# Full text of "A text-book of geometry"

## See other formats

GIFT OF MATHEMATICAL TEXT-BOOKS By G. A. WENTWORTH, A.M. Mental Arithmetic. Elementary Arithmetic. Practical Arithmetic. Primary Arithmetic. Grammar School Arithmetic. High School Arithmetic. High School Arithmetic (Abridged). First Steps in Algebra. School Algebra. College Algebra. Elements of Algebra. Complete Algebra. Shorter Course in Algebra. Higher Algebra. New Plane Geometry. New Plane and Solid Geometry. Syllabus of Geometry. Geometrical Exercises. Plane and Solid Geometry and Plane Trigonometry. New Plane Trigonometry. New Plane Trigonometry, with Tables. New Plane and Spherical Trigonometry. New Plane and Spherical Trig., with Tables. New Plane and Spherical Trig., Surv., and Nav. New Plane Trig, and Surv., with Tables. New Plane and Spherical Trig., Surv., with Tables. Analytic Geometry. TEXT-BOOK OF GEOMETBY REVISED EDITION. BY G. A. WENTWORTH, A.M., AUTHOR OF A SERIES OF TEXT-BOOKS IN MATHEMATICS. BOSTON, U.S.A.: PUBLISHED BY GINN & COMPANY. Entered, according to Act of Congress, in the year 1888, by G. A. WENT WORTH, in the Office ot the Librarian of Congress, at Washington, ALL RIGHTS RESERVED. y TYPOGBAPHT BY J. S. GUSHING & Co., BOSTON, U.S.A. PBESSWOBK BY GINN & Co., BOSTON, U.B.A. PREFACE. ~\ T~OST persons do not possess, and do not easily acquire, the power *** of abstraction requisite for apprehending geometrical concep tions, and for keeping in mind the successive steps of a continuous argument. Hence, with a very large proportion of beginners in Geom etry, it depends mainly upon the form in which the subject is pre sented whether they pursue the study with indifference, not to say aversion, or with increasing interest and pleasure. In compiling the present treatise, the author has kept this fact con stantly in view. All unnecessary discussions and scholia have been avoided ; and such methods have been adopted as experience and attentive observation, combined with repeated trials, have shown to be most readily comprehended. No attempt has been made to render more intelligible the simple notions of position, magnitude, and direc tion, which every child derives from observation ; but it is helieved that these notions have been limited and denned with mathematical precision. A few symbols, which stand for words and not for operations, have been used, but these are of so great utility in giving style and per spicuity to the demonstrations that no apology seems necessary for their introduction. Great pains have been taken to make the page attractive. The figures are large and distinct, and are placed in the middle of the page, so that they fall directly under the eye in immediate connec tion with the corresponding text. The given lines of the figures are full lines, the lines employed as aids in the demonstrations are short- dotted, and the resulting lines are long-dotted. 327374 iv PREFACE. In each proposition a concise statement of what is given is printed in one kind of type, of what is required in another, and the demon stration in still another. The reason for each step is indicated in email type between that step and the one following, thus preventing the necessity of interrupting the process of the argument by referring to a previous section. The number of the section, however, on which the reason depends is placed at the side of the page. The constituent parts of the propositions are carefully marked. Moreover, each distinct assertion in the demonstration and each particular direction in the construction of the figures, begins a new line; and in no case is it necessary to turn the page in reading a demonstration. This arrangement presents obvious advantages. The pupil perceives at once what is given and what is required, readily refers to the figure at every step, becomes perfectly familiar with the language of Geom etry, acquires facility in simple and accurate expression, rapidly learns to reason, and lays a foundation for completely establishing the science. Original exercises have been given, not so difficult as to discourage the beginner, but well adapted to afford an effectual test of the degree in which he is mastering the subjects of his reading. Some of these exercises have been placed in the early part of the work in order that the student may discover, at the outset, that to commit to mem ory a number of theorems and to reproduce them in an examination is a useless and pernicious labor; but to learn their uses and appli cations, and to acquire a readiness in exemplifying their utility is to derive the full benefit of that mathematical training which looks not eo much to the attainment oj information as to the discipline of the mental faculties. G. A. WENTWORTH. EXETER, N.H. 1878. PEEFACE. TO THE TEACHER. WHEN the pupil is reading each Book for the first time, it will be well to let him write his proofs on the blackboard in his own lan guage ; care being taken that his language be the simplest possible, that the arrangement of work be vertical (without side work), and that the figures be accurately constructed. This method will furnish a valuable exercise as a language lesson, will cultivate the habit of neat and orderly arrangement of work, and will allow a brief interval for deliberating on each step. After a Book has been read in this way, the pupil should review the Book, and should be required to draw the figures free-hand. He should state and prove the propositions orally, using a pointer to indicate on the figure every line and angle named. He should be encouraged, in reviewing each Book, to do the original exercises ; to state the converse of propositions ; to determine from the statement, if possible, whether the converse be true or false, and if the converse be true to demonstrate it ; and also to give well-considered answers to questions which may be. asked him on many propositions. The Teacher is strongly advised to illustrate, geometrically and arithmetically, the principles of limits. Thus a rectangle with a con stant base b, and a variable altitude x, will afford an obvious illus tration of the axiomatic truth that the product of a constant and a variable is also a variable ; and that the limit of the product of a constant and a variable is the product of the constant by the limit of the variable. If x increases and approaches the altitude a as a limit, the area of the rectangle increases and approaches the area of the rectangle ab as a limit; if, however, x decreases and approaches zero as a limit, the area of the rectangle decreases and approaches zero for a limit. An arithmetical illustration of this truth may be given by multiplying a constant into the approximate values of any repetend. If, for example, we take the constant 60 and the repetend 0.3333, etc., the approximate values of the repetend will be T 3 o, -f^, VI PREFACE. T 3 o 3 o 3 ff> rVtiW Q te-> an d these values multiplied by 60 give the series 18, 19.8, 19.98, 19.998, etc., which evidently approaches 20 as a limit; but the product of 60 into (the limit of the repetend 0.333, etc.) is also 20. Again, if we multiply 60 into the different values of the decreasing series ^, yfo, ^uW ^inr. etc., which approaches zero as a limit, we shall get the decreasing series 2, , ^, ^ 7 , etc.; and this series evi dently approaches zero as a limit. In this way the pupil may easily be led to a complete compre hension of the subject of limits. The Teacher is likewise advised to give frequent written examina tions. These should not be too difficult, and sufficient time should be allowed for accurately constructing the figures, for choosing the best language, and for determining the best arrangement. The time necessary for the reading of examination-books will be diminished by more than one-half, if the use of the symbols employed in this book be allowed. G. A. W. EXETER, N.H. 1879. PKEFACE. VI} NOTE TO REVISED EDITION. THE first edition of this Geometry was issued about nine years ago. The book was received with such general favor that it has been neces sary to print very large editions every year since, so that the plates are practically worn out. Taking advantage of the necessity for new plates, the author has re-written the whole work ; bat has retained all the distinguishing characteristics of the former edition. A few changes in the order of the subject-matter have been made, some of the demonstrations have been given in a more concise and simple form than before, and the treatment of Limits and of Loci has been made as easy of comprehension as possible. More than seven hundred exercises have been introduced into this edition. These exercises consist of theorems, loci, problems of con struction, and problems of computation, carefully graded and specially adapted to beginners. No geometry can now receive favor unless it provides exercises for independent investigation, which must be of such a kind as to interest the student as soon as he becomes acquainted with the methods and the spirit of geometrical reasoning. The author has observed with the greatest satisfaction the rapid growth of the demand for original exercises, and he invites particular attention to the systematic and progressive series of exercises in this edition. The part on Solid Geometry has been treated with much greater freedom than before, and the formal statement of the reasons for the separate steps has b efen in general omitted, for the purpose of giving a more elegant form tb the demonstrations. A brief treatise on Conic Sections (Book IX) has been prepared, and is issued in pamphlet form, at a very low price. It will also be bound with the Geometry if that arrangement is found to be gen erally desired. Vili PREFACE. The author takes this opportunity to express his grateful appre ciation of the generous reception given to the Geometry heretofore by the great body of teachers throughout the country, and he confidently anticipates the same generous judgment of his efforts to bring the work up to the standard required by the great advance of late in the sci ence and method of teaching. The author is indebted to many correspondents for valuable sug gestions ; and a special acknowledgment is due, for criticisms and careful reading of proofs, to Messrs. C. H. Judson, of Greenville, S.C. ; Samuel Hart, of Hartford, Conn. ; J. M. Taylor, of Hamilton, N.Y. ; W. Le Conte Stevens, of Brooklyn, N.Y. ; E. R. Offutt, of St. Louis, Mo.; J. L. Patterson, of Lawrenceville, N. J.; G. A. Hill, of Cam bridge, Mass. ; T. M. Blakslee, of Des Moines, la. ; G. W. Sawin, of Cain- bridge, Mass. ; Ira M. De Long, of Boulder, Col. ; and W. J. Lloyd, of New York, N.Y. Corrections or suggestions will be thankfully received. G. A. WENTWORTH. EXETER, N.H., 1888. CONTENTS. GEOMETRY. PAGE DEFINITIONS . . . . . "l STRAIGHT LINES . . . .- 6 PLANE ANGLES . . . 7 MAGNITUDE or ANGLES 9 ANGULAR UNITS .10 METHOD or SUPERPOSITION .- 11 SYMMETRY 13 MATHEMATICAL TERMS 14 POSTULATES 15 AXIOMS . . .16 SYMBOLS . . 16 PLANE GEOMETRY. BOOK I. THE STRAIGHT LINE. THE STRAIGHT LINE 17 PARALLEL LINES 22 PERPENDICULAR AND OBLIQUE LINES . . . . . 33 TRIANGLES 40 QUADRILATERALS 56 POLYGONS IN GENERAL 66 EXERCISES , , 72 I X CONTENTS. BOOK II. THE CIRCLE. PAGE DEFINITIONS 75 AKCS AND CHORDS 77 TANGENTS 89 MEASUREMENT. 92 THEORY or LIMITS 94 MEASURE or ANGLES .98 PROBLEMS OF CONSTRUCTION 106 EXERCISES 126 BOOK III. PROPORTIONAL LINES AND SIMILAR POLYGONS. THEORY OF PROPORTION 131 PROPORTIONAL LINES . 138 SIMILAR TRIANGLES 145 SIMILAR POLYGONS 153 NUMERICAL PROPERTIES OF FIGURES 156 PROBLEMS OF CONSTRUCTION 167 PROBLEMS OF COMPUTATION 173 EXERCISES 175 BOOK IV. AREAS OF POLYGONS. AREAS OF POLYGONS 180 COMPARISON OF POLYGONS 188 PROBLEMS OF CONSTRUCTION 192 PROBLEMS OF COMPUTATION 204 EXERCISES 205 BOOK V. REGULAR POLYGONS AND CIRCLES. REGULAR POLYGONS AND CIRCLES 209 PROBLEMS OF CONSTRUCTION . . . ... . 222 MAXIMA AND MINIMA . - . . . . . . .230 EXERCISES . . . . . . -. . ... 237 MISCELLANEOUS EXERCISES .... . 240 GEOMETRY. DEFINITIONS. 1, If a block of wood or stone be cut in the shape repre sented in Fig. 1, it will have six flat faces. Each face of the block is called a surface ; and if these faces are made D smooth by polishing, so that, when a straight-edge is applied to any one of them, the straight edge in every part will touch the surface, the faces are called plane surfaces, or planes. 2, The edge in which any two of these surfaces meet is called a line. 3, The corner at which any three of these lines meet is called a point. 4, For computing its volume, the block is measured in three principal directions : From left to right, A to B. From front to back, A to C. From bottom to top, A to D. These three measurements are called the dimensions of the block, and are named length, breadth (or width), thickness (height or depth). 2, . c >; :" ;* *;*: : .-..GEOMETRY. A solid, therefore, has three dimensions, length, breadth, and thickness. 5, The surface of a solid is no part of the solid. It is simply the boundary or limit of the solid. A surface, there fore, has only two dimensions, length and breadth. So that, if any number of flat surfaces be put together, they will coincide and form one surface. 6, A line is no part of a surface. It is simply a boundary or limit of the surface. A line, therefore, has only one dimen sion, length. So that, if any number of straight lines be put together, they will coincide and form one line. 7, A point is no part of a line. It is simply the limit of the line. A point, therefore, has no dimension, but denotes position simply. So that, if any number of points be put together, they will coincide and form a single point. 8, A solid, in common language, is a limited portion of space filled with matter; but in Geometry we have nothing to do with the matter of which a body is composed ; we study simply its shape and size; that is, we regard a solid as a limited portion of space which may be -occupied by a physical body, or marked out in some other way. Hence, A geometrical solid is a limited portion of space. 9, It must be distinctly understood at the outset that the points, lines, surfaces, and solids of Geometry are purely ideal, though they can be represented to the eye in only a material way. Lines, for example, drawn on paper or on the blackboard, will have some width and some thickness, and will so far fail of being true lines; yet, when they are used to help the mind in reasoning, it is assumed that they represent perfect lines, without breadth and without thickness. DEFINITIONS. 3 10, A point is represented to the eye by a fine dot, and named by a letter, as A (Fig. 2) ; a line is named by two letters, placed one at each end, as BF; a surface is represented and named by the lines which bound it, as BCDF; a solid is represented by the faces which bound it. FlQ - 2 - 11, By supposing a solid to diminish gradually until it vanishes we may consider the vanishing point, a point in space, independent of a line, having position but no extent. 12, If a point moves continuously in space, its path is a line. This line may be supposed to be of unlimited extent, and may be considered independent of the idea of a surface. 13, A surface may be conceived as generated by a line moving in space, and as of unlimited extent. A surface can then be considered independent of the idea of a solid. 14, A solid may be conceived as generated by a surface in motion. Thus, in the diagram, let the up- D H right surface ABCD move to the A ~"~ right to the position EFGH. The V | points A, B, C, and D will generate Q the lines AE, BF, CG, and DH, |A~" y respectively. The lines AB, BC, B " ~" Q ~" F CD, and AD will generate the sur faces AF, BG, CH, and AH, respectively. The surface ABCD will generate the solid AG. 15, Geometry is the science which treats of position, form, and magnitude. 16, Points, lines, surfaces, and solids, with their relations, constitute the subject-matter of Geometry. GEOMETRY. 17, A straight line, or right line, is a line which has the same direction throughout its . whole extent, as the line AB. 18, A curved line is a line no part of which is straight, as the line CD. 19, A broken line is a series of different successive straight lines, as the line ER FlG - 4 - 20, A mixed line is a line composed of straight and curved lines, as the line GH. A straight line is often called simply a line, and a curved line, a curve. 21, A plane surface, or a plane, is a surface in which, if any two points be taken, the straight line joining these points will lie wholly in the surface. 22, A curved surface is a surface no part of which is plane. 23, Figure or form depends upon the relative position of points. Thus, the figure or form of a line (straight or curved) depends upon the relative position of the points in that line ; the figure or form of a surface depends upon the relative position of the points in that surface. 24, With reference to form or shape, lines, surfaces, and solids are called figures. With reference to extent, lines, surfaces, and solids are called magnitudes. 25, A plane figure is a figure all points of which are in the same plane. 26, Plane figures formed by straight lines are called rec tilinear figures ; those formed by curved lines are called curvilinear figures ; and those formed by straight and curved lines are called mixtilinear figures. DEFINITIONS. 5 27, Figures which have the same shape are called similar figures. Figures which have the same size are called equiva lent figures. Figures which have the same shape and size are called equal or congruent figures. 28, Geometry is divided into two parts, Plane Geometry and Solid Geometry. Plane Geometry treats of figures all points of which are in the same plane. Solid Geometry treats of figures all points of which are not in the same plane. STRAIGHT LINES. 29, Through a point an indefinite number of straight lines may be drawn. These lines will have different directions. 30, If the direction of a straight line and a point in the line are known, the position of the line is known ; in other words, a straight line is determined if its direction and one of its points are known. Hence, All straight lines which pass through the same point in the same direction coincide, and form but one line. 31, Between two points one, and only one, straight line can be drawn ; in other words, a straight line is determined if two of the points are known. Hence, Two straight lines which have two points in common coincide throughout their whole extent, and form but one line. 32, Two straight lines can intersect (cut each other) in only one point ; for if they had two points common, they would coincide and not intersect. 33, Of all lines joining two points the shortest is the straight line, and the length of the straight line is called the distance between the two points. D GEOMETRY. 34, A straight line determined by two points is considered as prolonged indefinitely both ways. Such a line is called an indefinite straight line. 35, Often only the part of the line between two fixed points is considered. This part is then called a segment of the line. For brevity, we say "the line AB" to designate a segment of a line limited by the points A and B. 36, Sometimes, also, a line is considered as proceeding from a fixed point and extending in only one direction. This fixed point is then called the origin of the line. 37, If any point C be taken in a given straight line AB, the two parts CA and GB arc said to have opposite direc- ^ - fa & tions from the point C. FIG. 5. 38, Every straight line, as A B, may be considered as hav ing opposite directions, namely, from A towards B, which is expressed by saying "line AB"; and from B towards .4, which is expressed by saying "line BA" 39, If the magnitude of a given line is changed, it becomes longer or shorter. Thus (Fig. 5), by prolonging AC to B we add GB to AC, and AB = AC+ CB. By diminishing AB to C, we subtract CB from AB, and AC=AB- CB. If a given line increases so that it is prolonged by its own magnitude several times in A B c D E succession, the line is multi- H plied, and the resulting line is called a multiple of the given line. Thus (Fig. 6), if AC=2AB, AD = ZAB, and Hence, DEFINITIONS. Lines of given length may be added and subtracted; may also be multiplied and divided by a number. they FIG. 7. PLANE ANGLES. 40, The opening between two straight lines which meet is called a plane angle. The two lines are called the sides, and the point of meeting, the vertex, of the angle. 41. If there is but one angle at a given vertex, it is designated by a cap ital letter placed at the vertex, and is read by simply naming the letter ; as, angle A (Fig. 7). But when two or more angles have the same vertex, each angle is desig nated by three letters, as shown in Fig. 8, and is read by naming the three letters, the one at the vertex be tween the others. Thus, the angle DA Q means the angle formed by the sides AD and AC. It is often convenient to designate an angle by placing a small italic let ter between the sides and near the vertex, as in Fig. 9. 42, Two angles are equal if they can be made to coincide. FIG FIG. 43, If the line AD (Fig. 8) is drawn so as to divide the angle BAG into two equal parts, BAD and CAD, AD is called the bisector of the angle BAG. In general, a line that divides a geometrical magnitude into two equal parts is called a bisector of it. 8 GEOMETRY. 44. Two angles are called ad jacent when they have the same vertex and a common side be tween them ; as, the angles BOD and AOD (Fig. 10). 45, When one straight line stands upon another straight line and makes the adjacent angles equal, each of these angles is called a right angle. Thus, the equal angles DCA and DOB (Fig. 11) are each a right angle. O FIG. 10. C FIG. 11. 46, When the sides of an an gle extend in opposite directions, so as to be in the same straight line, the angle is called a straight angle. Thus, the angle formed at C (Fig. 11) with its sides CA and CB extending in opposite directions from C, is a straight angle. Hence a right angle may be defined as half a straight angle. 47, A perpendicular to a straight line is a straight line that makes a right angle with it. Thus, if the angle DCA (Fig. 11) is a right angle, DC is perpendicular to AB, and AB is per pendicular to DC. 48, The point (as C, Fig. 11) where a perpendicular meets another line is called the foot of the perpendicular. 49. Every angle less than a right an gle is called an acute angle; as, angle A. FIG 50, Every angle greater than a right angle and less than a straight angle is called an obtuse angle; as, angle C (Fig. 13). DEFINITIONS. 9 51, Every angle greater than a straight angle and less than two straight angles is called a reflex angle; as, angle (Fig. 14). FIG. 13. FIG. 14. 52, Acute, obtuse, and reflex angles, in distinction from right and straight angles, are called oblique angles ; and inter secting lines that are not perpendicular to each other are called oblique lines. 53, When two angles have the same vertex, and the sides of the one are prolongations of the sides of the other, they are called vertical angles. Thus, a and b (Fig. 15) are vertical an gles. 54, Two angles are called ^ FlQ complementary when their sum is equal to a right angle ; and each is called the complement of the other; as, angles DOB and DOC (Fig. 10). 55, Two angles are called supplementary when their sum is equal to a straight angle ; and each is called the supplement of the other ; as, angles DOB and DO A (Fig. 10). MAGNITUDE OF ANGLES. 56, The size of an angle depends upon the extent of opening of its sides, and not upon their length. Suppose the straight 10 GEOMETRY. line 00 to move in the plane of the paper from coincidence with OA, about the point as a pivot, to the position 0(7; then the line 00 describes or generates the angle AOC, and the magnitude of the angle AOC depends upon the amount of rotation of the line from the position OA to the position OC. If the rotating line moves from the position OA to the position OB, perpen dicular to OA, it generates the right angle AOB ; if it moves to the position OD, it generates the obtuse angle AOD ; if it moves to the posi tion OA 1 , it generates the straight angle AOA ; if it moves to the position OB 1 , it generates the reflex angle AOB 1 , indicated by the dotted line ; and if it continues its rotation to the posi tion OA, whence it started, it generates two straight angles. Hence the whole angular magnitude about a point in a plane is equal to two straight angles, or four right angles; and the angular magnitude about a point on one side of a straight line drawn through that point is equal to one straight angle, or two right angles. Angles are magnitudes that can be added and subtracted ; they may also be multiplied and divided by a number. ANGULAR UNITS. 57, If we suppose 00 (Fig. 17) to turn about from a position coinci dent with OA until it makes a com plete revolution and comes again into coincidence with OA, it will describe the whole angular magnitude about the point 0, while its end point O will describe a curve called a circum ference. DEFINITIONS. 11 58, By adopting a suitable unit of angles we are able to express the magnitudes of angles in numbers. If we suppose 00 (Fig. 17) to turn about from coinci dence with OA until it makes one three hundred and sixtieth of a revolution, it generates an angle at 0, which is taken as the unit for measuring angles. This unit is called a degree. The degree is subdivided into sixty equal parts called minutes, and the minute into sixty equal parts, called seconds. Degrees, minutes, and seconds are denoted by symbols. Thus, 5 degrees 13 minutes 12 seconds is written, 5 13 12". A right angle is generated when 00 has made one-fourth of a revolution and is an angle of 90; a straight angle is generated when 00 has made one-half of a revolution and is an angle of 180 ; and the whole angular magnitude about is generated when 00 has made a complete revolution, and contains 360. The natural angular unit is one complete revolution. But the adoption of this unit would require us to express the values of all angles by fractions. The advantage of using the degree as the unit consists in its convenient size, and in the fact that 360 is divisible by so many different integral numbers. METHOD OF SUPERPOSITION. 