateral edges of the pyramid are elements of the cone. An inscribed pyramid is wholly con- tained within the cone. 16. Definition. A pyramid is circum- scribed about a cone when its base is circumscribed about the base of the cone and its vertex co- incides with the vertex of the cone. Any lateral face, as SAB^ of the pyramid is tangent to the cone ; for, since it passes through a and S^ it contains the element Sa^ and it can- not cut the convex surface again without cutting the perimeter of the base again (YIII., Proposition 111.). The cone is then wholly contained within the pyramid. 17. Definition. A truncated cone is the portion of a cone included be- tween its base and a plane cutting its convex surface. When the cutting plane is par- allel to the base, the truncated cone is called a frustum of a cone; as ABCD-abcd. The altitude of a frus- tum is the perpendicular distance Tt between its bases. If a pyramid is inscribed in the cone, the cutting plane 296 ELEMENTS OF GEOMETRY. determines a truncated pyramid inscribed in the truncated cone ; and if a pyramid is circumscribed about the cone, the cutting plane determines a truncated pyramid circumscribed about the truncated cone. 18. Definition. In a cone of revolution all the elements are equal, and any ele- ment is called the slant height of the cone. In a cone of revolution the portion of an element included between the par- allel bases of a frustum, as Aa^ or Bh^ is called the slant height of the frustum. 19. Definition. Similar cones of revolu- tion are those which are generated by similar right triangles revolving about homologous sides. PROPOSITION IV.— THEOREM. 20. If a 'pyramid he inscribed in or circumscribed about a given cone, its volume will approach the volume of the cone as its limit, and its lateral surface will approach the convex surface of the cone as a limit, as the number of faces of the pyramid is indefinitely increased. The demonstration is precisely the same as that of Propo- sition I., substituting cone for cylinder and pyramids for prisms. 21. Corollary. A frustum of a cone is the limit of the in- scribed and circumscribed frustums of pyramids, the number of whose faces is indefinitely increased. BOOK IX. 297 PROPOSITION v.— THEOREM. 22. The lateral area of a cone of revolution is equal to the product of the circumference of its base by half its slant height. Suggestion. Circumscribe a regular pyramid about the cone, and then sup- pose the number of its faces to be indefinitely increased, (y. YII., Prop- osition XIY.) 23. CoROLLAEY I. Thc proposition may be formulated, S = tcRI/, where E is the radius of the base and L the slant height. 24. Corollary II. The lateral areas of similar cones of revolution are to each other as the squares of their slant heights, or as the squares of their altitudes, or as the squares of the radii of their bases. ^^u:;^ PROPOSITION VI.— THEOREM. 25. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multi* plied by its slant height. Suggestion. Circumscribe the frus- tum of a regular pyramid about the frustum of the cone (17), and sup- pose the number of its faces indefi- nitely increased, (y. YII., Proposi- tion XIY., Corollary.) 26. Corollary I. The proposition may be formulated, S = t:{B -\- r)L, if R and r are the radii of the bases and L is the slant height. 298 ELEMENTS OF GEOMETRY. 27. Corollary II. The lateral area of a frustum of a cone of revolution is equal to the circumfer- ence of a section equidistant from its ] bases multiplied by its slant height. jji Suggestion, IKz= ^(om + OM). (v. Exercise 24, Book I.) PROPOSITION VII.— THEOREM. 28. The volume of any cone is equal to one-third of the product of its base by its altitude. Suggestion. Inscribe a pyramid in the cone, and suppose the number of its faces to be indefinitely increased. (v. VII., Proposition XVIII.) 29. Corollary I. For a cone of revolution, the proposition may be for- mulated, V = InW^.H. 30. Corollary II. Similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. exercise. Theorem. — A frustum of any cone is equivalent to the sum of three cones whose common altitude is the altitude of the frus- tum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum, (v. VII., Proposition XIX., Corollary.) BOOK IX. 299 THE SPHERE. 