GIFT OF ROBBINS'S NEW PLANE GEOMETRY BY EDWARD RUTLEDGE ROBBINS, A.B. FORMERLY OF LAWRENCEVILLE SCHOOL AMERICAN BOOK COMPANY NEW YORK CINCINNATI CHICAGO COPYRIGHT, 1915, BY EDWARD RUTLEDGE ROBBINS. ROBBINS'S NEW PLANK GEOMETRY. W. P. I FOR THOSE WHOSE PRIVILEGE IT MAY BE TO ACQUIRE A KNOWLEDGE OF GEOMETRY THIS VOLUME HAS BEEN WRITTEN AND TO THE BOYS AND GIRLS WHO LEARN THE ANCIENT SCIENCE FROM THESE PAGES, AND WHO ESTEEM THE POWER OF CORRECT REASONING THE MORE BECAUSE OF THE LOGIC OF PURE GEOMETRY THIS VOLUME IS DEDICATED 459819 PREFACE THIS New Plane Geometry is not only the outgrowth, of the author's long experience in teaching geometry, but has profited further by suggestions from -teachers who have used Robbins's " Plane Geometry " and by many of the recommendations of the "National Committee of Fifteen." While many new and valu- able features have been added in the reconstruction, yet all the characteristics that met with widespread favor in the old book have been retained. Among the features of the book that make it sound and teach- able may be mentioned the following : 1. The book has been written for the pupil. The objects sought in the study of Geometry are (1) to train the mind to accept only those statements as truth for which convincing reasons can be provided, and (2) to cultivate a foresight that will appreci- ate both the purpose in making a statement and the process of reasoning by which the ultimate truth is established. Thus, the study of this formal science should develop in the pupil the ability to pursue argument coherently, and to establish geometric truths in logical order. To meet the requirements of the various degrees of intellectual capacity and maturity in every class, the reason for every statement is not printed in full but is indicated by a reference. The pupil who knows the reason need not con- sult the paragraph cited; while the pupil who does not know it may learn it by the reference. It is obvious that the greater progress an individual makes in assimilating the subject and in entering into its spirit, the less need there will be for the printed reference. 2. Every effort has been made to stimulate the mental activity of the pupil. To compel a young student, however, to supply his vi PREFACE own demonstrations frequently proves unprofitable as well as arduous, and engenders in the learner a distaste for a study in which he might otherwise take delight. This text does not aim to produce accomplished geometricians at the completion of the first book, but to aid the learner in his progress throughout the volume, wherever experience has shown that he is likely to require assistance. It is designed, under good instruction, to develop a clear conception of the geometric idea, and to produce at the end of the course a rational individual and a friend of this particular science. 3. The theorems and their demonstrations the real subject- matter of Geometry are introduced as early in the study as possible. 4. The simple fundamental truths are explained instead of being formally demonstrated. 5. The original exercises are distinguished by their abundance, their practical bearings upon the affairs of life, their careful gradation and classification, and their independence. Every ex- ercise can be solved or demonstrated without the use of any other exercise. Only the truths in the numbered paragraphs are nec- essary in working originals. 6. The exercises are introduced as near as practicable to the theorems to which they apply. 7. Emphasis is given to the discussion of original constructions. 8. The summaries will be found a valuable aid in reviews. 9. The historical notes give the pupil a knowledge of the devel- opment of the science of geometry and add interest to the study. 10. The attractive open page will appeal alike to pupils and to teachers. The author sincerely desires to extend his thanks to those friends and fellow teachers who, by suggestion and encourage- ment, have inspired him in the preparation of these pages. EDWARD K. BOBBINS. CONTENTS INTRODUCTION PAGE DEFINITIONS 1 ANGLES 2 TRIANGLES 4 CONGRUENCE ...... 5 SYMBOLS ............ 6 AXIOMS 6 POSTULATES . 7 EXERCISES 9 BOOK I. ANGLES, LINES, RECTILINEAR FIGURES PRELIMINARY THEOREMS . . . 13 THEOREMS AND DEMONSTRATIONS 15 TRIANGLES 15 PARALLEL LINES . .20 QUADRILATERALS 47 POLYGONS 60 SYMMETRY 65 CONCERNING ORIGINAL EXERCISES . . . .... 68 SUMMARY. GENERAL DIRECTIONS FOR ATTACKING ORIGINALS . 68 ORIGINAL EXERCISES 70 BOOK II. THE CIRCLE DEFINITIONS 75 PRELIMINARY THEOREMS 77 THEOREMS AND DEMONSTRATIONS 78 SUMMARY 94 ORIGINAL EXERCISES . 95 vii vi PREFACE own demonstrations frequently proves unprofitable as well as arduous, and engenders in the learner a distaste for a study in which he might otherwise take delight. This text does not aim to produce accomplished geometricians at the completion of the first book, but to aid the learner in his progress throughout the volume, wherever experience has shown that he is likely to require assistance. It is designed, under good instruction, to develop a clear conception of the geometric idea, and to produce at the end of the course a rational individual and a friend of this particular science. 3. The theorems and their demonstrations the real subject- matter of Geometry are introduced as early in the study as possible. 4. The simple fundamental truths are explained instead of being formally demonstrated. 5. The original exercises are distinguished by their abundance, their practical bearings upon the affairs of life, their careful gradation and classification, and their independence. Every ex- ercise can be solved or demonstrated without the use of any other exercise. Only the truths in the numbered paragraphs are nec- essary in working originals. 6. The exercises are introduced as near as practicable to the theorems to which they apply. 7. Emphasis is given to the discussion of original constructions. 8. The summaries will be found a valuable aid in reviews. 9. The historical notes give the pupil a knowledge of the devel- opment of the science of geometry and add interest to the study. 10. The attractive open page will appeal alike to pupils and to teachers. The author sincerely desires to extend his thanks to those friends and fellow teachers who, by suggestion and encourage- ment, have inspired him in the preparation of these pages. EDWARD R. ROBBINS. CONTENTS INTRODUCTION PAGE DEFINITIONS 1 ANGLES 2 TRIANGLES 4 CONGRUENCE ...... 5 SYMBOLS . 6 AXIOMS . 6 POSTULATES 7 EXERCISES 9 BOOK I. ANGLES, LINES, RECTILINEAR FIGURES PRELIMINARY THEOREMS 13 THEOREMS AND DEMONSTRATIONS 15 TRIANGLES 15 PARALLEL LINES . .20 QUADRILATERALS .......... 47 POLYGONS 60 SYMMETRY 65 CONCERNING ORIGINAL EXERCISES . . . .... 68 SUMMARY. GENERAL DIRECTIONS FOR ATTACKING ORIGINALS . 68 ORIGINAL EXERCISES 70 BOOK II. THE CIRCLE DEFINITIONS 75 PRELIMINARY THEOREMS 77 THEOREMS AND DEMONSTRATIONS 78 SUMMARY 94 ORIGINAL EXERCISES . 95 vii viii CONTENTS PAGE KINDS OF QUANTITIES. MEASUREMENT 96 ORIGINAL EXERCISES 109 Loci 114 ORIGINAL EXERCISES ON Loci 115 CONSTRUCTION PROBLEMS 117 ANALYSIS 131 ORIGINAL CONSTRUCTION PROBLEMS 132 BOOK III. PROPORTION. SIMILAR FIGURES DEFINITIONS 143 THEOREMS AND DEMONSTRATIONS . 144 CONCERNING ORIGINALS . 175 ORIGINAL EXERCISES (NUMERICAL) 176 SUMMARY 180 ORIGINAL EXERCISES (THEOREMS) 180 CONSTRUCTION PROBLEMS 186 ORIGINAL CONSTRUCTION PROBLEMS . . . . . . 190 BOOK IV. AREAS THEOREMS AND DEMONSTRATIONS 193 FORMULAS 207 ORIGINAL EXERCISES (NUMERICAL) 210 CONSTRUCTION PROBLEMS 213 ORIGINAL CONSTRUCTION PROBLEMS 221 BOOK V. REGULAR POLYGONS. CIRCLES THEOREMS AND DEMONSTRATIONS 225 ORIGINAL EXERCISES (THEOREMS) 239 CONSTRUCTION PROBLEMS 242 FORMULAS 245 ORIGINAL EXERCISES (NUMERICAL) 248 ORIGINAL CONSTRUCTION PROBLEMS 253 MAXIMA AND MINIMA 254 ORIGINAL EXERCISES 259 INDEX . . .261 PLANE GEOMETRY INTRODUCTION 1. Geometry is a science which treats of the measure- ment of magnitudes. 2. A point is that which has position but not magnitude. 3. A line is that which has length but no other magni- tude. 4. A straight line is a line which is determined (fixed in position) by any two of its points. That is, two lines that coincide entirely, if they coincide at any two points, are straight lines. 5. A rectilinear figure is a figure containing straight lines and no others. 6. A surface is that which has length and breadth but no other magnitude. 7. A plane is a surface in which if any two points are taken, the straight line connecting them lies wholly in that surface. 8. Plane Geometry is a science which treats of the proper- ties of magnitudes in a plane. 9. A solid is that which has length, breadth, and thick- ness. A solid is that which occupies space. 10. Boundaries. The boundaries (or boundary) of a solid are surfaces. The boundaries (or boundary) of a surface 1 PLANE GEOMETRY ;aJr,e' 'filler : -'T-he boundaries of a line are points. These boundaries can be no part of the things they limit. A sur- face is no part of a solid ; a line is no part of a surface ; a point is no part of a line. 11. Motion. If a point moves, its path is a line. Hence, if a point moves, it generates (describes or traces) a line ; if a line moves (except upon itself), it generates a surface ; if a surface moves (except upon itself), it generates a solid. NOTE. Unless otherwise specified the word " line " means straight line. ANGLES ANGLE ADJACENT VERTICAL ANGLES RIGHT ANGLES ANGLES PERPENDICULAR 12. A plane angle is the amount of divergence of two straight lines that meet. The lines are called the sides of the angle. The vertex of an angle is the point at which the lines meet. 13. Adjacent angles are two angles that have the same vertex and a common side between them. 14. Vertical angles are two angles- that have the same vertex, the sides of one being prolongations of the sides of the other. 16. If one straight line meets another and makes the ad- jacent angles equal, the angles are right angles. 16. One line is perpendicular to another if they meet at right angles. Either line is perpendicular to the other. The point at which the lines meet is the foot of the perpendicular. Oblique lines are lines that meet but are not perpendicular. INTRODUCTION . 3 17. A straight angle is an angle whose sides lie in the same straight line, but extend in opposite directions from the vertex. OBTUSE ACUTE COMPLEMENTARY SUPPLEMENTARY ANGLE ANGLE ANGLES ANGLES 18. An obtuse angle is an angle that is greater than a right angle. An acute angle is an angle that is less than a right angle. An oblique angle is any angle that is not a right angle. 19. Two angles are complementary if their sum is equal to one right angle. Two angles are supplementary if their sum is equal to two right angles. Thus, the complement of an angle is the difference between one right angle and the given angle. The supplement of an angle is the difference between .two right angles and the given angle. 20. A degree is one ninetieth of a right angle. The degree is the familiar unit used in measuring angles. It is evident that there are 90 in a right angle ; 180 in two right angles, or a straight angle ; 360 in four right angles. There are 60 minutes (60') in one degree, and 60 seconds (60") in one minute. 21. Parallel lines are straight lines that lie in the same plane and that never meet, however far they are extended in either direction. 22. Notation. A point is usually denoted by a capital letter, placed near it. A line is denoted by two capital letters, placed one at each end, or one at each of two of its points. Its length is sometimes represented advantageously by a small letter written near it. Thus, the line AB ; the line RS\ the line m. R S m 4 PLANE GEOMETRY There are various ways of naming angles. Sometimes three capital letters are used, one on each side of the angle and one at the vertex ; sometimes a small letter or a figure is placed within the angle. The symbol for angle is Z. M ZAMXo* /.XMA on AM X O C Z.a AND Z .BOO, NOT ZO In naming an angle by the use of three letters, the vertex letter is al- ways placed between the others. Thus the A above are Z.AMX or Z KM A, Z a, Z BOC, Z.x,/.APR,Z. APS, Z BPR, Z TPB, Z 5, etc. In the above figure Z x = Z 5. The size of an angle depends on the amount of divergence between its sides, and not upon their length. An angle is said to be included by its sides. An angle is bisected by a line drawn through the vertex and dividing the angle into two equal angles. TRIANGLES 23. A triangle is a portion of a plane bounded by three straight lines. These lines are the sides. The vertices of a triangle are the three points at which the sides intersect. The angles of a triangle are the three angles at the three vertices. Each side of a triangle has two angles adjoining it. The symbol for triangle is A. ISOSCELES A EQUILATERAL A BIGHT A OBTUSE A ACUTE A EQUIANGULAR A SCALENE & INTRODUCTION 5 The base of a triangle is the side on which the figure ap- pears to stand. The vertex of a triangle is the vertex op- posite the base. The vertex angle is the angle opposite the base. 24. Kinds of triangles : A scalene triangle is a triangle no two sides of which are equal. An isosceles triangle is a triangle two sides of which are equal. An equilateral triangle is a triangle all sides of which are equal. A right triangle is a triangle one angle of which is a right angle. An obtuse triangle is a triangle one angle of which is an obtuse angle, An acute triangle is a triangle all angles of which are acute angles. An equiangular triangle is a triangle all angles of which are equal. 25. The hypotenuse of a right triangle is the side opposite the right angle. The sides forming the right angle are called legs. CONGRUENCE 26. Two geometric figures are said to be equal if they have the same size or magnitude. Two geometric figures are said to be congruent if, when one is superposed upon the other, they coincide in all respects. The corresponding parts of congruent figures are equal, and are called homologous parts. 27. Homologous parts of congruent figures are equal. If the triangles DEF and HIJ are congruent, Z.D\$> homologous to and = to Z. H ; DE is homologous to and = to HI; Z E is homologous to and = to ^ /; EF is homologous to and = to IJ. NOTE. Congruent figures have the same shape as well as the same size, whereas equal figures do not necessarily have the same shape. Ex.1. What is the complement of an angle of 35? 48? 80? 75 50' ? 8 20' ? Ex. 2. What is the supplement of an angle of 100? 50? 148? 121 30'? 10 40'? 6 PLANE GEOMETRY 28. A curve or curved line, is a line no part of which is straight. A circle is a plane curve all points of which are equally distant from a point in the plane, called the center. An arc is any part of a circle. A radius is a straight line from the center to any point of the circle. A diameter is a straight line containing the center and having its extremities in the circle. The length of the circle is called the circumference. 29. Symbols. The usual symbols and abbreviations em- ployed in geometry are the following : + plus. minus. = equals, is equal to, equal. = does not equal. ^ congruent, or is con- gruent to. > is greater than. < is less than. .'. hence, therefore, consequently. JL perpendicular. Js perpendiculars. AXIOM, POSTULATE, AND THEOREM 30. An axiom is a statement admitted without proof to be true. It is a truth, received and assented to immediately. 31. AXIOMS. 1. Magnitudes that are equal to the same thing, or to equals, are equal to each other. 2. If equals are added to, or subtracted from, equals, the results are equal. 3. If equals are multiplied by, or divided by, equals, the results are equal. [Doubles of equals are equal; halves of equals are equal.] O circle. Ax. axiom. circles. Hyp. hypothesis. Z angle. comp. complementary. A angles. supp. supplementary. rt. Z right angle. Const. construction. rt. A right angles. Cor. corollary. A triangle. St. straight. & triangles. rt. right. rt. & right triangles. Def. definition. II parallel. alt. . alternate. Us parallels. int. interior. d parallelogram. ext. exterior. 17 parallelograms. INTRODUCTION 7 4. The whole is equal to the sum of all of its parts. 5. The whole is greater than any of its parts. 6. A magnitude may be displaced by its equal in any process. [Briefly called "substitution."] 7. If equals are added to, or subtracted from, unequals, the results are unequal in the same order. 8. If unequals are added to unequals in the same order, the results are unequal in that order. 9. If unequals are subtracted from equals, the results are un- equal in the opposite order. 10. Doubles or halves of unequals are unequal in the same order. Also, unequals multiplied by equals are unequal in the same order. 11. If the first of three magnitudes is greater than the second, and the second is greater than the third, the first is greater than the third. 12. A straight line is the shortest line that can be drawn be- tween two points. 13. Only one line can be drawn through a point parallel to a given line. 14. A geometrical figure may be moved from one position to another without any change in form or magnitude. 32. A postulate is something required to be done, the pos- sibility of which is admitted without proof. 33. POSTULATES. 1. It is possible to draw a straight line from any point to any other point. 2. It is possible to extend (prolong or produce) a straight line indefinitely, or to terminate it at any point. 8 PLANE GEOMETRY 34. A geometric proof or demonstration is a logical course of reasoning by which a truth becomes evident. 35. A theorem is a statement that requires proof. In the case of the preliminary theorems which follow, the proof is very simple ; but as these theorems are not admitted without proof they cannot be classified with the axioms. A corollary is a truth immediately evident, or readily es- tablished from some other truth or truths. A proposition, in geometry, is the statement of a theorem to be proved or a problem to be solved. Ex. 1. Draw an Z ABC. In /. ABC draw line BD. . What does Z ABD + Z. DEC equal ? What does Z ABC - ZABD equal? Ex. 2. In a rt. Z.ABC draw line BD. If ZABD = 25, how many degrees are there in Z DB C ? How many degrees are there in the complement of an angle of 38 ? How many degrees are there in the supplement ? Ex. 3. Draw a straight line AB and take a point X on it. What line does AX + BX equal? What line does AB - BX equal? Ex. 4. Draw a straight line AB and prolong it to X so ih&iBX = AB. Prolong it so that AB = AX. Historical Note. Probably as early as 3000 B.C. the Egyptians had some knowledge of geometric truths. The construction of the great pyramids required an acquaintance with the relations of geometry. This knowledge, however vague it may have been, was, according to Herodotus, employed in determining the amount of land washed away by the river Nile, during the reign of Rameses II (1400 B.C.). The Greeks, however, were the first to study geometry as a logical science. They enunciated theorems and demonstrated them, they pro- pounded problems and solved them as early as 300 B.C., and, in a crude way, two or three centuries earlier. To them belongs the credit of estab- lishing a logical system of geometry that has survived, practically un- changed, for twenty centuries. INTRODUCTION B EXERCISES EMPLOYING THE TWO INSTRUMENTS OF GEOMETRY Aside from pencil and paper, the only instruments necessary for the construction of geometrical dia- grams are the ruler and the compasses. Ex. 1. It is required to draw an equilateral tri- angle upon a given line as base. Suppose AB is the given base. Required to draw an equilateral A upon it. Using A as a center and AB as a radius, draw an arc. Using B as a center and AB as a radius, draw another arc cutting the first one at C. Draw AC and BC. The &ABC is an equi- lateral A, and AB is its base. Ex. 2. It is required to draw a triangle having its three sides each equal to a given line. Suppose the three given lines are a,b,c. Required to draw a A having for its sides lines equal to a, &, c-, respectively. Draw a line RS to a. Using R as a center and b as a radius, draw an arc. Using S as a center and c as a radius, draw another arc cutting the first arc at T. Draw straight lines R T and ST. A RST is the A whose three sides are equal to the lines a, b, c, respectively. Ex. 3. It is required to find the midpoint of a given straight line. Given the straight line AB. Required to find its midpoint. Using A and B as centers and a radius sufficiently / j \ long, draw two arcs, intersecting at P and Q. AI r^-j B Draw the straight line PQ cutting AB at M. \ I / Point M is the midpoint of A B. \ \ / Ex. 4. It is required to draw a perpendicular to a '"Q x line from a point within the line. Given the line CD and point P in it. ..K Required to construct a _L to CD, at P. Using P as center and any radius, draw two arcs ^ / I \ p cutting CD at E and F. Now using E and F as centers and a radius greater than before, draw two arcs intersecting at K. Draw KP. This line KP is _L to CD at P. BOBBINS'S NEW PLANE GEOM. 2 10 PLANE GEOMETRY P L Ex. 6. It is required to draw a perpendicular to a line from a point without the line. Given line AB and point P, without it. Required to draw a _L to A B from P. Using P as center and a sufficient radius, draw A- an arc cutting AB at C and D. Now using C and D as centers and a sufficient radius, draw two arcs intersecting at E. Draw PE, meeting AB at R. PR is the required _L to AB from P. Ex. 6. It is required to bisect a given angle. Given the Z ABC. Required to bisect it. Using vertex B as a center and any radius, draw arc DE cutting BC at D and BA at E. Using D and E as centers-and a sufficient radius, draw arcs intersect- ing at F. Draw straight line BF. BF bisects the Z ABC. Ex. 7. It is required to cpnstruct, at a given point on a given line, an angle equal to a given angle. Given line DE, point D in it, and Z B. Required to construct an Z at D, equal to Z 5. Using .B as a center and with any two distances as radii, draw an arc cutting AB at F and another cutting BC at G. Using D as a center and the same radii as before, draw one arc, and another arc cutting DE at /. Draw the straight line FG. Using / as a center and FG as a radius, draw an arc cutting a former arc at H. Draw the straight lines HJ and DHK. Now the Z KDE = Z B. Ex. 8. By the use of ruler and compasses, draw the following figures : G Ex. 9. Does it make any difference in these exercises, which lines are drawn first? In Ex. 7 and Ex. 8 explain the order of the lines drawn. INTRODUCTION 11 Ex. 10. Using the compasses only, draw the following figures: Ex. 11. Draw the following figures : Ex. 12. Draw the first of each of these three pairs of figures. Can you explain the construction of the second figure in each pair ? 12 PLANE GEOMETRY In this figure, ABCD is a square. On the sides are measured the equal dis- tances AE and BF, and CG and DH ; ( then the lines AG, BH, CE, and DF, are drawn intersecting at a, b, c, d. The figure abed is also .a square. This figure is the basis of an Arabic design used for parquet floors, etc. In this figure, which is the basis of a mosaic floor design, the radii of all com- plete circles equal one fourth of the side of the square ABCD. The radii of the semicircles GflJ, IKR, etc., equal one eighth of the side of the square. In this figure ABD is an equilateral arch ? and CD is its altitude. The several cen- ters used are A, of arc BD and arc CE ; 5, of arc AD and CF ; (7, of arcs AE and BF. This figure is the basis of a common Gothic window design. NOTE. The letters " Q.E.D." are often annexed at the end of a demon- stration and stand for "quod erat demonstrandum," which means, " which was to be proved." BOOK I ANGLES, LINES, RECTILINEAR FIGURES PRELIMINARY THEOREMS 36. A right angle is equal to half a straight angle. Because of the definition of a right angle. 37. A straight angle is equal to two right angles. (36.) 38. Two straight lines can intersect in only one point. Because they would coincide entirely if they had two common points. (4.) 39. Only one straight line can be drawn between two points. (40 40. A definite (limited or finite) straight line can have only one midpoint. Because the halves of a line are equal. 41. All straight angles are equal. Because they can be made to coincide. (26.) 42. \ All right angles are equal. ThejKare halves of straight angles. (36.) .. they are equal. (Ax. 3.) 43. Only one perpendicular to a line can be drawn from a point in the line. These right angles would not be equal if there were two perpendiculars. (42.) 13 14 BOOK I. "PLANE GEOMETRY 44. If two adjacent angles have their exterior sides in a straight line, they are supplementary. Because they together equal two rt. A. (19.) 45. If two adjacent angles are supplementary, their exte- rior sides are in the same _ straight line. Because their sum is two rt.^ (19) ; or a straight Z (37) . Hence the exterior sides are in the same straight line (17). 46. The sum of all the angles on one side of a straight line at a point equals two right angles. (Ax. 4 and 37.) 47. The sum of all the angles about a point in a plane is equal to four right angles. (46.) 48. Angles that have the same complement are equal. Or, complements of the same angle, or of equal angles, are equal. Because equal angles subtracted from equal right angles leave equal angles. (Ax. 2.) 49. Angles that have the same supplement are equal. Or, supplements of the same angle, or of equal angles, are equal. (Ax. 2.) 50. If two angles are equal and supplementary, they are right angles. Each is half a straight Z ; .'. each is a rt. XOTE. A reference number usually indicates only the statement in full face type in the section referred to. In giving demonstrations the pupil should quote the correct reason for each statement. Ex. The bisector^ of two supplementary adja- cent angles are perpendicular to each other. TRIANGLES 15 THEOREMS AND DEMONSTRATIONS PROPOSITION I. THEOREM 51. If two straight lines intersect, the vertical angles are equal. A Given : Lines AB and LM intersecting at o, A AOM and BOL, a pair of vertical A. To Prove : Z AOM = Z BOL. Proof : A AOM and MOB are supplementary (44). A MOB and BOL are supplementary (44). .'./-AOM = Z BOL. (49.) Q.E.D. PROPOSITION II. THEOREM 52. Two triangles are congruent if two sides and the in- cluded angle of one are equal respectively to two sides and the included angle of the other. A B R S Given : A ABC and RST ; AB = RS; AC=KT\ Z A = Z R. To Prove : A ABC is congruent to A RST. Proof : Place the A ABC upon the A RST so that Z A coin- cides with its equal Z R. AB falls upon RS and point B upon S (AB = RS~). AC falls upon RT and point C upon T (AC = RT). .*. BC coincides with 8T (39). /.the A coincide and are congruent (26). Q.E.D. 16 BOOK I. PLANE GEOMETRY 53. COROLLARY. Two right triangles are congruent if two legs of one are equal respectively to two legs of the other. This is a corollary following immediately from 52. B Ex. 1. If two triangles have two sides of one equal to two sides of the other, are the triangles necessarily congruent? Ex. 2. Illustrate your answer to Ex. 1 by draw- ing two triangles. Ex. 3. Find the distance AB if there is an ob- struction between A and B. Method. Take a convenient point P, from which A and B are accessible. Measure AP and in same straight line mark point R such that PR = A P. Similarly find point S. Show that the two & ABP and SRP are congruent, therefore the length of AB may be found by measuring RS. Ex. 4. Prove that a point, P, in the perpendicular bisector MC of a line AB is equally distant from the ends of the line AB. (Show that the & AMP and BMP are congruent, (1) by using 52; and (2) by using 53.) Ex.6. If the line AC bisects ZBAD, and BA = AD, prove that the triangles ABC and ADC are congruent, the line A C bisects Z BCD, and EC equals CD. Ex. 6. Prove that if a line from a vertex of a triangle perpendicular to the opposite side bisects that side, the triangle is isosceles. Ex. 7. Draw two angles that are adjacent and not supplementary; adjacent and not complementary. Ex. 8. Of two unequal angles which has the greater supplement? Ex. 9. The complement of a certain angle added to the supplement of the same angle is 176. Find the angle. Ex. 10. What angle added to one fifth of its supplement equals a right angle ? Ex. 11. In the figure of 51, if Z A OM is 100, how many degrees are there in each of the other angles at ? TRIANGLES 17 \ = to PB. 3. 4. 5. 6. Construction, Identical. / Quote 53. Quote ^7, PROPOSITION III. THEOREM 54. Only one perpendicular can be drawn to a line from an external point. Given: PR _L to AB from P, and PD any other line from P to AB. To Prove : PD is not J_ to AB ; that is, PR is the only-L from P to AB. Proof : 1. Extend PR to S making RS 2. Draw DS. 3. In rt. A PDR and SDR, PR = RS. 4. DR = DR. 5. .'.A PDR is congruent to A SDR. 6. .-. ZPD = Z SDR. 7. That is, Z PDR = half of Z PDS. 8. Now line PRS is a st. line. 9. .*. line PDS is not a st. line. 10. /. Z PDE, half of Z PDS, is wo a rt. Z. 11. /. PD is not J_ to ^1#. That is, PR is the only J_ from P to AB. Q.E.D. The preceding form of demonstration will serve to illustrate an ex- cellent plan of writing the proofs. It will be observed that the state- ments appear at the left of the page and their reasons at the right. This arrangement will be found of great value in the saving of time, both for the pupil who writes the proofs and for the teacher who reads them. 8. Construction, 9. Quote 39. 10. Qoote 36. 11. \Quotef Historical Note. The proof of the following theorem as given in the fifth proposition of Euclid's " The Elements," the most famous geometry that was ever written, was considered by the beginners as pre- senting great difficulties. The theorem was therefore called by the ancient teachers, the pans asinorum, or the bridge of the asses. Euclid discussed only magnitudes, not their numerical measures. Another note (p. 45) will tell more of the author of this renowned book. 18 BOOK I. PLANE GEOMETRY PROPOSITION IV. THEOREM 55. The angles opposite the equal sides of an isosceles triangle are equal. Given : A ABC, AB = AC. To Prove : Z # = Z c. B x c Proof: Suppose AX is drawn dividing Z BAG into two equal A and meeting BC at X. In the A ABX and ACX, AX = AX (Identical) . AB = AC (Given). /.EAX /.GAX (Const.). .'.A A EX is congruent to A ACX (52). .'. Z.B = Z.c (27). Q.E.D. 56. COROLLARY. An equilateral triangle is equiangular. PROPOSITION V. THEOREM 57. Two right triangles are congruent if the hypotenuse and an adjoining angle of one are equal respectively to the hypote- nuse and an adjoining angle of the other. A CD F Given : Rt. A ABC and DEF ; AB = DE ; and Z A = D. To Prove : A ABC is congruent to A DEF. Proof : Place A ABC upon A DEF so that Z A coincides with its equal, Z D, and AC falls along DF. Then AB coincides with DE and point B falls exactly on E (AB = DE). TRIANGLES 19 Now, from point E, BC and EF are both _L to DF .'. BC coincides with EF .-.A ABC is congruent to A DEF (16). (54). (26). Q.E.D. Ex. 1. In the adjoining diagram, if Z 1 = Z 2, prove the right triangles congruent. Ex. 2. By use of 27 prove PA = PC. Ex. 3. Prove that the line, EM, from the vertex of an isosceles triangle, ABC, and perpendicular to the base, bisects the vertex angle and also bisects the base. (First prove the two right & congruent, and then use 27.) Ex. 4. Prove that the perpendiculars, SM and TP, upon the equal sides of an isosceles triangle, RST, from the opposite vertices, S and T, are equal. Two ways : (a) Show rt. A PST and MST are con- gruent; or (b) show rt. &RSM and RPT are con- gruent. Ex. 6. Prove that the bisector of the vertex angle of an isosceles tri- angle bisects the base at right angles. 