IN MEMORIAM FLORIAN CAJOF ELEMENTARY SYNTHETIC GEOMETRY. BY GEORGE BRUCE HALSTED, A.B.^ A.M., and ex-Fellow of Princeton College ; Ph.D. and ex-Fellow of Johns Hopkin University ; Professor of Mathematics in the University of Texas. NEW YORK: JOHN WILEY & SONS, 53 East Tenth Street, 1892. Copyright, iSgz, BV GEORGE BRUCE HALSTED. CAJORI ]FEMU8 B»03., ROBERT Dromhont>, Pnnters, Eleotrotvper, ^^^ g^.'-eets tU&m Pearl Street. * ^^^ ^^^^ New York. PREFACE. My conception of Comparative Geometry dates back to 1877, when, having had the good fortune in BerHn to meet a copy of Lobatschewsky, then rare and unappreciated, I was enjoying a comparison of it with my beloved EucHd. I have at last mustered courage to put my Pure Spherics where it belongs. But as my first book is geometry with one parallel geodesic, my second book geometry with no parallel geodesic, my third book should be geometry with more than one parallel geodesic through the same point (the Lobat- schewsky-Bolyai elementary geometry). So it shall be, if, after another decade, I write still another geometry. George Bruce Halsted. 2407 Guadalupe Street, Austin, Texas. ill 918190 Digitized by tine Internet Arciiive in 2008 witin funding from IVIicrosoft Corporation littp://www.arcliive.org/details/elementarysyntlieOOIialsricli CONTENTS. BOOK I. PAGB Symmetry, Symcentry, and Congruence, i CHAPTER SKCTIONS I. The Primary Concepts of Geometry, . . . 1-54 II. The Circle, 55-io9 III. The Fundamental Problems 110-119 IV. Symmetry and Symcentry 120-152 V. Tangents, 1 53-171 VI. Chords 172-183 VII. Two Circles 184-186 VIII. Parallels 187-205 IX. The Triangle 206-227 X. Polygons, 228-243 XI. Periphery Angles, 244-261 XII. The Symmetrical Triangle, 262-268 XIII. The Symcentral Quadrilaterals, 269-280 XIV. Symmetrical Quadrilateral, 281-299 XV. Congruence of Triangles, 300-313 BOOK II. Pure Spherics 60 CHAPTER SECTIONS I. Primary Concepts, 314-340 II. Symcentry on the Sphere 34^-343 III. Symmetr)' on the Sphere, 344-383 IV. The Symcentral Quadrilateral : 384-389 •V. Spherical Triangles, 390-428 VI CONTENTS. BOOK III. PAGE Equivalence 85 BOOK IV. Proportion 92 BOOK V. Similarity 95 BOOK VI. Mensuration 104. BOOK VII. Modern Geometry, 117- CHAPTER SECTIONS I. Transversals 522-533 II. Harmonic Ranges and Pencils 534-55° III. Principle of Duality 551-564 IV. Complete Quadrilateral and Quadrangle, . 565-572 V. Inversion 573-581 VI, Pole and Polar with respect to a Circle, . . 582-599 VII. Cross Ratio 600-623 BOOK VIII. Recent Geometry 146 CHAPTER SECTIONS I. Anti-parallels, Isogonals, Symmedians, . . 624-660 II. The Brocard Points 661-675 TABLE OF SYMBOLS. bi' bisector. e.g exempli gratia [for example]. p't point. p'ts points. quad' quadrilateral. r't right. r't bi' right bisector [perpendicular bisector]. sq' square. s't straight. s'ts straights. O circle. • surface of circle. O* circles. A. triangle. A" triangles. A spherical triangle. .• therefore. ~ similar. ~C center of similitude. = equivalent. ^ congruent. symcentral. •|' symmetrical. II parallel. IP parallels. I! g'm parallelogram. J_ perpendicular. J_^ perpendiculars. -f- plus. — minus. < less than. > greater than. ^ angle. ^ABC. . . .angle from ray BA to ray BC. '^ab angle from ray a to ray b. QC{r) circle with center C and radius r. A perspective. A\C . : center of perspective. A projective. ELEMENTARY SYNTHETIC GEOMETRY. BOOK I. SYMMETRY, SYMCENTRY, AND CONGRUENCE. CHAPTER I. THE PRIMARY CONCEPTS OF GEOMETRY. 1. A natural object, say a crystal, is bounded ; and this boundary divides it from the air around it, but is not a thin film of the crystal itself. It is where the crystal ends and the air begins. It is also a boundary of the air where it joins the crystal, but it is not air. It is the boundary between the two, and is common to the crystal and the air. 2. A boundary of the sort capable of wholly enclosing a solid, so that nothing could get into the solid except through this boundary, but itself no solid, is called a S7irface. 3. Surface is an ideal or imaginary concept drawn from the apparent (not real) boundaries of physical objects. We natu- rally associate the surface with the limited solid, not with the Fig. 2 .^LEMBA'tA^HY SYNTHETIC GEOMETRY. surrounding air. Thus we think of the colored surface of the crystal as belonging to the crystal ; and if yellow oil lies on the water in a glass, we think of the under surface of the oil as yellow and belonging only to the oil : while a mathematical surface pertains equally to the two solids that it separates. 4. These ideal mathematical surfaces may be dealt with as existing by themselves, and as movable. In illustration of this, think of a shadow. 5. A surface may be finite yet unbounded in the sense of having no abrupt or natural stopping-place on it, no visible break or obvious limit in it. Such is the surface of an egg. Set it up in an egg-cup, and run a pencil-mark around it. Then you may think of the surface of the egg as divided into three parts, the white surface within the cup, the ribbon-like black surface of the pencil-mark, and the white surface above this black ribbon. 6. Between the black surface and the two white surfaces are two boundaries which are neither black nor white. These boundaries are not thin strips of surface any more than the surface is a thin layer of solid. Where a white surface meets a black there is a common boundary of both, dividing each from the other, and belonging to both. 7. A boundary of the sort capable of wholly enclosing a piece of surface so that nothing mov- ing in the surface could enter this piece of surface except through this boundary, but itself no sur- face, is called a line. Fig. 2. 8. A boundary between two adjacent pieces of a line, and common to both pieces, but itself no line, is called 2i point. 9. Two lines cross or intersect in a point. Fig, 3. PRIMARY CONCEPTS. 10. Two surfaces intersect in a line. Fig. 11. When a chalk mark is drawn across a blackboard, each of the two edges, neither white nor black, is a line. 12. When one chalk mark crosses another, four points are fixed by the crossing of the four edges. Fig. s. 13. Surfaces, lines, or points, or any combinations of them, are called figures. 14. Any figure may be looked upon as two coincident figures; Mathematical figures wholly lack impenetrability. 15. If we imagine a figure to move, we may also suppose it to leave behind its outline or trace. „ ^ Fig. 6. 16. Two coincident figures cannot be distinguished from one another unless they be separated by moving one. 17. Assumption I. Figures may be moved about, with- out any other change. 18. Figures which can be made to coincide are called congruent. 19. If a solid has, as part of its boundary, a piece of sur- face which appears the same from within the solid as from without, and if any two of three such soHds will fit each other all over these surfaces, then each of these surfaces is called plane. Such a surface' unbounded is called 3. plane. 20. Any piece of a plane will slide in the plane, and after being turned over will fit the plane. ^ Fig. 7. ELEMENTARY SYNTHETIC GEOMETRY. 21. The intersection of two planes is called a straight line, or simply a straight. 22. A straight is a line in a plane which appears the same , from both the regions bounded by it in the „ „ plane. Fig. 8. ^ (~-\s\ 23. A piece of the plane with part of the straight as one of its boundaries would fit all along the straight from both sides. Fig. 9. 24. Assumption II. If two straights have two points in common they coincide throughout. 25. A straight with two points in a plane lies wholly in that plane. For it lies in a plane, and if this is another plane the two intersect in a straight which has two points in com- mon with the given straight. Fig. 10. 26. Assumed Construction I. A straight can be drawn through any two points. , ^, 27. A sect is the piece of a straight between two definite points. Fig. II. 28. A curve is a line no part of which is straight. Fig, 12. 29. Assumption III. A figure with two points fixed can still be moved, and the whole figure partakes of the motion, except the straight through the two fixed points. Such motion is called revolution about this straight as axis. It may be continued until each point of the figure coincides with its trace. Such a turning is spoken of as one complete revolution, or simply, a revolution. 30. If a third point, not on the axis, be fixed, all motion of a rigid figure is prevented. PRIMARY CONCEPTS. Fig. 13. Fig. 14. 31. Three points not on a straight are necessary and suf- ficient to determine a plane. 32. Any straight in a plane cuts it into two parts called Jiemiplanes. 33. By a half-revolution of their plane about the coTimon straight, either of two hemiplanes may be brought into coinci- dence with the trace of the other. Thus one hemiplane may be thought of as made to coincide with the other by folding over along the common axis. 34. Any point in a straight cuts it into two a o b parts called rays. 35. The figure so formed is a special case of the figure formed by two rays going out from the same point, called a bi-radial. 36. A bi-radial lies wholly in one plane. 37. One ray, a, of a bi-radial, may be brought into coin- cidence with the other ray, b, by a turning in the plane, or rotation, about the common point, or vertex O ; and this turning may be in the sense indicated by the arrow in Fig. 14, or in the opposite sense. 38. To fix that sense of rotation which is to be considered as positive (which kind is meant if nothing else is stated), we take the turning of a ray in the sense opposite to that of the hands of a watch as positive. The watch hands, then, turn in the negative sense. Clockwise is minus [ — ]. Counter-clockwise is plus [-|-]. 39. A bi-radial looked at with special reference to the magnitude and sense-of-turning of a ray's rotation from one of its rays into the other, is called an angle. Thus, though we consider no turning beyond one complete rotation, yet the same bi-radial is four different angles, 6 ELEMENTARY SYNTHETIC GEOMETRY. T^A^ ab, ^ — ab, ^ -f~ ^^y t- — ba, where the turning is always from the first-mentioned ray into the second. 40. If O (Fig. 14) is the origin or initial point of ray a and of ray b, and A any other point oh a, and B on b, then ^ ■\- ab may be written -a^-^^ OA/OB, or even -4.-^- AOB^\\\\^x& the order of the letters denotes that the angle is generated by a ray rotating about O from OA to OB, and the sign fixes the sense of that rotation. /'' 'x 41- If a ray, a, is turned about the initial ih c a\ point, C, until it coincides with the continu- ation, b, of its trace beyond C, the angle ab is called a straight angle. /" X 42. If we turn still more, until the c moving ray has made a complete rota- \^^^y tion, and coincides with its trace, the ^"^- ^7- angle is called Si perigon. 43. If ^ab equals a perigon, then the ray a coincides with the ray b. 44. When a bi-radial is looked upon as an angle, its two rays are called the arjns of the angle. 45. Two angles are eqtialM they can be so placed that their arms and therefore their ver- tices coincide, and that both are described simultaneously by the turning of the same ray about their common vertex. Fig. 18. 46. Equality implies that both angles have the same sense. 47. Two angles which can be made equal by changing the sign of one, are said to be equal in magni- tude but opposite in sense. 48. Since turning the plane of a bi-radial through half a revolution changes the '°' '^" sense of each of its four angles, therefore, if one angle by folding over along an axis is made equal to PRIMARY CONCEPTS. another, then the angles were equal in magnitude but opposite in sense. 49. Assumption IV. All straight angles are equal in magnitude. 50. As a consequence, all perigons are equal in magnitude. 51. If two angles have a vertex and an arm in common, they are called adjacent angles. Fig. 20. 52. When two adjacent angles are of the same sense, and so situated that they cannot be simultaneously described, even Fig. 21. Fig. 22. in part, by the same ray rotating, their sum is the angle of like sense whose arms are their two non-coincident arms. 53. When the sum of any two angles \^^ _^ is a straight angle, each is said to be ^>\ \ the supplement of the other. Fig. 23. 54. When the sum of any two angles is a perigon, each is said to be the explement of the other. Thus -4.-^- ab and i. -j- ba are exple- mental. p,^ ^^ CHAPTER II. THE CIRCLE. 55. If, in a plane, a sect turns about one of its end points the other end point describes a curve called a circle. 56. The fixed end point is called the center of the circle. 57. Any sect from the center to a point on the curve is called a radius. 58. All radii are equal, being equal to the generating sect. Fig. 26. 59. Since the moving sect, after rotating through a perigon, returns to its trace, therefore the moving end point describes a closed curve. 60. This curve divides the plane into two parts, one of which is finite and is swept over by the moving sect. 61. This finite plane surface is called the surface of the circle. Any point in this finite FioTay. plane is said to lie within the circle. 62. Assumed Construction II. A circle may be described from any given point as center with any given sect as radius. 63. A theorem is a statement usually capable of being inferred from other statements previously accepted as true. THE CIRCLE. g t 64. A corollary to a theorem is a statement whose truth follows readily from that of the theorem. 65. A theorem consists of two parts, the hypothesis (that which is assumed), and the conclusion (that which is asserted to follow therefrom). 66. A problem is a proposition in which something is required to be done by a process of construction. 67. The treatment of a problem in elementary geometry ■consists, — [i] Construction. In indicating how the ruler and compasses are to be used in effecting what is required. [2] Proof. In showing that the construction so given is correct. [3] Determination. In fixing whether there is only a single solution, or suitable result of the indicated construction ; or more than one ; and in discussing the limitations, which some- times exist, within which alone the solution is possible. 68. Our assumed constructions allow the use of the straight- edge not marked with divisions, for drawing and producing ^e^ts, and the use of compasses for drawing circles and the transference of sects. It is important to note the implied restriction, namely, that we work in the plane, and that no construction in elementary geometry is allowable which cannot be effected by combinations of these two primary construc- tions. 69. Theorem. The sect to a point, from the center of a circle, is less than, equal to, or greater than the radius, according as the point is within, on, or without the circle. Proof. For any point Q, within the cir- cle, lies on some radius, OQR. If ^ is without the circle, then the sect OS con- """""■fIgTIs. tains the radius OR. 70. Inverse. A point is within, on, or without the circle. lO ELEMENTARY SYNTHETIC GEOMETRY. according as its sect from the center is less than, equal to, or greater than the radius. 71. Theorem. Circles of equal radii are congruent. Proof. For, if put in the same plane, with centers in coin- cidence, every point of each is on the other, because of the equality of their radii. Thus Q)C\r\ ^ Q)0 [r]. 72. Corollary. A circle turned about its center slides on its trace. This fundamental property of this curve enables us to turn any figure connected with the circle about the center without changing the relation to the circle. 73. Circles which have its same center are called concentric. 74. Concentric circles with a point in common coincide. 75. A sect whose end points are on the circle is called a chord. "jG. Any chord through the center is called a diameter. jj. All diameters are equal, each being equal to two radii. 78. Every diameter is bisected by the center of the circle. 79. No circle can have more than one center. For, if it had two, the diameter through Fig. 30. them would have two mid points. Fig. 31. Fig. 32. 80. Any ray from the center of a circle cuts the circle in one, and only one, point. THE CIRCLE. II Fig. 33. 81. Any straight through its center cuts the circle in two and only two points. 82. Any piece of a circle is called an arc. 83. When the end point of a radius de- scribes an arc, the radius rotates through an angle having its vertex at the center. This angle is called the angle at the center, and is said to be subtended by the arc simultaneously described, or to stand upon that arc. 84. An arc, being described by the end point of a rotating radius, is said to have the same sense as the angle through which that radius rotates. 85. Arcs congruent and of the same sense are called equal. 86. The sum of two arcs, of the same circle, or of equal circles, is the arc which subtends an angle at the center equal to the sum of the angles subtended by those arcs separately. 87. Theorem. Equal arcs subtend equal angles at the center, and, inversely, equal angles at the center stand upon equal arcs. Proof. For, if arc AB equal arc CD, we may slide the arc AB, together with the radii OA and OB, along the circle until A coincides with C\ then will B coincide with D, since arc CD equals arc AB. Therefore 4- A OB will coincide with ^ COD, and will be equal to it in magnitude and sense. 88. It follows, that if y^, B, C, etc., denote points on the circle and a, b, c, etc., the radii drawn to those points, then every equation between arcs AB, BC, etc., will carry v/ith it c' an equation between the corresponding angles ab, be, etc. ; and inversely. p^^ ,g Fig. 35. 12 ELEMENTARY SYNTHETIC GEOMETRY. 89. Theorem. In the same or equal circles, of two unequal arcs, the greater subtends the greater angle at the center. Proof. If the first arc is greater than the second, it equals the second plus a third arc, and so the angle which the first subtends is ^^^^"^ greater than the angle which the second sub- tends by the angle which the third arc subtends at the center. 90. Inversely : Of two unequal angles at the center, the greater intercepts the greater arc. 91. Two arcs which together equal the whole circle are called explemental. Thus the explemental angles at the center of a circle, whose arms are the same radii, are said to stand upon the explemental arcs which would be described simultaneously Avith the angles, the greater angle upon the greater arc. 92. Explemental arcs equal in magnitude are called semi- circles. 93. A semicircle subtends a straight angle. For two sub- tend a perigon, and are equal. 94. Any straight through the center cuts the circle into two semicircles. For it makes at the center straight angles which together are subtended by the whole circle. 95. If we fold over about a straight through the center of a circle, the semicircles it makes are brought into coinci- dence. For every point on the turned semicircle must fall on some point of the other, as its sect from the center is a radius. 96. Two arcs which together equal a semi- circle are called supplemental. 97. Half a straight angle is called a right angle. Fic?3&. 98. All right angles are equal in magnitude. THE CIRCLE. 13 99. The arc subtending a right angle is called a quadrant. It is one quarter of a circle. 100. Two straights which make a right angle are said to be perpendicular to one another. 10 1. Two angles whose sum is a right angle are caW^d complement al. Fig. 40. 102. An angle less than a right angle is called acute. 103. An angle greater than a right angle, but less than a straight angle, is called obtuse. 104. An angle greater than a straight angle, but less than a perigon, is called reflex. 105. An angle which is either acute, right, or obtuse, is called a minor angle. Fig. 42. 106. An arc less than a semicircle is called a minor arc. 107. An arc less than a circle, but greater than a semi- circle, is called a major arc. 108. Theorem. If two circles have one common point not on the straight through their centers, they have also another such point. Proof. Let O C and O O have the the point A in common. Fold the figure over along the straight through their centers, CO. Then the semi- circles which have A in common are brought into coincidence with the other semicircles, fore these also have a common point. A'. There- 14 ELEMENTARY SYNTHETIC GEOMETRY. 109. Theorem. If two circles have a common point not on the straight through their centers, and therefore another such point, then the center-straight bisects the angles made at the centers by the radii to these two common points, and is the perpendicular bisector of the common chord. Proof. For by folding over along CO we bring A into coin- cidence with A'. Therefore sect AM = sect A' M. ^ OMA = t A' MO. t MCA =: ^ A' CM. 4- AOM = :^ MOA'. Fig. 44. CHAPTER III. THE FUNDAMENTAL PROBLEMS. Fig. I lo. Problem. To bisect any given sect. Construction. With its end points, A and A', as centers, and itself as radius, describe two circles. They will have one common point not on their center straight, and therefore a second such. Join these two common points, C and O. Then CO bisects the given sect AA'. Proof. For A A' is J_ to CO, and if from C and O as centers, with radii equal to AA' , two circles were described, then A A' would be a common chord,' bisected by the center-straight CO. III. Theorem. The straight through the mid point of a chord, and the center of the circle, is per- pendicular to the chord, and bisects the exple- mental arcs, and their angles at the center. Proof. A and B are any points on qO. Turn the whole figure over and apply it to its trace, so that O falls on O, but A on the trace of B, and B on the trace of A. Then the bisection point, C, of the chord AB falls on its own trace, and consequently the whole change amounts only to half a revolution of the figure about the straight CO. Fig. 46. 15 i6 ELEMENTARY SYNTHETIC GEOMETRY. l\2. Problem. To bisect any given angle. Construction. With its vertex, C, as- center, and any sect, r, as radius, describe a circle cutting the arms of the angle at A and A'. Bisect the chord AA' , and Fig. 47. join its mid point, M, to the center C Then MC bisects t ACA'. 113. Problem. At a given point on a given straight, to- draw a perpendicular to that straight. Construction. Bisect the straight angle at the point. 114. Problem. Through a given point, not in a given: straight, to draw a perpendicular to that straight. Construction. In the hemiplane not contain- ing the given point, C, take any point D. Call A and A' points where (dC\CD~\ cuts the given straight. Bisect the chord A A' at M. Then is CM ]^\.o AA'. Determination. Through a given point only one perpendicular can be drawn to a given straight. For, if the plane were folded over along the given straight, the given point would fall on the production of any perpendicular from it to the straight. 