31. Definition. A spherical segment is a portion of a sphere included between two parallel planes. The sections of the sphere made by the parallel planes are the bases of the segment ; the distance between the planes is the altitude of the segment. Let the sphere be generated by the revo- lution of the semicircle EBF about the axis EF; and let Aa and Bb be two parallels, perpendicular to the axis. The solid gener- ated by the figure ABba is a spherical seg- ment ; the circles generated by Aa and Bb are its bases ; and ab is its altitude. If two parallels Aa and TE are taken, one of which is a tangent at E^ the solid generated by the figure EAa is a spherical segment having but one base, which is the section generated by Aa. The segment is still included be- tween two parallel planes, one of which is the tangent plane at E, generated by the line ET. 32. Definition. A zone is a portion of the surface of a sphere included between two parallel planes. The circumferences of the sections of the sphere made by the parallel planes are the bases of the zone ; the distance between the planes is its altitude. A zone is the curved surface of a spherical segment. In the revolution of the semicircle EBF about EFy an arc AB generates a zone ; the points A and B generate the bases of the zone ; and the altitude of the zone is ab. An arc, EA, one extremity of which is in the axis, gener- ates a zone of one base, which is the circumference described by the extremity A. 300 ELEMENTS OF GEOMETEY. 33. Definition. When a semicircle revolves about its diam- eter, the solid generated by any sector of the semicircle is called a spherical sector. Thus, when the semicircle EBF revolves about EFj the circular sector COD generates a spherical sector. The spherical sector is bounded by three curved surfaces; namely, the two conical surfaces generated by the radii 00 and OD, and the zone generated by the arc CD. This zone is called the base of the spherical sector. OD may, however, coincide with OF, in which case the spherical sector is bounded by a conical surface and a zone of one base. -Again, 00 may be perpendicular to OF, in which case the spherical sector is bounded by a plane, a conical surface, and a zone. V PROPOSITION VIII.— LEMMA. 34. The area of the surface generated by a straight line re- volving about an axis in its plane, is equal to the projection of the line on the axis multiplied by the circumference of the circle whose radius is the perpendicular erected at the middle of the line and terminated by the axis. Let AB be the straight line revolving about the axis J^Y; ab its projection on the axis ; 01 the perpendicular to it, at its middle point 7, terminating in the axis; then area AB = ab X circ. 01. For, draw IK perpendicular and AH parallel to the axis. The area generated BOOK IX. 301 by AB is that of the frustum of a cone ; hence (Proposition " VI., Corollary II.) area AB = AB X circ. IK. The triangles ABH and lOK are sinxjlar, being mutually equiangular, and we have rCj-^] / but an4 AH^IK ab_^IK, AB or AB OV circ^ = ^(Y., Proposition YIIL), ah circ. IK A B^ circ.OI' ah X circ. 01 = AB X circ. IK. au ;x^ cue. ui = jljd ;<, circ. in.. q Therefore _^--^ ~~^ Jp V'-'x area AB = ah X circ. 01 AN \ ' ' If AB meets XY^ the surface generated is a conical sur- face ; but the proposition still holds, as may be easily proved. (v. Proposition Y.) If AB is parallel to the axis, the result is the same. (y. Proposition II., Corollary I.) PEOPOSITION IX.— THEOREM. ^ 35. The area of a zone is equal to the product of its altitude by the circumference of a great circle. E Let the sphere be generated by the revo- ^^.^^ lution of the semicircle EBF about the axis /(i EF ; and let the arc AD generate the zone b/^-— X^ whose area is required. I Jj^t the arc AD be divided into any num- ^V' ber of equal parts. AB^ BG, CD, and draw ^^— the chords AB, BC, etc. These chords are 26 302 ELEMENTS OF GEOMETRY. all equal, since they subtend equal ares ; and the perpendic- ulars at their middle points all pass through the centre of the semicircle, and are equal (II., Proposition YII.). Let abj he, etc., be the projections of these chords on the axis. Then, by Proposition YIII., area AB = ab X cire. 01, area BC == be X cire. 01, area CD =z ed X cire. 01. Hence the sum of these areas, which is the area generated by the broken line ABCD, is equal to (ab + 6c + cd) X cire. 01; that is, to ad X cire. 01. Calling the area generated by the broken inscribed line, 8y we have S = ad X cire. 01, no matter what the number of the equal parts into which the arc AD is divided. If, now, we increase the number of parts indefinitely, 01 will approach the radius of the sphere, and cire. 01 the circumference of a great circle as its limit, and S will approach the surface of the zone as its limit. There* fore surfaee of zone = ad X cireumferenee of great eirele. 36. Corollary. The proposition may be formulated, S = 2tzR.H, where R is the radius of the sphere and II Ihe altitude of the zone. A'- cJ-r K'J+ t BOOK IX. 303 PEOPOSITION X.