59, The test of the equality of two geometrical magnitudes is that they coincide throughout their whole extent. Thus, two straight lines are equal, if they can be so placed that the points at their extremities coincide. Two angles are equal, if they can be so placed that they coincide. In applying this test of equality, we assume that a line may be moved from one place to another without altering its length; that an angle may be taken up, turned over, and put down, without altering the difference in direction of its sides. 12 GEOMETEY. This method enables us to compare magnitudes of the same kind. Suppose we have two angles, ABC and DEF. Let the side ED be placed on the side BA, so that the vertex E shall fall on B; then, if the side EF falls on BO, the angle DEF equals the angle ABC; if the side EF falls between EG and BA in the direction BG, the angle DEF is less than ABO; but if the side EF falls in the direction BH, the angle DEF is greater than ABO. This method enables us to add magnitudes of the same kind. Thus, if we have two straight lines BC AB and CD, by placing the point Q D on B, and keeping CD in the ^ # same direction with AB, we shall FlQ - 19 - have one continuous straight line AD equal to the sum of the lines AB and CD. C / FIG. 20. B FIG. 21. Again : if we have the angles ABC and DEF, and place the vertex E on B and the side ED in the direction of BC, the angle DEF will take the position CBH, and the angles DEF and ABC will together equal the angle ABU. If the vertex J is placed on B, and the side ED on J:L4, the angle DEFwitt take the position ABF, and the angle FBC will be the difference between the angles ABC and DEF, DEFINITIONS. 13 SYMMETRY. 60, Two points are said to be symmetrical with respect to a third point, called the centre of sym- metry, if this third point bisects the p> \- p straight line which joins them. Thus, FlQ - 22 - P and P are symmetrical with respect to as a centre, if C bisects the straight line PP 1 . 61, Two points are said to be sym metrical with respect to a straight line, called the axis of symmetry, if this straight line bisects at right angles the straight line which joins them. Thus, P and P are symmet rical with respect to XX 1 as an axis, if XX 1 bisects PP at right angles. 62, Two figures are said to be sym metrical with respect to a centre or an axis if every point of one has a corresponding symmetrical point in the other. Thus, if every point in the figure A B C* has a symmetrical point in ABO, with respect to D as a centre, the figure A B C is sym metrical to ABO with respect to D as a centre. 63, If every point in the figure A B C has a symmetrical point in ABO, with respect to XX 1 as an axis, the figure A JB C 1 is symmetri cal to AB with respect to XX 1 as an axis. 14 GEOMETRY. 64, A figure is symmetrical with re spect to a point, if the point bisects every straight line drawn through it and terminated by the boundary of the figure. 65, A plane figure is symmetrical with respect to a straight line, if the line divides it into two parts, which are sym metrical with respect to this straight line. MATHEMATICAL TERMS. FIG. 27. 66, A proof or demonstration is a course of reasoning by which the truth or falsity of any statement is logically established. 67, A theorem is a statement to be proved. 68, A theorem consists of two parts: the hypothesis, or that which is assumed ; and the conclusion, or that which is asserted to follow from the hypothesis. 69, An axiom is a statement the truth of which is admitted without proof. 70, A construction is a graphical representation of a geo metrical figure. 71, A problem is a question to be solved. 72, The solution of a problem consists of four parts : (1) The analysis, or course of thought by which the con struction of the required figure is discovered ; (2) The construction of the figure with the aid of ruler and compasses ; (3) The proof that the figure satisfies all the given condi tions; DEFINITIONS. 15 (4) The discussion of the limitations, which often exist, within which the solution is possible. 73, A postulate is a construction admitted to be possible. 74, A proposition is a general term for either a theorem or a problem. 75, A corollary is a truth easily deduced from the propo sition to which it is attached. 76, A scholium is a remark upon some particular feature of a proposition. 77, The converse of a theorem is formed by interchanging its hypothesis and conclusion. Thus, If A is equal to B, C is equal to D. (Direct.) If is equal to D, A is equal to B. (Converse.) 78, The opposite of a proposition is formed by stating the negative of its hypothesis and its conclusion. Thus, If A is equal to B, C is equal to D. (Direct.) If A is not equal to B, C is not equal to D. (Opposite.) 79, The converse of a truth is not necessarily true. Thus, Every horse is a quadruped is a true proposition, but the converse, Every quadruped is a horse, is not true. 80, If a direct proposition and its converse are true, the opposite proposition is true ; and if a direct proposition and its opposite are true, the converse proposition is true. 81, POSTULATES. Let it be granted 1. That a straight line can be drawn from any one point to any other point. 2. That a straight line can be produced to any distance, or can be terminated at any point. 3. That a circumference may be described about any point as a centre with a radius of given length, 16 GEOMETRY. 82. AXIOMS. 1. Things which are equal to the same thing are equal to each other. 2. If equals are added to equals the sums are equal. 3. If equals are taken from equals the remainders are equal. 4. If equals are added to unequals the sums are unequal, and the greater sum is obtained from the greater magnitude. 5. If equals are taken from unequals the remainders are unequal, and the greater remainder is obtained from the greater magnitude. 6. Things which are double the same thing, or equal things, are equal to each other. 7. Things which are halves of the same thing, or of equal things, are equal to each other. 8. The whole is greater than any of its parts. 9. The whole is equal to all its parts taken together. 83, SYMBOLS AND ABBREVIATIONS. + increased by. O circle. circles. diminished by. Def. . . . definition. X multiplied by. Ax. ... axiom. -f- divided by. Hyp. . . hypothesis. = is (or are) equal to. Cor. . . . corollary. =:= is (or are) equivalent to. Adj. . . . adjacent. > is (or are) greater than. Iden. . . identical. < is (or are) less than. Cons. . . construction. .-. therefore. Sup. . . . supplementary. angle. Sup. -adj. supplementary. Bangles. Ext. -int. exterior-interior. _L perpendicular. Alt.-int. alternate-interior. Jl perpendiculars. Ex. ... exercise. II parallel. rt right. lie parallels. st straight. A triangle. Q.E.D. . . quod erat demonstrandum, A triangles. which was to be proved. O parallelogram. Q.E.F. . , quod erat faciendum, 17 parallelograms. which was to be done. PLANE GEOMETRY. BOOK I. THE STRAIGHT LINE. PROPOSITION I. THEOREM. 84i All straight angles are equal. A - B D * F Let Z.BCA and /.FED be any two straight angles. To prove ABCA = FED. Proof, Apply the Z EC A to the Z. FED, so that the vertex C shall fall on the vertex E, and the side GB on the side EF. Then GA will coincide with ED, (because BOA and FED are straight lines and have two points common). Therefore the Z EGA is equal to the Z FED. 59 Q. E. D. 85, COR. 1. All right angles are equal. Ax. 7. 86, COR. 2. The angular units have constant values. 87, COR. 3. The complements of equal angles are equal. Ax. 3. 88, COR. 4. The supplements of equal angles are equal. Ax. 3. 89, COR. 5. At a given point in a given straight line one perpendicular, and only one, can be erected. HINT. Consider the given point as the vertex of a straight angle, and draw the bisector of the angle. 18 PLANE GEOMETRY. BOOK I. PROPOSITION II. THEOREM. 90. If two adjacent angles have their exterior sides in a straight line, these angles are supplements of each other. Let the exterior sides OA and OB of the adjacent A AOD and BOD be in the straight line A3. To prove A AOD and BOD supplementary. Proof. AOB is a straight line. Hyp. . . the Z AOB is a st. Z. 46 But theZ^KXD + ZmD^thest. /.AOB. Ax. 9 /. the A AOD and BOD are supplementary. 55 Q. E. D. 91. SCHOLIUM. Adjacent angles that are supplements of each other are called supplementary-adjacent angles. 92. COR. Since the angular magnitude about a point is neither increased nor diminished by the number of lines which radiate from the point, it follows that, The sum of all the angles about a point in a plane is equal to two straight angles, or four right angles. The sum of all the angles about a point on the same side of a straight line passing through the point is equal to a straight angle, or two right angles. THE STRAIGHT LINE. 19 PROPOSITION III. THEOREM. 93. CONVERSELY : If two adjacent angles are supple ments of each other, their exterior sides lie in the same straight line. o AC B F Let the adjacent A OCA + OCB = 2 rt. A. To prove A C and CB in the same straight line. Proof, Suppose CF to be in the same line with.-4Cl 81 Then Z OCA + Z OCF= 2 rt. A, 90 (being sup.-adj. A). But Z OCA + Z OCB = 2 rt. A. Hyp. . . Z OCA + Z OCr= Z OCA -f Z OCB. Ax. 1 Take away from each of these equals the common Z OCA. Then Z 0CF= Z 0C5. Ax. 3 .*. CB and CF coincide. . . A C and CB are in the same straight line. Q.E. D. 94. SCHOLIUM. Since Propositions II. and III. are true, their opposites are true ; namely, 80 If the exterior sides of two adjacent angles are not in a straight line, these angles are not supplements of each other. If two adjacent angles are not supplements of each other, their exterior sides are not in the same straight line. 20 PLANE GEOMETRY. BOOK I. PROPOSITION IV. THEOREM. 95, If one straight line intersects another straight line, the vertical angles are equal. Let line OP cut AS at C. To prove Z OCB = Z ACP. Proof, Z OCA + Z OCB = 2 rt. A, (being sup.-adj. A)- AACP=2rt. A, (being sup.-adj. ). 90 90 Ax. 1 Take away from each of these equals the common Z OCA. Then Z OCB -Z ACP. Ax. 3 In like manner we may prove Q. E. D. 96. COR. If one of the four angles formed by the intersection of two straight lines is a right angle, the other three angles are right angles. THE STRAIGHT LINE. 21 PROPOSITION V. THEOREM. 97. From a point without a straight line one per pendicular, and only one, can be drawn to this line. JT / D\ C \ V Let P be the point and AB the line. To prove that one perpendicular, and only one, can be drawn from Pto AB. Proof, Turn the part of the plane above AB about AB as an axis until it falls upon the part below AB, and denote by P 1 the position that P takes. Turn the revolved plane about AB to its original position, and draw the straight line PP , cutting AB at C. Take any other point D in AB, and draw PD and P D, Since POP is a straight line, PDP is not a straight line. (Between two points only one straight line can be drawn.) . . Z PCP is a st. Z, and Z PDP is not a st. Z. Turn the figure POD about AB until P falls upon P. Then CP will coincide with OP, and DP with DP. . . Z PCD = Z. POD, and Z PDO= Z PDC. 59 .-. Z POZ), the half of st. Z PC/*, is a rt. Z ; and Z PZ>C, the half of Z PZ>^, is not a rt. Z. . . PC is to ^15, and PD is not _L to AB. 47 .*. one _L, and only one, can be drawn from P to AB. Q.E.D, 22 PLANE GEOMETRY. BOOK! I. PARALLEL LINES. 98, DEF. Parallel lines are lines which lie in the same plane and do not meet however far they are prolonged in both directions. 99, Parallel lines are said to lie in the same direction when they are on the same side of the straight line joining their ori gins, and in opposite directions when they are on opposite sides of the straight line joining their origins. PROPOSITION VI. 100, Two straight lines in the same plane perpen dicular to the same straight line are parallel. -B Let AB and CD be perpendicular to AC. To prove AB and CD parallel. Proof. If AB and CD are not parallel, they will meet if sufficiently prolonged, and we shall have two perpendicular lines from their point of meeting to the same straight line ; but this is impossible. 97 (From a given point without a straight line, one perpendicular, and only one, can be drawn to the straight line.} . . AB and CD are parallel. Q.E.D. REMARK. Here the supposition that AB and CD are not parallel leads to the conclusion that two perpendiculars can be drawn from a given point to a straight line. The conclusion is false, therefore the supposi tion is false; but if it is false that AB and CD are not parallel, it is true that they are parallel. This method of proof is called the indirect method. 101, COR. Through a given point, one straight line, and only (me, can be drawn parallel to a given straight line. PAEALLEL LINES. 23 PROPOSITION VII. THEOREM. 102, If a straight line is perpendicular to one of two parallel lines, it is perpendicular to the other. H M- E ar ar K Let AB and EF be two parallel lines, and let HK be perpendicular to AB, and cut EF at C. To prove HK\_EF. Proof, Suppose MN drawn through (7J_ to HK. Then MN \a\\toAB, 100 (two lines in the same plane _L to a given line are parallel). But EFia \\ to AB. Hyp. /. EF coincides with MN, 101 (through the same point only one line can be drawn \\ to a given line). that is, .ffiTis J_ to EF. Q.E.D. 103, If two straight lines AB and CD are cut by a third line EF, called a transversal, the eight angles formed, are named as follows : The angles a, d, f, g are called interior ; b, c, e, h are called ex terior angles. The angles d and /, or a and g, are called alt. -int. angles. The angles b and h, or c and e, are called alt. -ext. angles. The angles b and /, c and g, a and e, or d and h, are called ext. -int. angles. 24 PLANE GEOMETRY. BOOK I. PROPOSITION VIII. THEOREM. 104, If two parallel straight lines are cut by a third straight line, the alternate-interior angles are equal. E Let EF and GH be two parallel straight lines cut by the line BQ. To prove Z.B = /.Q. Proof. Through 0, the middle point of BC, suppose AD drawn J_ to GH. Then AD is likewise _L to EF, 102 (a straight line _L to one of two Us is _L to the other), that is, CD and BA are both JL to AD. Apply figure COD to figure BOA, so that OD shall fall on OA. Then 00 will fall on OB, 95 (since /. COD = Z BOA, being vertical A) ; and the point C will fall upon B, (since 00 = OB by construction). Then the J_ CD will coincide with the _L BA, 97 (from a point without a straight line only one JL to that line can be drawn}. . . /. OCD coincides with Z OB A, and is equal to it. 59 Q. E. D. Ex. 1. Find the value of an angle if it is double its complement ; if it is one-fourth of its complement. Ex. 2. Find the value of an angle if it is double its supplement ; if it is one-third of its supplement. PAKALLEL LINES./ 25 PROPOSITION IX. THEOREM. 105. CONVERSELY : When two straight lines are cut by a third straight line, if the alternate-interior an gles are equal, the two straight lines are parallel. MA \f/ D N Let EF cut the straight lines AB and CD in the points E and K, and let the To prove AB II to CD. Proof, Suppose MN drawn through H II to CD \ 101 then Z MHK= Z HKD, 104 (being alt.-int. A of II lines). But Z AHK= Z HKD. Hyp. /. Z MHK= Z AHK. Ax. 1 /. the lines JIfJVand AB coincide. But MNis II to CD. Cons. .*. AB, which coincides with JOT, is II to QD. Q.E. D. Ex. 3. How many degrees in the angle formed by the hands of a clock at 2 o clock ? 3 o clock ? 4 o clock ? 6 o clock ? 26 PLANE GEOMETRY. BOOK I. PROPOSITION X. THEOREM. 106, // two parallel lines are cutlby a third straight line, the exterior-interior angles are equal. Let AB and CD be two parallel lines cut by the straight line EF, in the points H and K, To prove Z EHB = Z HKD. Proof, Z EHB = Z AHK, 95 (being vertical A). . But Z AHK= Z HKD, 104 (being alt.-int. Aof\\ lines). Ax. 1 In like manner we may prove Z E HA = Z HKC. Q. E. D. 107, COR. The alternate-exterior angles EHB and CKF, and also AHE and DKF, are equal. Ex. 4. If an angle is bisected, and if a line is drawn through the vertex perpendicular to the bisector, this line forms equal angles with the sides of the given angle. Ex. 5. If the bisectors of two adjacent angles are perpendicular to each other, the adjacent angles are supplementary. PARALLEL LINES. 27 PROPOSITION XI. THEOREM. 108, CONVERSELY : When two straight lines are cut ly a third straight line, if the exterior-interior an gles are equal, these two straight lines are parallel. B Let EF cut the straight lines AB and CD in the points H and K, and let the To prove AB \\ to CD. Proof, Suppose MN drawn through 5" II to CD. 101 Then Z EHN= Z HKD, 106 (being ext.-int. A of II lines). But Z EHB = Z HKD. Hyp. . . Z EHB = Z EHN. Ax. 1 . .the lines JOT" and AB coincide. But MNia II to CD. Cons. . .AB, which coincides with MN, is II to CD. Q. E. D. Ex. 6. The bisector of one of two vertical angles bisects the other. Ex. 7. The bisectors of the two pairs of vertical angles formed by two intersecting lines are perpendicular to each other. 28 PLANE GEOMETRY. BOOK I. PROPOSITION XII. THEOREM. 109. If two parallel lines are cut by a third straight line, the sum of the two interior angles on the same side of the transversal is equal to two right angles. -B Let AB and CD be two parallel lines cut by the straight line EF in the points H and K. To prove Z BIIK+ Z HKD = 2 rt. A Proof, Z EHB + Z EHK = 2 rt. 4 90 (being sup.-adj. zt). But Z EHB = Z HKD, 106 (being ext.-int. A of II lines). Substitute Z HKD for Z EHB in the first equality ; then Z BHK+ Z HKD - 2 rt. A. Q. E. D. Ex. 8. If the angle AHE is an angle of 135, find the number of degrees in each of the other angles formed at the points -ffand K. Ex. 9. Find the angle between the bisectors of adjacent complemen tary angles. PARALLEL LINES. 29 PROPOSITION XIII. THEOREM. 110. CONVERSELY : When two straight lines are cut by a third straight line, if the two interior angles on the same side of the transversal are together equal to two right angles, then the two straight lines are parallel. Let EF cut the straight lines AB and CD in the points H and K, and let the BHK+^HKD equal two right angles. To prove AB II to CD. Proof, Suppose MN drawn through H \\ to CD. Then Z NHK+ Z. HKD = 2 rt. A, 109 (being two interior Aof\\son the same side of the transversal}. But Z.BHK+HKL = 2rt.A. Hyp. . .Z.NHK+Z.HKD = Z.BHK+Z.HKD. Ax. 1 Take away from each of these equals the common Z. HKD ; then Z. NHK= Z. BHK. Ax. 3 . . the lines AB and MN coincide. But MNis II to CD. Cons. . . AB, which coincides with MN, is II to CD. Q.E.O. 30 PLANE GEOMETRY. BOOK I. PROPOSITION XIV. THEOREM. Ill, Two straight lines which are parallel to a third straight line are parallel to each other. K Let AB and CD be parallel to EF. To prove AB II to CD. Proof, Suppose HK drawn _L to EF. 97 Since CD and EF are II, HKis _L to CD, 102 (if a straight line is to one of two Us, it is J_ to the other also). Since AB and EF are II, fflTis also _L to AB. 102 (each being a rt. Z). .-. AB is II to CD, 108 (when two straight lines are cut by a third straight line, if the ext.-int. A are equal, the two lines are parallel). Q. E D Ex 10. It has been shown that if two parallels are cut by a trans* versal! the alternate-interior angles are equal, the exterior-interior angles are equal the two interior angles on the same side of the transversal are Bupplementary. State the opposite theorems. State the converse theo rems. PARALLEL LINES. 31 PKOPOSITION XV. THEOREM. 112, Two angles whose sides are parallel, each to each, are either equal or supplementary. L r M -fcr- N F Let AB be parallel to EF, and BC to MN. To prove Z ABO equal to Z EHN, and to Z MHF, and supplementary to Z EHM and to Z NHF. Proof, Prolong (if necessary) BO and FE until they inter sect at D. 81 (2) Then Z B = Z EDO, 106 and Z DHN= Z ^D(7. 106 (being ext.-int. A of II ines), and Z.B = Z. MHF (the vert. Z of DHN). Now Z DHN is the supplement of Z EHM and Z ^V^F. . . Z ^, which is equal to Z DJ77V, is the supplement of Z EHM and of Z NHF. Q. E. D. REMARK. The angles are equal when both pairs of parallel sides extend in the same direction, or in opposite directions, from their ver tices ; the angles are supplementary when two of the parallel sides extend in the same direction, and the other two in opposite directions, from their vertices. 32 PLANE GEOMETKY. BOOK I. PROPOSITION XVI. THEOREM. 113, Two angles whose sides are perpendicular, each to each, are either equal or supplementary. O K \ F G Let AB be perpendicular to FD, and AC to GI. To prove Z BAG equal to Z DFG, and supplementary to /.DPI. Proof, Suppose AK drawn _L to AB, and AH A- to AC. Then AKis \\ to FD, arid AJI to IG, 100 (two lines J_ to the same line are parallel). 112 (two angles are equal. whose sides are II and extend in the same direction from their vertices). The Z BAK is a right angle by construction. . . Z BAH is the complement of Z KAH. The Z CAJTis a right angle by construction. . . Z HAHis the complement of Z BAG. 87 (complements of equal angles are equal). .\/.DFG = ^BAO. Ax. 1 . . Z DFI, the supplement of Z DFG, is also the supplement tf/.BAC. Q.E.D. REMARK. The angles are equal if both are acute or both obtuse ; they are supplementary if one is acute and the other obtuse. PEKPENDICULAR AND OBLIQUE LINES. 33 PERPENDICULAR AND OBLIQUE LINES. PROPOSITION XVII. THEOREM. 114, The perpendicular is the shortest Line that can be drawn from a point to a straight Line. D\ i Let AB be the given straight line, P the given point, PC the perpendicular, and PD any other line drawn from P to AB. To prove PC < PD. Proof, Produce PC to P , making CP = PC; and draw DP . On AB as an axis, fold over CPD until it conies into the plane of CP D. The line CP will take the direction of CP , (since Z PCD - Z P>CD t each being a rt. Z by hyp.). The point P will fall upon the point P , (since PC= P Cby cons.). . . line PD = line P D, and PC +CP =2 PC. Cons. But PC + CP <PD + DP 1 , (a straight line is the shortest distance between two points). . .2PC<2PD, or PC<PD. Q. E . D . 34 PLANE GEOMETRY. BOOK X, 115, SCHOLIUM. The distance of a point from a line is under stood to mean the length of the perpendicular from the point to the line. PROPOSITION XVIII. THEOREM. 116, Two oblique lines drawn from a point in a perpendicular to a given line, cutting off equal dis tances from the foot of the perpendicular, are equal. A F Let FC be the perpendicular, and CA and CO two oblique lines cutting off equal distances from F. To prove CA = CO. Proof, Fold over CFA, on CF&s an axis, until it comes into the plane of CFO. FA will take the direction of FO, (since Z CFA = Z CFO, each being a rt. Z by hyp.). Point A will fall upon point 0, (since FA = FO by hyp.). . .line C4 = line CO, (their extremities being the same points). Q. E. o. 117, COR. Two oblique lines drawn from a point in a per pendicular to a given line, cutting off equal distances from the foot of the perpendicular, make equal angles with the given line, and also with the perpendicular. PEEPENDICULAE AND OBLIQUE LINES. 35 PROPOSITION XIX. THEOREM. 118, The sum of two lines drawn from a point to the extremities of a straight line is greater than the sum of two other lines similarly drawn, but included by them. C A B Let CA and CB be two lines drawn from the point C to the extremities of the straight line AB. Let OA and OB be two lines similarly drawn, but included by CA and CB. To prove CA+CB>OA + OB. Proof. Produce AO to meet the line CB at E. Then AC+ CE > OA + OE, (a straight line is the shortest distance between two points), and BE+OE>BO. Add these inequalities, and we have CA+CE+BE+OE>OA + OE+ OB. Substitute for CE+ BE its equal CB, and take away OE from each side of the inequality. We have CA+CB>OA + OB. Ax. 5 aEilx 36 PLANE GEOMETRY. BOOK I. PROPOSITION XX. THEOREM. 119, Of two oblique lines drawn from the same point in a perpendicular, cutting off unequal dis tances from the foot of the perpendicular, the more remote is the greater. b Let OC be perpendicular to AB, OG and OE two oblique lines to AB, and GE greater than CG. To prove OE > OG. Proof. Take CF equal to CG, and draw OF. Then 0F= OG, 116 (two oblique lines drawn from a point in a _L, cutting off equal distances from the foot of the JL, are equal). Prolong OC to D, making CD =00. Draw ED and FD. Since AB is _L to OD at its middle point, FO = FD, and EO = JED, 116 But OE+ED> OF+ FD, 118 (the sum of two oblique lines drawn from a point to the extremities of a straight line is greater than the sum of two other lines similarly drawn, but included by them). . . 20E>20F, or OE> OF. But OF= OG. Hence OE > OG. a E . D . 120, COR. Only two equal straight lines can be drawn from a point to a straight line ; and of two unequal lines, the greater cuts off the greater distance from the foot of the perpendicular. PEKPENDICULAE, AND OBLIQUE LINES. 37 PROPOSITION XXI. THEOREM. 121, Two equal oblique lines, drawn from, the same point in a perpendicular, cut off equal distances from tine foot of the perpendicular. E F K Let CF be the perpendicular, and CE and CK be two equal oblique lines drawn from the point C to AB. To prove FE=FK. Proof, Fold over GFA on OF as an axis, until it conies into the plane of CFB. The line FE will take the direction FK, (since Z CFE= Z. CFK, each being a rt. Z by hyp). Then the point E must fall upon the point K, Otherwise one of these oblique lines must be more remote from the perpendicular, and therefore greater than the other ; which is contrary to the hypothesis that they are equal. 