58. Homologous parts. Triangles are proved congruent in order that their homologous sides or homologous angles may be proved equal. 59. Auxiliary lines. Often it is impossible to give a simple demonstration without drawing lines not described in the hypothesis. Such lines, used only for the proof, are usually dotted in order to distinguish them from the lines mentioned in the hypothesis and the conclusion. 60. Elements of a theorem. Every theorem contains two parts. The one is assumed to be true ; the other results from this assumption. The one part contains the given con- ditions ; the other part states the resulting truth. The assumed part of a theorem is called the hypothesis. 20 BOOK I. PLANE GEOMETRY The part of the theorem which is to be proved true is the conclusion. Often the hypothesis is a clause introduced by the word "if." When this conjunction is omitted, the subject of the sentence is known and its qualities, described in the quali- fying words, constitute the "given conditions." Thus, in the theorem of 52, the assumed part follows the word " if," and the truth to be proved is : " Two triangles are equal." The converse of a theorem is the theorem obtained by inter- changing the hypothesis and the conclusion of the original theorem. Consult 44 and 45, 55 and 114. NOTE. Every theorem having a simple hypothesis and a simple con- clusion has a converse, but only a few of these converse theorems are true. 61. Elements of a demonstration. All correct demonstra- tions should consist of certain distinct parts, namely : 1. Full statement of the given conditions as applied to a particular figure. 2. Full statement of the truth which it is required to prove. 3. The proof a series of successive statements, for each of which a valid reason should be quoted. (The drawing of auxiliary lines is sometimes essential.) 4. The conclusion declared to be true. PROPOSITION VI. THEOREM 62. Two lines in the same plane and perpendicular to the same line are parallel. Given: CD and EF in same plane c ~ and both _L to AB. To Prove : CD and EF II. Proof. If CD and EF were not II, they would meet if sufficiently prolonged. PARALLEL LINES 21 CD and EF are both J_ to AB (Given). .*. there would be two lines _L to AB from the point of meeting. But this is impossible ..CD and EF do not meet and are || Q.E.D. PROPOSITION VII. THEOREM 63. Two lines in the same plane and parallel to the same line are parallel. Given : AB II to BS, and CD II to BS, ^ p in the same plane. r> ~ To Prove : AB II to CD. Proof : If AB and CD were not II, they would meet if suf- ficiently prolonged. AB and CD are both II to RS (Given). .'. there would be 'two lines II to BS through the point of meeting. But this is impossible (Ax. 13). .'. AB and CD do not meet and are || (21). Q.E.D. PROPOSITION VIII. THEOREM 64. If a line is perpendicular to one of two parallels, it is perpendicular to the other also. Given: LM _L to AB and AB II to A ~~ CD. X To Prove : LM to CD. Proof: Suppose XY is drawn through M _L to LM XT is II to AB 9 (62). But CD is II to AB (Given). And CD and XY both contain point M (Const.). .*. CD and XY coincide (Ax. 13). But LM is J_ to XY (Const.). That is, LM is to CD Q.E.D. M Y 22 BOOK I. PLANE GEOMETRY 65. If one line cuts other lines, it is called a transversal Angles are formed at the several intersections, as follows : b, c, e, h are interior angles, a, ?,/, g are exterior angles. b and A, c and e, a and g, d and/ (on opposite sides of the transversal) are alternate angles. b and h, c and e are alternate interior angles. a and g, d and /are alternate exterior angles. a and e, d and h, b and / c and g are corresponding angles. PROPOSITION IX. THEOREM 66. If a transversal intersects two E parallels, the alternate interior angles R \\/ are equal. A ~ Given: AB II to CD; transversal EF c . cutting the Us at H and K. To Prove : Z a = Z i and Z x = Z v. Proof: Suppose through 3f, the midpoint of HK, ES is drawn J_ to AB. Then S is _L to CD. (64). In rt. A EMH and KMS, HM = JOf (Const.). .'.A EMH is congruent to A KM8 (57). .-.Za = Z; (27). Again Z ^ is the supplement of Z a (44). Also Z v is the supplement of Z e (44). .-.Z* = Z (49). Q.E.D. Ex. 1. If a line through the vertex of an isosceles triangle is parallel to the base, it makes equal angles with the sides of the triangle. Ex. 2. If from each point at which a transversal intersects two parallels a perpendicular to the other parallel is drawn, two congruent right triangles are formed. PARALLEL LINES PROPOSITION X. THEOREM 67. If a transversal intersects two parallels, the corresponding angles are equal. A Given : AB II to CD ; transversal EF cutting the Us and forming the 8 A. To Prove : Z s = Z i; Z f \ \E and BS are II to DF. Prove that AR = BS. ~/\ ^^F Ex. 3. In the accompanying figure, AB is II to DC and AD is II to BC. Prove that the A are congruent. / A TRIANGLES 29 PROPOSITION XX. THEOREM 78. Two triangles are congruent, if the three sides of one are equal respectively to the three sides of the other. Given: &ABC and RST; AB = RS-, AC=RT-, BC=*ST. To Prove: A RST is congruent to A ABC. Proof: Place A ABC in the position of A AST, so that the longest equal sides, BC and ST coincide, and A is opposite 8T from JR. Draw RA. Now Also Also Adding, RS = AS '. AASR is an isosceles A TR= TA '.AATR is an isosceles A .'.Z.SRA = /.SAR Z TEA = Z TAR (Hyp.)- (Def.). /-SRT = /-SAT .'. A RST is congruent to A AST That is, by substituting, A RST is congruent to A ABC (Ax. 6) . Q.E.D (Def.). (55). (55). (Ax. 2). (52). Ex. In the figure of Ex. 3 on the opposite page, if the opposite sides are equal, prove them parallel. 79. The distance from one point to another is the length of the straight line joining the two points. 30 BOOK I. PLANE GEOMETRY PROPOSITION XXI. THEOREM 80. Any point in the perpendicular bisector of a line is equally distant from the extremities of the line. Given : AE J_ to CD at its midpoint, B ; P any point in AE ; PC and PD. To Prove : PC = PD. Proof : In the rt. A PBC and PBD, PB = PB BC = BD .'.A PBC is congruent to A PBD .'. PC = PD. PROPOSITION XXII. THEOREM 81. Any point not in the perpen- dicular bisector of a line is not equally distant from the extremities of the line. Given : AB J_ bisector of CD ; P any point not in AB ; PC and PD. To Prove : PC = PD. (Identical). (Hyp.). (53). (27). Q.E.D. Proof: Either PC or PD will cut AB. AB at O. Draw OD. DO + OP > PD But CO = DO Substituting, CO 4- OP > PD That is, PC > PD or PC = PD. Suppose PC cuts (Ax. 12). (80). (Ax. 6). Q.E.D. 82. COROLLARY. If a point is equally distant from the ex- tremities of a line, it is in the perpendicular bisector of the line. (80 and 81.) 83. COROLLARY. Two points each equally distant from the extremities of a line determine the perpendicular bisector of the line. (82 and 4.) TRIANGLES 31 PROPOSITION XXIII. THEOREM 84. Two right triangles are equal if the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other. K R Given : rt. A UK and LMR ; Kl = RM ; KJ = RL. To Prove : A UK = A LMR. Proof : Place A UK in the position of A jrjR so that the equal sides, KJ and RL, coincide, and I is at X, opposite RL from M. A RLM and RLX are supplementary .*. -ZX.M" is a straight line .'. figure XMR is a A RX = R M .'. A XMR is isosceles Now Now (19). (45). (23). (Hyp.)- (Def.). (55). (57). That is, A UK is congruent to A LMR (Ax. 6). Q.E.D. 85. COROLLARY. The perpendicular from the vertex of an isosceles triangle to the base bisects the base, and bisects the vertex angle. In the congruent right A of 84, XL = LM (27). Also Z. XRL = MRL (27). Q.E.D. .A XLR is congruent to A LMR A UK is congruent to A LMR Ex. In the figure of 84, if XL LM, prove by two methods that XR = RM. 32 BOOK I. PLANE GEOMETRY PROPOSITION XXIV. THEOREM 86. The sum of two sides of a triangle is greater than the sum of two lines drawn to the ex- tremities of the third side, from any point within the triangle. Given: P, any point in A ABC; lines PA and PC. To Prove : AB + BC> AP + PC. Proof : Extend AP to meet BC at X. IU ^ Subtract AB + BX > AP + PX (Ax. 12). CX 4- PX > PC (Ax. 12). + BX + CX + PX > AP + PC + PX (Ax. 8). PX = PX .'. AB H- BC> AP + PC (Ax. 7). Q.E.D. Ex. If from any point within a triangle lines are drawn to the three vertices : (1) their sum is less than the sum of the sides of the triangle. (2) their sum is greater than half the sum of the sides of the triangle. PROPOSITION XXV. THEOREM 87. The perpendicular is the shortest line that can be drawn from a point to a A straight line. Given : PR -L to AB ; PC not _L. To Prove : PR < PC. Proof : Extend PR to X, making RX = to PR. (1) PR + RX < PC + ex But AR is -L bisector of PX .-.CX= PC Also RX = PR X Draw CX. (Ax. 12). (Const.). (80). (Const.). TRIANGLES 33 .-. Substituting in (1), PR + PB DT. A To Prove : PR>PT. Proof : Because DR > DT, we may take DS (on XXR) = to DT. Draw PS. Extend PD to X, making DX = to PD. Draw EX and SX. Now AD is the J_ bisector of PX and CD is the _L bisector of ST (Const.). PE + RX>PS+ sx (86). But EX = PR, and SX = PS = PT (80). Substituting, PR -h Pfl > PS + PS (Ax. 6). That is, 2 PR > 2 PS .'. PR>PS (Ax. 10). Substituting, PR > PT (Ax. 6). Q.E.D. IV. Given: [Use no dotted lines.] CD_Lto AB ; oblique lines PR and PT ; P# > PT. To Prove : DR > DT. Proof : It is evident that DR < DT, or DR DT, or DR > DT. IST: If DR < Dr, PE < PT (By III). But PR>PT (Hyp.). .'. DR is not
PT (Hyp.). .'. DR is not = DT 3D: Therefore, the only possibility is that DR > DT. Q.E.D. 89. COROLLARY. From an external point it is not possible to draw three equal straight lines to a given straight line. TRIANGLES 35 90. The method of demonstration employed in 88, IV, is called the method of exclusion. It consists in making all possible suppositions, leaving the probable one last, and then proving all these suppositions impossible, except the last, which must necessarily be true. The method of proving the individual steps is called re- ductio ad absurdum (reduction to an absurd or impossible conclusion). This method consists in assuming as false the truth to be proved and then showing that this assumption leads to a conclusion altogether contrary to known truth or the given hypothesis. (Examine the last preceding proof.) This is sometimes called the indirect method. The theorems of 62 and 63 are demonstrated by a single use of this method. PROPOSITION XXVII. THEOREM 91. If two triangles have two sides of one equal to two sides of the other, but the included angle in the first greater than the included angle in the second, the third side of the first is greater than the third side of the second. Given: A ABC, DEF; AB = DE', BC=EF; Z ABC > Z.E. To Prove : AC> DF. Proof : Place the A DEF upon A ABC so that side DE coin- cides with its equal AB, A DEF taking the position of A ABH. There remains an /. HBC. (Hyp.). Draw HC and suppose its JL bisector, MX, to be erected, meeting AC at X. Draw HX. Now HX = XC (80). Also AX + xn > AH (Ax. 12). Substituting, AX + XC, or AC > AH (Ax. 6). Substituting, AC> DF (Ax. 6). Q.E.D. 36 BOOK I. PLANE GEOMETRY PROPOSITION XXVIII. THEOREM 92. If two triangles have two sides of one equal to two sides of the other, but the third side of the first greater than the third side of the second, the included angle of the first is greater than the included r> angle of the second. [Con- verse.] Given : A ABC and RST ; AB RS ; BC = 8T ; A* ^C R 4 AC > RT. To Prove : Z B > Z s. Proof : Z.B < Z s, or Z B = Z s, or Z B > Zs. 1. IfZ.B RT (Hyp.). .*. Z B is not < Z s. 2. If Z B = Z 5, the A are congruent (52). AT -PT f f ) r l\ . . AL Kl (/> But AC > RT (Hyp.). 3. .*. the only possibility is that Z B > Z s. Q.E.D. 93. The distance from a point to a line is the length of the perpendicular from the point to the line. If the per- pendiculars from a point to two lines are equal, the point is said to be equally distant from the lines. PROPOSITION XXIX. THEOREM 94. Every point in the bisector of an angle is equally distant from the sides of the angle. Given: /.ACE; bisector CQ ; point P in CQ; distances PB and PD. To Prove : PB = PD. TRIANGLES 37 Proof: A PEC and PDC are rt. A. In rt. A PBC and PDC, PC = PC (Iden.). Z.PCB = ^PCD (Hyp.). /.A PBC is congruent to A PDC (57). PB = PD (27). Q.E.D. PROPOSITION XXX. THEOREM 95. Every point equally distant from the sides of an angle is in the bisector of the angle. Given : Z. A CE ; P, a point, such that PB = PD (distances) ; CQ, a line from vertex of the angle, and containing P. To Prove : Z ACQ = Z ECQ. Proof : A PC and PDC are rt. A (93). In rt. A PBC and PDC, PC = PC (Iden.). PB = PD (Hyp.) .*. A PBC is congruent to A PDC (84). .' . Z. ACQ = /. ECQ (27). Q.E.D. 96. COROLLARY. If a point is not equally distant from the sides of an angle, it is not in the bisector of the angle. (If it were in the bisector, it would be equally distant.) 97. COROLLARY. The vertex of an angle and a point equally distant from its sides determine the bisector of the angle. 98. The altitude of a triangle is the perpendicular from any vertex to the opposite side (prolonged if necessary). A triangle has three altitudes. The bisecljgj of an angle of a triangle is the line dividing any angle into' two equal angles. A triangle has three bisec- tors of its angles. The median of a triangle is the line drawn from any vertex to the midpoint of the opposite side. A triangle has three medians. THE THREE MEDIANS 38 BOOK I. PLANE GEOMETRY PROPOSITION XXXI. THEOREM 99. The bisectors of the angles of a triangle meet in a point which is equally distant from the sides. Given : A ABC, AX bisecting /. A, BY and CZ the other bisectors. To Prove : AX, BY, CZ meet in a point equally distant from AB, AC, and BC. B C Proof : Suppose that AX and B T intersect at O. O, in AX, is equally distant from AB and AC (94). O, in BY, is equally distant from AB and BC (?). .'. point O is equally distant from AC and BC (Ax. 1). /. O is in bisector CZ (95). That is, all three bisectors meet at O, and O is equally dis- tant from the three sides. Q.E.D. PROPOSITION XXXII. THEOREM 100. The three perpendicular bisec- tors of the sides of a triangle meet in a point which is equally distant from the vertices. Given: A ABC-, LB, MS, NT, the three J_ bisectors. To Prove: LE, MS, NT meet in a point equally distant from A, B, C. N C Proof : Suppose that LB and MS intersect at O. O, in LB, is equally distant from A and B (80). O, in MS, is equally distant from A and C (?). /. point O is equally distant from B and C (Ax. 1). .'. O is in _L bisector NT (82). That is, all three _L bisectors meet at O, and O is equally distant from A and B and C. Q.E.D. TRIANGLES 39 101. An exterior angle of a triangle is an angle formed outside the triangle, between one side of the triangle and another side prolonged. [Z.ABX.~] The angles within the triangle at the other vertices are the opposite interior ^ angles. [Z.4 and Zc.] PROPOSITION XXXIII. THEOREM 102. An exterior angle of a triangle is D equal to the sum of the opposite interior angles. Given : A ABC ; exterior Z ABD. To Prove : Z ABD = Z A + Z c. Proof : Suppose BP to be drawn through B II to A C. Z ABD = Z ABP + Z PBD (Ax. 4). But ZABP=Z.4 (66). Also Z PBD = ZC (67). = Z.A + ZC (Ax. 6). Q.E.D, 103. COROLLARY. An exterior angle of a triangle is greater than either of the opposite interior angles. (Ax. 5.) Ex. 1. If lines are drawn from any point within a triangle to two vertices of the triangle, they in- clude an angle greater than the third angle of the triangle. [Notice that Z 1 is an ext. Z of A CDP and Z 2 is B- an ext. Z of &ABD.~] Ex. 2. If the side LM, of equilateral triangle LMN, is produced to P, and PN is drawn, /.P AC. c To Prove : Z ACB > Z.B. Proof : On AB take AR = to AC. Draw CR and let Z ARC = x. Z. ARC is an ext. Z of ACBR (Def. 101). .vZa?> z# (103). But Z.ACR = ^X (55). Substituting, Z.ACR > Z# (Ax. 6). Again, Z ACB > Z ACR (Ax. 5). ..Z^>Z* (Ax. 11). Q.E.D. PROPOSITION XXXVII. THEOREM 117. If two angles of a triangle are unequal, the side opposite the greater angle is longer than the side opposite the less angle. [Converse.] Given : A ABC ; Z ACB > Z B. To Prove: AB > AC. Proof: In Z ACB, suppose Z BCR constructed = to Then CR = BR (H4). Also AR + CR > AC (Ax. 12). Substituting, AR + BR > AC (Ax. 6). That is, AB > AC. Q.E.D. 118. COROLLARY. The hypotenuse is the longest side of a right triangle. (117.) QUADRILATERALS 47 QUADRILATERALS 119. A quadrilateral is a portion of a plane bounded by four straight lines. These four lines are called the sides. The vertices of a quadrilateral are the four points at which the sides intersect. The angles of a quadrilateral are the four angles at the vertices. The diagonal of a rectilinear figure is a line joining two vertices, not in the same side. 120. A trapezium is a quadrilateral having no two sides parallel. A trapezoid is a quadrilateral having two and only two sides parallel. A parallelogram is a quadrilateral having its opposite sides parallel (O). NOTE. In the first figure below, angles A, Z>, or B, C are opposite angles, angles A, C, or B, Z>, or A, B, or (7, D are consecutive angles. C D PARALLELOGRAM RHOMBOID SQUARE RECTANGLE TRAPEZOID 121. A rectangle is a parallelogram whose angles ar_e right angles. A rhomboid is a parallelogram whose angles are not right angles. 122. A square is an equilateral rectangle. an equilateral rhomboid. A rhombus is 123. The side upon which a figure appears to stand is called its base. A trapezoid and all parallelograms have two bases, the actual base and the side parallel to it. The non- parallel sides of a trapezoid are sometimes called the legs. An isosceles trapezoid is a trapezoid whose legs are equal. The median of a trapezoid is the line connecting the midpoints of the legs. The altitude of a trapezoid and of all parallelograms is the perpendicular distance between the bases. 48 BOOK I. PLANE GEOMETRY PROPOSITION XXXVIII. 124. The opposite sides of a paral- lelogram are equal. Given : O LMOP. To Prove : LM = PO and LP = MO. Proof : Draw diagonal PM. In A LMP and OM P, PM = PM '.A LMP is congruent to A OMP .'. LM = PO and LP = MO Ex. 1.' If two perpendiculars are drawn to the R upper base of a parallelogram from the extremities of the lower base, two congruent right triangles are formed. (Iden.). (66). CO- (76). (27). Q.E.D. Ex. 2. The angles adjoining each base of an isosceles trapezoid are equal. Ex. 3. If the angles at the base of a trapezoid are equal, the figure is isosceles. Ex. 4. The diagonals of a rectangle are equal. Ex. 6. If the diagonals of a parallelogram are equal, the figure is a rectangle. Ex. 6. Any line terminated in a pair of oppo- R site sides of a parallelogram and passing through the midpoint of a diagonal is bisected by this point. A D Ex. 7. If one angle of a parallelogram is a right angle, the figure is a rectangle. Ex. 8. The bisectors of the angles of a trapezoid form a quadrilateral, two of whose angles are right angles. QUADRILATERALS 49 Ex. 9. The bisectors of the four interior angles formed by a transversal cutting two parallels form a rectangle. [Prove each Z of LMPQ a rt. Z.] M Ex. 10. The bisectors of the angles of a parallelogram form a rec- tangle. Ex. 11. The bisectors of the angles of a rectangle form a square. [In order to prove EFGH equilat- eral, the /& A HB and CDF are proved congruent and isosceles ; similarly &BGC and AED.~\ Ex. 12. The perpendiculars upon a diagonal of a parallelogram from the opposite vertices are equal. Ex. 13. If on diagonal BD, of square A BCD, BE is taken equal to a side of the square, and EP is drawn perpendicular to BD meeting AD at P, AP = PE = ED. Ex. 14. If AB CD is a square and E, F, G, H are points on the sides, such that AE = BF=CG=DH, EFGH is a square. [First, prove EFGH equilateral ; then one Z a rt. Z.] Ex. 16. The diagonals of an isosceles trapezoid are equal. 125. COROLLARY. Parallel lines included between parallel lines are equal. (124.) 126. COROLLARY. The diagonal of a parallelogram divides it into two congruent triangles. 127. COROLLARY. The opposite angles of a parallelogram are equal. (27.) 50 BOOK I. PLANE GEOMETRY PROPOSITION XXXIX. THEOREM 128. If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. [Converse of 124.] Given: Quadrilateral ABCD ; AB = DC', AD = BC. To Prove : ABCD is a O. Proof : Draw diagonal BD. In A ABD and CBD, BD = BD AB == DC and AD = BC .'.A ABD is congruent to A CBD .'. AB is II to DC Also Z y = Z x .'. AD is II to BC Hence ABCD is a O PROPOSITION XL. THEOREM 129. If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram. Given: Quadrilateral ABCD ; AB = DC and AB II to DC. To Prove : ABCD is a O. Proof: Draw diagonal BD. In A ABD and CBD, BD = BD (Iden.). AB = DC (Given). Za = Zi (66). .-.A ABD is congruent to A CBD (52). Hence Z.y=.x (27). .'. AD is II to BC (70). .'. ABCD is a O (Def.). Q.E.D. 130. COROLLARY. Any pah" of consecutive angles of a paral- lelogram are supplementary. (69.) QUADRILATERALS 51 PROPOSITION XL I. THEOREM 131. The diagonals of a parallelo- p gram bisect each other. Given : O EFGH ; diagonals EG and FH intersecting at X. To Prove : FX = XH and GX = XE. E H Proof : In A FXG and EXH, FG = EH (124). Za = Zo and ^c = /_r (66). .'. A FXG is congruent to A EXH (76). .'. FX = XII and GX = XE (27). Q.E.D. PROPOSITION XLII. THEOREM 132. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Given: Quadrilateral EFGH, diagonals EG and FH, FX = XH, EX = GX. To Prove : EFGH is a O. Proof : In A FXG and EXH, FX = XH and EX = XG (Hyp. ). Z FXG = Z EXH (?). .*. A FXG is congruent to A EXH (?). .-. FG = EH and Z f of AB ; P, the midpoint of BC ; line \ MP. To Prove: MP II to AC; A and MP = ^ ^1C. Proof : Suppose ^1R is drawn through 4, II to BC and meeting 3fP produced, at E. In A ABM and BPJtf, -4lf = BM (Hyp.)- Z^ = Ze (51). Z o = Z.B (66). .'. A ^.Rjtf is congruent to A BPM (76). .'.AR = BP (27). But BP = PC (Hyp.)- .'.AR = PC (Ax. 1). .-.^ICPBisaO (129). Hence UP, or JfP, is II to AC (Def.). Q.E.D. Also RP=AC (124). But MP = BM (27). /. MP, the half of HP, = % AC (Ax. 6). Q.E.D. 54 BOOK I. 'PLANE GEOMETRY PROPOSITION XLVI. THEOREM 137. The line bisecting one side of a triangle and parallel to a second side, bisects also the third side. Given: A ABC; MP bisecting AB and II to AC. To Prove : MP bisects BC also. Proof: Suppose MX is drawn from Jf, the midpoint of AB to X, the midpoint of BC. MX is II to AC (136). But MP is II to AC (Hyp.). .*. MX and MP coincide (Ax. 13). That is, MP bisects BC. Q.E.D. Ex. 1. The lines joining the midpoints of the sides of a triangle divide the triangle into four congruent triangles. Ex. 2. The lines joining (in order) the midpoints of the sides quadrilateral form a parallelogram the sum of whose sides is equal to the sum of the diagonals of the quadrilateral. Ex. 3. The lines joining (in order) the mid- points of the sides of a rectangle form a rhombus. [Draw the diagonals.] Ex. 4. If the four midpoints of the four halves of the diagonals of a parallelogram are joined in order, another parallelogram is formed. Prove this in four ways. Ex. 6. If lines are drawn from a pair of opposite vertices of a parallelogram to the midpoints of a pair of opposite sides, they trisect the diagonal joining the other two vertices. QUADRILATERALS 55 Ex. 6. If two medians are drawn from two vertices of a triangle and produced their own length beyond the opposite sides, and if these ex- tremities are joined to the third vertex, these two lines are equal, and in the same straight line. Ex. 7. The line joining the midpoints of one pair of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals bisect each other. PROPOSITION XL VII. THEOREM 138. The line bisecting one leg of a trapezoid and parallel to the base bisects the other leg, is the median, and is equal to half the sum of the bases. Given: Trapezoid ABCD; Jtf, the midpoint of AB; MP II to AD, meeting CD at P. B C To Prove : /\T I. P is the midpoint of CD. M / R II. MP is the median. Y III. MP = 1(AD + BC). *- a D Proof: I. Draw diagonal BD, meeting MP at R. MP is II to BC (63). InA^iBD, MR bisects BD (137). In A BbC, RP bisects CD (?). That is, P is the midpoint of CD. Q.E.D. II. MP is the median (Def. 123). Q.E.D. III. IllA^BD, MR = I AD (136). Also, in A BDC, RP = ^ BC (?). Adding, .-. MP= \(AD+ BC) (Ax. 2). Q.E.D. 56 BOOK I. PLANE GEOMETRY 139. COROLLARY. The median of a trapezoid is parallel to the bases and equal to half their sum. Ex. 1. In a trapezoid one of whose bases is double the other, the diagonals intersect at a point two thirds of the distance from each end of the longer base to the opposite vertex. Proof : Take M, the midpoint of A (136), etc. Ex. 2. If one angle of a triangle is double an- other, the line from the third vertex, making with the longer adjacent side an angle equal to the less given angle, divides the triangle into two isosceles triangles. NOTE. The verb " to intersect " means merely " to cut." In geometry, the verb " to intercept " means " to include between" Thus the statement " AB and CD intercept XY on the line EF" really means, "AB and CD intersect EF&nd include XY, a part of EF, between them." / f PROPOSITION XL VIII. THEOREM 140. Parallels intercepting equal parts on one transversal intercept equal parts on any transversal. Given : I Is AB, CD, EF, GH, U inter- cepting equal parts AC, CE, EG, GI, on the transversal AI, and cutting trans- versal BJ. To Prove : BD = DF = FH = HJ. Proof: The figure ABFE is a trapezoid CD bisects AE and is || to EF .-. D is midpoint of BF That is, BD = DF. Similarly, CDHG is a trapezoid and DF= FH. Similarly, FH=HJ. . ' . BD = DF= FH= HJ (?), (Hyp.). (?). (Ax. 1). Q.E.D. QUADRILATERALS 57 PROPOSITION XLIX. THEOREM 141. The midpoint of the hypotenuse of a right triangle is equally distant from the three vertices. Given : Rt. A ABC ; Jf, the midpoint of the hypotenuse AB. To Prove : AM = CM = BM. Proof : Suppose MX drawn n to BC, meeting AC at X. X is the midpoint of AC MX is .L to AC (64). .'. AM = CM (80). But AM=BM (Hyp.). .*. AM = CM = BM (Ax. 1). Q.E.D. Ex. 1. Any right triangle can be divided by one line into two isosceles triangles. Ex. 2. If through the vertex of the right p Q i angle of a right triangle a line is drawn parallel to the hypotenuse, the legs of the right triangle bisect the angles formed by this parallel and the median drawn to the hypotenuse. Ex. 3. If one leg of a trapezoid is perpendicular to the bases, the mid- point of the other leg is equally distant from the ends of the first leg. [Draw the median.] Ex. 4. The median of a trapezoid bisects both the diagonals. Ex. 5. The line joining the midpoints of the diagonals of a trapezoid is a part of the median, is parallel to the bases, and is equal to half their difference. Ex. 6. If the median of a triangle is equal to half the side to which it is drawn, it is a right triangle. Ex. 7. The line (prolonged if necessary) joining the midpoints of two sides of a triangle, bisects the altitude drawn to the third side. ROBBINS'S NEW PLANE GEOM. 6 M 58 BOOK I. PLANE GEOMETRY Ex. 8. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter leg. Proof: Use fig. of 141. Denote A by x, then Z B = 2 x and x = 30 (?). AM=MC = BM (141). .-. Z BCM = Z B = 60 (55). .-. Z MC = 60 (104). .-. MB = BC (115). .-. AB, or 2 x M, = 2 x C (Ax. 6). PROPOSITION L. THEOREM 142. The perpendiculars from the vertices of a triangle to the opposite sides meet in a point. R,- C A . Z Given: A ABC, AX J_ to BC, F-L to ^ic, and cz_L to AB. To Prove : These three J meet in a point. Proof : Through A suppose BS drawn II to BC ; through B, TS II to AC; through C, ET II to AB, forming AEST. The figure ABCR is a O (Const). and ABTC is a O (?). BC = AB and CT = AB (124) . .'.BC=CT (Ax. 1). Now CZ is _L to #r (64). That is, CZ is _L bisector of BT. Similarly, AX is J_ bisector of E-S. And BY is _L bisector of TS. .-. in A BST, AX, BY, CZ meet at a point (100). Q.E.D. QUADRILATERALS 59 PROPOSITION LI. THEOREM 143. The three medians of a triangle meet in a point which is two thirds the distance from any vertex to the midpoint of the opposite side. Given : A ABC, medians AF, BD and CE, the latter two meeting at O. (Fig. 1.) To Prove : BO = f BD ; co = | CE ; AO = | AF and that all three medians B F c meet at o. FlG< L Proof : Suppose H is midpoint of BO and z is the midpoint of CO. Draw ED, DI, IH, HE. In A ABC, ED is II to BC and = \ BC (136). In A OBC, HI is II to BC and = BC (136). .-. ED = HI (Ax. 1); and ED is II to HI (63). .-. EDIH is a O (129). .-. HO = OD and IO = OE (131). .-. BH = HO = OD and CI = IO = OE (Ax. 1). That is, BO = f BD and CO = f CE. Suppose AF meets BD at O r . (Fig. 2.) Then BO=%BD (Proved above). And BO r =%BD (Proved similarly). .-. BO =BO f (Ax. 1). That is, O and o' are the same point, and the three medians meet at O, which is J the distance from any vertex to the midpoint of the opposite side. Q.E.D. F FIG. 2. Ex. 1. If two lines (AB and CD) are equal and parallel, the lines connecting their op- posite ends bisect each other. Ex. 2. If the bisector of one angle of a tri- angle is perpendicular to the opposite side, the triangle is isosceles. B 60 BOOK I. PLANE GEOMETRY Ex. 3. The median of an isosceles triangle is perpen- dicular to the base. Ex. 4. The medians from the ends of the base of an isosceles triangle are equal. Ex. 6. If a triangle has two equal medians, it is isosceles. A B Ex. 6. If ABC is an equilateral triangle and D, E, F are points on the sides, such that AD = BE = CF, triangle DEF is also equilateral. B Ex. 7. Any two vertices of a triangle are equally distant from the median from the third vertex. f\ " \ Y Ex. 8. The lines bisecting two interior angles that a transversal makes with one of two parallels cut off equal segments on the other parallel from the point at which the transversal meets it. [The & formed are isosceles.] POLYGONS 144. A polygon is a portion of a plane bounded by straight lines. The lines are called the sides. The points of intersection of the sides are the vertices. The angles of a polygon are the angles at the vertices. 145. The number of sides of a polygon is the same as the number of its vertices or the number of its angles. An exterior angle of a polygon is an angle without the polygon, between one side of the polygon and another side prolonged. 146. An equilateral polygon has all its sides equal to one another. An equiangular polygon has all its angles equal to one another. POLYGONS 61 147. A convex polygon is a polygon no side of which if produced will enter the surface bounded by the sides of the polygon. A concave polygon is a polygon at least two sides of which if produced will enter the polygon. EQUILATERAL EQUIANGULAR CONCAVE, OR CONVEX POLYGONS REENTRANT NOTE. A polygon may be equilateral and not be equiangular; or it may be equiangular and not be equilateral. The word "polygon" usually signifies convex figures. 148. Two polygons are mutually equiangular if for every angle of the one there is an equal angle in the other and similarly placed. Two polygons are mutually equilateral, if for every side of the one there is an equal side in the other, and similarly placed. 149. Homologous angles in two mutually equiangular polygons are the pairs of equal angles. Homologous sides in two polygons are the sides between two pairs of homolo- gous angles. 150. Two polygons are congruent if they are mutually equi- angular and their homologous sides are equal ; or if they are composed of triangles, equal each to each and similarly placed. (Because in either case the polygons can be made to coincide.) Ex. 1. If a quadrilateral has three equal sides, is it necessarily a parallelogram ? a trapezoid ? Ex. 2. Two quadrilaterals are congruent if three sides and the two included angles of one are equal respectively to three corresponding sides and the two included angles of the other. 