— THEOREM. ^ Sfr. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle% This follows directly from Proposition IX., since the sur« face of the whole sphere may be regarded as a zone whose altitude is the diameter of the sphere. 38. Corollary I. This may be formulated^ S = 27: B X2R = 4:nR\ ._ Hence the surface of a sphere is equivalent to four great circles. ' 39. Corollary II. The surfaces of two spheres are to each other as the squares of their diameters^ or as the squares of their radii. 40. Scholium. The area of a spherical degree on a sphere whose radius is R is -^ (^IH., 69), and, by the aid of this value, we may readily reduce the area of a spherical polygon to ordinary square measure. . PROPOSITION XI.— THEOREM. 41. The volume of a sphere is equal to the area of its surface multiplied by one-third of its radius. Circumscribe a polyedron about the sphere. This may be done by taking at pleasure points on the sur- face of the sphere, and drawing tangent planes at these points. The circumscribed polyedron wholly con- tains the sphere, and is greater than the sphere. Join all the vertices of the polyedron with the centre of the sphere, and pass planes through the edges of the polyedron and these lines, thus dividing the polyedron 304 ELEMENTS OF GEOMETRY. into pyramids, each of which has its vertex at the centre of the sphere, and has a face of the poly- edron as its base, and has, therefore, the radius of the sphere for its alti- tude (YIII., Proposition YII.). The volume of any one of these pyramids ^ J^ \pJh is then one-third of the product of a face of the polyedron by the radius of the sphere, and the sum of the volumes of the pyramids, or the whole volume of the poly- edron, is one-third of the product of the sum of the faces of the polyedron by the radius of its sphere ; that is, one-third of the product of the whole surface of the polyedron by the radius of the sphere. Representing the surface of the poly- edron by s, and its volume by u, we have V = IBs, and this equation holds no matter what the number of the faces of the polyedron. If, now, we increase the number of faces of the polyedron by drawing additional tangent planes to the sphere, we de- crease the volume v, for each new tangent plane cuts off a corner of the polyedron. We may carry on indefinitely this process of shaving down the polyedron, and may thus make the difference between its volume and the volume of the sphere as small as we please ; but we cannot make the two volumes absolutely coincide. As the two volumes approach coincidence, the two surfaces also approach coincidence. If, now, S is the surface and V the volume of the sphere, S is the limit of s and Y the limit of v, as the number of faces of the circumscribed polyedron is indefinitely increased. Therefore, by III., Theorem of Limits, BOOK IX. 305 42. Corollary I. The result of this proposition may be for- mulatedj F = p' 43. Corollary II. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. PROPOSITION XII.— THEOREM. 44. The volume of a spherical sector is equal to the area of the zone which forms its base multiplied by one-third the radius of the sphere. The proof is analogous to the proof of Proposition XI. The form of the circumscribed polyedron is, however, some- what more complicated, as it will be bounded by a surface made up of plane faces tangent to the zone of the spherical sector, and by two pyramidal faces tangent to, or inscribed in, the two conical surfaces of the spherical sector. 45. Definition. A spherical pyramid is a f. ^^ solid bounded by a spherical polygon and y^ \ /\ the planes of the sides of the polygon ; as ^ \ /\,''' 0-ABCD. The centre of the sphere is the \ // \ / vertex of the pyramid j the spherical poly- • '^~ Y gon is its base. EXERCISE. Theorem. — The volume of a spherical pyramid is equal to the area of its base multiplied by one-third of the radius of the sphere. u 26* 306 ELEMENTS OF GEOMETRY. y PROPOSITION XIII.— PROBLEM. 46. To find the volume of a spherical segment. Any spherical segment may be obtained from a spherical sector by adding to it, or subtracting from it, cones having as bases the bases of the segment. For example, let us consider a segment of two bases which does not contain the centre of the sphere. The segment generated by the revolution of ABGD about OC may be ' obtained by taking the cone generated by OAD from the sum of the cone generated by OBG and the spherical sector generated by OAB. Call OC /, ODp, DC h, AD r, BC r', and OA E, and the volume of the segment F. Then we have the simple relations h r' + p' P —Pi K*, r'^ + p"" B\ The area of the zone of the segment is 27ri2.