115. Since the perpendicular from the center to a chord of a circle bisects that chord, and also the explemental arcs and the explemental angles pertaining to that chord, therefore the r't bi' of any chord passes through the A center ; and the straight which possesses any two of these seven properties possesses also the other five. 116. Problem. To bisect any given arc. Construction. Join its extremities, and Fig. 49. \ X Fig. 50. draw the r't bi' of this chord. THE FUNDAMENTAL PROBLEMS. 17 117. Theorem. A straight cannot have more than two points in common with a circle. Proof. For, if it had a third, then, since the r't bi' of a chord contains the center, there would be three perpendiculars from the center to the same straight. 118. Theorem. Every point which joined to two points gives equal sects is on the per- pendicular bisector of the sect joining those two points. Proof. The r't bi' of the chord contains the center. 119. Corollary. Circles with three points in common coincide. CHAPTER IV. iM A Fig. 52. SYMMETRY AND SYMCENTRY. 120. Two points are said to be sym- metrical with regard to a given straiglit, called the Axis of Symmetry, when the axis bisects at right angles the sect joining the tw'o points. 121. Two points have always one, and only one, symmetry axis. 122. A point has, with regard to a given axis of symmetry, always one, and only one, symmetrical point ; namely, the one on the ray from the given point perpendicular to the axis, which ends the sect bisected by the axis. 123. Two figures have an axis of symmetry when, with regard to this straight, every point of each has its symmetrical point on the other. 124. Two figures are symmetrical when they can be placed so as to have an axis of symmetry. 125. One figure has an axis of symmetry when, with regard to this straight, every point of the figure has its symmetrical point on the figure. 126. One figure is symmetrical when it has an axis of symmetry. 127. Any figure has, with regard to any given straight as axis, always one, and only one, sym- metrical figure. 128. One figure is symmetrical when it has an axis with re- gard to which its symmetrical figure coincides with itself. 18 Fig. 53. ^3\ Fig. s4. SYMMETRY AND SYM GENTRY. 19 'C\ Fig. 56. T29, Every point in the axis is symmetrical to itself. 130. The axis is symmetrical with regard to itself. 131. Two points are said lo be symcentral . with regard to the mid point of their joining ~~~ ' ' sect. ^"^- 55- 132. A point has, with regard to a given symcenter, always one, and only one, symcentral point ; namely, the one on the ray from the given point through the symcenter, which ends the sect bisected by the symcenter. 133. Two figures have a sym- center when, with regard to this point, every point of each has its pi^_/_ .\ be a point not in the straight s, and OC X. to s : then s is tangent to O^ \PC^ at C. Any second circle concentric with the first, but of lesser radius, lies wholly within the first. A third concentric circle, with radius > C'^T, lies wholly without the OO \PC\ and cuts s in D-\-D', axis OC; .'. CD = fig. 75. CD'. A fourth concentric circle, with radius > OD, lies wholly without the third ; .*. its intersections with s lie without the sect DD'. Hence the four following theorems : 162. A straight will be a secant, a tangent, or not meet the circle, according as the perpen- dicular to it from the center is less than, equal to, or greater than the radius. 163. The perpendicular is the least sect between a given point and a given straight. f,^. yg. 24 ELEMENTARY SYNTHETIC GEOMETRY Fig. 77. 164. Except the perpendicular, any sect from a poirit to a straight is called an obligtie. 165. Two obliques from a point to a straight, making equal sects from the foot of the perpendicular, are equal. 166. Of any two obliques between a point and a straight, that which makes the greater sect from the foot of the perpen- dicular is the greater. 167. Problem. From a given point without the circle to dravv^ a tangent to the circle. Construction, Join the given point A to the center C, cutting the circle in B. Draw BDX_ to CB, and cutting in D the Q)C[CA]. Join DC, cutting qC {CB'] in F. Then AF \s tangent \.o Q)C ICB\ Proof. Radius CA, _L to chord HD, bisects arc HD\ .'. if we rotate the figure until H comes upon the trace of A, then A is on the trace of D, .*. tangent HB on trace oi AF. Determination. Always two and only two tangents. 168. Corollary. By symmetry the straight through C, the center of a circle, and A an exter- nal point, bisects the angle be- tween the two tangents from A to OC, and also the angle between the radii to the points of contact, F and F' ; and sect AF = sect AF'. 169. Corollary. Any point from which the perpendiculars on two intersecting straights are equal, is on one of their angle bisectors. It is center of a circle to which they .are tangent. 170. Corollary. The centers of all circles tangent to two intersecting straights are in their angle bisectors. 171. Inversely. From any point on a bisector of an angle Fig. 78. TANGENTS. 2$ made by two straights, the perpendiculars to those straights are equal. For the bisector is a symmetry axis for the two straights : so when we fold along it, the foot of the perpendicular to one straight falls on the other straight, and there is only one per- pendicular from a point to a straight. CHAPTER VI. CHORDS. 172. Take AB any chord in oO. The qA [AB] cuts OO- in two points, B \ B' , axis AO [the center- straight]. But the end points of all sects from A which are equal to AB must lie on Q)A ^° {_AB], ; whence : Theorem. Chords from any point of a circle are equal in pairs, one on each side of the Fig' 79/ diameter from that point. 173. A circle = ©(9, and containing a chord = AB, can be superimposed upon OO, and then rotated until one end of the chord comes at A. The other end of this chord then lies on both QO and QA [AB], and so falls on B ox B' \ and the chord coincides with AB or AB' . Hence the theorem: In the same or equal circles, to equal chords pertain equal minor arcs. 174. Corollary. In the same or equal circles of arcs pertain- ing to equal chords any two are either equal or explemental. 175. If with center A and radius AC <. AB we describe a second circle, it will lie wholly within qA [AB]. Conse- quently it cuts QO in points C and C on the arc BAB' \ .'. arc AC < arc AB. Thus if the chord decreases, so does the minor arc ; and inversely, of two unequal minor arcs, the greater has the greater chord. 176. If the chord increases, its major arc decreases, since its major and minor arcs are always explemental. Inversely, if a major arc decreases, its chord increases. 177. A diameter is the greatest chord. Every other chord equals a piece of the diameter. 26 CHORDS. 27 178. Equal arcs, being congruent, have equal chords. Therefore, also, explemental arcs have equal chords. 179. Equal chords, having equal minor arcs, which may be brought into coincidence by rotation about the center, have' also equal perpendiculars from the center. 180. In the same or equal circles, chords which have equal perpendiculars from the center, since by rotation one may be put upon the other, are equal. S >.b 181. Since the end point 6^ of a chord AC < AB lies on the minor arc AB, :. it is on the side of AB remote from the center 0\ .'. the mid point of chord AC is on the side oi AB remote from O. fig. 80. Theorem. In the same or equal circles, the greater chord has the lesser perpendicular from the center. 182. Inversely. The chord with the greater perpendicular from the center is the lesser; for [181] it cannot be the greater chord, nor [179] can they be equal. 183. Problem. At a given point G, in a given straight s, to make an angle equal to a given ^v\g\e ACB. Construction. With any radius draw OC[r], cutting CA at Z?, and CB at F. Join DF. Draw OC [r], cutting the given straight at H. Draw QH \r' = BF], cutting QG [r] at K. Join GK. Then ^B'GK= ^ ACB. Proof. 06^ [r] = OA [r], and chord //'TT^ chord DF. .•. minor arc HK = minor arc BF. .'. minor ^ HGK^=. minor 4- BCF. Determination. The construction will give four minor angles, at G, all equal, namely, Fig. 81. H,GK,; K,GH,; KJGH,-, HfiK,. CHAPTER VII. TWO CIRCLES. 184. A figure formed by two circles is symmetrical with regard to their center-straight as axis. Every chord perpendicular to this axis is bisected by it. If the circles have a common point on this straight, they cannot ^"^- ^^- have any other point in common, for any point in each has its symmetrical point with regard to this axis, and circles with three points in common coincide. 185. Two circles with only one point in common are called tangent, are said to touch ; and the common point is called the point of tangency or contact. 186. If two circles touch, then, since there is only one common point, this point of contact lies on the center-straight, and a per- pendicular to the center-straight ^"^' ^^' through the point of contact is a common tangent to the two circles. CHAPTER VIII. PARALLELS. 187. A straight cutting across other straights is called a. transversal. [In plane geometry, all are in one plane.] 188. If, in a plane, two straights are cut in two distinct points by a transversal, at each of these points four positive minor lo.z^. angles are made. Of these eight angles, four are between the two straights^ [namely, 3, 4, a, ^], and are called Interior \ Angles : the other four lie outside the two J^ straights, and are called Exterior Angles. \ ^ Angles, one at each point, which lie <^ d on the same side of the transversal, the one exterior and the other interior, are called Corresponding Angles [e.g., 1 and a']. Two non adjacent angles on opposite sides of the trans- versal, and both interior or both exterior, are called Alternate Angles [e.g., 3 and «]. Two angles on the same side of the transversal, and both interior or both exterior, are called Conjugate Angles [e.g., 4 and a\ 189. Theorem. If two corresponding or two alternate angles are equal, or two conjugate angles are supplemental, then every angle is equal to its corresponding and to its alternate, and supplemental to its conjugate. [Use vertical angles and supplemental adjacent angles.} 29 30 ELEMENTARY SYNTHETIC GEOMETRY. Fig. 86. 190. Parallels are straights in the same plane which nowhere meet. [Note. As we are working on a plane, the clause " in the same plane" would be understood even if not mentioned.] 191. Assumption V. Two coplanar straights are parallel if a transversal makes equal alternate angles. 192. Assumption VU If two coplanar straights cut by a transversal have a pair of alternate interior angles unequal, they meet on that side of the transversal where lies the smaller angle. 193. Theorem. If two straights cut by a transversal have corresponding angles equal, or conjugate angles supplemental, they are parallel. For either hypothesis makes the alternate angles equal. 194. If two straights cut by a transversal have conjugate angles not supplemental, they meet. For the alternate angles are unequal. :^ 195. Problem. Through a given point to draw a parallel to a given straight. Construction. Join the given point P to any point C, of the given straight CB. Then at P make an angle CPD alternate and equal to tPCB. Determination. There is only one solution. 196. Corollary. Two coplanar straights parallel to the same straight are parallel to one another. For they cannot meet. 197. Theorem. If a transversal cuts two parallels, the alternate angles are equal. PARALLELS. 31 Fig. 91. Proof. For if they were unequal, the straights would meet. 198. Theorem. Any two parallels c x^ are symcentral with regard to the mid \3fl point of the sect which they intercept ^ ° on any transversal. fig. 90. Proof. Rotating the figure about M through a straight angle brings A into coincidence with the trace of B and :4. CAM into coincidence with the trace of the equal alternate t DBM. 199. Two angles with their arms parallel are either equal or supplemental [189 and 197.] 200. If two angles have their arms respectively perpendic- ular, they are either equal or supple- mental. For rotating one of the angles through a r't 4- around its vertex, its arms become _L to their traces, and .'. II to the arms of the other %.. 201. Points all in the same straight are called costraight. 202. Problem. To pass a circle through any three points not costraight. Construction. Join the three points by three sects ; to these sects erect r't bisectors ; of these every two will meet, since they make an angle = or , supplemental to that to whose arms they are J_. Suppose two to meet at C. This point joined to the three points gives three equal sects. Therefore it is the center of a circle containing the three -given points. 203. Corollary. The center of any O through the three points must lie on all three r't bi's. Fig. 92. Fig. 93. 32 ELEMENTARY SYNTHETIC GEOMETRY. .'. the third r't bi' passes through O. 204. Problem. To describe a circle touching three givei* intersecting straights not all through the same point. Construction. At each of two intersection points draw the two angle-bisectors. Every pair of these meet, since they make conjugate angles which are not supplemental. [Two of the four different angles bisected are together less than a straight angle ; the other two each less than a straight angle, and the angle between bisectors of supplemental adjacent angles is right.] From any point, as /, on a bisector through A and one through B, drop a perpendicular upon one of the given straights, as AB. A circle described with this perpendicular as radius is tangent to AB\ but it also touches the second given straight BC [/ lies on the bisector of an angle between AB and ^C], and the third CA [/ is on a bisector of an angle between AB and AC]. Determination. Every intersection point of two angle- bisectors has thus equal perpendiculars to the three given straights. It is therefore on a third angle-bisector. Thus the four intersection points of the two bisectors through A with the two through B are the eight intersection points of the two bisectors through C with the other four.. PARALLELS. 33 Thus the two bisectors through the third point give no new intersections, and there are just four solutions. 205. Problem. To draw a common tangent to two given circles. Construction. A and B are the points where QC [CA] and OOIOBI are cut by CO. Sup- pose CA > OB. From AC or AO cut ofi AD = OB. Describe OC [CD^. To it, from 6?, draw tangent OP. Let CP cut qC [CA] in Q. Through O, on the same side of OP as Q, draw OR \\ to CP, cutting qO [OB] in R. Then QR is a common tangent. Fig. 95. Fig. 96. Proof. Radii CQ = CA, CP=CD', .-. PQ = AD - OB = OR. But OR I to PQ and ^ OPQ a r't ^ ; .-. ^ /'C>y? is a r't ^. .-. e -I- y?, axis _L to OP, .'. 0P\\ to QR, .'. t PQR = 4. QRO = a r't ^. CHAPTER IX. THE TRIANGLE. 206. Three points A, B, C, not co-straight, and the three straights they determine, form a figure called a triangle. 207. The three points of intersection are the three vertices of the triangle \_A, B, C, 208. The circle through the vertices of a triangle is called its circumcircle, O O [R\, and the center O of the circumcircle is called the circumcenter of the triangle ; its radius, R, the circumradius. 209. The three sects joining the vertices are the sides of the triangle. The side opposite the angle A is called a ; opposite :4- B \s> side b ; opposite C, c. 210. Straights which all intersect in the same point are called concurrent. 21 r. The three perpendicular bisectors of the sides of a triangle are concurrent in its circumcenter. Fig. 98. Fig. 99. THE TRIANGLE. 35 212. The circle tangent to the three sides of a triangle is called its in-circle, ©/ [r], and its center /, the triangle's in- center [r, in-radius]. 213. The three internal bisectors of the angles of a triangle are concurrent in its in-center. 214. A circle touching one side of a triangle and the other two sides produced is called an escribed cxrcle, or ex-Q. The three centers /, , /, , /, of the escribed circles O /, r^i]» O ^1 [^j]' O h [''3] of ^ triangle are called its ex-centers. 215. The sum of two sects is the sect obtained by placing them on the same straight, with one end point of each in coin- cidence, but no other point in common. 216. An exterior aiigle of a triangle is one between a side and the continuation of another side. 217. Through the vertex B of a. A draw BD \\ to AC. The exterior ^ ABE is made up of ^ ABD = 4. c- BAC [alternate], and ^ DBE = 4. ^C5 [corresponding]. Therefore: Theorem. In every triangle any exterior angle equals the sum of the two interior angles not adjacent to it. Therefore : 218. Theorem. The sum of the angles in any plane triangle is a straight angle. 219. Corollary. In a triangle, at least two angles are acute. The third angle may be acute, right, or obtuse ; and the tri- angle is called acute-angled, right-angled, or obtuse-angled, ac- cordingly. 220. In a right-angled triangle the side opposite the right angle is called the hypothenuse. 221. A triangle with two sides equal is called isosceles. 222. Theorem. If one side of a triangle be greater than a second, the angle opposite the first must be greater than the angle opposite the second. Fig. ioi. Fig. 102. 36 ELEMENTARY SYNTHETIC GEOMETRY. Given BA > BC. Draw bisector BD of ^ B, and' fold over along this axis. Then C falls on BA at C between B and A. Then 4(; (7 now appears as an exterior 2^ to A AC'D, and .". > 4- -^ ^ot adjacent. If one angle of a triangle is greater than a second, the side opposite the first must be greater than the side opposite the second. Proof. Given i^ C > 4. A. Draw the bisector BD of ^ B. Then is :^ ADB [= ^ {C -\- ^B)\ > ^ BDC [= ^ {A -\- ^B)] ; therefore on folding over along the axis BD, ^ BDC will fall within ^ ADB, and therefore C must fall be- tween A and B. 224. Corollary I. In an isosceles triangle, the angles op- posite the equal sides are equal. 225. Corollary II. If two angles of a triangle are equal, the triangle is isosceles. 226. If we join CC in the preceding figure then ^ DCC = ^ CCD, since A DCC is isosceles; .'. 2j^ ACC < 4C F?G. .04. " CCA ; .-. AC < AC. But AC = AB ~ BC. Therefore : Theorem. In every triangle the difference of two sides is less than the third side. AB -BC if within it [257]. 260. Corollary I. The vertices of all right-angled triangles on the same hypothenuse are concyclic. 261. Cor. II. If two opposite angles of a quadrilateral are supplemental, it is cyclic. Fig. 122. CHAPTER XII. THE SYMMETRICAL TRIANGLE. 262. The figure consisting of three points can only be sym- central if they are in the same straight : consequently no tri- angle has a symcenter. 263. In any triangle a sect joining a vertex to the mid point of the opposite side is called a median. Fig. 123. Fig. 124. 264. A perpendicular from a vertex to the opposite side is called an altitude. 265. The figure consisting of three points can only be sym- metrical with regard to an axis passing through one and bisect- ing at right angles the sect joining the other two ; consequently, every symmetrical triangle is isosceles, and has a median which is an altitude- and an angle-bisector. 266. If with the intersection of the equal sides of any isosceles triangle as center, and one of the sides as radius, we describe a circle, it will pass through the other two vertices. Therefore in every isosceles triangle the Fig. 125. median concurrent with the equal sides is an altitude- and an angle-bisector. So every isosceles triangle is symmetrical. 43 44 ELEMENTARY SYNTHETIC GEOMETRY. 267. Theorem. A triangle having a median which is an angle-bisector is isosceles. Proof. Produce this median BD to F, making DF=BD. ]o\n AF. A^/^/^ is symcentral to A aDBC; .: ^F= 4.CBD, and FA = BC. But ^CBD = ^DBA ; .-. :^F= ^DBA ; .-. FA = AB ; .-. AB = BC. F'G. 126. 268(«). Theorem. A triangle is symmetrical if two angle-bisectors are equal. p^ Proof. If ^OBC is not = ^OCB, suppose ^OBC> 4-OCB ; .: CD> BE. (304.) ^CBF=^ECB; tBCF= tEBC\ .: BF= CE, CF=BE. F.G.,.6W. ]o\n DF. Then, since BF ■= BD, .'. ^BFD = ^BDF; ^OCD < 4.OBE, ^COD = 4-BOE', .'. to DO ^OEB, .'. to DC > t-BFC. and by hypoth., and and THE SYMMETRICAL TRIANGLE. 45; Hence, subtracting ^BDF — 4-BFD, .'.4-FDC> tDFC\ ,:CF>CD, .'. BE > CD, .-. BE > and < CD; which is absurd. .'. :^OBC = ^OCB. 268 {b). If any triangle has one of the following properties, it has all : [i] Symmetry. [2] Two equal sides. [3] Two equal angles. [4] A median which is an altitude. [5] A median which is an angle-bisector. [6] An altitude which is an angle-bisector. [7] A perpendicular side-bisector which contains a vertex.. [8] Two equal angle-bisectors. CHAPTER XIII. THE SYMCENTRAL QUADRILATERAL. 269. A quadrilateral with a symcenter is called a parallelo- gram. (Ig'm). 270. Because it has symcentry, every parallelogram has its opposite sides parallel and equal, its opposite angles equal, and Fig. 127. ^ diagonals which bisect each other. Also, ■every straight through the symcenter cuts the parallelogram into congruent parts. 271. Theorem. A quadrilateral with 'O \ each side parallel to its opposite is a Fig. 128. parallelogram. Proof. Since for any two ||s the mid point of the sect they intercept on any transversal is a symcenter, .'. the mid point of a diagonal, being a symcenter for both pairs, is a symcenter for the quad. 272. Theorem. A quadrilateral with a pair of sides equal and parallel is a parallelogram. Proof. The mid point of a diagonal is a symcenter for the four vertices. 273. Theorem. A quadrilateral with each side equal to its c _!|B_ opposite is a parallelogram. Proof. Any vertex, B, is the only inter- \ section point of OA {^AE] with oC\CW\ on ° Fig. 129. that side of the center-straight A C. But a straight through A \\ to Z>C meets a straight through C \\ to DA at that point, since opposite sides of a ||g'm are equal. 46 THE SYMCENTRAL QUADRILATERAL. 47 Fig. 130. 274. Theorem. A quadrilateral with a pair of opposite sides equal and each greater than a diagonal \b' making equal alternate angles with the other sides, is a parallelogram. Prooif. For the mid point C of this diagonal is the symcenter of its end points ; and also of the other two vertices, since one of these, B, is the one intersection point of a semicircle, whose center O is one end point of this diagonal, with a ray starting in this diameter ; and the other, B', is the one intersection point of a semicircle and ray symcentral to those with regard to this diagonal's mid point. 275. If the sides given equal were less than the diagonal mak- ing equal angles with the other sides, then the first ray would start from without the first semi- circle and meet it twice [looking at a tangent as a secant through two coincident points]. 