62 BOOK I. PLANE GEOMETRY 161. Two polygons may be mutually equiangular without being mutually equilateral ; also, they may be mutually equilateral without being mutually equiangular except in the case of triangles. The first two figures are mutually equilateral, but not mutually equi- angular. The last two figures are mutually equiangular, but not mutu- ally equilateral. 152. A 3-sided polygon is a triangle. A 4-sided polygon is a quadrilateral. A 5-sided polygon is a pentagon. A 6-sided polygon is a hexagon. A 7-sided polygon is a heptagon. An 8-sided polygon is an octagon. A 10-sided polygon is a decagon. A 12-sided polygon is a dodecagon. A 15-sided polygon is a pentadecagon. An w-sided polygon is called an n-gon. PROPOSITION LII. THEOREM 153. The sum of the interior angles of an n-gon is equal to (n 2) times 180. Given : A polygon having n sides. To Prove : The sum of its interior A = (n-2). 180. Proof : If all possible diagonals are drawn from any vertex it is evident that there are formed (n 2) triangles. The sum of the A of one A = 180 (104). ANGLES 63 /. the sum of the A of (n - 2) A = (n - 2) 180 (Ax. 3). The sum of A of the A = sum of A of n-gon (Ax. 4). /. the sum of A of the n-gon (n 2) 180 (Ax. 1). Q.E.D. 154. COROLLARY. The sum of the interior angles of an n- gon is equal to 180 n 360. 155. COROLLARY. Each angle of an equiangular n-gon = Qi - 2) 180 n 156. COROLLARY. The sum of the angles of any quadrilat- eral is equal to four right angles. 157. COROLLARY. If three angles of a quadrilateral are right angles, the figure is a rectangle. PROPOSITION LIII. THEOREM 158. If the sides of a polygon are produced hi order, one at each vertex, the sum of the exterior angles of the polygon equals four right angles, that is, 360. Given: A polygon with sides prolonged in succession forming the several exterior angles &, 6, , etc. Then Z^t + Zs + Zc+ZD + Z^-f etc. = 4 rt. A (47). But Z^t = Za, Z = Z, Zc = Z Z o. To Prove : Arc LM > arc AB. Proof : Superpose O O upon O C, making sector AOB fall in position of sector XCM, OB coinciding with CM. CX is within the angle LCM. (Because Z LCM > Z o.) Arc AB falls upon LM, in the position XM (179). .'. arc LM > arc XM (Ax. 5). That is, arc LM > arc AB. Q.E.D. II. Given: (?). To Prove : Z LCM > ^ o. Proof : The pupil may employ either superposition, as in I, or the method of exclusion, as in 92. NOTE. Unless otherwise specified, the arc of a chord always refers to the lesser of the two arcs. If two arcs (in the same or equal circles) are concerned, it is understood either that each is less than a semicircle, or each is greater. Ex. 1. Are equal circles also congruent? Why? Ex. 2. Is there a geometrical figure that is both sector and segment? 80 BOOK II. PLANE GEOMETRY PROPOSITION IV. THEOREM 196. In the same circle (or in equal circles) equal chords subtend equal arcs. Given : O o = O C ; chord AB = chord LM. To Prove : Arc AB = arc LM. Proof : Draw the several radii to the ends of the chords. In A OAB and CLM, OA = CL and OB = CM (188). Chord AB = chord LM (Hyp.). .-. A OAB is congruent to A CLM (?). Hence Zo = Zc (?). .-. arc AB = arc LM (193). Q.E.D. PROPOSITION V. THEOREM 197. In the same circle (or hi equal circles) equal arcs are subtended by equal chords. [Converse.] Given : O o = O C ; arc AB = arc LM. To Prove: Chord AB = chord LM. Proof : Draw the several radii to the ends of the chords. In A OAB and CLM, OA = CL (188). OB= CM (80). Zo = ZC (194). . . A AOB is congruent to A CLM (?). .-. chord AB = chord LM (?). Q.E.D. THE CIRCLE 81 PROPOSITION VI. THEOREM 198. In the same circle (or in equal circles) : I. If two chords are unequal, the greater chord subtends the greater arc. II. If two arcs are unequal, the greater arc is subtended by the greater chord. [Converse.] I. Given : O = O C ; chord AB > chord RS. To Prove : Arc AB > arc RS. Proof : Draw the several radii to the ends of the chords. In A AOB and RCS, AO = EC and BO = SC (?). Chord AB > chord B8 (Hyp.). .-. Z O > Z C (92). .. arc AB > arc RS (195, I). Q.E.D. II. Given : O o = O C ; arc AB > arc RS. To Prove : Chord AB > chord RS. Proof : Draw the several radii. In A AOB and RCS, AO = EC and BO = SC (?). But Z o > Z c (195, II). .-. chord AB > chord RS (91). Q.E.D. Ex. Can either part of Proposition VI be proved by the method of ex- clusion ? Can Proposition IV or V be proved by that method ? 82 BOOK II. PLANE GEOMETRY PROPOSITION VII. THEOREM 199. The diameter perpendicular to a chord bisects the chord and both the sub- tended arcs. Given : Diameter DR _L to chord AB in 00. To Prove : I. AM = MB ; II. Arc AB = arc RB, arc AD = arc DB. Proof : Draw radii to ends of the chord. I. In rt. A OAM and OBM, OA = OB OM= OM A OAM is congruent to A OBM .'. AM= BM II. Also Z AOM = Z BOM arc AR = arc RB Z AOD = Z BOD arc -4D = arc DB (84). CO- Q.E.D. (27). (193). (49). (193). Q.E.D. 200. COROLLARY. The line from the center of a circle perpendicular to a chord bisects the chord. 201. COROLLARY. The perpendicular bisector of a chord passes through the center of the circle. Proof : The center is equally distant from the extremities of the chord (187). .. the center is in the JL bisector of the chord (82). Ex. 1. The perpendicular bisectors of all chords in a circle pass through a common point. THE CIRCLE 88 Ex. 2. A diameter bisecting a chord is perpendicular to the chord and bisects the subtended arcs. Ex. 3. A diameter bisecting an arc is the perpendicular bisector of the chord of the arc. Ex. 4. A line bisecting a chord and its arc is the B perpendicular bisector of the chord. Ex. 6. If a circle is described on the hypotenuse of a right triangle as diameter, it passes through the vertex of the right angle (141). Ex. 6. If any number of parallel chords are drawn in a circle, their midpoints all lie on the same straight line. Ex. 7. If two perpendicular diameters of a circle are drawn and their extremities are joined in order, these chords form a square. Ex. 8. If any two diameters of a circle are drawn and their extremities are joined in order, the figure is a parallelogram. PROPOSITION VIII. THEOREM 202. The line perpendicular to a radius at its extremity is tangent to the circle. Given: Radius OA of O O, and ET JL to OA at A. To Prove: ET tangent to the circle. Proof : Take any point P in ET (except A) and draw OP. OP > OA (87). . . P lies without the O (Because OP > radius). That is, eve'ry point (except A) in ET is without the O. .. ET is a tangent to the O O (Def.). Q.E.D. 84 BOOK II. PLANE GEOMETRY PROPOSITION IX. THEOREM 203. If a line is tangent to a circle, the radius drawn to the point of contact is perpendicular to the tangent. [Converse.] Given: ET tangent to O O at A ; radius OA. To Prove : OA J_ to ET. Proof : Every point (except A) in ET is without the O (181). ..a line from O to any point in ET (except^) is > OA. (Because it is > a radius.) That is, OA is the shortest line from O to ET. .-. OA is_L to ET (87). 204. COROLLARY. The perpendicular to a tangent at the point of contact passes through the center of the circle. (43.) Q.E.D. PROPOSITION X. THEOREM 205. If two circles are tangent to each other, the line joining their centers passes through then* point of contact. Given : O and C tangent to a line at A, and line OC. To Prove : OC passes through A. Proof: Draw radii OA and CA. (203). (48). (39). Q.E.D. Let the pupil supply the proof if the circles are tangent internally. OA is _L to the tangent and CA is J_ to the tangent .-. OAC is a straight line .*. OAC and OC coincide, and OC passes through A THE CIRCLE 85 PROPOSITION XL THEOREM 206. Two tangents drawn to a circle from an external point are equal. NOTE. In this theorem the word "tangent" signifies the distance between the external point and the point of contact. Given : O O and tangents PA, PB. To Prove : Distance PA = distance PB. Proof : Draw radii to points of contact, and join OP. Z OAP and OBP are right A (203). In rt. A OAP and OBP, OP = OP (?) ; OA = OB (?). .*. A O^iP is congruent to A OBP (84). .'.PA = PB (?). Q.E.D. Historical Note. Pythagoras, a Greek philosopher, born probably at Samos, in the sixth century B.C., had the reputation of being an " assiduous inquirer," and of having a great fund of general knowledge. He was a moral reformer as well as a scientific teacher. He was the head of a secret society the members of which were pledged to the severest discipline, and to the practice of temper- ance, purity, and obedience. The study of mathe- matics in Greece was magnified by him and ad- vanced to the rank of a science. The invincible proof given on page 204 of the theorem that the square on the hypotenuse of a right triangle is equal in area to the sum of the squares on the legs, has been attributed to Pythagoras, and is often referred to as the Pythagorean proposition. The discovery of incommensurable magnitudes (p. 96) and of many other theorems and problems is ascribed to him. PYTHAGORAS 86 BOOK II. PLANE GEOMETRY PROPOSITION XII.. THEOREM 207. If from an external point tangents are drawn to a circle, and radii are drawn to the points of contact, the line joining the center and the external point bisects : I. The angle formed by the tangents. n. The angle formed by the radii. HI. The chord joining the points of contact. IV. The arc intercepted by the tangents. Given: Tangents AP and BP from point P and radii OA and OB. To Prove : Line OP bisects : I. Z APB, II. Z AOB, III. Chord AB, IV. Arc AXE. Proof : A OAP and OBP are rt. A (?). They are congruent. (Explain.) II. III. O is equidistant from A and B (?). P is also equidistant from A and B (206). /. OP is -L to AB at its midpoint (83). IV. Arc AX = arc BX (193). Q.E.D. Ex. 1. Tangents drawn to a circle at the extremities of a diameter are parallel. Ex. 2. Tangents drawn to a circle at the extremities of a chord form, with the chord, an isosceles triangle. Ex. 3. The bisector of the angle between two tangents to a circle passes through the center. THE CIRCLE Ex. 4. The sum of one pair of opposite sides of a circumscribed quadrilateral is equal to the sum of the other pair. A Ex. 5. A circumscribed parallelogram is equilateral. Ex. 6. A circumscribed rectangle is a square. Ex. 7. If a circle is inscribed in a right triangle, the sum of the diameter and the hypotenuse is equal to the sum of the legs. Ex. 8. If two parallel tangents meet a third tangent, and lines are drawn from the points of intersection to the center, they are perpendicular. Ex. 9. Tangents drawn to two tangent circles from any point in their common interior tangent are equal. Ex. 10. The common interior tangent of two tan- gent circles bisects their common exterior tangent. Ex. 11. Do the theorems of Ex. 9 and 10 apply if the circles are tangent internally? If so, prove. Ex. 12. In the adjoining figure, if AE and AD are secants, AE passing through the center, and the ex- ternal part of AD being equal to a radius, the angle DCE = Z /.A. [Draw BC. /. DEC = ext. Z of &ABC = 2^A . = D. (Explain.) Z DCE = an ext. Z, etc.] Ex. 13. The two common interior tangents of two circles are equal. Ex. 14. The common exterior tangents to two circles are equal. [Produce them to in- % tersection.] Ex. 15. In the preceding figure, prove that RH SF. Proof: AR + RB = CS + SD ; XD' 88 BOOK II. PLANE GEOMETKY ... AR + (RH + HF} = (SF + HF) + SD. .-. RH + RH+ HF= SF + HF + SF; .-. 2 RH = 2 SF, etc. Give reasons and explain. Ex. 16. The common exterior tangents to two circles intercept on a common interior tangent (produced), a line equal to a common exterior tangent. To Prove : RS = AB. Ex. 17. AB and AC are two tangents from A ; in the less arc BC a point D is taken and a tangent drawn at D, meeting AB at J and AC at F. Prove that AE + EF + AF remains a uniform length for all posi- tions of D in arc BC. Ex. 18. If perpendiculars are drawn upon a tangent from the ends of any diameter : (1) The point of tangency bisects the line between the feet of the perpendiculars. [Draw CP.] (2) The sum of the perpendiculars equals the diameter. (3) The center of the circle is equally distant from the feet of the perpendiculars. PROPOSITION XIII. THEOREM 208. In the same circle (or in equal circles) equal chords are equally distant from the center. Given : O o ; chord AB = chord CD, and distances OE and OF. To Prove : OE = OF. Proof : Draw radii OA and OC. In the rt. A AOE and COF, AE= \AB and CF= But AB = CD .'. AE= CF Also AO = co .-. A AOE is congruent to A COF .-. OE = OF CD (200) (HypO (Ax. 3) (?) (84) (? Q.E.D THE CIRCLE 89 PROPOSITION XIV. THEOREM 209. In the same circle (or in equal circles) chords which are equally distant from the center are equal. [Converse.] Given : O O ; chords AB and CD ; distance OE = distance OF. To Prove : Chord AB = chord CD. Proof : Draw radii OA and OC. In rt. A AOE and COF, AO = CO (?). Also EO=OF (Hyp.). .*. A AOE is congruent to A COF (84). .'.AE=CF (?). Now AB is twice AE and CD is twice CF .'.AB=CD (200). (Ax. 3). Q.E.D. Ex. 1. If two circles are concentric, all chords of the greater that are tangent to the less are equal. Ex. 2. If at the midpoint of an arc a tangent is drawn, it is parallel to the chord of the arc. Ex. 3. If two equal chords intersect on the circle, the radius drawn to their point of intersection bisects their angle. Ex. 4. If the line joining the point of intersection of two chords and the center bisects the angle formed by the chords, they are equal. Ex. 5. The radius of the circle inscribed in an equi- lateral triangle is half the radius of the circle circum- scribed about it. [Use 143.] Ex. 6. If the inscribed and circumscribed circles of a triangle are concentric, the triangle is equilateral. Ex. 7. If two circles are concentric and a secant cuts them both, the portions of the secant intercepted between the circumferences are equal. Ex. 8. Of all secants that can be drawn to a circumference from a fixed external point, the longest passes through the center. p ^ OV-" 1 -^ c- Ex. 9. The shortest line from an external point to a circumference is that which, if produced, would pass through the center. ROBBINS'S NEW PLANE GEOM. 7 90 BOOK II. PLANE GEOMETRY PROPOSITION XV. THEOREM 210. In the same circle (or in equal circles) if two chords are unequal, the greater chord is at the less distance from the center. Given: O O ; chord AB > chord CD, and distances OE and OF. To Prove : OE < OF. Proof: Arc AB > arc CD (198,1). Suppose arc AH is taken on arc AB, equal to arc CD. Draw chord AH. Draw OK _L to J.H, cutting AB at I. Now chord AH = chord CD .'. distance OK = distance OF But OE < oi Also 01 < OK .*. OE < OK Substituting, OE < OF (197). (208). (87). (Ax. 5). (Ax. 11). (Ax. 6). Q.E.D. PROPOSITION XVI. THEOREM 211. In the same circle (or in equal circles) if two chords are unequally distant from the center, the chord at the less distance is the greater. [Converse.] Given : O O ; chords AB and CD ; distance OE < distance OF. To Prove : Chord AB > chord CD. Proof: It is evident that chord AB < CD, or = chord CD, or > chord CD. Proceed by the method of exclusion. Another Proof : On OF take OX = to OE. At X draw a chord RS J_ to OX. Then chord RS is II to chord CD. (62). .-. arc RS > arc CD (Ax. 5). THE CIRCLE 91 .-. chord ES > chord CD (198, II). But chord AB = chord R8 (209). Substituting, chord AB > chord CD (Ax. 6). Q.E.D. 212. COROLLARY. The diameter of a circle is longer than any other chord. Ex. 1. What is the longest chord that can be drawn through a given point within a circle ? Ex. 2. Of all chords that can be drawn through a given point within a circle, the chord perpendicular to the diameter through the given point is the shortest. Given : P, the point ; BOC the diam. ; LS to EC at P ; GR, any other chord through P. To Prove: (?) Proof: Draw OA to GR. Etc. PROPOSITION XVII. THEOREM 213. Through three points, not in the same straight line, one circle can be drawn, and only one. Given : Points A and B and c. To Prove: I. (?). II. (?). Proof: I. Draw lines AB, BC, AC. Suppose their _L bisec- tors, OZ, OX, OF, are drawn. These J will meet at a point (100). With O as a center and OA or OB or OC as a radius, a circle can be described through A, JB, and C (100). II. These Ji meet at only one point (100). That is, there is only one center. The distances O^L, OB, OC are all equal (100). That is, there is only one radius. .*. there can be only one circle (192). Q.E.D. 92 BOOK II. PLANE GEOMETRY 214. COROLLARY. One circle, and only one, can be drawn through the vertices of a triangle. 215. COROLLARY. A circle is determined by three points. 216. COROLLARY. A circle cannot be drawn through three points which are in the same straight line. [The J would be II.] ,217. COROLLARY. A straight line can intersect a circle in only two points. (216.) 218. COROLLARY. Two circles can intersect in only two points. PROPOSITION XVIII. 219. If two circles intersect, the line joining their centers is the per- pendicular bisector of their common chord. Given: (?). To Prove: (?). THEOREM Proof Draw radii in each O to ends of AB. Point O is equally distant from A and B Point C is equally distant from A and B .-. OC is the _L bisector of AB (1ST). CO- (83). Q.E.D. Ex. 1. Illustrate the five corollaries on this page by diagrams. Ex. 2. On an island six miles from the mainland is a gun having a range of ten miles. Draw a diagram, using a scale of f in. to the mile, showing the range of the gun. Ex. 3. On the opposite sides of the entrance to a harbor are two forts, twelve miles apart. In each there is a gun with a range of nine miles. Draw a diagram showing the region exposed to the fire of each gun, and to the fire of both guns. Ex. 4. Make a similar problem using three forts, and guns of different ranges, and draw the diagram, showing regions exposed to one gun only and to all three guns (if any). THE CIRCLE 93 PROPOSITION XIX. THEOREM 220. Parallel lines intercept equal arcs on a circle. M Vc N Given: A circle and a pair of parallels intercepting two arcs. To Prove : The intercepted arcs are equal. There may be three cases : I. If the Us are a tangent (^1J?, tangent at P) and a secant (CD, cutting the circle at E and F). Proof : Draw diameter to point of contact, P. This diameter is _L to AB (203). PP' is also _L to EF (64). .-. arc EP = arc l^P (199). II. If the Us are two tangents, points of contact being M and N. Proof : Suppose a secant is drawn II to one of the tangents, cutting the O at R and 8. RS will be II to the other tangent (63). .-. arc MR = arc M8 and arc RN arc SN (I). Adding, arc MRN = arc MSN (Ax, 2). III. If the Us are two secants, one cutting the O at A and B, the other at C and D. Proof : Suppose a tangent is drawn, II to AB and touching the O at P. This tangent is II to CD (63). Arc PC = arc PD and arc PA = arc PB (I). Subtracting, arc AC= arc BD (Ax. 2). Q.E.D. 94 BOOK II. PLANE GEOMETRY 221. A polygon is inscribed ] if the vertices of the poly- in a circle, or a circle is cir- cumscribed about a polygon A polygon is circumscribed about a circle, or a circle is gon are in the circle, and its sides are chords. if the sides of the polygon ., are all tangent to the circle. inscribed m a polygon The perimeter of a figure is the sum of all its bounding lines. EXERCISES IN DRAWING CIRCLES 1. Draw two unequal intersecting circles. Show that the line joining their centers is less than the sum of their radii. 2. Draw two circles externally (not tangent) and show that the line joining their centers is greater than the sum of their radii. 3. Draw two circles tangent externally. Discuss these lines similarly. 4. Draw two circles tangent internally. Discuss these lines similarly. 6. Draw two circles so that they can have only one common tangent. 6. Draw two circles so that they can have two common tangents. 7. Draw two circles so that they can have three common tangents. 8. Draw two circles so that they can have four common tangents. 9. Draw two circles so that they can have no common tangent. SUMMARY 222. The following is a summary of the truths relating to magnitudes, which have been already established in Book II. I. Arcs are equal if they are : (1) Intercepted by equal central angles. (2) Subtended by equal chords. (3) Intercepted by parallel lines. (4) Halves of the same arc, or of equal arcs. II. Lines are equal if they are : (1) Radii of the same or of equal circles. (2) Diameters of the same or of equal circles. (3) Chords that subtend equal arcs. (4) Chords that are equally distant from the center. (5) Tangents to one circle from the same point. III. Unequal arcs and unequal chords have like relations. ORIGINAL EXERCISES 95 ORIGINAL EXERCISES 1. Show that an inscribed trapezoid is isosceles. 2. In the figure of 242, show that arc DBX = arc DB + arc A C. 3. In the figure of 243, show that arc CA=arc CMB-arc CNB. 4. In the figure of 244, show that arc CX - arc CE - arc BD. 5. In the figure of 245, show that arc BX = arc BE arc BD. 6. Show that the perpendiculars to the sides of a circumscribed poly- gon at the points of contact meet at a common point. 7. Show that the bisectors of the angles of a circumscribed polygon meet at a common point. 8. If two circles intersect and the four radii are drawn to the points of intersection, prove that the line joining the centers of the circles bisects the central angles formed by these radii. 9. If two chords of a circle are equal, but not parallel, and their mid- points are joined by a line, prove that the line from the center of the circle to the midpoint of the other line is perpendicular to it. 10. If a hexagon is circumscribed about a circle, prove that the sum of three alternate sides equals the sum of the other three sides. 11. Draw two circles which can have neither a common chord nor a common tangent. 12. Two perpendicular radii are prolonged to meet a tangent to a circle, and from the two points of intersection two other tangents are drawn to this circle. Prove that these two tangents are parallel. (Hint. Draw radii to the three points of contact.) 13. If from the midpoint of an arc perpendiculars are drawn to the radii drawn to the ends of the arc, prove that these perpendiculars are equal. 14. If through the extremities of a diameter two equal chords are drawn, one on each side of the diameter, prove that they are parallel. 15. Prove the theorem of 210 by the accompanying figure. [Hint, in &AEK show that two sides are unequal, hence two A are unequal, hence two angles in A EKO are unequal, etc.] 16. Prove the theorem of 211 by this figure, and a method similar to that employed in Ex. 15. 96 BOOK II. PLANE GEOMETRY KINDS OF QUANTITIES MEASUREMENT 223. A ratio is the quotient of one quantity divided by another both being of the same kind. 224. To measure a quantity is to find the number of times it contains another quantity of the same kind, called the unit. This number is the ratio of the quantity to the unit. 225. Two quantities are called commensurable if there exists a common unit of measure which is contained in each a whole (integral) number of times. Two quantities are called incommensurable if there does not exist a common unit of measure which is contained in each a whole number of times. Thus, $17 and $35 are commensurable, but $ 17 and $V35 are incom- mensurable. Two lines 18 ft. and 13 yd. are commensurable, but 18 ft. and V13 yd. are incommensurable. 226. A constant quantity is a quantity the value of which does not change during a discussion. A constant may have only one value. A variable is a quantity that has different successive values during a discussion. It may have an unlimited number of values. 227. The limit of a variable is a constant, to which the variable cannot be equal, but from which the variable can be made to differ by less than any mentionable quantity. 228. Illustrative. The ratio of 15 yd. to 25 yd. is written either J$ or 15 -4- 25 and is equal to three fifths. If we state that a son is two thirds as old as his father, we mean that the son's age divided by the father's, equals two thirds. A ratio is a fraction. The statement that a certain distance is 400 yd. signifies that the unit (the yard), if applied to this distance, will be contained exactly 400 times. Are $ 7.50 and $3.58 commensurable if the unit is $ 1 ? 1 dime ? 1 cent ? Are 10 ft. and vT9 ft. commensurable ? The height of a steeple is a constant; the length of its shadow made by the sun is a variable. The distance a train goes varies with the time it travels. Our ages are variables. The length of a standard yard, mile, MEASUREMENT 97 or meter, etc., is a constant. The height of a growing plant or a child is a variable. The limit of a variable may be illustrated by considering a right tri- angle ABC, and supposing the vertex A to move farther and farther from the vertex of the right angle. It. is evident that the hypotenuse becomes longer, that AC increases, but EC remains the same length. The angle A decreases, the angle B increases, but the angle C remains constantly a right angle. If we carry vertex A toward the left indefinitely, the ^ A becomes less and less but cannot become zero. [Because then there could be no A.] Hence the limit of the decreasing ZA is zero. Likewise, the Z B becomes larger and larger but cannot become equal to a right angle. [Because then two sides of the triangle would be parallel, which is impossible.] But it may be made as nearly equal to a right angle as we choose. Hence the limit of Z. B is a right angle. To these limits we cannot make the variables equal, but from these limits we can make them differ by less than any mentionable angle, how- ever small. The following supplies another illustration of the limit of a variable. The sum of the series 1 + $ + \ + i + & + -h + -fa + T$T + etc -> win always be less than 2, no matter how many terms are collected. But by taking more and more terms we can make the actual difference between this sum and 2 less than any conceivable fraction, however small. Hence 2 is the limit of the sum of the series. The limit is not 3 or 4, because the difference between the sum and 3 cannot be made less than any assigned fraction. Neither is the limit 1. (Why not?) Similarly, the limit of the value of .333333 ad infinitum is %. Certain variables become equal to a fixed magnitude ; but this fixed magnitude is not a limit. Thus, the length of the shadow of a tower really becomes equal to a fixed distance (at noon). A man's age really attains to a definite number of years and then ceases to vary (at death). Hence if a variable approaches a constant, and the differ- ence between the two can be made indefinitely small while the variable cannot become equal to the constant, the con- stant is the limit of the variable. This is merely another definition of a limit. 98 BOOK II. PLANE GEOMETRY PROPOSITION XX. THEOREM OF LIMITS 229. If two variables are always equal and each approaches a limit, their limits are equal. Given : Two variables v and v f ; v always = v 1 ; also v ap- proaching the limit I ; v' approaching the limit I'. To Prove: l = V. Proof: v is always = to v' (Hyp.). Hence they may be considered as a single variable. Now a single variable can approach only one limit (228). Hence 1 = 1'. Q.E.D. 230. (1) Algebraic principles concerning variables. If v is a variable and k is a constant : I. v + k is a variable. IV. kv is a variable. II. v k is a variable. V. - is a variable. k III. k v is a variable. VI. - is a variable. v These six statements are obvious. (2) Algebraic principles concerning limits. If v is a variable whose limit is Z, and k is a constant : I. v k will approach I k as a limit. II. k v will approach k I as a limit. III. kv will approach kl as a limit. IV. - will approach - as a limit. K K k k V. - will approach - as a limit. v I NOTE. A variable, as applied to Plane Geometry, is not added to, sub- tracted from, multiplied by, or divided by another variable. Proofs: I. v cannot = I (227). .. v k cannot = I k. Also, v I approaches zero (227) . .-. (v k) (/ k) approaches zero. (Because it reduces to v /.) Hence v k approaches I k (227). MEASUREMENT OF ANGLES 99 II. Demonstrated similarly. III. If kv = kl t then v = I (Ax. 3). But this is impossible (227). .. kv cannot = kl. Also v I approaches zero (227) . .. k(v 1) or kv kl approaches zero. Therefore kv approaches kl (227). IV and V. Demonstrated similarly. PROPOSITION XXI. THEOREM 231. In the same circle (or in equal circles) the ratio of two central angles is equal to the ratio of their intercepted arcs. Given : To Prove : O o = O c ; central O arc AB o and c ; arcs AB and XT. C arc XY Proof: I. If the arcs are commensurable. There exists a common unit of measure of AB and XY (225). Suppose this unit, when applied to the arcs, is contained 5 times in AB and 7 times in XY. arc AB __ 5 "7 (Ax. 3). arc XY Draw radii to the several points of division of the arcs. O is divided into 5 parts and Z c into 7 parts. These 12 parts are all equal (194). (Ax. 3). Zo Zo ZC 5 7 arc AB ' arc XY (Ax. 1). Q.E.D. 100 BOOK II. PLANE GEOMETRY II. If the arcs are incommensurable. There does not exist a common unit (225). Suppose arc AB is divided into equal parts (any number of them). Apply one of these as a unit of measure to arc XT. There is a remainder PT. (Because AB and XT are incommensurable.) Draw CP. Now -- = ^. (Case I). Z XCP arc XP Indefinitely increase the number of subdivisions of arc AB. Then each part, that is, our unit or divisor, is indefinitely decreased. Hence PF, the remainder, is indefinitely de- creased. (Because the remainder < the divisor.) That is, arc PT approaches zero as a limit, and Z. PCT approaches zero as a limit. .*. arc XP approaches arc XT as a limit (227). and Z. XCP approaches Z. XCT as a limit (227). Z O . Z.O ^' -4. approaches - as a limit ZXCP ZXCT and arcAB approaches arc AB as a limit. &TCXP arc XT (229) . XCT arc XT Ex. 1. If you double an arc do you double its central angle ? its chord ? Ex. 2. If in two equal circles, an arc in one is taken three times as long as an arc in the other, how do their central angles compare? Is there any similar law that you know, applying to their chords? Ex. 3. Two arcs of a circle contain 80 and 120 respectively. What is the ratio of their central angles ? MEASUREMENT OF ANGLES 101 PROPOSITION XXII. THEOREM 232. A central angle is measured by its intercepted arc. Given: Oo; Z.AOY; arcJ.F. To Prove : Z. AOY is measured by the arc AY, that is, they contain the same number of units. Proof : The sum of all A about O = 4 rt. A = 360 (47). If this O is. divided into 360 equal parts and radii are drawn to the several points of division, there will be 360 equal central A (194). Each of these 360 central angles will be a degree of angle (20). Each of the 360 equal arcs is called a degree of arc. Take Z AOT, one of these degrees of angle, and arc AT, one of the ,. rp, Z.AOY a. /^OQ1\ degrees of arc. Then = - (231). Z AOT arc AT Z AOY _ ^ AOY -L- a unit = the number of units in Z AOY (224). - = arc AY-s- a unit = the number of units in arc AY we AT .-. the number of units in Z AOY= the number of units in arc AY (Ax. 1). That is, /. A OY is measured by arc AY. Q.E.D. 233. COROLLARY. A central right angle intercepts a quadrant of arc. (Because each contains 90 units.) 234. COROLLARY. A right angle is measured by half a semicircle, that is, by a quadrant. 235. An angle is inscribed in a segment if its vertex is on the arc and its sides are drawn to the ends of the arc of the segment. Thus, ABCD is a segment and Z ABD is inscribed in it. -102 BOOK II. PLANE GEOMETRY PROPOSITION XXIII. THEOREM 236. An inscribed angle is measured by half its intercepted arc. Given : O O ; inscribed Z A ; arc CD. To Prove : Z A is measured by | arc CD. Proof : I. If one side of the Z is a diameter. Draw radius CO. A ^ioc is isosceles That is, But . Z COD = ^A + ^ A = i Z COD = Z ^ Z COD is meas. by arc CD *. JZ COD is meas. by \ arc CD .