^ (Proposition IX., Corollary). Hence , ^ ^^/{ V = ^T:hB^ + i7r/r"-— Upr" (Proposition XII., and Proposi- ^ - .T^~--^:;r2r:^^ ^ tion YII., C orollayg I.), y = !;:(/ -"^ + %^^^i§^'^i - i^p(^ -%^ V = (/ - p)7:R' - UG" -^ p% [1] a convenient formula when the distances of the bases of the segment from the centre of the sphere are given. Another convenient formula can be obtained by introducing in [1] A, r, and / in place of j? and p'. We have F=(/-;>)|[3iJ' (y + /j' + i'0]- BOOK IX. Now Hence and 2/i> + ;>'- ,;^it+Jt^\ = f (iP - /' + iJ' - r") - 1 = 3JP - Kr" + /^ - 1', and we have [2] This formula is convenient when the areas of the bases of the segment are given, and it may be put into words as follows : The volume of a spherical segment is equal to the half sum of its bases multiplied by its altitude plus the volume of a sphere of which that altitude is the diameter, . V^^ \ -.^^; EXERCISES ON BOOK IX. X THEOREMS. 1. Give a strict proof of Proposition I. and Proposition IV. for the volumes of cylinder and cone, by showing that the difference between the volumes of the inscribed and circumscribed figures can be decreased at pleasure. 2. Assuming that if a solid has a plane face the area of that face is less than the rest of the surface of the solid, prove, first, that if two convex solids have a plane face in common, and one solid is wholly included by the other, its surface is less than that of the other {v. V., 13), and then give a strict proof of Proposition I. and Proposition IV. for the surfaces of cylinder and cone. 8. The volumes of a cone of revolution, a sphere, and a cylinder of revolution are proportional to the numbers 1, 2, 3 if the bases of the cone and cylinder are each equal to a great circle of the sphere, and their altitudes are each equal to a diameter of the sphere. 4. An equilateral cylinder (of revolution) is one a section of which through the axis is a square. An equilateral cone (of revolution) is one a section of which through the axis is an equi- lateral triangle. These definitions premised, prove the following theorems : I. The total area of the equilateral cylinder inscribed in a sphere is a mean proportional between the area of the sphere and the total area of the inscribed equilateral cone. The same is true of the volumes of these three bodies. II. The total area of the equilateral cylinder circumscribed about a sphere is a mean proportional between the area of the sphere and the total area of the circumscribed equilateral cone. The same is true of the volumes of these three bodies. 5. If h is the altitude of a segment of one base in a sphere whose radius is r, the volume of the segment is equal to Trh'^H — ^h), 6. The volumes of polyedrons circumscribed about the same sphere are proportional to their surfaces. 308 . 7^ MISCELLANEOUS EXERCISES ON THE GEOMETKY OF SPACE. 1. A PERPENDICULAR let fall from the middle point of a line upon any plane not cutting the line is equal to one-half the sum of the perpendiculars let fall from the ends of the line upon the same plane. 2. The perpendicular let fall from the point of intersection of the medial lines of a given triangle upon any plane not cutting the triangle is equal to one-third the sum of the perpendiculars from the vertices of the triangle upon the same plane. 3. The perpendicular from the centre of gravity of a tetraedron upon any plane not cutting the tetraedron is equal to one-fourth the sum of the perpendiculars from the vertices of the tetraedron upon the same plane. 4. The volume of a truncated triangular prism is equal to the product of the area of its lower base by the perpendicular upon the lower base let fall from the intersection of the medial lines of the upper base. 6. The volume of a truncated parallelopiped is equal to the product of the area of its lower base by the perpendicular from the centre of the upper base upon the lower base. 6. If ABCD is any tetraedron, and O any point within it, and if the straight lines AO^ BO^ CO^ DO, are produced to meet the faces in the points a, 6, c, d, respectively, then Aa'^ Bb~^ Cc "^ Dd 809 310 - ELEMENTS OF GEOMETRY. /7. If the three face angles of the vertical triedral angle of a tetraedron are right angles, and the lengths of the lateral edges are represented by a, 6, and c, and of the altitude by p, then 8. If the three face angles of the vertical triedral angle of a tetraedron are right angles, the square of the area of the base is equal to the sum of the squares of the areas of the lateral faces. 