119 Q. E. D. Ex. 11. Show that the bisectors of two supplementary-adjacent angles are perpendicular to each other. Ex. 12. Show that the bisectors of two vertical angles form one straight line. Ex. 13. Find the complement of an angle containing 26 52 37". Find the supplement of the same angle. 38 PLANE GEOMETRY. BOOK I. PROPOSITION XXII. THEOREM. 122, Every point in the perpendicular, erected at the middle of a given straight line, is equidistant from the extremities of the line, and every point not in the perpendicular is unequally distant from the extremities of the line. R Let PR be a perpendicular erected at the middle of the straight line AB, any point in PR, and G any point without PR. Draw OA and OB, CA and CB. To prove OA and OB equal, CA and CB unequal. Proof, PA = PB. Hyp. 116 (two oblique lines drawn from the same point in a _L, cutting off" equal dis tances from the foot of the _L, are equal). Since C is without the perpendicular, one of the lines, CA or CB, will cut the perpendicular. Let CA cut the J_ at D, and draw DB. Then DB = DA, 116 (two oblique lines drawn from the same point in a _L, cutting off equal dis tances from the foot of the _L, are equal). But CB<CD+DB, (a straight line is the shortest distance between two points). Substitute in this inequality DA for DB, and we have CB<CD + DA. That is, CB<CA. PERPENDICULAR AND OBLIQUE LINES. 39 123, Since two points determine the position of a straight line, two points equidistant from the extremities of a line deter mine the perpendicular at the middle of that line. THE Locus OF A POINT. 124, If it is required to find a point which shall fulfil a single geometric condition, the point will have an unlimited number of positions, but will be confined to a particular line, or group of lines. Thus, if it is required to find a point equidistant from the extremities of a given straight line, it is obvious from the last proposition that any point in the perpendicular to the given line at its middle point does fulfil the condition, and that no other point does ; that is, the required point is confined to this perpendicular. Again, if it is required to find a point at a given distance from a fixed straight line of indefinite length, it is evident that the point must lie in one of two straight lines, so drawn as to be everywhere at the given distance from the fixed line, one on one side of the fixed line, and the other on the other side. The locus of a point under a given condition is the line, or group of lines, which contains all the points that fulfil the given condition, and no other points. 125, SCHOLIUM. In order to prove completely that a certain line is the locus of a point under a given condition, it is neces sary to prove that every point in the line satisfies the given condition; and secondly, that every point which satisfies the given condition lies in the line (the converse proposition), or that every point not in the line does not satisfy the given condi tion (the opposite proposition). 126, Con. The locus of a point equidistant from the extrem ities of a straight line is the perpendicular bisector of that line. 122, 123 40 PLANE GEOMETEY. BOOK I. D TRIANGLES. 127. A triangle is a portion of a plane bounded by three straight lines; as, ABC. The bounding lines are called the sides of the triangle, and their sum is called its perimeter ; the angles formed by the sides are called the angles of the triangle, and the vertices of these an gles, the vertices of the triangle. 128, An exterior angle of a triangle is an angle formed between a side and the prolongation of another side ; as, ACD. The interior angle ACE is adjacent to the exterior angle ; the " FIG. 2. other two interior angles, A and B, are called opposite- interior angles. FIG. 1. Scalene. Isosceles. Equilateral. 129, A triangle is called, with reference to its sides, a scalene triangle when no two of its sides are equal ; an isos celes triangle, when two of its sides are equal ; an equilateral triangle, when its three sides are equal. Right. Obtuse. Acute. Equiangular. 130, A triangle is called, with reference to its angles, a right triangle, when one of its angles is a right angle ; an obtuse TEIANGLES. 41 triangle, when one of its angles is an obtuse angle ; an acute triangle, when all three of its angles are acute angles ; an equiangular triangle, when its three angles are equal. 131, In a right triangle, the side opposite the right angle is called the hypotenuse, and the other two sides the legs, of the triangle. 132, The side on which a triangle is supposed to stand is called the base of the triangle. Any one of the sides may be taken as the base. In the isosceles triangle, the equal sides are generally called the legs, and the other side, the base. 133, The angle opposite the base of a triangle is called the vertical angle, and its vertex the vertex of the triangle. 134, The altitude of a triangle is the perpendicular distance from the vertex to the base, or to the base produced ; as, AD. 135, The three perpendiculars from the vertices of a tri angle to the opposite sides (produced if necessary) are called the altitudes; the three bisectors of the angles are called tha bisectors; and the three lines from the vertices to the middle points of the opposite sides are called the medians of the triangle. 136, If two triangles have the angles of the one equal respec tively to the angles of the other, the equal angles are called homologous angles, and the sides opposite the equal angles are called homologous sides. In general, points, lines, and angles, similarly situated in equal or similar figures, are called homologous. 137, THEOREM . The sum of two sides of a triangle is greater than the third side, and their difference is less than the third side. In the A ABC(Yi%. 1), AB + BC>AC, for a straight line is the shortest distance between two points ; and by taking away BCiiQm both sides, AB>AC-BC, or AC-BC<AB. 42 PLANE GEOMETRY. BOOK I. PROPOSITION XXIII. THEOREM. 138, The sum of the three angles of a triangle is equal to two right angles. A Let ABC be a triangle. To prove Z B + Z BOA + Z A = 2 rt. A. Proof, Suppose CE drawn II to AB, and prolong A to F. Then Z ECF+ Z ECE + Z J5O4 = 2 rt. Z, 92 (/ie swra o/ a^ the A about a oin on the same side of a straight line = 2 rt. A). But Z A - Z J57CF, 106 (Jetn^r ext.-int. A o/\\ lines). w&/.B = Z.BCE, 104 (being alt.-int. A of II lines ). Substitute for Z EOF and Z .SC^ the equal A A and 5. Then Z ^. + Z B + Z .SO4 - 2 rt. Zs. Q. E. D. 139, Con. 1. If the sum of two angles of a triangle is sub tracted from two right angles, the remainder is equal to the third angle. 140, COR. 2. If two triangles have two angles of the one equal to two angles of the other, the third angles are equal. 141, COR. 3. If two right triangles have an acute angle of the one equal to an acute angle of the other, the other acute angles are equal. TRIANGLES. 43 142, COR. 4. In a triangle there can be but one right angle, or one obtuse angle. 143, COR. 5. In a right triangle the two acute angles are complements of each other. 144, COR. 6. In an equiangular triangle, each angle is one- third of two right angles, or two-thirds of one right angle. PROPOSITION XXIV. THEOREM. 145, The exterior angle of a triangle is equal to the sum of the two opposite interior angles. A C* -*"* -*i U Let BCH be an exterior angle of the triangle ABC. To prove Z BCH Z A -f Z B. Proof. Z B CH+ ^ACB = 2rt.A y (being sup.-adj. A}, (the sum of the three A of a A = 2 rt. A). Take away from each of these equals the common Z ACB ; then Z.BCH=Z.A + B. Ax. 3 Q. E. D. 146, COR. The exterior angle of a triangle is greater than either of the opposite interior angles. 44 PLANE GEOMETRY. BOOK I. 7 PROPOSITION XXV. THEOREM. 147, Two triangles are equal if a side and two ad jacent angles of the one are equal respectively to a side and two adjacent angles of the other. A CD ff In the triangles ABC and DEF, let AB = DE, Z.A = Z.D, /LB = ^E. To prove A ABC= A DEF. Proof, Apply the A AB C to the A DEF so that AB shall coincide with DE. A C will take the direction of DF, (for ZA = ZD,by hyp.) ; the extremity C of AC will fall upon DF or DF produced. EC will take the direction of EF, (for ZB = ZE,by hyp.) the extremity C of .#(7 will fall upon EF or EF produced. . .the point C, falling upon both the lines DF and EF, must fall upon the point common to the two lines, namely, F. /.the two A coincide, and are equal. Q.E.D. 148, COR. 1. Two right triangles are equal if the hypotenuse and an acute angle of the one are equal respectively to the hypote nuse and an acute angle of the other. 149, COR. 2. Two right triangles are equal if a side and an acute angle of the one are equal respectively to a side and homologous acute angle of the other, TRIANGLES. 45 PROPOSITION XXVI. THEOREM. 150, Two triangles are equal if two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. A B D E In the triangles ABC and DEF, let AB = DE, AG-DF, A = ^D. To prove AAC-=A DEF. Proof, Apply the A ABC to the A DEF so that AB shall coincide with DE. Then A will take the direction of DF, (for ZA = ZD,by hyp} ; the point C will fall upon the point F, (for AC= DF, by hyp.). (thdr extremities being the same points). .the two A coincide, and are equal. Q.E. D. 151, COR. Two right triangles are equal if their legs are equal, each to each. 46 PLANE GEOMETRY. BOOK I. PROPOSITION XXVII. THEOREM. 152, If two triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second. -y E In the triangles ABC and ABE, let AB = AB, BC=BE; but ZABC greater than /.ABE. To prove AC > AE. Proof, Place the A so that AB of the one shall coincide with AB of the other. Suppose BF drawn so as to bisect Z EBG. Draw EF. In the A EBF and CBF EB = J30, Hyp. BF=BF, Iden. Z EBF= Z. CBF. Cons. .*. the A EBF&K& CBFwQ equal, 150 (having two sides and the included /. of one equal respectively to two sides and the included Z. of the other). .\EF=FC, (being homologous sides of equal A). Now AF+ FE > AE, 137 (the sum of two sides of a A is greater than the third side). . . AF+FO AE-, or, AC>AE. Q.E.D. TRIANGLES. 47 PROPOSITION XXVIII. THEOREM. 153. CONVERSELY. If two sides of a triangle are equal respectively to two sides of another, but the third side of the first triangle is greater than the third side of the second, then the angle opposite the third side of the first triangle is greater than the angle opposite the third side of the second. D A B C E In the triangles ABO and DEF, let AB = DE, AC = DF; but let BG be greater than EF. To prove Z A greater than Z D. Proof. Now Z A is equal to Z D, or less than Z D, or greater than Z D. But Z A is not equal to Z D, for then A ABC would be equal to A DEF, 150 (having two sides and the included Z of the one, respectively equal to two sides and the included /. of the other), and BG would be equal to EF. And Z A is not less than Z D, for then BC would be less than EF. 152 /. Z A is greater than Z D. Q.E.D. PLANE GEOMETRY. BOOK I. PROPOSITION XXIX. THEOREM. 154, In an isosceles triangle the angles opposite the equal sides are equal. A B D C Let ABO be an isosceles triangle, having the sides AB and AC equal. To prove Z B = Z C. Proof, Suppose AD drawn so as to bisect the /.BAG. In the A ADB and ADC, AJ3 = AC. Hyp. AD = AD, Iden. Z. BAD = Z. CAD. Cons. . .AAD = &ADC, 150 (two & are equal if two sides and the included Z of the one are equal respectively to two sides and the included Z of the other). .\Z.B = C. Q.E.D. 155, COR. An equilateral triangle is equiangular, and each angle contains 60. Ex. 14. The bisector of the vertical angle of an isosceles triangle bisects the base, and is perpendicular to the base. Ex. 15. The perpendicular bisector of the base of an isosceles triangle passes through the vertex and bisects the angle at the vertex. TKIANGLES. 49 PEOPOSITION XXX. THEOREM. 156, If two angles of a triangle are equal, the sides opposite the equal angles are equal, and the triangle is isosceles. B D c In the triangle ABC, let the Z = ZC. To prove AB = AC. Proof, Suppose AD drawn J_ to BO. In the rt. A ALB and ADC, AD = AD, 4.B = Z.Q. .*. rt. A ALB = rt. A ADC, Iden. Hyp. 149 (having a side and an acute Z of the one equal respectively to a side and an homologous acute Z of the other). (being homologous sides of equal &). Q.E. D. 157, COR. An equiangular triangle is also equilateral. Ex. 16. The perpendicular from the vertex to the base of an isosceles triangle is an axis of symmetry. 50 PLANE GEOMETRY. BOOK X, PROPOSITION XXXI. THEOREM. 158, If two sides of a triangle are unequal, the an gles opposite are unequal, and the greater angle is opposite the greater side. C B In the triangle ACB let AB be greater than AC. To prove Z A CB greater than Z B. Proof. Take AE equal to AQ. Draw EC. AAEC=Z.ACE, 154 (being A opposite equal sides). But Z AEG is greater than Z. B, 146 (an exterior /. of a A is greater than either opposite interior Z). and Z A CB is greater than Z ACE. Ax. 8 Substitute for Z ACE its equal Z ^(7, then Z.ACB\* greater than Z AEC. Much more, then, is the Z ACB greater than Z. B. Ex. 17. If the angles ^Cand ACB, at the base of an isosceles tri angle, be bisected by the straight lines BD, CD, show that DBCmll be an isosceles triangle. TKI ANGLES. 51 PROPOSITION XXXII. THEOREM. 159, CONVERSELY : If two angles of a triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle. In. the triangle AGE, let angle ACE be greater than angle E. To prove AB > AC. Proof, Now AB is equal to AC, or less than AC, or greater than AC. But AB is not equal to AC, for then the /. C would be equal to the Z B, 154 (being A opposite equal sides ). And AB is not less than AC, for then the Z. C would be less than the Z B, 158 (if two sides of a A are unequal, the A opposite are unequal, and the greater Z is opposite the greater side). . . AB is greater than AC. Q. E. D. Ex. 18. ABC and ABD are two triangles on the same base AB, and on the same side of it, the vertex of each triangle being without the other. If AC equal AD, show that BC cannot equal BD. Ex. 19. The sum of the lines which join a point within a triangle to the three vertices is less than the perimeter, but greater than half the perimeter. 52 PLANE GEOMETRY. BOOK: I. PROPOSITION XXXIII. THEOREM. 160, Two triangles are equal if the three sides of the one are equal respectively to the three sides of the other. In the triangles ABC and A B C , let AB = A B , AC=A C t , BC=B C . To prove A ABC = A A B C . Proof. Place A A B C in the position AB C, having its greatest side A C in coincidence with its equal AC, and its vertex at , opposite B ; and draw BB 1 . Since AB = AB , Hyp. Z ABB = Z AB B, 154 (in an isosceles A the A opposite the equal sides are equal). Since CB = CB\ Hyp. ZCBB = ZCB R 154 Hence, ^ ABC= Z AB C, Ax. 2 /. A ABG= A AB t O= A A B C 150 (two & are equal if two sides and included Z of one are equal to two sides and included Z of the other). TRIANGLES. 53 PROPOSITION XXXIV. THEOREM. 161. Two right triangles are equal if a side and the hypotenuse of the one are equal respectively to a side and the hypotenuse of the other. In the right triangles ABC and A B C , let AB^A Bf, and AC=A C . To prove A AC= A A C ? . Proof, Apply the A ABC to the A A B C t so that AE shall coincide with A , A falling upon A 1 , E upon E\ and and C l upon the same side of A -B r . Then BO will take the direction of B &, (for Z ABC= Z A B C f , each being a rt. Z). Since AC=A C 9 , the point C will fall upon <7 , 121 (two equal oblique lines from a point in a _L cut off" equal distances from the foot of the JL). .*. the two A coincide, and are equal. Q.E. 0. 54 PLANE GEOMETRY. BOOK I. PROPOSITION XXXV. THEOREM. 162, Every point in the bisector of an angle is equi distant from the sides of the angle. Let AD be the bisector of the angle BAG, and let O be any point in AD. To prove that is equidistant from AB and AC. Proof, Draw O^and OG J_ to AB and A C respectively. In the rt. A ^O^Fand AOO AO=AO, Iden. JBAO = CAO. Hyp. .-.AAOF=AAOG, 148 (two rt. A are equal if the hypotenuse and an acute Z of the one are equal respectively to the hypotenuse and an acute Z of the other). . . OF= OG, (homologous sides of equal &). .*. is equidistant from AB and AC. Q. E. D. What is the locus of a point : Ex. 20. At a given distance from a fixed point ? \ 57. Ex. 21. Equidistant from two fixed points? 119. Ex. 22. At a given distance from a fixed straight line of indefinite length ? Ex. 23. Equidistant from two given parallel lines ? Ex. 24. Equidistant from the extremities of a given line ? TRIANGLES. 55 PROPOSITION XXXVI. THEOREM. 163, Every point within an angle, and equidistant from its sides, is in the bisector of the angle. A G C Let be equidistant from the sides of the angle BAG, and let AO join the vertex A and the point 0. To prove that AO is the bisector of Z. BAG. Proof. Suppose OF and OG drawn J. to AB and AC, respectively. In the rt. A ^O^and AOG OF= OG, Hyp. AO^AO. Iden. .-. AAOF=AAOG, 161 (two rt. & are equalif the hypotenuse and a side of the one are equal to the hypotenuse and a side of the other). /. Z FAO = /. GAO, (homologous A of equal A). .*. AO is the bisector of Z BAG. Q.E. D. 164. COR. The locus of a point within an angle, and equi distant from its sides, is the bisector of the angle. 56 PLANE GEOMETRY. BOOK I. QUADRILATERALS. 165. A quadrilateral is a portion of a plane bounded by four straight lines. The bounding lines are the sides, the angles formed by these sides are the angles, and the vertices of these angles are the vertices, of the quadrilateral. 166. A trapezium, is a quadrilateral which has no two sides parallel. 167. A trapezoid is a quadrilateral which has two sides, and only two sides, parallel. 168. A parallelogram is a quadrilateral which has its oppo site sides parallel. Trapezium. Trapezoid. Parallelogram. 169. A rectangle is a parallelogram which has its angl< right angles. 170. A rhomboid is a parallelogram which has its angl< oblique angles. 171, A square is a rectangle which has its sides equal. 172, A rhombus is a rhomboid which has its sides equal. Square. Rectangle. Rhombus. Rhomboid. 173, The side upon which a parallelogram stands, and the opposite side, are called its lower and upper bases. QUADETLATEEALS. 57 174, The parallel sides of a trapezoid are called its bases, the other two sides its legs, and the line joining the middle points of the legs is called the median. 175, A trapezoid is called an isosceles trapezoid when its legs are equal. 176, The altitude of a parallelogram or trapezoid is the perpendicular distance between its bases. 177, The diagonal of a quadrilateral is a straight line joining two opposite vertices. PROPOSITION XXXVII. THEOREM. 178, The diagonal of a parallelogram divides the figure into two equal triangles. B C A. E Let ABCE "be a, parallelogram and AC its diagonal. To prove A ABC= A AEQ. In the A ABC and AEQ, AC=AC, Iden. Z.ACB = CAE, 104 and Z.CAB = Z.ACE, (being alt.-int. A of II lines). .-.AAC=AAEG, 147 (having a side and two adj. A of the one equal respectively to a side and two adj. A of the other). Q.E.D. 58 PLANE GEOMETRY. BOOK I. PROPOSITION XXXVIII. THEOREM. 179, In a parallelogram the opposite sides are equal, and the opposite angles are equal. Let the figure ABCE be a parallelogram. To prove BC= AE, and AB = EC, also, Z B = /. E, and Z BAE^Z. BCE. Proof. Draw AC. AABC^AAEC, 178 (the diagonal of a O divides the figure into two equal &). .-. BC= AE, and AB= CE, (being homologous sides of equal A). Also, /.B = Z.E,mdL/.BAE=BCE, 112 (having their sides II and extending in opposite directions from their vertices). Q. E. D. 180, Con. 1. Parallel lines comprehended between parallel lines are equal. -L. 181. COR. 2. Two parallel lines are everywhere equally distant. For if AB and DC are parallel, D C J dropped from any points in AB to DC, measure the distances of these points from DO. But these J are equal, by 180; hence, all points in AB are equidistant from DO. QUADRILATERALS. 59 PROPOSITION XXXIX. THEOREM. 182, If two sides of a quadrilateral are equal and parallel, then the other two sides are equal and par allel, and the figure is a parallelogram. B O Let the figure ABCE be a quadrilateral, having the side AE equal and parallel to BO. To prove AB equal and II to EC. Proof. Draw AC. In the A ABC and AEC BC= AE, Hyp. AC=AC, Iden. BCA = Z.CAE, 104 (being alt.-int. A of II lines). 150 (having two sides and the included Z. of the one equal respectively to two sides and the included Z of the other). \ . . AB = EC, (being homologous sides of equal A). Also, Z.BAC=/-ACE, (being homologous A of equal &). .-.AJBis II to J^C; 105 (when two straight lines are cut by a third straight line, if the alt.-int. A are equal, the lines are parallel). . . the figure ABCE is a O, 168 (the opposite sides being parallel). Q. E. n 60 PLANE GEOMETRY. BOOK I. PROPOSITION XL. THEOREM. 183, If the opposite sides of a quadrilateral are equal, the figure is a parallelogram. Let the figure ABCE be a quadrilateral having BG = AE and AB = EC. To prove figure ABCE a O. Proof. Draw AC. In the A AB and AEQ 0= AE, Hyp. AB=CE, Hyp. AC= AC. Hen. .:AABC=AAEC, 160 (having three sides of the one equal respectively to three sides of the other). and (being homologous A of equal &). .-.BO is II ioAJE, and AB is II to EC, 105 (when two straight lines lying in the same plane are cut by a third straight line, if the alt.-int. A are equal, the lines are parallel). .-. the figure ABCE is a O, 168 (having its opposite sides parallel). Q.E. D. QUADRILATERALS. 61 PROPOSITION XLI. THEOREM. 184, The diagonals of a parallelogram bisect each other. B O Let the figure ABCE be a parallelogram, and let the diagonals AC and BE cut each other at 0. To prove A0= 00, and BO = OK In the A AOE and BOO AE=BC, 179 (being opposite sides of a CJ). ZOAE=ZOC, 104 and OEA=QBC, (being alt.-int. A of II lines). .-.AAOE = AOC, 147 (having a side and two adj. A of the one equal respectively to a side and two adj. A of the other). (being homologous sides of equal A). Q.E. D. Ex. 25. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Ex. 26. The diagonals of a rectangle are equal. Ex. 27. If the diagonals of a parallelogram are equal, the figure is a rectangle. Ex. 28. The diagonals of a rhombus are perpendicular to each other, and bisect the angles of the rhombus. Ex. 29. The diagonals of a square are perpendicular to each other, and bisect the angles of the square. 62 PLANE GEOMETRY. BOOK I. PROPOSITION XLIL THEOREM. 185. Two parallelograms, having two sides and the included angle of the one equal respectively to two sides and the included angle of the other, are equal. B CD In the parallelograms ADCD and A B C D , let AB = A B , AD = A D , and Z.A = Z.A>. To prove that the UJ are equal. Apply H ABCD to O A C D f , so that AD will fall on and coincide with A D . Then AB will fall on A B\ and the point B will fall on B\ (for AB = A B , by hyp.). Now, BO and C are both II to A D and are drawn through point . . . the lines BO and B C* coincide, 101 and C falls on C or C produced. In like manner, Z>(7and D C are II to A and are drawn through the point D f . .-. Z)(7and DC 1 coincide. 101 .-. the point falls on D C , or D C produced. /. C falls on both B C and D C 1 . . . (7 must fall on the point common to both, namely, C . . . the two UJ coincide, and are equal. Q. E. O. 186, COR. Two rectangles having equal bases and equal altitudes are equal. QUADRILATERALS. 63 PROPOSITION XLIII. THEOREM. 187. If three or more parallels intercept equal parts on any transversal, they intercept equal parts on every transversal. E M Let the parallels AH, BK, CM, DP intercept equal parts HK, KM, MP on the transversal HP. To prove that they intercept equal parts AB, EG, CD on the transversal AD. Proof, From A, B, and tf suppose AE, BF, and CO drawn II to HP. Then AE = HK, BF= KM, CO = MP, 180 (parallels comprehended between parallels are equal). . .AE=BF=CO. Ax. 1 Also Z.A = AB = Z.C, 106 (being ext.-int. A of II lines) ; and E=Z.F=/.O, 112 (having their sides II and directed the same way from the vertices). /. A ABE= A BCF= A CDG, 147 (each having a side and two adj. A respectively equal to a, side and two adj. A of the others). /. AB mm BC= CD, (homologous sides of equal &). Q. E. o. 64 PLANE GEOMETRY. BOOK I. 188, COR. 1. The line parallel to the base of a triangle and bisecting one side bisects the other side also. For, let DE be II to BC and bisect AB. Draw through A a line II to BC. Then this line is II to DE, by 111. The three parallels by hypothesis intercept equal parts on the transversal AB, and there fore, by 187, they intercept equal parts on the transversal AC-, that is, the line DE bisects AC. 189, COR. 2. The line which joins the middle points of two sides of a triangle is parallel to the third side, and is equal to half the third side. For, a line drawn through D, the middle point of AB, II to BC, passes through E, the middle point of AC, by 188. Therefore, the line joining D and j7 coincides with this parallel and is II to BC. Also, since EF drawn H to AB bisects AC, it bisects BC, by 188 ; that is, BF= FG = \BG. But BDEF is a O by construction, and therefore DE=BF=\BG. 190, COR. 3. The line which is parallel to the bases of a trap ezoid and bisects one leg of the trap ezoid bisects the other leg also. For rx; ? if parallels intercept equal parts on / \ x ^ \ any transversal, they intercept equal j JP\ \ parts on every transversal by 187. / 191, COR. 4. The median of a trapezoid is parallel to the bases, and is equal to half the sum of the bases. For, draw the diagonal DB. In the A ADB join E, the middle point of AD, to F, the middle point of DB. Then, by 189, EF is II to AB sui& = %AB. In the ADBC join Fto G, the middle point of BC. Then FG is II to DC and = \DG. AB and FG, being II to DC, are II to each other. But only one line can be drawn through F II to AB. There fore FG is the prolongation of EF. Hence EFG is II to AB and DC, and = * (AB + DC). EXERCISES. 65 EXERCISES. 30. The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle. HINT. Let the bisectors AD and BE intersect at 0. Then being in AD is equidistant from AC and AB. (Why ?) And being in BE is equidistant from BC and AB. Hence is equidistant from AC and BC, and therefore is in the bisector CF. (Why ?) 