-. Z 4 is meas. by J arc CD (?). (102). (?), (Ax. 6). (Ax. 3). (232). (Ax. 3). (Ax. 6), II. If the center is within the angle. Z CAX is measured by ^- arc CX Z D4X is measured by J arc . Z C^ID is measured by J arc CD III. If the center is without the angle. Z O4X is measured by J arc CX Z D^JT is measured by | arc DX . Z CAD is measured by J arc CD Draw diameter AX. (i). (I). Adding, (Ax. 2). Draw diameter AX. a)- (I). Subtracting, (Ax. 2). Q.E.D. MEASUREMENT OF ANGLES 103 237. COROLLARY. Angles measured by half the same arc, or halves of equal arcs, are equal. 238. COROLLARY. In the same circle (or in equal circles), equal angles are measured by equal arcs. PROPOSITION XXIV. THEOREM 239. All angles inscribed in the same segment are equal. Given: The several A A inscribed in segment BAG. To Prove : These angles all equal. Proof : Each Z. BAG is measured by J arc BC .-. these angles are all equal. 240. COROLLARY. All angles inscribed hi a semicircle are right angles. Proof : Each Z K is measured by J a semicircle .-. each Z K = a rt. Z (236). (237). Q.E.D. (236). (234). Q.E.D. Historical Note. Thales, a Greek from Asia Minor, studied geometry from the Egyptians in the sixth century B.C. He discovered the truth of 240 as well as a number of very important theorems. For example : Book I, Propositions I, II, IV and XXXIV, and Book III, Proposition XX. Thales was one of the "Seven Wise Men of Greece," and made important contributions to astronomy and philosophy as well as to geometry. He regarded water as the principle of all things. THALES 104 BOOK II. PLANE GEOMETKY PROPOSITION XXV. THEOREM 241. The angle formed by a tangent and a chord is measured by half the intercepted arc. Given : O o, tangent TN ; chord AP ; Z TPA ; arc PA. To Prove : Z TPA is measured by J arc PA. Proof : Through A suppose AX drawn II to TN. Now Z A is meas. by J arc PX But Z ^4 = z TP4 Also arc PX = arc P^4 Substituting, Z TP^4 is meas. by J arc P^l (236). (66). (220). (Ax. 6). Q.E.D. Ex. 1. A chord divides a circle into two arcs, one containing 100, the other, 260. An angle is inscribed in each segment. How many degrees are there in each angle? Ex. 2. In a circle, an inscribed angle and a central angle intercept the same arc, which contains 140. How many degrees are there in each angle ? Ex. 3. A chord subtends an arc of 74. How many degrees are there in the angle between the chord and a tangent at one extremity ? Ex. 4. The circumference of a circle is divided into four arcs, 40, 70, 100, and x. Find x and the angles of the quadrilateral formed by the chords of these arcs. Ex. 6. In a segment of a circle whose arc contains 210 is inscribed an angle. How many degrees are there in this angle? Ex. 6. An inscribed angle contains 35. How many degrees are there in its intercepted arc ? Ex. 7. The line bisecting an inscribed angle bisects also its inter- cepted arc. Ex. 8. State and prove the converse of Ex. 7. Ex. 9. The line bisecting the angle between a tan- gent and a chord bisects the intercepted arc. Ex. 10. State and prove the converse of Ex. 9. A v MEASUREMENT OF ANGLES 105 Ex. 11. The angle between a tangent and a chord is half the angle between the radii drawn to the ends of the chord. Ex. 12. If a triangle is inscribed in a circle and a tangent is drawn at one of the vertices, the angles formed between the tangent and the sides equals the other two angles of the triangle. Ex. 13. By the figure of Ex. 12 prove that the sum of the three angles of a triangle equals two right angles. Ex. 14. Tf one pair of opposite sides of an inscribed quadrilateral are equal, the other pair are parallel. Proof: Draw Js BX, CY; arc A B = arc CD (?). .-. arc ABC = arc BCD (Ax. 2). Hence prove rt. &.ABX and DCY equal. Ex. 16. If any pair of diameters is drawn, the lines joining their extremities (in order) form a rectangle. Ex. 16. The opposite angles of an inscribed quadri- lateral are supplementary. Ex. 17. If a tangent and a chord are parallel, and the chords of the two intercepted arcs are drawn, they make equal angles with the tangent. Ex. 18. The circle described on one of the equal sides of an isosceles triangle as a diameter bisects the base. Proof : Draw line EM. The A .are rt. & (?) and congruent (?). Ex. 19. If the circle, described on a side of a triangle as diameter, bisects another side, the triangle is isosceles. Ex. 20. All angles that are inscribed in a segment greater than a semicircle are acute, and all angles inscribed in a segment less than a semicircle are obtuse. Ex. 21. An inscribed parallelogram is a rectangle. Prove arc ABC = arc A DC, etc. Ex. 22. The diagonal of an inscribed rectangle is a diameter Ex. 23. A circle described on the hypotenuse of a right triangle as a diameter passes through the vertices of all the right triangles having the same hypotenuse. ROBBINS'S NEW PLANE GEOM. 8 10(3 BOOK II. PLANE GEOMETRY Ex. 24. If from one end of a diameter a chord is drawn, a perpen- dicular to it drawn from the other end of the diameter intersects the first chord on the circum- ference. Ex. 25. If two circles intersect and a diameter is drawn in each circle through one of the points of intersection, the line joining the ends of these diameters the other point of intersection. [Draw chord ABJ] Ex. 26. If a tangent is drawn at one end of a chord, the midpoint of the intercepted arc is equally distant from the chord and the tangent. [Draw chord A M and prove the rt. A congruent.] Ex. 27. If two circles are tangent at A and a common tangent touches them at B and C, the angle BA C is a right angle. [Draw tangent at A. Use 206, 240.] Ex. 28. The bisectors of all the angles inscribed in the same segment of a circle pass through a common point Ex. 29. If two circles intersect and a line is drawn through each point of intersection terminat- ing in the circles, the chords joining these extremi- ties are parallel. [Draw RS. Z A is supp. of ^RSC (?). Finally use 73.] Ex. 30. A circle described on the radius of another circle as diameter bisects all chords of the larger circle drawn from their point of contact. To Prove: AB is bisected at C. Proof: Draw chord OC. (Use 240, 200). Ex. 31. If two equal chords intersect within a circle, the segments of one are equal to the segments of the other, each to each. Ex. 32. Prove Proposition XXV by drawing a diameter to the point of tangency, instead of a chord parallel to the tangent. Ex. 33. If in figure of Ex. 25 above, line CD met the two circles at M and N instead of at a single point B, what could be said of the lines AM and AN? Ex. 34. If a circle is divided into four equal arcs and if chords of these arcs are drawn, the inscribed figure is a square. MEASUREMENT OF ANGLES 107 PROPOSITION XXVI. THEOREM 242. The angle formed by two chords intersecting within the circle is measured by half the sum of the intercepted arcs. (The arcs are those intercepted by the given angle and by its vertical angle.) Given: Chords AB and CD intersect- ing at P ; Z APC ; arcs AC and DB. To Prove: Z APC is measured by J (arc AC+ arc DB). Proof: Suppose CX drawn through c II to AB. Now Z c is measured by ^ arc DX (236). That is, Z C is measured by J (arc BX + arc DB). But Z c = Z ^IPO (66). Also arc BX = arc AC (220). .-. Z.4PO is meas. by J (arc ^1(7 -f- arc DB) (Ax. 6). Q.E.D. PROPOSITION XXVII. THEOREM 243. The angle formed by two tangents is measured by hah 5 the difference of the inter- cepted arcs. Given: The two tangents AC and AB ; Z A ; arcs CMB and CNB. / N\ /M To Prove : Z ^4 is measured by | (arc CMB arc CT.B). Proof : Suppose CX drawn II to AB. Now Z DOT is meas. by J arc CX (241). That is, Z DCX is meas. by | (arc CMB arc .BJT). But ZDCX = ZA (67). Also arc BX arc OZV5 (220). . . Z 4 is meas. by (arc C3f arc ora) (Ax. 6). Q.E.D. 108 BOOK 11. PLANE GEOMETRY PROPOSITION XXVIII. THEOREM 244. The angle formed by two secants which intersect without the circle is measured by half the differ- ence of the intercepted arcs. Given: (?). To Prove: (?). Proof : Suppose BX drawn. Where ? How ? Z CBX is meas. by J arc CX That is, ^ CBX is meas. by | (arc CE arc XE) But Z CBX Z A And arc XE = arc BD Substituting, Z A is meas. by (arc CE arc BD) CO. (67). CO- (Ax. 6). Q.E.D. PROPOSITION XXIX. THEOREM 245. The angle formed by a tangent and a secant which intersect without the circle is measured by half the dif- ference of the intercepted arcs. Given: (?). To Prove: (?). Proof : Suppose BX drawn, etc. Z CBX is meas. by \ arc BX (241). That is, Z CBX is meas. by J (arc BXE arc XE). But Z.CBX=^A (?). And arc XE arc BD (?). Substituting, Z A is meas. by J (arc BXE arc .BD) (?). Q.E.D. Ex. 1. Where is the vertex of an angle that is measured by one arc? by half an arc? by half the sum of two arcs? by half the difference of two arcs? Ex. 2. State these truths all in a single theorem of your own. ORIGINAL EXERCISES 109 ORIGINAL EXERCISES 1. The arcs intercepted by two secants intersecting without a circle contain 20 and 140 respectively. How many degrees are there in the angle formed by the secants? 2. If in Ex. 1 the intersecting lines were chords, how many degrees would there be in their angle ? 3. One of the arcs intercepted by two intersecting tangents is 72. Find the angle formed by the tangents. 4. Two intersecting chords intercept opposite arcs of 28 and 80. How many degrees are there in the angle formed by the chords? 6. The angle between a tangent and a chord contains 27. How many degrees are there in the intercepted arc ? 6. The angle between two chords is 30 ; one of the arcs intercepted is 40. Find the other arc. [Denote the arc by #.] 7. If in figure of 241, arc AP contains 124, how many degrees are there in Z TPA ? in Z 1 NPA ? in arc AX ? 8. If in figure of 242, arc A C is 85, Z A PC is 47, find arc DB. 9. If the arcs intercepted by two tangents contain 80 and 280, find the angle formed by the tangents. 10. If the arcs intercepted by two secants contain 35 and 185, find the angle formed by the secants. 11. If in figure of 243, arc CB is 135, find the angle A. 12. If in figure of 244, angle A = 42 and arc BD = 70, find arc CE. 13. If in figure of 245, angle A = 18, arc BXE = 190, find arc BD. 14. If the angle between two tangents is 80, find the number of degrees in each intercepted arc. [Denote the arcs by # and 360 x.~\ 15. Three of the intercepted arcs of a circumscribed quadrilateral are 68, 98, 114. Find the angles of the quadrilateral. If the chords are drawn connecting (in order) the four points of contact, find the angles of this inscribed quadrilateral. Also find the angles between the diagonals of the two quadrilaterals. 16. If the angle between two tangents to a circle is 40, find the other angles of the triangle formed by drawing the chord joining the points of contact. 110 BOOK II. PLANE GEOMETKY 17. The circumference of a circle is divided into four arcs, three of which are, RS = 62, ST = 142, TU = 98. Find : (1) Arc UR. (2) The three angles at R ; at S\ at T; at U. (3) The angles A, B, C, D of circumscribed quadrilateral. (4) The angles between the diagonals R T and SU. (5) The angle between RL7 and ST 1 at their point of intersection (if produced). (6) The angle between RS and TU at their intersection. (7) The angle between AD and BC at their intersection. (8) The angle between AB and DC at their intersection. (9) The angle between RS and DC at their intersection. (10) The angle between AD and S^at their intersection. 18. If in the figure of Ex. 17, Z A =96; ZB = 112 ; and /. C = 68, find the angles of the quadrilateral RSTU. [Denote arc RU by x. .. in A ARU y 96 + i x + \ x = 180. .-. x = etc.] 19. If two circles are tangent externally and any line through their point of contact intersects them at B and C, the tangents at B and C are par- allel. [Draw common tangent at A. Prove : Z A CT 20. Prove the same theorem if the circles are tangent internally. 21. If two circles are tangent externally and any line is drawn through their point of contact terminating in the circles, the two diame- ters drawn to the extremities are parallel. 22. Prove the same theorem if the circles are tangent internally. 23. If two circles are tangent externally and any two lines are drawn through their point of contact intersecting the circles, the chords joining these points of intersection are parallel. [Draw common tangent at 0. Prove : Z C = /. D.~] 24. Prove the same theorem if the circles are tangent internally. 25. Prove the theorem of 242 by drawing AD instead of CX, and using Z.APC as an exterior angle of A APD. 26. Prove the theorem of 213 by drawing BC and using /.DCB as an exterior angle of A ABC. ORIGINAL EXERCISES 111 27. Prove the theorem of 244 by drawing CD and using angle CDE as an exterior angle of triangle A CD. 28. Prove the same theorem by drawing BE. 29. Prove the theorem of 245 by drawing BE. 30. If the opposite angles of a quadrilateral are supplementary, a circle can be drawn circumscribing it. p To Prove : A O can be drawn through A, B, C, P. Proof: A O can be drawn through A, B, C (?). It is required to prove that it will contain point P. /_ P + Z B are supp. (?) .-. they must be meas. by half the entire circle. Z B is meas. by | arc ADC (?). Hence /.P is meas. by \ arc ABC. If ZP is within or without the circle, it is not meas. by % arc ABC. (Why not?) 31. The circle described on the side of a square, or of a rhombus, as a diameter passes through the point of intersection of the diagonals. [Use 135, 141.] 32. The line joining the vertex of the right angle of a right triangle to the point of intersection of the diagonals of the square constructed upon the hypotenuse as a side, bisects the right angle of the triangle. Proof : Describe a O upon the hypotenuse as diameter and use 141, 196, 237. 33. If two secants, PAR and PCD, meet a circle at A, B, C, and D, respectively, the triangles PBC and PAD are mutually equiangular. 34. If PAB is a secant and PM is a tangent to a circle from P, the triangles PAM and PBM are mutually equiangular. 35. If two equal chords intersect within a circle, (1) One pair of intercepted arcs are equal. (2) Corresponding parts of the chords are equal. (3) The lines joining their extremities (in order) form an isosceles trapezoid. (4) The radius drawn to their intersection bisects their angle. 112 BOOK II. PLANE GEOMETRY 36. If a secant intersects a circle at D and E, PC is a parallel chord, and PR a tangent at P meeting the secant at R, the triangles PCD and PRD are mutually equiangular. [/. R and /. CDP are measured by arc PC. (Explain.) Etc.] 37. If a circle is described upon one leg of a right triangle as diameter and a tangent is drawn at the point of its intersec- tion with the hypotenuse, this tangent bisects the other leg. [Draw OP and OD. CD is tangent (?). OD bi- sects arc PC (207). ^COD=ZA (237). .-. OD is II to AB (?). Etc.] 38. If an equilateral triangle ABC is inscribed in a circle and P is any point of arc AC, AP + PC = BP. [Take PN = PA ; draw AN. A ANP is equilateral. (Explain.) AANB = &APC(?). Etc.] 39. If two circles are tangent internally at C, and a chord AB of the larger circle is tangent to the less circle at M, the line CM bisects the angle A CB. [Draw tangent CX and chord RS. (Explain.) .-. AB is || to RS (?). Etc.] V 40. If two circles intersect at A and C and lines are drawn from any point P, in one circle, through A and C terminating in the other at points B and Z>, chord BD will be of constant length for all posi- tions of point P. [Draw EC. Prove /.BCD, the ext. Z of APBC, = a constant. Etc.] 41. The perpendiculars from the vertices of a triangle to the opposite sides are the bisectors of the angles of the triangle formed by joining the feet of these perpendiculars. To Prove : BS bisects Z. RST, etc. Proof: If a circle is described on AO as diarn., it will pass through T and S (141). If a circle is described on OC as diam., it will pass through R and ,S (?). .-. ZBAR = BST (?) ; and BCT = Z. BSR (?). But Z. BAR = Z BCT. (Each is the comp. of ABC.) .-. Etc. ORIGINAL EXERCISES 113 42. If ABC is a triangle inscribed in a circle, BD is the bisector of angle ABC, meeting A C at O and the circle at D, the triangles A OB and COD are mutually equiangular. Also triangles BOC and AOD. Also triangles BOC and ABD. Also triangles A OD and Also triangles BCD and COZ). 43. If two circles intersect at A and J3, and from P, any point on one of them, lines A P and BP are drawn cutting the other circle again at C and D respectively, CD is parallel to the tangent at P. 44. If two circles intersect at A and B, and through B a line is drawn meeting the circles at R and 5 respectively, the angle RAS is constant for all positions of the line RS. [Prove /. R + Z. S is constant. .-, Z. RAS is also constant.] 45. Two circles intersect at A, and through A any secant is drawn meet- ing the circles again at M and N. Prove that the tangents at M and N meet at an angle which remains constant for all positions of the secant. [Prove the angle between these tangents equal to the angle between the tangents to the circles at A .] 46. Two equal circles intersect at A and B, and through A any straight line MAN is drawn, meeting the circles at M and N respectively. Prove chord BM = chord BN. 47. If the midpoint of the arc subtended by any chord is joined to the extremities of any other chord, (1) The triangles formed are mutually equiangular. (2) The op- posite angles of the quadrilateral thus formed are supplementary. 48. Two circles meet at A and B and a tangent to each circle is drawn at A, meeting the circles at R and S respectively. Prove that the triangles ABR and ABS are mutually equiangular. 49. Two chords intersecting within a circle divide the circumfer- ence into parts that bear the relation to each other of 1, 2, 3, 4. Find the angles made by the chords. [Denote the arcs by ar, 2 x, 3 x, 4 #.] 50. If A BCD is an inscribed quadrilateral, AB and DC produced to meet at E, AD and EC produced to meet at F t the bisectors of angles E and F are perpendicular. [The difference of one pair of arcs = difference of a second pair ; the difference of a third pair = difference of a fourth pair. (Explain.) Transpose negative terms and add correctly, rioting that the sum of 4 arcs = sum of 4 others, and hence = 180. Half the sum of these 4 arcs measures the angle between the bisectors. (Explain.) Etc.] 114 BOOK II. PLANE GEOMETRY LOCI 246. The locus of a point is the series of positions the point must occupy in order that it may satisfy a given con- dition. It is the path of a point whose positions are limited or denned by a given condition, or given conditions. 247. Explanatory. I. If a point is moving so that it is always one inch from a given point, the moving point may occupy any position in a circle whose center is the fixed point and whose radius is one inch. Furthermore, this moving point cannot occupy any position outside of the circle, or its position will not fulfill the given condition. THEOREM. The locus of points in a plane a given dis- tance from a given point is a circle the center of which is the given point and the radius of which is the given distance. II. If a point is moving so that it is always equally dis- tant from the ends of a straight line, it must move in the perpendicular bisector of the line. THEOREM. The locus of points equally distant from the ends of a line is the perpendicular bisector of the line. Proof: Every point in the _L bisector is equally distant from the ends of the line. (80.) No point without this _L fulfills that condition. (81.) .-. the JL bisector is the locus. (246.) III. THEOREM. The locus of points equally distant from the sides of an angle is the bisector of the angle. Proof : Any point within the bisector of an angle is equally distant from the sides. (94.) Any point equally distant from the sides of an angle lies in the bisector. (95.) Hence all the points in the bisector fulfill the condition and there are no other points that fulfill it. "That is, the bisector is the locus, etc. Q.E.D. ORIGINAL EXERCISES ON LOCI 115 IV. The method of proving that a certain line or a group of lines is the locus of points satisfying a given condition, consists in proving that every point in the line fulfills the given requirement, and that there is no other point that ful- fills it. In the above illustrations it is evident that every point in the lines that were called the "locus," fulfilled the conditions of the case. It is evident also that there is no point outside these "loci" that does so fulfill the con- ditions. That is, these u loci" contain all the points de- scribed. ORIGINAL EXERCISES ON LOCI 1. What is the locus of a point so moving that it is always two feet away from a given line ? 2. What is the locus of a point so moving that it is always equally distant from two parallel lines ? 3. What is the locus of points equally distant from two given fixed points ? 4. If all the radii of a circle were drawn, what would be the locus of their midpoints V 5. If all possible lines were drawn from a vertex of a triangle and terminating in the opposite side, what would be the locus of their mid- points ? 6. What is the locus of the midpoints of a series of parallel chords in a circle? Prove. 7. What is the locus of the midpoints of all chords of the same length in a given circle? Prove. 8. What is the locus of all points from which two equal tangents can be drawn to two circles which are tangent to each other? 9. What is the locus of all points at a given distance from a given circumference ? Discuss if the distance is > radius. If it is less. 10. What is the locus of the vertices of the right angles of all the right triangles that can be constructed on a given hypotenuse ? Prove. 11. What is the locus of the vertices of all the triangles which have a given acute angle (at that vertex) and have a given base ? Prove. 116 BOOK II. PLANE GEOMETRY 12. A line of given length moves so that its ends are in two perpen- dicular lines. What is the locus of its midpoint? Prove. [Suppose AB represents one of the positions of J the moving line. Draw OP to its midpoint. In all the positions of AB, OP = | A B - a constant (141). .*. P is always a fixed distance from 0. Etc.] 13. What is the locus of the midpoints of all the chords that can be drawn through a fixed point on a given circle? [Suppose AB represents one of the chords from B in circle 0, with radius OB ; and P is the midpoint of A B. Draw OP. Z P is a rt. Z (?). That 'is, where- ever the chord may be drawn, Z P is a rt. Z. .'. locus of P is, etc.] 14. A definite line which is always parallel to a given line moves so that one of its extremities is on a given circle. Find the locus of the other extremity. [Suppose CP represents one position of ^ ^xC p the moving line CP. Draw OQ = and II to CP from center 0. Join OC and PQ. Wherever CP is, this figure is a O (?). Its sides are of constant length (?). That is, P ^ is always a fixed distance from Q, etc.] A B 15. What is the locus of the centers of all circles tangent to a given line at a given point ? to a given circle at a given point ? 16. A parallelogram, ABCD, is hinged at the vertices, and AB only is fixed in position. What is the locus of vertex C? of vertex D't of the midpoint of BC1 of the midpoint of CDt 248. Heretofore only a few of the simplest exercises in construction have been given* (pages 8-12), and formal proofs of these were not required. The following methods for constructing lines are given so that mathematical precision may be employed in drawing accurate diagrams of a complex nature. No construction is considered valid unless a proof of its correctness can be given. The pupil should be familiar with the use of the ruler and compasses. CONSTRUCTION PROBLEMS 117 CONSTRUCTION PROBLEMS 249. A geometrical construction is a diagram made of points and lines. 250. A geometrical problem is the statement of a required construction. Thus, " It is required to bisect a line " is a problem. A problem is sometimes defined as " a question to be solved" and includes other varieties besides those in- volved in geometry. 251. The word proposition is used to include both theorem and problem. 252. The complete solution of a problem consists of five parts: I. The Given data are to be described. II. The Required construction is to be stated. III. The Construction is to be outlined. This part usually contains the verb only in the imperative. The only limitation in this part of the process is that every construction demanded shall have been shown possible by previous constructions or postulates. (See 32, 33, 186.) IV. The Statement that the required construction has been completed. V. The Proof of this declaration. Frequently a discussion of ambiguous or impossible in- stances is necessary. 253. NOTES. (1) A straight line is determined by two points. (2) A circle is determined by three points. (3) A circle is determined by its center and its radius. Whenever a circumference, or even an arc, is to be drawn, it is essential that the center and the radius be mentioned. (4) "Q.E.F. " = Quod erat faciendum "which was to be done." These letters follow the statement that the construc- tion which was required has been accomplished. 118 BOOK II. PLANE GEOMETRY PROPOSITION XXX. PROBLEM 254. To bisect a given line. Given : The definite line AB. Required : To bisect AB. Construction: Using A and B as centers and one radius, sufficiently A | long to make the circumferences in- tersect, describe two arcs meeting at B and T. Draw ET meeting AB at M. Statement : Point M bisects AB. Proof : B is equally distant from A and B T is equally distant from A and B Hence ET is the _L bisector of AB That is, M bisects AB. M ; "side adjoining." Second, If "side opposite" = " side adjoining." Third, If " side opposite " < " side adjoining." Construction for all : Draw an indefinite line, CX, and at one extremity construct an Z = to Z. C. Take on the side of this angle a distance from the vertex equal to the " side adjoining." Using the end of this side as a center and the " side opposite " as a radius, describe an arc intersecting CX. Draw radius to the intersection just found. 130 BOOK II. PLANE GEOMETRY If the known angle is obtuse or right. Given : Z c, side c opposite y it, and side b adjoining Z C. Construction: As above. Discussion : Case I. c > b. The A is always possible. Case II. c = b. The A is never possible (106). Case III. c b. The A is always possible. Case II. c = b. The A is always possible and isosceles. Case III. c < b. First, c <.the J_ from A to CX. The A is never possible. Second, c the -L from A to CX. The A is possible and a right A. Third, c > the _L from A to CX. There are two A, ACB and ACE' ; both contain the three given parts. ,Y ANALYSIS 131 ANALYSIS Many constructions are so simple that their correct solu- tion will readily occur to the pupil. Sometimes, in the case of complicated constructions, the pupil must have the ability to put the given parts together, one by one. The following outline may be found helpful if employed intelligently. It is called the method of analysis. I. Suppose the construction made, that is, suppose the figure drawn. II. Study this figure in search of truths by which the order of the lines that have been drawn can be determined. III. One or more auxiliary lines may be necessary. IV. Finally, construct the figure and prove it correct. EXERCISE. Given the base of a triangle, an adjacent acute angle, and the difference of the other sides, to construct the triangle. Given : Base AB ; Z A' ; difference d. Required : To construct the A. [Analysis: Suppose A ABC is the re- quired A. It is evident if CD = CB, they may be sides of an isos. A and AD = d.~\ Construction : At A on AB construct Z BAX = toZA r and on AX, take AD = to d. Join DB. At M, midpoint of DB, draw M Y _L to DB meeting AX at Draw CB. C. Statement: Proof: (?). Discussion: Historical Note. The philosopher Plato and his school flourished at Athens in the fourth century B.C. Plato brought to geometry exact definitions and axioms as well as the method of analysis, which is helpful in discovering difficult proofs and constructions. 132 BOOK II. PLANE GEOMETRY ORIGINAL CONSTRUCTIONS I. Construct an isosceles triangle, having given: 1. The base and one of the equal sides. 2. The base and one of the equal angles. 3. One of the equal sides and the vertex angle. 4. One of the equal sides and one of the equal angles. 5. The base and the altitude upon it. 6. The base and the radius of the inscribed circle. [Bisect the base ; erect a JL = to the radius ; describe O, etc.] 7. The base and the radius of the circumscribed circle. 8. The altitude and the vertex angle. 9. The base and the vertex angle. [Find the supplement of the given base construct an Z = to this half ; etc.] 10. The perimeter and the altitude. Given : Perimeter = AB ; alt. = h. Re- quired: (?). Construction: Bisect AB; erect at M _L = to h ; draw 4P and BP. Bisect these ; erect Js SC and RE ; etc. bisect this; at each end of C M E II. Construct a right triangle, having given: 11. The two legs. 12. One leg and the adjoining acute angle. 13. One leg and the opposite acute angle. The hypotenuse and an acute angle. 14. 16. The hypotenuse and the altitude upon it. 16. The median and the altitude upon the hypotenuse 17. The radius of the circumscribed circle and a leg. 18. The radius of the inscribed circle and a leg. Given: Radius = r; leg = CA. Required: (?). Analysis : Consider that ABC is the completed fig- ure ; CNOM is a square, whose vertex O is the center of the circle, and side ON is the given radius. AB is tangent from A . Construction: On CA take CN = to r and construct square, CNOM. Prolong CM in- definitely. Describe O, etc. ORIGINAL CONSTRUCTIONS 133 B .- 19. One leg and the altitude upon the hypotenuse. 20. An acute angle and the sum of the legs. Given: AD = sum; Z K. Required: (?). Construction : At A construct /. A = to Z K ; at D construct Z D = to 45, etc. ^ C 21. The hypotenuse and the sum of the legs. [Use A as center, hypotenuse as radius, etc.] 22. The radius of the circumscribed circle and an acute angle. 23. The radius of the inscribed circle and an acute angle. Construction : Take CS = to r, on indefinite line, ZA. On CS construct square CSOM. At O construct Z MOX = to Z K. Draw radius OT _L to OX. Draw tangent at T, etc. III. Construct an equilateral triangle, having given : 24. One side. 25. The altitude. 26. The perimeter. 27. A median. 28. The radius of the inscribed circle. [Draw circle and radius ; at center construct = to 120 and Z ROT = to 120 ; etc.] 29. The radius of the circumscribed circle. IV. Construct a triangle, having given : 30. The base, an angle adjoining it, the altitude upon it. 31. The midpoints of the three sides. [Draw RS, RT, ST, etc.] 32. One side, altitude upon it, and the radius of the circumscribed circle. Construction: Draw O with given radius and any center. Take chord = to given side ; etc. 33. One side, an adjoining angle, and the radius of the circumscribed circle. 34. Two sides and the altitude from the same vertex. Construction: Erect JL equal to altitude, upon an indefinite line. Use the end of this altitude as center, and the given sides as radii, etc. 134 BOOK II. PLANE GEOMETRY C / KX::=:V*C L.L yt\ lih : \i^__ A v c.. 36. One side, an angle adjoining it, and the sum of the other two sides. Construction : At A construct /.BAX = to given ZK. On AX take AD = to s; draw DB ; bisect DB at M, etc. 36. Two sides and the median to the third side. Given : a, b, m . Construction : Con struct A A BR with three sides, A B = to a, BR = to b, AR = to 2 m. Draw A C II to BR and .RC II to ,4 meeting atC. Draw^C. Statement: (?). Proof: (?). 