9. The perpendicular from the middle point of the diagonal of a rectangular parallelepiped upon a lateral edge bisects the edge, and is equal to one-half of the projection of the diagonal upon the base. 10. A straight line of a given length moves so that its extremi- ties are constantly upon two given perpendicular but non-inter- secting straight lines : what is the locus of the middle point of the moving line ? PROBLEMS. 11. To cut a given polyedral angle of four fac^s by a plane so that the section shall be a parallelogram. ^\ 12. To cut a cube by a plane so that the section shall be a regular hexagon. 13. To find the ratio of the volumes generated by a rectangle revolving successively about its two adjacent sides. SYLLABUS OP PROPOSITIONS IN SOLID GEOMETRY. BOOK VI. THEOREMS. Proposition I. Through any given straight line a plane may be passed, but the line will not determine the plane. Proposition II. A plane is determined, 1st, by a straight line and a point with- out that line ; 2d, by two intersecting straight lines ; 3d, by three points not in the same straight line ; 4th, by two parallel straight lines. Corollary, The intersection of two planes is a straight line. Proposition III. From a given point without a plane one perpendicular to the plane can be drawn, and but one ; and the perpendicular is the shortest line that can be drawn from the point to the plane. Corollary. At a given point in a plane one perpendicular can be erected to the plane, and but one. Proposition IV. If a straight line is perpendicular to each of two straight lines at their point of intersection, it is perpendicular to the plane of those lines. Corollary I. At a given point of a straight line, one plane can be drawn perpendicular to the line, and but one. Corollary II. Through a given point without a straight line, one plane can be drawn perpendicular to the line, and but one. 311 312 ELEMENTS OF GEOMETRY. Proposition V. Two lines in space having the same direction are parallel. Corollary, Two lines parallel to the same line are parallel to each other. Proposition VI. If two straight lines are parallel, every plane passed through one of them and not coincident with the plane of the parallels is parallel to the other. Corollary I. Through any given straight line a plane can be passed parallel to any other given straight line. Corollary II. Through any given point a plane can be passed parallel to any two given straight lines in space. Proposition VII. Planes perpendicular to the same straight line are parallel to each other. Proposition VIII. The intersections of two parallel planes with any third plane are parallel. Proposition IX. A straight line perpendicular to one of two parallel planes is perpendicular to the other. Corollary. Through any given point one plane can be passed parallel to a given plane, and but one. Proposition X. If two angles, not in the same plane, have their sides respec- tively parallel and lying in the same direction, they are equal and their planes are parallel. Proposition XI. If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to that plane. Corollary. Two straight lines perpendicular to the same plane are parallel to each other. Proposition XII. Two diedral angles are equal if their plane angles are equal. Proposition XIII. Two diedral angles are in the same ratio as their plane angles. SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 313 Proposition XIV. If a straight line is perpendicular to a plane, every plane passed through the line is perpendicular to the plane. Proposition XV. If two planes are perpendicular to each other, a straight line drawn in one of them, perpendicular to their intersection, is perpendicular to the other. Corollary I. If two planes are perpendicular to each other, a straight line drawn through any point of their intersection per- pendicular to one of the planes will lie in the other. Corollary II. If two planes are perpendicular, a straight line let fall from any point of one plane perpendicular to the other will lie in the first plane. Proposition XVI. If two intersecting planes are each perpendicular to a third plane, their intersection is also perpendicular to that plane. Proposition XVII. Through any given straight line a plane can be passed perpen- dicular to any given plane. Proposition XVIII. The projection of a straight line upon a plane is a straight line. Proposition XIX. The acute angle which a straight line makes with its own pro- jection upon a plane is the least angle it makes with any line of that plane. Proposition XX. The sum of any two face angles of a triedral angle is greater than the third. Proposition XXI. The sum of the face angles of any convex polyedral angle is less than four right angles. Proposition XXII. If two triedral angles have the three face angles of the one re- spectively equal to the three face angles of the other, the corre- sponding diedral angles are equal. 