31. The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle. HINT. Let the _L bisectors EEf and DD intersect at 0. Then being in EE / is equidistant from A ^ and C. (Why ?) And being in DD / is equidistant F from A and B. Hence is equidistant from B and C, and therefore is in the JL bisector FF . (Why?) 32. The perpendiculars from the vertices of a* triangle to the opposite eides meet in a point. HINT. Let the JL be AH, BP, and CK. Through A, B, C suppose B C , A f C f , A B drawn II to BC, AC, AB, respectively. Then AH is JL to B C . (Why ?) Now ABCB and ACBff are Hf (why?), and AB = BC, and ACT ^ = BC. (Why ?) That is, A is the middle point of B &. In the same way, B and C are the middle points of A C and A B , respectively. There fore, AH, BP, and C!2Tare the _L bisectors of the sides of the A A B C . Hence they meet in a point. (Why ?) 33. The medians of a triangle meet in a point which is two-thirds of the distance from each vertex to the middle of the opposite side. HINT. Let the two medians AD and CE meet in 0. Take ,Fthe middle point of OA, and G of OC. Join OF, FE, ED, and DO. In A AOC, OF i* II to AC and equal to ,} AC. (Why ?) DE is \\ioAC and equal to %AC. (Why ?) Hence DOPE is a O. (Why ?) Hence AF= FO = OD, and CG=GO= OE. (Why ?) Hence, any median cuts off on any other median two-thirds of the dis tance from the vertex to the middle of the opposite side. Therefore the median from B will cut off AO, two-thirds of AD; that is, will pass through 0. 66 PLANE GEOMETRY. BOOK I. POLYGONS IN GENERAL. 192, A polygon is a plane figure bounded by straight lines. The bounding lines are the sides of the polygon, and their sum is the perimeter of the polygon. The angles which the adjacent sides make with each other are the angles of the polygon, and their vertices are the ver tices of the polygon. The number of sides of a polygon is evidently equal to the number of its angles. 193, A diagonal of a polygon is a line joining the vertices of two angles not adjacent ; as AC, Fig. 1. B JO D E Fio. 1. FIG. 2. FIG. 3. 194, An equilateral polygon is a polygon which has all its sides equal. 195, An equiangular polygon is a polygon wh^ch has all its angles equal. 196. A convex polygon is a polygon of which no side, when produced, will enter the surface bounded by the perimeter. 197. Each angle of such a polygon is called a salient angle, and is less than a straight angle. 198. A concave polygon is a polygon of which two or more sides, when produced, will enter the surface bounded by the perimeter. Fig. 3. 199, The angle FDE is called a re-entrant angle, and is greater than a straight angle. If the term polygon is used, a convex polygon is meant. POLYGONS. 67 200. Two polygons are equal when they can be divided by diagonals into the same number of triangles, equal each to each, and similarly placed ; for the polygons can be applied to each other, and the corresponding triangles will evidently coincide. 201. Two polygons are mutually equiangular, if the angles of the one are equal to the angles of the other, each to each, when taken in the same order. Figs. 1 and 2. 202, The equal angles in mutually equiangular polygons are called homologous angles ; and the sides which lie between equal angles are called homologous sides. 203, Two polygons are mutually equilateral, if the sides of the one are equal to the sides of the other, each to each, when taken in the same order. Figs. 1 and 2. FIG. 4. FIG. 5. FIG. 6. FIG. 7. Two polygons may be mutually equiangular without being mutually equilateral ; as, Figs. 4 and 5. And, except in the case of triangles, two polygons may be mutually equilateral without being mutually equiangular ; as, Figs. 6 and 7. If two polygons are mutually equilateral and equiangular, they are equal, for they may be applied the one to the other ao as to coincide. 204, A polygon of three sides is called a trigon or triangle; one of four sides, a tetragon or quadrilateral ; one of five sides, a pentagon; one of six sides, a hexagon; one of seven sides, a heptagon; one of eight sides, an octagon; one of ten sides, a decagon ; one of twelve sides, a dodecagon. 68 PLANE GEOMETRY. BOOK I. PROPOSITION XLIV. THEOREM. 205, The sum of the interior angles of a polygon is equal to two right angles, taken as many times less two as the figure has sides. A D Let the figure ABCDEF be a polygon having n sides. To prove Z.A+AB + AC, etc. = (w-2) 2 rt.A. Proof, From the vertex A draw the diagonals AC, AD, and AE. The sum of the A of the A = the sum of the A of the polygon. Now there are (n 2) A, and the sum of the A of each A = 2 rt. A. 138 .-. the sum of the A of the A, that is, the sum of the A of the polygon = (n 2) 2 rt. A. a E. D. 206. COR. The sum of the angles of a quadrilateral equals two right angles taken (4 2) times, i.e., equals 4 right angles; and if the angles are all equal, each angle is a right angle. In general, each angle of an equiangular polygon of n sides is o /~ 2^ equal to ^ * right angles. n POLYGONS. 69 PROPOSITION XLV. THEOREM. 207, The exterior angles of a polygon, made by pro ducing each of its sides in succession, are together equal to four right angles. \ Let the figure ABODE be a polygon, having its sides produced in succession, To prove the sum of the ext. A = 4 rt. A. Proof, Denote the int. A of the polygon by A, B, C, D, JS, and the ext. A by a, b, c, d, e. AA + Za = 2rt.A, 90 and Z B -f Z b = 2 rt. A, (being sup.-adj. A). In like manner each pair of adj. A = 2 rt. A. . . the sum of the interior and exterior A=2 rt. A taken as many times as the figure has sides, or, 2 n rt. A. But the interior A = 2 rt. A taken as many times as the figure has sides less two, = (n 2) 2 rt. A, or, 2 n rt. A 4 rt. A. .*. the exterior A = 4 rt. A. Q.E.D. 70 PLANE GEOMETRY. BOOK I. PROPOSITION XL VI. THEOREM. 208, A quadrilateral which has two adjacent sides equal, and the other two sides equal, is symmetrical with respect to the diagonal joining the vertices of the angles formed by the equal sides, and the diago nals intersect at right angles. Let ABCD be a quadrilateral, having AB = AD, and CB = CD, and having the diagonals AC and BD. To prove that the diagonal AC is an aoris of symmetry, and is J_ to the diagonal BD. Proof, In the A ABC ^d, ADC AB = AD, and B0= DC, Hyp. and AC=AO. Hen. . .AABC^AADC, 160 (having three sides of the one equal to three sides of the other). .-. Z BAQ= Z DAO, and Z EGA = /. DCA, (homologous A of equal A). Hence, if ABC is turned on AC as an axis, AB will fall upon AD, CB on CD, and OB on OD. Hence Ada an axis of symmetry, 65, and is J_ to BD^ ^ POLYGONS. 71 PROPOSITION XLVII. THEOREM. 209, // a figure is symmetrical with respect to two axes perpendicular to each other, it is symmetrical with respect to their intersection as a centre. r D E Let the figure ABCDEFGH be symmetrical with respect to the two axes XX , YY , which intersect at right angles at 0. To prove the centre of symmetry of the figure. Proof. Let N be any point in the perimeter of the figure. Draw NMIL to YY and IKL to XX . Join LO, ON, and KM. Now KI= KL, 61 (the figure being symmetrical with respect to XX ). But KI= OM, 180 (Us comprehended between Us are equal). .-. KL = OM, and KLOMia a O, 182 (having two sides equal and parallel). . . LO is equal a,nd parallel to KM. 179 In like manner we may prove ON equal and parallel to KM. Hence the points L, 0, and JVare in the same straight line drawn through the point II to KM; and LO=ON, since each is equal to KM. . . any straight line LON, drawn through 0, is bisected at 0. . . is the centre of symmetry of the figure. 64 Q. E. D. 72 PLANE GEOMETRY. BOOK I. EXERCISES. 34. The median from the vertex to the base of an isosceles triangle is perpendicular to the base, and bisects the vertical angle. 35. State and prove the converse. 36. The bisector of an exterior angle of an isosceles triangle, formed by producing one of the legs through the vertex, is parallel to the base. 37. State and prove the converse. 38. The altitudes upon the legs of an isosceles triangle are equal. 39. State and prove the converse. 40. The medians drawn to the legs of an isosceles triangle are equal. 41. State and prove the converse. (See Ex. 33.) 42. The bisectors of the base angles of an isosceles triangle are equal. 43. State the converse and the opposite theorems. 44? The perpendiculars dropped from the middle point of the base of an isosceles triangle upon the legs are equal. 45. State and prove the converse. /- 46. If one of the legs of an isosceles triangle is produced through the vertex by its own length, the line joining the end of the leg produced to the nearer end of the base is perpendicular to the base. 47. Show that the sum of the interior angles of a hexagon is equal to eight right angles. 48. Show that each angle of an equiangular pentagon is f of a right angle. 49. How many sides has an equiangular polygon, four of whose angles are together equal to seven right angles ? 50. How many sides has a polygon, the sum of whose interior angles is equal to the sum of its exterior angles ? 51. How many sides has a polygon, the sum of whose interior angles is double that of its exterior angles ? 52. How many sides has a polygon, the sum of whose exterior angles is double that of its interior angles? EXERCISES. 73 53. BAG is a triangle having the angle B double the angle A. If BD bisect the angle , and meet AQ in D, show that BD is equal to AD. 54. If from any point in the base of an isosceles triangle parallels to the legs are drawn, show that a parallelogram is formed whose perimeter is constant, and equal to the sum of the legs of the triangle. 55. The lines joining the middle points of the sides of a triangle divide the triangle into four equal triangles. 56. The lines joining the middle points of the side of a square, taken in order, enclose a square. 57. The lines joining the middle points of the sides of a rectangle, taken in order, enclose a rhombus. 58. The lines joining the middle points of the sides of a rhombus, taken in order, enclose a rectangle. 59. The lines joining the middle points of the sides of an isosceles trapezoid, taken in order, enclose a rhombus or a square. 60. The lines joining the middle points of the sides of any quadri lateral, taken in order, enclose a parallelogram. 61. The median of a trapezoid passes through the middle points of the two diagonals. 62. The line joining the middle points of the diagonals of a trapezoid is equal to half the difference of the bases. 63. In an isosceles trapezoid each base makes Q P equal angles with the legs. / \ \ HINT. Draw CE \\ DB. / \ \ 64. In an isosceles trapezoid the opposite angles ^ are supplementary. 65. If the angles at the base of a trapezoid are equal, the other angles are equal, and the trapezoid is isosceles. 66. The diagonals of an isosceles trapezoid are equal. 67. If the diagonals of a trapezoid are equal, the trapezoid is isosceles. HINT. Draw CE and DF _L to CD. Show that & ADF and BCE are equal, that & COD and AOB are isosceles, and that A AOC and BOD are equal. 74 PLANE GEOMETRY. BOOK I. 68. ABOD is a parallelogram, E and F the middle points of AD and BC respectively: show that BE sand DJ^will trisect the diagonal AC. 69. If from the diagonal BD of a square ABCD, BE is cut off equal to BO, and EF is drawn perpendicular to BD to meet DC at F, show that DE is equal to EF, and also to FC. 70. The bisector of the vertical angle A of a triangle ABC, and the bisectors of the exterior angles at the base formed by producing the sides AB and AC, meet in a point which is equidistant from the base and the sides produced. 71. If the two angles at the base of a triangle are bisected, and through the point of meeting of the bisectors a line is drawn parallel to the base, the length of this parallel between the sides is equal to the sum of the segments of the sides between the parallel and the base. 72. If one of the acute angles of a right triangle is double the other, the hypotenuse is double the shortest side. 73. The sum of the perpendiculars dropped from any point in the base of an isosceles triangle to the legs is constant, and equal to the altitude upon one of the legs. HINT. Let PD and PE be the two Js, BF the altitude upon AC. Draw PG to BF, and prove the A PBQ and PBD equal. .. p 74. The sum of the perpendiculars dropped from any point within an equilateral triangle to the three sides is constant, and equal to the altitude. HINT. Draw through the point a line II to the base, and apply Ex. 73. 75. What is the locus of all points equidistant from a pair of inter secting lines ? 76. In the triangle CAB the bisector of the angle C makes with the perpendicular from C to AB an angle equal to half the difference of the angles A and E. 77. If one angle of an isosceles triangle is equal to 60, the triangle is equilateral. BOOK II. THE CIRCLE. DEFINITIONS. 210, A circle is a portion of a plane bounded by a curved line called a circumference, all points of which are equally dis tant from a point within called the centre. 211, A radius is a straight line drawn from the centre to the circumference ; and a diameter is a straight line drawn through the centre, having its extremities in the circumference. By the definition of a circle, all its radii are equal. All its diameters are equal, since the diameter is equal to two radii. 212, A secant is a straight line which intersects the circum ference in two points ; as, AD, Fig. 1. 213, A tangent is a straight line which touches the circum ference but does not intersect it ; as, 0, Fig. 1. The point in which the tangent touches the circumference is called the point of contact, or point of tangency. 214, Two circumferences are tangent to each other when they are both tan- Fia l gent to a straight line at the same point; and are tangent internally or externally, according as one circumference lies wholly within or without the other. 76 PLANE GEOMETRY. BOOK II. 215, An arc of a circle is any portion of the circumference. An arc equal to one-half the circumference is called a semi- circumference. 216, A chord is a straight line having its extremities in the circumference. Every cliord subtends two arcs whose sum is the circum ference; thus, the chord AB (Fig. 3) subtends the smaller arc AB and the larger arc BCDEA. If a chord and its arc are spoken of, the less arc is meant unless it is otherwise stated. 217, A segment of a circle is a portion of a circle bounded by an arc and its chord. A segment equal to one-half the circle is called a semicircle. 218, A sector of a circle is a portion of the circle bounded by two radii and the arc which they intercept. A sector equal to one-fourth of the circle is called a quadrant. 219, A straight line is inscribed in a circle if it is a chord. 220, An angle is inscribed in a circle if its vertex is in the circumference and its sides are chords. 221, An angle is inscribed in a segment if its vertex is on the arc of the segment and its sides pass through the extrem ities of the arc. 222, A polygon is inscribed in a circle if its sides are chords of the circle. 223, A circle is inscribed in a polygon if the circumference touches the sides of the polygon but does not intersect them. ARCS AND CHORDS. 77 224, A polygon is circumscribed about a circle if all the sides of the polygon are tangents to the circle. 225, A circle is circumscribed about a polygon if the circum ference passes through all the vertices of the polygon. 226, Two circles are equal if they have equal radii ; for they will coincide if one is applied to the other; conversely, two equal circles have equal radii. Two circles are concentric if they have the same centre. PROPOSITION I. THEOREM. 227, The diameter of a circle is greater than any other chord; and bisects the circle and the circum ference. p Let AB be the diameter of the circle AMBP, and AE any other chord. To prove AB > AE, and AB bisects the circle and the circumference. Proof, I. From C, the centre of the O, draw OR CE^CB, (being radii of the same circle). But AC+CE>AE, 137 (the sum of two sides of a A is > t he third side). Then AC+ CB > AE, or AB > AE. Ax. 9 II. Fold over the segment A MB on AB as an axis until it falls upon APB, 59. The points A and B will remain fixed; therefore the arc AMB will coincide with the arc APB ; because all points in each are equally distant from the centre C. 210 Hence the two figures coincide throughout and are equal. 59 Q.E.O. 78 PLANE GEOMETRY. BOOK II. PROPOSITION II. THEOREM. 228, A straight line cannot intersect the circum ference of a circle in more than two points* LetHKbe any line cutting the circumference AMP. To prove that HK can intersect the circumference in only two points. Proof, If possible, let HK intersect the circumference in three points If, P, and K, From 0, the centre of the O, draw OH, OP, and OK. Then OH, OP, and OJiTare equal, (being radii of the same circle). Hence, we have three equal straight lines OH, OP, and OK drawn from the same point to a given straight line. But this is impossible, 120 (only two equal straight lines can be drawn from a point to a straight line). Therefore, HK can intersect the circumference in only two points. a E. a ARCS AND CHORDS. 79 PROPOSITION III. THEOREM. v^ 229, In the same circle, or equal circles, equal an gles at the centre intercept equal arcs; CONVERSELY, equal arcs subtend equal angles at the centre. p P In the equal circles ABP and A B P let Z O -= Z V. To prove arc BS = arc IPS . Proof, Apply O ABP to O A &P, so that Z. shall coincide with Z 0*. B will fall upon B , and 8 upon 8 , 226 (for OE = O R , and 08= O S , being radii of equal ). Then the arc BS will coincide with the arc JR /S , since all points in the arcs are equidistant from the centre. 210 . .arc BS=&rc B &. CONVERSELY : Let arc RS arc R S . To prove ZO = ZO . Proof, Apply O ABP to O A E F, so that arc BS shall fall upon arc -R S , B falling upon B , S upon S 9 , and upon . Then BO will coincide with B 0\ and SO with S O . , ,<0 and O 1 coincide and are equal. Q. e. a 80 PLANE GEOMETRY. BOOK II. PROPOSITION IV. THEOREM. 230. In the same circle, or equal circles, if two chords are equal, the arcs which they subtend are equal; CONVERSELY, if two arcs are equal, the chords which subtend them are equal. p p In. the equal circles ABP and A B P , let chord RS = chord R S . To prove arc 8 arc B 8 . Proof. Draw the radii OB, OS, 0>B t and In the A OBS and O JZ /S" B8=B 8 t Hyp. the radii OR and 08= the radii O B and 8 . 226 . .AB08=AB O>8 , 160 (three sides of the one being equal to three sides of the other). 229 (in equal , equal A at the centre intercept equal arcs). CONVEESELY : Let arc RS= arc R S 1 . To prove chord S = chord B 8 1 . Proof. Z = Z , 229 (equal arcs in equal subtend equal A at the centre), and OB and OS= O B and 8 , respectively. 226 . .AOBS=AO B 8 , 150 (having two sidos equal each to each and the included A equal). . . chord B8 = chord B 8 . a E . D. ARCS AND CHORDS. - 81 PROPOSITION V. THEOREM. 231, In the same circle, or equal circles, if two arcs are unequal, and each is less than a sewii-circumfer ence, the greater arc is subtended by the greater chord; CONVERSELY, the greater chord subtends the greater are. JC. In the circle whose centime is 0, let the arc AMB be greater than the arc AMF. To prove chord AB greater than chord AF. Proof. Draw the radii OA, OF, and OB. Since Fis between A and B, OF will fall between OA and OB, and Z AOB be greater than Z A OF. Hence, in the A AOB and AOF, the radii OA and OB = the radii OA and OF, but Z AOB is greater than Z AOF. /. AB > AF, 152 (the & having two sides equal each to each, but the included A unequal). CONVERSELY : Let AB be greater than AF. To prove arc AB greater than arc AF. In the A AOB and AOF, OA and OB= OA and OF respectively. But AB is greater than AF. Hyp- /. Z AOB is greater than Z AOF, ^- 153 (the A having two sides equal each to each, but the third sides unequal}. . . OB falls without OF. . . arc AB is greater than arc AF. Q.E.D. 82 PLANE GEOMETRY. BOOK II. PROPOSITION VI. THEOREM. 232, The radius perpendicular to a chord bisects the chord and the arc subtended by it. E M * ^_ S Let AB be the chord, and let the radius OS be per pendicular to AB at M. To prove AM= BM, and arc AS = arc BS. Proof, Draw OA and OB from 0, the centre of the circle. In the rt. A OA M and OB M the radius OA = the radius OB, and OM = OM. Hen. .\AOAM=AOBM, 161 (having the hypotenuse and a side of one equal to the hypotenuse and a side of the other). /. AM= BM, . .*,TGAS=&Tc8, 229 (equal A at the centre intercept equal arcs on the circumference). Q.E.D. 233, COR. 1. The perpendicular erected at the middle of a chord passes through the centre of the circle. For the centre is equidistant from the extremities of a chord, and is therefore in the perpendicular erected at the middle of the chord. 122 234, COR. 2. The perpendicular erected at the middle of a chord bisects the arcs of the chord. 235, COR. 3. The locus of the middle points of a system of parallel chords is the diameter perpendicular to them. ARCS AND CHORDS. 83 PROPOSITION VII. THEOREM. 236, In the same circle, or equal circles, equal chords are equally dislant from the centre ; AND CONVERSELY. B Let AB and OF be equal chords of the circle ABFC. To prove A 13 and CF equidistant from the centre 0. Proof, Draw OP to AB, OH. to OF, and join OA and OC. OP and OH bisect AB and CF, 232 (a radius _l_ to a chord bisects it). Hence, in the rt. A OP A and OHO AP=CH, Ax. 7 the radius OA = the radius 00. .-. A OP A - A OHC, 161 (having a side and hypotenuse of the one equal to a side and hypotenuse of the other). . . OP = OH. .*. AB and OF are equidistant from 0. COJTVEBSELY : Let OP = OH. To prove AB = CF. Proof, In the rt. A OP A and OHO the radius OA = the radius 0(7, and OP= OH, Hyp. /. A OP A and OJffC are equal. 161 /. AP= OH. CF. Ax C. 84 PLANE GEOMETRY. BOOK II. PROPOSITION VIII. THEOREM. 237, In the same circle, or equal circles, if two chords are unequal, they are unequally distant from the. centre, and the greater is at the less distance. In the circle whose centre is 0, let the chords AB and CD be unequal, and AB the greater; and let OE and OF be perpendicular to AB and CD respectively. To prove OE < OF. Proof, Suppose AO drawn equal to CD, and OR JL to AG. Then OH= OF, 236 (in the same O two equal chords are equidistant from the centre). Join Elf. OE and OH bisect AB and AG, respectively, 232 (a radius A. to a chord bisects it). Since, by hypothesis, AB is greater than CD or its equal A G, AE, the half of AB, is greater than AH, the half of AG. . . the Z AHE is greater than the Z AEH, 158 (the greater of two sides of a A has the greater Z opposite to it). Therefore, the Z ORE, the complement of the Z AHE, is less than the Z OEH, the complement of the Z AEH. . . OE < OH, 159 (the greater of two A of a A has the greater side opposite to it). . .OE< OF, the equal of OH. Q.E.Q ARCS AND CHORDS. 85 PROPOSITION IX. THEOREM. 238, CONVERSELY : In the same circle, or equal cir cles, if two chords are unequally distant from the centre, they are unequal, and the chord at the less distance is the greater. In the circle whose centre is 0, let AB and CD be unequally distant from 0; and let OE perpendicular to AB be less than OF perpendicular to CD. To prove AB > CD. Proof, Suppose AG drawn equal to CD, and OH _L to AG. Then OH= OF, 236 (in the sameQ two equal chords are equidistant from the centre). Hence, OE < OH. Join EH. In the A OEHfhe Z OHE is less than the Z OEH, 158 (the greater of two sides of a A has the greater Z. opposite to it). Therefore, the Z A HE, the complement of the Z OHE, is greater than the Z A EH, the complement of the Z OEH . . AE > AH, 159 (the greater of two A of a A has the greater side opposite to it). But AE=\AB, and AH=%AQ. .\AB>AG; hence AB > CD, the equal of AG. Q. E. D. 86 PLANE GEOMETRY. BOOK II. PROPOSITION X. THEOREM. 239, A. straight line perpendicular to a radius at its extremity is a tangent to the circle. M- H A Let MB be perpendicular to the radius OA at A. To prove MB tangent to the circle. Proof, From draw any other line to MB, as OCH. OH>OA, 114 (a JL is the shortest line from a point to a straight line). . . the point ZTis without the circle. Hence, every point, except A, of the line MB is without the circle, and therefore MB is a tangent to the circle at A. 213 Q. E. D. 240, COR. 1. A tangent to a circle is perpendicular to the radius drawn to the point of contact. For, if MB is tangent to the circle at A, every point of MB, except A, is without the circle. Hence, OA is the shortest line from to MB, and is therefore perpendicular to MB ( 114) ; that is, MB is per pendicular to OA. 241, COR. 2. A perpendicular to a tangent at the point of contact passes through the centre of the circle. For a radius is perpendicular to a tangent at the point of contact, and there fore, by 89, a perpendicular erected at the point of contact coincides with this radius and passes through the centre. 242, COR. 3. A perpendicular let fall from the centre of a circle upon a tangent to the circle passes through the point of contact. ARCS AND CHORDS. 87 PROPOSITION XI. THEOREM. 243, Parallels intercept equal arcs on a circum ference. F FIG. 1. FIG. 2. Let AB and CD be the two parallels. CASE I. When AB is a tangent, and CD a secant. Fig. 1. Suppose AB touches the circle at F. To prove arc OF arc DF. Proof. Suppose FF drawn J_ to AB. This J. to AB at F is a diameter of the circle. 241 It is also _L to CD. 102 .% arc CF= arc DF, 232 (a radius A. to a chord bisects the chord and its subtended arc). Also, arc FCF 1 = arc FDF , 227 . .a,Tc(FCF -FC) = a,rc(FDF -FI>), 82 that is, arc OF = arc DF . CASE II. When AB and CD are secants. Fig. 2. Suppose JEF drawn 11 to CD and tangent to the circle at M. Then arc AM = arc BM and arc CM = arc DM- Case I. /.by subtraction, arc AC = arc BD. CASE III. When AB and CD are tangents. Fig. 3. Suppose AB tangent at E, CD at F, and GH II to AB. Then arc OE = arc EH Case I. and arc OF = arc .