276. Theorem. A quadrilateral with a side equal to its opposite and less than a diagonal opposite €qual angles is a parallelogram. ' *^^* Proof. For the mid point of the diagonal is the symcenter Fig. 132. of its end points ; and also of the other two vertices, since one of them, B, is the one intersection point of an arc on this 48 ELEMENTARY SYNTHETIC GEOMETRY. diagonal as chord with a semicircle whose center is one end' point of this diagonal and radius a side less than it ; and the other vertex, B' , is the one intersection point of an arc and semicircle symcentral to those with regard to this diagonal's mid point. 277. A circle OO {f), on a ray from whose center a chord 00' is, can meet that chord only once ; but if it cuts the arc of that chord twice before meeting the chord, it never- meets the chord. 278. If the given equal sides, r, were greater than the diagonal OO' op- posite the equal angles, then the first semicircle would not meet the chord of the first arc, and so would intersect that arc twice. 279. Theorem. A quadrilateral with each angle equal to its opposite is a parallelogram. Proof. For then any two of the angles, not opposite equal the other two, and there- fore are supplemental. So each side is || to A . . Fig. 134. its opposite. 280. Theorem. A quadrilateral- ^_ \o__Tni^^3'B whose diagonals bisect each other is a parallelogram. Fig. 135. Proof. Their intersjection is then a symcenter for the four vertices. Fig. 133. CHAPTER XIV. SYMMETRICAL QUADRILATERALS. 281. A symmetrical quadrilateral with a diagonal as axis is called a deltoid. 282. A sect joining the mid points of the opposite sides of a quadrilateral is called a median. 283. A symmetrical quadrilateral with a median as axis is called a symtra. 284. Theorem. Every symmetrical quadrilateral not a deltoid is a symtra. Proof. For to every vertex corresponds a vertex, hence the number of vertices not on the axis must be even, — here four ; and the sects joining corresponding vertices are bisected at right angles by the axis, hence parallel, hence sides ; for the two diagonals of an undivided tetragon can never be parallel, since not every pair of conjugate angles made 49 Fig. 137. 50 ELEMENTARY SYNTHETIC GEOMETRY. by the diagonals with the four sides can be as great as a straight angle. 285. In any deltoid, since a diagonal is axis of symmetry, ^ therefore : [i] One diagonal [the axis] is the perpen- dicular bisector of the other. [2] One diagonal [the axis] bisects the angles at the two vertices. [3] Sides which meet on one diagonal [the axis] are equal ; so each side is equal to one of its adjacent sides. [4] One diagonal [not the axis] joins the vertices of equal angles, and makes equal angles with the equal sides. [5] The triangles made by one diagonal [the axis] are congruent, and their equal sides meet. [6] One diagonal [not the axis] makes two isosceles triangles. CONDITIONS SUFFICIENT TO MAKE A QUADRILATERAL A DELTOID. 286. Any quadrilateral which has one of the six preceding pairs of properties is a deltoid ; for from [i] that diagonal is an axis of symmetry ; from [2] that diagonal is axis ; from [3] a AB = AD and CB = CD, then the isosceles triangles ABD, CBD have a common axis of symmetry, AC. This follows also from [6] ; from [4] the perpendicular bisector of that diagonal must be axis of symmetry for the two equal angles, and their corresponding sides must intersect on it, hence it is a diagonal ; from [5] taking two adjacent sides equal, and the angle contained by them bisected by a diagonal, then the ends of these equal sides are corresponding points with regard to this diagonal as axis of symmetry. 287. Theorem. A quadrilateral with a diagonal which bi- sects the angle made by two sides, and is less than each of the S YMME TRICAL Q UADRILA TERALS. 51 Fig. 141. other two sides, and these sides equal, is a deltoid with this diagonal as axis. Proof. One of the two vertices not on this diagonal is the one intersection point of a semi- circle whose center is one end point of that BA'. 304. Theorem. If two triangles have two sides of the one respectively equal to two sides of the other, but the included angles unequal, then that third side is the greater which is op- posite the greater angle. Proof. Slide the triangles in the plane until a pair of equal sides coincide, and the other pair of equal sides have a common end point. Bisect the angle made by these equal sides. This axis cuts the third side which is opposite the greater angle. \ 305. Inverse. If two triangles have two sides Fig. 150a. qJ ^YiQ one respectively equal to two sides of the other, but the third sides unequal, then, of the angles opposite these third sides, that is the greater which is opposite the greater third side. 306. Theorem. If three parallels intercept equal sects on one transversal they intercept equal sects on every transversal. Proof. If, on a straight, AB = BC, and f through A, B, and C intersect another st' in A', B\ C, then st's through B' and C, drawn j| to BC, make ^ A^ 307. Corollary I. The intercept made on the mid of three parallels by two transversals differs from the intercepts on the others by equal sects. 308. Cor. II. If through the mid point of one side of a triangle a straight be drawn parallel to a second side, it will bisect the third side. Fig. 151. CONGRUENCE OF TRIANGLES. 57 309. Cor. III. If a straight parallel to one side of a trian- -gle cuts off any fractional part of a side, it cuts off the same fraction of the other side. 310. Inverse. The sect joining the mid points of any two sides of a triangle is parallel to the third side, and equal to half of it. 311. Corollary. The sect joining points which bound with any vertex of a triangle the same fractional parts of two sides is parallel to the third side and is that fractional part of it. ROTATION-CENTER. 312. Theorem. In a plane, the result of sliding any poly- gon is the same as of a rotation about a fixed point. Fig. 152. Proof. Join vertex A' with its trace A'\ and B' with B". The perpendicular bisectors of A' A" and B' B" intersect in the rotation-center O. For A A' OB' ^ A A" OB" [having three sides respectively equal]. Consequently t A'OA" = t B'OB". 313. The altitudes of a triangle are l b concurrent, and the point is called the \ triangle*s orthocenter. They must cointersect, since each con- tains the circumcenter of a triangle made by drawing through the vertices of the given triangle parallels to its sides. Fig. 153. 58 ELEMENTARY SYNTHETIC GEOMETRY. EXERCISES ON BOOK I. 1. If, with the vertex of an angle as center, two circles be described, and the points in which they cut its arms be joined, the joins are || or intersect on the angle's bisector. 2. In a -I- A two sides, two altitudes, two medians, two -^-bisectors are =, and cross on the axis. 3. Intersecting equal circles are -I- with regard to their common chord. 4. If about two given points as centers pairs of equal intersecting Gs be described, all the pairs have their common points on one straight. 5. If of two convex polygons one is wholly within the other, then the outer has the greater perimeter. 6. An interior angle of a regular dodecagon is what fraction of a r't 4. ? 7. If two sides of a A be produced through their common vertex until each is doubled, the join of the ends is || to the third side. 8. The join of the points of contact of || tangents to a O is a diam- eter. 9. If in a -I- A we decrease one of the equal sides and increase the other equally, the join of the points so obtained is bisected by the third side. 10. In A ABC, if r't bi' oi a cuts the st' d in D and c in E, then is 4 ABD = 4- ACE = dif between ^s B and C. 11. If from two p'ts of a st', ±s to another st' are =, either the st's are ||, or the sects from their cross to the p'ts are =. 12. The sum of the ±s to the = sides from any p't in the third side of a -I- A equals one of the = altitudes. 13. All equal sects between two ||s belong to two sets of \\s. 14. If from the vertices in the same sense on the sides of a ]|g'm a given sect be taken, the points so obtained are vertices of a ||g'm cosym- central with the first. 15. Find the bisector of an 4 without using its vertex. 16. A quad' with two sides || and the others = is either a ||g'm or a symtra. 17. If two sides of a quad' have a common r't bi', it is a symtra. 18. The r't bi's of the non-|| sides of a symtra cross on the r't bi' of the other sides. EXERCISES ON BOOK I. 59 19. If the diagonals of a quad' are =, its medians are ±. 20. If in a trapezoid three sides are =, then the angles adjacent to the fourth side are bisected by the diagonals. 21. The sects to the intersection points of a secant from the ± pro- jections of ends of a diameter on it are =. 22. A quad' is fixed by 5 given magnitudes. 23. An «-gon is fixed by 2« — 3 given magnitudes. 24. The bisector of an ^ of a a and the r't bi' of the opposite side cross on the circum-O. 25. The cross of an altitude (produced through its foot) with the cir- cum-0 is -I- to the orthocenter with respect to that side of the A. 26. Whether their vertex be on or within the O, a pair of vertical angles together intercept the same part of a O. 27. Vertical r't ^s with vertex on or within a intercept half of it. 28. Joining one common p't of two = intersecting Os to the crosses, of a secant through the other common point gives = sects. 29. If from one intersection p't of two = Os as center we describe any third circle cutting them, then the four intersection p'ts are vertices of a symtra whose non-|| sides go through the other intersection p't of the = Os. 30. A on the common chord of .two = Os as diameter bisects all sects through an intersection p't of the Os and ending in them. 31. A symtra is cyclic. 32. A deltoid is a circumscribed quad'. 33. The four ^-bisectors of a quad' make a cyclic quad'. 34. The four crosses of the inner with the outer common tangents to- two Os lie on a circle with their center-sect as diameter, 35. The sect of an outer between the inner tangents equals the sect of an inner between its points of contact. 36. Each side of a A is, by the p'ts of contact of the in-Oand an ex-O. divided into three sects, of which the outer two are =. 37. If a polygon has a circum-O and a concentric in-O, it is regular. 38. To make a regular hexagon, trisect the sides of a regular trigon and join the points next its vertices. 39. To make a regular octagon, about each vertex of a square, with half the diagonal as radius, describe a O and join the crosses next its vertices. 40. If a p't of its circum-O be joined to the vertices of a regular A^ the greatest sect equals the sum of the other two. BOOK 11. PURE SPHERICS. CHAPTER I. PRIMARY CONCEPTS. 314. A circle is a closed line that will slide in its trace. Though in itself unbounded and everywhere alike, yet it is finite. On it two points starting from coincidence and moving in opposite senses will meet. 315. Every point on a circle has one other on it such that the two bisect the circle. Two such are called opposite points. 316. If a pair of opposite points can be kept fixed while a circle moves, it describes a surface called a sphere. 317. A sphere is a closed surface which will slide in its trace. Though in itself unbounded and everywhere alike, yet it is finite, being generated completely by one finite motion of a finite line. 318. Assumption I. Any figure drawn on the sphere may be moved about in the sphere without any other change. 319. Assumed Construction I. Through any two points, in a sphere, can be passed a line congruent with the generating^ line of the sphere. In Book II. g-line will always mean such a line, and sect will mean a piece of it less than half. 60 PRIMARY CONCEPTS. 61 320. Assumption II. Two sects cannot meet twice on the sphere. If two sects have two points in common, their g-Iines coin- cide throughout. Through two points, not opposite points of a g-hne, only one distinct g-Hne can pass. 321. A piece of the sphere with part of a g-Hne as one of its boundaries, would fit all along the g-line from either side. 322. Because of the symmetry in its generation, the sphere is cut by any g-line on it into two equal parts, called hemi^ spheres. ' 323. If one end point of a sect be kept fixed, the other end point moving in the sphere describes what is called an arc, and the sect is said to rotate in the sphere about the fixed end point. The arc is greater as the amount of rotation is greater. 324. Two sects from the same point, when looked at with special reference to the amount of rotation necessary to bring their g-lines into coincidence, are said to form a spherical angle. The spherical angle is greater as the amount of rotation is greater. 325. When a sect has rotated just suf- ficiently to fall again into the same g-line, the angle described is called a straight angle, and the arc described is called a semicircle. 326. Half a straight angle is called a right angle. 327. The whole angle about a point in the sphere, that is, the angle described by a sect rotating until it coincides with its trace, is called a perigon ; the whole arc is called a circle. Fig. 155. 62 ELEMENTARY SYNTHETIC GEOMETRY. The fixed end point is called a pole of the circle, and the sect is called a spherical radius of the circle. 328. Assumed Construction II. A circle can be described from any pole, with any sect as radius. 329. Assumption III. All straight angles are equal. 330. Corollary I. All perigons are equal. 331. Corollary II. The two angles on the same side of a g-line, made by a sect with one extremity in that g-line, are together k straight angle. 332. Corollary III. Vertical angles are equal, being supplements of the same angle. 333. Theorem. Every g-line in the sphere cuts every other in two opposite points. Proof. Let BB' and CC be any two g-lines. Since each bisects the sphere, therefore the second cannot lie wholly in one of the hemispheres made by the first, therefore they intersect at two Fig. 157. points, which are therefore opposite. 334. On a sphere, every circle has two poles, which are opposite points, and its« spherical radius to one pole is the supple- ment of that to the other. 335. A spherical figure made by two half g-lines intersecting in opposite points, is called a lune. PRIMARY CONCEPTS. 63 336. Theorem. The angle contained by the sides of a lune at one of their points of intersection equals the angle contained at the other. Proof. Slide the lune, in the sphere, until each of the two intersection points falls on the trace of the other, and one of the half g-lines on the trace of the other. If the angles were unequal, the smaller could thus be brought within the trace of the greater, and its second half g-line would start between the traces, and since it could meet neither again until it reached the opposite inter- section point, we would find the surface of the lune less than its trace. 337. One quarter of a g-line is called a quadrant. 338. A spherical polygon is a closed figure, in the sphere, bounded by sects, no two of which cross. 339. A spherical triangle is a three- sided spherical polygon, with no interior angle greater than a straight angle. 340. A spherical triangle is positive [-(-] if a sect with one end pivoted within it and rotating counter-clockwise, after passing through the vertex of the greatest angle goes next over the vertex of the least. Fig. 161. CHAPTER II. SYMCENTRY ON THE SPHERE. 341. On a sphere a point has, with regard to a given sym- center, always one and only one symcentral point, namely, the one which ends the sect from the given point bisected by the symcenter. 342. Two figures are symcentral when they can be placed so as to have a sym- center. One figure is symcentral when it has a symcenter, that is, a point with respect to which every point of the figure has its- symcentral point on the figure. A lune is symcentral with regard to the cross of the g-line bisecting its angles, with the g-line bisecting its sides. 343. Symcentral figures on a sphere have precisely the same properties as in the plane, including congruence. 64 Fig. 162. CHAPTER III. SYMMETRY ON THE SPHERE. 344. Two points on a sphere are symmetrical with respect to a g-line, when it bisects at right angles the sect joining them. This g-line is called their axis of sym- metry. 345. Two points on a sphere have always one, and only one, symmetry axis on that sphere. 346. A point has, with regard to a given axis of symmetry, always one, and only one, symmetrical point, namely, the one which ends the sect from the given point perpendicular to the axis and bisected by the axis. 347. Two figures on the sphere have an axis of symmetry when, with regard to this g-line, every point of each has its symmetrical point on the other. 348. Two figures are symmetrical when they can be placed so as to have an axis of symmetry. 349. One spherical figure has an axis of symmetry when, with regard to this g-line, every point of the figure has its sym- metrical point on the figure. 350. One figure is symmetrical when it has an axis of sym- metry. 351. Any figure on the sphere has, with regard to any g-line on the sphere as axis, always one, and only one, symmetrical figure. 65 Fig. 163. 66 ELEMENTARY SYNTHETIC GEOMETRY. 352. One figure is symmetrical when it has an axis with regard to which its sym- metrical figure coincides with itself. 353. Every point on an axis is sym- metrical to itself. Fig. 164, 354. Assumption IV. The figure sym- metrical to a sect is an equal sect ; to a spherical angle, is a spherical angle equal in magnitude but opposite in sense. 355. Corollary. A sect, or g-line, or spherical angle, in one of two symmetrical figures, has a symmetrical sect, or g-line, '°' * ^' or spherical angle, in the other. 356. The intersection point of two sects is symmetrical to the intersection of two symmetrical to those. 357. The intersection points of two symmetrical g-lines are on the axis. 358. The bisector of a spherical angle is symmetrical to the bisector of the symmetrical spherical angle. 359. The angle between two symmetrical g-lines is bisected iby the axis. 360. Two g-lines are symmetrical with regard to. either of their angle-bisectors. For there is a g-llne symmetrical to the first with regard to that angle-bisector, and the angle between the two symmetri- cal g-lines is bisected by the axis. 361. Any g-line is symmetrical with regard to any of its perpendiculars. 362. Any circle is symmetrical with Fig. 166. regard to any of its spherical diameters. SYMMETRY ON THE SPHERE. 67 363. Every point on the perpendicular bisector of a sect is the pole of a circle passing through its end points. For A-\'B\ axis CD ; .-. CA = CB. Thus sects from any point on its per- pendicular bisector to the end points of the sect are equal. 364. The perpendicular bisector of a spherical chord contains the poles of the circle. For the end points of the chord ^'<^- ^^^• are symmetrical with regard to this perpendicular, and also with regard to the perpendicular from a pole. 365. Two circles with three points in common coincide. 366. One spherical radius, of every circle on the sphere, is less than a quadrant. Call its pole the q-pole, and it the q-radius. 367. If the q-pole-sect of two circles equals the sum of their q-radii, they have a common point on their q-pole-sect, and by symmetry no other common point. Such circles are said to be tangent ex- ternally. Neither has a point in common with a circle concentric with the other, but of lesser q-radius. 368. If the q-pole-sect equals the difference of the q-radii, the two circles have a common point on their pole-g-line, and by symmetry, no other common point. Such circles are said to be tangent in- ternally. Neither has a point in common with a circle concentric with the lesser and of lesser q-radius. Fig. 168. Fig. 169. 68 ELEMENTARY SYNTHETIC GEOMETRY. 369. While the q-pole-sect is growing, from equality with the difference of the q-radii, up to equality with their sum, the two circles have always two common points, symmetrical with regard to their pole-g-line. 370. Problem. To make a spherical triangle, given its sides. Construction. If two of its sides are each less than a quadrant, then with these as q-radii, and the end points of the third side as poles, describe two circles. Their two common points will be the third vertices of two symmetrical triangles with the three given sides. If two of the given sides are each greater than a quadrant, take, in the above, their supplements with the given third side. Then in the two triangles obtained, produce these two supple- ments until they meet. These two meeting points will be the third vertices of two symmetrical triangles with the three given sides. 371. Corollary I. Any two sides of a spherical triangle are together greater than the third. For if two be each less than a quadrant, and together equal to the third, the construction circles will be tangent extenally. If two be each greater than a quadrant, their difference is that of their supplements, which is less than the third side ; for if equal to it, the construction circles would be tangent inter- nally. 372. Corollary II. The sum of the three sides of any spherical triangle is less than a g-line. 373. Since any chord is bisected by the perpendicular from a pole, .*. a g-line J_ to a diameter at an end point has only this point in common with the circle. Fig. 170. This point of the circle is symmetrical to itself with regard to this diameter as axis. But if we draw through this point B any g-line BF not_L to SYMMETRY ON THE SPHERE. 69 the spherical radius AB, then the perpendicular from a pole A will meet this g-hne BF at some other point E. Hence the g-line BF cuts the circle again at B' •{• B, axis AE; .-. Theorem. At every point on the circle one, and only one, tangent can be drawn, namely, the perpendicular to a radius at that point. 374. Let P be a point not in the g-line g, 2iX\dPC^_tog : then^ is tangent to 0P[/'C] at C. If PC is less than a quadrant, any second circle with q-radius < PC, and q-pole P, lies wholly within © P [_PC\ Therefore: Theorem. If less than a quadrant, the perpendicular is the least sect between a point and a g-line. 375. The poles of all circles tangent to two intersecting g-lines are in their angle- bisectors. 376. From any point on an angle-bisec- tor the perpendiculars to the g-lines are equal. 377. Theorem. If two angles of a triangle be equal, the triangles is isos- celes. Proof. The perpendicular bisector of the side joining the equal angles is the symmetry axis for that side and its end points, and so for angles made with that side at those points which are equal in magnitude and opposite in sense. 378. Theorem. If one angle of a spherical triangle is greater than a second, the side opposite the first must be greater than the side opposite the second. Fig. 17a. Fig. 173. 70 ELEMENTARY SYNTHETIC GEOMETRY. Fig. 174. Proof. Given the 4.C> 4. A. If i. DCA = 4A, then DC=DA. But DC+ DB > BC; .-. DA + DB > BC. 379. Inverse. If one side of a spheri- cal triangle is greater than a second, the angle opposite the first must be greater than the angle opposite the second. Proof. For the angle opposite the second cannot be the greater, nor can they be equal. 