37. A side, the altitude upon it, and the angle opposite it. Given : Side = to AB, alt. = to h ; opposite Z C = to Z C". Construction: Upon AB construct segment ACB which contains Z C = to Z C' (by 267). At A erect ^4/2 _L to ^ and = to h ; etc. 38. A side, the median to it, the angle oppo- site it. [Statement : A ABC is the required A.] 39. One side and the altitude from its extrem- ities to the other sides. Given: Side = AB, altitudes x and y. Construction: Bisect AB] describe a semicircle. Using A as center and x as radius, describe arc cutting the semicircle at R ; etc. 40. Two sides and the altitude upon one of them. [Given : Sides = to AB and EC ; alt. on EC = to x.~\ 41. One side, an angle adjoining it, and the radius of the inscribed circle. Construction : Describe O with given radius, any center. Construct central Z = to given Z. Draw two tangents || to these radii; etc. V. Construct a square, having given : 42. One side. 43. The diagonal. 44. The perimeter. 45. The sum of a diagonal and a side. ORIGINAL CONSTRUCTIONS 135 VI. Construct a rhombus, having given : 46. One side and an angle adjoining it. 47. One side and the altitude. 48. The diagonals. 49. One side and one diagonal. [Use 269.] 60. An angle and the diagonal to the same vertex. 51. An angle and the diagonal between two other vertices 52. One side and the radius of the inscribed circle. VII. Construct a rectangle, having given : 53. Two adjoining sides. 54. A diagonal and a side. 55. One side and the angle formed by the diagonals. 56. A diagonal and the sum of two adjoining sides. [See Ex. 21.] 67. A diagonal and the perimeter. 58. The perimeter and the angle formed by the diagonals. Construction : Bisect the perimeter and V.. take AB = to half of it. Bisect Z K. ^ - ^K At A construct ZBAX = to half /.K. " Etc. VIII. Construct a parallelogram, having given : 69. One side and the diagonals. 60. The diagonals and the angle between them. 61. One side, an angle, and the diagonal not to the same vertex. 62. One side, an angle, and the diagonal to the same vertex. 63. One side, an angle, and the altitude upon that side. 64. Two adjoining sides and the altitude. IX. Construct an isosceles trapezoid, having given : 65. The bases and an angle adjoining the larger base. 66. The bases and an angle adjoining the less base. 67. The bases and the diagonal. 68. The bases and the altitude. 136 BOOK II. PLANE GEOMETRY 69. The bases and one of the equal sides. 70. One base, an angle adjoining it, and one of the equal sides. 71. One base, the altitude, and one of the equal sides. 72. One base, the radius of the circumscribed circle, and one of the equal sides. [First, describe a O.] 73. One base, an angle adjoining it, and the radius of the circum- scribed circle. 74. The bases and the radius of the circumscribed circle. 75. One base and the radius of the inscribed circle. Construction : Bisect the base and erect a _L = to radius ; etc. X. Construct a trapezoid,* having given : 76. The bases and the angles adjoining one of them. Construction : Take EC = to longer base, and on it take ED = to less base. Construct A DBC (by 271). 77. The four sides. 78. A base, the altitude, and the non-parallel sides. Construction: Construct a A two sides of which equal the given non-|| sides of the trapezoid, and the altitude from same vertex of which equals the given altitude. (See Ex. 34.) 79. The bases, an angle, and the altitude. Construction: Construct 7 on ED, having given altitude and . 80. A base, the angles adjoining it, and the altitude. 81. The longer base, an angle adjoining it, and the non-parallel sides. 82. The shorter base, an angle not adjoining ;t, and the non-parallel sides. XI. Construct the locus of a point which shall be : 83. At a given distance from a given point. 84. At a given distance from a given line. * NOTE. It is evident that every trapezoid may ^ B be divided into a parallelogram and a triangle by drawing one line (as BD] II to one of the non-11 sides Hence the construction of a trapezoid is often merely constructing a triangle and a parallelogram. :r/\ ORIGINAL CONSTRUCTIONS 137 86. At a given distance from a given circle : () If the given radius is < the given distance ; (ii) If the given radius is > the given distance. 86. Equally distant from two given points. 87. Equally distant from two intersecting lines. XII. Find (by intersecting loci) * the point P, which shall be : 88- At two given distances from two given points. f 89. Equally distant from three given points. 90. In a given line and equally distant from two given points. 91. In a given line and equally distant from two given intersecting lines. 92. In a given circle and equally distant from two given points.f 93. In a given circle and equally distant from two intersecting lines, f 94. Equally distant from two given intersecting lines and equally distant from two given points.f 95. At a given distance from a given line and equally distant from two given points.f 96. At a given distance from a given line and equally distant from two other intersecting lines. f 97. Equally distant from two given points and at a given distance from one of them.f 98. Equally distant from two given intersecting lines and at a given distance from one of them.f 99. At a given distance from a point and equally distant from two other points. f 100. At given distances from two given intersecting lines. f 101. At given distances from a given line and from a given circle.f * It is well to draw the loci concerned as dotted lines. (See Ex. 105.) t In the Discussion, include the answers to questions like these : (1) Is this ever impossible ? (i.e. must there always be such a point ?) (2) Are there ever two such points ? when ? (3) Are there ever more than two ? when ? (4) Is there ever only one ? when ? ROBBINS'S NEW PLANE GEOM. 10 138 BOOK II. PLANE GEOMETRY 102. At given distances from a given line and from a given point.* 103. Equally distant from two parallels and equally distant from two intersecting lines.* 104. At a given distance from a given point and equally distant from two given parallels.* , 105. At a given distance from a given point and equally distant from two given intersecting p^v^x^ "\ p lines. Can C be so taken that there will be no point? Can C be so taken that there will be only one \ \/ point? only two? only three? more than four? ''-:- XIII. Find (by intersecting loci) the center of a circle which shall : 106. Pass through three given points. f 107. Pass through a given point and touch a given line at a given point, f 108. Have a given radius and be tangent to a given line at a given point.f 109. Have a given radius, touch a given line, and pass through a given point.f 110. Pass through a given point and touch two given parallel lines. f 111. Touch two given parallels, one of them at a given point, f 112. Have a given radius and touch two given intersecting lines, t 113. Have a given radius and pass through two given points. f 114. Touch three given indefinite lines, no two of them being parallel. :{: 116. Touch three given lines, only two of them being parallel. XIV. Construct a circle which shall : 116. Pass through a given point and touch a given line at a given point. 117. Touch two given parallel lines, one of them at a given point. 118. Pass through a given point and touch two given parallels. * See note (t) on preceding page. t Discussion : Is this ever impossible ? Are there ever two circles and hence two centers ? Are there ever more than two ? Etc. \ Four solutions. One is in 265. ORIGINAL CONSTRUCTIONS 139 119. Have a given radius, touch a given line, and pass through a given point. 120. Have its center in one line, touch another line, and have a given radius. 121. Have a given radius and touch two given intersecting lines. 122. Have a given radius and pass through two given points. 123. Have a given radius and touch a given circle at a given point. [Draw tangent to the given O at the given point.] 124. Have a given radius and touch two given circles. 125. Touch three indefinite intersecting lines.* 126. Touch two given intersecting lines, one of them at a given point. 127. Touch a given line and a given circle at a given point. Given: Line AB; O C; point P. Construction : Draw radius CP. Draw tan- gent at P meeting AB at R. Bisect /.PRB, meeting CP produced at 0; etc. 128. Be inscribed in a given sector. Construction : Produce the radii to meet the tangent at the midpoint of the arc. In this A inscribe a O. 129. Have a given radius and touch two given circles. 130. Have a given radius, touch a given line, and a given circle, 131. Touch a given line at a given point and touch a given circle. Given: Line AB] point P; O C. Construction : At P erect PX to AB, and extend it below AB, so PR = radius O C. Draw CR and bisect it at M. Erect MY to CR at M, meeting PX at ; etc. Y 132. What is the locus of the vertices of all right triangles having the same hypotenuse? 133. Through a given point on a given circumference draw two equal chords perpendicular to each other. * Four solutions. One is in 265. 140 BOOK II. PLANE GEOMETRY 134. Draw a line of given length through a given point and terminat- ing in two given parallels. Construction: Use any point of one of the ll s as center and the given length as radius to describe an arc meeting the other II. Join these two points. Through the given point draw a line II, etc. 135. Draw a line, terminating in the sides Q of an angle, which shall be equal to one line x / and parallel to another. / ./^ / Statement : RS = a, and is II to x. 136. Draw a line through a given point within an angle, which is terminated by the sides of the angle and bisected by the point. Construction : Through P draw PD II to AC. Take on AB, DE = AD. Draw EPF; etc. 137. Circumscribe a circle about a rec- tangle. 138. Construct three circles having the vertices of a given triangle as centers so that each touches the other two. Construction : Inscribe a O in the A ; etc. 139. Construct within a circle three equal circles each of which touches the given circle and the other A two. Construction: Draw a radius, OA, and construct Z A OB = to 120 and Z. A OC = to 120. In these sectors inscribe, etc. 140. Through a point without a circle draw a secant having a given distance from the center. 141. Draw a diameter to a circle at a given dis- tance from a given point. 142. Through two given points within a circle draw two equal and parallel chords. Construction: Bisect the line joining the given points and draw a diameter, etc. 143. Draw a parallel to side EC of triangle ABC, meeting AB in X and AC in F, such that XY = YC. l ORIGINAL CONSTRUCTIONS 141 144. Find the locus of the points of contact of the tangents drawn to a series of concentric circles from an external point. 146. Given : Line AB and points C and D on the same side of it ; find point X in AB such Construction: Draw CE _L to AB and pro- duce to F so that EF = CE. Draw FD meeting AB in X. Draw CX. 146. Draw from one given point to another the shortest path which has one point in common with a given line. Statement : CX + XD is < CR + RD. Another statement of this exercise: If C is an object before a mirror ER, and D is an eye, draw a diagram showing the path of a ray of light, from C, reflected to D. 147. Draw a line parallel to side BC of triangle ABC meeting AB at X and A C at F, so that XY = BX + YC. Construction : Draw bisectors of A B and C, meeting at 0, etc. 148. Draw in a circle, through a given point of an arc, a chord that is bisected by the chord of the arc. Construction : Draw radius OP meeting chord at C. Prolong PO to X so that CX = OP. Draw XM II to AB meeting O at M. Draw PM cutting AB at Z>; etc. Is there any other chord from P bisected by AB ? 149. Inscribe in a given circle a triangle whose angles are given. Construction : Construct 3 central A, doubles of the given A. Etc. 160. Circumscribe about a given circle a triangle whose angles are given. Construction : Inscribe A (like Ex. 149) first, and draw tangents II to the sides. 151. Three lines meet in a point. Draw a line terminating in the outer two and bisected by the inner one. Construction : Through any point P, of OB, draw Us to the outer lines. Draw diagonal RS ; etc. 142 BOOK II. PLANE GEOMETRY 152. Draw through a given point P, a line that is terminated by a given circle and a given line and is bisected by P. Construction: Draw any line DX meeting AB at D. Draw PE II to AB meeting DX at E. Take EF = ED ; etc. 153. Through a given point without a circle draw a secant to the circle which is bisected by the circle. Construction: Draw arc at J", using P as center and diam. of O O as radius. Using T as center and same radius as before, describe circle touching O at C and passing through P. Draw PC meet- ing O at M. 154. Inscribe a square in a given rhombus. [Bisect the four A formed by the diagonals.] 155. Bisect the angle formed by two lines without producing them to their point of intersection. Construction : At P, any point in RS, draw PA II to XY\ bisect ZAPS by PB. At any point in P.B erect ML _L to PB, meeting the given lines in M and L. Bisect ML at D and erect DC ML, etc. 156. Construct a common external tangent to two circles. Construction: Using as a center and a radius equal to the difference of the given radii, construct (dotted) circle. Draw QA tangent to this O from Q; draw radius OA and produce it to meet given O at B. Draw radius QC II to OB. Join BC. Statement: BC is tangent to both (D. Proof: AB = CQ (Const.). AB is II to CQ (?). /. ABCQ is a O (?). But Z OAQ is a rt. Z; etc. 157. Construct a common internal tan- gent to two circles. Construction : Using as a center and a radius equal to the sum of the given radii, construct (dotted) circle. Draw QA tangent to this from Q. Draw radius OA meeting given O at B, etc., as above. BOOK III PROPORTION. SIMILAR FIGURES 274. A ratio is the quotient of one quantity divided by another, both being of the same kind. 275. A proportion is an equation whose members are ratios. 276. The extremes of a proportion are the first and the last terms. The means of a proportion are the second and the third terms. 277. The antecedents are the first and the third terms. The consequents are the second and the fourth terms. 278. A mean proportional is the second or the third term of a proportion in which the means are identical. A third proportional is the last term of a proportion in which the means are identical. A fourth proportional is the last term of a proportion in which the means are not identical. 279. A series of equal ratios is the equality of more than two ratios. A continued proportion is a series of equal ratios in which the consequent of any ratio is the antecedent of the next following ratio. EXPLANATORY. A ratio is written as a fraction or as an indicated division ; - , or a -f- b, or a : b. A proportion is usually written - ?, or b by a:b = x:y, and is read : " a is to b as x is to y" In this proportion the extremes are a and y\ the means are b and x ; the antecedents are a and a:; the consequents are b and #; and y is a fourth proportional to a, b, x. In the proportion a : m = m : z, the mean proportional is m, and z is the third proportional. 143 144 BOOK III. PLANE GEOMETRY THEOREMS AND DEMONSTRATIONS PROPOSITION I. THEOREM 280. In a proportion the product of the extremes is equal to the product of the means. Given : | = - or a : b = x : y. To Prove : ay = bx. & Proof: 7 = - (Hyp.). Multiply by the common denomi- t/ nator by and obtain, ay = bx (Ax. 3). Q.E.D. PROPOSITION II. THEOREM 281. If the product of two quantities is equal to the product of two others, one pair may be made the extremes of a propor- tion and the other pair the means. Given : ay = bx. To Prove : These eight proportions : 1. a : b = x : y, 5. x : y = a : b, 2. a : x = b : y, 6. x : a = y i b, 3. b : a = y : a/, 7. y : x = b : a, 4. b : y = a : x, 8. y : b = x : a. Proof. 1. ay = lx (Hyp.). Divide each member by 5v, = (Ax. 3). by by .. j- = - QY a:b = xiy Q.E.D. 2. ay = bx (Hyp.). Divide by xy, etc. 3. ay = bx (?). Divide by ax, etc. Etc., etc. NUMERICAL ILLUSTRATION. Suppose in this paragraph a = 4, b = 14, x = 6, y = 21 ; the truth of the above proportions can be clearly seen by writing these equivalents. 4 x 21 = 14 x 6 (True). 1. 4 : 14 = 6 : 21 (True) ; 2. 4 : 6 = 14 : 21 (True) ; etc. They will all be recognized as true proportions. THEORY OF PROPORTION 145 PROPOSITION III. THEOREM 282. In any proportion the terms are also in proportion by alternation (that is, the first term is to the third as the sec- ond is to the fourth). Given : a : b = x : y. To Prove : a:x=b:y. Proof : a:b = xiy (Hyp.). .:ay = bx (280). .:a:x=b:i/ (281). Q.E.D. PROPOSITION IV. THEOREM 283. In any proportion the terms are also in proportion by inversion (that is, the second term is to the first as the fourth term is to the third). [The proof is similar to the proof of 282.] PROPOSITION V. THEOREM 284. In any proportion the terms are also in proportion by composition (that is, the sum of the first two terms is to the first, or the second, as the sum of the last two terms is to the third, or the fourth). I a + b : a=x+ V : X, or Given: a:b = x:y. To Prove: Proof : a:b = x:y (Hyp.). . -. ay = bx (280). Add ax to each, and obtain, ax + ay = ax + bx (Ax. 2). That is, a (x + y) = x(a + b). Hence a+b : a = x + y : x (281). Similarly, by adding by, a + b : b = x -f y : y. Q.E.D. Ex. 1. Is-the equation 12 : 9 = 28 : 21 a true proportion ? Ex. 2. Apply Proposition I to the above proportion. Ex. 3. Apply Proposition III to the above proportion. Ex. 4. Apply Proposition IV to the above proportion. Ex. 6. Apply Proposition V to the above proportion. 146 BOOK III. PLANE GEOMETRY PROPOSITION VI. THEOREM 285. In any proportion the terms are also in proportion by division (that is, the difference between the first two terms is to the first, or the second, as the difference between the last" two terms is to the third, or the foiirth). Given: a: b = x :y. ToProve: {-'-*- = . a o : o = x y - y> Proof : a:b = x:y (Hyp.). . . ay = bx (280). Subtracting each side from ax, ax ay = ax bx (Ax. 2). That is, a(x y) = x(a b). Hence a'b-.a x y-.x (281). Similarly, a b:b = x y:y. Q.E.D. NOTE 1. The proportions of 284 and 285 may be written in many different forms (282, 283) . Thus, (1) a b:a = x y\x\ (2) ab:b = xy:y; (3) ab:xy = a:x, etc. NOTE 2. In any proportion the sum of the antecedents is to the sum of the consequents as either antecedent is to its consequent. Also, in any proportion the difference of the antecedents is to the difference of the consequents as either antecedent is to its consequent. (Explain.) Thus : a + x:b + y = a:b = x:y. Also, a x:b y = a:b = x:y. PROPOSITION VII. THEOREM 286. In any proportion the terms are also in proportion by composition and division (that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference). Given: a:b=x:y. ToProve: ^-^ = ^1. a b x y Proof: fLi = ^^ (284). Divide the first by the second, ^- = ?-l (Ax. 3). 7* x - y Q.E.D. THEORY OF PROPORTION 147 PROPOSITION VIII. THEOREM 287. In any proportion, like powers of the terms are also in proportion, and like roots of the terms are in proportion. Given : a : b = x : y. To Prove : a n : b n = x n : y n ; and -\/a : -\/b = ^/x : ^Jy. Proof: [Write the hypothesis in fractional form, etc.] PROPOSITION IX. THEOREM 288. In two or more proportions the products of the corre- sponding terms are also in proportion. Given : a\~b x\y^ and c : d = I : m, and e : f = r : s. To Prove : ace : bdf = xlr : yms. Proof: [Write in fractional form and multiply.] PROPOSITION X. THEOREM 289. A mean proportional is equal to the square root of the product of the extremes. Given : a : x = x : b. To Prove : x = Va6. Proof: [Use 280.] PROPOSITION XI. THEOREM 290. If three terms of one proportion are equal to the corre- sponding three terms of another proportion, each to each, the remaining terms are also equal. Given: ' . ToProve:m = r. \ a : b = c : r. Proof : am = be and ar = be (280). ..am = ar (Ax. 1). .-. m = r (Ax. 3). Q.E.D. 148 BOOK III. PLANE GEOMETRY PROPOSITION XII. THEOREM 291. In a series of equal ratios the sum of all the ante- cedents is to the sum of all the consequents as any antecedent is to its consequent. Given: f= = f = f- b d f h To Prove: * + ' + e + f = f=^ = etc. b+d+f+h b d Proof : Set each given ratio = to m ; thus, a c e a - = m; 3 = w ; -r = w ; ^ = m. b d f h . . a = bm, c = dm, e =/m, g = Jim (Ax. 3). TT a + ++F : AC. To Prove : The A similar. Proof : Superpose A ABC upon A DEF so that Z. A coincides with its equal, Z D, and A ABC takes the position of A DBS. Now DE:DB = DF:DS (Hyp.). .-. US is II to EF (296). .-. A DBS is similar to A DEF (305). Substituting, A ABC is similar to A DEF (Ax. 6). Q.E.D. 158 BOOK III. PLANE GEOMETRY 307. COROLLARY. If two triangles are similar to the same triangle, they are similar to each other. Proof: The three angles of each of the first two triangles are respectively equal to the three angles of the third (301). Hence the first two A are mutually equiangular (Ax. 1). Therefore they are similar (302). Q.E.D. PROPOSITION XXIII. THEOREM 308. If two triangles have their homologous sides propor- tional) they are similar. D B C E F Given : A ABC and DEF\ DE : AB = DF : AC = EF : EC. To Prove : A ABC similar to A DEF. Proof : On DE take DK = to AB ; and on DF take DL = to AC. Draw KL. 1st Now DE : AB = DF : AC (Hyp.). DE : DK = DF : DL (Ax. 6). .-. EL is 11 to EF (296). A DKL is similar to A DEF (305). 2d DE : DK = EF : KL (Def. sim. A) (301). Substituting, DE : AB = EF : KL (Ax. 6). But DE:AB = EF:BC (Hyp.). .'.BC = KL (290). 3d A ABC is congruent to A DKL (78). But A DKL has been proved similar to A DEF. Substituting, A ABC is similar to A DEF (Ax. 6). Q.E.D. SIMILAR POLYGONS 159 PROPOSITION XXIV. THEOREM 309. If two triangles have their homologous sides parallel, they are similar. A D E F Given : A ABO and DEF ; AB II to DE ; AC II to DF ; and BC II to EF. To Prove : A ABC similar to A DEF. Proof : Produce BC of A ABC until it intersects two sides of A DEF at E and s. Now Z B = Z DBS, and Z DBS = Z E (67). .-. ZB = /.E (Ax. 1). Similarly, we may prove Z ACB = to Z F. . . A ^BC is similar to A DEF (303). Q.E.D. Ex.1. Draw the figure and prove Proposition XXIV: (a) If one triangle is entirely within the other ; (6) If no side of either, when prolonged, meets any side of the other ; (c) If one side of one intersects two sides of the other, without pro- longation ; (d) If a vertex of one is upon a side of the other. Ex. 2. If in a right triangle, a perpendicular is drawn from the vertex of the right angle upon the hypotenuse, the two new right triangles are similar to the original triangle and to each other. Ex. 3. Are all equilateral triangles similar? Why? Ex. 4. Are all squares similar? all rectangles? Why? Ex. 5. Do a square and a rectangle fulfill either of the conditions for similar polygons? Ex. 6. If from any point in a leg of a right triangle a line is drawn perpendicular to the hypotenuse, the triangles are similar. 160 BOOK III. PLANE GEOMETRY PROPOSITION XXV. THEOREM 310. If two triangles have their homologous sides perpen- dicular, they are similar. M Given: A ABC and DEF -, AB _L to VE\ AC . to DF\ etc. To Prove : A ABC similar to A DEF. Proof: Through P, any point in EF, construct PR I! to AC, meeting DF at M. At R, any point in PM, draw RS II to AB, meeting ED at N. Draw PS II to BC, meeting NR at 8, form- ing the A PRS. Now PM is J_ to DF and RN is _L to DE (64) . In quadrilateral DMRN, Z D + Z 3f + Z MRN + Z zV = 4 rt. /4 (156). But Zjtf + Z TV = 2 rt. ^ (16). .-. ZD +/.MRN = 2rt. A (Ax. 2). But Z a + Z ^f## = 2 rt. A (46). .-. ZD-Za (49). Similarly, by quadrilateral ^P5JV, it may be proved that Z# = Z6. . . A DEF is similar to A pis (303). But A ABC is similar to A PRS (309). . . A ABC is similar to A DEF (307). Q.E.D. Historical Note. Thales, as early as the sixth century B.C., used simi- lar triangles to determine the height of the great Egyptian pyramid. He measured the shadow of a pole of known height, and the shadow of the pyramid at the same time, to obtain homologous sides of similar right triangles. SIMILAR POLYGONS 161 PROPOSITION XXVI. THEOREM 311. Two homologous altitudes of two similar triangles are proportional to any two homologous sides. Given: (?). To Prove BL AB AC BC B'L' A'B' A'C' B'C' A L C A' Proof : A ABC is similar to A A' B'C' .: A ABL is similar to A A 1 B'L' BL AB C But BL B'L' AB A'B' AB BC A'B' AC = A'C'~ B'C' AC BC A'B' A'C' B'C' (Hyp.)- (301). (304). (301). (301). (Ax. 1). Q.E.D. 312. In similar figures, homologous angles are equal (Def.). 313. In similar figures, homologous sides are proportional (Def.). The antecedents of this proportion belong to one of the similar figures, and the consequents to the other. 314. In similar triangles, homologous sides are opposite homologous angles. Shortest sides are homologous. (Opposite smallest A}. Medium sides are homologous. (Opposite medium A}. Longest sides are homologous. (Opposite largest A). 162 BOOK III. PLANE GEOMETRY PROPOSITION XXVII. THEOREM 315. If two parallel lines are cut by three or more transversals that meet at a point, the corresponding segments of the parallels are proportional. Given: (?). To Prove: ^ = ^ = **. CG GH HD Proof : In A OCG, AE is II to CG .'. A OAE is similar to A OCG Likewise, A OEF is similar to A OGH, and A OFB to A OHD L \\ G CG OG GH OG AE _EF ' CG ~ GH Likewise = , and = GH OH OH HD AE = EF ^FB ' CG GH~ HD PROPOSITION XXVIII. THEOREM 316. If three or more non-parallel trans- versals intercept proportional segments on two parallels, they meet at a point. [Con- verse.] Given: Transversals AB, CD, EF; par- allels AE and BF; proportion, AC: BD = CE : DF. To Prove : AB, CD, EF meet at a point. H D (Hyp.)- (305). is similar (305). (313). (Ax. 1). (313). (Ax. 1). Q.E.D. kC B Proof : Produce AB and CD until they meet, at o. Draw OF cutting AE at X. Now AC:BD=CX:DF (315). But AC:BD=CE:DF (Hyp.)- SIMILAR POLYGONS 163 .-. CX=CE (290). .-. FE and FX coincide (39). That is, FE prolonged, passes through o. Q.E.D. Ex. 1. Show the truth of Proposition XXVIII, by an accurate diagram. Ex. 2. In the figure of Proposition XXVIII, what is point XI PROPOSITION XXIX. THEOREM 317. The perimeters of two similar polygons are to each other as any two homologous sides. D' D Given: Polygon R whose perimeter = P and similar polygon 8 whose perimeter = P f . To Prove : p : p' = AB : A'B' = BC : B'C' = etc. Proof : AB : A'B' = BC : B'C' = CD : C'D' = etc. (313). . AB + BC+CD+ ... = AB = BC = ^ .^ ' A'B' + B'C' + C'D' + ... A'B' B'C' Substituting, y = ^ = - = etc. (Ax. 6). Q.E.D. Ex. 1. The sides of a certain polygon are 4, 10, 12, 15, and 18. The shortest side of a similar polygon is 6. Find the other four sides. Ex. 2. The perimeters of two similar polygons are 125 and 275 respectively. The longest side of the smaller polygon is 40. Find the longest side of the larger polygon. Ex. 3. Enumerate the ways of proving two triangles similar. Which is the easiest of these ways? 164 BOOK III. PLANE GEOMETRY PROPOSITION XXX. THEOREM 318. If two polygons are similar, they may be decomposed into the same number of triangles, similar each to each and similarly placed. B' B ED E' D 1 Given : Similar polygons BE and B'E'. To Prove : A ABC similar to A A' B'C' ; A ACD similar to A A'C'D' ; A AED similar to A A'E'D' '. Proof: First. AB : A'B' = BC : B'C' (313). Also Z B = Z B' (312). Therefore A ABC is similar to A A' B'C' (306). Second. In A ABC and A' B'C', ^- = -^- (313). B'c' A'C' T>/"f /^7~i In the similar polygons, ^ = -7-7 (313). B C CD Consequently -=- (Ax. 1). A C 1 C 1 D In the polygons, Z BCD = Z 5'c'D' 1 In the A ^IBC and A'B'C', Z. BCA= ^ B' C* A 1 } Hence, by subtraction, Z ACD = Z. A'C'D' (Ax. 2). Therefore A ACD is similar to A A'C'D' (306). Third. A AED is proved similar to A A'E'D' in like manner. Q.E.D. Ex. A map is drawn on a scale of 1 in. to 400 miles. How far is Paris from Antwerp if they are 1| in. apart on the map? SIMILAR POLYGONS 165 PROPOSITION XXXI. THEOREM 319. If two polygons are composed of triangles similar each to each and similarly placed, the polygons are similar. [Con- verse.] H' H J K' Given: A GHI similar to A G'H'I' ; A GIJ similar to A G'I'J' ; A GJK similar to A G'J'K'. To Prove : The polygons HK and H' K' similar. Proof : First. In A HGI and H' G' l' , Z H = Z H' (312). Also in these A Z HIG = Z H'I'G' (312). In A GIJ and G'I'J', Z GIJ = Z G'I'J' (312). Adding, ZniJ=^H f I f J f (Ax. 2). Likewise Z u^r = Z I'J'K' ; etc. That is, the polygons are mutually equiangular. Second. In A GHI and G' H GH G'H' HI H'I' In A GIJ and eW, -- = = G'l' I'J' In A GJK and 6?' J'K', = G'J' GH HI IJ G'J 1 KG ~ K'G> JK KG K'G' That is, the homologous sides are proportional. .. The polygons are similar (Ax. 1). (301). Q.E.D. Ex. If two parallelograms are similar, and a diagonal in each is drawn, are the resulting triangles necessarily similar in pairs? 166 BOOK 111. PLANE GEOMETRY PROPOSITION XXXII. THEOREM 320. If through a fixed point within a circle two chords are drawn, the product of the segments of one equals the product of the segments of the other. Given : Point O in circle C ; chords AB and BS intersecting at O. (Review the note, p. 149.) To Prove : AO OB = RO - os. Proof : Draw AS and RB. In A AOS and BOB, Z s = /- B And Z^=ZE .. these A are similar .-. AO : BO = OS : OB .'. AO- OB = EO OS (239). ' (303). (313). (280). Q.E.D. 321. COROLLARY. The product of the segments of any chord drawn through a fixed point within a circle is constant for all chords through this point. 322. Direct proportion and reciprocal (or inverse) proportion. Illustrations. I. If a man earns $4| each day, in 8 days he earns $36. In 12 days he earns $54. Hence, 8 da. : 12 da. = $36: $54 is a proportion in which the antecedents belong to the same condition or circumstance, and the consequents belong to some other condition or circumstance. This is called a direct proportion. II. If one man can build a certain wall in 120 days, 8 men can build it in 15 days ; or 12 men in 10 days. Hence, 8 men : 12 men = 10 da. : 15 da. is a proportion in which the means belong to the same condition or circumstance, and the extremes belong to some other condition or cir- cumstance. This is called a reciprocal (or inverse) proportion. Definitions. A direct proportion is a proportion in which the antecedents belong to the same circumstance or figure, and the consequents belong to some other circumstance or figure. Thus, the ordinary proportions derived from similar figures are direct proportions. CIRCLES 167 A reciprocal (or inverse) proportion is a proportion in which the means belong to the same cir- cumstance or figure, and the extremes belong to some other circumstance or figure. Thus, in the adjoining figure, a b = x >y (320). .*. a : x y : b (281). This is a reciprocal pro- portion because the means are parts of one chord, and the extremes are parts of the other chord. 323. THEOREM. If through a fixed point within a circle two chords are drawn, their four segments are reciprocally (or inversely) proportional. Proof: [Identical with 320; omitting the last step.] PROPOSITION XXXIII. THEOREM 324. If from a fixed point without a circle a secant and a tangent are drawn, the product of the whole secant and the external segment equals the square of the tangent. Given : O C ; secant PAB ; tan- gent PT. To Prove : PB PA = PT 2 . Proof: Draw AT and BT. In A PAT and PBT, Z P = Z P Z.PTA is measured by ^ arc AT Z B is measured by ^ arc ^.T Therefore Hence A PAT is similar to A PBT PB : PT = PT : PA (Iden.). (241). (236). (237). (303). (313). (280). Q.E.D. 168 BOOK III. PLANE GEOMETKY 325. COROLLARY. If from a fixed point without a circle any secant is drawn, the product of the secant and its external segment is constant for all secants. Proof: Any secant x ext. seg. = (tan.) 2 = constant. 