27 314 ELEMENTS OF GEOMETRY. BOOK VII. THEOREMS. Proposition I. The sections of a prism made by parallel planes are equal poly- gons. Corollary. Any section of a prism made by a plane parallel to the base is equal to the base. Proposition II. The lateral area of a prism is equal to the product of the perim- eter of a right section of the prism by a lateral edge. Corollary. The lateral area of a right prism is equal to the product of the perimeter of its base by its altitude. Proposition III. Two prisms are equal, if three faces including a triedral angle of the one are respectively equal to three faces similarly placed including a triedral angle of the other. Corollary I. Two truncated prisms are equal, if three faces in- cluding a triedral angle of the one are respectively equal to three faces similarly placed including a triedral angle of the other. Corollary II. Two ri^ht prisms are equal if they have equal bases and equal altitudes. Proposition IV. Any oblique prism is equivalent to a right prism whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. Proposition V. Any parallelopiped is equivalent to a rectangular parallelopiped of the same altitude and an equivalent base. Proposition VI. The plane passed through two diagonally opposite edges of a parallelopii^ed divides it into two equivalent triangular prisms. Proposition VII. Two rectangular parallelepipeds having equal bases are to each other as their altitudes. SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 315 Proposition VIII. Two rectangular parallelepipeds having equal altitudes are to each other as their bases. Proposition IX. Any two rectangular parallelopipeds are to each other as the products of their three dimensions. Proposition X. The volume of a rectangular parallelopiped is equal to tne product of its three dimensions, the unit of volume being the cube whose edge is the linear unit. Proposition XI. The volume of any parallelopiped is equal to the product of the area of its base by its altitude. Proposition XII. The volume of a triangular prism is equal to the product of its base by its altitude. Corollary. The volume of any prism is equal to the product of its base by its altitude. Proposition XIII. If a pyramid is cut by a plane parallel to its base, 1st, the edges and the altitude are divided proportionally ; 2d, the section is a polygon similar to the base. Corollary I. If a pyramid is cut by a plane parallel to its base, the area of the section is to the area of the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. Corollary II. If two pyramids have equal altitudes and equiva- lent bases, sections made by planes parallel to their bases and at equal distances from their vertices are equivalent. Proposition XIV. The lateral area of a regular pyramid is equal to the product of the perimeter of its base by half its slant height. Corollary. The lateral area of the frustum of a regular pyra- mid is equal to the half sum of the perimeters of its bases multi- plied by the slant height of the frustum. 316 ELEMENTS OF GEOMETRY. Proposition XV. If the altitude of any given triangular pyramid is divided into equal parts, and through the points of division planes are passed parallel to the base of the pyramid, and on the sections made by these planes as upper bases prisms are described having their edges parallel to an edge of the pyramid and their altitudes equal to one of the equal parts into which the altitude of the pyramid is divided, the total volume of these prisms will approach the volume of the pyramid as its limit as the number of parts into which the altitude of the pyramid is divided is indefinitely increased. Proposition XVI. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. Proposition XVII. A triangular pyramid is one-third of a triangular prism of the same base and altitude. Corollary. The volume of a triangular pyramid is equal to one- third of the product of its base by its altitude. Proposition XVIII. The volume of any pyramid is equal to one-third of the product of its base by its altitude. Proposition XIX. A frustum of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frus- tum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum. Corollary. A frustum of any pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum. Proposition XX. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism and whose vertices are the three vertices of the inclined section. Proposition XXI. Only five regular (convex) polyedrons are possible. SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 317 BOOK Till. THEOREMS. Proposition I. Every section of a cylinder made by a plane passing through an element is a parallelogram. Corollary. Every section of a right cylinder made by a plane perpendicular to its base is a rectangle. Proposition II. The bases of a cylinder are equal. Corollary I. Any two parallel sections of a cylindrical surface are equal. Corollary II. All the sections of a circular cylinder parallel to its bases are equal circles, and the straight line joining the centres of the bases passes through the centres of all the parallel sections. Proposition III. Every section of a cone made by a plane passing through its vertex is a triangle. Proposition IV. If the base of a cone is a circle, every section made by a plane parallel to the base is a circle. Corollary. The axis of a circular cone passes through the centres of all the sections parallel to the base. Proposition V. Every section of a sphere made by a plane is a circle. Corollary I. The axis of a circle on a sphere passes through the centre of the circle. Corollary II. All great circles of the same sphere are equal. Corollary III. Every great circle divides the sphere into two equal parts. Corollary IV. Any two great circles on the same sphere bisect each other. Corollary V. An arc of a great circle may be drawn through any two given points on the surface of a sphere, and, unless the points are the opposite extremities of a diameter, only one such arc can be drawn. Corollary VI. An arc of a circle may be drawn through any three given points on the surface of a sphere. 27* 318 ELEMENTS OF GEOMETRY. Proposition VI. All the points in the circumference of a circle on a sphere are equally distant from either of its poles. Corollary I. All the arcs of great circles drawn from a pole of a circle to points in its circumference are equal. Corollary II. The polar distance of a great circle is a quadrant. Corollary III. If a point on the surface of a sphere is at a quad- rant's distance from each of two given points of the surface, which are not opposite extremities of a diameter, it is the pole of the great circle passing through them. Proposition VII. A plane tangent to a sphere is perpendicular to the radius drawn to the point of contact. Corollary. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere. Proposition VIII. The intersection of two spheres is a circle whose plane is per-^ pendicular to the straight line joining their centres, and whose centre is in that line. Proposition IX. The angle of two arcs of great circles is equal to the angle of their planes, and is measured by the arc of a great circle described from its vertex as a pole and included between its sides (produced if necessary). Corollary. All arcs of great circles drawn through the pole of a given great circle are perpendicular to its circumference. Proposition X. If the first of two spherical triangles is the polar triangle of the second, then, reciprocally, the second is the polar triangle of the first. Proposition XI. In two polar triangles, each angle of one is measured by the supplement of the side lying opposite to it in the other. Proposition XII. Two triangles on the same sphere are either equal or symmet- rical when two sides and the included angle of one are respectively equal to two sides and the included angle of the other. SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 319 Proposition XIII. Two triangles on the same sphere are either equal or symmet- rical when a side and the two adjacent angles of one are respec- tively equal to a side and the two adjacent angles of the other. Proposition XIV. Two triangles on the same sphere are either equal or symmet- rical when the three sides of one are respectively equal to the three sides of the other. Proposition XV. If two triangles on the same sphere are mutually equiangular, they are mutually equilateral, and are either equal or symmet- rical. Proposition XVI. Any side of a spherical triangle is less than the sum of the other two. Proposition XVII. The sum of the sides of a convex spherical polygon is less than the circumference of a great circle. Proposition XVIII. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles. Proposition XIX. Two symmetrical spherical triangles are equivalent. Proposition XX. If two arcs of great circles intersect on the surface of a hemi- sphere, the sum of the opposite spherical triangles which they form is equivalent to a lune whose angle is the angle between the arcs in question. Proposition XXI. A lune is to the surface of the sphere as the angle of the lune is to four right angles. Corollary. The area of a lune is expressed by twice its angle, the angular unit being the degree, and the unit of surface the spherical degree. 