380. Theorem. In an isosceles triangle the angles opposite the equal sides are equal. Proof. The bisector of the angle between the equal sides is a symmetry axis for those sides and their end points, hence for the triangle. 381. Corollary. In an isosceles triangle the bisector of the angle between the equal sides is perpendicular to the third side. 382. If the vertices of a polygon are concyclic, the polygon may be called cyclic. 383. In a cyclic quadrilateral, the sum of one pair of opposite angles equals the sum of the other pair. Proof. Join the circumcenter E with A, B, C, D, the vertices. By isosceles tri- FiG. X7S. angles, ^ ABC = ^ BAE -\- ^ ECB, and CDA = t DCE + ^ DAE. CHAPTER IV. THE SYMCENTRAL QUADRILATERAL. diagonal making 384. A symcentral spherical quadrilateral, or cenquad, has its opposite sides equal, its opposite angles equal, and diagonals which bisect each other. Also, every g-line through the symcenter cuts the cenquad into congruent parts. 385. Theorem. A quadrilateral with a with each side an angle equal to its alternate, is. a cenquad. Proof. The mid point of this diagonal is a symcenter for both pairs of opposite sides. 386. Theorem. A quadrilateral with a pair of opposite sides equal and making equal alternate angles with a diagonal, is a cenquad. Proof. The mid point of the diagonal is a symcenter for the four vertices. 387. Theorem. A quadrilateral with a pair of opposite sides equal, and a diagonal making equal alternate angles with the other sides and opposite angles not supplemental, is a cenquad. Proof. The mid point of this diagonal is the symcenter of its end points; and also of the other two vertices, since one of these is an intersection point of'a semicircle, of which a diame- ter is bisected by one end point of this diagonal, with a g-line through its other end ; and the other is the symcentral inter- 71 Fig. 176. 72 ELEMENTARY SYNTHETIC GEOMETRY. section point of a semicircle and g-line symcentral to those with regard to this diagonal's mid point. 388. Theorem. A quadrilateral with each side equal to its opposite is a cenquad. Proof. Any vertex, B, is the only inter- section point of QA [AB'] with OC [CB] on that side of their pole-g-line, AC. But the fourth vertex of a cenquad with Fig. 177. sides CD = AB and DA = CB, and symcen- ter the mid point of AC, is that point B, 389. Theorem. A quadrilateral whose diagonals bisect each other is a cenquad. Proof. Their intersection is then a symcenter for the four vertices. CHAPTER V. SPHERICAL TRIANGLES. 390. Theorem. Spherical triangles of the same sense are congruent if they have a side and two angles adjacent to it equal ; or two sides and the included angle equal ; or two sides and the angles opposite one pair equal, opposite the other pair not supplemental; or three sides equal. Proof. Slide the two triangles in the sphere until a pair of equal sides coincide, but beyond this common side are no co- incident points. The triangles are then ^'°- ^'''^• symcentral with regard to the mid point of the common side. 391. Triangles which would be con- gruent, but that they differ in sense, are symmetrical. Symmetrical triangles are of opposite sign. 392. Corollary. Symmetrical isosceles spherical triangles are congruent. For the equality of two angles in a tri- angle obliterates the distinction of sense or sign. 73 Fig. i8o. 74 ELEMENTARY SYNTHETIC GEOMETRY. 393. Theorem. An exterior angle of a spherical triangle is greater than, equal to, or less than either of the interior opposite angles, according as the median from the other interior opposite angle is less than, equal to, or greater than a quadrant. Proof. Let A CD be an exterior angle "f^^T"^ of the A ABC. Bisect AC dit E. Join. BE, and produce to F, making EF ^= BE. Join EC. 2 ABE ^ 2 CFE. [Spherical triangles of the same sense having two sides- and the included angle equal are congruent.] .-. 4.BAE = ^FCE. If, now, the median BE be a quadrant, BEF is a half-g-line, and F lies on BD; .-. 4.DCE coincides with :^FCE^ .'. tDCE = 4.BAE. If the median BE be less than a quadrant, BEF is less than a half-g-Hne, and F lies between CD and AC\ .'. ^DCA > ^FCE, .-. DCA > ^BAC And if BE be greater than a quadrant, BEF is greater than a half-g-line, and F lies between CD and AC produced through C; .: ^DCA < 4.FCE, .: DCA < ^BAC. Thus, according as BE is greater than, equal to, or less than a quadrant, the exterior ^ACD is less than, equal to, or greater than, the interior opposite ^BAC. 394. Inversely, according as the exterior angle A CD is greater than, equal to, or less than the interior opposite angle BAC, the median BE is less than, equal to, or greater than a quadrant. SPHERICAL TRIANGLES. n Fig. 182. 395. Theorem. Any two perpendiculars to a g-line intersect in two points, from either of which all sects drawn to that g-line are quadrants perpendicular to it. Proof. Let AB and CB, drawn at right angles to AC, intersect at B, and meet AC again at A' and C, respectively. Then ^BA'C'=^BAC' and i^BCA' = ^BCA'. [The angles contained by the sides of a lune, at their two points of intersection, are equal.] Moreover, AC = AC, for they have the common supple- ment A C. Hence, keeping A and C on the line A C, slide ABC until ^6" comes into coincidence with A'C. Then the angles 2itA,C,A', C being all right, AB will lie along A'B, and CB along CB, and hence the figures ABC 2iV\d A' BC coincide. Therefore each of the half-lines y^^^' and CBC is bisected at^. In like manner, any other line drawn at right angles to AC passes through B, the mid point of ABA'. Hence every sect from AC to B \s3i quadrant ^ to AC. 396. Corollary I. A g-line is a circle whose spherical radius is a quadrant. 397. Corollary H. A point which is a quadrant from two points in a g-line, and not in the g-line, is its pole. 398. Corollary HI. Equal angles at the poles of lines inter- cept equal sects on those lines. 399. T^iQ polar of any point is the g-line of which that point is a pole. 400. If an angle be a fraction of a perigon, it intercepts on the polar of its vertex that fraction of a g-line. 401. Theorem. If a median is a quadrant, it is an angle- bisector, and the sides of the bisected angle are supple- mental. v^ ELEMENTARY SYNTHETIC GEOMETRY. Proof. The quadrant and the sides BA, BC, all produced, are concurrent in B' opposite to B. .'. ABCB'A, is a cenquad [its diagonals AC and BB' bisect each other]. .-. AB=CB', and AB, BC are supplemental. Also, the complements of AB and CB' are equal, i.e., AH ^^ CF \ also, the supplements of 4. BAD and ^ BCD, or t DAH = ^DCF; .'. 2AD//^ a CD F[t\vo sides and included angle]; .-. ND = DF; .-. ^ ABC is bisected by BD. 402. Corollary. If two sides of a triangle are supplemental, the opposite angles are supplemental. 403. Theorem. Two spherical triangles, of the same sense, having two angles of the one equal to two angles of the other, •the sides opposite one pair of equal angles equal, and those opposite the other pair not supplemental, are congruent. Proof. Given 4B=:^E; 4- C — F ; AB = DE; AC not supplemental to FD. Make DE coincide with AB : then EF will lie along BC, and FD must coincide with AC; else would it make a a A GC v^'xth. Fig 184.- exterior 4. AGB = interior opposite 4 C, and .'. with median a quadrant, and .*. with ^6^ supplemental to AG, that is, with AC supplemental to FD. 404. Theorem. Two spherical triangles of the same sense, having in each one, and \ only one, right angle, equal hypothenuses, and another side or angle equal, are con- gruent. Proof. If 4C=4H=x't4, and c =-- h, Pig. i8s. ^nd ^ =/, then if AC > g, make CD — g; .\ BD = h = c, and the bisector of 4DBA is X to CD A, ,\ Bis pole to CDA, .'. 4 A is also r't. SPHERICAL TRIANGLES. 77 Fig. i86. If t C = t H — x\ t, and c = //, and t A =4- F, then if ^ ABC > ^ G, make 4. ABD = 4. G, .'. 4- BDA = ^ H ^^ C = r't ^, .: B is pole to CD A. 405. Theorem. Of sects joining two sym- metrical points to a third, that cutting the axis is the greater. Proof. BA=BC-^ CA = BC + CA'>BA'. 406. Theorem. If two spherical triangles have two sides of the one equal to two sides of the other, but the included angles unequal, then that third side is the greater which is opposite the greater angle. Proof. Slide the triangles in the sphere until a pair of equal sides coincide and the other pair of equal sides have a common end point. Bisect the angle made by these equal sides. This axis cuts the third side, which is opposite the greater angle. 407. Inverse. If two triangles have two sides of the one equal to two sides of the other, but the third sides unequal, then of the angles opposite these third sides that is the greater which is opposite the greater third side. 408. Theorem. The g-line through the poles of two g-lines is the polar of their intersection points. Proof. If A and B are poles of the g-lines a and b, which intersect in P, then AP and BP diXQ quadrants ; .•. AB is the polar of P. 409. Corollary I. The g-line through the poles of two g-lines cuts both at right angles. 410. Corollary II. If three g-lines are concurrent, their poles are collinear. 411. Of the sides of a spherical angle, we may call those Fig. 187. Fig. 188. 78 ELEMENTARY SYNTHETIC GEOMETRY. poles positive from which in the figure these sides would be described from the vertex by a quadrant rotating positively. 412. Theorem. The sect which an angle intercepts on the polar of its vertex equals the ^sect between the positive poles of its sides. Proof. Slide the quadrant BF along the polar of A until B comes to C. The -|- pole F oi.AB will then coincide with the -f- pole G oiAC. 413. The sect joining any point to one pole of a g-line is less than a quadrant if the two points are in the same one of that g-line's hemispheres ; greater than a quadrant if they are in different hemispheres. By a pole's hemisphere we mean that one of its g-line's hemispheres in which the pole is. 414. Of a given spherical triangle ABC, the polar is a new triangle A'B' C\ where A' is that pole of 56^ which has A in its hemisphere, and B' that pole oi AC which has B in its hemisphere, and C that pole of AB which has C in its hemisphere. 415. Theorem If of two spherical tri- angles the second is the polar of the first, then the first is the polar of the second. Hypothesis. Let ABC be the polar of A'B'C. Conclusion. Then A'B'C is the polar of ABC. Proof. Join A'B and A'C. Since B is pole of A' C, therefore BA' is a quad- rant ; and since C is pole of A 'B', therefore CA' is a quadrant; .*. A' is pole of BC. In like manner, B' is pole of AC, and F.G. X9». C of AB. Fig. 191. SPHERICAL TRIANGLES. 79 Moreover, since A \\a.s A' in its hemisphere, ,-. the sect ^^' is less than a quadrant, .*. A' has A in its hemisphere. 416. Theorem. In a pair of polar triangles, any angle of either intercepts, on the side of the other which lies opposite to it, a sect which is the supplement of that side. Proof. Let ABC and A'B'C be two polar triangles. Produce ^'^' and A' C to meet BC at D and E, respectively. Since B is the pole of A'C, therefore BE is a quadrant ; and since C is the pole of A'B', therefore Fig. 193. CD is a quadrant; therefore BE -\- CD =^ hdM-g-Wwe-, but BE -^ CD =z BC -\- DE, Therefore DE, the sect of BC which A' intercepts, is the supplement of BC, 417. Theorem. Two spherical triangles of the same sense, having three angles of the one equal respectively to three angles of the other, are congruent. Proof. Since the given triangles are respectively equiangu- lar their polars are respectively equilateral. For equal angles at the poles of g-lines intercept equal sects on those lines ; and these equal sects are the supplements of corresponding sides. Hence these polars, having three sides equal, are respectively equiangular, and therefore the original triangles are respectively equilateral. 418. Of a convex spherical polygon ABCD is a new spherical polygon A'B'C'D' ..., where A' is that pole of BC which has A in its hemisphere, etc. 419. Theorem. The polar of a cenquad is a concentric cenquad. Proof. The g-line HK through the sym- center O and _L to ^^ is also _L to CD ; and OH =: OK are the complements of the sects from O to poles D' and B' of the sides AB and CD. the polar Fig. 194. 8o ELEMENTARY SYNTHETIC GEOMETRY:. Hence O is symcenter for B' and D'. In the same way prove O symcenter for A' and C 420. Theorem. The opposite sides of a cenquad intersect on the polar of its symcenter. Proof. O is symcenter for F and F'. 421. Theorem. Any two consecutive vertices of a cenquad' p^. ~-^c and the opposites of the other two are con- cyclic. Proof. The perpendiculars, to the g-Iine through O and bisecting BC and DA, from A Fig. 195. and B in one of its hemispheres and C and D in the other, are equal. So also the perpendiculars from their opposites D' and C in the first hemisphere and A' and B' in the second. So A, B, C't D' and A\ B\ C, D are on equal circles with opposite q-poles. Such circles are called parallels ; the co-polar g-line, equator. 422. The perpendiculars erected at the mid points of the sides of a spherical triangle are concurrent in its circumcenter.^ 423. Theorem. The g-line bisecting two sides of a triangle intersects the third side at a quadrant from its mid point. Proof. AL, BM, CN are X to the g-Hne through A\ B\ the mid points of two sides BC, CA, D' and meeting the third side pro- duced at n and B'. .-. ALB' = CNB' [having the right angle, hypothenuse, and one oblique Q angle equal], :.AL = CN. Fig. ,96. Similarly, BM = CN. .'. ALD = BMD' [having two angles and an opposite side- equal, and the other pair of opposite sides not supplemental]. .-. AD- BD', .-. DC a quadrant. SPHERICAL TRIANGLES. 8 1 424. Corollary I. The altitudes of a spherical triangle are concurrent in its orthocenter. For, regarding A'B'C as the triangle, the perpendicular to DC at C is the polar of D, and .*. J. to A'B'. Similarly, the perpendicular to BA' at A' is _L to B'C, etc. So the three altitudes of A'B'C are concurrent in the cir- cumcenter of ABC. 425. Cor, II. The vertices of spherical triangles of the same angle sum on the same base are on a circle co-polar with the g-line bisecting their sides. For AO = BO, ^ OAB = 4. OBA, ^ LAB = ^ MBA = i[A + B-{-Cl Hence AOB is fixed, and .-. (9C [supplemental to OA]. 426. Theorem. The g-lines through the corresponding ver- tices of a triangle and its polar are con- current in the common orthocenter of the two triangles. Proof. For A A' is _L to BC and B'C, b'< since it passes through their poles. 427. Theorem. The sides of a triangle ^'*^- 's?- intersect the corresponding sides of its polar on the polar of their orthocenter. Proof. For AA' is the polar of the intersection points of BC and B'C ; similarly, BB' is the polar of the intersection points of CA and CA', etc. Sects from the orthocenter to these intersection points are all quadrants. a' 428. Theorem. A triangle's in-center is also its polar's circumcenter ; and R is complemental to r. Proof. ID _L to BC contains A'. .'. /A' is b" the complement of r. So is IB' and IC. ^"'- '^- 82 ELEMENTARY SYNTHETIC GEOMETRY. ■ I EXERCISES ON BOOK II. I. Explemental -^s at the q-poles of = Os intercept explemental arcs. Fig a. Fig. B. 2. Explemental arcs of equal circles have equal spherical chords. 3. As a spherical chord increases its major arc decreases. 4. If Os pass through 2 given p'ts their cen- ters all lie on the r't bi' of the join of the 2 p'ts. 5. If 2 Os touch internally, a 1 to the di- ameter through the p't of contact has equal pieces between the 20 s. 6. The g-lines on which ±s from a fixed p't are equal envelop a O with this p't for center. 7. The centers of Os touching two given g-lines all lie on the bisectors of the 2(^s made by these g-lines. 8. The centers of Os touching 3 given g-lines lie on the bisectors of the ^s made by these g-lines. 9. If a quad' is cyclic, the r't bi's of its sides and of its two diagonals are concurrent. 10. ABCD is a cyclic quad' ; AD, ^Cmeet in F. Where does tan at F to circum-o CDF ^G- D' meet AB ? EXERCISES ON BOOK II. 83 II, One convex polygon wholly contained within another has the lesser perimeter. Fig. E. Fig. F. 12. The perimeter of any a is less than a g-line. The perimeter of any convex spherical polygon is less than a g-line. 13. If 2 Os touch, and through the p't of contact a g-line be drawn to cut the Os again, where will the tangents at these crosses meet.? 14. If 2 Os touch, and through the p't of contact 2 g-lines be drawn cutting the Os again, where will the joins of these crosses meet.? 15. If the common chord of 2 intersecting ©s be produced to any p't, the tangents to the 2 os from this p't are =; and inversely. 16. If the common chord of 2 intersecting os be produced to cut a common tangent, it bisects it. 17. The 3 common chords of 3 Os which intersect each other are concurrent. 18. How do the in-, circum-, and ex-radii of a regular a^ compare in size? 19. If a quad' can have a circle inscribed in it, the sums of the oppo. site sides are equal. 20. If two equal Os intersect, each contains the orthocenters of As inscribed in the other on the common chord as base. 21. Three equal Os intersect at a p't H, their other points of inter- section being A, B, C. Show that H is orthocenter of a ABC; and that the triangle formed by joining the centers of the circles is ^ to A ABC. 22. The feet of Is from A ol A ABC on the external and internal bi's of ^s B and C are co-st' with the mid p'ts of 6 and c. Does this hold for the sphere ? 23. If two opposite sides of a quad' are =, they make = ^s with the median of the other sides. Prove for the plane, then extend to the sphere. 84 ELEMENTARY SYNTHETIC GEOMETRY. 24. (Bordage.) The ceniroids of the 4 As determined by 4 concyclic p'ts are concyclic. 25. The orthocenters of the 4 As determined by 4 concyclic p'ts A, B, C, D are the vertices of a quad' ^ to ABCD. The in-centers are vertices of an equiangular quad'. 26. CBrahmegupta.) If the diagonals of a cyclic quad' are 1, the J- from their cross on one side bisects the opposite side. 27. If the diagonals of a cyclic quad' are J-, the feet of the Is from their cross on the sides and the mid p'ts of the sides are concyclic. 28. It tangents be drawn at the ends of any two diameters, what sort of a quad' is circumscribed } 29. In any equiangular polygon inscribed in a ©, each side is equal to the next but one to it. Hence, if an equiangular polygon inscribed in a G have an odd number of sides it must be equilateral. Any equilateral polygon inscribed in a ©is equiangular. 30. In any equilateral polygon circumscribed about a O, each 4^ is = to the next but one to it. Hence, if an equilateral polygon circumscribed about a circle have an odd number of sides, it must be equiangular. Any equiangular polygon described about a O is equilateral. 31. The circle through any 3 vertices of a regular polj'gon contains the remaining vertices. 32. If one of 2 equal chords of a bisects the other, then each bisects the other. 33. Given 2 symcentral g-lines and their symcenter. Find the g-line symcentral to a third given g-line with respect to this symcenter. 34. All =A' on the same side of the same base have their sides bisected by the same g-line. BOOK III. EQUIVALENCE. df2C). Magnitudes are equivalent which can be cut into parts congruent in pairs. 430. Problem. To describe a rectangle, given two consecu- tive sides. Construction. Draw a straight, erect to it a perpendicular. From the vertex of the right angle lay off one given sect on the straight, the other on the perpendicular. Through their second end points draw parallels, one to the straight, one to the perpendicular. 431. Corollary, A rectangle is completely deterrhined by two consecutive sides ; so if two sects, a and b, are given, we may speak of the rectangle of a and b, or we may call it the rectangle ab. Thus, when a and b are actual sects, we mean ^ by ab a definite plane figure with four right Fig. 199. angles, four sides, and an enclosed surface. 432. The sum of two polygons is any polygon equivalent to them. 433. Theorem. In any right-angled triangle, the square on the hypothenuse is equivalent to the sum of g' the squares on the other two sides. y^ \ Hypothesis, a ABC, r't angled at B. Conclusion. Square on AB-\-sq^ on BC = sq' on AC. Proof. By 430, on hypothenuse AC, on ^ the side toward the A ABC describe the sq' ADFC. 85 86 ELEMENTARY SYNTHETIC GEOMETRY, On the greater of the other two sides, as BC, lay off CG = AB. Join FG. Then, by construction, CA = FC, and AB = CG, and ^ CAB = 4. FCG, since each is the complement of ACB; .: a ABC ^ a CGF. Rotate the a ABC about A through a minus r't ^ ; this brings^ to B'. Likewise rotate CGF about F through a + r't ^ ; this brings G to G\ The sum of the angles a.t I) = st' ^. .*. G'D and DB' are in one straight. Produce GB to meet this straight at H; then BC = GF = FG'; and r't ^ 6^ = ^ GFG' = 4 FGH\ .: GFG'H equals, square on BC Again, BA = AB' , and r't ^ B' = ^ B'AB = ^ ABH\ .'. ABHB is the sq' on AB. .-. sq' o{AC= sq' of AB + sq' of BC. 434. An altitude of a parallelogram is a perpendicular from a point in one side to the straight of the opposite side, which is then called the base. 435. Theorem. A parallelogram is equivalent to the rect- G D F c angle of either altitude and its base. I -J — -I f' ° j / I / Proof. If CD, the side of the |g'm opposite j / I / the base AB, contains F, a vertex of the rect- i / j / angle, then ABFD ^ ABFD, and A BCF ^ 1/ / A ADG. A B Fig. 201, If the sides AD, BF intersect in H, then, G F_ D c by continued bisection, cut BF into equal I / / parts each less than BH. Through these y( / points draw straights \ to the base, so ~/\ ~~~y dividing the rectangle into congruent rect- ./- — v-^ angles, each as above, equivalent to the ^ corresponding parallelogram. Fig. 202- 436. Corollary. All parallelograms hav- ing equal altitudes and equal bases are equivalent. EQUIVALENCE. 87 /\ \ ^F 437; Theorem. A triangle is equivalent to the 1 bctangle of its base and half-alti- tude. Procf . Join the mid points D, M of the Fig. 203. sides CB, BA of A ABC, and produce MD' = MD. Then aAD'M^ aBDM. 438. Corollary. Triangles of equal bases and altitudes are equivalent. 439. Theorem. A trapezoid is equivalent to a triangle of equal altitude, whose base is the sum of the parallel sides. Proof. Join AC, BD. To DA produced ^ ^^ dr2iw BF\to CA. F.G.204. A BCD = A BCA ^ A AFB. 440. Theorem. The sect cut out, on a parallel to the base of a triangle by the sides, is bisected by the corresponding median. Proof. Let AI be the mid point of PQ \ to AB. t PMC = t QMC\ also trapezoid AC'MP= trapezoid C'BQM; .'. AC MCA = CBCMC. Were M not in CC, but on Q's sivle, then AC MCA > A ACC> CBCMC. 441. Theorem. Sects joining intersections of the sides of a parallelogram with straights drawn parallel d p rg c to the sides through a point on one diagonal, if they cut that diagonal, are parallel to the other. Proof. Through O draw QR \\ to BC, cutting HK in 5. Since DR = RC, .: MS = S£, and HM = EK=FB', .: HMBF is a ||g'm. Again, since MK = HE = DG, .