326. COROLLARY. If from a fixed point without a circle two secants are drawn, these secants and their external segments are reciprocally (or inversely) proportional. Proof : PB - PA = PY - PX (325). .-. PB : PY = PX : PA (281). Q.E.D. 327. THEOREM. If from a fixed point without a circle a secant and a tangent are drawn, the tangent is a mean proportional between the secant and its external segment. Proof : [Identical with proof of 324 ; omitting the last step.] Ex. 1. If PA = 3 in., and PB = 12 in., find the length of PT. Ex. 2. If PB = 21 in., PY = 15 in., and PA = 5 in., find PX. Ex. 3. Two altitudes of a triangle are reciprocally proportional to the bases to which they are drawn. Ex. 4. If AD and BE are two altitudes of a triangle, and DE is drawn, the triangle ABC is similar to the triangle CED. [Use 306.] Ex. 5. The four segments of the diagonals of a trapezoid are propor- tional. CIRCLES 169 PROPOSITION XXXIV. THEOREM 328. In any triangle the product of two sides is equal to the diameter of the cir- cumscribed circle multiplied by the alti- tude upon the third side. Given : A ABC ; circumscribed Q o ; altitude BK. To Prove : a c = d h. Proof : Draw chord CD. Z BCD = rt. Z In rt. A ABK and BCD, Z A = Z D .. these A are similar (240). (239). (304). (313). (280). Q.E.D. PROPOSITION XXXV. THEOREM 329. In any triangle the product of two sides is square of the bisector of their included angle, plus the product of the seg- ments of the third side formed by the bisector. Given: A ABC, CO bisector of Z ACS. To Prove : a - b = t 2 -f- n r. Proof : Circumscribe a O about the A ABC. Produce CO to meet O at D. Draw BD. In A AGO and BCD, Z AGO = Z BCD Also Z A = Z D .-. &ACO and BCD are similar Now Substituting, AB and CD are chords a b = t 2 + n r equal to the ........... - C /o A I / I (Ax. (Hyp.). (239). (303). (313). (280). (Const.). (320). 6). Q.E.D. BOBBINS'S NEW PLANE GEOM. 12 170 BOOK III. PLANE GEOMETRY 330. The projection of a point upon a line is the foot of the perpendicular from the point to the line. Thus, the projection of P is J. B S L ^1 /\ \ I i i / = ;J_ J C D R T NM The projection of a definite line upon an indefinite line is the part of the indefinite line between the feet of the two perpendiculars to it, from the extremities of the definite line. The projection of AB is CD ; of RS is ET ; of LM is NM. PROPOSITION XXXVI. THEOREM 331. If in a right triangle a perpendicular is drawn from the vertex of the right angle upon the hypotenuse : I. The triangles formed are similar to the given triangle and similar to each other. n. The perpendicular is a mean proportional between the segments of the hypotenuse. r* Given : Rt. A ABC ; CP _L to AB from C. To Prove : A I. A APC, ABC, and BPC similar. /_ II. AP : CP = CP : PB. A Proof : I. In rt. A APC and ABC, ^-A = /.A (Iden.). . . A APC is similar to A ABC (304). In rt. A BPC and ABC, ^ B = Z B (Iden.). . . A BPC is similar to A ABC (304). Therefore A APC, ABC, and BPC are all similar (307). II. In the A APC and BPC, AP : CP = CP : PB (313). Q.E.D. PROJECTIONS 171 332. COROLLARY. If from any point in a circumference a perpendicular is drawn to a diameter, p it is a mean proportional between the segments of the diameter. Given: (?). To Prove: (?). Proof : Draw chords AP and BP. A APB is a rt. A .-. AD : PD = PD :DB (240). (331, II). Q.E.D. PROPOSITION XXXVII. THEOREM 333. The square of a leg of a right triangle is equal to the product of the hypotenuse and the projection of this leg upon the hypotenuse. Given : Rt. A ABC ; AC and BC the legs. I. AC? = AB - AP. AB- BP. To Prove II. Proof: I. The rt. A ABC and APC are similar .'. AB :_AC = AC : AP *. AC^ = AB AP The rt. A ABC and BCP are similar .'. AB : BC= BC : BP 2J(?2 __ ^jj . -gp II. (331, I). (313). (280). (313)! (280). Q.E.D. Ex. 1. If, in 331, AP = 3, PB = 27, find CP. Ex. 2. If, in 333, AP = 4, PB = 21, find A C and BC. Ex. 3. If, in 333, AB = 2Q,AC = 10, find AP, BP, CP, and BC. Ex. 4. If, in 331, AP = 10 and CP = 20, find BP. Ex. 5. If, in 333, AB = 45 and AC = 15, find AP, BP, CP, and BC. Ex. 6. If, in 333, AP = 9 and BP = 16, find AC, PC, and BC. Ex. 7. A stone arch in the shape of the arc of a circle is 4 ft. high. The chord of half the arch is 1.0 ft. Find the diameter. 172 BOOK 111. PLANE GEOMETRY PROPOSITION XXXVIII. THEOREM 334. The sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Given : Rt. A ABC. To Prove : a 2 + V* = c 2 . Proof : Draw CP _1_ to the hypotenuse AB. Denote AP by p and PB by p'. Now b 2 = c - p Also a 2 = c p' Adding, a = c-p + c .p f = c(p +p f ) = C C = C 2 . (333). (?). (Ax. 2). Q.E.D. 335. COROLLARY. The square of either leg of a right triangle is equal to the square of the hypotenuse minus the square of the other leg. That is, a 2 = c 2 - b* ; and b 2 = c* - a 2 (Ax. 2). Ex. 1. If AC = 28 and EC = 45, find AB. Ex. 2. If AC = 21 and AB = 29, find EC. Ex. 3. The square of the altitude of an equilateral triangle equals three fourths the square of a side. Ex. 4. In any triangle the difference of the squares of two sides is equal to the 'difference of the squares of their projections on the third side. [ZS 2 = (?) ; BC 2 = (?). Subtract, etc.] Ex. 6. If the altitudes of triangle ABC meet at 0,.AB* - AC 2 = B0 2 - CO 2 . A ~~ [Consult Ex. 4 and substitute.] Ex. 6. If lines are drawn from any point in a circle to the four vertices of an inscribed square, the sum of the squares of these four lines is equal to twice the square of the diameter. Proof: &APC, DPB are vt, A; etc. SQUARES OF SIDES OF TRIANGLES 173 PROPOSITION XXXIX. THEOREM 336. In an obtuse triangle the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides plus twice the product of one of these two sides and the projection of the other side upon that one. Given : Obtuse A ABC ; etc. ^- - C M To Prove : c 2 = a 2 + b 2 + 2 bp. Proof : In right A CBM, h 2 +p 2 = a 2 (334). In right A ABM, (? = h* + (p + b) 2 (334). = h 2 + p 2 + b 2 + 2 bp = ~a?~^ + b 2 + 2bp (Ax. 6). Q.E.D. PROPOSITION XL. THEOREM 337. In any triangle the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides minus twice the product of one of these two sides and the pro- jection of the other side upon that one. Given: (?). To Prove : c? = a 2 + b 2 2 bp. ^ t>-t> I P Proof : In one rt. A, h 2 + p 2 = a 2 (334) . In the other rt. A, c 2 = W + (b-p)* (334). (Ax. 6). Q.E.D. NOTE. This theorem is equally true in the case of an obtuse triangle. Ex. If lines are drawn from any external point to the vertices of a rectangle ABCD, the sum of the squares of two of them which are drawn to a pair of opposite vertices is equal to the sum of the squares of the other two. To Prove : PA 2 + PC 2 = PB 2 + P&. 174 BOOK III. PLANE GEOMETRY PROPOSITION XLI. THEOREM 338. I. The sum of the squares of two sides of a triangle is equal to twice the square of hah" the third side increased by twice the square of the median upon that side. II. The difference of the squares of two sides of a triangle is equal to twice the product of the third side by the projection of the median upon that side. Given : A ABC ; median = m ; its projection = p ; and side b > side a. To Prove : Proof : In A ARC and BRC, AR = BE (Hyp.). And CR = CR (Iden.). Also AC > BC (Hyp.). .-. Z ARC > Z BRC (92). That is, Z.ARC is obtuse and Z BRC is acute. 5 2 = (<^) 2 + w 2 + ^ (336). a?= (^e) 2 + m*-cp (337). I. Adding, 6 2 + a 2 = 2(-|V) 2 +2w 2 (Ax. 2). II. Subtracting, b 2 a 2 = 2cp (Ax. 2). Q.E.D. 339. Formulas. If the vertices of a triangle are denoted by A, -B, C, the lengths of the sides opposite are denoted by a, 5, 0, respectively ; the altitude upon these sides by A , h b h c , respectively ; the bisectors of the angles by t a , t b , t c , respec- tively ; the medians by m a , m b , m^ respectively ; the seg- ments of the sides formed by the bisectors of the opposite angles by n a and r , n b and r b , n c and r c \ and the projections as follows: the projection of side a upon side 5, by a p b \ of side a upon side 2 rt 2 7> 2 r 2 rfi W 1. If /. C is obtuse, a ^ 6 = C-jL- ; 6 p a = |_A; etc. 2 i ;.2 ,.2 ^2 i 1,2 ~2 2. If Z C is acute, a p b = - + ^~-; ^ = - + ^ g ~ ; etc. IT. Altitudes. h b = Va 2 - a p b 2 ; A = V& 2 - 6 p 2 ; etc. III. Medians. m c = iV2(' 2 + b 2 ) - c 2 ; m = ^V2(& 2 + c 2 ) - a 2 ; etc. IV. Bisectors. t c = Vab - n c r* ; t a = V6c - n a r a * ; f 6 = Vac - n b r b * V. Diameter of circumscribed circle = ; = ; = h b h c h a VI. Largest Angle. Denote largest angle by C. 1. If c 2 = a 2 + 6 2 , Z C is right (334). 2. If c 2 = a 2 f b 2 plus something, Z C is obtuse (336). 3. If c 2 = a 2 + 6 2 minus something, Z. C is acute (337). Ex. 1. If the sides of a triangle are a = 7, b = 10, c = 12, find the nature of Z C. Ex. 2. In the same triangle find ra a . Find m b . Find m c . Ex. 3. In the same triangle find a p b . Find b p a . Find a jt> c . Find fr j9 c . Find c p a . Find ./>&. 1 Ex. 4. Find /*. Find h b . Find ^ c . Ex. 5. Find the diameter of the circumscribed circle. Ex. 6. Find n a and r a : Find n b and r b . Find n c and r c . Ex. 7. Find * . Find t b . Find < c . CONCERNING ORIGINALS 340. First determine from the nature of each numerical exercise upon which theorem it depends, and then apply that theorem. * The segments n and r can be found by 297 ; n b : r b = c : a, etc. 176 BOOK III. PLANE GEOMETRY ORIGINAL EXERCISES (NUMERICAL) 1. The legs of a right triangle are 12 and 16 inches. Find the hypote- nuse. 2. The side of a square is 6 feet. What is the diagonal? 3. The base of an isosceles triangle is 16 and the altitude is 15. Find the equal sides. 4. The tangent to a circle from a point is 12 inches and the radius of the circle is 5 inches. Find the length of the line joining the point to the center. 5. In a circle whose radius is 13 inches, what is the length of a chord 5 inches from the center? 6. The length of a chord is 2 feet and its distance from the center is 35 inches. Find the radius of the circle. 7. The hypotenuse of a right triangle is 2 feet 2 inches, and one leg is 10 inches. Find the other. 8. The base of an isosceles triangle is 90 inches and the equal sides are each 53 inches. Find the altitude. 9. The radius of a circle is 4 feet 7 inches. Find the length of the tangent drawn from a point 6 feet 1 inch from the center. 10. How long is a chord 21 yards from the center of a circle whose radius is 35 yards ? 11. Each side of an equilateral triangle is 4 feet. Find the altitude. 12. The altitude of an equilateral triangle is 8 feet. Find the side. 13. Each side of an isosceles right triangle is a. Find the hypotenuse. 14. If the length of the common chord of two intersecting circles is 16, and their radii are 10 and 17, what is the distance between their centers? 15. The diagonal of a rectangle is 82 and one side is 80. Find the other. 16. The length of a tangent to a circle whose diameter is 20, from an external point, is 26. What is the distance from this point to the center ? 17. The diagonal of a square is 10. Find each side. 18. Find the length of a chord 2 feet from the center of a circle whose diameter is 5 feet. 19. A flagpole was broken 16 feet from the ground, and the top struck the ground 63 feet from the foot of the pole. How long was the pole ? ORIGINAL EXERCISES 177 20. The top of a ladder 17 feet long reaches a point on a wall 15 feet from the ground. How far is the lower end of the ladder from the wall? 21. A chord 2 feet long is 5 inches from the center of a circle. How far from the center is a chord 10 inches long? [Find the radius.] 22. The diameters of two concentric circles are 1 foot 10 inches and 10 feet 2 inches. Find the length of a chord of the larger which is tan- gent to the less. 23. The lower ends of a post and a flagpole are 42 feet apart; the post is 8 feet high and the pole, 48 feet. How far is it from the top of one to the top of the other? 24. The radii of two circles are 8 inches and 17 inches, and their cen- ters are 41 inches apart. Find the lengths of their common external tangents ; of their common internal tangents. 25. A ladder 65 feet long stands in a street. If it inclines toward one side, it will touch a house at a point 16 feet above the pavement ; if to the other side, it will touch a house at a point 56 feet above the pave- ment. How wide is the street? 26. Two parallel chords of a circle on opposite sides of the center are 4 feet, and 40 inches long, respectively, and the distance between them is 22 inches. Find the radius of the circle. [Draw the radii to ends of chords ; these = hypotenuses = R ; the distances from the center = x and 22 #.] 27. The legs of an isosceles trapezoid are each 2 ft. 1 in. long, and one of the bases is 3 ft. 4 in. longer than the other. Find the altitude. 28. One of the non-parallel sides of a trapezoid is perpendicular to both bases, and is 63 feet long ; the bases are 41 feet and 25 feet long. Find the length of the remaining side. 29. If a = 10, h = 6, find p, c, p', b. 30. If h = 8, p' = 4, find b, c, p, a. 31. If a = 10, p' = 15, find p, c, h, b. 32. If a = 9, b = 12, find c, p,p', h. . 33. Ifp= 3, p' = 12, find h, a, b. 34. The line joining the midpoint of a chord to the midpoint of its arc is 5 inches. If the chord is 2 feet long, what is the diameter? [A ACE is rt. A (?). .-. AC 2 = CE . CD (?).] 35. If the chord of an arc is 60 and the chord of its half is 34, what is the diameter? 178 BOOK III. PLANE GEOMETRY 36. The line joining the midpoint of a chord to the midpoint of its arc is 6 inches. The chord of half this arc is 18 inches. Find the diameter. Find the length of the original chord. 37. To a circle whose radius is 10 inches, two tangents each 2 feet long are drawn from a point. Find the length of the chord joining their points of contact. 38. The sides of a triangle are 6, 9, 11. Find the segments of the shortest side made by the bisector of the opposite angle. 39. Find the segments of the longest side made by the bisector of the largest angle in Ex. 38. 40. The sides of a triangle are 5, 9, 12. Find the segments of the shortest side made by the bisector of the opposite exterior angle ; also of the medium side made by the bisector of its opposite exterior angle. 41. In the figure of 295, if AC = 3, CE = 5, EG = 8, BD = 4, find DF and FH. 42. If the sides of a triangle are 6, 8, 12 and the shortest side of a similar triangle is 15, find its other sides. 43. If the homologous altitudes of two similar triangles are 9 and 15 and the base of the former is 21, what is the base of the latter? 44. In the figure of 315, AE = 4, EF = 6, FB = 9, GH = 15. Find CG and CD. 45. The sides of a pentagon are 5, 6, 8, 9, 18, and the longest side of a similar pentagon is 78. Find the other sides. 46. A pair of homologous sides of two similar polygons are 9 and 16. If the perimeter of the first is 117, what is the perimeter of the second? 47. The perimeters of two similar polygons are 72 and 120. The shortest side of the former is 4. What is the shortest side of the latter? 48. Two similar triangles have homologous bases 20 and 48. If the altitude of the latter is 36, find the altitude of the former. 49. The segments of a chord, made by a second chord, are 4 and 27. One segment of the second chord is 6. Find the other. 50. One of two intersecting chords is 19 inches long and the segments of the other are 5 inches and 12 inches. Find the segments of the first chord. 51. Two secants are drawn to a circle from a point ; their lengths are 15 inches and 10| inches. The external segment of the latter is 10 inches. Find the external segment of the former. ORIGINAL EXERCISES 179 52. The tangent to a circle is 1 foot long and the secant from the same point is 1 foot 6 inches. Find the chord part of the secant. 53. The internal segment of a secant 25 inches long is 16 inches. Find the tangent from the same point to the same circle. 54. Two secants to a circle from a point are 1| feet and 2 feet long ; the tangent from the same point is 12 inches. Find the external segments of the two secants. 55. If the sides of a triangle are 5, 6, 8, is the angle opposite 8 right, acute, or obtuse ? if the sides are 8, 7, 4 ? 56. If the sides of a triangle are 8, 9, 12, is the largest angle right, acute, or obtuse? if the sides are 13, 7, 11 ? 57. The sides of a triangle are x, y, z. If z is the greatest side, when will the angle opposite be right? obtuse? acute? 58. The sides of a triangle are 6, 8, 9. Find the length of the projec- tion of side 6 upon side 8 ; of side 8 upon side 9 ; of side 9 upon side 6. 59. The sides of a triangle are 5, 6, 9. Find the length of the pro- jection of side 6 upon side 5 ; of side 9 upon side 6. 60. Find the three altitudes in a triangle with sides 9, 10, 17. 61. Find the three altitudes in a triangle with sides 11, 13, 20. 62. Find the diameter of a circle circumscribed about a triangle with sides 17, 25, 26. 63. Find the length of the bisector of the least angle of a triangle with sides 7, 15, 20; also of the largest angle. 64. Find the length of the bisector of the largest angle of a triangle with sides 12, 32, 33 ; also of the other angles. 65. Find the three medians in a triangle with sides 4, 7, 9. 66. Find the product of the segments of every chord drawn through a point 4 units from the center of a circle whose radius is 10 units. 67. The bases of a trapezoid are 12 and 20, and the altitude is 8. The other sides are produced to meet. Find the altitude of the larger tri- angle formed. 68. The shadow of a yardstick perpendicular to the ground is 4| feet. Find the height of a tree whose shadow at the same time is 100 yards. 69. There are two belt-wheels 3 feet 8 inches and 1 foot 2 inches in diameter, respectively. Their centers are 9 feet 5 inches apart. Find the length of the belt suspended between the wheels if the belt does not cross itself ; also the length of the belt if it does cross. 180 BOOK III. PLANE GEOMETRY SUMMARY 341. Triangles are proved similar b}^ showing that they have : (1) Two angles of one equal to two angles of the other. (2) An acute angle of one equal to an acute angle of the other. [In right triangles.] (3) Homologous sides proportional. (4) An angle of one equal to an angle of the other and the including sides proportional. (5) Their sides respectively parallel or perpendicular. Four lines are proved proportional by showing that they are : (1) Homologous sides of similar triangles. (2) Homologous sides of similar polygons. (3) Homologous lines of similar figures. The product of two lines is proved equal to the product of two others, by proving these four lines proportional and making the product of the extremes equal to the product of the means. One line is proved a mean proportional between two others by proving that two triangles containing this line in common are similar, and obtaining the proportion from their sides. In cases dealing with the square of a line, one uses : (1) Similar triangles having this line in common, or, (2) A right triangle containing this line as a part. ORIGINAL EXERCISES (THEOREMS) 1. In any right triangle the product of the hypotenuse and the altitude upon it is equal to the product of the legs. 2. If two circles intersect at A and J5, and AC and AD are drawn, each a tangent to one circle and a chord of the other, the common chord AB is a mean proportional between BC and BD. 3. If AB is a diameter and BC a, tangent, and A C meets the circumference at D, the diameter is a mean proportional between AC and AD. ORIGINAL EXERCISES 181 4. If two circles are tangent externally, the chords formed by a straight line drawn through their point of contact have the same ratio as the diameters of the circles. 6. If a tangent is drawn from one extremity of a diameter, meeting secants from the other extremity, these secants and their internal segments are recip- rocally proportional. To Prove: AC : AD = AS : AR. Proof : Draw RS. In&ARS and A CD, ^A=ZAai\ (Explain.) Etc. 6. If AB is a chord and CE, another chord, drawn from C, the midpoint of arc AB, meeting chord AB at D, A C is a mean proportional between CD and CE. Prove the above theorem and deduce that, CE CD is constant for all positions of the point E on arc AEB 7. If chord AD is drawn from vertex A of in- scribed isosceles triangle ABC, cutting BC at E, AB is a mean proportional between AD' and AE. Prove the above theorem and deduce that, AD AE is constant for all positions of the point D on arc BDC. 8. If a square is inscribed in a right triangle so that one vertex is on each leg of the triangle and the other two vertices on the hypotenuse, the side of the square is a mean proportional between the other segments of the hypotenuse. To Prove : AD : DE = DE : EB. First prove A ADG and BEF similar. 9. If the sides of two unequal triangles are respectively parallel, the lines joining homologous vertices meet in a point. (These lines to be produced if necessary.) 10. AB is any chord; AC is a tangent and CDE is a secant parallel to AB cutting the circle at D and E. Prove that A C : AE = DC : BE. 11. Prove theorem of 320, by drawing two other auxiliary lines. 12. Prove theorem of 316 if point is between the parallels. 13. Prove theorem of 327 by drawing auxiliary lines A Y and BX. 182 BOOK III. PLANE GEOMETRY 14. If one leg of a right triangle is double the other, its projection upon the hypotenuse is four times the projection of the other. Proof: (2 a) 2 = cp-, a?=cp< ; p'=^L. (AX. 3). C C 15. If the bisector of an angle of a triangle bisects the opposite side, the triangle is isosceles. 16. The tangents to two intersecting circles from any point in their common chord produced are equal. [Use 324.] 17. If two circles intersect, their common chord, produced, bisects their common tangents. 18. If AB and A C are tangents to a circle from A ; CD is perpendicular to diameter BOX from C; then AB CD = BD BO. 19. If the altitude of an equilateral tri- angle is h, find the side. X" 20. If one side of a triangle is divided by a point into segments which are proportional to the other sides, a line from this point to the opposite angle bisects that angle. To Prove : Z n = Z m in figure of 297. Proof: Produce CB to P, making BP - AB; draw AP; etc. 21. State and prove the converse of 298. 22. Two rhombuses are similar if an angle of one is equal to an angle of the other. 23. If two circles are tangent internally and any two chords of the greater are drawn from their point of contact, they are divided propor- tionally by "the less circle. [Draw diameter to point of contact and prove the right & similar.] 24. The non-parallel sides of a trapezoid and the line joining the mid- points of the bases, if produced, meet at a point. [Use Ax. 3 and 316.] 25. The diagonals of a trapezoid and the line joining the midpoint of the bases meet at a point. 26. If one chord bisects another, either segment of the latter is a mean proportional between the segments of the other. 27. Two parallelograms are similar if they have an angle of the one equal to an angle of the other and the including sides proportional. ORIGINAL EXERCISES 183 28. Two rectangles are similar if two adjoining pairs of homologous sides are proportional. 29. If two circles are tangent externally, the common exterior tangent is a mean proportional between the diameters. [Draw chords PA, PC, PB, PD. Prove Z APE a rt, Z. Then prove APD and BPC straight lines. Then prove A ABC and ABD similar.] 30. In any rhombus the sum of the squares of the diagonals is equal to the square of half the perimeter. 31. If in an angle a series of parallel lines are drawn having their ends in the sides of the angle, their midpoints lie in one straight line. 32. If ABC is an isosceles triangle and BX is the altitude upon AC (one of the legs), BC 2 = 2 A C CX. [Use 337.] 33. In an isosceles triangle the square of one leg is equal to the square of the line drawn from the vertex to any point of the base, plus the product of the seg- ments of the base. Proof: Circumscribe a O; use method of 329. 34. If a line is drawn in a trapezoid parallel to the bases, the segments between the diagonals and the non-parallel sides are equal. Proof: & AH I and ABC are similar (?) ; & DKJ and DCB also. Efcc - 35. A line through the point of intersection of the diagonals of a trapezoid, and parallel to the bases, is bisected by that point. 36. If M is the midpoint of hypotenuse AB of right triangle ABC, AB 2 + BC 2 + AC* = 8 CM 2 . 37. The squares of the legs of a right triangle have the same ratio as their projections upon the hypotenuse. 184 BOOK III. PLANE GEOMETRY 38. If the diagonals of a quadrilateral are perpendicular to each other, the sum of the squares of one pair of opposite sides is equal to the sum of the squares of the other pair. 39. The sum of the squares of the four sides of a parallelogram is equal to the sum of the squares of the diagonals. [Use 338, I.] 40. If DE is drawn parallel to the hypotenuse AB of right triangle ABC, meeting AC at D and CB at E, AE* + BE? = AB 2 + DE 2 . [Use 4 rt. & having vertex C.~\ 41. If between two parallel tangents a third tangent is drawn, the radius of the circle is a mean proportional between the segments of the third tangent. To Prove : BP : OP = OP : PD. Proof: A BOD is a rt A (?). Etc. B 42. If ABCD is a parallelogram, BD a / diagonal, A G any line from A meeting BD at / E, CD at F, and EC (produced) at G, AE is a mean proportional between EF and EG. ^ Proof: A ABE and EDF are similar (?) ; also & ADE and BEG (?). Obtain two ratios equal to BE : ED and then apply Ax. 1. 43. An interior common tangent of two circles divides the line join- ing their centers into segments proportional to the radii. 44. An exterior common tangent of two circles divides the line join- ing their centers (externally) into segments proportional to the radii. 46. The common internal tangents of two circles and the common external tangents meet on the line deter- mined by the centers of the circles. 46. If from the midpoint P, of an arc subtended by a given chord, chords are drawn cutting the given chord, the product of each whole chord from P and its segment adjacent to P is constant. Proof: Take two such chords, PA and PC; draw diameter PX; etc. Rt. & PST and PCX are similar. (Explain.) 47. If from any point within a triangle ABC, perpendiculars to the sides are drawn OR to A B, OS to BC, OT to A C, At? + BS* + CT 2 = BR 2 + CS 2 + AT 2 . [Draw OA, OB, OC.] ORIGINAL EXERCISES 185 APT and PBS 48. If two chords intersect within a circle and at right angles, the sum of the squares of their four seg- ments equals the square of the diameter. To Prove : AP* + BP 2 + CP 2 + DP 2 = AR 2 . Proof: Draw EC, AD, RD. Chord BR is _L to AB (240). .-. CD is II to BR (?) .-. arc EC = arc RD .-. chord EC = chord RD (?). Also A ARD is rt. A Now AR 2 = AD* + DR 2 But AD 2 = AP 2 + PD 2 and DR 2 = etc. 49. The perpendicular from v any point flf an arc upon its chord is a mean proportional between the perpendiculars from the same point to the tangents at the ends of the chord. To Prove : PR : PT = PT : PS. Proof: Prove & ARP and BTP similar; also Thus, get two ratios each = to PA : PB. 50. If each of three circles intersects the other two, the three common chords meet in a point. To Prove: AB, LM, RS meet at 0. Proof: Suppose AB and LM meet at O. Draw R O and produce it to meet the at X and X'. Prove OX=OX' (by 320). .-. X, X', S are coincident. 61. In an inscribed quadrilateral the sum of the products of the two pairs of opposite sides is equal to the product of the diagonals. Proof : Draw DX making Z CDX = ZADB; &ADB and CDX are sim. (?) ; also &BCD and ADX (?). Hence AB DC = DB - XC (?), Also AD EC = DB . AX (?). Adding, etc. 52. If AB is a diameter, EC and AD tangents, meeting chords AF and BF (produced) at C and D respectively, AB is a mean proportional between the tangents EC and AD. 53. If from a point A on the circumference of a circle two chords are drawn and a line parallel to the tangent at A meet them, the chords and their segments nearer to A are inversely proportional. ROBBINS'S NEW PLANE GEOM. 13 186 BOOK III. PLANE GEOMETRY CONSTRUCTION PROBLEMS PROPOSITION XLII. PROBLEM 342. To find a fourth proportional to three given lines. I ? v B oxK Given : Three lines , b, c. R \ a \ c \ nz Required : To find a fourth propor- A- s w tional to a, 5, c. Construction: Take two indefinite lines, AB and AC, meeting at A. On AB take AE = to a, RV= to b. On AC take AS = to c. Draw ES. From v draw VW II to BS, meeting AC at W. Statement: SJF is the fourth proportional required. Q.E.F. Proof: In AAVW, RS is II to VW (Const.). .-. a: b = c: SW (294). Q.E.D. PROPOSITION XLIII. PROBLEM 343. To find a third proportional to two given lines. Given: (?). Required: (?). Construction : . Like that for 342. ^( Statement: (?). Proof: (?) A"" ::: T - : - - ........ :C PROPOSITION XLIV. PROBLEM 344. To divide a given line into segments proportional to any number of given lines. A t H G B Given: AB; a, b, c, d. Required : To divide AB into parts a - b -,(.. d which shall be proportional to a, 6, c _ o, d. * CONSTRUCTION PROBLEMS 187 Construction: Draw AX oblique to AB from A. On AX take AC = to #, CD = to 6, DE to c, J.F = to d. Draw FB also through ", D, and (7, lines II to FB, as #, DH, and CI. Statement: Ai, IH, HG, GB are the required parts. Q.E.F. Proof : Ai: a = IH: b = HG : c = GB : d (295). Q.E.D. PROPOSITION XLV. PROBLEM 345. To construct a triangle similar to a given triangle and having a given side homologous to a side of the given triangle. Given : A ABC and RS c yv i ' homologous to ^4#. /\. Required : To con- struct a A on RS similar Construction: At R construct ^SRX=toZAi at s con- struct Z RST= to Z J5, the sides of these angles meeting at T. Statement: (?). Proof: (?). (303). PROPOSITION XLVI. PROBLEM 346. To construct a polygon similar to a given polygon and having a given side homologous to a side of the given polygon. Given: Polygon EB\ line A'B' homologous to AB. Required: To con- struct a polygon upon A*!*', similar to poly- gon EB. A B Construction: From A draw diagonals AC and AD. On A'B f construct A A'B'C' similar to A ABC (by 345). On A r B' construct A A'C'D' similar to A ACD. Etc. Statement: (?). Proof: (?). (319). 188 BOOK III. PLANE GEOMETRY PROPOSITION XLVII. PROBLEM 347 . To find the mean proportional D , between two given lines. / t Given : Lines a and b. Required: To find the mean pro- A 1 :"" A" portional between them. b Construction : On an indefinite line, AX, take AB = to a and BC = to b. Using o, the midpoint of AC, as center, and AO as radius, describe the semicircle, ADC. At B erect BD _L to AC, meeting the arc at D. Draw AD and CD. Statement : BD, or m, is the mean proportional required. Q.E.F. Proof: a:m = m:b. (?) Q.E.D. 348. A line is divided into extreme and mean ratio if one segment is a mean proportional between the whole line and the other segment ; in other words, if a line is to one of its parts as that part is to the other part. (See 292.) PROPOSITION XLVIII. PROBLEM ...?... 349. To divide a line into ex- treme and mean ratio. / ... Given: Line AB = a. \ ...--'" Required: To divide AB into ...-*'" \\ extreme and mean ratio; that is, - F ' a _ x ~ B so that AB : AF = AF: FB. a Construction : At B erect BR, -L to AB and = to AB. Using C, the midpoint of BR, as center, and CB as radius, describe a O. Draw AC meeting O at D and E. On AB take AF = AD ; let AF= x. Statement : F divides AB so that AB : AF AF: FB. Q.E.F. Proof: AB is tangent to O C (202). .-. AE- AD=AB 2 (324). CONSTRUCTION PROBLEMS 189 Now AD = x (187) and DE=a (190). .-. AE= a + x (Ax. 4). Substituting, (a +x)x = a 2 (Ax. 6). Or ax + x 2 = a 2 .-. a :x = x:a-x (281). That is, AB : AF=AF:FB. Q.E.D. PROPOSITION XLIX. PROBLEM. 