320 ELEMENTS OF GEOMETRY. Proposition XXII. The area of a spherical triangle is equal to the excess of the sum of its angles over two right angles. Proposition XXIII. The shortest line that can be drawn on the surface of a sphere between two points is the arc of a great circle, not greater than a semi-circumference, joining the two points. BOOK IX. THEOREMS. Proposition I. If a prism whose base is a regular polygon be inscribed in or circumscribed about a given cylinder, its volume will approach the volume of the cylinder as its limit, and its lateral surface will approach the lateral surface of the cylinder as its limit as the number of sides of its base is indefinitely increased. Proposition II. The lateral area of a cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of the surface. Corollary I. The lateral area of a cylinder of revolution is equal to the product of the circumference of its base by its altitude. This may be formulated, Corollary II. The lateral areas of similar cylinders of revolu- tion are to each other as the squares of their altitudes, or as the squares of the radii of their bases. Proposition III. The volume of a cylinder is equal to the product of its base by its altitude. Corollary I. For a cylinder of revolution this may be formu- lated, Corollary II. The volumes of similar cylinders of revolution are to each other as the cubes of their altitudes, or as the cubes of their radii. SYLLABUS OF PROPOSITIONS IN SOLID GEOMETRY. 321 Proposition IV. If a pyramid be inscribed in or circumscribed about a given cone, its volume will approach the volume of the cone as its limit, and its lateral surface will approach the convex surface of the cone as its limit as the number of faces of the pyramid is indefi- nitely increased. Corollary, A frustum of a cone is the limit of the inscribed and circumscribed frustums of pyramids, the number of whose faces is indefinitely increased. Proposition V. The lateral area of a cone of revolution is equal to the product of the circumference of its base by half its slant height. Corollary I. This proposition may be formulated, S = ^RL. Corollary II. The lateral areas of similar cones of revolution are to each other as the squares of their slant heights, or as the squares of their altitudes, or as the squares of the radii of their bases. Proposition VI. The lateral area of a frustum of a cone of revolution is equal to the half sum of the circumferences of its bases multiplied by its slant height. Corollary I. This proposition may be formulated, S=Tr{R-{.r)L. Corollary II. The lateral area of a frustum of a cone of revolu- tion is equal to the circumference of a section equidistant from its bases multiplied by its slant height. Proposition VII. The volume of any cone is equal to one-third the product of its base by its altitude. Corollary I. For a cone of revolution this proposition may be formulated, V= ITER'S. Corollary II. Similar cones of revolution are to each other aa the cubes of their altitudes, or as the cubes of the radii of their 322 ELEMENTS OF GEOMETRY. Proposition VIII. The area of the surface generated by a straight line revolving about an axis in its plane is equal to the projection of the line on the axis multiplied by the circumference of the circle whose radius is the perpendicular erected at the middle of the line and terminated by the axis. Proposition IX. The area of a zone is equal to the product of its altitude by the circumference of a great circle. Corollary. This proposition may be formulated, Proposition X. The area of the surface of a sphere is equal to the product of^ its diameter by the circumference of a great circle. Corollary I. This may be formulated, S = 27ri2 X2B = ^TzB^,- Hence the surface of a sphere is equivalent to four great circles. Corollary II. The surfaces of two spheres are to each other aa the squares of their diameters, or as the squares of their radii. Proposition XI. The volume of a sphere is equal to the area of its surface multi- plied by one-third of its radius. Corollary I. This proposition may be formulated, Corollary II. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. Proposition XII. The volume of a spherical sector is equal to the area of the zone which forms its base multiplied by one-third the radius of the sphere. ^__ " Of THB THE END. T ^ I T^ F ({ THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. JEP 27 1934 i .'^-. 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