-. by addition, arc EQF= arc ^.ZTF. a E . a 88 PLANE GEOMETRY. BOOK II. PROPOSITION XII. THEOREM. 244, Through three points not in a straight line, one circumference, and only one, can be drawn. Let A, B, C be three points not in a straight line. To prove that a circumference can be drawn through A, B, and (7, and only one. Proof, Join AB and BO. At the middle points of AB and BC suppose _I erected. Since BC is not the prolongation of AB, these J will inter sect in some point 0. The point 0, being in the J_ to AB at its middle point, is equidistant from A and B\ and being in the J_ to J5(7at its middle point, is equidistant from B and C, 122 (every point in the perpendicular -bisector of a straight line is equidistant from the extremities of the straight line). Therefore is equidistant from A, B, and C; and a cir cumference described from as a centre, with a radius OA, will pass through the three given points. Only one circumference can be made to pass through these points. For the centre of a circumference passing through the three points must be in both perpendiculars, and hence at their intersection. As two straight lines can inter sect in only one point, is the centre of the only circumfer ence that can pass through the three given points. Q . E . D . 245, COR. Two circumferences can intersect in only two points. For, if two circumferences have three points common, they coincide and form one circumference. TANGENTS, 89 PROPOSITION XIII. THEOREM. 246, The tangents to a circle drawn from an exte rior point are equal, and make equal angles with the line joining the point to the centre. B. c Let AB and AC be tangents from A to the circle whose centre is 0, and AO the line joining A to 0. To prove AB = AC, and Z BAO = Z CAO. Proof, Draw OB and 00. AB is _L to OB, and AC _L to OC, 240 (a tangent to a circle is _L to the radius drawn to the point of contact). In the rt. A OAB and OAC OB=OC, (radii of the same circle). OA = OA. Iden. .-.A OAB = A OAC t 161 (having a side and hypotenuse of the one equal to a side and hypotenuse of thz other). and Z BAO = Z CAO. Q. E. D. 247, DEF. The line joining the centres of two -circles is called the line of centres. 248, DEF. A common tangent to two circles is called a common exterior tangent when it does not cut the line of cen tres, and a common interior tangent when it cuts the line of centres. 90 PLANE GEOMETRY. BOOK II. PROPOSITION XIV. THEOREM. 249, If two circumferences intersect each other, the line of centres is perpendicular to their common chord at its middle point. Let C and C be the centres of two circumferences which intersect at A and B. Let AB be their common chord, and CC join their centres. To prove CO _L to AB at its middle point. Proof. A _L drawn through the middle of the chord AB passes through the centres C and (7 f , 233 (a _L erected at the middle of a chord passes through the centre of the O). /. the line (7(7 , having two points in common with this _L, must coincide with it. /. CO is _L to AB at its middle point. Q.E. D. Ex. 78. Describe the relative position of two circles if the line of centres : \ (i.) is greater than ^he sum of the radii ; (ii.) is equal to the sum of the radii ; (iii.) is less than the sum but greater than the difference of the radii ; (iv.) is equal to the difference of the radii ; (v.) is less than the difference of the radii. Illustrate each case by a figure. TANGENTS. 91 PROPOSITION XV. THEOREM. 250, If two circumferences cure tangent to each other, the line of centres passes through the point of contact. Let the two circumferences, whose centres are C and C , touch each other at 0, in the straight line AB, and let CO be the straight line joining their centres. To prove is in the straight line CO . Proof, A _L to AB, drawn through the point 0, passes through the centres C and C , 241 (a JL to a tangent at the point of contact passes through the centre of the circle). . . the line CO , having two points in common with this _L must coincide with it. .*. is in the straight line CO 1 . Q. E. D. Ex. 79. The line joining the centre of a circle to the middle of a chord is perpendicular to the chord. Ex. 80. The tangents drawn through the extremities of a diameter are parallel. Ex. 81. The perimeter of an inscribed equilateral triangle is equal to half the perimeter of the circumscribed equilateral triangle. Ex. 82. The sum of two opposite sides of a circumscribed quadri lateral is equal to the sum of the other two sides. 92 PLANE GEOMETRY. BOOK II. MEASUREMENT. 251, To measure a quantity of any kind is to find how many times it contains another known quantity of the same kind. Thus, to measure a line is to find how many times it con tains another known line, called the linear unit. The number which expresses how many times a quantity contains the unit-quantity, is called the numerical measure of that quantity ; as, 5 in 5 yards. 252, The magnitude of a quantity is always relative to the magnitude of another quantity of the same kind. No quantity is great or small except by comparison. This relative magni tude is called their ratio, and is expressed by the indicated quotient of their numerical measures when the same unit of measure is applied to both. The ratio of a to b is written -, or a : b. b 253, Two quantities that can be expressed in integers in terms of a common unit are said to be commensurable. The common unit is called a common measure, and each quantity is called a multiple of this common measure. Thus, a common measure of 2J feet and 3f feet is -J- of a foot, which is contained 15 times in 2-J- feet, and 22 times in 3| feet. Hence, 2^ feet and 3f feet are multiples of -J- of a foot, 2J feet being obtained by taking % of a foot 15 times, and 3f feet by taking % of a foot 22 times. 254, When two quantities are incommensurable, that is, have no common unit in terms of which both quantities can be expressed in integers, it is impossible to find a fraction that will indicate the exact value of the ratio of the given quanti ties. It is possible, however, by taking the unit sufficiently small, to find a fraction that shall differ from the true value of the ratio by as little as we please. RATIO. 93 Thus, suppose a and b to denote two lines, such that a - a .- Now V5= 1.41421356 ..... , a value greater than 1.414213, but less than 1.414214. If, then, a millionth part of b be taken as the unit, the value of the ratio lies between and and there- fore differs from either of these fractions by less than By carrying the decimal farther, a fraction may be found that will differ from the true value of the ratio by less than a billionth, a trillionth, or any other assigned value whatever. Expressed generally, when a and b are incommensurable, and b is divided into any integral number (ri) of equal parts, if one of these parts is contained in a more than m times, but less than m -f 1 times, then n that is, the value of % lies between and w + . b n n The error, therefore, in taking either of these values for - is less than -. But by increasing n indefinitely. - can be b n n made to decrease indefinitely, and to become less than any assigned value, however small, though it cannot be made absolutely equal to zero. Hence, the ratio of two incommensurable quantities cannot be expressed exactly by figures, but it may be expressed ap proximately within any assigned measure of precision. 255, The ratio of two incommensurable quantities is called an incommensurable ratio ; and is a fixed value toward which its successive approximate values constantly tend. 94 PLANE GEOMETRY. BOOK II. 256. THEOKEM. Two incommensurable ratios are equal if, when the unit of measure is indefinitely diminished, their ap proximate values constantly remain equal. Let a : b and a : b be two incommensurable ratios whose true values lie between the approximate values and n n when the unit of measure is indefinitely diminished. Then thev cannot differ so much as - n Now the difference (if any) between the fixed values a : b and a : b , is a fixed value. Let d denote this difference. Then d<-. n But if d has any value, however small, -, which by hypoth- n esis can be indefinitely diminished, can be made less than d. Therefore d cannot have any value; that is, d 0, and there is no difference between the ratios a : b and a : b ; there fore a : b = a 1 : b 1 . THE THEORY OF LIMITS. 257, When a quantity is regarded as having a/zec? value throughout the same discussion, it is called a constant; but when it is regarded, under the conditions imposed upon it, as having different successive values, it is called a variable. When it can be shown that the value of a variable, measured at a series of definite intervals, can by continuing the series be made to differ from a given constant by less than any assigned quantity, however small, but cannot be made abso lutely equal to the constant, that constant is called the limit of the variable, and the variable is said to approach indefi nitely to its limit. If the variable is increasing, its limit is called a superior limit ; if decreasing, an inferior limit. THEORY OF LIMITS. 95 Suppose a point to move from A toward B, under the con ditions that the first ^ # tr M? B second it shall move one-half the distance from A to B, that is, to M; the next second, one-half the remaining distance, that is, to M ; the next second, one-half the remaining distance, that is, to M" ; and so on indefinitely. Then it is evident that the moving point may approach as near to as we please, but will never arrive at B. For, how ever near it may be to B at any instant, the next second it will pass over one-half the interval still remaining ; it must, therefore, approach nearer to B, since half the interval still remaining is some distance, but will not reach B, since half the interval still remaining is not the whole distance. Hence, the distance from A to the moving point is an in creasing variable, which indefinitely approaches the constant AB as its limit ; and the distance from the moving point to B is a decreasing variable, which indefinitely approaches the constant zero as its limit. If the length of AB is two inches, and the variable is denoted by x, and the difference between the variable and its limit, by v : after one second, x = 1, v = 1 ; after two seconds, a? = 1 + -J, v = ^\ after three seconds, a? = 1 -f -J- -f -J-, ^ = ^ ; after four seconds, #=l-}-^-f-^-|--J-, v = %\ and so on indefinitely. Now the sum of the series 1 + J + + -j-, etc., is less than 2 ; but by taking a great number of terms, the sum can be made to differ from 2 by as little as we please. Hence 2 is the limit of the sum of the series, when the number of the terms is increased indefinitely ; and is the limit of the dif ference between this variable sum and 2. 96 PLANE GEOMETRY. BOOK II. Consider the repetend 0.33333 , which may be written TG + dhr + nflnr + ydhnr + However great the number of terms of this series we take, the sum of these terms will be less than ; but the more terms we take the nearer does the sum approach . Hence the sum of the series, as the number of terms is increased, approaches indefinitely the constant as a limit. 258. In the right triangle ACE, if the vertex A approaches indefinitely the base BO, the angle B diminishes, and approaches zero indefi nitely ; if the vertex A moves away from the base indefinitely, the angle B increases and approaches a right angle indefinitely ; but B cannot become zero or a right angle, so long as A OB is a triangle ; for if B be comes zero, the triangle becomes the straight line BC t and if B becomes a right angle, the triangle becomes two parallel lines AC&ud AB perpendicular to BO. Hence the value of B must lie between and 90 as limits. 259. Again, suppose a square A BOD inscribed in a circle, and E, F, H, K the middle points of the arcs subtended by the sides of the square. If we draw the straight lines AE, EB, BF, etc., we shall have an inscribed polygon of double the number of sides of the square. K[ The length of the perimeter of this polygon, represented by the dotted lines, is greater than that of the square, since two sides replace each side of the square and form with it a triangle, and two sides of a triangle are together greater than the third side; but less than the length of the circumference, for it is made up of THEORY OF LIMITS. 97 straight lines, each one of which is less than the part of the circumference between its extremities. By continually repeating the process of doubling the num ber of sides of each resulting inscribed figure, the length of the perimeter will increase with the increase of the number of sides ; but it cannot become equal to the length of the cir cumference, for the perimeter will continue to be made up of straight lines, each one of which is less than the part of the circumference between its extremities. The length of the circumference is therefore the limit of the length of the perimeter as the number of sides of the inscribed figure is indefinitely increased. 260, THEOREM. If two variables are constantly equal and each approaches a limit, their limits are equal. A: ________^^ N ~~ ~C Let AM and AN be two variables which are con stantly equal and which approach indefinitely AB and AC respectively as limits. To prove AB = AC. Proof, If possible, suppose AB > AC, and take AD = AC. Then the variable AMm&y assume values between AD and AB, while t*he variable AN must always be less than AD. But this is contrary to the hypothesis that the variables should continue equal. .*. AB cannot be > AC. In the same way it may be proved that AC cannot be >AB. . . AB and AC are two values neither of which is greater than the other. Hence AB = AC. 98 PLANE GEOMETRY. BOOK II. MEASURE OF ANGLES. PROPOSITION XVI. THEOREM. 261. In the same circle, or equal circles, two angles at the centre have the same ratio as their intercepted arcs. CASE I. When the arcs are commensurable. In the circles whose centres are G and D, let AC B and EDF be the angles, AB and EF the intercepted arcs. Z AOB = arc AB ~ arc EF To prove Z.EDF Proof. Let m be a common measure of AB and EF. Suppose m to be contained in AB seven times, and in EF four times. arc AB __1 arc EF 4 Then (1) At the several points of division on AB and EF draw radii. These radii will divide /. ACB into seven parts, and . EDF into four parts, equal each to each, 229 (in the same O, or equal <D, equal arcs subtend equal 4 at the centre). (2) EDF 4 From (1) and (2), EDF arc EF MEASURE OF ANGLES. 99 CASE II. When the arcs are incommensurable. p P In the equal circles ABP and A B P* let the angles ACB and A C S 1 intercept the incommensurable arcs AB and A B . Z ACB arc AB To prove ______ Proof, Divide AB into any number of equal parts, and apply one of these parts as a unit of measure to A B as many times as it will be contained in A B 1 . Since AB and AB are incommensurable, a certain number of these parts will extend from A 1 to some point, as D, leav ing a remainder DB less than one of these parts. Draw C D. Since AB and A D are commensurable, Z ACB arc AB A C D arc A D Case I. If the unit of measure is indefinitely diminished, these ratios continue equal, and approach indefinitely the limiting ratios Z AGE and arc AB Z. A C B arc A B ^,-^fr . 1260 (If two variables are constantly equal, and each approaches a limit, their limits are equal.) Q.I.Q. 100 PLANE GEOMETRY. BOOK II. 262, The circumference, like the angular magnitude about a point, is divided into 360 equal parts, called degrees. The arc-degree is subdivided into 60 equal parts, called minutes ; and the minute into 60 equal parts, called seconds. Since an angle at the centre has the same number of angle- degrees, minutes, and seconds as the intercepted arc has of arc- degrees, minutes, and seconds, we say : An angle at the centre is measured by its intercepted arc ; meaning, An angle at the centre is such a part of the whole angular magnitude about the centre as its intercepted arc is of the whole circumference. PROPOSITION XVII. THEOREM. 263, An inscribed angle is measured by one-half the arc intercepted between its sides. CASE I. When one side of the angle is a diameter. In the circle PAB (Fig. 1), let the centre C be in one of the sides of the inscribed angle B. To prove /. B is measured by J arc PA. Proof. Draw CA. Eadius CA = radius CB. .\Z.B = A, 154 (being opposite equal sides of the A CAB). PCA = Z + Z.A, .145 equal to the sum of the two opposite interior A). But (the exterior Z of a A But Z PC A is measured by PA, (the at the centre is measured by the intercepted arc). . ./. B is measured by PA. 262 MEASURE OF AJKiLIJSI ; >, [ \i \>>\ J.Q1 CASE II. When the centre is within the angle. In the circle BAE (Fig. 2), let the centre C fall within the angle EBA. To prove Z EBA is measured by -J arc EA. Proof, Draw the diameter EG P. Z PEA is measured by \ arc PA, Case I. Z PBE is measured by arc PE, Case I. /. Z PEA + Z PEE is measured by \ (arc PA + arc PE), or Z EBA is measured by -| arc -ZL4. CASE III. When the centre is without the angle. In the circle BFP (Fig. 3), let the centre C fall without the angle ABF. To prove Z ABF is measured by -J arc AF. Proof, Draw the diameter BOP. Z PEF is measured by \ arc PF, Case I. Z PEA is measured by \ arc PA. Case I. , ./. PEF-/. PEA is measured by (arc PF~ arc P^4), or Z AEF is measured by -J- arc AF. Q.E.D. B FIG. 1. FIG. 2. FIG. 3. 264, COR. 1. An angle inscribed in a semicircle is a right angle. For it is measured by one-half a semi-circumference. 265, COR. 2. An angle inscribed in a segment greater than a semicircle is an acute angle. For it is measured by an arc less than half a semi-circumference ; as, Z CAD. Fig. 2. 266, COR. 3. An angle inscribed in a segment less than a semicircle is an obtuse angle. For it is measured by an arc greater than half a semi-circumference ; as, Z GBD. Fig. 2. 267, COR. 4. All angles inscribed in the sam.e segment are equal. For they are measured by half the same arc. Fig. 3. 102 *LANE GfEQMETRY. BOOK II. PROPOSITION XVIII. THEOREM. 268, An angle formed ~by two chords intersecting within the circumference is measured by one-half the sum of the intercepted arcs. Let the angle AOC be formed by the chords AB and CD. To prove Z AOC is measured by %(AC+ BD). Proof, Draw AD. Z.COA = /.D + ^A, 145 (the exterior Z of a A is equal to the sum of the two opposite interior A}. But Z D is measured by % arc AC, 263 and Z A is measured by % arc BD, (an inscribed Z. is measured by % the intercepted arc). ,-.Z CO A is measured by \ (AC+ BD). Q.E.D. Ex. 83. The opposite angles of an inscribed quadrilateral are sup plements of each other. Ex. 84. If through a point within a circle two perpendicular chords are drawn, the sum of the opposite arcs which they intercept is equal to a semi-circumference. Ex. 85. The line joining the centre of the square described upon the hypotenuse of a rt. A, to the vertex of the rt. Z, bisects the right angle. HINT. Describe a circle upon the hypotenuse as diameter. MEASURE OF ANGLES. 103 PROPOSITION XIX. THEOREM. 269, An angle formed by a tangent and a chord is measured by one-half the intercepted arc. ill- MAH fee the angle formed by the tangent MO and chord AH. To prove Z MAH is measured by J arc A EH. Proof. Draw the diameter ACF. Z MAFiszrt. Z, 240 (the radius drawn to a tangent at the point of contact is JL to it). Z MAF being a rt. Z, is measured by the semi-circum ference AEF. But Z HAF is measured by J arc HF, 263 (an inscribed Z -is measured by % the intercepted arc). .% Z MAF-/. HAF is measured by (AEF- HF) ; or Z MAH is measured by A EH. Q. E. D. Ex. 86. If two circles touch each other and two secants are drawn through the point of contact, the chords joining their extremities are parallel. HINT. Draw the common tangent. 104 PLANE GEOMETRY. BOOK II. PROPOSITION XX. THEOREM. 270, An angle formed by two secants, two tangents, or a tangent and a secant, intersecting without the circumference, is measured by one-half the difference of the intercepted arcs. FIG. 1. CASE I. Angle formed by two secants. Let the angle (Fig. 1) be formed by the two se cants OA and OB. To prove Z is measured by ( AB EC). Proof, Draw CB. ^ACB = Z.O + Z.B, 145 (the exterior Z of a A is equal to the sum of the two opposite interior A). By taking away Z B from both sides, But and Z AGE is measured by \ AB, Z B is measured by \ OE, (an inscribed Z. is measured by $ the intercepted arc). .-. Z is measured by %(AB CE\ 263 MEASURE OF ANGLES. 105 CASE II. Angle formed by two tangents. Let the angle (Fig. 2) be formed by the two tan gents OA and OB. To prove Z is measured by -| (AMJ3 ASH). Proof. Draw AB. OAB, 145 (the exterior Z of a A is equal to the sum of the two opposite interior A}. By taking away Z OAB from both sides, But Z ABO is measured by -J- AMB, 269 and Z OAB is measured by % ASB, (an Z formed by a tangent and a chord is measured by I the intercepted arc}. . . Z is measured by 1 (A MB ASB). CASE III. Angle formed by a tangent and a secant. Let the angle (Fig. 3) be formed by the tangent OB and the secant OA. To prove Z is measured by % (ADS CES). Proof. Draw OS. ZACS=ZO + ZCSO, 145 (the exterior Z of a A is equal to the sum of the two opposite interior A}. By taking away Z GSO from both sides, But Z ACS is measured by | ADS, 263 (being an inscribed Z), and Z CSO is measured by $CES, 269 (being an /.formed by a tangent and a chord). . . Z is measured by %(ADS CES). Q.E.D. 106 PLANE GEOMETRY. BOOK II. PROBLEMS OF CONSTRUCTION. PROPOSITION XXL PROBLEM. 271. At a given point in a straight line, to erect a perpendicular to that line. A HOB JS\ ________ ,-- B FIG. 1. FIG. 2. I. Let be the given point in AC. (Fig. 1). To erect a A. to the line AC at the point 0. Construction, From as a centre, with any radius OB, describe an arc intersecting AC in two points 77 and B. From .ZTand B as centres, with equal radii greater than OB, describe two arcs intersecting at R. Join OR. Then the line OR is the _L required. Proof. Since and R are two points at equal distances from 7?" and B, they determine the position of a perpendicular to the line HE at its middle point 0. 123 II. When the given point is at the end of the line. Let B be the given point. (Fig. 2). To erect a J_ to the line AB at B. Construction, Take any point O without AB ; and from C as a centre, with the distance CB as a radius, describe an arc intersecting AB at E. Draw EC, and prolong it to meet the arc again at D. Join BD, and BD is the J_ required. Proof. The Z. B is inscribed in a semicircle, and is therefore a right angle. 264 Hence BD is _L to AB. o. E . F . PROBLEMS. 107 PKOPOSITION XXII. PROBLEM. 272, From a point without a straight line, to let fail a perpendicular upon that line. H --. M ,., K / B Let AB be the given straight line, and C the given point without the line. To let fall a _L to the line AB from the point Q. Construction, From C as a centre, with a radius sufficiently great, describe an arc cutting AB in two points, .5" and K. From .ZTand K as centres, with equal radii greater than \HK, describe two arcs intersecting at 0. Draw CO, and produce it to meet A B at M. CM is the _L required. Proof, Since C and are two points equidistant from J?"and K) they determine a _L to HK at its middle point. 123 ______ Q. E. F. NOTE. Given lines of the figures are full lines, resulting lines are long-dotted, and auxiliary lines are short-dotted. 108 PLANE GEOMETRY. BOOK II. PROPOSITION XXIII. PROBLEM. 273, To bisect a given straight line. Let AB be the given straight line. To bisect the line AB. Construction, From A and B as centres, with equal radii greater than AB, describe arcs intersecting at C and E. Join GE. Then the line GE bisects AB. Proof, G and E are two points equidistant from A and B. Hence they determine a J_ to the middle point of AB. 123 Q.E. F. Ex. 87. To find in a given line a point X which shall be equidis tant from two given points. Ex. 88. To find a point X which shall be equidistant from two given points and at a given distance from a third given point. Ex. 89. To find a point X which shall be at given distances from two given points. Ex. 90. To find a point X which shall be equidistant from three given points. PKOBLEMS. 109 PROPOSITION XXIV. PROBLEM. 274, To bisect a given arc. Let ACS be the given arc. To bisect the arc ACB. Construction, Draw the chord AB. From A and B as centres, with equal radii greater than AB, describe arcs intersecting at D and E, Draw DE. DE bisects the arc ACB. Proof, Since D and E are two points equidistant from A and B, they determine a _L erected at the middle of chord AB. 123 And a _L erected at the middle of a chord passes through the centre of the O, and bisects the arc of the chord. 234 Q.E.F. Ex. 91. To construct a circle having a given radius and passing through two given points. Ex. 92. To construct a circle having its centre in a given line and passing through two given points. 110 PLANE GEOMETRY. BOOK II. PKOPOSITION XXV. PROBLEM. 275, To bisect a given angle. Let AEB be the given angle. To Used Z AEB. Construction. From E as a centre, with any radius, as EA, describe an arc cutting the sides of the Z E at A and B. From A and B as centres, with equal radii greater than one-half the distance from A to B, describe two arcs inter secting at C. Join EC, AC, and EG. EC bisects the Z E. Proof, In the A AEC an& BEG AE = BE, and AG= BC, Cons. and EC = EC. Men. .-.A AEC^&BEC, 160 (having three sides equal each to each). .-. Z AEG =Z. BEG. Q. E. F. Ex. 93. To divide a right angle into three equal parts. Ex. 94. To construct an equilateral triangle, having given one side. Ex. 95. To find a point X which shall be equidistant from two given points and also equidistant from two given intersecting lines. PROBLEMS. Ill PROPOSITION XXVI. PROBLEM. 