-. MKGD is a ||g'm. 442. Corollary. If through any point on a diagonal of a parallelogram straights be drawn parallel to the sides, the two 88 ELEMENTARY SYNTHETIC GEOMETRY. parallelograms, one on each side of this diagonal, will be equivalent. For through E drawing NP \ to BD, we get a FBL ^ A HMD, and a ENK ^ a PEG, and Ig'm BNEL = ||g'm DMEP. 443. Theorem. Any angle made with a side of a spherical triangle by joining its extremity to the circumcenter, equals half the angle-sum less the opposite angle of the triangle. Proof. For ^.A-^^B^tC— 2 4- OCA +2 4. OCB ± 2 ^ OAB .'. ^ OCA =\ltA^^B-\-tC\- [^ OCB ± ^ OAB] = ^[^ A -^ 4 B F,o..o,. +^C]-^B. 444. Corollary. Symmetrical spherical triangles are equiva- lent. For the three pairs of isosceles triangles formed by joining the vertices to the circumcenters, having respectively a side and two adjacent angles equal, are congruent. 445. Theorem. When three g-lines mutually intersect, the two triangle^ on opposite sides of any vertex are together equivalent to the lune with that vertical angle. Proof. A ABC -{- A ADF= lune ABHCA. For DF — BC, having the common Id supplement CD ; and FA = CH, hav- ing the common supplement HF\ and AD = BH, having the common supple- ^ ment ND; .: 2 ADF = 2 BCH , .: Fig. 208 a. A ABC + A ADF = 2 ABC 4- A BCH = \\xnQ ABHCA. 446. The spherical excess, e, of a spherical triangle is the ex- cess of the sum of its angles over a straight angle. In general. : EQUIVALENCE. 89 the spherical excess of a spherical polygon is the excess of the sum of its angles over as many straight angles as it has sides, less two. 447. Theorem. A spherical triangle is equivalent to a lune whose angle is half the triangle's spher- ical excess. Proof. Produce the sides of the ^ ABC until they meet again two and b| two at D, F, and H. The ^ ABC now forms part of three lunes, whose angles are A, B, and C, respectively. But, by 436, lune with ^ A = £\ fTg. 2o87. ABC-\-2ADF. Therefore the lunes whose angles are A, B, and C are to- gether equal to a hemisphere plus twice a ABC. But a hemisphere is a lune whose angle is a straight angle; .*. 2 a ABC = lune whose ^ is [A -\- B -\- C — st. ^ ] = lune whose ^ is e. 448. Corollary I. The sum of the ^s of a a is > a st' ^ ^nd < 3 st' ^ s. 449. Cor. II. Every :^ of a a is > ^e. 450. Cor. III. A spherical polygon is equivalent to a lune whose angle is half the polygon's spherical excess. 451. Cor. IV. To construct a lune equivalent to any spheri- cal polygon, add its angles, subtract [w — 2] st' ^s, halve the remainder, and produce the arms of a half until they meet again. 90 ELEMENTARY SYNTHETIC GEOMETRY. EXERCISES ON BOOK III. I. The joins of the centroid and vertices of a triangle trisect it. Proof. A ABM = A MBC, A AGM = A MGC\ .-. A ABO = A GBC. I. Make a ||g'm triple a given ||g'm. 3. Make a A triple a given A. 4. Make a symtra triple a given symtra. 5. Trisect a given symtra. 6. {a + by t: a' + zab + ^». 7. {a — by = a' - 2ab + P. 8. {a + b){a — b) = a" - b\ 10. On each side of a quad' describe a sq' outwardly. Of the four as made by joining their neighboring corners, two opposite equal the other two and equal the quad. II. (Pappus.) Describe on two sides AB, AC of a A any ||g'ms (both outwardly or both inwardly). Designate the cross of the sides opposite b and c by F. On the st' FA from a cut off A'// = AF. Con- struct a II g'm on a whose opposite side goes through //. It equals the sum of the other two. 12. If from an ^ a we cut two = as, one •!• , the sq' of one of the = sides of the -I- A equals the rectangle of the sides of the other a on the arms of the ^ a. 13. Transform a given A into an = -j- a. 14. Transform a given a into an = regular a. 15. If a vertex of a A moves on a ± to the opposite side, the differ- ence of the squares of the other sides is constant. 16. The ^ bisectors of a rectangle make a sq', which is half the sq' on the difference of the sides of the rectangle. 17. The bisectors of the exterior 2(1 s of a rectangle make a sq' which is half the square on the sum of the sides of the rectangle. 18. The sura of the squares made by the bisectors of the interior and EXERCISES ON BOOK III. 9r exterior^ ^of a rectangle equals the square on its diagonal ; their differ- ence is double the rectangle, 19. If on the hypothenuse we lay off from each end its consecutive side, the sq' of the mid sect is double the rectangle of the others. 20. If in A ABC, the foot of altitude from A be D, from C be ^, then rectangles BD . a = BF . c. (Hint. From 4- B are two r't as cut off. Turn one about the bi- sector of 4(! B.) 21. In a trapezoid, the sum of the squares on the diagonals equals the sum of the squares on the non-J sides plus twice the rectangle of the H sides. BOOK IV. PROPORTION. 452. A greater magnitude is said to be a multiple of a lesser magnitude when the greater is the sum of a number of parts each equal to the lesser; that is, when the greater contains the lesser an exact number of times. The lesser is then called a submultiple of the greater. 453. Any multiple of any submultiple of a magnitude is called d, fraction of that magnitude. 454. Two magnitudes of which neither is a fraction of the other are called incominensiirable ; for example i and 4^2. 455. That definite mmterical relation of any magnitude to any magnitude of the same kind, in virtue of which the former is either a fraction of the latter or is greater than one and less than the other of two fractions of the latter differing by less than any given fraction however small, is called the ratio of the former to the latter. 456. If the first of two magnitudes is a fraction of the second, the ratio of the former to the latter is expressed by the numerical fraction whose denominator is the number indicating the submultiple of the second, and whose numerator is the number indicating the multiple of that submultiple. Thus the ratio of a foot to 8 inches is 3/2. 457. The ratio of the first of two magnitudes to the second is said to be greater than a numerical fraction expressing the ratio, to the second, of any magnitude less than the first. 458. Two ratios are equal if no numerical fraction is greater than one and less than the other. 459. When the ratio of two magnitudes A and B, which 92 PROPORTION. 95. may be written A/B, equals that of the other two a aiid b, the. four are said to form a proportion ; which may be written A/B = a/b. 460. Theorem. If to every one of a series of magnitudes A, B, C, . . . there corresponds one of a second series. «, b, c, . . . in such manner that, 1. If the magnitudes A and B are equal, so are also the. corresponding magnitudes a and b ; and, II. The sum vS" of two magnitudes A and B corresponds to the sum s of the corresponding magnitudes a and b', Then two magnitudes of the first series have the same ratio, as the corresponding magnitudes of the second series. Proof. I. If B corresponds to b, and 7t is any integer, then tiB corresponds to 7ib ; for the sum of ti equal parts B must [by II] correspond to the sum of 11 equal parts b. 2. Also the wth part of B corresponds to the «th part of b; for a magnitude which, taken 11 times gives B must correspond to that which taken 71 times gives b. First Case. When ^ is a fraction of B. Then A = {7i/d)B = 7t . {B/d). Now [by 2] the magnitude B/d corresponds to b/d, an(\ [by i] the magnitude 7i.{B/d) corresponds to 7i{b/d). Consequently a = 71. [b/d) = {7i/d)b. Second Case. When A is no fraction of B. Then U A> {7i/d)B, .: A/B > {n/d)B/B, .'. A/B > 7t/d.. 71 71 But since A > -B, .-. [by II] a^ —b, .-. a/b > 7t/d. d d 461. Corollary I. If parallels cut two straights, the inter-, cepts on one have the same ratio as the cor- responding intercepts on the other. For to sects a,b,c,... on one, the parallels give corresponding sects a', b', c' , . . . on the other, such that \{ a =^ b, then a' =^ b' \ and to the sum a-\-b corresponds the sum a' -\- b' , etc. ' jne. 309. •94 ELEMENTARY SYNTHETIC GEOMETRY, 462. Cor. II. Parallelograms with an angle and a side in one equals to an angle and side in the other have the same ratio as their other sides. For this other side and the ||g'm are then corresponding magni- tudes, such that if of sides a, b, c, . . . and ||g'ms A, B, C, . . . /I z= d, .'. A = B, also \.o a -\- b corresponds A ■\- B. 463. Cor. III. In the same circle or in equal circles, angles at the center have the same ratio as their arcs. For these angles A, B, C, . . . and arcs a, b, c, . . . so corre- spond that if A = B, then a = b ; and to A -\- B corresponds ^i-\-b. 464. Chords are not proportional to their arcs. For if arcs A, B correspond to chords a, b, then arc A -\- B does not correspond to a chord equal to a-\- b. BOOK V. SIMILARITY. 465. If from a point we draw rays to all the points of a given figure, and take on each of these rays another point, these latter points determine a second figure which we may call a perspective of the given figure. Two figures are called perspective when each point of one can be so paired with a point of the other that the joins of all the pairs concur in one point called the center of perspective. Two figures are called projective if they can be moved so as to be perspective. Two figures are called similar when they can be so placed that on any straight whatever through one point sects from it to the perimeters of the figures have always the same ratio. Figures are similar which, being projective, when made per- spective have sects from the center of perspective to the pairs of points always in the same ratio. 466. The sect from the center of perspective to any point is called that Rami's perspective sect. 467. A point in the perimeter of one figure and a point in the perimeter of the other are said to correspond if their perspective sects are co-straight when the figures are in perspective position. Should the perimeters then have four points co-straight, of the two on one figure, that whose per- spective sect is the lesser corresponds to that of the two on the other figure, whose perspective sect is the lesser. 468. The ratio of corresponding perspective sects is called the ratio of similitude of similar figures ; the perspective center, their center of similitude. 469. The center of perspective is called internal when corresponding perspective sects lie on opposite sides of it ; otherwise, external. 95 96 ELEMENTARY SYNTHETIC GEOMETRY. Fig. Fig. A symccnter is that special case of an internal center of perspective where the corresponding perspective sects are; equal in magnitude. 470. Theorem. Any two circles are similar figures. Proof. When concentric, their center is a center of perspec- tive, and the ratio of corresponding perspective sects is constant, being the ratio of the radii. 471. Corollary. The intersection point of two straights; is '^ {7 for the arcs they intercept on circles with that point as center. For the ratio of the radii gives a constant; ratio of similitude. "S 472. The intersection point of two straight, is ~ (7 for the sects they cut out, on any two parallels. For a parallel through this point shows a constant ratio of similitude. 473. Theorem. If points on two straights are made corresponding which end sects from their intersection having the same ratio, the straights through cor- responding points [projection-straights] are parallel. Proof. By hypothesis CA/CA' = CB/CB'\ .-. CA/CB =-. CA'/CB'. But the parallel to A A' drawn through B gives- CA/CB = CA'/CB". .-. B" coincides with B' . A'B.' 474. Theorem. Similar sects have the ratio of similitude. Proof. When in perspec b' tive position since CA/CB = Fig. 214. Fig. 215. O^'/C^',.'. thesectsareparallel. Now slide CBB' until B' comes on A' and BB' contains A. Thus A' becomes ~ C for the parallels AC and. BC '„ ,'. B'B/A'A = B'C'/A'C. Fig. 213. SIMILARITY. 97 475. Problem. To three given sects to find a fourth propor- tional. Construction. On one arm of any ^ C cut off CD ■=. a, and DF =^ b\ on the other arm make CH =. c. Join DH. Draw FK \ to DH. a/b = c/ {_HK\ 476. We say that by a ^ ^ point Pox\ the straight AB, but not on the sect AB, \\\ ^ this sect is divided extcr- nally; and y^/* and ^/^ are fig. 216. called exter7ial segments of the sect AB. If the point Pis on the sect AB, this is said to be divided internally. 477. Problem. To divide a given sect AB in a given ratio, AS/BT. Construction. On parallels, from A and B take on opposite sides of the straight AB [or the same side] sects AS and BT. Join 57; cutting AB in P. Then AP/PB = AS/BT. 478. When a sect is divided internally and externally into segments having the same ratio, it is said to be divided /lar- tnonically. 479. Theorem. If a sect ^^ is divided harmonically by the points Pand Q, the sect PQ will be divided harmonically by the points A and B. Proof. Since AP/BP = AQ/BQ, .:BP/AP=BQ/AQ; .-.BP/BQ = AP/AQ. 480. The points A, B, and P, Q, of which each pair divides harmonically the sect terminated by the ■' ^ A P B Q, other pair, are called four harmonic points / • Fig. 218. or a harmomc range. 481. Problem. Given a point P on the straight AB, to de- termine the fourth harmonic point. Construction. Through A and B draw parallels, and by a 98 ELEMENTARY SYNTHETIC GEOMETRY. straight through P, cut them in vS and T. On straight BT take BT = BT. The fourth point is on the straight ST'. 482, Corollary. With ABP only one point forms a har- monic range, for if S be any point without the straight in which is the harmonic range APBQ, and if through B we draw BT II to AS, meeting SPm T, SQ in T, then BT = BT . 483. Theorem. If, of two parallels, those points which end proportional sects be joined, these projection-straights are concurrent. Proof. Let the projection-straights A A' and BB' meet in C. Then let CD cut A' B' in F. By hypothesis AB/BD = A'B'/B'D'. By 472 AB/BD = A'B'/B'F. Hence F coincides with D'. 484. Theorem. In any trapezoid the mid points of the parallel sides and the intersection point of the non-parallel sides and the intersection point of the diagonals form a har- monic range. p „ Proof. P divides the median AB in the ratio of the parallel sides. 485. Theorem. The medians of a triangle are concurrent in that trisection A point of each remote from its vertex. Proof. The sect joining the mid points of two sides of a triangle is |I to and ^ of the third side ; .-. the intersection point of any two medians, since it divides each median in the ratio of these \s, is that trisection point of each remote from its vertex. 486. The intersection point of its medians is called the tri- angle's centroid. 487. Theorem. The bisector of an exterior angle or in- terior angle of a triangle divides the opposite side externally or internally in the ratio of the other two sides of the triangle. Proof. ABC any A, BD the bisector of ^ at B. Draw Fig. 221. SIMILARITY. 99 AF II BD. Then of the two angles at B given equal by- hypothesis, one equals the correspond- ing interior angle at F, and the other the corresponding alternate angle at A^ .'. AB = BF [sides opposite equal ^s]. But BF/BC = AD/ DC. [If parallels cut two straights, their intercepts are proportional.] .-. AB/BC = AD/DC. 488. Inverse. If one side of a triangle is divided inter- nally or externally in the ratio of the other sides, the straight from the point of division to the opposite vertex bisects the interior or exterior angle. 489. Corollary. The bisectors of an interior and exterior angle at one vertex of a triangle divide the opposite side har- monically. 490. Theorem. Two triangles are similar if they have two angles respectively equal, or two sides proportional and the included angles equal, or two sides proportional and the angles opposite the greater equal, or their three sides proportional. Proof. Put one angle upon its equal, and then the common vertex is ~ C The A with three sides proportional to those of a given A is ^ to the A made by a straight || to one side of the given a , and cutting off from a second side a sect equal to the corre- sponding side of the other a. 491. Theorem. In a right triangle the altitude to the hypothenuse is a mean proportional between the segments of the hypothenuse, and each side is a mean proportional between the hypothenuse and its adjacent segment. Proof. R't /\ABC '^ £,ACD ^ aCBD. 492. Corollary. To find a mean proportional to two given sects, put a semicircle on their sum as diameter, and produce lOO ELEMENTARY SYNTHETIC GEOMETRY. to this semicircle the perpendicular erected at their common' point. 493. Theorem. If four sects are proportional, the rectangle contained by the extremes is equivalent to the rectangle con- tained by the means. Proof. Let the four sects a, b, c, d be proportional. On a and on b construct rectangles with altitude c. On c a ac c a nd d. ac a c i k 1 6 ■C Fig. 223. and on d construct rectangles of altitude a. Then a/b = ac/bc^ and c/d = ac/ad. [Rectangles of equal altitudes are to each other as their bases.] But by hypothesis, a/b = c/d. .'. ac/bc = ac/ad; .'. be = ad. 494. Theorem. The rectangle of the segments into which a given point divides chords of a given circle is constant. p Fig. 224. Fig. 225. Hypothesis. Let chords AB and CD intersect in P. Conclusion. Rectangle AP.PB = rectangle CP.PD. Proof, i. PAC — ^ PDB [inscribed angles on the same SIMILARITY. lOI arc], and t APC = ^ BPD- angular triangles]. t.APC'^ A BPD [equian- FlG. .-. AP/CP = PD/PB- .-. AP.PB = CP.PD. 495, Corollary. Let the point P be without the circle, and suppose DCP to revolve about P until C and D coin- cide: then the secant DCP becomes a tangent, and the rectangle CP.PD becomes the square on PC. There- fore, if the point is without the circle, the rectangle is equivalent to the square of the tangent ; if within, to the square on half the smallest chord. 496, Theorem, If a triangle have two sides each equal to c, and from their intersection a sect d cut the third side into segments /"and ^, then e = d' -^fg. Proof. ^" = //^ + [K/+^)-/r; ,.d^ = h^-^\k{g-f)-\\ ■ -^ .: But ^ = >^' + [K^ + /)]'. 497, Theorem. A point without a circle, and its chord of •contact, divide harmonically any chord whose straight contains the point. Proof. AP.PB = €' = PQ' -\-CQ.QD ■^PQ^AQ.QB. But AP.PB = AP' + AP.AQ + AP.QB: and PQ' -\-AQ.QB = AP" -\- AP.AQ -\-AQ.PQ + AQ.QB. .:AP.QB = AQ.PB; .: AP/PB = AQ/QB. Fig. 228. I02 ELEMENTARY SYNTHETIC GEOMETRY. 498. Theorem. The rectangles of opposite sides of a noiv cyclic quadrilateral are together greater than the rectangle of its diagonals. ^ Proof. Make 4. BAF — ^ CAD, and tABF=4.ACD. ^"^^^ ^^ I Join FD. Then aABF^^ aACD, Fig. 229^. .-. BA/AC = FA/AD. But this shows (since 14. BAC = t F^D), ABAC -^ aFAD. From aABF -^ aACD, .'. AB.CD = BF.AC. From ABAC '^ aFAD, .'. BC.AD= FD.AC. \< ab.cd-\-bc.ad = bf.ac-\-fd.ac>bd.ac: 499. Corollary (Ptolemy). The rectangle of the diagonals. of a cyclic quad' equals the sum of the rectangles of opposite sides. For then F falls on BD. EXERCISES ON BOOK V- I03 EXERCISES ON BOOK V. 1. The joins of the vertex of one ^ of a A to the ends of that diam- eter of the circum O which is J. to the opposite side are the bisectors of that t- 2. If 2 AS have a common base, they are as the segments into which the join of the vertices is divided by the common base. 3. The 3 external bisectors of the ^s of a A meet the sides co- straightly. 4. Given one side of a A, and the ratio of the other sides, find the path of its movable vertex. 5. The sect || to one side of a quad' from the cross of 2 diagonals and bisected by the opposite side ends where .-* 6. If equiangular as have a common vertex and second vertices co- st', so are the third vertices. 7. If c be the center-sect of the in- and circum- ©s of a A, then R + c R —c = I. 8. The OS on the sides of a a as diameters cross on the sides of the A . 9. If a o be described on one of the X sides of a r't A as diameter, the tangent at the p't where it divides the hypothenuse bisects the other ± side. 10. The mid p'ts of concurrent chords are concyclic. 11. A A', BB', CC are || chords of a ©. Show A ABC \ A A'B*C'. 12. AB is trisected in C and D ; CPD is a regular A ; show that D is circumcenter of BPC, and AP the tangent at P to the circum- 0. 13. Two AS on opposite sides of the same base have the ^s opposite it supplemental. Show that the join of their supplemental ^s is j to the join of their orthocenters. 14. If the ± projections of any vertex of a quad' on the other sides and diagonal of the quad' are co-straight, so are the like projections of any other vertex. BOOK VI. MENSURATION. 500. In practical science, every quantity is expressed by another of the same kind preceded by a number. From our knowledge of the number and the quantity it multiplies, we get knowledge of the quantity to be expressed. So in each kind of magnitude we select one convenient quantity as a standard or unit, to be known familiarly by us, and then to be used in expressing every other magnitude of the same kind. The measurement of a magnitude consists in finding its ratio to its unit. 501. For sects, the unit is the centimeter ['^'"•], which is the hundredth part of the sect between two marked points on a special bar of platinum at Paris, when the bar is at the tem- perature of melting ice. The length of any sect is its ratio to the centimeter. 502. An accessible sect may be approximately measured by the direct application to it of a centimeter, or a sum of centi- meters, such as the edge of a ruler suitably divided. But because of incommensurability, even were our senses perfect, any direct measurement must be usually imperfect and merely approximate. 503. For the measurement of surfaces the standard is the square centimeter [*"""], the square on the linear unit. 504. The area of any surface is its ratio to this square. 104 MEN SURA TION. 105 505. Theorem. The area of a rectangle equals the product ■of the length of its base by the length of its altitude. Proof. If the altitude of the rectangle R is a, and its base b, then its ratio to a rectangle of altitude i*^™ and base tF^ is a', but the ratio of this rectangle to the square centimeter is h ; .-. R — ab'''^\ 506. Corollary. The area of a square is the second power of the number denoting the length of its side. 507. Cor. From the area of a square, to find the length of its side : extract the square root of its area. 508. To find the length of the other side, from the length of the hypothenuse and of one side of a right triangle, multiply the sum of the lengths by the difference, and extract the square root. C — a- \c - a\\c -^ d"^ = b\ 509. Given the chord of an arc and the radius of the circle, to find the chord of half g the arc. BC =k' = Vbd' + DC = '/[py+:^». Fig. 229. But and DO = iOC - ODY = [r - ODY, OD = ^OB" - 30-" = Vr' - -. 4 .