350. To divide a ...8... line externally into extreme and mean / ratio. '\ f *- Given: (?). ,""*\ Required: (?). F"" A B % Construction : The same as in 349, except that AF r is taken on BA produced, = to AE. Statement : AB : AF' = AF r : BF'. Q. E. F. Proof: AB is tangent to O C (202). . . AE : AB = AB : AD (325). . '. AE + AB : AE = AB + AD : AB (284). Now AE+AB = BF f (Ax. 4). Also AB + AD = AE=AF' (Const.). Substituting, BF 1 : AF 1 = AF f : AB (Ax. 6). That is, AB:AF f = AF r : BF f . Q.E.D. 351. The lengths of the several lines of 349 and 350 may be found by algebra, if the length of AB is known. Thus if AB = a, we know in 349, a : x = x : a x. Hence x 2 = a 2 ax. Solving this quadratic, x = AF=^ a(V5 1) ; also, a x = BF= % a(3 V5). Likewise in 350, if AB = a, AF' = y, a\y = y : a -f- y. Solving for y,y = AF' = | a( V5 + 1). Also a + y = BF' = J a(3 + V5). 190 BOOK III. PLANE GEOMETKY ORIGINAL CONSTRUCTIONS It is required : 1. To construct a fourth proportional to lines that are exactly 3 in., 5 in., and 6 in. long. How long should this constructed line be? 2. To construct a mean proportional between lines that are exactly 4 in. and 9 in. How long should this constructed line be? 3. To construct a fourth proportional to three lines 5 in., 8 in., and 10 in. will this be the same length as a fourth proportional to 5 in., 10 in., and 8 in.? to 8 in., 10 in., and 5 in. ? to 10 in., 5 in., and 8 in. ? 4. To construct a third proportional to lines 3 in. and 6 in. long. 5. To produce a given line AB to point P, such that AB : AP =3:5. [Divide AB into three equal parts, etc.] 6. To divide a line 8 in. long into two parts in the ratio of 5:7. [Divide the given line into 12 equal parts.] 7. To solve Ex. 6 by constructing a triangle. [See 297.] 8. To divide one side of a triangle into segments proportional to the other two sides. 9. To divide one side of a triangle externally into segments propor- tional to the other sides. 10. To construct two straight lines having given their sum. and ratio. [Consult Ex. 6.] 11. To construct two straight lines having given their difference and ratio. [Consult Ex. 5.] 12. To construct a triangle similar to a given triangle and having a given perimeter. [First, use 344.] 13. To construct a right triangle having given its perimeter and an acute angle. [Constr. a rt. A having the given acute Z. Etc.] 14. To construct a triangle having given its perimeter and two angles. [Constr. a A having the two given A. Etc.] 15. To construct a triangle similar to a given triangle and having a given altitude. 16. To construct a rectangle similar to a given rectangle and having a given base. 17. To construct a rectangle similar to a given rectangle and having a given perimeter. 18. To construct a parallelogram similar to a given parallelogram and having a given base. ORIGINAL CONSTRUCTIONS 191 19. To construct a parallelogram similar to a given parallelogram and having a given perimeter. 20. To construct a parallelogram similar to a given parallelogram, and having a given altitude. 21. To draw through a given point another line, which is termi- nated by the outer two of three lines meet- ing in a point and bisected by the inner one. Construction: From E on BD draw Us. Etc. Through P draw RT II to GF. Statement: RS = ST. T C 22. To inscribe in a given circle a triangle similar to a given triangle. Construction: Circumscribe a O about the given A; draw radii to the vertices ; at center of given O construct 3 A = to the other central angles. 23. To circumscribe about a given circle a triangle similar to a given triangle. Construction : First, inscribe a A similar to the given A. 24. To construct a right triangle, having given one leg and its projection upon the hypotenuse. 25. To inscribe a square in a given semicircle. Construction: At B erect BD JL to AB and = to AB-, draw DC, meeting O at R ; draw R U II to BD. Etc. Statement: (?). Proof: RSTU is a rectangle (?). A CRU is similar to A CDB (?). /. CU : CB = UR:BD (?). But CB = \BD (?). :.CU=%UR (?). Etc. 26. To inscribe in a given semicircle a rectangle similar to a given rectangle. Construction : From the midpoint of the base draw line to one of the opposite vertices. At given center construct an Z = to the Z at the mid- point. Proceed as in Ex. 25. 27. To inscribe a square in a given triangle. Construction : Draw altitude A D ; construct the square ADEF upon AD as a side; draw BF meeting A C at 72. Draw RU II to AD ; RS II to BC. Etc. 192 BOOK III. PLANE GEOMETRY 28. To inscribe in a given triangle a rectangle similar to a given rectangle. Construction : Draw the altitude. On this construct a rectangle similar to the given rectangle. Proceed as in Ex. 27. 29. To construct a circle which shall pass through two given points and touch a given line. Given : Points A and B ; line CD. f* . P R Construction: Draw line AB meeting CD at P. Construct a mean proportional between PA and PR (by 347). On PD take PR = to this mean. Erect OR _L to CD at R, meeting _L bisector of AB at 0. Use as center, etc. 30. To construct a line = to V2 in. [Diag. of square the side of which is 1 in.] 31. To construct a line = to A/5 in. [Hyp. of a rt. A, whose legs are 1 in. and 2 in. respectively.] 32. To divide a line into segments in the ratio of 1 : A/2. 33. To divide a line into segments in the ratio of 1 : A/5. 34. To construct a line x, if x = , and a, 6, c are lengths of three c given lines. [That is, to construct x, if c : a = b : x (281) .] 35. To construct a line x, if x = . [3 c : a = b : a:.] 3c 36. To construct a line x, if x = Vab. [a : x = x : &.] 37. To construct a line x, if x = . c 38. To construct a line a;, if x = Va 2 6 2 . [a + b : x x : a &.] 39. To construct a line x, if x = . c 40. To construct a line y, if ay = f 6 2 . . 41. To construct a line = to A/10 in. 42. To construct a line = to 2 A/6 in. 43. To construct a line = to Va 2 + 6 2 , if a and & are given lines. BOOK IV AREAS OF POLYGONS THEOREMS AND DEMONSTRATIONS 352. The unit of surface is a square each side of which is a unit of length. 353. The area of a surface is the number of units of surface it contains. The area of a surface is the ratio of that surface to the unit of surface. UNIT OF LENGTH NOTE. It is often convenient to speak of "triangle," "rectangle," etc., when one really means "the area of a triangle," or "the area of a rectangle," etc. PROPOSITION I. THEOREM 354. If two rectangles have equal altitudes, they are to each other as their bases. Given: Rectangles AC and EG having equal altitudes, with bases AB and EF. To Prove : AC:EG = ABiEF. Proof: I. ABE F If AB and EF are commensurable. There is a common unit of measure of AB and EF (225). Suppose this unit is contained 3 times in AB and 5 times in EF. Hence AB:EF=3:5 (Ax. 3). Draw lines through these points of division J_ to the bases. These divide rectangle AC into three' parts and EG into 5 parts, and all of these eight parts are equal (134). Hence A C : EG = 3 : 5 (Ax. 3). .'. AC:EG = AB'.EF (Ax. 1). Q.E.D. 193 194 BOOK IV. PLANE GEOMETRY II. If AB and EF are incom- mensurable. There does not exist a com- mon unit (225). Divide AB into several equal parts. Apply one of these as a unit of measure to EF. There is a remainder, JtF Draw RS _L to EF. C H SG AB (Hyp.)- (Case I). Now = ES ER Indefinitely increase the number of equal parts of AB ; that is, indefinitely decrease each part, or the unit or di- visor. Then the remainder, RF, is indefinitely decreased. That is, RF approaches zero as a limit, Also RFGS approaches zero as a limit. Hence ER approaches EF as a limit, Also ES approaches EG as a limit. A C 1 AC* Therefore approaches as a limit, ES EG A 7? A 7? Also - approaches as a limit. ER EF (229). Q.E.D. AC _AB ' EG ~~ EF 355. COROLLARY. Two rectangles having equal bases are to each other as their altitudes. (Explain.) PROPOSITION II. THEOREM 356. Any two rectangles are to each other as the products of their bases by their altitudes. 1 X B A 2 i d AREAS 195 Given: Rectangles A and B the altitudes of which are a and c, and the bases b and c?, respectively. To Prove : A-.B = a-b:c- d. Proof : Construct a third rectangle X, whose base is b and whose altitude is c. Then Also Multiplying, - = - B d A = a^b B c a (355). (354). (Ax. 3). Q.E.D. PROPOSITION III. THEOREM 357. The area of a rectangle is equal to the product of its base by its altitude. Given: Rectangle #, with i! U ! base b and altitude h. u To Prove : Area of R = b h. Proof: Draw a square tr, each side of which is a unit of length. This square is a unit of surface (352). Now |=^4 = 6. A (356). But - = the area of R U the area of R = b h 358. COROLLARY. square of its side. (353). (Ax. 1). Q.E.D. The area of a square is equal to the (357.) . Ex. I have enough material to build 1000 yards of fence. If I put this around a square field, how many square yards will the field contain ? If I put it around a rectangular field that is four times as long as it is wide, how many square yards will the field contain ? 196 BOOK IV. PLANE GEOMETRY PROPOSITION IV. THEOREM 359. The area of a parallelogram is equal to the product of its base by its altitude. F D EC Given: HJ ABCD, with base b and altitude h. To Prove : Area of ABCD b h. A B Proof: From A and B, the extremities of the base, draw Js to the upper base meeting it in F and E respectively. In rt. A ADF and BCE, AF=BE (124). AD=BC (124). . . A ADF is congruent to A BCE (84). Now from the whole figure subtract A ADF and the parallelogram ABCD remains. And from the whole figure subtract A BCE and rectangle ABEF remains. .*. O ABCD = rectangle ABEF (Ax. 2). But rectangle ABEF = b h (357). . . H ABCD = b - h (Ax. 1). Q.E.D. 360. COROLLARY. All parallelograms having equal bases and equal altitudes are equal in area. 361. COROLLARY. Two parallelograms having equal altitudes are to each other as their bases. Proof : P = b - h and P 1 = b' . h (359). Dividing, p=^j=4 ( Ax ' 3 > Q.E.D. 362. COROLLARY. Two parallelograms having equal bases are to each other as their altitudes. Proof : (?). 363. COROLLARY. Any two parallelograms are to each other as the products of their bases by their altitudes. Proof : (?). AREAS 197 PROPOSITION V. THEOREM 364. The area of a triangle is equal to half the product of its base by its altitude. Given: A^i?C,with base b and alti- tude h. To Prove : Area of A ABC = | b h. Proof : Through A draw AR II to BC and through C draw CR || to AB, meeting AR at R. Now ABCR is a O (120). The area of O ^CE = b - h (359). Dividing by 2, | O ^tfOR = | b A (Ax. 3). Also EJABCR = AABC (126). .*. the area of A ^ijBC = ^ 6 ^ (Ax. 1). Q.E.D. 365. COROLLARY. A triangle having the same base and alti- tude as a parallelogram equals half the parallelogram. 366. COROLLARY. All triangles having equal bases and equal altitudes are equal in area. 367. COROLLARY. All triangles having the same base and whose vertices are in a line parallel to the base are equal. 368. COROLLARY. Two triangles having equal altitudes are to each other as their bases. Proof: Ar=i& . h ; andAr' = J&'A (364). 369. COROLLARY. Two triangles having equal bases are to each other as their altitudes. Proof : (?). 370. COROLLARY. Any two triangles are to each other as the products of their bases by their altitudes. Proof : (?). 371. COROLLARY. The area of a right triangle is equal to half the product of the legs. Proof : (?). 198 BOOK IV. PLANE GEOMETEY PROPOSITION VI. THEOREM 372. The area of a trapezoid is equal to half the product of the altitude by the sum of the bases. Given: Trapezoid ABCD, with alti- tude h and bases b and c. To Prove : Area = \ h (6 + 0- A Proof: Draw diagonal AC. The A ABC and A DC have the same altitude, 7i, and their bases are b and A DEF DE 2 ^> F 2 EF * Proof : Denote a pair of homologous altitudes by h and h 1 ', and the corresponding bases by & and &'. AABC_ b - h _b h_ ~v'n Now But Substituting, That is, A DEF b' - h' *L = L h'~b' AABC = b b_ ADEF~b' b 1 b* or-So*. ADEF EF * D E* (311). (Ax. 6). Q.E.D. DF PROPOSITION IX. THEOREM 376. Two similar polygons are to each other as the squares of any two homologous sides and as the squares of their perimeters. Given: Similar polygons ABCDE and A'B'C'D'E', with per- imeters P and P f respectively. AREAS 203 TV D-r x "Ayu" ABCDE A!? -, P 2 To Prove : - ^ = = = etc., and = Polygon A'B'C'D'E' ~A r B r P 12 Proof : Draw from homologous vertices, A and A', all the pairs of homologous diagonals, dividing the polygons into A. These A are similar in pairs. (318). An A T?2 "> ^iJj xrrrr-\ A S CJ? ^011^. (313). (287). (Ax. 1). (291). (Ax. 6). (Above). (Ax. 6). Q.E.D. (317). (287). (Ax. 1). Q.E.D. AS' C'D'* A T Z>2? A T' D< E >* g j. AB BC CD DE A'B' B'C' C'D' D'E' A T> T> n fiT\ T\ I/ -/1-O t>(J \jU JJJL A R AS AT ' A R' As' AT' t AR + AS + AT AR ' A R' + A s' + A T' AR f Substituting Polygon ABCDE A R Polygon A'B'C'D'E' AR' But AR __AB 2 Polygon ABCDE AB 2 Polygon A'B'C'D'E' J/^ 2 Also P : P' = AB : A' B f = etc. . -. P 2 : P' 2 = AB* : A' B r * = etc. Polygon ABCDE P 2 ' Polygon A'B'C'D'E' p' 2 204 BOOK IV. PLANE GEOMETRY PROPOSITION X. THEOREM 377. The square described upon the hypotenuse of a right triangle is equal in area to the sum of the squares described upon the legs. Given: (?). To Prove: (?). L D Proof: Draw CL J_ to AB, meeting AB at K and ED at L. Draw BF and CE. Now A ACS, ACG, and BCR are all rt.A (Hyp.) Hence ACH and BCG are straight lines. (45.) Also AELK and BDLK are rectangles. (Def.) In A ABF and ACE, AB = AE, AF = AC (122). Z BAF= Z CAE (Each is composed of a rt. Z plus Z CAB). .-. AABF^AACE (52). Also A ABF and square AG have the same base, AF, and the same altitude, AC. . -. square AG = 2 A ABF (365). Similarly, rectangle AKLE = 2 A ACE (365). .'. rectangle AKLE = square AG (Ax. 1). By drawing AI and CD, it may be proved in the same manner that rectangle BDLK = square BH. By adding, square AD = square AG -f- square BH (Ax. 2). Q.E.D. AEEAS 205 378. COROLLARY. The square described upon one of the legs of a right triangle is equal in area to the square described upon the hypotenuse minus the square described upon the other leg. PROPOSITION XI. THEOREM 379. If the three sides of aright triangle are the homologous sides of three similar polygons, the polygon described upon the hypotenuse is equal in area to the sum of the two polygons described upon the legs. Given: (?). Proof : And To Prove: (?). S Z? AB Adding, &+ T = AC 2 + BC 2 ^_AB 2 = ^ * AB 2 ~ AB*~~ Clearing of fractions, B= 8 + T (Ax. 3). Q.E.D. 380. COROLLARY. If the three sides of a right triangle are the homologous sides of three similar polygons, the polygon described upon one of the legs is equal in area to the polygon described upon the hypotenuse minus the polygon described upon the other leg. 381. COROLLARY. The two squares described upon the legs of a right triangle are to each other as the projections of the legs upon the hypotenuse. A Proof : Square S = A c 2 Square T = i?c 2 Square S = AC 2 = AB AP _ AP Square T J^ 2 AB - BP BP (Ax. 3; 333). 206 BOOK IV. PLANE GEOMETRY 382. COROLLARY. If two similar poly- gons are described upon the legs of a right triangle as homologous sides, they are to each other as the projections of the legs upon the hypotenuse. Polygon 8 _ A(? _AB Polygon T Be 2 AB Proof: = (376; 333). BP BP ^ J Ex. 1. In the figure of 377, prove that : (t) Points /, C, and F are in a straight line. (ii) CE and BF are perpendicular. (iii) A G and BH are parallel. (iv) &AEF=A CGH =-- A BDI = &ABC. Ex. 2. The sum of the squares described upon the four segments of two perpendicular chords in a circle is equal to the square described upon the diameter. (Fig. is on page 185.) Ex. 3. The square described upon the sum of two lines is equal to the sum of the squares described upon the two lines, plus twice the rectangle of these lines. To Prove : Square AE = m* + n 2 + 2 mn. Ex. 4. The square described upon the difference of H two lines is equal to the sum of the squares described upon the two lines minus twice the rectangle of these G lines. To Prove : Square AD = m* + n 2 - 2 mn. Ex. 5. A and B are the extremities of a diameter ^ of a circle; C and D are the points of intersection of any third tangent to this circle with the tangents at A arid B respectively. Prove that the area of ABDC is equal to \AB CD. Ex. 6. If the four points midway between the center and vertices of a parallelogram are joined in order, a parallelogram is formed similar to the original parallelogram ; its perimeter is half of the perimeter of the original figure ; and its area is one quarter of the area of the original figure. R- n n C n m-n m-n m-n m-n m-n n 1 5 n n m FORMULAS 207 Ex. 7. If two triangles of equal area have the same base and lie on opposite sides of it, the line joining their vertices is bisected by the line of the base. Ex. 8. What part of a right triangle is the quadrilateral which is cut from the triangle by a line joining the midpoints of the legs ? Ex. 9. From M, a vertex of parallelogram LMNO, a line MPX is drawn meeting NO at P and LO produced, at X. LP and NX are also drawn. Prove that triangles LOP and XNP are equal in area. FORMULAS PROPOSITION XII. PROBLEM 383. To derive the formulas for the altitude and the area of an equilateral triangle, in terms of its side. Solution: Let each side = a, and alti- tude = h. h 2 = a 2_^ = 3as 4 ' 4 |/> Now Also CD= b p a = i & AD 2 = AC? - //2 [ ^2 /?^V h * = b *-( 2^ J B (339). (335). (Ax. 6). 208 BOOK IV. PLANE GEOMETRY Hence W= h+*=* Hi- za ) I Xa Alsc fl^aft+ga + ff-cg 2a6-a 2 -6 2 + (by factoring). 4 a 2 Then it is evident that a+ b ) is diminished by x so that the area remains unchanged. Find x. 28. The projections of the legs of a right triangle upon the hypote- nuse are 8 and 18. Find the area of the triangle. 29. In triangle A B C, AB is 5, B C is 8, and AB is produced to P, mak- ing BP = 6. BC is produced (through B) to Q and PQ is drawn so that triangle BPQ is equal in area to triangle ABC. Find the length of BQ. 212 BOOK IV. PLANE GEOMETRY 30. The angle C of triangle ABC is right; AC = 5; BC = 12. BA is produced through A, to D making AD = 4; CA is produced through ^4, to E so that triangle A ED is equal in area to triangle ABC. Find AE. 31. Find the area of a square inscribed in a circle whose radius is 6. 32. Find the side of an equilateral triangle whose area is 25 V3. 33. Two sides of a triangle are 12 and 18. What is the ratio of the two triangles formed by the bisector of the angle between these sides ? 34. The perimeter of a rectangle is 28 m. and one side is 5 m. Find the area. 35. The perimeter of a polygon is 5 ft. and the radius of the inscribed circle is 5 in. Find the area of the polygon. In the following triangles, find the area, the three altitudes, the radius of the inscribed circle, the radius of circumscribed circle : 36. a = 13, b = 14, c = 15. 38. 20, 37, 51. 37. a = 15, b = 41, c = 52. 39. 140, 143, 157. 40. The sides of a triangle are 15, 41, 52. Find the areas of the two triangles into which this triangle is divided by the bisector of the largest angle. 41. Find the area of the quadrilateral ABCD if AE = 78 m., BC = 104 m., CD = 50 m., AD = 120 m., and AC = 130 m. 42. One diagonal of a rhombus is T V of the other and the difference of the diagonals is 14. Find the area and the perimeter of the rhombus. 43. A trapezoid is composed of a rhombus and an equilateral triangle. Each side of each figure is 16 inches. Find the area of the trapezoid. 44. Find the side of an equilateral triangle equal in area to the square whose diagonal is 15\/2. 46. Which of the figures in Ex. 44 has the smaller perimeter? 46. In a triangle whose base is 20 and whose altitude is 12, a line is drawn parallel to the base, bisecting the area of the triangle. Find the distance from the base to this parallel. 47. Two lines are drawn parallel to the base of a triangle whose base is 30 and altitude 18. These lines divide the area of the triangle into three equal parts. Find their distances from the vertex. 48. Around a rectangular lawn 30 yards x 20 yards is a drive 16 feet wide. How many square yards are there in the drive ? CONSTRUCTION PROBLEMS 213 CONSTRUCTION PROBLEMS PROPOSITION XVII. PROBLEM 388. To construct a square equal to the sum of two squares. Given : (?). Required : (?). Construction : Construct a rt. Z.E, with sides EX and EY. On EX take EF = to AB ; on ET take EG = to CD. Draw FG. On FG construct square T. Statement : Proof : T = B + s. T = R+ S PROPOSITION XVIII. PROBLEM 389. To construct a square equal to the sum of several squares. Given: Squares whose sides are a, 6, c,d. Required : To construct a square = to Construction : Construct a rt. Z. whose sides are equal to a and b. Draw hypotenuse EC. At B erect a J_ = to c and draw hypotenuse DC. At D erect a _L = to c?, etc. Statement : The square constructed on EC = the sum of the several given squares. Q.E.F. (334). (?>. = a 2 + i 2 + (294). a ^1D ==2 (287). <* 9 But 2^ = square S, and a 2 = square R (358). Substituting, square S : square R = h : g (Ax. 6). Q.E.D. PROPOSITION XXIV. PROBLEM. 400. To construct a polygon similar to a given polygon and having a given ratio to it. .. D Given: (?). Required: (?). Construction and Statement are the same as for Proposition XXIII. Proof: Very much like the proof of Proposition XXIII. 220 BOOK IV. PLANE GEOMETRY PROPOSITION XXV. PROBLEM 401. To construct a polygon similar to one given polygon and equal in area to another. Given: Polygons R and s. Required: (?). Construction : Construct squares R r = R, and s 1 = 8 (by 396). Find a fourth proportional to a, b, and AB. This is CD. Upon CD, homologous to AB, construct T similar to R. Statement: T = 8. Q.E.F. Proof: f = ^ ( 3T6 > - = (Const.). b CD t = (287). * 2 c& -!= (Ax.l). Now a 2 = B' = R ; a 2 = s' = S (358 & Const.). Substituting, -=- (Ax. 6). .-. T=S (Ax. 3). Q.E.D. NOTE. By means of this proposition we are able to maintain the size of a rectilinear figure, but change its shape to any desired form. Thus we can construct an equilateral triangle equal to a given square. The pupil will explain. ORIGINAL CONSTRUCTIONS 221 ORIGINAL CONSTRUCTIONS It is required : 1. To construct a right triangle equal to a given parallelogram. 2. To construct a square equal to the sum of two given parallelograms. 3. To construct a square equal to the difference of two given paral- lelograms. 4. To construct a square equal to the sum of several given right triangles. 6. To construct a square equal to the sum of several given paral- lelograms. 6. To construct a square equal to the sum of several given triangles. 7. To construct a square equal to the sum of several given polygons. 8. To construct a square equal to the difference of two given polygons. 9. To construct a square equal to three times a given square ; seven times a given square. 10. To construct a right triangle equal to the sum of several given triangles. 11. To construct a right triangle equal to the difference of any two given triangles ; of any two given parallelograms. 12. To construct a square equal to a given trapezoid ; equal to a given trapezium. 13. To construct a square equal to a given hexagon. 14. To construct a rectangle equal to a given triangle, having given its perimeter. 15. To construct an isosceles right triangle equal to a given triangle. 16. To construct a square equal to a given rhombus. 17. To construct a rectangle equal to a given trapezium, and having its perimeter given. 18. To find a line whose length shall be \/2 units. [See 388.] 19. To find a line whose length shall be V3 units. 20. To find a line whose length shall be VTi units. 21. To find a line whose length shall be V7 units. 22. To find a line whose length shall be V30 units. 222 BOOK IV. PLANE GEOMETRY 23. To construct a square which shall be f of a given square. 24. To construct a square which shall be f of a given square. 25. To construct a polygon which shall be f of a given polygon, and similar to it. 26. To construct a square which shall have to a given square the ratio V3 : 4 ; the ratio 4 : \/3. 27. To draw through a given point, within a parallelogram, a line which shall bisect the parallelogram. 28. To construct a rectangle equal to a given trapezoid, having given the difference of its base and altitude. 29. To construct a triangle similar to two given similar triangles and equal to their sum. 30. To construct a triangle similar to a given triangle and equal to a given square. [See 401.] 31. To construct a triangle similar to a given triangle and equal to a given parallelogram. 32. To construct a square having twice the area of a given square. [Two methods.] 33. To construct a square having 3| times the area of a given square. 34. To construct an isosceles triangle equal to a given triangle and upon the same base. 35. To construct a triangle equal to a given triangle, having the same base, and also having a given angle adjoining this base. 36. To construct a parallelogram equal to a given parallelogram having the same base and also having a given angle adjoining the base. 37. To draw a line that shall be perpendicular to the bases of a parallelogram and that shall bisect the parallelogram. 38. To construct an equilateral triangle equal to a given triangle. [See 401.] 39. To trisect (divide into three equal parts) the area of a triangle, by lines drawn from one vertex. 40. To construct a square equal to f of a given pentagon. 41. To construct an isosceles trapezoid equal to a given trapezoid. 42. To construct an equilateral triangle equal to the sum of two given equilateral triangles. ORIGINAL CONSTRUCTIONS 223 43. To construct an equilateral triangle equal to the difference of two given equilateral triangles. 44. To construct upon a given base a rectangle that shall be equal to a given rectangle. Analysis : Let us call the unknown altitude x. Then b h = V x (?). Hence, b':b = h:x (?). A b That is, the unknown altitude is a fourth pro- portional to the given base, the base of the given rec- b' tangle, and the altitude of the given rectangle. Construction : Find a fourth proportional, x, to b', b and h. Construct a rectangle having base = b' and alt. = x. Statement: This rectangle, B = A. x B Proof: b f : b = h : x (Const.). .-. b'x= bh (?). p But Vx = the area of B (?), etc. 45. To construct a rectangle that shall have a given altitude and be equal to a given rectangle. 46. To construct a triangle upon a given base that shall be equal to a given triangle. 47. To construct a triangle that shall have a given altitude and be equal to a given triangle. 48. To construct a rectangle that shall have a given base, and shall be equal to a given triangle. 49. To construct a triangle that shall have a given base, and be equal to a given rectangle. 50. To construct a triangle that shall have a given base, and be equal to a given polygon. 51. To construct the problems 45, 46, 47, 48, 49, substituting "parallel- ogram " in each case for the figure to be constructed. 52. To construct upon a given hypotenuse, a right triangle equal to a given triangle. 53. To construct upon a given hypotenuse, a right triangle equal to a given square. 54. To construct (a) a triangle which shall have a given base, a given adjoining angle, and be equal to a given triangle ; (&) a triangle equal to a given 'square ; (c) a triangle equal to a given polygon. 56. To construct (a) a parallelogram which shall have a given base, a given adjoining angle, and be equal to a given parallelogram ; (5) a par- allelogram equal to a given triangle ; (c) a parallelogram equal to a given polygon. 224 BOOK IV. PLANE GEOMETRY 56. To construct a line, DE, from D, a given point in AB of triangle ABC, so that DE bisects the triangle. Analysis: After DE is drawn, A ABC = 2 A ADE (Hyp.). But A ABC : A ADE = AB AC : AD AE (?). Hence, AB AC = 2 (A D - AE) (Ax. 6). .-. 2AD-.AB = AC: x (?). Thus a: (that is, AE) is a fourth proportional to three given lines. 67. To draw a line meeting two sides of a triangle and forming an isosceles triangle equal to the given triangle. Analysis: Suppose A X is a leg of the required isos- celes A. .-. A ABC :&AXX = AB - AC : AX AX'. But the A are equal and A X = AX' (Hyp.). y Hence, AB AC AX 2 . .-. AX is a mean proper-^- \ c tional between AB and A C. x - 58. To draw a line parallel to the base of a triangle which shall bisect the triangle, 59. To draw a line meeting two sides of a triangle forming an isosceles triangle equal to half the given triangle. 60. To draw a line parallel to the base of a triangle forming a tri- angle equal to one third of the original triangle. 61. To draw a line parallel to the base of a trapezoid so that the area is bisected. /' \ Analysis : A OXX' = J (A OAD + A OBC) and is similar to A OB C. 62. To construct two lines parallel to the base of a triangle, that shall trisect the area of the triangle. 63. To construct a triangle, having given its angles and its area. Analysis : The required A is similar to any A containing the given A. The given area may be a square. This reduces the problem to 401. 64. To find two straight lines in the ratio of two given polygons. BOOK V REGULAR POLYGONS. CIRCLES THEOREMS AND DEMONSTRATIONS 402. A regular polygon is a polygon which is equilateral and equiangular. PROPOSITION I. THEOREM 403. An equilateral polygon inscribed in a circle is regular. Given: AG, an equilateral inscribed polygon. To Prove: AG is regular. K^=:_ j Proof : chord AB = chord EC = chord CD = etc. (Hyp.). .'. arc AB = arc BC= arc CD = etc. (196). .'. arc AC = arc BD =.arc CE = etc. (Ax. 3). ..Z^LBC=ZCD = Z CDE=etc. (239). That is, the polygon is equiangular. .'. the polygon is regular (402). Q. E. D. 404. COROLLARY. If the circumference of a circle is divided into any number of equal arcs, and the chords of these arcs are drawn, they form an inscribed polygon. Given : Arc AB = arc BC = arc CD = etc. and chords AB, etc. To Prove : Polygon AG is a, regular polygon. Proof: Chords AB, BC, CD, etc. are all equal. (197). .'. the polygon is regular (403). Q. E. D. 405. COROLLARY. If chords are drawn joining the alternate vertices of an inscribed regular polygon (having an even num- ber of sides), another inscribed regular polygon is formed. 225 226 BOOK V. PLANE GEOMETRY PROPOSITION II. THEOREM 406. If the circumference of a circle is divided into any number of equal parts, and tangents are drawn, at the several points of division, they form a circum- scribed regular polygon. Given: Arcs AB, BC, CD, etc. all equal ; and GH, HI, H, etc. tangents at A, B, C, etc. To Prove : Polygon HK is regular. Proof : Draw chords AB, BC, CD, etc. In A ABH, BCI, CDJ, etc. AB = BC = CD, etc. . (197). Z HAB = Z HBA = Z IBC = Z ICB =Z JCD, etc. .*. these A are congruent and isosceles That is, Also That is, (237). (76; 114). .-. Ztf=Zi=Z Hence approaches 1. That is, approaches 1, or P and P 1 approach equality ; that is, they approach the same constant as a limit. But P>C and C>P f and C is constant. Hence P approaches C and P' approaches C. Q.E.D. II. = (420). If the number of sides of these polygons is indefinitely in- - o creased, off approaches OE 2 , and thus : - approaches unity. OD (The argument continues the same as in I.) BOBBINS' s NEW PLANE GEOM. 16 234 BOOK V. PLANE GEOMETRY NOTE. The theorems of 423 and 424 are considered so evident, and rigorous proofs (as in the case of the demonstrations for many funda- mental principles in mathematics) are so difficult for young students to comprehend, that it is advisable to omit the profound demonstrations and insert only simple explanations. PROPOSITION IX. THEOREM 425. The circumferences of two circles are to each other as their radii. Given : Two whose radii are R and E f and circumfer- ences, C and cf respectively. To Prove: C: C f = R: it'. ^ I >? ?v I .J Proof: Circumscribe regular polygons (having the same number of sides) about these CD and let P and p' denote their perimeters. Then P : P r =R: R r (419). Hence P R ! = P r R (?). Now suppose the number of sides of these polygons to be indefinitely increased, P approaches C (424). p' approaches C f (?). .