276, At a given point in a given straight line, to construct an angle equal to a given angle. Let C be the given point in the given line CM, and A the given angle. To construct an /. at equal to the Z. A. Construction, From A as a centre, with any radius, as AE, describe an arc cutting the sides of the Z A at E and F. From O as a centre, with a radius equal to AE, describe an arc cutting CM at H. From H as a centre, with a radius equal to the distance EF, describe an arc intersecting the arc HG at m. Draw Cm, and HCm is the required angle. Proof, The chords jEFand Hm are equal. Cons. /. arc EF= arc Hm, 230 (in equal equal chords subtend equal arcs). . .ZC=ZA, 229 (in equal equal arcs subtend equal A at the centre}. Q. E. F. Ex. 96. In a triangle ABC, draw DE parallel to the base BC, cut ting the sides of the triangle in D and E t so that DE shall equal DB + EC. Ex. 97. If an interior point of a triangle ABC is joined to the ver tices B and C, the angle BOG is greater than the angle BAG of the triangle. 112 PLANE GEOMETRY. BOOK II. PROPOSITION XXVII. PROBLEM. 277, Two angles of a triangle being given, to find the third angle. E i F Let A and B be the two given angles of a triangle. To find the third Z of the A. Construction, Take any straight line, as JEF, and at any point, as If, construct Z a equal to Z A, 276 and Z b equal to Z B. Then Z. c is the Z required. Proof. Since the sum of the three A of a A = 2 rt. A, 138 and the sum of the three A a, b, and c, 2 rt. A\ 92 and since two A of the A are equal to the A a and b, the third Z of the A will be equal to the Z c. Ax. 3. Q. E. F. Ex. 98. In a triangle ABC, given angles A and B, equal respectively to 37 13 32" and 41 17 56". Find the value of angle C. PROBLEMS. H3 PROPOSITION XXVIII. PROBLEM. 278, Through a given point, to draw a straight line parallel to a given straight line. n D Let AB be the given line, and C the given point. To draw through the point a line parallel to the line AB. Construction, Draw DOE, making the Z EDB. At the point C construct Z. ECF= Z EDB. 276 Then the line FCHis \\ to AB. Proof. Z ECF= Z EDB. Cons. .\HFia II to A3, 108 (when two straight lines, lying in the same plane, are cut by a third straight line, if the ext.-int. A are equal, the lines are parallel). Q.E.F. Ex. 99. To find a point X equidistant from two given points and also equidistant from two given parallel lines. Ex. 100. To find a point X equidistant from two given intersecting lines and also equidistant from two given parallels. 114 PLANE GEOMETRY. BOOK II. PROPOSITION XXIX. PROBLEM. 279, To divide a given straight line into equal parts. -0 Let AB be the given straight line. To divide AB into equal parts. Construction, From A draw the line AO. Take any convenient length, and apply it to AO as many times as the line AB is to be divided into parts. From the last point thus found on AO, as (7, draw CB. Through the several points of division on AO draw lines II to CB, and these lines divide AB into equal parts. Proof. Since A C is divided into equal parts, AB is also, 187 (if three or more \\s intercept equal parts on any transversal, they intercept equal parts on every transversal). Q. E. F. Ex. 101. To divide a line into four equal parts by two different methods. Ex. 102. To find a point X in one side of a given triangle and equi distant from the other two sides. Ex. 103. Through a given point to draw a line which shall make equal angles with the two aides of a given angle. PROBLEMS. 115 PROPOSITION XXX. PROBLEM. 280, Two sides and the included angle of a trian gle being given, to construct the triangle. D 7) Let the two sides of the triangle be b and c, and the included angle A. To construct a A having two sides equal to b and c respec tively, and the included Z. = /. A. Construction, Take AB equal to the side c. At A, the extremity of AB, construct an angle equal to the given Z A. 276 On^Dtake A C equal to b. Draw CB. Then A AGE is the A required. Q. E. F. Ex. 104. To construct an angle of 45. Ex. 105. To find a point X which shall be equidistant from two given intersecting lines and at a given distance from a given point. Ex. 106. To draw through two sides of a triangle a line || to the third side so that the part intercepted between the sides shall have a given length. PLANE GEOMETRY. BOOK [I. PROPOSITION XXXI. PROBLEM. 281, A side and two angles of a triangle being given, to construct the triangle. Let c be the given side, A and B the given angles. To construct the triangle. Construction, Take EC equal to c. At the point E construct the Z.CEH equal to Z A. 276 At the point C construct the /. EQK equal to Z B. Let the sides EH and GK intersect at 0. Then A COE is the A required. Q. E. F. REMARK. If one of the given angles is opposite to the given side, find the third angle by g 277, and proceed as above. Discussion, The problem is impossible when the two given angles are together equal to or greater than two right angles. Ex. 107. To construct an angle of 150. Ex. 108. A straight railway passes two miles from a town. A place is four miles from the town and one mile from the railway. To find by construction how many places answer this description. Ex. 109. If in a circle two equal chords intersect, the segments of one chord are equal to the segments of the other, each to each. Ex. 110. AB is any chord and AC is tangent to a circle at A, CDE a line cutting the circumference in D and E and parallel to AB; show that the triangles ACD and EAB are mutually equiangular. PROBLEMS. 117 PROPOSITION XXXII. PROBLEM. 282, The three sides of a triangle being given> to construct the triangle. 7 \ A*-- ^B A o B Let the three sides be m, ?i, and o. To construct the triangle. Construction. Draw AB equal to o. From A as a centre, with a radius equal to n, describe an arc ; and from B as a centre, with a radius equal to m, describe an arc intersecting the former arc at C. Draw GA and GB. Then A GAB is the A required. Q.E.F. Discussion, The problem is impossible when one side is equal to or greater than the sum of the other two. Ex. 111. The base, the altitude, and an angle at the base, of a tri angle being given, to construct the triangle. Ex. 112. Show that the bisectors of the angles contained by the oppo site sides (produced) of an inscribed quadrilateral intersect at right angles. Ex. 113. Given two perpendiculars, AB and CD, intersecting in 0, and a straight line intersecting these perpendiculars in E and F\ to construct a square, one of whose angles shall coincide with one of the right angles at O, and the vertex of the opposite angle of the square shall lie in EF. (Two solutions.) 118 PLANE GEOMETRY. BOOK II. PROPOSITION XXXIII. PROBLEM. 283, Two sides of a triangle and the angle opposite one of them being given, to construct the triangle. CASE I. If the side opposite the given angle is less than the other given side. Let b be greater than a, and A the given angle. To construct the triangle. Construction. Construct Z DAE to the given Z A. 276 On AD take AB = b. From B as a centre, with a radius equal to a, describe an arc intersecting the line AE at Cand C . Draw BO and C . Then both the A ABC and ABC 1 ,D fulfil the conditions, and hence we .- JD / have two constructions. This is called the ambiguous case. * / \ a Discussion, If the side a is equal ^L. x --... i -- E to the JL BH, the arc described from B will touch AE, and there will be but one construction, the right tri- ,D angle ABH. B^- If the given side a is less than the ^/ \ a JL from B, the arc described from B / i-- will not intersect or touch AE, and -. E hence the problem is impossible, THE CIKCLE. 119 If the Z A is right or obtuse, the problem is impossible ; for the side opposite a right or obtuse angle is the greatest side. 159 CASE II. If a is equal to b. If the Z. A is acute, and a = b, the arc described from B as a centre, and with a radius equal to a, will cut the line AE at the points A and O. B ,P There is therefore but one solution : the i>/ Vx - isosceles A AEG. J[ C ~^c~ E Discussion, If the Z A is right or obtuse, the problem is impossible ; for equal sides of a A have equal A opposite them, and a A cannot have two right A or two obtuse A. CASE III. If a is greater than b. If the given Z. A is acute, the arc described from B will cut the line ED on opposite sides of A, at O and C 1 . The A ABO answers the required conditions, but the B / A ABC* does not, for it does not contain the acute Z. A. There is then only one E y^ solution ; namely, the A ABC. ^ , * If the Z A is right, the arc described from B cuts the line ED on opposite /?\ sides of A, and we have two equal right a / ^\ A which fulfil the required conditions. E ^ * ^ D C ~~-rA If the /. A is obtuse, the arc described from B cuts the line ED on opposite \ B sides of A, at the points C and <? . The <j/ V v<1 A ABC answers the required conditions, ^_ v -" ^ - but the AA.BC 1 does not, for it does c "~ ..... 4 " c not contain the obtuse Z A. There is then only one solu tion ; namely, the A A BO. Q.E. F. 120 PLANE GEOMETEY. BOOK II. PROPOSITION XXXIV. PROBLEM. 284, Two sides and an included angle of a paral lelogram being given, to construct the parallelogram. / / / / Let m and o be the two sides, and C the included angle. To construct a parallelogram. Construction, Draw AB equal to o. At A construct the Z. A equal to Z (7, 276 and take AH equal to m- From T&s a centre, with a radius equal to o, describe an arc. From B as a centre, with a radius equal to m, describe an arc, intersecting the former arc at E. Draw EH and EB. The quadrilateral ABEH\$> the O required. Proof. AB = HE, Cons. ^#= 5^7. Cons. /. the figure ABEHis a O, 183 (having its opposite sides equal). Q. E. F. PEOBLEMS. 121 PROPOSITION XXXV. PROBLEM. 285, To circumscribe a circle about a given tri angle. . Let ABC be the given triangle. . To circumscribe a circle about ABC, Construction, Bisect AB and BO. 273 At the points of bisection erect Js. 271 Since BO is not the prolongation of AB, these J will in tersect at some point 0. From 0, with a radius equal to OB, describe a circle. A B C is the required. Proof. The point is equidistant from A and B, . and also is equidistant from B and (7, 122 (every point in the _L erected at the middle of a, straight line is equidistant from the extremities of that line). . . the point is equidistant from A, B, and O, and a described from as a centre, with a radius equal to OB, will pass through the vertices A, B, and C. ae<F . 286, SCHOLIUM. The same construction serves to describe a circumference which shall pass through the three points not in the same straight line ; also to find the centre of a given circle or of a given arc. 122 PLANE GEOMETRY. BOOK II. PROPOSITION XXXVI. PROBLEM. 287, Through a given point, to draw a tangent to a given circle. >iJ CASE I. When the given point is on the circle. Let C be the given, point on the circle. To draw a tangent to the circle at C. Construction, From the centre draw the radius 00. Through C draw AM to 00. 271 Then AM is the tangent required. Proof, A straight line _!_ to a radius at its extremity is tan gent to the circle. 239 CASE II. When the given point is without the circle. Let be the centre of the given circle, E the given point without the circle. To draw a tangent to the given circle from the point E. Construction, Join OE. On OE as a diameter, describe a circumference intersecting the given circumference at the points M and H. Draw OM and EM. Then EM is the tangent required. Proof, Z OME is a right angle, 264 (being inscribed in a semicircle). . . EM is tangent to the circle at M. 239 In like manner, we may prove HE tangent to the given O. Q. E. F. PROBLEMS. 123 PROPOSITION XXXVII. PROBLEM. 288, To inscribe a circle in a given triangle. Let ABC be the given triangle. To inscribe a circle in the A ABC. Construction, Bisect A A and C. 275 From E, the intersection of these bisectors, draw EH J. to the line AC. 272 From E, with radius EH, describe the O KMH. The O KHMis the O required. Proof, Since E is in the bisector of the Z A, it is equidis tant from the sides AB and AC; and since E is in the bisector of the Z (7, it is equidistant from the sides AC and BC, 162 (every point in the bisector of an /. is equidistant from the sides of the Z). .-. a O described from E as centre, with a radius equal to EH will touch the sides of the A and be inscribed in it. SCHOLIUM. The intersec tions of the bisectors of exterior angles of a triangle, formed by- producing the sides of the tri angle, are the centres of three circles, each of which will touch one side of the triangle, and the two other sides produced. These three circles are called escribed circles. 124 PLANE GEOMETRY. BOOK II. PKOPOSITION XXXVIII. PROBLEM. 290, Upon a given straight line, to describe a seg ment of a circle which shall contain a given angle. Let AB be the given line, and M the given angle. To describe a segment upon AB which shall contain Z M. Construction, Con struct Z ABE equal to Z.M. 276 Bisect the line AB by the _L FO. 273 From the point B draw BO _i_ to EB. 271 From 0, the point of intersection of FO and BO, as a cen tre, with a radius equal to OB, describe a circumference. The segment AKB is the segment required. Proof, The point is equidistant from A and B, 122 (every point in a JL erected at the middle of a straight line is equidistant from the extremities of that line). . . the circumference will pass through A. But BE is -L to OB. Cons. /. BE is tangent to the O, 239 (a straight line to a radius at its extremity is tangent to the O). /. Z ABE is measured by arc AB, 269 (being an ^.formed by a tangent and a chord). An Z inscribed in the segment AKB is measured by 263 t *. segment AKB contains Z M. Ax. 1 0. E. F. PROBLEMS. 125 PROPOSITION XXXIX. PROBLEM. 291, To find the ratio of two commensurable straight lines. i % I I r F Let AB and CD "be two straight lines. To find the ratio of AB and CD. Apply CD to AB as many timesjas possible. Suppose twice, with a remainder EB. Then apply EB to CD as many times as possible. Suppose three times, with a remainder FD. Then apply FD to EB as many times as possible. Suppose once, with a remainder HB. Then apply HB to FD as many times as possible. Suppose once, with a remainder KD. Then apply KD to HB as many times as possible. Suppose KD is contained just twice in HB. The measure of each line, referred to KD as a unit, will then be as follows : HB = 2KD-, EB = FD+HB = 5KD- CD = 3 EB -f FD = 18 KD ; AB = 2 CD + EB = 41 CD 18 /.the ratio CD 18 Q.E.F. 126 PLANE GEOMETRY. BOOK II. THEOREMS. 114. The shortest line and the longest line which can be drawn from a given exterior point to a given circumference pass through the centre. 115. If through a point within a circle a diameter and a chord _L to the diameter are drawn, the chord is the shortest cord that can be drawn through the given point. 116. In the same circle, or in equal circles, if two arcs are each greater than a semi-circumference, the greater arc subtends the less chord, and conversely. 117. If ABC is an inscribed equilateral triangle, and Pis any point in the arc BC, then PA = PB + PC. HINT. On PA take PIT equal to PB, and join BM. "118. In what kinds of parallelograms can a circle be inscribed? Prove your answer. 119. The radius of the circle inscribed in an equilateral triangle is equal to one- third of the altitude of the triangle. 120. A circle can be circumscribed about a rectangle. 121. A circle can be circumscribed about an isosceles trapezoid. -- 122. The tangents drawn through the vertices of an inscribed rec tangle enclose a rhombus. -f^ 123. The diameter of the circle inscribed in a rt. A is equal to the difference between the sum of the legs and the hypotenuse. 124. From a point A without a circle, a straight line AOB is drawn through the centre, and also a secant ACD, so that the part A C without the circle is equal to the radius. Prove that Z. DAB equals one-third the Z DOB. 125. All chords of a circle which touch an interior concentric circle are equal, and are bisected at the points of contact. 126. If two circles intersect, and a secant is drawn through each point of intersection, the chords which join the extremities of the secants are parallel. HINT. By drawing the common chord, two inscribed quadrilaterals are obtained. 127. If an equilateral triangle is inscribed in a circle, the distance of each side from the centre of the circle is equal to half the radius. 128. Through one of the points of intersection of two circles a diameter of each circle is drawn. Prove that the straight line joining the ends of the diameters passes through the other point of intersection. EXERCISES. 127 129. A circle touches two sides of an angle BAG at B, C\ through any point D in the arc BC a tangent is drawn, meeting AB at J^and AQ at F. Prove (i.) that the perimeter of the triangle ^.EFis constant for all positions of D in BC\ (ii.) that the angle EOF\& also constant. Loci. 130. Find the locus of a point at three inches from a given point. 131. Find the locus of a point at a given distance from a given circumference. 132. Prove that the locus of the vertex of a right triangle, having a given hypotenuse as base, is the circumference described upon the given hypotenuse as diameter. 133. Prove that the locus of the vertex of a triangle, having a given r base and a given angle at the vertex, is the arc which forms with the base a segment capable of containing the given angle. 134. Find the locus of the middle points of all chords of a given length that can be drawn in a given circle. 135. Find the locus of the middle points of all chords that can be drawn through a given point A in a given circumference. jv 136. Find the locus of the middle points of all straight lines that ca.n t>e drawn from a given exterior point A to a given circumference. 137. A straight line moves so that it remains parallel to a given line, and touches at one end a given circumference. Find the locus of the other end. 1 138. A straight rod moves so that its ends constantly touch two fi&ed rods which are _L to each other. Find the locus of its middle point. 139. In a given circle let AOB be a diameter, 00 any radius, CD the perpendicular from C to AB. Upon 00 take OM=CD. Find the locus of the point J/as 00 turns about 0. CONSTRUCTION OF POLYGONS. To construct an equilateral A, having given : 140. The perimeter. 141. The radius of the circumscribed circle. 142. The altitude. 143. The radius of the inscribed circle. To construct an isosceles triangle, having given: 144. The angle at the vertex and the base. 128 PLANE GEOMETRY. BOOK II. 145. The angle at the vertex and the altitude. 146. The base and the radius of the circumscribed circle. 147. The base and the radius of the inscribed circle. 148. The perimeter and the alti tude. HINTS. Let ABC be the A re quired, and EF the given perimeter. The altitude CD passes through the ,. middle of EF, and the A AJSG, .- FO&Te isosceles. To construct a right triangle, having given : 149. The hypotenuse and one leg. 150. The hypotenuse and the altitude upon the hypotenuse. 151. One leg and the altitude upon the hypotenuse as base. l|| %152. The median and the altitude drawn from the vertex of the rt. Z.. 153. The radius of the inscribed circle and one leg. 154. The radius of the inscribed circle and an acute angle. 155. An acute angle and the sum of the legs. 156. An acute angle and the difference of the legs. To construct a triangle, having given : V157. The base, the altitude, and the Z at the vertex. V 158. The base, the corresponding median, and the Z at the vertex. 159. The perimeter and the angles. vQ - 160. One side, an adjacent Z, and the sum of the other sides. 161. One side, an adjacent Z, and the difference of the other sides/- ** 162. The sum of two sides and the angles. 163. One side, an adjacent Z, and radius of circumscribed O. 164. The angles and the radius of the circumscribed O. 165. The angles and the radius of the inscribed O. 166. An angle, the bisector, and the altitude drawn from the vertex. 167. Two sides and the median corresponding to the other side. 168. The three medians. To construct a square, having given: 169. The diagonal. 170^The sum of the diagonal and one side. EXERCISES. 129 To construct a rectangle, having given: 171. One side and the Z formed by the diagonals. 172. The perimeter and the diagonal. 173. The perimeter and the Z "of the diagonals. 174. The difference of the two adjacent sides and the Z of the diagonals. To construct a rhombus, having given : 175. The two diagonals. 176. One side and the radius of the inscribed circle. 177. One angle and the radius of the inscribed circle. ... 178. One angle and one of the diagonals. To construct a rhomboid, having given: 179. One side and the two diagonals. 180. The diagonals and the Z formed by them. 181. One side, one Z, and one diagonal. 182. The base, the altitude, and one angle. To construct an isosceles trapezoid, having given: 183. The bases and one angle. 184. The bases and the altitude. 185. The bases and the diagonal. 186. The bases and th radius of the circumscribed circle. To construct a trapezoid, having given : 187. The four sides. 188. The two bases and the two diagonals. The bases, one diagonal, and the Z formed by the diagonals. CONSTRUCTION OF CIRCLES. Find the locus of the centre of a circle : 190. Which has a given radius r and passes through a given point P. 191. Which has a given radius r and touches a given straight line AE. 192. Which passes through two given points P and Q. 193. Which touches a given straight line AB at a given point P. 194. Which touches each of two given parallels. 195. Which touches each of two given intersecting lines. 130 PLANE GEOMETRY. BOOK II. To construct a circle which has the radius r and which also : 196. Touches each of two intersecting lines AB and CD. 197. Touches a given line AB and a given circle K. 198. Passes through a given point -P and touches a given line AB. 199. Passes through a given point P and touches a given circle K. To construct a circle which shall : 200. Touch two given parallels and pass through a given point P. 201. Touch three given lines two of which are parallel. 202. Touch a given line A B at P and pass through a given point Q. 203. Touch a given circle at P and pass through a given point Q. 204. Touch two given lines and touch one of them at a given point P. 205. Touch a given line and touch a given circle at a point P. 206. Touch a given line AB at P and also touch a given circle. 207. To inscribe a circle in a given sector. * 208. To construct within a given circle three equal circles, so that each shall touch the other two and also the given circle. X209. To describe circles about the vertices of a given triangle as centres, so that each shall touch the two others. CONSTRUCTION OF STRAIGHT LINES. 210. To draw a common tangent to two given circles. 211. To bisect the angle formed by two lines, without producing the lines to their point of intersection. j 212. To draw a line through a given point, so that it shall form with the sides of a given angle an isosceles triangle. 213. Given a point P between the sides of an angle BAC. To draw through P a line terminated by the sides of the angle and bisected at P. ^214. Given two points P, Q, and a line AB; .to draw lines from P and Q which shall meet on AB and make equal angles with AB. HINT. Make use of the point which forms with P a pair of points symmetrical with respect to AB. 7 215. To find the shortest path from Pto Q which shall touch a line AB. , 216. To draw a tangent to a given circle, so that it shall be parallel to a given straight line. BOOK III. PROPORTIONAL LINES AND SIMILAR POLYGONS, THE THEORY OF PROPORTION. 292, A proportion is an expression of equality between two equal ratios. A proportion may be expressed in any one of the follow ing forms : -=-; a:b = c:d; a:b ::c:d; b d and is read, " the ratio of a to b equals the ratio of c to d" 293, The terms of a proportion are the four quantities com pared ; the first and third terms are called the antecedents, the second and fourth terms, the consequents ; the first and fourth terms are called the extremes, the second and third terms, the means. 294, In the proportion a : b = c : d, d is a fourth propor tional to a, b, and c. In the proportion a:b = b:c, c is a third proportional to a and b. In the proportion a:b = b :c, b is a mean proportional between a and c, 132 PLANE GEOMETRY. BOOK III. PROPOSITION I. 295, In every proportion the product of the extremes is equal to the product of the means. Let a:b = c-.d. To prove ad be. AT a C Now 7 = -y b d whence, by multiplying both sides by bd, ad = be. o. E. D. PROPOSITION II. 296, A mean -proportional between two quantities is equal to the square root of their product. In the proportion a : b = b : c, V = ac, 295 (the product of the extremes is equal to the product of the means). Whence, extracting the square root, 1) = Va<?. a E. D. PROPOSITION III. 297, // the product of two quantities is equal to the product of two others, either two may be made the extremes of a proportion in which the other two are made the -means. Let ad=bc. To prove a:b = c:d. Divide both members of the given equation by bd. a c Then = Q.E.D. THEORY OF PROPORTION. 133 PROPOSITION IV. 298, If four quantities of the same kind are in pro portion, they will be in proportion by alternation ; that is, the first term will be to the third as the sec ond to the fourth. Let a:b = c:d. To prove a:c = b:d. f-j Multiply each member of the equation by -. Then = * c d or, a:c = b:d. Q.E.D. PROPOSITION V. 299, If four quantities are in proportion, they will be in proportion by inversion ; that is, the second term will be to the first as the fourth to the third. Let a:b = c:d. To prove b:a = d:c. Now bo = ad. 295 Divide each member of the equation by ac. Then *= a c or, b : a = d : c. at.