-. k' = \/--\-{r- \^r^ - -S = V^2r' - 2r/; 4 4^ 510. Since the angle at the center subtended by the side of a regular inscribed hexagon is one third a straight angle, and io6 ELEMENTARY SYNTHETIC GEOMETRY. SO this side equals the radius, therefore the length of a side of a regular dodecagon inscribed in a circle whose radius is i<='" is k' = ^2 — 2Vi—i — - 0.517638C9+. The length of one side of a regular inscribed polygon of 24 sides is r' = /rryi-(ii7Q38°9+)'^ 0.26,05238 +. 4 511. Given the radius of a circle and the side of a regular inscribed polygon, to find the side of a similar circumscribed polygon. Suppose AB the given side k. Draw the tangent at the mid point C of the arc AB, and produce it both ways to the points D and F, where it meets the Fig. 230, j.^(jii Q^ ^LU^ QB produced. DF is the side required, t. t.OFC ^ /\OBM, .'. OC/OM=CF/BM=t/k. kr But t = 0M = OM' VOB' - BM' = Vr' - \k\ kr V'i 512. Corollary. When r = i"", the side of a regular circum* scribed hexagon I _ 2 vr^~ 4/3' / = MENSURA TION. 107 The side of a regular circumscribed dodecagon . 0.51763809 „ o , t' = ^ y ^ ^ = .535898 +. Vi i(. 5 1 763809)' 513. Since no part of a circle can be congruent to any sect, so no part of a circle can be equivalent to any sect in ac- cordance with our definition of equivalent magnitudes as such as can be cut into pieces congruent in pairs. Hence we assume-: [i] No arc is less than its chord. [2] No arc is greater than the sum of the tangents at its extremities. As a consequence of these paradoxical assumptions, an approximate value of a semicircle is given by the semiperime- ter of every polygon inscribed or circumscribed. Moreover, the semicircle cannot be less than the inscribed semiperime- ter nor greater than the circumscribed. 514. Calculating the length of a side in the regular inscribed and circumscribed polygons of 6, 12, 24, 48, 96, etc, sides, radius i*^"", and in each case multiplying the length of one side by half the number of sides, we get the following table of semiperimeters : « i»K i"^n 6 3.0000000 3.4641016 12 3.1058285 3 2153903 24 3 1326286 3 1596599 48 3.1393502 3 1460862 96 3.1410319 3 1427146 192 3.1415424 3 1418730 384 3-1415576 3 1416627 768 3-1415838 3 1416101 1536 3.1415904 3 1415970 3072 3.1415921 3 1415937 6144 3-1415925 3 141 5929 12288 3.1415926 3 141 5927 393216 3.1415926535 3 1415926537 i.e., 6 X 2i« I08 ELEMENTARY SYNTHETIC GEOMETRY. 515. Since a regular polygon of any number of sides, say 393216, is similar to any other regular polygon of that number of sides, therefore their sides have the same ratio as the radii of their circum-circles, or their in-circles. So 3.141592653 is not only an expression, exact to nine places of decimals, for the length of the semicircle whose radius is i*^", but also for the ratio of any semicircle to its radius. 516. This ratio of any circle to its diameter Euler desig- nated by 7t. The Bible [i Kings vii. 23] gives for its value 3. The Egyp- tians twenty-two centuries before Christ gave [4/3]* = 3- 16. Archimedes, from the perimeters of the regular inscribed and circumscribed polygons of 96 sides, placed it between 31^ and 3|. Ptolemy used n = |^^ = 3.1416. The Hindoos gave 3927/1250 = 3.1416. Adriaan Anthoniszoon, father of Adriaan Metius [before 1589] gave 355/113 = 3. 14.15929. Ludolf van Ceulen [1540- 1610] gave Tt = 3. 141 59265358979323846264338327950288. Lambert in 1761 demonstrated the irrationaHty of tt. In June 1882 Professor Lindemann proved that tt is a transcen- dental irrational, that is, rt cannot be a root of a rational algebraic equation of any degree. Hence the rectification of the circle is proved insoluble by •compassiand ruler. CIRCULAR MEASURE OF AN ANGLE. 517. When its vertex is at the center of the circle, any ^ its intercepted arc arc any ^ _ arc st' ^ ~ semicircle ~ r;r ' " (i/7r)st'^ ~ r' st' ^ So, adopting as unit angle , that is, the angle sub- tended at the center of every circle by the arc equal to its MEN SURA TION. 109 radius, and hence called a radian, then the ratio of any angle to the radian equals the ratio of its arc to the radius. If u denote the number of radians in an angle, and / its intercepted arc, then u = l/r. The quotient are/radius, or u, is called the circular measure of an angle. 518. Since a triangle is half the rectangle of either of its sides and the altitude to that side, therefore the area of a triangle is half the product of the length of a side by the length of its altitude. 519. Theorem. In any triangle, the square on a side oppo- site any acute angle is less than the sum of the squares on the other two sides by twice the rectangle contained by either of those sides and a sect from the foot of that side's altitude to the vertex of the acute angle. Proof. Let a, b, c denote the lengths of the sides, and It denote <^'s altitude, and j the sect from its foot to the acute angle A. a^ _ h^ = (b -jy z=b^ - 2bj-\-f' = b'' - 2bj^ c^ - h\ ... a^^b''-2bj-^c\ 520. (Heron.) If A denote the area of any triangle and s — \\a-^b-\- c), then A = \/ s\s — d\\s — b\ \s — c\ F -\- c" - a" Proof, j = 2b .'. h' = c' -j' = c' - (^' -f g' - aj 4b' .'. 4/i'b' = 4b'c' - lb' -\- c' - a')\ no ELEMENTARY SYNTHETIC GEOMETRY. .'. 2hb = V4dV' - {b' -\-c' - dj, .: 4 A = V{2bc + b' -[- c' - a') {2bc - b' - c' -{- a'), A = i i/(a-]-b^c) {b-\-c-a){a-{-b — c) {a — b-{-c). 521. The area of a regular polygon is half the product of its perimeter by the radius of the inscribed circle. For sects from the center to the vertices divide the polygon into congruent isosceles triangles whose altitude is the radius, r, of the inscribed circle, and the sum of whose bases is the perimeter, p, of the polygon. .*. N = ap/2. 522. The area of any circle 0= rV. For if a regular polygon of 393216 sides be circumscribed about the circle its area is \rp. But \p is rit ; therefore its area is rV. EXERCISES ON BOOK VI. Ill EXERCISES ON BOOK VI. 1. In a regular triangle the side {b) = \ perimeter (/) = 1/3 circum- radius {R)= 2 1/3 in-radius (r) = '- — - altitude {K) = -y 3 V's'area (v). 2. The area of a tangent- polygon (circum-polygon) is half perimeter by in-radius (ipr). 3. How many times greater does a quad' become if we magnify it until a diagonal is tripled ? 4. Lengthening through A the side ^ of a a by r and c by d, they be- come diagonals of a symtra which is to the A as (^ + c)' to dc. 5. One vertex of a ||g'm and the mid points of the other two sides determine a A. What is its ratio to the ig'm ? 6. The squares of chords from the same point are as their ± projec- tions on the diameter from that p't. 7. Make a sq' equal to ^ a given sq'. 8. Make a sq' = | a given sq'. 9. If from an ^ or from supplemental ^s we cut + as, they are as the sq's on one of th^e~=r-sides^ 10. Trisect a •!• A by |s. 11. Their cross divides the non-| sides of a trapezoid externally, the diagonals internally, in the ratio of the || sides. 12. (Circle of Apollonius.) If a sect is cut in a given ratio, and the interior and exterior points of division are taken as ends of a diameter, this circle contains the vertices of all triangles on the given sect whose other two sides have the given ratio. • 13. If AD and BE are altitudes of a ABC, then a/i> = -A, / 7^. AlJ/ BE 112 ELEMENTARY SYNTHETIC GEOMETRY, MISCELLANEOUS EXERCISES ON THE FIRST SIX BOOKS. 1. Describe a having center in a given st' and containing two given points. 2. A may be described which shall contain two p'ts, and have r = «; {a> \AB). 3. a ■\- b ■\- c > 2a. 4. a + b + c < 2a + 2b. 5. If the sides of a regular polygon be produced to meet, their inter- section points are the vertices of a similar polygon. 6. Trisect a st' ^. 7. From a -I- a, cut a synitra with three sides =. Hint. Join extremities of the two equal angle-bisectors. 8. Two external ^ bi's of a •!• A are || to a side. 9. A median, a', is >, =, < a, according as ^ ^ is acute, r't, or ob- tuse. 10. AS having a r't 4- common, and = hy's, have mid's of hy's on a quadrant. 11. The angles made by productions of the sides of a reg' pentagon are together a st' 4-. 12. The angles made by productions of the sides of a reg' hex' are together a perigon. 13. Any two llg'ms on two sides of a A are together = to a ||g'm on the third side, whose consecutive side is = and || to the sect joining the intersection of two sides produced of the other llg'ms to their common vertex. 14. The squares on the sides of a A are together triple the squares on the sects joining the vertices to the centroid. 15. Triple the squares of the sides of a a is quadruple the squares on the medians. 16. The sum of the sides of a a is greater than the sum of its medians. 17. From the vertices, equal sects taken in order on the sides of a sq' give the vertices of a sq'. 18. With a vertex on a vertex, inscribe in a sq' a reg' A. MISCELLANEOUS EXERCISES ON THE FIRST SIX BOOKS. 1 1 3 19. Ilg'ms inscribed in a IJg'm have common sC. 20. If either diag' of a ||g'm be = to a side, the other diag' > any side. 21. Sects from a point in a diag' of a ||g'm to vertices give As = in pairs. 22. One median of a trapezoid bisects it. 23. Sects from any p't in a ||g'm to its vertices bisect it. 24. Sects from the mid p't of a non-|| side of a trapezoid to opposite vertices bisect it. 25. Medians of a quad' bisect. 26. The sum of sq's of diag's of a trap' = sq's of non-|| sides + two rect' of I sides. 27. Draw a chord bisected by a given p't within a given o. 28. Any chords which intersect on a diam', and make = 4-^ with it, are =. 29. Describe a o with given r, center in given st', and tan' to another given St'. 30. The opposite sides of a circum-quad' subtend suppl' ^s at the center. 31. HD produced to circum-o is doubled. 32. In an inscribed even polygon, non-consecutive angles make half the angle-sum. 33. On a given sect as chord describe a segment which will contain a given 4-. 34. Find a curvilinear figure equivalent to a regular even polygon, 35. In a regular even polygon, any vertex and the center are co-st' with another vertex. 36. II chords are sides of a symtra. 37. The sum of the squares of the segments of two 1 chords = d"^. 38. The ± projections of opposite p'ts of a © on any st" are on a con- centric 0. 39. If through a p't on a common chord pass two chords, their four ejctremities are concyclic. 40. y^s" = kt" + y^io". 41. U\ d:: d: if,. 42. OA' -H OB' + OC = R + r. 43. Any rectangle is half the rectangle of the diagonals of squares on its sides. 44. If two = chords intersect they make equal segments [they are diag's of a symtra]. 114 ELEMENTARY SYNTHETIC GEOMETRY. 45. A II through the center is \ perimeter 6f a circum-symtra, ifi. b : c :: 1. ir. A' on c : 1. fr. A' on b. 47. The sum of the diag's of a quad' is less than the sum of any other four sects from a p't to the vertices. 48. Through a given p't draw a st' on which is from two given p'ts shall be =. 49. The 4- bi's of a |ig'm make a rectangle. 50. From the r't 4C. the median and altitude of the r't A contain 2(1 = dif of the acute angles. 51. An angle-bi' and median contain 2(! = to dif of other 4-^ of the a. 52. If of the four As into which the diag's divide a quad', two oppo- site are =, it is a trap'. 53. If two circles cut, the intercepts on any two ||s through the points of section are =. 54. Chords all drawn from a p't on a have their mid p'ts concyclic. 55. If from one common p't of two equal intersecting ©s as center a O be drawn, two of the points in which it cuts them, and their other common p't, are cost'. 56. If two = OS cut, the part of a st' through a common p't inter- cepted between them is bi'd by the O on their common chord as diame- ter. 57. If two OS are tangent, two st's through the p't of contact inter- cept arcs whose chords are ||. 58. If two OS touch externally, and || d's be drawn, a st' joining their extremities will contain the p't of contact. 59. In a st' through the center determine a p't from which a tan' shall be = d. 60. The 4- made by tan's from a p't to a o is double the 4- of chord of contact and diam' through a p't of contact. 61. Through a given p't to draw a st' which shall make equal 4-^ with two given st's. 62. From two given p'ts to draw two = sects which shall meet on a given st'. 63. From two given p'ts on the same side of a given st' to draw two st's which shall cross on that st' and make = 4-^ with it. 64. If a tan' be || to a chord, the p't of contact will be the mid p't of the chord's arc. 65. Of st's drawn from two given p'ts to meet on a O, the sum of those two will be least which make = ^s with the tan' at the point of concourse. MISCELLANEOUS EXERCISES ON THE FIRST SIX BOOKS. 1 1 5 66. If two Os cut, and from either common p't diam's be drawn, their extremities and the other common p't are co-st'. d"]. If a be described on the r of another © as d, any sect from the common p't to the greater is bisected by the lesser. 68. The st's joining to the centre the intersections of a tan' with two II tan's are 1. 69. St's from two p'ts in a O to a p't in a tan' make the greatest ^ ■when drawn to the p't of contact. 70. If any chord be bisected by another, and produced to meet the tan's drawn at the extremities of this other, the parts between the tan's -ana the are =. 71. If one chord bisect another, and tan's at the extremities of each be produced to meet, the join of their points of intersection is || to the bisected chord. 72. If from the extremities of a diameter chords be drawn intersect- ing, two and two, on a ± to that d', the joins of the extremities of the pairs are concurrent. 73. If from any p't in the base of •}• a st's making equal 4-'^ with the base be drawn to the sides, the as formed by joining the intersections to the opposite vertices are =. 74. Which st' through a given p't within a given "4- will cut off the least A ? 75. The diag's of a trap' cross on a median. 76. A st' bisecting a side of a A is cut harmonically by the three sides and a || to the bisected side through the opposite vertex. "J"]. A st' from a vertex of a A is cut harmonically by the opposite side, a median, and a || to either of the other sides through the opposite vertex. 78. If from the ends of a side of a A st's be drawn intersecting in the altitude to that side, the straights joining the points where they cross the other sides to the foot of that altitude make equal angles with it. 79. If from any 2(^ of a rectangle a sect be drawn to a side, and a ± to it from the adjacent 4- of the quad' so formed, their rect' = the given rect'. 80. The two spherical tan's from a p't to a are =. 81. If the g-lines joining the corresponding vertices of two A'scon- <:ur, the crosses of opposite sides are colli near ; and inversely. 82. If a spherical quad' is inscribed, and another circumscribed touch- ing at the vertices of the first, the crosses of the opposite sides of these •quad's are collinear. Il6 ELEMENTARY SYNTHETIC GEOMETRY. [The crosses of the opposite sides of the inscribed are on the diago- nals of the circumscribed.] 83. The crosses of the sides of an inscr' a with the spherical tan's at the opposite vertices are collinear. 84. If from the greater of two sides of a A a portion be cut ofT equal to the lesser, the join of the p't of section and the opposite ^ makes an ^ = ^ dif of ^s adjacent to third side of a. 85. Every © passing through a given p't and centered in a given st', passes also through another fixed p't. 86. The rectangles of opposite sides are together double a cyclic quad' whose diag's are ±. 87. If through the mid p't of any chord two chords be drawn, the joins of their extremities will cut off equal sects on the first chord. 88. In a r't a the dif between the hy' and the sum of the other sides equals the ^ of the in- O- 89. If from the extremity of the radius of its circum-O bisecting one side of a A a J. be drawn to the larger of the other two sides, one of the segments made is half the sum, the other half the difference, of these sides. 90. The center of O touching semicircles described outwardly on the two sides of a r't a is the mid p't of the hypothenuse. 91. An angle-bi' of a a cuts the circum m the center of a O con- taining the other two vertices and the in center. 92. If from a vertex of a A inscribed in a st's be drawn || to the tangents at the extremities of the opposite side, they cut oflf ~ as. 93. The joins of the vertices and the points of contact of the in-o of a A concur. 94. If from the ends of a side of a square 0s be described, one with the side, the other with the diagonal, as radius, the lune formed equals the square. 95. If the diam' of a semi-0 be cut in pieces and on ;hem semi-0s be described, these together equal the given semicircle. 96. In a given st' determine the p't at which st's from two given p'ls on the same side of it will contain the greatest ^ . 97. If the rectangles of the segments of two intersecting sects are tqual, their extremities are concyclic. 98. If two altitudes are equal, is the A isosceles ? 99. If two medians are equal, is the A isosceles? 100. A ABC ~ A A'B'C [where A' bisects a; B',b; C, c.\ BOOK VII. MODERN GEOMETRY. CHAPTER I. TRANSVERSALS, 522. [Menelaus.] If the sides of the triangle ABC, or the sides produced, be cut by any transversal in the points : then Bc/cA = BD/Ab, and Ca/aB = bC/BD ; therefore [Ca/aBJ_Bc/cA'\ = \bC/BD\BD/Ab'\ = bC/Ab = \/{^Ab/bC\ Inversely, if the straight ab meet AB in c\ then by Menelaus, lAb/bCWCa/aB-WBc'/c'AI = i. But by hypothesis, [Ab/bCllCa/aBjlBc/cA] = i. Therefore c and c' coincide. "7 'O- Il8 ELEMENTARY SYNTHETIC GEOMETRY. 523. Corollary. If a traversal intersects the sides AB, BC, CD, etc., of any polygon in the points a, b, c, etc., in order, then [Aa/aB][Bd/dC\\lCc/cD][nd/dE] . . . etc. = i. Proof. Divide the polygon into triangles by straights through one vertex, apply Menelaus to each triangle, and com- bine the results. 524. [Ceva.] If the sides of triangle ABC are cut by A O, BO, CO in a, b, c, then lAb/bC^iCa/aB\Bc/cA'\ = i. 'b ^ Inversely, given this relation, the straights iG. 233. ^^^ j^^^ Q^ ^jjj i^g concurrent. Proof. By the transversal ^/5 to the t.AaCy^Q have [Mene- laus] {_Ab/bC\CB/Ba\aO/OA'\ = 1 ; and by the transversal Cc to the AaB, [Bc/cA][AO/Oa][aC/CB] = i. Multiply these equations together. Inverse as in Menelaus. 525. Corollary. If transversals through O from the vertices of any odd polygon meet the sides AB, BC, CD, etc., in the points a, b, c, etc., in order, then {_Aa/aB\Bb/bC\Cc/cD\Dd/dE\ . . . etc. = i. 526. Theorem. If any transversal cuts the sides of a triangle and their three intersectors AO, BO, CO, in the points A' , B', C, a', b' , c' , respectively, then \A'b' /b' C\C' a' /a' B'\B' c' /c' A'^ = t. Proof. Each side forms a triangle with its intersector and the transversal. Take the four remaining straights in succes- TRANSVERSALS. 119 Pig. 234. sion for transversals to each triangle, applying Menelaus symmetrically, and combine the twelve equations. 527. Theorem. If the vertices of two triangles join concurrently, the pairs of corresponding sides in- tersect co-straightly, and inversely. Proof. Take be, ca, ab, trans- versals respectively to the triangles OBC, OCA, OAB\ apply Mene- laus, and the product of the three equations shows that P, Q, R lie on a transversal to ABC. 528. Corollary. If the vertices of any two polygons join concurrently, the pairs ot corresponding sides intersect co- straightly. 529. The straight on which the pairs of sides cross is called the axis of perspective. 530. The ]_ projection of a point on a sect is the foot of the perpendicular from the point to the straight of the sect. 531. The J_ projection of a sect on a straight is the piece between the perpendiculars dropped upon the straight from the ends of the sect. 532. Theorem. The _L projections on the sides of a triangle of any point on its circumcircle are co-straight. [This straight is called the Simson's straight of the triangle with respect to the given point.] Proof. Let O be any point on circumcircle of A ABC. Join its _L projections GF, GH. Join OA, OC. Since 2^ OGC and ^ OHC are r't, therefore C, H, G, O are concyclic. Simi- larly, G, B, F, O are concyclic. .'. •4. OGF = 4- OBF, inscribed angles on same arc of circle OGBF. But ^ OBF— 4 OCA, being supplemental to ^ OBA. .-. 40GH-\-4 0GF= ^.OGH^ 4 0CH=st4.. Fig. 235. 120 ELEMENTARY SYNTHETIC GEOMETRY. 533. Inverse. If the projections on the sides of a triangle of a point be co-straight, that point is on the triangle's circum- circle. Proof. Let G, H, F, _L projections of O on a, b, c be co- straight. Since (9, C, H, G are concyclic, .•. ^ OCH = ^ OGF, being supplements of ^ OGH. But -^ OGF = ^ OBF, inscribed angles on same arc of circle OGBF .: ^ OCA + 4- OB A = s't ^. .'. O, C, A, B are concyclic. CHAPTER II. HARMONIC RANGES AND PENCILS. 534. A system of co-straight points is called a range, of which the straight is the bearer. 535. A system of concurrent straights is called a pencil, of which the intersection point is the vertex or the bearer. 536. Straights all parallel form a pencil of parallels or a parallel-pencil. 537. Thus straights with equal perpendiculars from two given points form two pencils, one parallel to their join, and the other bisecting it. 538. A range and a pencil are called perspective when each point of the range lies on a straight of the pencil. 539. Two ranges are called perspective when their points lie in pairs on the straights of a pencil. The bearer of the pencil is called XJae. perspective-center. 540. Two pencils are called perspective when their straights cross in pairs in the points of a range. The bearer of the range is called the projection axis. 541. Ranges and pencils are CdiW^d projective if they can be put in perspective position. 542. If A, B be two points, and C, D, two co-straight with them, be so taken that AC/BC — AD/DB, then the points A, C, B, D form a harmonic range; C and D are harmonic conjugates with respect to A and B\ AC, AB, AD are said to be in harmonic progression ; and AB is said to be a harmonic mean between AC and AD. Thus we have proved (479) that if C and D are harmonic conjugates with respect to A and B, then A and B are harmonic conjugates with respect to Cand D. 121 122 ELEMENTARY SYNTHETIC GEOMETRY. 543. If A, C, B, D form a harmonic range, and O be the mid point of AB, then OA'' = OB' = OC . OD. For AD/DB = AC/BC. .'. (AD + DB)/{AD - DB) = {AC-{- CB)/{A C - CB)\ :. 2OD/2OB = 2OB/2OC, .'. OB' = OC. OD. 544. Theorem. If four concurrent straights cut any trans- s versal in a harmonic range, they will cut / \\\ / every transversal in a harmonic range. a'./_\_V^'X T, Proof. Through B and B' draw BT / Ti'(c \b/ X and j5' 7^ II to AA'S, and meeting SC in \ / 7", , T/ and SD in Zj, , TJ . Then since v(, BT, = BT,, .'. B'T/ = B'TJ. F"'- ^36. A'C'B'D' is a harmonic range. 545. If A, C, B, D is a harmonic range, 5^, SC. SB, SD' is a harmonic pencil, and 5(7, SD are harmonic conjugates of SA, SB. 546. We have shown that the arms of any angle form with its internal bisector and its external bisector a harmonic pencil. I 547. If, in a harmonic pencil, one element bisect the angle between two conjugates, then it is perpendicular to its con- jugate. 