*. P R f approaches C R r . Also P r - R approaches C f R. Hence C R f = C r - R (229). Therefore C : C r = B : B r (281). Q.E.D. 426. THEOREM. The ratio of any circumference to its diam- eter is constant for all circles. That is, any circumference divided by its diameter is the same as any other circumference divided by its diameter. Proof: =, (425). But CIKCLES 235 That is, .-.- = , D D r circumference = ft congtant for (Ax. 1). (282). Q>E ^ diameter 427. Definition of IT (pi). The constant ratio of a circurn- y-^ ference to its diameter is called TT. That is, = TT. D The numerical value of TT is 3.141592, or 3^, approximately. (This is computed in 453.) 428. FORMULA. Let c = the circumference of a circle with radius 1?. Then -- = TT. (427). (Ax. 3). Ex. 1. Find the circumference of a circle the radius of which is 12 in. Ex. 2. Find the radius of a circle the circumference of which is 66 feet. Historical Note. Gottfried Wilhelm von Leibnitz, a German phi- losopher, mathematician, and man of affairs, was born in 1646 and died in 1716. He could read Latin easily at 12, and wrote some Latin verse, While at the University of Leipsic he became acquainted with Francis Bacon, Kepler, Galileo, and Descartes, modern thinkers who had revolu- tionized science and philosophy. He resolved to study mathematics, but not until he had reached his majority did he throw himself into deep mathemati- cal research. It was while he was living in Paris and Mainz that he announced his imposing discoveries in natural philosophy, mathematics, mechanics, optics, hydrostatics, pneumatics, and nautical science. In mathematics, he was the discoverer of the differ- ential and integral calculus. He possessed a marvel- LEIBNITZ ous ability for rapid and continuous work. Even in traveling his time was employed in solving mathematical problems. He is described as moderate in habits, quick of temper, charitable in judgment of others, tolerant of differences of opinion, but impatient of contradiction on small matters and desirous of honor. 236 BOOK V. PLANE GEOMETRY PROPOSITION X. THEOKEM 429. The area of a circle is equal to half the product of its circumference by its radius. .'**> P Given : O with circumference C, area -S, radius R. To Prove : s = c - R. Proof: Circumscribe a regular poly- gon about the circle ; denote its area by JET and perimeter by P. Now E = P.R (421). Suppose the number of sides of the polygon is indefinitely increased, .HT approaches S, and P approaches C (424). J P R approaches J C R as a limit. Hence s = c - R (229) . Q. E. D. 430. FORMULA. Let -S = the area of a circle whose circum- ference = C, and whose radius = R. Then S=^C-R (429). But c = 2irR (428). Substituting, S = irll* (Ax. 6). Ex. 1. Find the area of a circle the radius of which is 8 in. Ex. 2. Find the radius of a circle the area of which is 500 sq. ft. 431. COROLLARY. The areas of two circles are to each other as the squares of their radii, and as the squares of their di- ameters. To Prove : s : s f = E 2 : fi' 2 = D 2 : D' 2 . Proof : s = 7TB 2 , and s f = TTR'* (430). CIRCLES 237 432. COROLLARY. The area of a sector is the same part of the circle as its central angle is of 360. (Ax. 1.) 433. FORMULA. An arc : circum. = central Z : 360 (231). . . arc : 2 irJB = Z : 360. NOTE. If any two of the three quantities, arc, R, Z, are known, the remaining one can be found by this proportion. 434. FORMULA. Sector : area of O = central Z : 360 (432). .-. sector : IT J 2 = Z: 360. NOTE. If any two of the three quantities, sector, R, Z, are known, the remaining one can be found by this proportion. 435. FORMULA. Sector : area of O = arc : circum. .'. sector : TrR 2 = arc : 27rR . sector = \ R arc 436. Similar arcs, similar sectors, and similar segments are those which corre- spond to equal central angles, in unequal circles. Thus, AB, A r B f , A !I B" are similar arcs ; AOB, A' OB', and A" OB 1 ' are similar sec- tors ; and the shaded segments are simi- lar segments. 437. THEOREM. Similar arcs are to each other as their radii. Given: Arcs whose lengths are a and a', radii E and R f . To Prove : a : a' = E : R r . Proof: (Ax. 1) R Q.E.D. 238 BOOK V. PLANE GEOMETRY 438. THEOREM. Similar sectors are to each other as the squares of their radii. Given : Sectors whose areas are T and r', radii R and JB'.. To Prove : T : T f = R 2 : R' 2 . -y-sS-S < 282 > Q.E.D. PROPOSITION XI. THEOREM 439. Similar segments are to each other as the squares of their radii. o' Given : Similar segments ABC and A'B'C'. ent A'B'C' = R 2 : R f2 . ^^ ftT^m Proof : A AOB and A'O'B' are similar A AOB R 2 (306). (375). Ax. 1). (282). f285\ Also sector sector OACB R 2 sector O'A'C'B' R' 2 sector CMC A AOB ^ ' sector O'A'C'B' ~ A A'O'B' sector OACB sector O'A'C'B' AAOB A A' O'B' OACB A AOB sector O'A'C'B' A ^'o'j?' That is segment ABC _ segment A'B'C' AAOB A A' O'B' segment ABC _ AAOB _ R 2 * segment A'B'C' ~~ A A'O'B' ~~ R' 2 ORIGINAL EXERCISES 239 ORIGINAL EXERCISES (THEOREMS) 1. The central angle of a regular polygon is the supplement of the angle of the polygon. 2. An equiangular polygon inscribed in a circle is regular (if the number of its sides is odd). 3. An equiangular polygon circumscribed about a circle is regular. [Draw radii and apothems.] 4. The sides of a circumscribed regular polygon are bisected at the points of contact. 5. The diagonals of a regular pentagon are equal. 6. The diagonals drawn from any vertex of a regular n-gon divide the angle at that vertex into n 2 equal parts. 7. If a regular polygon is inscribed in a circle, and another regular polygon having the same number of sides is circumscribed about it, the radius of the circle is a mean proportional between the apothem of the inner and the radius of the outer polygon. 8. The area of the square inscribed in a sector the central angle of which is a right angle, is equal to half the square of the radius. [Find x 2 , the area of OEDC.~] 9. The apothem of an equilateral triangle is one third the altitude of the triangle. 10. The chord that bisects a radius of a circle at right angles is the side of the inscribed equilateral triangle. [Prove that the central Z subtended is 120.] 11. If ABODE is a regular pentagon, and diagonals A C and BD are drawn, meeting at : (a) A0 = AB. (l>) AO is II to ED. (c) A BOC is similar to A BDC. (d) ZACB = 36. (e) AC is divided into mean and extreme ratio at O. 12. The altitude of an equilateral triangle is three fourths the diameter of the circumscribed circle. 13. The apothem of an inscribed regular hexagon equals half the side of an inscribed equilateral triangle. 240 BOOK V. PLANE GEOMETRY 14. The area of a circle is four times the area of another circle described upon its radius as a diameter. 15. The area of an inscribed square is half the area of the circum- scribed square. 16. An equilateral polygon circumscribed about a circle is regular (if the number of its sides is odd) . 17. The sum of the circles described upon the legs of a right triangle as diameters is equal to the circle described upon the hypotenuse as a diameter. 18. A circular ring (the area between two con- centric circles) is equal to the circle described upon the chord of the larger circle, which is tangent to the less, as a diameter. Proof : Draw radiiOB, OC. A OBC is rt. A (?) ; and OC 2 - OB 2 = BC 2 (?). Etc. 19. If semicircles are described upon the three sides of a right triangle (on the same side of the hypotenuse), the sum of the two crescents thus formed is equal to the area of the triangle. Proof: Entire figure = \wAJi + crescent BDC + crescent A EC (?) Entire figure = irAC 2 + | TrBC 2 + A AB C (?) Now use Ax. 1 ; etc. 20. Show that the theorem of Ex. 19 is true in the case of a right triangle whose legs are 18 and 24. 21. If from any point in a semicircle a line is drawn perpendicular to the diameter and if semicircles are described on the two segments of the original diameter as diameters, the area of the surface bounded by these three semicircles equals the area of a circle whose diameter is the perpendicular first drawn. Proof: - J,r(f) 2 -i,r(f) 2 = 22. Show that the theorem of Ex. 21" is true in the case of a circle with diameter AB equal to 25 and AD equal to 5. ORIGINAL EXERCISES 241 23. If the sides of a circumscribed regular polygon are tangent to the circle at the vertices of an inscribed regular polygon, each vertex of the outer lies on the prolongation of the apothems of the inner polygon, drawn perpendicular to the several sides. 24. The sum of the perpendiculars drawn from any point within a regular n-gon to the several sides is constant [= n apothem]. 25. The area of a circumscribed equilateral triangle is four times the area of the inscribed equilateral triangle. 26. If a point is taken dividing the diameter of a circle into two parts and circles are described upon these parts as diameters, the sum of the circumferences of these two circles equals the circumference of the original circle. 27. Show that the theorem of Ex. 26 is true in the case of a circle the segments of the diameter of which are 7 and 12. 28. The area of an inscribed regular octagon is equal to the product of the diameter by the side of the inscribed square. 29. If squares are described on the six sides of a regular hexagon (externally), the twelve exterior vertices of these squares are the vertices of a regular 12-gon. 30. If the alternate vertices of a regular hexagon are joined by draw- ing diagonals^ another regular hexagon is formed. Also its area is one third of the original hexagon. 31. Show that the theorem of Ex. 18 is true in the case of two con- centric circles whose radii are 34 and 16. 32. In the same or equal circles two sectors are to each other as their central angles. 33. If the diameter of a circle is 10 in. and a point is taken dividing the diameter into segments with lengths 4 in. and 6 in., and on these segments as diameters semicircles are described on opposite sides of the diameter, these arcs form a curved line which divides the original circle into two parts in the ratio of 2 : 3. 34. If the diameter of a circle is d and a point is taken dividing the diameter into segments with lengths a and d a, and on these segments as diameters semicircles are described on opposite sides of the diameter, these arcs form a curved line which divides the original circle into two parts in the ratio of a : d a. 242 BOOK V. PLANE GEOMETRY CONSTRUCTION PROBLEMS PROPOSITION XII. PROBLEM 440. To inscribe a square in a given circle. Given: The circle o. Required : To inscribe a square. Construction : Draw any diameter, AB, and another diameter, CD, JL to AB. Draw AC, BC, BD, AD. Statement: ACBD is an inscribed square. Q.E.F. Proof : The central A are all equal .-. arcs AC, CB, BD, DA are equal .. ACBD is an inscribed square (42). (193). (404). Q.B.D. PROPOSITION XIII. PROBLEM 441. To inscribe a regular hexagon in a given circle. Given: (?). Required: (?). Construction: Draw any radius, A O. At A, with radius = to AO, describe arc intersecting the given O at B. Draw AB. Statement: AB is the side of an in- scribed regular hexagon. Proof: Draw BO. A ABO is equilateral (Const.). .. A ABO is equiangular (56). .-. Z AOB = 60 (109). That is, arc AB = i of the circumference ; and if arc AB is used as a unit, it divides the circumference into 6 equal arcs. If the chords are drawn, an inscribed regular hexa- gon is formed (404). Q.E.D. CONSTRUCTION PROBLEMS 243 PROPOSITION XIV. PROBLEM 442. To inscribe a regular decagon in a given circle. Given: (?). Required: (?). Construction: Draw any radius AO. Divide it into mean and extreme ratio (by 349), having the larger segment next to the center. Taking A as a center and OB as a radius, draw an arc cutting O at c. Draw AC, EC, OC. Statement : AC is a side of the inscribed regular decagon. Proof: AO:BO = BO:AB Substituting, AO : AC = AC: AB .-. A ABC and AOC are similar .-. First, Z.ACB = ZO Second, A ABC is isosceles .'. AC BC But AC=BO .'. BC = BO .'. z. BCO = Z o Now Z AGO == 2 Z O Q.E.F. (Const.). (Ax. 6). (306). (312). (similar to A AOC). (24). (Const.). (Ax. 1). (55). And Adding, (Ax. 4). (55). Z o = 1 Z o the A of &ACO^~5ZTd (Ax. 2). /. 5Z = 180 (104). .'. Z o = 36 = iV of 360 (Ax. 3). .'. arc AC= -^Q of the circumference (193). That is, if arc AC is used as a unit it divides the circum- ference into ten equal arcs ; and if the chords are drawn, an inscribed regular decagon is formed. (404). Q.E.P. 244 BOOK V. PLANE GEOMETRY PROPOSITION XV. PROBLEM 443. To inscribe a regular i5-gon (pentadecagon) in a given circle. Given: (?). Required: (?). Construction: Draw AB, the side of an inscribed hexagon, and AC, the side of an inscribed decagon. Draw BC. Statement : BC is the side of an inscribed regular 15-gon. Q.E.F. Proof : Arc BC = arc AB arc AC = 1 y 1 ^- of circumference (Const.). = ^ of circumference. That is, if arc BC is used as a unit, it divides the circum- ference into fifteen equal arcs; and if the chords are drawn, an inscribed regular pentadecagon is formed (404). Q.E.D. 444. To inscribe in a given circle: I. A regular 8-gon, a regular i6-gon, a regular 32-gon, etc. II. A regular i2-gon, 24-gon, etc. III. A regular ao-gon, 6o-gon, etc. Construction : I. Inscribe a square. (440.) Bisect the arcs and draw the chords. Proof : (404). II. Inscribe a regular hexagon, etc. III. Inscribe a regular pentadecagon, etc. 445. To inscribe an equilateral triangle in a circle. Construction: Join the alternate vertices of an inscribed regular hexagon. Proof : (?) (405). 446. To inscribe a regular pentagon in a given circle. 447. To circumscribe a regular polygon about a circle. Construction : Inscribe a polygon having the same number of sides. At the several vertices draw tangents. Statement: (?). Proof: (?) (406). CONSTRUCTION PROBLEMS 245 FORMULAS 448. Sides of inscribed polygons. 1. The side of inscribed equilateral triangle = B V3. Proof: Z ACB is a rt. Z (240). AB = 2R (189). BC=R (441). z 2 =(2B) 2 -E 2 (335). x= R V3. Q.E.D. 2. The side of an inscribed square = z?V2. Proof : In fig. of 440 Ic 2 = B 2 4- 2 (334). .'. AC=RV2. Q.E.D. 3. The side of an inscribed regular hexagon = R (441). 4. The side of an inscribed regular decagon = | R ( V5 1) (352; 442). D B 449. Sides of circumscribed poly- gons. 1. The side of a circumscribed equilateral A = 2 R V3. Proof: ^DAB = Z DBA = ^D=60 (?). .-. A ABD is equilateral. AD^AB^B^ (448). .'. DF= 2BV3. Q.E.D. 2. The side of a circumscribed square ~ 2 R (?) 246 BOOK V. PLANE GEOMETRY 3. The side of a circumscribed regular hexagon = f R V3. Proof : This side = a side of an equilateral A with altitude R. Let x = this side. 450. In an equilateral triangle, apothem = J R. Proof : Fig. of 449. Bisect arc AC at H. Draw OA, OC, AH and CH. Figure AOCH is a rhombus (448, 3). ..Otf = *OH=iJS ^ (135). Q.E.D. PROPOSITION XVI. PROBLEM 451. In a circle whose radius is R is inscribed a regular polygon whose side is s ; to find the formula for the side of an inscribed regular polygon having double the number of sides. Given : AB = s, a side of an inscribed regular polygon in O whose radius is R ; C, the midpoint of arc AB ; chord AC. Required: To find the value of AC, the side of a regular polygon having double the number of sides and inscribed in the same circle. Construction : Draw radii OA and OC. Computation : OC bisects AB at right A In rt. A AON, O is an acute Z. Hence in A AOC, A(?=~OA 2 +W?-2>OC' ON (337). (83). But And Substituting, AO = OC = R (187). (335). (Ax. 6). Q.E.F. ON= Vfl 2 -(is) 2 = Ic 2 =2fl 2 -2 J R. JV< |V4 E 2 s 2 * 2 -* 2 .-. AC = \/2 R 2 - R V4 R' J_ j2 p 452. FORMULA. If R = 1, and given side = s, the side of a regular polygon having twice as many sides = \/2 V4 s 2 . CONSTRUCTION PROBLEMS 247 PROPOSITION XVII. PROBLEM 453. To find the approximate numerical value of IT, Given : A circle whose diameter = D ; circumference = C. Required : The value of TT, that is . Method: 1. We may select a O of any diameter. (426.) 2. We can compute the side and the perimeter of some in- scribed regular polygon. (448.) 3. We can compute the side and perimeter of another inscribed regular polygon having double the number of sides. (451.) 4. We can now compute the side and perimeter of a third inscribed regular polygon, having still double the number of sides. (451.) 5. By continuing this process until the consecutive perim- eters differ very slightly, we can find the approximate value of the circumference. /* 6. Thus, knowing both C and D, we know or TT. Computation : 1. For simplicity, take D = 2, and R = 1. 2. We will select the regular hexagon as the first polygon. .*. S 6 = 1, and P 6 = 6. 3. .'. 5 12 = V2 _ V4 - S 6 2 = V2 _ V3 = .5176381 (452). and P 12 = 6.2116572. 4. Hence S 24 = ^2 _ V4 - S 12 2 = V2 _ V4^(751763S1) 2 = .2610524. Also P 24 = 6.2652576. 5. By continuing, S 3072 = .002045. Also P 3072 = 6.283184. 6. It now appears that, approximately, C = 6.283184. Hence TT = 6 ' 288184 = 3.141592+. Q. E.F. 248 BOOK V. PLANE GEOMETRY This calculation is tabulated for reference. * 6 = 1, .-. P 6 = 6. Sw2 = 0.032723, .-. P 192 = 6.282904. s n = 0.517638, .-. P 12 = 6.211657. , 384 = 0.016362, .-. P 884 = 6.283115. * 24 = 0.261052, .-. P 24 = 6.265257. s m = 0.008181, .-. P 768 = 6.283169. S48 = 0.130806, .-. P 48 = 6.278700. 15 , 6 = 0.004091, .-. P 1536 = 6.283180. s 96 = 0.065438, .-. P 96 = 6.282053. s 3072 = 0.002045, .-. P 8072 = 6.283184. ORIGINAL EXERCISES (NUMERICAL) MENSURATION OF REGULAR POLYGONS AND THE CIRCLE 1. Find the angle and the central angle of : () a regular pentagon ; (ii) a regular octagon ; (Hi) a regular do- decagon ; (iv) a regular 20-gon. 2. Find the area of a regular hexagon whose side is 8. 3. Find the area of a regular hexagon whose apothem is 4. 4. In a circle whose radius is 10 are inscribed an equilateral triangle, a square, and a regular hexagon. Find the perimeter, the apothem, and the area of each. 6. About a circle whose radius is 10 are circumscribed an equilateral triangle, a square, and a regular hexagon. Find the perimeter and the area of each. 6. Find the circumference and the area of a circle whose radius is 5 inches. [Use * = 3}.] 7. Find the circumference and the area of a circle whose diameter is 42 centimeters. 8. The radius of a certain circle is 9 meters. What is the radius of a second circle whose circumference is twice as long as the first ? of a third circle whose area is twice as great as the first ? 9. If the circumference of a circle is 55 yards, what is its diameter ? 10. If the area of a circle is 113f square meters, what is its radius ? 11. In a circle whose radius is 35 there is a sector whose angle is 40. Find the length of the arc and the area of the sector. 12. The area of a circle is 6 times the area of another. If the radius of the smaller circle is 12, what is the radius of the larger circle? 13. If the angle of a sector is 72 and its arc is 44 inches, what is the radius of the circle? What is the area of the sector? 14. In a circle whose radius is 7 find the area of the segment whose central angle is 120. [Required area = 1 (area of O area eq. A.] ORIGINAL EXERCISES 249 15. If the radius of a circle is 4 feet, what is the area of a segment whose arc is 60? of a segment whose arc is a quadrant? 16. Find the area of a circle inscribed in a square whose area is 75. 17. Find the area of an equilateral triangle inscribed in a circle whose area is 441 TT square meters. 18. If the length of a quadrant is 8 inches, what is the radius ? 19. Find the length of an arc subtended by the side of an inscribed regular 15-gon if the radius is 4| inches. 20. The side of an equilateral triangle is 10. Find the areas of its in- scribed and circumscribed circles. 21. Find the perimeter and the area of a segment whose chord is the side of an inscribed regular hexagon, if the radius of a circle is 5^. 22. A circular lake 9 rods in diameter is surrounded by a walk \ rod wide. What is the area of the walk ? 23. A locomotive driving wheel is 7 feet in diameter. How many revolutions will it make in running a mile ? 24. What is the number of degrees in the central angle whose arc is as long as the radius ? 25. Find the side of the square equal to a circle whose diameter is 4.2 meters. 26. Find the radius of that circle equal to a square whose side is 5.5 inches. 27. Find the radius of the circle which divides a given circle whose radius is 10 into two equal parts. 28. Three equal circles are each tangent to the other two and the diameter of each is 40 feet. Find the area between these circles. [Required area = area of an eq. A minus area of three sectors.] 29. Find the area of the three segments of a circle whose radius is 5\/3, formed by the sides of the inscribed equilateral triangle. 30. If a cistern can be emptied in 5 hours by a 2-inch pipe, how long will be required to empty it by a 1-inch pipe ? 31. Find the side, the apothem, and the area of a regular decagon in- scribed in a circle whose radius is 6 feet. 32. What is the area of the circle circumscribed about an equilateral triangle whose area is 48 V3 ? 33. The circumferences of two concentric circles are 40 inches and 50 inches. Find the area of the circular ring between them. ROBBINS'S NEW PLANE GEOM. 17 250 BOOK V. PLANE GEOMETRY 34. A circle has an area of 80 square feet. Find the length of an arc of 80. 36. Find the angle of a sector whose perimeter equals the circum- ference. 36. Find the angle of a sector whose area is equal to the square of the radius. 37. Find the area of a regular octagon inscribed in a circle whose radius is 20. [Inscribe square, then octagon. Draw radii of octagon. Find area of one isosceles A formed, whose altitude is half the side of the square.] 38. A rectangle whose length is double its width, a square, an equi- lateral triangle, and a circle all have the same perimeter, namely, 132 meters. Which has the greatest area? the least? 39. Through a point without a circle whose radius is 35 inches two tangents are drawn, forming an angle of 60. Find the perimeter and the area of the figure bounded by the tangents and their smaller intercepted arc. 40. In a circle whose radius is 12 are two parallel chords which sub- tend arcs of 60 and 90 respectively. Find the perimeter and the area of the figure bounded by these chords and their intercepted arcs. 41. A quarter mile race track is to be laid out, having parallel sides but semicircular ends with radius 105 feet. Find the length of the parallel sides. 42. If the diameter of the earth is 7920 miles, how far at sea can the light from a lighthouse 150 feet high be seen ? 43. The diameter of a circle is 18 inches. Find the area of the figure between this circle and the circumscribed equi- lateral triangle. 44. How far does the end of the minute hand of a clock move in 20 minutes, if the hand is 3 inches long ? 46. The diameter of a circle is 16 inches. What is the area of that portion of the circle outside the inscribed regular hexagon? 46. Using the vertices of a square whose side is 12, as centers, and radii equal to 4, four quadrants are described within the square. Find the perimeter and the area of the figure thus formed. ORIGINAL EXERCISES 251 47. Using the four vertices of a square whose side is 12 as centers, and radii equal to 6, four arcs are described without the square (see figure). Find the perimeter and the area of the figure bounded by these four arcs. 48. Using the vertices of an equilateral triangle whose side is 16 as centers and radii equal to 8, three arcs are described within the triangle. Find the perimeter and the area of the figure bounded by these arcs. Do the same if the three arcs are described without the triangle. 49. Using the vertices of a regular hexagon, whose side is 20, as centers and radii equal to 10, six arcs are described within the hexagon. Find the perimeter and the area of the figure bounded by these arcs. Do the same if the six are described without the hexagon. 60. If semicircumferences are described with- in a square, with side 8 inches, upon the four sides as diameters, find the areas of the four lobes bounded by the eight quadrants. Find the area of any one. In the following exercises let n = number of sides of the regular polygon ; s = length of side ; r = apothem ; R = radius ; K = area. 61. If n = 3, show that s = U V3 ; r = \ # ; K = 3 R *^ = 3 r 2 V3. 4 62. If n = 4, show that s = R\/2 = 2r; K = 2 R 2 = 4 r 2 . 53. If = 6, show that, = R=-; K = 3 =2r* 64. If n = 8, show that s = R \/2 - v/2 = 2 r( V2- 1) ; r= V2+\/2~; 65. If n = 10, show that s=~ ( V5 - 1) ; r = VlO + 2V5. 66. Ifn = 5, show that s= VlO -2V5; r = (V5 + 1). 57. If n = 12, show that s = R~Vz - V3 = 2 r (2 - V3) ; 72 = 2rV2-V3 i; r= - V2+V3; K = 12 r 2 (2 - V3) = 252 BOOK V. PLANE GEOMETRY 58. The apothem of a regular hexagon is 18 V3 inches. Find its side and area. Find the area of the circle circumscribed about it. 69. What is the radius of a circle whose area is doubled by increas- ing the radius 10 feet? 60. If an 8-inch pipe will fill a cistern in 3 hours 20 minutes, how long will it require a 2-inch pipe to fill it ? 61. The radius of a circle is 12 meters. Find : (a) The area of the inscribed square. (&) The area of the inscribed equilateral triangle. (c) The area of the inscribed regular hexagon. (d) The area of the inscribed regular dodecagon. (e) The area of the circumscribed square. (/) The area of the circumscribed equilateral triangle. (AABD. Proof : Draw altitude DE. Now AD>DE .'. AODE Multiply each member by J AB. (87). (Ax. 6). Then | AB AC> \ AB DE (Ax. 10). Now | AB AC = area A ABC (364), And J AB - DE = area ABD (?). Therefore A ABC > ABD (Ax. 6). Q.E.D. This theorem may be stated thus : Of all triangles having two given sides, that triangle whose third side is the diameter of the circle which circumscribes it is the maximum. 456. COROLLARY. Of all n-gons having n i sides given, that polygon whose nth side is the diameter of a circle which circumscribes the polygon is the maximum. MAXIMA AND MINIMA 255 PROPOSITION XIX. THEOREM 457. Of all isoperimetric triangles having the same base, the isosceles triangle is the maximum. Given : A ABC and ABD isoper- imetric, having the same base, AB, and A ABC isosceles. To Prove : A ABC > A ABD. Proof: Prolong AC to E, making CE = to -4(7, and draw BE. Using D as a center and BD as a radius, describe an arc cutting EB prolonged, at F. Draw CG and DH II to AB, meet- ing EF at G and H respectively. Draw AF. With C as a center and AC, BC or EC as a radius, the Q described will pass through A, B, and E (Hyp. and Const.). .-. ^ABE=rt. Z (240). That is, AB is J_ to EF. Hence CG and DH are J_ to EF (64). AC + CE = AC + CB = AD + DB = AD + DF (Hyp. and Const.). That is, AE= AD + DF (Ax. 1). But AD + DF > AF (Ax. 12). .'. AE > AF (Ax. 6). .'.BE>BF (88, IV). And i BE > J BF (Ax. 10). Now BG = \ BE, and BH = I BF (85). .'. BG > BH (Ax. 6). Mult, by AB, % AB > BG> % AB BH (Ax. 10). But AB-BG = area A ABC (364). And I AB BH= area A ABD (?). Substituting, A ABC > A ^D (Ax. 6). Q.E.D. 256 BOOK V. PLANE GEOMETRY 458. COROLLARY. Of isoperimetric triangles the equilateral triangle is the maximum. [Any side may be considered the base.] PROPOSITION XX. THEOREM 459. Of isoperimetric polygons having the same number of sides the maximum is equilateral. ^^^^^^" ^^^^\ K/l Given: Polygon AD, the maximum of all polygons having the same perimeter and the same number of sides. To Prove : AB = BC = CD = DE = etc. Proof : Draw AC and suppose AB not = to BC. On AC as base, construct A ACM isoperimetric with A ABC and isosceles ; that is, make AM = CM. Then A ACM > A ABC (457). Add to each member, the polygon ACDEF. .*. polygon AMCDEF > polygon AD (Ax. 7). But the polygon AD is maximum (Hyp.). .-. AB cannot be unequal to BC as we supposed (because that results in an impossible conclusion). Hence AB = BC. Likewise it is proved that .8(7= CD = etc. Q.E.D. 460. COROLLARY. Of isoperimetric polygons having the same number of sides the regular polygon is maximum. Proof : Only one such polygon is maximum, and the maxi- mum is equilateral. (459.) It can also be inscribed in a circle and is therefore regular. (403.) MAXIMA AND MINIMA 257 PROPOSITION XXI. THEOREM 461. Of isoperimetric regular polygons, the polygon having the greatest number of sides is maximum. Given: Equilateral A ABC and square S, having the same perimeter. To Prove : Square s> A ABC AT~ ~C Proof: Take D, any point in BC, and draw AD. On AD as base, construct isosceles A ADE, isoperimetric with A ABD. Now A AED> A ABD (457). Adding A ADC to each member, AEDC> A ABC (Ax. 7). AEDCis isoperimetric with A ABC and S (Hyp. and Const.). Hence S>AEDC (460). .-. 8>AABC (Ax. 11). Similarly, we may prove that an isoperimetric regular pen- tagon is greater than S ; and an isoperimetric regular hexa- gon is greater than this pentagon, etc. Therefore the regular polygon having the greatest num- ber of sides is maximum. Q.E.D. 462. COROLLARY. Of all isoperimetric plane figures the circle is the maximum. Ex. 1. Of isoperimetric triangles, the maximum is equilateral. Ex. 2. Of all right triangles that can be constructed upon a given hypotenuse, which is maximum? Why? Ex. 3. Of all triangles having a given base and a given vertex angle, the isosceles is the maximum. Ex. 4. Of all mutually equilateral polygons, that which can be in- scribed in a circle is the maximum. 258 BOOK V. PLANE GEOMETRY PROPOSITION XXII. THEOREM 463. Of equal regular polygons the perimeter of the polygon having the greatest number of sides is the minimum. Given : Any two equal regular polygons, A and B, A hav- ing the greater number of sides. To Prove : the perimeter of A < the perimeter of B. Proof : Construct regular polygon 8, similar to B and iso- perimetric with A. Then A>8 (460). But A = B (Hyp.)- .-. B>s (Ax. 6). Hence the perimeter of B > perimeter of 8 (376). But the perimeter of s = perimeter of A (Const.). .*. perimeter of JS> perimeter of A (Ax. 6). That is, the perimeter of A< the perimeter of B. Q.E.D. 464. COROLLARY. Of all equal plane figures the circle has the minimum perimeter. Historical Note. Rene* Descartes was born near Tours, France, in 1596. He was a man of wonderful intellect. He simpli- fied and generalized the notation of algebra and in- troduced the use of exponents as now employed. The restriction of final letters of the alphabet to represent unknown quantities is also due to him. Descartes was the first to adapt algebra to geom- etry, showing that geometrical figures can be repre- sented by algebraic equations. On this general truth he based the development of analytical geom- DESCARTES etry which is known by his own name, as Cartesian geometry. He gave a large part of his life to original and creative work in mathematics, philosophy, physics and astronomy. ORIGINAL EXEKCISES 259 ORIGINAL EXERCISES 1. Of all equal parallelograms having equal bases, the rectangle has the minimum perimeter. 2. Of all lines drawn between two given parallels (terminating both ways in the parallels), which is the minimum? Prove. 3. Of all straight lines that can be drawn on the ceiling of a room 12 feet long and 9 feet wide, what is the length of the maximum? 4. Find the areas of an equilateral triangle, a square, a regular hexagon, and a circle, the perimeter of each being 264 inches. Which is maximum ? What theorem does this exercise illustrate? 5. Find the perimeters of an equilateral triangle, a square, a regular hexagon, and a circle, if the area of each is 1386 square feet. Which perimeter is the minimum? What theorem does this exercise illustrate? 6. Of isoperimetric rectangles which is maximum? 7. Divide a given line into two parts such that their product (rectangle) is maximum. 8. Of all equal triangles having the same base, the isosceles triangle has the minimum , ^ .^..^i. > perimeter. To Prove: The perimeter of A ABC < the perimeter of A AB'C. Proof: AD