* 134 PLANE GEOMETRY. BOOK III. PROPOSITION VI. 300, If four quantities are in proportion, they will be in proportion by composition ; that is, the sum of the first two terms will be to the second term as the sum of the last two terms to the fourth term. Let a:b = c:d. To prove a + b :b = c + d:d. Now T ~y b d Add 1 to each member of the equation. Then f +lr= | +1 a + b _c -f- d that is, 7 ~~d~ or, a + b : b = c + d : d. In like manner, a + b :a=c + d:c. ^^ PROPOSITION VII. 301, // four quantities are in proportion, they will be in proportion by division ; that is, the difference of the first two terms will be to the second term as the difference of the last two terms to the fourth term. , Let a:b = c:d. To prove a b:b = c d:d. a_c Now h~"f7 Subtract 1 from each member of the equation. Then - 1=1-1; b a a-b__c-d that is, 7 -5 In like manner, a b : a=c d .c. THEORY OF PROPORTION. 135 PROPOSITION VIII. 302, In any proportion the terms are in proportion by composition and division ; that is, the sum of the ftrst two terms is to their difference as the sum of the last two terms to their difference. Let a-.b = c: d. Then, by 300, And, by 301, v ,. . . By division, = - a b cd or. a-\-b : a b c-}-d: c d. Q.E. D. PEOPOSITION IX. 303. In a series of equal ratios, the sum of the an tecedents is to the sum of the consequents as any antecedent is to its consequent. Let a:b = c:d = e :f= g : h. To prove a + c + e + g : b + d+f+h = a : b. Denote each ratio by r. Then , = ? = = i = f. b d f h Whence, a = br, c = dr, e =fr, g = hr. Add these equations. Then a + c + e + g = (b + d+f+ h)r. Divide by (b + d+f+h). Then a or, Q. E. D. 136 PLANE GEOMETRY. BOOK III. PROPOSITION X. 304, The products of the corresponding terms of two or more proportions are in proportion. Leta:b = c:d, e:f=g:h, k:l = m-.n. To prove aek : bfl = cgm : dhn. vr a c e a Ic m NOW T = T 7 i 7 = -- b d f h I n Whence, by multiplication, aek _ cgm bfl ~~ dhn or, aek : bfl = cgm : dhn. Q.E.D. PROPOSITION XI. 305, Like powers, or like roots, of the terms of a proportion are in proportion. Let a-.b = c:d. To prove a n :b n = c n :d n , I I 1 i and Q> n : b = C* : d*. NT a c NOW 7 = -y b d By raising to the nth power, ^-=~ } or a b n d n By extracting the nth root, a i ,1 i ,1 = ; or, a : o = c : a Q.E.O. 306, Equimultiples of two quantities are the products ob tained by multiplying each of them by the same number. Thus, ma and mb are equimultiples of a and b. THEORY OF PROPORTION. 137 PROPOSITION XII. 307, Equimultiples of two quantities are in the same ratio as the quantities themselves. Let a and b be any two quantities. To prove ma :mb a:b. H- Multiply both terms of first fraction by m. mi ma a Then - = - mo b or, ma :mb = a:b. Q.E.D. SCHOLIUM. In the treatment of proportion it is as sumed that fractions may be found which will represent the ratios. It is evident that the ratio of two quantities may be represented by a fraction when the two quantities compared can be expressed in integers in terms of a common unit. But when there is no unit in terms of which both quantities can be expressed in integers, it is possible to find a fraction that will represent the ratio to any required degree of accuracy. (See 251-256.) Hence, in speaking of the product of two quantities, as for instance, the product of two lines, we mean simply the product of the numbers which represent them when referred to a com mon unit. An interpretation of this kind must be given to the product of any two quantities throughout the Geometry. 138 PLANE GEOMETRY. BOOK III. PROPORTIONAL LINES. PROPOSITION I. THEOREM. 309, If a line is drawn through two sides of a tri angle parallel to the third side, it divides those sides proportio nally. B FIG. 1. C FIG. 2. In the triangle ABC let EF be drawn parallel to BC. EB FC Toprove ZS = Zp CASE I. When AE and EB (Fig. 1) are commensumble. Find a common measure of AE and EB, as BM. Suppose BMiQ be contained in BE three times, and in AE four times. Then (1) AE 4 At the several points of division on BE and AE draw straight lines II to BC. These lines will divide AC mto seven equal parts, of which FC will contain three, and A. ill contain four, 187 (if parallels intercept equal parts on any transversal, they intercept equal parts on every transversal). FC AF Compare (1) and (2), AE (2) Ax. 1. PROPORTIONAL LINES. 139 CASE II. When AE and EB (Fig. 2) are incommensurable. Divide AE into any number of equal parts, and apply one of these parts as a unit of measure to EB as many times as it will be contained in EB. Since AE and EB are incommensurable, a certain number of these parts will extend from E to a point K, leaving a remainder KB less than the unit of measure. Draw KH II to BO. Suppose the unit of measure indefinitely diminished, the ra tios =?== and =-== t continue equal ; and approach indefi- AE AJb Tt 1 T-? nitely the limiting ratios - and -, respectively. Therefore Jf = 26 310, COR. 1. One side of a triangle is to either part cut off by a straight line parallel to the base as the other side is to the corresponding part. For EB : AE= FC: AF, by the theorem. /. EB + AE .AE=FC+AF:AF, 300 or AB:AE=AO:AF. 311. COR. 2. If two lines are cut by any number of parallels, the corresponding intercepts are proportional. Let the lines be AB and CD. Draw AN II to CD, cutting the Us at L, M, and N. Then AL= GG, LM= GK, MN= KD. 187 _ By the theorem, B N AIT: AM= AF:AL = FIT: LM= HB : MN. That is, AF:CG= FH: GK= HB : KD. If the two lines AB and CD were parallel, the correspond ing intercepts would be equal, and the above proportion be true. 140 PLANE GEOMETRY. BOOK III. PROPOSITION II. THEOREM > 312, // a straight line divides two sides of a tri angle proportionally, it is parallel to the third side. In the triangle ABC let EF be drawn so that AB = AC t AE AF To prove EF \\toBG. Proof, From E draw EH II to BO. Then AB : AE-= AC: AH, 310 (one side of a A is to either part cut off by a line II to the base, as the other side is to the corresponding part). But AB : AE = AC : AF. Hyp. The last two proportions have the first three terms equal, each to each ; therefore the fourth terms are equal ; that is, AF=AH. .*. .KFand EH coincide. But EH\s II to BC. Cons. /. EF, which coincides with EH, is || to BC. Q.E.Q. PROPORTIONAL LINES. 141 PROPOSITION III. THEOREM. 313, The bisector of an angle of a triangle divides the opposite side into segments proportional to other two sides. A M B Let CM "bisect the angle C of the triangle GAB. To prove MA : MB = CA CB. Proof, Draw AE II to MC to meet BC produced at E. Since MC is II to AE of the A BAE, we have 309 MA:MB=CE:CB. (1) Since MC is II to AE, ZACM=ZCAE, 104 (being alt.-int. of\\ lines) ; and Z BCM= Z CEA, 106 (being ext.-int. A of II lines). But the Z. A CM= Z BCM. Hyp. .-. the Z CAE = Z CEA. Ax. 1 /. CE= CA, 156 (if two A of a A are equal, the opposite sides are equal), Putting CA for CE in (1), we have MA:MB=CA: CB. Q.E.D. 142 PLANE GEOMETRY. BOOK III. PROPOSITION IV. THEOREM. 314, The bisector of an exterior angle of a triangle meets the opposite side produced at a point the dis tances of which from the extremities of this side are proportional to the other two sides. A Let CM bisect the exterior angle ACE of the tri angle CAB, and meet BA produced at M f . To prove M A : M B = CA : OB. Proof, Draw AF II to CM to meet BC&t F. Since AFis II to OH 1 of the A BOM 1 , we have 309 WA\WB=CF\CB. (1) Since AF is II to CM\ theZM CE = ZAFC, 106 (being ext.-int. A of II lines) \ and the Z M CA = Z CAF, 104 (being alt.-int, A of I! lines). Since CM 1 bisects the Z ECA, :. the Z AFC = Z CAF. Ax. 1 /. CA = CF, 156 (if two A of a A are equal, the opposite sides are equal). Putting CA for CF in (1), we have WA : M B = CA : CB. Q.E.D. PROPORTIONAL LINES. 143 315, SCHOLIUM. If a given line AB is divided at M, a point between the extremities A and B, it is said to be divided internally into the segments MA and MB ; and if it is divided at M , a point in the prolongation of AB, it is said to be divided externally into the segments M A and M f B. , - B A M In either case the segments are the distances from the point of division to the extremities of the line. If the line is divided internally, the sum of the segments is equal to the line ; and if the line is divided externally, the difference of the segments is equal to the line. Suppose it is required to divide the given line AB inter nally and externally in the same ratio; as, for example, the ratio of the two numbers 3 and 5. cc. _ , _ , _ .i.i...... - ; M A M B y We divide AB into 5 + 3, or 8, equal parts, and take 8 parts from A ; we then have the point J/j such that MA: MB = 3: 5. (1) Secondly, we divide AB into two equal parts, and lay off on the prolongation of AB, to the left of A, three of these equal parts ; we then have the point M f , such that M A : M B -3:5. (2) Comparing (1) and (2), 316, If a given straight line is divided internally and externally into segments having the same ratio, the line is said to be divided harmonically. 144 PLANE GEOMETRY. BOOK III. 317. COR. 1. The bisectors of an interior angle and an exte rior angle at one vertex of a triangle divide the opposite side harmoni cally. For, by 313 and 314, each \o bisector divides the opposite side into segments proportional to the o^her two sides of the triangle. M M D 318. COR. 2. If the points M and M divide the line AB harmonically, the points A and B divide the line MM har monically. For, if MA : MB = M A : M B, by alternation, MA : M A = MB : M . 298 That is, the ratio of the distances of A from M and M is equal to the ratio of the distances of B from M and M . The four points A, B, M, and M 1 are called harmonic points, and the two pairs, A, B, and M, M\ are called con jugate harmonic points. SIMILAR POLYGONS. 319. Similar polygons are polygons that have their homol ogous angles equal, and their homologous sides proportional. D and ED E D f Thus, if the polygons ABODE and A B C D E are similar the A A, B, C, etc., are equal to A A , , C , etc. CD A JB etc. 320. In two similar polygons, the ratio of any two homol ogous sides is called the ratio of similitude of the polygons. SIMILAR TRIANGLES. 145 SIMILAR TRIANGLES. PROPOSITION V. THEOREM. 321, Two mutually equiangular triangles are sim ilar. A In the triangles* ABC and A B C let angles A, B, G be equal to angles A , B , C r respectively. To prove A ABC and A B C similar. Proof, Apply the A A B C* to the A AEG, so that Z A shall coincide with Z A. Then the A A B C 1 will take the position of A AEH. Now Z AEJI(ssime as Z B 1 ) = Z B. :. Elf is II ioBC, 108 (when two straight lines, lying in the same plane, are cut by a third straight Line, if the ext.-int. 4 are equal the lines are parallel). /. AB : AE= AC: AH, 310 or AB:A B f = AC:A C . In like manner, by applying A A B C to A ABC, so that Z B 1 shall coincide with Z B, we may prove that AB : A B = BC: J3 <7 . Therefore the two A are similar. 319 Q. E. D. 322, COR. 1. Two triangles are similar if two angles of the one are equal respectively to two angles of the other. 323, COR. 2. Two right triangles are similar if an acute angle of the one is equal to an acute angle of the other. 146 PLANE GEOMETRY. BOOK III. PROPOSITION VI. THEOREM. 324, If two triangles have their sides respectively proportional, they are similar. In the triangles ABC and A B f C> let AB = AC = BC A B A C B C To prove A ABC and A B C similar. Proof, Take AE= A B , and AH= A O . Draw EH. Then from the given proportion, AB = AC AE AH :. Effis II to BC, 312 (if a line divide two sides of a A proportionally, it is II to the third side). Hence in the A ABC&nd. AEH 106 and (being ext.-int. A of II lines). .-. A ABC and AEH are similar, 322 (two A are similar if two A of one are equal respectively to two A of the other). .:AB\ AE= that is, AB A B 1 = SIMILAR TRIANGLES. But by hypothesis, 147 The last two proportions have the first three terms equal, each to each ; therefore the fourth terms are equal ; that is, Hence in the A AE=A B t 160 (having three sides of the one equal respectively to three sides of the other). But A AEH is similar to A ABC. .-. A A 3 is similar to A ABO. aE . D . 325, SCHOLIUM. The primary idea of similarity is likeness of form ; and the two conditions necessary to similarity are : I. For every angle in one of the figures there must be an equal angle in the other, and II. The homologous sides must be in proportion. In the case of triangles, either condition involves the other, but in the case of other polygons, it does not follow that if one condition exist the other does also. Q Thus in the quadrilaterals Q and Q 1 , the homologous sides are proportional, but the homologous angles are not equal. In the quadrilaterals R and It 1 the homologous angles are equal, but the sides are not proportional. 148 PLANE GEOMETKY. BOOK III. PROPOSITION VII. THEOREM. 326, If two triangles have an angle of the one equal to an angle of the other, and the including sides pro portional, they are similar. In the triangles ABC and A B C , let AB = AC A B ~ A C To prove A ABC and A B C similar. Proof, Apply the A A B C 1 to the A ABC, so that Z A 1 shall coincide with Z A. Then the A A B C will take the position of A AEH. AB AC Now That is, A B 1 A C AB__ AC^ AE AH Therefore the line EH divides the sides AB and A C pro portionally ; toC, 312 (if a line divide two sides of a A proportionally, it is \\ to the third side}. Hence the A ABC and AEH are mutually equiangular and similar. .-. A A B C 1 is similar to A ABO. Q. E. D. SIMILAR TRIANGLES. 149 PROPOSITION VIII. THEOREM. 327, // two triangles have their sides respectively parallel, or respectively perpendicular, they are sim ilar. B C In the triangles A B C and ABC let A B , A C , B <y be respectively parallel, or respectively perpendicular, to AB, AC, BC. To prove A A B C 1 and ABC similar. jf Proof, The corresponding A are either equal or supplements of each other, 112, 113 (if two A have their sides II, or _L, they are equal or supplementary). Hence we may make three suppositions : 1st. A + A = 2rt.A, B + B = 2rt.A 2d. A = A , B + B = 2rt.A 3d. A - A , B= B\ . . C= C . 140 Since the sum of the A of the two A cannot exceed four right angles, the third supposition only is admissible. 138 . . the two A ABC and A B C are similar, 321 (two mutually equiangular & are similar). 150 PLANE GEOMETRY. BOOK III. PROPOSITION IX. THEOREM. 328, The homologous altitudes of two similar tri angles have the same ratio as any two homologous sides. A o A In the two similar triangles ABC and A B C , let the altitudes be CO and C O . To prove CO AC AB C O A C A B Proof, In the rt. A CO A and C O A , Z A = Z A , 319 (being homologous A of the similar & ABC and A B C 1 }. .-. A CO A and C O A are similar, 323 (two rt. & having an acute Z of the one equal to an acute Z of the other are similar). CO AC C O A C In the similar A AB C and A B C , AC _ = AB A C ~ A B , CO AC AB Therefore, =_ = _. 319 Q. E. O. SIMILAR TRIANGLES. 151 PROPOSITION X. THEOREM. 329, Straight lines drawn through the same point intercept proportional segments upon two parallels. /B /C" \D \E ABC D E Let the two parallels AE and A E cut the straight lines OA, OB, OC, OD t and OE. AE BO CD DE Toprove _ = ---_= . Proof, Since A E 1 is II to AE, the pairs of A OAB and OA B , OBC and OB C , etc., are mutually equiangular and similar, AS " A ff OB 1 (homologous sides of similar & are proportional). Ax 1 In a similar way it may be shown that BC _ CD 1 CD DE a B O C D C D Q. E. D. REMARK. A condensed form of writing the above is _ _^ = _ ==== ^ A B \OB j B C \00 J C D 1 \OD ) D E^ where a parenthesis about a ratio signifies that this ratio is used to prove the equality of the ratios immediately preceding and following it. 152 PLANE GEOMETRY. BOOK III. PBOPOSITION XI. THEOREM. CONVERSELY: If three or more non-parallel straight lines intercept proportional segments upon two parallels, they pass through a wmrrwn, point. AC E Let AB, CD, EF, cut the parallels AE and BF so that AC : BD=CE : DF. To prove that AB, CD, EF prolonged meet in a point. Proof, Prolong AB and CD until they meet in 0. Join OE, If we designate by F the point where OE cuts BF, we shall have by 329, AC:BD=CE:DF . But by hypothesis AC:BD = CE.DF. These proportions have the first three terms equal, each to each ; therefore the fourth terms are equal ; that is, .. F coincides with F. .-. .EF prolonged passes through O. . . AB, CD, and EF prolonged meet in the point 0. SIMILAR POLYGONS. 153 SIMILAR POLYGONS. PROPOSITION XII. THEOREM. 331, If two polygons cure composed of the same num ber of triangles, similar each to each, and- similarly placed, the polygons are similar* E B C B V In the two polygons ABODE and A B C D E , let the triangles AEB, BEG, CED be similar respectively to the triangles A E B , B E C , C E D*. To prove ABODE similar to A B C D E 1 . Proof, Z A = Z A 1 , 319 (being homologous A of similar A). Also, Z ABE = Z A B E , 319 and Z EEC = Z E B C . By adding, Z ABC = Z A B C . In like manner we may prove Z BCD = Z B C D , etc. Hence the two polygons are mutually equiangular. Now AE = AB ^( EB\ BO ^( EC\ CD = ED A E AB \E &) B C \E &) C D E D (the homologous sides of similar A are proportional). Hence the homologous sides of the polygons are proportional. Therefore the polygons are similar, 319 (having their homologous A equal, and their homologous sides proportional). Q. E. D, 154 PLANE GEOMETRY. BOOK III. PROPOSITION XIII. THEOREM. 332, If two polygons cure similar, they are composed of the same number of triangles, similar each to each, and similarly placed. B C B f Let the polygons ABODE and A B C D E be similar. From two homologous vertices, as E and E 1 , draw diagonals EB, EC, and E B , E C . To prove A EAB, EEC, ECD similar respectively to A E A B , E B C , E C D . Proof, In the A EAB and E A B , Z.A=/.A , (being homologous A of similar polygons) ; AE AB A B and 319 319 A E (being homologous sides of similar polygons ). . . A EAB and E A B are similar, 326 (having an of the one equal to an /. of the other, and the including sides proportional). Also, ZABC=ZA B C , (1) (being homologous A of similar polygons). And Z ABE= Z A B E , (2) (being homologous A of similar A). Subtract (2) from (1), /. E B C 1 . Ax. 3 Now And SIMILAR POLYGONS. 155 EB _ AB (being homologous sides of similar A). BO = AB B C A ^ (being homologous sides of similar polygons). , A v 1 * * -r~ttT~\i *~T\I ~7f" XIA. JL . . A EBCw& E B C are similar, 326 (having an Z of the one equal to an Z of the other, and the including sides proportional). In like manner we may prove A ECD and E C D similar. Q. E. D. PROPOSITION XIV. THEOREM. 333, The perimeters of two similar polygons have the same ratio as any two homologous sides. E B c B c Let the two similar polygons be ABODE and A B C D E 1 , and let P and P represent their perimeters. To prove P : P = A B : A B . AB : A B = EG: = CD : C J D , etc., 319 (the homologous sides of similar polygons are proportional). .-. AB+BC,etc. : A B +B C , etc. =AB: A B 1 , 303 (in a series of equal ratios the sum of the antecedents is to the sum of the " n equents as any antecedent is to its consequent). conse That is, P:f t = AB:A B . a E 156 PLANE GEOMETRY. BOOK III. NUMERICAL PROPERTIES OF FIGURES. PROPOSITION XV. THEOREM. 334, If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse : I. The perpendicular is a mean proportional be tween the segments of the hypotenuse. II. Each leg of the right triangle is a mean pro portional between the hypotenuse and its adjacent segment. F In the right triangle ABC, let BF be drawn from the vertex of the right angle B, perpendicular to AC. I. To prove AF: BF= BF: FO. Proof, In the rt. A BAFwd BAC the acute /. A is common. Hence the A are similar. 323 In the rt. A OF and BCA the acute /. (7 is common. Hence the A are similar. . 323 Now as the rt. A ABF and CBF&re both similar to ABO, they are similar to each other. In the similar A ABFsiul CBF, AF, the shortest side of the one, : BF, the shortest side of the other, : : BF, the medium side of the one, : FO, the medium side of the other. II. To prove AC : AB = AB : AF, and NUMERICAL PROPERTIES OF FIGURES. 157 In the similar A ABO and ABF, AC, the longest side of the one, : AB, the longest side of the other, : : AB, the shortest side of the one, : A F, the shortest side of the other. Also in the similar A AfiCand FBC, AC, the longest side of the one, : BC, the longest side of the other, : : EC, the medium side of the one, : FCj the medium side of the other. Q . E . o. 335, COR. 1. The squares of the two legs of a right triangle are proportional to the adjacent segments of the hypotenuse. The proportions in II. give, by 295, AB* = ACxAF, and BC* = ACxCF. By dividing one by the other, we have AGx CF OF 336, COR. 2. The squares of the hypotenuse and either leg are proportional to the hypotenuse and the adjacent segment. -r^ -^L O _/lL/ /\ ^J_ O _/T_ O l^ = I77>o4> = ZF 337, COR. 3. An angle inscribed in a semicircle is a right angle ( 264). Therefore, I. The perpendicular from any point in the circumference to the diameter of a circle is a mean proportional between the segments ^_ _^ ^ of the diameter. & D II. The chord drawn from the point to either extremity of the diameter is a mean proportional between the diameter and the adjacent segment. REMARK. The pairs of corresponding sides in similar triangles may be called longest, shortest, medium, to enable the beginner to see quickly these pairs ; but he must not forget that two sides are homologous, not because they appear to be the longest or the shortest sides, but because they lie opposite corresponding equal angles. 158 PLANE GEOMETRY. BOOK III. PROPOSITION XVI. THEOREM. 338, The sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. B Let ABC fee a right triangle with its right angle at B. To prove AB* + B& = AC*- Proof, Draw BF to AC. Then A& = ACxAF 334 and BC* = ACX OF By adding, AB* + SO* = AC(AF+ CF) = AC\ a E. D. 339, COR. The square of either leg of a right triangle is equal to the difference of the squares of the hypotenuse and the other leg. 340, SCHOLIUM. The ratio of the diagonal of a D square to the side is the incommensurable num ber V2. For if AC is the diagonal of the square ABCD, then AC* = IB* + BC\ or AC 2 = 2 AC* AC Divide by AJ3\ we have ~ = 2, or -^-g =-- V2. Since the square root of 2 is incommensurable, the diagonal and side of a square are two incommensurable lines. 341. The projection of a line CD upon a straight line AB is that part of the line AB comprised between the perpendiculars CP and DR let fall from the extremities of CD. Thus, PR is the projection of CD upon AB. A ~ NUMERICAL PROPERTIES OF FIGURES. 159 PROPOSITION XVII. THEOREM. 342, In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides diminished Ijy twice the product of one of those sides and the projection of the other upon that side. A Let G be an acute angle of the triangle ABC, and DC the projection of AC upon BC. To prove AB* ^ BC* + AC* -2BCx DO. Proof, If D fall upon the base (Fig. 1), DB = BC- DC- If D fall upon the base produced (Fig. 2), DB = DC- BC. In either case, JOB* = BC* + DC* -2BCx DC. Add AD to both sides of this equality, and we have AD* + DB* = BC* + AD*+DC*-2BCx DC. But AD* + DB 2 = AB\ 338 and AD* + DC* = AC*, (the sum of the squares of the two legs of a rt. A is equal to the square of the hypotenuse}. Put AH* and AC* for their equals in the above equality, BC* -}- AC*- 2BCx DC. Q. E. D. - 160 PLANE GEOMETRY. BOOK III. PROPOSITION XVIII. THEOREM. 343, In any obtuse triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other upon that side. A Let C be the obtuse angle of the triangle ABC, and CD be the projection of AC upon BC produced. To prove A3* = C 2 + Iff + 2Cx DO. Proof, DB = BC+DC. Squaring, DB* = C* + DO* + 2Cx DO. Add Alf to both sides, and we have AD* + 52? = W*+AD i +D But AD* + ~DS = AB\ 338 and (the sum of the squares of the two legs of a rt. A is equal to the square of the hypotenuse). Put AS and AC? for their equals in the above equality, AW = C 2 + AC* + 2Cx DC. Q.E.O. NOTE. The last three theorems enable us to compute the lengths of the altitudes if the lengths of the three sides of a triangle are known. NUMERICAL PROPERTIES OF FIGURES. 161 ^ PROPOSITION XIX. THEOREM. 344, I. The sum of the squares of two sides of a tri angle is equal to twice the square of half the third side increased ~by twice the square of the median upon that side. II. The difference of the squares of two sides of a triangle is equal to twice the product of the third side by the projection of the median upon that side. A M u In. the triangle ABC let AM "be the median, and MD the projection of AM upon the side BO. Also let AS be greater than AC. Toprove I. A& + AC* = 2 3M* + 2 AM\ II. A?-A(T = 2CxMD. Proof, Since AB>AC, the Z AME will be obtuse, and the Z AMC will be acute. 152 Then A =M + AM + 2JBMx MD, 343 (in any obtuse A the square of the side opposite the obtuse /. is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other on that side) ; and AC = MC + AM-2MCx MD, 342 (in any A the square of the side opposite an acute Z is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other upon that side). Add these two equalities, and observe that BM=- MC. Then AS + AC 2 = 2 BM i + 2AM\ Subtract the second equality from the first. Then Z&-AO* = 23Cx MD. aE . D . NOTE. This theorem enables us to compute the lengths of the medians if the lengths of the three sides of the triangle are known. 162 PLANE GEOMETRY. BOOK III. PROPOSITION XX. THEOREM. 345, If any chord is drawn through a fixed point within a circle, the product of its segments is con- stant in whatever direction the chord is drawn. Let any two chords AB and CD intersect at 0. To prove OAxOB=ODx 00. Proof. Draw A O and ED. In the A AOC and BOD, ^C=Z.B, 263 (each being measured by j arc AD}. Z A - Z D, 263 (each being measured by % arc BC). /. the A are similar, 322 (two & are similar when two A of the one are equal to two A of the other). Whence OA, the longest side of the one, : OD, the longest side of the other, : : 00, the shortest side of the one, : OB, the shortest side of the other. .-. OAxOB = ODxOO. 