548. If in a harmonic pencil one pair of conjugates be at right angles, then these are the internal and external bisectors of the angle between the other pair. 549. Theorem. If two harmonic ranges are taken, one in each of two straights, and if three of the four projection- straights are concurrent, then so are the four. Proof. For the three concurrent projection-straights and the straight from their bearer through one of the fourth points form a harmonic pencil ; so this latter straight contains also the other fourth point. 550. Corollary. If two corresponding points coincide in the cross of the two straights, then one projection-straight being free, the other three are always concurrent. CHAPTER III. PRINCIPLE OF DUALITY. 551. Not only the sect joining two points, but the whole straight, may be called the join of the two points. 552. The point common to two straights may be called the cross of the two straights. 553. In a pencil consisting of straights through one fixed point, any one of the straights may be called an element of the pencil, or a line on the fixed point or bearer. In this sense, we say not only that points may lie on a straight, their bearer, but also that straights may lie on a point, their bearer, meaning that the straights pass through this point. 554. In most cases we can, when one figure is given, con- struct another, such that straights take the place of points in the first, and points the place of straights. Thus from a definition or a theorem we can obtain another by interchanging point and straight, cross and join, range and pencil, or by similar interchanges. 555. A figure regarded as consisting of a system of straights crossing in points will thus give a figure which may be regarded as a system of points joined by straights ; and in general with any figure coexists another having the same genesis from these elements, point and straight, but that these elements are inter- changed. Any descriptive theorem or theorem of position concerning 123 124 ELEMENTARY SYNTHETIC GEOMETRY, one, thus gives rise to a corresponding theorem concerning the other figure, 556. Figures or theorems related in this manner are called reciprocal figures or reciprocal theorems. 557. This correlation of point and straight is termed a principle of duality. 558. Each of two descriptive theorems so correlated is said to be tJie dual of the other ; and it will be found that if any descriptive property is demonstrated, its dual also holds. 559. Since capitals mean points, and two fix a straight, their join ; so small letters may denote straights, and two will fix a point, their cross. Thus AB denotes the straight which is the join of the points A and B ; while ab denotes the point which is the cross of the straights a and b. 560. In plane geometry to all points on a straight the recip- rocal figure is all straights on a point. 561. A sect, AB, is that piece of a range containing the initial point A of the sect, its final point B, and all intermediate successive positions of the generating point. 562. The figure reciprocal to sect AB is ^ ab, that piece of a pencil containing the initial straight a of the angle, its final straight b, and all intermediate positions of the generating straight. RECIPROCAL THEOREMS. 563,. If two harmonic 563'. If two harmonic pen- ranges are taken, one in each cils are such that the three •of two straights, and if three crosses of three pairs of corre- of the four projection-straights sponding straights are co- are concurrent, then so are the straight, then this straight four. contains the cross of the fourth pair. PRINCIPLE OF DUALITY. 125 564,. If two harmonic ranges are taken one in each of two straights, and two corresponding points coincide in the cross of the straights, then the other three projec- tion-straights are concurrent. 564'. If two straights, one in each of two harmonic pen^ cils, are coincident, then the three crosses of the other three pairs of straights are costraight. CHAPTER IV. COMPLETE QUADRILATERAL AND QUADRANGLE. 565,. A system of four straights, no three concurrent, and their six crosses is called a complete quadrilateral, or co7n- quad. 566j. The four straights are called the "sides " of the quad- rilateral ; and the six crosses, the vertices. 567,. Two vertices which •do not lie on the same " side " are called opposite vertices. There are three pairs. 5681. The three straights joining opposite vertices are called diagonal straights, and the triangle formed by the diagonal straights is called the diagonal triangle of the com- plete quadrilateral. 565'. A system of four points, no three costraight, and their six joins is called a quadrangle. 566'. The four points are called the: summits of the quad- rangle, and their six joins the " sides." 567'. Two sides which do not pass through the same summit are called opposite sides. There are three pairs. 568'. The three crosses of opposite sides are called diag- onal points, and the , triangle determined by the diagonal points is called the diagonal triangle of the quadrangle. 126 COMPLETE QUADRILATERAL AND QUADRANGLE. 1 2/ Fig. 237. Fig. 238. 569,. In this complete 569'. In this quadrangle quadrilateral a, b, c, d, are the a, B, C, D, are the summits, sides. The vertices are i, 2, 3, 4, The sides are i, 2, 3, 4, 5^ 5, 6. I and 6 are opposite 6. i and 6 are opposite sides, vertices. So are 2 and 5. So are 2 and 5. Al.so 3 and 4. Also 3 and 4- Fig. 239. 570,. In the above com- plete quadrilateral, \{ f be the joni of I and 6, ^ of 2 and 5, h of 3 and 4, then fgh is the diagonal triangle. 571,. Theorem. In a com- plete quadrilateral each pair of opposite vertices forms with two of the angular points of Ihe diagonal triangle a har- monic ranore. Fig. 240. 570'. In the above quad- rangle if F be the cross of i and 6, 6^ of 2 and 5, .^ of 3 and 4, then FGH is the diag- onal triangle. 571'. In a quadrangle, each pair of opposite sides forms with two of the sides of the diagonal triangle a harmonic pencil. 128 ELEMENTARY SYNTHETIC GEOMETRt W Fig. 241. Fig. 242. Proof. The range q\ASC^^ Proof. The pencil Q\asc] is. is perspective with the range perspective with the pencil r\^ESF'] to projection center B ; R{esf\ to projection axis b ; on .*. on a straight through i5must lie the harmonic conjugates R and ^ to 5" of the.se ranges. But also q\ASC^ is per- spective with rlFSEA^ to pro- jection center/?; .-.also on a straight through D must lie R and Q. Hence they must lie on s, the join of B and D. b must cross the harmonic conjugates r and ^ to .y of these pencils. But also 0,\asc\ is projec- tive to R\^fse\ to projection axis d\ .'. on d also must cros? r and g. Hence they must gc through S, the cross of b and d. 572. Problem. To draw a pair of tangents to a given circle from a given external point by means of a ruler only. Construction. From the given point S, draw SCA, SEF, cutting the given circle in A, C and E, F. Join AE, CF, crossing at B. Join AF, CE, and produce to meet at D. The st' DB contains the chord of contact of S. For we have proved in [497] that the chord of contact of 5 contains the harmonic conjugates R, Q of S on EF 2iWdiAC, and we have just proved in [571,] the opposite ver- tices BD of the complete quadrilateral abed costraight with R, Q, these harmonic conjugates of 5. , diagonals of a sym- c_ _ivi B tra. The mid points O, P^ , P, O' . /"^X' / '^v.,^-^^ ^''^ always costraight, and the Oy^'"' Pj>^^ p' \ rectangle OP^ . OP', constant. // >\ \ / ^^"---..N. For if M, H be the mid points D^^ -"h ^-^A of the symtra's || sides, then Fig. 249- OM = OH ^nd MP, = MP' = HP' = HP, . [The sect joining the mid points of two sides of a triangle is half the third] ; so the relative position of the points OP,P' is kept the same as in Peaucellier's Cell. So if O is fixed and P, describes any line, then P' must describe its inverse. CHAPTER VI. POLE AND POLAR WITH RESPECT TO A CIRCLE. 582. Theorem. If the cross of two tangents glides on a straight, their chord of contact rotates about a point ; and inversely. Fig. 250. Proof. Draw CR _L to /, meeting TT' \x\ P. Since CG is r't bi' of TT, .'. ^ PDG is r't. .-. P, R, D, G are concyclic ; .-. CR .CP = CD.CG = CT' [since TD is ± to hy' of r't A CTG]. But CR and CT dixe. fixed; .'. also CP. Inversely, draw GR _L to CP. Since, in the inverse, CP and CT are fixed, .'. so also is CR. P is called the pole of p, and / the polar of P with respect to the given circle. 583. Since R and P are inverse points, with respect to the center C, and radius CT, therefore the perpendicular to their 133 134 ELEMENTARY SYNTHETIC GEOMETRY. Straight through either of two inverse points is the polar of the other, which is the pole of the perpendicular. 584. To get the pole of P with respect to a circle with center C, join PC cutting the circle in A and B, and through R, the harmonic conjugate of P with regard to AB, draw a perpendicular/. 585. Inversely, the pole of/ with respect to a circle, center C, is the harmonic conjugate of the foot of the perpendicular from C on p with regard to the intersection points of the per- pendicular with the circle. 586,. If a straight passes 586'. If a point lie on the through the pole of a second polar of a second point, so straight, so does the second does the second point lie on straight pass through the pole the polar of the first, of the first. A a' Proof. If A lie on AR the polar of B, then CB _L to AR. Draw BD J_ to CA, then A, R, B, D are concyclic ; .-. CA . CD = CR.CB = r'', .'. BD is the polar of A. Fig. 252. 587,. Corollary. The join 587'. The cross of the polars of the poles of two straights is of two given points is the pole the polar of their cross. of their join. 588. Theorem. Any secant through a pole is cut harmoni- cally by the circle, pole, and polar. We have proved this in [497] if the pole be without the circle. If the pole be within the circle, we may substitute for it the intersection of its polar and the secant, since [586] the polar of that point contains the given pole. 589. A triangle of which each side is the polar of the oppo- site vertex with regard to a circle is said to be self -conjugate with respect to the circle. POLE AND POLAR WITH RESPECT TO A CIRCLE. 1 35 590. The diagonal triangle of a quadrangle inscribed in a circle is self-conjugate. Proof. Ranges QAD, QBC are perspective from center S\ .'. the harmonic conjugates to Q are on a straight through 5. But ranges QAD, QCB are perspective from center R ; .". the harmonic con- jugates to Q are on a straight through R. .'. SR is the polar of Q. In the same way QR is the polar of S. polar of R. 591. Corollary. With ruler only, draw the polar of a given point, or find the pole of a given straight, with respect to a given circle. Fig. 253. OS is the Fig. 2SS. 592,. The polars of the 592'. The poles of the four four points of a harmonic straights of a harmonic pencil range form a harmonic pencil, form a harmonic range. ' Proof. Let P be the pole of the straight ACBD with respect to ©a Oi A, C, B, D, the polars PA', PC, PB' , PD' , are _L to OA, OC, OB, OD. Thus the angles between the straights PA' , PC, PB', PD' are respectively equal to the angles of the harmonic pencil OA, OC, OB, OD. 593. If with respect to a given fixed circle be taken the 136 ELEMENTARY SYNTHETIC GEOMETRY. pole of each straight, the polar of each point, of a figure F^, we obtain a dual figure F' . This method is called polarization or reciprocation, and either of the figures is termed the polar reciprocal of the other, and any geometrical property of the one has its correlative for the other. 594j. The pole of each 594'. The straight through a point lies point on a each on the polar of this point. 595,. The join of the poles of two straights is the polar of their cross. 596,. The poles of the straights of a pencil form a range whose bearer is the polar of the pencil's vertex. the polar of straight goes pole of this cross of points is the the through straight. 595'. The polars of two pole of their join. 596'. The polars of the points of a range form a pen- cil whose vertex is the pole of the range's bearer. 597. Thus in reciprocal polars correspond in /^ a jom, a pencil, parallels, angle between two straights; \xvF' a cross, a range, points co-straight with the center of reciprocation, ^ [or its supplement] sub- tended by two points at center of reciprocation ; and vice versa. 598. A self-conjugate triangle is its own reciprocal polar. 599. The diagonal triangle of a cyclic quadrangle is also the diagonal triangle of the complete quadrilateral whose sides touch the circle at the vertices of the quadrangle. Fig. 256. CHAPTER VII. CROSS RATIO. 600. If in a range consisting of four points, A, B, C, D, we take A and B, called conjugate points, as the extremities of a sect, this is divided internally or externally by C ; and dis- tinguishing the " step'' AC from CA as of opposite " sense," so Vhat AC ^ — CA, the ratio AC/BC is never the same for two positions of C. The like is true of the positive or negative number AD/BD. 137 N R. :u2 ii^ m ' 73]) 138 ELEMENTARY SYNTHETIC GEOMETRY. _JBJ5 D.. 1, -Tl The ratio [^(7^(r]/[^Z>/^Z>] is called the cr^j^ r«//^ of . . » ■ ■ — ^ — """ the range, and is written {^ABf.CD\ 601. Four elements may be arranged in twenty-four different ways: ^ {_AbCD\ \bAdC\ \CbAB\ {_DCBA\ \ [ABDCl [BACD], [DCABl [CDBAl ^ [[ACBD], [CADB], [BDAC], [DBCA], \,\[aCDBI \CABD\ IDBAC\ \_BDCA\ y lADBC], ibACBl IBCAD\ {CBDA\ ' \^ADCB\ \_bABC\ \CBAD\ \BCDA\, ft;5 but four cross ratios in each of these six rows are equal, as may be readily proved by writing out any two in a row. 602. If in a cross ratio the two points belonging to one of the two groups be interchanged, the cross ratio changes to its reciprocal. [Proved by writing out their values.] Thus the ratio in the second row is reciprocal to that in the tirst, fourth to third, sixth to fifth. 603. If in a cross ratio the two middle letters be inter- changed, the cross ratio changes to its complement. {^ABCD'\ = I - [ACBDl For we have, taking account of sense or sign, BC -\- C A-^- AB^o; BC. AD -^ CA . AD -^ AB . AD = o; BC.AD^CA.[_BD+AB^+AB.ICD-CA] =^0; BC.AD-{-CA.BD-\- AB.CD = o; I + [CA .BDyiBC.AD] + [AB. CDyiBCAD] = 0; I -[AB. CD]/[CB. AD] = IAC. BD]/[BC. AD] ; ;/■ AB /AD __AC/Ap ~CD " BC' Bb ' CB I - \ACBD] __AC 1. "Bc' ' [AbCD].. CHOSS RATIO. 139 604. By 603 [AbCEX = I - [ACDB] = [by 602] i - l/[ACBD] = [by 603] I - i/[i_- {ABCno Thus if the cross ratio {ABCB] = A, then the six cross I ratios derivable from these four co-straight points are A, A - I A A' ^ ~ '^' I - A' A ' A - r 605. Theorem. If 5 be a point without the range ABCD, and if through C a straight be drawn parallel to SD, meeting SA, SB in G, H, respectively, then GC/HC- [ABCD\ Fig. 257. Proof. GC/SD = CA/DA. SD/HC=DB/CB. GC Sp _CA DB CA iCB SD' HC .: GC/HC I LB 'DA'~CB ~DA/^B' ICDAB^ = [ABCD]. 606. If two transversals meet the straights of the pencil 5 \^abcd\ in A, B,C, a D and in A', B\ C, D\ then [AbCD] ;g = [A'B'CB']. Proof. Through C and C draw GH and ^ G'H' II to SD. Then , GC/HC = G' C'/H C. • njh Fig. 258. 607. The cross ratio of the pencil 5 [abed'] means the cross ratio of the four points ABCD on any transversal, and is. written 5 [ABCD]. 140 ELEMENTARY SYNTHETIC GEOMETRY. 608. If two ranges or pencils have equal cross ratios they ■are said to be eqiii-cross. 609. Mutually equiangular pencils are equi-cross. Fig. 260. 610'. The crosses of cor- responding straights of two equi-cross pencils which have two corresponding straights coincident are co-straight. Fig. 259. 610,. The joins of corre- sponding points of two equi- cross ranges which have two corresponding points coinci- dent are concurrent. Proof. Let the join of the two crosses B and C cut the common straight in A, and cut d in D. Then is D also on d' , since by hypothesis d' cuts ABC in a point D' such that IAbCD'\ = [ABCDf]. 611. Corollary. Equi-cross ranges or pencils are projective. 612,. Pencils whose straights 612'. Ranges whose points pass through four fixed points lie on four fixed tangents to a on a circle, and whose vertices lie on the circle, are equi-cross. Proof. The pencils are mu- tually equiangular. 6i3i. [Pascal.] In a cyclic hexagon the crosses of oppo- site sides are co-straight. circle and whose bearers are tangent to the circle are equi- cross. Proof. Polarization from 612,. 613'. [Brianchon.] In a cir- cumscribed hexagon the joins of opposite vertices are con- current. CROSS RATIO. I4I: Fig. 261. Proof. By6i2„ the pencils B.ACDE, and F.ACDE are equi-cross ; .-. iRHDE] = VQCDL] ; .*. by (610,) RQ, HQ EL are concurrent. 614. If the figure formed by joining the six concyclic points, by consecutive sects in any order be called a hexagram, there are 60, and Pascal holds for each. 615. Pascal holds for six points, three co-straight and also, the other three. INVOLUTION. 616. If a system of pairs of co-straight points AA\ BB\ CC, etc., be so situated with regard to a point, O on the same, straight that OA . OA' = OB . OB' = OC . OC, etc., they are said to be in invohition. The point O is called the center, and AA' , BB', CC, etc., are called conjugate points of the involution. The points E, F, situated on the range, on opposite sides of C, such that OE" = OF' = OA . OA' are called the double points of the involution. If straights be drawn from a point .S outside the range to A, A', B, B', C, C, etc., they form a pencil in involution, and SE, SF are called the double straights of the pencil. [Observing sense or sign, the double points and double straights are real only when conjugate points of the involution; are on the same side of the center.] 142 ELEMENTARY SYNTHETIC GEOMETRY. 617. Theorem. The two double points and any pair of con- jugate points of an involution form a harmonic range. For the sect between two double points is diameter of the circle with regard to which the conjugate points are inverses. 618. In a system of points or straights in involution the cross ratio of any four points or straights is equal to that of their conjugates. 619. If two pairs of conjugate straights of a pencil in invo- lution be at right angles, then every pair of conjugate straights are at right angles. RADICAL AXIS. 620. Corollary. The straights of a series of right angles at the same vertex form a system in involution. 621. Points from which tangents to two given circles are equal lie on a perpendicular to the center-sect which so divides it that the difference of the squares on the segments is equal to the difference of the squares on the radii of the two circles. 622. The bearer of the points from each of which tangents drawn to two given circles are equal is called the radical axis of the two circles. 623. If two circles intersect, their radical axis contains their common chord. MILNE'S SYMMETRY THEORY OF MAXIMUM AND MINIMUM. 623,. If a continuously varying magnitude, changing in accordance with some definite law, first increases until it attains a certain value and then decreases, that specific value, both preceded and followed by lesser values, is called a inaxiimiin value of the varying quantity. Similarly, a value immediately preceded and followed by greater values of the variable is called a minimum value. 623,. Just so the form of a geometrical figure, varying in a definite way, may approach symmetry, may attain symmetry, may immediately become unsymmetrical. 6233 . The positions which give the maximum and minimum CROSS RATIO. 143 values of a continuously varying geometrical magnitude in any figure are positions of symmetry with regard to other parts of the figure which are fixed in position. For example, the _L from a chord to its arc is a maximum when on the axis of symmetry of the figure. Again, the perimeter of a triangle of fixed surface on a given base is a minimum when the A is -l- . 623^. Thus every varying geometrical magnitude may be considered to have two properties, one metrical, one positional or descriptive. When the magnitude has a symmetrical position it has a maximum [minimum] value, and inversely. So we may reduce the problem of finding the maximum [min'] values of any varying geometrical quantity to the much simpler one of find- ing its positions of symmetry. Ex. I. The minimum [max'] sect between two Os is on their axis of -I- [is on their center-st']. Ex. 2. A and B are two fixed p'ts without a given O O. Find a p't P on the o such that AF" -\- BP^ may be a mini- mum. Bisect AB in C. Then AP^ -[- BP'' = 2AO^ iCP^ . Now AC is constant, .*. CP must be a minimum ; .-. the re- quired p't is where CO cuts the ©. [For max' take its other -cross.] Ex. 3. In Ex' 2 substitute a st' for the Q. Ex. 4. Through a given p't draw a chord which shall cut off a minimum surface. [The p't must be on the -I- axis, .-. it bisects the chord.] Ex. 5. Substitute in Ex' 4 two intersecting st's for the arc. [Same solution.] Ex. 6. Through a given p't within a Q draw the minimum ■chord. [Same solution.] Ex, 7. If two sides of a a be given in magnitude, the sur- c ■' >' 144 ELEMENTARY SYNTHETIC GEOMETRY. face is a maximum when each is -I- axis for the st' of the other [when they are _L]. Ex. 8. To cut a sect so that the rectangle of the 2 pieces^ may be a maximum. [The mid p't.] [For the same p't, the sum of the sq's on the segments is min.] Ex. 9. The p't within a sq' such that sq's on the _Ls from it to the sides are together a minimum is its symcenter. Ex, 10. If two sects cut J_, the sum of the rectangles of the segments of each is a maximum when they mutually bisect. Ex. II. In a given square inscribe the minimum sq. Ex. 12. The p't within a a such that the sum of the squares of its sects from the vertices is a minimum, is the centroid. Ex. 13. Within a A find a p't such that the sum of the squares on the _Ls from it to the sides may be a minimum. fA _ A _ A _ 2A \ \a~ b~ c~ a^ A-b" A- c'J' [For development of this theory see Milne's Companion to Problem Papers.] EXERCISES ON BOOK VII. I45 EXERCISES ON BOOK VII. 1. The Simson-line of a p't bisects the join of that p't and the or- thocenter of the A. 2. Circles described on any 3 chords from one p't of a O as diameters have their other 3 p'ts of intersection co-straight. 3. The circum-Os of the 4 As formed by 4 intersecting straights concur. 4. The diameter of the in-o of a r't A and the hypothenuse together equal the other sides. 5. If A, B, C, D are concyclic, show that the Simson linesof /?, B, C, D with respect to as BCD, CDA, DAB, ABC, and the nine-point-circles of those AS, all pass through the same point. 6. The bisectors of the ^ s in a segment of a O form 2 pencils, whose bearers are the ends of the diameter bisecting the segment. 7. If from a p't within a A is be drawn to the sides, then is the sum of the sq's of the three segments of the sides which have no common end point equal to the sum of the squares of the other 3. [True when the p't is on or without the perimeter of the A.] 8. Inverse of the preceding. 9. If from a p't within a A ±s be drawn to the sides, then is the sum of the 3 rectangles of side-segments having no common end p't each with its A-side equal the sum for the other 3, and equal half the sum of the squares of the sides of the A. 10. The 3 internal and 3 external bisectors of the ^s of a a meet the opposite sides in 6 p'ts, which are 3 by 3 in 4 st's. 11. If of 4 p'ts one is the orthocenter of the other 3, then every one is the orthocenter of the other 3. 12. A, B, Cand their orthocenter //"are the centers of the 4 ©s which touch DEF, the orthocentric a. -\ 1 BOOK VIII. RECENT GEOMETRY, [The Lemoine-Brocard Geometry.] CHAPTER I. ANTI-PARALLELS, ISOGONALS, SYMMEDIANS. 624. Two mutually equiangular polygons are co-sensal when rays pivoted within them and containing the vertices of equal angles, rotate in the same sense to pass through the vertices of the consecutive equal angles. 625. If two transversals cross an ^ , so as to make with its arms two As equiangular, but not co-sensal, then either of these transversals is said to be anti-parallel to the other with regard to the angle. Thus, if ABK, ACHhe straights, and ^ AKH - 4 ACB, then KH anti-| to BC with respect to :2f A. 626. KB and HC are anti-|| with regard to the ■4- of BC with KH. 627. B, C H, K are concyclic ; and inversely, if a quad' is cyclic, either opposite pair of its sides are anti-|| with regard to the ^ between the othei pair. • 628. Any St' II to the tan' to circum-Q of A ABC at A is anti-|| to BC 629. Anti-||s to 2 sides of a a make the same ^s with the 3d side. Thus anti-IJs to the sides a and /5 of A ABC make each with <: an ^ = C. 146 Fig. 263. ANTI-PARALLELS, ISOGOXALS, SYMMEDIANS. ^A7 If the join of their 2 ends not in the 3d side is | to it, they are =, since their 4 ends are then vertices of a symtra. Inversely, if 2 anti-Ils are =, their 4 B' A . Fig. 264. ends are vertices of a symtra. In each case the center of the circle circumscribing the symtra is in the bisector of the ^ between the anti-fs. 630. The joins of the feet D, E, F of the altitudes of a a ABC are anti-jl to its sides. A ABC is called the original tri- angle, and A DEF is called the orthocentric triangle. 631. Given any anti-!| E'F \.o a within A ABC; two anti-[|', FD, D'E, to b, c, can always be found within the A. [By drawing FD' \ to b and E' D \ to <;.] 632. Any two straights symmetrical with regard to an angle-bisector are called isogonals with reference to that angle. 633. If from two points, one on each of two isogonals with respect to a given angle, perpendiculars be drawn to its arms, then, I. The rectangle of the perpendiculars to one arm equals that of those to the other ; II. The feet of the perpendiculars are concyclic ; III. The join of the feet of perpendiculars from the point on either isogonal is perpendicular to the other isogonal. Proof. By ~ as, GM/GA = KQ/KA ; AG/GP = AK/KN\ .'.GM/GP= KQ/KN; .'.GM.KN= GP.KQ. II. By ~ AS, F.0..65. AM/AG=AQ/AK; AG/AP = AK/AN\ .'. AM/AP = AQ/AN; .\AM.AN=AP.AQ, •. M, N, P, Q are concyclic. ' 148 ELEMENTARY SYNTHETIC GEOMETRY. III. Since -4.?, AMG, APG are r't, .-. AMGP is cyclic; .-. ^ MAG = ^ AfPG; but ^ G^y^y]/= ^ A'^Q- .-. ^ MPC; = ^ A'^S; but also PG ±to AQ; .: PM± to AK. 634. Inverse. If the rectangle of the JLs from 2 given p'ts on one of the arms of a given ^ equal the rectangle of the J_& on the other arm, the joins of the vertex and the p'ts are isog- onal with respect to the ^ . 635. If 3 st's through the vertices of a A concur, so do their isogonals. '^ Proof. Let AG, BG be the isogonals ^M of AK, BK. Then by 633, AA' = /s A' > A A' =P%P'%\ hence /,// — P^Px^ /. by 634, GO and KG are isogonals. • 636. Two points so related to a triangle that the three joins of one to the vertices are isogonal to the joins of the other, are called isogonal conjugates. 637. Theorem. The six _L projections of 2 isogonal conju- gates on the sides of the triangle are concyclic • and the center of this circle bisects their join. Proof. By ~ AS, BQ'/BM = BK/BG = BN/BF ; .'.BM.BN=^BP'.BQ'', .'. M, N, P', Q! are concyclic. And as the center of © round them lies in the r't bi's of MA^and Q P\ it is the mid point of GK. Similarly, Q, Pare on the same ©. ( ANTI-PARALLELS, ISOGONALS, SYMMEDIANS. 149 638. Corollary I. The circumcenter and orthocenter of a A are isog' conj's; .'. a O passes through A\ B', C, D, E, F, and the mid points of AH, BH, CH , with center, TV, bisecting OH, and diameter R. This is called the nine-point circle. Its center is called the triangle's medio-center. 639. Corollary. NI = ^R - r. ,*. the ninepoint-O touches the in-O and each ex-O (Feuer- bach's Theorem). 640. The isogonal conjugate to the centroid of a A is called the Lemoinc point of the A . 641. The isogonals to the medians of a A are called its sytnmedians. 642. Since a median bisects all |Is to its side of the A, .•. by symmetry its symmedian bisects all anti-||s to the side. 643. The Lemoine point bisects 3 anti-||s, which are equal, since the two halves going to a side make with it ^ s each =r to the opposite ^ of the A. Thus the ends of any 2 of these are vertices of a rectangle. 644. The circle through the 6 p'ts in which anti-|jS to the sides of a A, through its Lemoine p't, meet the sides, is called the 2d Lemoine O of that A. 645. Since the sides of a A and of its orthocentric A are anti- II , .'. the sides of the orthocentric are bisected by the symmedians. .'. join each angle of a a to the mid point of that side of the orthocentric which ends in its arms ; the joins concur in the Lemoine p't. 646. The J_s from the Lemoine p't to the sides of a a are proportional to the sides. From B' bisecting b, and K, the Lemoine p't, draw J_s. Then by~ as, B' P, . KP,' = B'P, . KP,' . But since a ABB' = A B'BC, .-. B'P, . a = B'P^ . c. .'. KP,'/a = KP:/c. 150 ELEMENTARY SYNTHETIC GEOMETRY. 647. \{ k^, k^, k^ are _Ls from K on a, b, c, then a A a'^b'' + c'' 648. [Grebe.] Describe sq's APQB, BUVC, CXYA on the sides of A y^^^' [all externally or all internally], and let QP^ ^Fmeet in a ; PQ, VU m yS ; UV, YX in y ; then aA, fiB, yC concur in K. 649. [Mathieu.] The Lemoine point of a triangle is the center of perspective of that triangle and its polar triangle with respect to any circle. Proof. With respect to circum-O of A ABC, the pole of BC is P, of CA is Q, of AB is R. LPM \ to QR is anti- \ to BC. .'. PL = PB =PC = PM; .'. AP is a symmedian of a ABC, .'. the Lemoine point is in AP. In same way, it is in BQ and CR ; .*. it is A C of As ABC, PQR; and .•. of a ABC and any other of its polar As. 650. The joins of the points of contact of the in-O of a a with its opposite vertices concur in the Lemoine p't of the A formed by joining the points of contact. This is called the Gergonne point of the first A . 651. [Schlomilch.] The three joins of the mid point of each side of a triangle, to the mid point of the corresponding altitude, concur in the Lemoine point. ANTI-PARALLELS, ISOGONALS, SYMMEDIANS. \\\ In A ABC let a, a' be p'ts in BC', ft, ft' in CA, y, y' in AB ; such that oiKft', ftKy', yKa' are the respective anti- jjs through K [the Lemoine p't], to AB, BC, CA. :. K\^ the mid p't of these sects and aftft'y' is a rect'. The median BB' [of a ABC^ cuts ay' in its mid p't J/; and if MK meets ^C in iV^, A'J/ = KNy^nd MKN is || to aft, and .-. to ^^, the alt' from B. :. B'K is median of a B'BE and .'. bisects BE. ■ 652. [R. F. Davis.] If of six points a pair is in each side of a triangle and concyclic with each other such pair, then the six are concyclic. Proof. Ay, . Ay, = Aft,. Aft, ; .-. A is on the radical axis of ©s a,a,y,y, , o(,ot^ft,ft,. But if these circles are distinct, yet intersect, their radical axis contains their common chord a,a, . .". They coincide. Fig. 270. 653. If a,ft, anti-|| to a,ft, , and a^y^ anti-ll to a,y, , and ft,y, anti-|| to ft,y, , then the six points ^if '^^y fti, ft,, Vi, 7, are concyclic. 654. [Tucker.] The six ends of three equal anti-parallels in a triangle are concyclic. 152 ELEMENTARY SYNTHETIC GEOMETRY. J I Nr- V L Fig. 271. Proof. Let /' be the in-center of A LMN. Since E' F, F'D make ^ E^FA = 4 BF'D, .-. aFNF' is •]• ; .-. a bisector of ^ iV is r't bi' of FF^ ; .\ the r't bi' of FF' passes through /'. Similarly, the r't bi' of DD' passes through /'. But since E'F= D'E, :. D'F \ to AC,.: D\ F, F' , D are concyclic ; .•. /' is the center of a O through D' , F, F', D. Similarly, /' is center of O through E' , D, D', E. 655. The circles got by varying the size of the three equal anti-parallels are all called Tucker's circles. 656. Corollary. K is cost' with O and O' the circumcenters of aABC ?ind Aa^y; and/' bisects the join 00'. For, since aE'AF is a II g'm, .*. K is /^ C oi A ABC and Aafty. Also OC, JL to tan' at C, is _L to i?'/;anti-|| to ^; and .'. O'y, || to OC, is ± to D'E', .'. through J/ (mid p't P'°-'7«. of yC and of BE) a || to OC will bisect 00' in T and contain /'. Similarly for E'F', .'. TisI'. ANTI-PARALLELS, ISOGONALS, SYMMEDIANS. 1 53 657. Straights through the Lemoine point of a triangle parallel to its sides, are called the Lemoine parallels of that triangle. 658. The crosses of the Lemoine parallels and the sides of a triangle are concyclic. For K being now the common vertex of 3 ||g'ms, its joins to the vertices of the a bisect the other 3 diagonals, joins of these crosses, which are thus 3 anti-||^ and all equal, being non-|j sides of 3 symtras. 659. This Tucker's circle through the ends of the Lemoine "N parallels is called the First Lemoine Circle. 660. The center of the First Lemoine Circle is the mid point of the sect KO. For A a-/?;/ has become the point A"; so /', bisecting OO' , now bisects OK. 660 (b). [H. M. Taylor.] The six J. projections of the feet of its altitudes on the other sides of a A are on a Tucker's 0, x:alled its Taylor's O. ^lUf ^^ h^5> '^5^? < CHAPTER II. THE BROCARD POINTS. 66i. Problem. To draw a circle which shall pass through a given point and touch a given straight at a given point. Construction. The J_ to the st' at the given p't and the r't bi' of the join of the two given points cross in the center. 662. Problem. To find within a a a p't which its sides will contain after equal positive rotations about their vertices. Construction. Describe a © passing through one vertex,, as C, and touching the opposite side c at the next counter-clockwise vertex, A. Draw the chord AP \ to BC, Join BP, cutting the O in/2. Proof. Then [periphery ^^ on the same arc CLA^ ^ AC^ — -2^ BAD. — •4. APn = 4 CBa [its alternate ^]. Determination. Only one solution. 663. This p't fL is called the positive Brocard p't. Its isogonal conjugate, the negative Brocard p't fl' , is given by substi- tuting negative for positive in the preceding problem. \ ' 664. D. and/2' are isogonal conjugates. 665. The magnitude of the angle of rotation for £1 and for /2' is designated by 00, and ^ -j- co is called the Brocard ^ of the A. 154 Fig. 273. THE BROCARD POINTS. I 55 666. The 3 O® each passing through a vertex and touching the opposite side at the next counter-clockwise vertex concur in the positive Brocard p't. [If next clockwise vertex, in /^'.] 667. At A draw AX \ to BC. At C make ^ XCA = ^ CBA. Then ^ CBX is the Brocard ^ of A ABC. 668. From ^y, the Brocard ^ , ^ -|- &?, is the same for all ~ As. 669. In the construction 662 we may keep ^ B constant and increase :^ &?, by sliding BC \ to itself until it touches the O; .-. if one ^ of a A is fixed, co is greatest when the A is •!• ; .'. an equilateral A has the greatest of all Brocard angles, which is ^ r't ^ . 670. The arcs A^IC, AflB, BflA, Bft'C, CflB, Cfl' A are called the Brocard arcs of the triangle. 671. [Brocard.] If O is circumcenter and K Lemoine of A ABC; and if O on diameter OK cuts the Lemoine jj* to BCt CA, AB, in a, /?, y, respectively ; then I. Ay, Bar, C/3 concur on the O; and their cross is the positive Brocard point D,. II. A^, By, Coc concur on the circle, and their cross is the negative Brocard point /2'. Proof. Let £KaF\ FK/3D', DyKE' be the Lemoine f. Then ^ OaK is r't. .-. aO _L to BC bisects BC in A' [since O is circumcenter]. ,In same way /5(9, yO meet b, c in their mid points B' , C. Let Ba, Cft cross in £1\ then aA'/BA' = twice i. from K on BC/BC', = twice i. from X on d/d [646] ; =r /3ByCB\ .'. A aBA' -^ A/3CB'. .-. ^ BaA' = ^ C/3B' = ^ 0/3n. \ 156 ELEMENTARY SYNTHETIC GEOMETRY. .'. £1 is concyclic with a, fS, O, and .•. also with y. Similarly, Ay, Ba concur on this ©; i.e., Ay, Ba, C^ concur in £1 on the © whose diam' is OK* Similarly, AjS, By, Cot concur in a p't DJ on the O. Moreover, :^nBC = ^nCA = :fnAB[{rom'^ AsaBA', §CB' , yAC'\ .'. O is the positive Brocard p't. Similarly, /2' is the negative Brocard p't. 672. The O on the joint of the circumcenter and Lemoine p't of a A as diameter is called the Bro card Q of the A , from the name of its discoverer. 673. The A whose vertices are the _L projections of the circumcenter on the Lemoine ||s is called Brocard's first a. 674. The A whose vertices are the X projections of the circumcenter on the symmedians is called Brocard's second A. 675. In A ABC to inscribe a A ~ to a given a. In AB take any p't D, and draw any sect JDF to another side, as AC; and at the points D and i^ make ^^ FJDE, DFE equal to 2 of the ^* of the given a ; .-. ^ ^5" = 3** angle. Join AE and produce it to meet the side a\n G ', from G draw GH, GI, respectively |1 to ED, EF. THE BROCARD POINTS l^T Join HI. A HIE is the required A. ', Jl H J- $ • For ^ E = 7f G ; ■ ^ also BE : HG ::AE: AG ::EF:GI; .-. A HGI'^ A BEF. 676. A ABC<^ A A'B'C. 677. Corollary. The sect from any angle of a a to // is. twice the _L from O on opposite side, 678. Theorem. Of any a, O, >^C, and //are co-st'. Proof. Join OH, meeting A A' in G. Bisect AG in X and AH in V. XV \\ to and = ^GH, .: ^ AXY = ^ AGH = ^ OG^ and^F= OA'. \_e77\ A\so AD W OA'; .\XY= OG = ^GH, and AX = XG = GA'. .'. G is ■'^C. CO-SYMMEDIAN AND CO-BROCARDAL TRIANGLES. 679. Let the symmedians A,K, B^K, C,K of a A^B.C, be produced to meet the circum-O in A^, B^, C^, and let the op- posite sides j?,^', , -5^^, of the quad' i?, 6^,^52 Cj be produced to meet in L. Then the polar of L passes through K, the cross of B^B^ . But since A^A^ is a symmedian of the a A^B^C,, the tan- gents at B^ and C^ cross on A^KA^ produced, .•. A^KA^ is the polar of L, and the tangents at B^ , C^ must cross on this st , which is consequently a symmedian of the a A^BC^ also ; similarly for the straights B,KB^ , C^KC^. These as A^B^C^, A,B^C, having the same symmedians are called co-symmedian triangles. They have the same Lemoine p't, and the same circum- center, consequently the same Brocard ©. Also the same 158 ELEMENTARY SYNTHETIC GEOMETRY. Brocard ^ , and the same Brocard p'ts, the same first Lemoine © , and the same second Lemoine O , also the Tucker Os of one are Tucker Qs of the other, though a par- ticular Tucker O of one is not always the same Tucker O of the other. Thus the Taylor Q of one is not the Taylor © of the other. 680. If 2 As are co-symmedian, the sides of one are propor- tional to the medians of the other. For ^s C,A^B, = C,A,A, + B.A^A, = KC,A, + KB,A, = GCjBj -[- GB^C^, since G and AT are isogonal conjugates. Hence QA,B, = BfiC; = A/C/'G, where C/' is the cross of C,G with a i to B,G through A/ . Similarly, ^ A,B,C, = i-C/'GA/. Thus, A A^B^C^ ~ A C^"GA^' , each of whose sides is -^ the corresponding median of a A^B^C^. 681. To show that any A y^j^jC, has corresponding to it not only the co-syn^edian a A^B^C^, but an infinity of others having the same Brocard p'ts, Lemoine p't, Brocard ©, ist Lemoine 0, 2d Lemoine ©, we should study the triangle's Brocard ellipse, but this would carry us beyond strictly elementary geometry. A system of As thus connected are called co-Brocardal AS. 682. If a a A,B,C, be inscribed m a given a ABC, the ©s AB,C, , BA,C, , CA,B, concur. For let ©s AB,C, , BA,C, meet in O. Then since B,OC, = st' ^-A, and C,OA, = sV ^-B, we have B,OA, = 2 st' ^s - [sf ^ - A] - [sf ^ - B] = A-^B = st' 4 -C; .'. the quad' A,OB,C is cyclic. 683. If the A A.B^C, inscribed in A ABC is ~ and co-sensal to it, and A, falls on a, then B0C = A-\~A, = 2A; THE BROCARD POINTS. I59 simKarly, CO A =28, and A0B = 2C\ therefore is the circumcenter. 684. If A^ falls on c, the Os concur in D,. If A^ falls on d, the Os concur in fi'. In the first case, the a and its inscribed A have the same positive Brocard p't. In the second cdse, the same negative Brocard p't. 685. Similar as have the same Brocard ^ a?. 686. O is the circumcenter of A ABC; OA^ , OB^ , OC, are drawn to a, b, c, so that ^OA^A = yf OB^B = ^ OC,C. Show that A A^B^C^ is ~ and co-sensal to a ABC. Show also that O is the orthocenter of A A^B^C, . I60 ELEMENTARY SYNTHETIC GEOMETRY. EXERCISES ON BOOK VIII. 1. Through the mid point of each side of a A are drawn ±8 to the other 2 sides. Show that the 2 as thus formed have the same Lemoine p't. 2. Show that the Lemoine p't of a A is the centroid of its 1 projec- tions on the sides ; and inversely. 3. Find a p't within a A such that the sum of the squares of its Xs on the sides is a minimum. [The p't must be the centroid of its 1 projections on the sides, and .*. the Lemoine p't.] 4. The joins of the circumcenter of a A to its vertices are J. to the sides of its orthocentric A. 5. Brocard's first A is in perspective with its original A. 6. Brocard's first A is ~ but not cosensal to its original. 7. The join of the circumcenter and Lemoine p't of a A is the r't bi' of the join of its Brocard p'ts. [ ^ aaK = ^ ChBC =00 = a'CB = n'a/C] 8. If T is the center of Brocard's O, then ^ flTfl' = 2ilTK = 4cj. 9. If AK, BK, CK meet the Brocard © in a', /J', y' , then a' is the cross of the Brocard arcs AilC, Afl'B ; ff is the cross of BJCIA, BLl'C; and y' is the cross of CD.B, Cfl'A. 10. [Dewulf.J If through the Brocard p't £1 three 0» be described each passing through two vertices of a ABC, the A formed by their centers has the circumcenter of ABC for one of its Brocard p'ts. 11. The join of any two p'ts, and the join of their isogonal conjugates, with respect to a A, subtend at any vertex of the a ^ either = or sup- plemental. 12. If three st's through the v^nices of a A meet the opposite sides co-st'ly, so do their isogonals. 13. The joins of the ± projections of the Lemoine p't on the sides of the A are X to the medians. 14. If on a given sect, and on the same side of it, be described six. A^ ~ to a given A, the vertices are concyclic. 15. In Fig. 275, Aa, Bfi, Cy concur. INDEX. (The numbers refer to .he pages.) Acute 13 Adjacent 7 Adrian ' 108 Altitude 43 Angle 5 Angle, Brocard 154 inscribed 40 periphery 40 tanchord 29 Angles, alternate 40 Anti-parallels 146 Apollonius Ill Archimedes 108 Area 104 Arms of angle 6 Axis of perspective 119 of symmetry 18 radical 142 Base 86 Bearer I2I Bi-radial 5 Bisectors of angles 34 Brianchon 140 Brocard angle 154 arcs 155 circle I55 points 154 first triangle 156 second triangle 156 Broken line 37 Cenquad 71 Center 8 Center of inversion 129 of involution 141 of perspective 95 of similitude • . - 95 Centimeter 104 Centroid 98 Ceva 118 Chord 10 Circle S Circle, Apollonius 1 1 1 Brocard i55 circum 34 ex 35 in 34 Lemoine's, 1st 153 2d 149 Nine-point 149 Taylor i53 Circles, Tucker's 152 Circular measure 108 Circum-center 34 -circle 34 -radius 34 Clockwise 5 Co-Brocardal.triangles I57 Complement of angle 13 161 l62 INDEX. Complete quadrilateral 126 Comquad 126 Conclusion g Concurrent 34 Concyclic , 41 Congruent 3 Conjugate points 137 Conjugates, harmonic 121 isogonal 148 Construction 9 Contraparallelogram 132 Convex 38 Corresponding points 95 Co-sensal 146 Co-straight 31 Cosymmedian triangles 157 Cross 123 Cross-ratio 137 Curve 4 Cyclic 41 polygon 70 Davis 151 Deltoid 49 Determination 9 Diagonal 38 points 126 triangle 126 Diameter 10 Dual .... 124 Egyptians 108 Element 123 Equal 6 Equi-cross 140 Equivalent 85 Escribed circle 35 Ex-center 35 Ex-circle 35 Explement 7 Explemental arcs 12 Exterior angle ". .. 35 Euler 108 Feuerbach 149 Figures 3 reciprocal 124 Fraction .... 92 G-Iine 60 Gergonne 150 Grebe 150 Harmonic conjugates 121 pencil 122 progression 121 range 97 Harmonically divided 97 Hart , 132 Hemiplanes 5 Heron 109 Hexagon 38 Brianchon's 140 Pascal's 140 Hexagram 141 Hindoos 108 Hypothenuse 35 Hypothesis 9 In-center 34 -circle <3^ '\^ -radius 34 Incommensurable 92 Inscribed angle 40 Inverse points 129 Inversion 129 Involution 141 Isogonal conjugates 148 Isogonals 146 Isosceles 35 Join 123 K in Lemoine point Lambert 108 Lemoine's first circle 153 second circle 149 parallels..., 153 point 149 Length 104 Lindemann 108 Line 2 INDEX. 163 Line, Simson 119 Linkage 130 Ludolf. 108 Lune 62 Major 13 Maximum 142 Measurement 104 Median 43 Medio-center 149 Menelaus 117 Milne 142 Minimum 142 Minor , 13 Multiple 92 Nine-point circle 149 Oblique 24 Obtuse 13 Opposite points 60 Orthocenter 57 Orthocentric triangle 147 Parallelogram 46 Parallels '. 30 Pascal 140 Peaucellier 130 Pencil 121 Perigon 6 Perimeter 37 Periphery angle 40 Perspective 95 Perspective-center 121 Pi (tt) 108 Plane 3 Point 2 of Gergonne 150 Lemoine 149 Points, Brocard 154 conjugate 121 corresponding 95 double 141 Polar 75, 133 Polar reciprocal 136 Polar triangle 78 Polar polygon 79 polarization 136 Pole of circle 62 Pole and Polar 133 Polygon 37 convex 30 star 33 undivided 37 Problem 9 Projection 1 119 Projection-axis 11 1 Projection-straights 96 Projective 95 Proportion 92 Ptolemy 102 q- radius 67 q-pole 67 Quadrangle 126 Quadrant 13 Quadrilateral 33 complete 126 Radian 109 Radical axis 142 Radius 8 Radius of inversion 129. Range 121 harmonic 97 Ratio 92 cross 137 Rays 5 Reciprocal 124 Rectangle 52 Reflex 13 Regular 38 Revolution 4 Rhombus 52 Rotation 5 -centre 57 Schlomilch , 150 Secant 22 Sect 4 Segments 97 164 INDEX. Sense 5 Similarity 95 Similitude, ratio of 95 Simson IIO Sphere .• 60 Spherical excess 88 Square 53 Straight 4 angle 6 Submultiple 92 Sum 7 Summit 126 Supplement 7 Surface I Symmedians 146 Symcenter 19 Symcentry 19 Symmetry 18 Symtra 49 Tanchord angle 40 Tangent 22 Tangent circles 28 Theorem 9 Bordage 84 Brahmegupla 84 Brianchon 140 Theorem of Brocard 157 Ceva 118 Davis 151 Dewulf 160 Feuerbach i^g Grebe 150 Hart 132 Mathieu 150 Menelaus 117 Pappus 90 Pascal 140 Ptolemy 102 Schlomilch.. 150 Tucker 151 Trace , Transversal 29 Trapezoid 91 Triangle, Brocard's first 157 second. . . ... 157 diagonal 126 orthocentric 147 self-conjugate 134 Tucker's circles > . . . 152 Unit 104 Vertex 121 UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which bonowed This book is DUE on the last date stamped below. Oct 11 '48T!\ri 0ctl2'48J5 l2Mlar'49^^ i9S1 l9Aug'60Er§ REC D uD MG22mo ^^^^^ IN STACKS APR 61954 REC'D LD APR 2 6 '64 -4 PM LD 21-100ot-9,'47(A5702s16)476 9181iJ0 H^4 THE UNIVERSITY OF CALIFORNIA LIBRARY \ ^%^\1