295 Q.E. D. 346, SCHOLIUM. This proportion may be written OA = 00 OA 1 OD OB r OD~OB 00 that is, the ratio of two corresponding segments is equal to the reciprocal of the ratio of the other two corresponding segments. In this case the segments are said to be reciprocally proportional. NUMERICAL PROPERTIES OF FIGURES. 163 PROPOSITION XXL THEOREM. 347, If from a fixed point without a circle a secant is drawn, the product of the secant and its external segment is constant in whatever direction the secant is drawn. Let OA and OB be two secants drawn from point 0. To prove OA X 00= OB X OD. Proof. Draw EC and AD. In the A OAD and OBC Z is common, A = Z.B, 263 (each being measured by % arc CD). . . the two A are similar, 322 (two A are similar when two A of the one are equal to two A of the other). Whence OA, the longest side of the one, : OB, the longest side of the other, : : OD, the shortest side of the one, : 0(7, the shortest side of the other. /. OA x 00= OB x OD. 295 Q. E. D. EEMABK. The above proportion continues true if the secant OB turns about until B and D approach each other indefinitely. Therefore, by the theory of limits, it is true when B and D coincide at H. Whence, OA x 00= OH\ This truth is demonstrated directly in the next theorem. 164 PLANE GEOMETRY. BOOK III. PROPOSITION XXII. THEOREM. 348, If from a point without a circle a secant and a tangent are drawn, the tangent is a mean propor tional between the whole secant and the external segment* Let OB "be a tangent and OG a secant drawn from the point to the circle MBC. To prove OG : OB = OB : M. Proof. Draw EM and BO. In the A OB M and OBO /. is common. Z. OEM is measured by % arc MB, 269 (being an Z. formed by a tangent and a chord). Z C is measured by -J- arc B M, 263 (being an inscribed Z). .\/.OBM=Z. O. .-. A OBC and OEM are similar, 322 (having two A of the one equal to two A of the other). Whence OC, the longest side of the one, : OB, the longest side of the other, : : OB, the shortest side of the one, : OM, the shortest side of the other. Q. E. D. NUMERICAL PROPERTIES OF FIGURES. 165 PROPOSITION XXIII. THEOREM. 349, The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments determined by the bisector upon the third side of the triangle. Let AD bisect the angle BAG of the triangle ABC. To prove AD* = ABxAC~DBx DC. Proof, Circumscribe the O ABO about the A ABO. 285 Produce AD to meet the circumference in E, and draw EG. Then in the A ABD and AEC, ^BAD = ZCAE, Hyp. Z.B = E, 263 (each being measured by ^ the arc AC}. .-. A ABD and AEC are similar, 322 (two A are similar if two A of the one are equal respectively to two A of the other). Whence AB, the longest side of the one, : AE, the longest side of the other, : : AD, the shortest side of the one, : A 0, the shortest side of the other. .-. AB x A0= ADxAE 295 = AD(AD+DE) = AI?+ADxDE. But ADxDE=DBxDC, 345 (the product of the segments of a chord drawn through a fixed point in a Q is constant). .-.ABxAC=AD i + DBxDC. Whence AD* = AB X AC DB x DC. Q . E . D . NOTE. This theorem enables us to compute the lengths of the bisectors of the angles of a triangle if the lengths of the sides are known. 166 PLANE GEOMETRY. BOOK III. PROPOSITION XXIV. THEOREM. 350, In any triangle the product of two sides is equal to the product of the diameter of the circum scribed circle by the altitude upon the third side. Let ABC be a triangle, AD the altitude, and ABO the circle circumscribed about the triangle ABC. Draw the diameter AE, and draw EG. To prove AB xAC=AEx AD. Proof, In the A ABD and A EC, Z BDA is a rt. Z, Cons. Z EGA is a rt. Z, 264 (being inscribed in a semicircle), and Z B - Z E. 263 .-. A ABD and AEG are similar, 323 (two rt. & having cm acute /. of the one equal to an acute Z. of the other are similar). Whence AB, the longest side of the one, : AE, the longest side of the other, : : AD, the shortest side of the one, : AC, the shortest side of the other. .:. ABxAC^AExAD. 295 Q.E. D. NOTE. This theorem enables us to compute the length of the radius of a circle circumscribed about a triangle, if the lengths of the three sides of the triangle are known. PROBLEMS OF CONSTRUCTION. 167 PROBLEMS OF CONSTRUCTION. PROPOSITION XXV. PROBLEM. 351, To divide a given straight line into parts pro portional to any number of given lines. H K 7? Let AB, m, n, and p, be given straight lines. To divide AB into parts proportional to m, n, and p. Construction, Draw AX, making an acute Z with AB. On AX take AC=m, CE=n, EX^p. Draw EX. From E and draw UK and GH II to BX. JTand .5" are the division points required. (a line drawn through two sides of a A II to the third side divides those sides proportionally). /. AH : UK : KB = AC : CE : EX. Substitute m, n, and p for their equals AC, CE, and EX. Then AIT : HK : KB - m n : p. Q. E. F. 168 PLANE GEOMETRY. BOOK III. PROPOSITION XXVI. PROBLEM. 352, To find a fourth proportional to three given straight lines. Let the three given lines be m, n, andp. To find a fourth proportional to m, n, andp. Draw Ax and Ay containing any acute angle. Construction, On Ax take AB equal to m, BQn. On Ay take AD=p. Draw BD. From C draw CF \\ to BD, to meet Ay at F. DF is the fourth proportional required. Proof, A B : BG = AD : DF, 309 (a line drawn through two sides of a A II to the third side divides those sides proportionally}. Substitute m, n, and^> for their equals AB, JBC, and AD. Then m : n = p : DF. Q. E. F. PROBLEMS OF CONSTRUCTION. 169 PROPOSITION XXVH. PROBLEM. 353, To -find a third proportional to two given straight lines. A Let m and n be the two given straight lines. To find a third proportional to m and n. Construction, Construct any acute angle A, and take AB m, AC=n. Produce AB to D, making ED =-- AC. Join EC. Through D draw DE II to 0to meet AC produced at E. CE is the third proportional to AB and AC. Proof, A E : ED = A C : CE. 309 (a line drawn through two sides of a A II to the third side divides those sides proportionally}. Substitute, in the above proportion, AC for its equal ED. Then AB : AC = AC : CE. That is, m:n= n : CE. Q.E. F. Ex. 217. Construct x, if (1) x = , (2) x = c c Special Cases : (1) a = 2, b = 3, c = 4; (2) a = 3, 6 = 7, c=ll;(3) a = 2, c - 3 ; (4) a = 3, c = 5 ; (5) a = 2c. 170 PLANE GEOMETRY. BOOK III. PROPOSITION XXVIII. PROBLEM. 354, To find a mean proportional between two given straight lines. H m A~-- 1 _ -^ -TT 1 ra c n B Let the two given lines be m and n. To find a mean proportional between m and n. Construction, On the straight line AE take AC= m, and OB = n. On AB as a diameter describe a semi-circumference. At erect the _L CH to meet the circumference at H. CHis a mean proportional between in and n. Proof, .-. AC : CH = CH CB, 337 (the _L let fall from a point in a circumference to the diameter of a circle is a mean proportional between the segments of the diameter). Substitute for A O and OB their equals m and n. Then m : CH = CH : n. Q. E. F. 355, A straight line is said to be divided in extreme and mean ratio, when the whole line is to the greater segment as the greater segment is to the less. Ex. 21 8. Construct x if a; = Special Cases : (1) a = 2, 6 = 3; (2)a=l, b = 5- (3)a = 3, 6 = 7. PROBLEMS OF CONSTRUCTION. 171 v PROPOSITION XXIX. PROBLEM. 356, To divide a given line in extreme and mean ratio. .-- " X. .%""" A C B Let AB be the given line. To divide AB in extreme and mean ratio. Construction, At B erect a J_ BE equal to one-half of AB. From E as a centre, with a radius equal to EB, describe a 0. Draw AE, meeting the circumference in .Fand G. On AB take A C = AF. On BA produced take AC 1 = AG. Then AB is divided internally at C and externally at C in extreme and mean ratio. Proof, AG : AB - AB : AF, 348 (if from a point without a O a secant and a tangent are drawn, the tan gent is a mean proportional between the whole secant and the external segment). Then by 301 and 300, AG-AB:AB = AB-AF : AF, (1) AG + AB: AG=AB+AF : AB. (2) By construction FG = 2 EB == AB. .-. AG-AB = AG-FG--=AF=AC. Hence (1) becomes AC : AB = BC : AC; or, by inversion, AB : AC= AC : BC. 299 Again, since C A = AG = AB + AF, (2) becomes C B : C A == C A : AB. Q.E.F. 172 PLANE GEOMETRY. BOOK III. PROPOSITION XXX. PROBLEM. 357, Upon a given line homologous to a given side of a given polygon, to construct a polygon similar to the given polygon. E Let A E be the given line homologous to AE of the given polygon ABODE. To construct on A E 1 a polygon similar to the given polygon. Construction, From E draw the diagonals EE and EC. From E 1 draw E B , E C , and E D\ making AA E B , B E C , and C E D 1 equal respectively to A AEB, BEG, and OED. From A draw A t making Z E A B =^EAB, and meeting E B 1 at Bf . From Bf draw , making Z E B C = Z EBC, and meeting E C at C . From C draw C D , making Z E C D = ZECD, and meeting j5"Z> at D . Then A B C D E is the required polygon. Proof, The corresponding A A BE and A B E , EBC and E B C , ECD and ^"(7 Z) are similar, 322 (two A are similar if they have two A of the one equal respectively to two A of the other}. Then the two polygons are similar, 331 (two polygons. composed of the same number of A similar to each other and similarly placed, are similar). PROBLEMS OF COMPUTATION. 173 PROBLEMS OF COMPUTATION. 219. To compute the altitudes of a triangle in terms of its sides. O At least one of the angles A or B is acute. Suppose it is the angle B. In the A CDB, In the A ABC, Whence, Hfin^p li? n? h 2 = a 2 ^ 2 = a 2 -BD\ + c*-2cxBD. ^338 342 (a 2 + c 2 -6 2c 2 ) 2 4 a 2 c 2 (a 2 + c 2 6 2 ) 2 4 c 2 4 c 2 -6 2 )(2ac-a 2 -c 2 4c 3 = {(a + c) 2 - b 2 } {b* -(a- c) 2 } 4c 2 _ (a + b + c) (a + c 5) (6 + a c) (6 a + c) 4c 2 Let a-f6 + c = 2s. Then a -f c - b = 2(s- 6), 6 a + c = 2(s a). Hence A 2 = 2s X 2 ( s ~ a ) X 2(s - 6) x 2(s -c) 4c 2 By simplifying, and extracting the square root, 220. To compute the medians of a triangle in terms of its sides. By? 344, a 2 + 6 2 = 2m 2 -h2/ r -Y- Whence 4 m 2 = 2 (a 2 + 6 2 ) - c 2 . By? 344, a 2 + 6 2 = 2m 2 -h2-- (Fig. 2) 174 PLANE GEOMETRY. BOOK III. 221. To compute the bisectors of a triangle in terms of the sides. By I 349, P - ab - AD x BD. ~ * J?Z> c a+b By 313, ^1=^ = a + b be a + b Whence , and BD = abc 2 (a + 6) 2 c 2 /!_ ab (a + b + c) (a + b c) (a + Z>) 2 a5x2sx2(s-c) (a + by Whence 222. To compute the radius of the circle circumscribed about a tri angle in terms of the sides of the triangle. By 350, ABxAC=AExAD t be = 2 R x AD. But Whence (s-a)(s-b}(s-c). abc E 223. If the sides of a triangle are 3, 4, and 5, is the angle opposite 5 right, acute, or obtuse ? 224. If the sides of a triangle are 7, 9, and 12, is the angle opposite 12 right, acute, or obtuse ? 225. If the sides of a triangle are 7, 9, and 11, is the angle opposite 11 right, acute, or obtuse ? 226. The legs of a right triangle are 8 inches and 12 inches ; find the lengths of the projections of these legs upon the hypotenuse, and the dis tance of the vertex of the right angle from the hypotenuse. - 227. If the sides of a triangle are 6 inches, 9 inches, and 12 inches, find the lengths (1) of the altitudes ; (2) of the medians ; (3) of the bisec tors ; (4) of the radius of the circumscribed circle. EXERCISES. 175 THEOREMS. 228. Any two altitudes of a triangle are inversely proportional to the corresponding bases. 229. Two circles touch at P. Through P three lines are drawn, meet ing one circle in A, B, C, and the other in A } B / , C , respectively. Prove that the triangles ABC, A B C are similar. ^230. Two chords AB, CD intersect at M, and A is the middle point of the arc CD. Prove that the product AB X AM remains the same if the chord AB is made to turn about the fixed point A. HINT. Draw the diameter AE, join BE/qpd. compare the triangles thus formed. 231. The sum of the squares of the segments of two perpendiculai chords is equal to the square of the diameter of the circle. HINT. If AB, CD are the chords, draw the diameter BE, join AC, ED, BD, and prove that AC = ED. Apply \ 338. 232. In a parallelogram ABCD, a line DE is drawn, meeting the diagonal AC in F, the side BO in G, and the side AB produced in E. Prove that DF* = FGx FE. 233. The tangents to two intersecting circles drawn from any point in their common chord produced, are equal, (g 348.) 234. The common chord of two intersecting circles, if produced, will bisect their common tangents. ($ 348.) 235. If two circles touch each other, their common tangent is a mean proportional between their diameters. HINT. Let AB be the common tangent. Draw the diameters AC, BD. Join the point of contact P to A, B, C, and D. Show that APD and BPC are straight lines _L to each other, and compare A ABC, ABD. "" 236. If three circles intersect one another, the common chords all pass flhrough the same point. HINT. Let two of the chords AB and CD meet at 0. Join the point of intersection E to 0, and suppose that EO produced meets the same two circles at two different points P and Q. Then prove that OP- OQ; hence, that the points P and Q coincide, 176 PLANE GEOMETRY. BOOK TTI. 237. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. HINT. Draw any two of the chords, join the points where they meet the circumferences, and prove that the A thus formed are similar. 238. In an inscribed quadrilateral, the product of the diagonals is equal to th.e sum of the products of the opposite sides. HINT. Draw DE, making Z CDE= /.ADB. The & ABD and CDE are similar. Also the & BCD and ADE are similar. 239. The sum of the squares of the four sides of any quadrilateral is equal to the sum of the squares of the diagonals, increased by four-iimes the square of the line joining the middle points of the diagonals. HINT. Join the middle points F, E, of the diag onals. Draw EB and ED. Apply \ 344 to the A ABC and ADC, add the results, and eliminate BE 1 + DE 1 by applying 343 to the A BDE. 240. The square of the bisector ofan exterior angle of a triangle is equal to the product of the external segments deter mined by the bisector upon one of the sides.dimin- ished by the product of the other two sides. HINT. Let CD bisect the exterior Z BCH of the A ABC. Circumscribe a about the A, pro- II duce DC to meet the circumference in F, and draw BF. BCF similar. Apply 347. Prove &ACD, < 7 241. If a point is joined to the vertices of a triangle ABC, and /through any point A / in OA a line parallel to AB is drawn, meeting OB at B , and then through B f a line parallel to BC, meeting OC at C , and C / is joined to A , the triangle A B C will be similar to the tri angle ABC. 242. If the line of centres of two circles meets the circumferences at the points A, B, C, D, and meets the common exterior tangent at P, then PAxPD = PBxPC. 243. The line of centres of two circles meets the common exterior tangent at P, and a secant is drawn from P, cutting the circles at the consecutive points E, F, G, H. Prove that PExPH = PFx PQ. t EXERCISES. 177 NUMERICAL EXERCISES. *P 244. A line is drawn parallel to a side AB of a triangle ABC, and Cutting AC in D, BC in E. HAD -. DC= 2 : 3, and AB= 20 inches, id DE. i 245. The sides of a triangle are 9, 12, 15. Find the segments made by bisecting the angles, (g 313.) i-.2246. A tree casts a shadow 90 feet long, when a vertical rod 6 feet high casts a shadow 4 feet long. How high is the tree ? 247. The bases of a trapezoid are represented by a, b, and the altitude h. Find the altitudes of the two triangles formed by producing the legs till they meet. s \248. The sides of a triangle are 6, 7, 8. In a similar triangle the side homologous to 8 is equal to 40. Find the other two sides. -j 249. The perimeters of two similar polygons are 200 feet and 300 feet. If a side of the first polygon is 24 feet, find the homologous side of the second polygon. **/ 250. How long must a ladder be to reach a window 24 feet high, if line lower end of the ladder is 10 feet from the side of the house ? 251. If the side of an equilateral triangle = a, find the altitude. ^, 252. If the altitude of an equilateral triangle = h, find the side. ? 253. Find the lengths of the longest and the shortest chord that can bo drawn through a point 6 inches from the centre of a circle whose radius is equal to 10 inches. / 7 254. The distance from the centre of a circle to a chord 10 inches long is 12 inches. Find the distance from the centre to a chord 24 inches long. 255. The radius of a circle is 5 inches. Through a point 3 inches from the centre a diameter is drawn, and also a chord perpendicular to the diameter. Find the length of this chord, and the distance from one end of the chord to the ends of the diameter. 256. The radius of a circle is 6 inches. Through a point 10 inches from the centre tangents are drawn. Find the lengths of the tangents, and also of the chord joining the points of contact. V 257. If a chord 8 inches long is 3 inches distant from the centre of the circle, find the radius and the distances from the end of the chord to the ends of the diameter which bisects the chord. 178 PLANE GEOMETRY. BOOK ITI. 258. The radius of a circle is 13 inches. Through a point 5 inches fif6m the centre any chord is drawn. What is the product of the two seg ments of the chord ? What is the length of the shortest chord that can be drawn through the point ? 259. From the end of a tangent 20 inches long a secant is drawn through the centre of the circle. If the exterior segment of this secant is 8 inches, find the radius of the circle. ^ 260. The radius of a circle is 9 inches ; the length of a tangent is 12 inches. Find the length of a secant drawn from the extremity of the tangent to the centre of the circle. 261. The radii of two circles are 8 inches and 3 inches, and the dis tance between their centres is 15 inches. Find the lengths of their com mon tangents. - ^262. Find the segments of a line 10 inches long divided in exirerne and mean ratio. 263. The sides of a triangle are 4, 5, 6. Is the largest angle acute, right, or obtuse ? PROBLEMS. 264. To divide one side of a given triangle into segments proportional to the adjacent sides. ( 313.) 265. To produce a line AB to a point C so that AB -. AC= 3 : 5. 266. To find in one side of a given triangle a point whose distances from the other sides shall be to each other in a given ratio. 267. Given an obtuse triangle ; to draw a line from the vertex of the obtuse angle to the opposite side which shall be a mean proportional between the segments of that side. 268. Through a given point P within a given circle to draw a chord ABso that AP:5P=2:3. 269. To draw through a given point P in the arc subtended by a chord AB a chord which shall be bisected by AB. 270. To draw through a point P, exterior to a given circle, a secant PAB so that PA: AB = 4: 3. 271. To draw through a point P, exterior to a given circle, a secant PAB so that ZB 3 = PA X PB. 272. To find a point P in the arc subtended by a given chord AB so EXEECISES. 179 273. To draw through one of the points of intersection of two circles a secant so that the two chords that are formed shall be to each other in the ratio of 3 : 5. 274. To divide a line into three parts proportional to 2, f, $. 275. Having given the greater segment of a line divided in extreme and mean ratio, to construct the line. 276. To construct a circle which shall pass through two given points and touch a given straight line. 277. To construct a circle which shall pass through a given point and touch two given straight lines. 278. To inscribe a square in a semicircle. 279. To inscribe a square in a given triangle. HINT. Suppose the problem solved, and DEFQ the inscribed square. Draw CM II to AB, and let AF produced meet CM in M. Draw Gffand MN to AB, and produce AB to meet MN at N. The & ACM, AOF are similar; also the A AMN, AFE are similar. By these triangles show that the figure CMNH is a square. By construct ing this square, the point F can be found. A. D H E B 280. To inscribe in a given triangle a rectangle similar to a given rectangle. 281. To inscribe in a circle a triangle similar to a given triangle. 282. To inscribe in a given semicircle a rectangle similar to a given rectangle. 283. To circumscribe about a circle a triangle similar to a given triangle. 284. To construct the expression, x = ^^ ; that is ~ x - de d e 285. To construct two straight lines, having given their sum and their ratio. 286. To construct two straight lines, having given their difference and their ratio. 287. Having given two circles, with cef*W6s and , and a point A in their plane, to draw through the point A a straight line, meeting the circumferences at B and C, so that AB : AC= 1 : 2. HINT. Suppose the problem solved, join OA and produce it to D, making OA -. AD - I ; 2. Join DC; & OA 5, ADC are similar. BOOK IV. AREAS OF POLYGONS. 358, The area of a surface is the numerical measure of the surface referred to the unit of surface. The unit of surface is a square whose side is a unit of length ; as the square inch, the square foot, etc. 359, Equivalent figures are figures having equal areas. PKOPOSITION I. THEOREM. 360, The areas of two rectangles having equal alti tudes are to each other as their bases. D B F E O Let the two rectangles be AC and AF, having the same altitude AD. Toprme ^t^=4|- lect.AF AE Proof, CASE I. When AB and AE are commensurable. Suppose AB and AE have a common measure, as AO, which is contained in AB seven times and in AE four times. AB ^7 AE 4 Apply this measure to AB and AE, and at the several points of division erect Js. The rect. A C will be divided into seven rectangles, and the rect. AF will be divided into four rectangles. Then (1) AREAS OF POLYGONS. These rectangles are all equal. Hence From (1) and (2) AE 181 186 l Ax . l CASE II. When AB and AE are incommensurable. Dl B F Divide AB into any number of equal parts, and apply one of them to AE as often as it will be contained in AE. Since AB and AE are incommensurable, a certain number of these parts will extend from A to a point K, leaving a remainder KE less than one of the parts. Draw KH II to EF. Since AB and AK &TQ commensurable, Case I. These ratios continue equal, as the unit of measure is indefi nitely diminished, and approach indefinitely the limiting ratios rect.AF , AE .. , and -_ respectively. AB * * . rect. AF AE (if two variables are constantly equal, and each approaches a limit, the limits are equal). Q E D 361, COB. The areas of two rectangles having equal bases are to each other as their altitudes. For AB and AE may be con sidered as the altitudes, AD and AD as the bases. NOTE. In propositions relating to areas, the words "rectangle," " triangle," etc., are often used for " area of rectangle," " area of tri angle," etc. 182 PLANE GEOMETRY. BOOK IV. PROPOSITION II. THEOREM. 362, The areas of two rectangles are to each other as the products of their bases l)y their altitudes. b b b Let R and E be two rectangles, having for their bases b and b , and for their altitudes a and a . m E aXb To prove = -- E 1 a X b 1 Proof, Construct the rectangle S, with its base the same as that of E, and its altitude the same as that of E f . E a Then 8 a 1 361 and rectangles having equal bases are to each other as their altitudes) ; Sb 360 (rectangles having equal altitudes are to each other as their bases) By multiplying these two equalities, E aXb E a X b Q. E. O. Ex. 288. Find the ratio of a rectangular lawn 72 yards by 49 yards to a grass turf 18 inches by 14 inches. Ex. 289. Find the ratio of a rectangular courtyard 18 yards by 15 yards to a flagstone 31 inches by 18 inches. Ex. 290. A square and a rectangle have the same perimeter, 100 yards. The length of the rectangle is 4 times its breadth. Compare their areas. Ex. 291. On a certain map the linear scale is 1 inch to 5 miles. How many acres are represented on this map by a square the perimeter of which is 1 inch ? AREAS OF POLYGONS. 183 PROPOSITION III. THEOREM. 363, The area of a rectangle is equal to the product of its base and altitude. a Let E be the rectangle, b the base, and a the alti tude; and let U be a square whose side is equal to the linear unit. To prove the area of R = a X b. Jii, a X o i - = aXb, IXl c (two rectangles are to each other as the product of their bases and altitudes). But = the area of R. .. the area of R = a X b. 358 Q. E. D. 364. SCHOLIUM. When the base and altitude each contain the linear unit an integral number of times, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of measure. Thus, if the base contain seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of measure ; and the area of the figure equals 7x4 units of surface. 184 PLANE GEOMETRY. BOOK IV. PROPOSITION IV. THEOREM. 365, The area of a parallelogram is equal to the product of its base and altitude. BE C F A b D A b D Let AEFD be a parallelogram, AD its base, and CD its altitude. To prove the area of the H AEFD = ADx CD. Proof, From A draw AB II to DC to meet FE produced. Then the figure ABCD will be a rectangle, with the same base and altitude as the 7 AEFD. In the rt. A ABE and DCF AB = CD and A E = DF, 179 (being opposite sides of a ZZ7). .-.AAJBE=ADCF, 161 (two rt. A are equal when tlic, hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other). Take away the A DCF, and we have left the rect. ABCD. Take away the A ABE, and we have left the O AEFD. .-. rect. ABCD =c= O AEFD. Ax. 3 But the area of the rect. ABCD = axb, 363 .-. the area of the O AEFD = axb. Ax. 1 Q. E. D. 366, COR. 1. Parallelograms having equal bases and equal altitudes are equivalent. 367, COR. 2. Parallelograms having equal bases are to each other as their altitudes ; parallelograms having equal altitudes are to each other as their bases ; any two parallelograms are to each other as the products of their bases by their altitudes. AREAS OF POLYGONS. 185 PROPOSITION V. THEOREM. 368, The area of a triangle is equal to one-half the product of its base by its altitude. Let ABC be a triangle, AB its base, and DC its altitude. To prove the area of the &ABC=%ABx DC. Proof, From C draw CH II to BA. From A draw AH II to BO. The figure ABCHis a parallelogram, (hawing its opposite sides parallel), and AC is its diagonal. 168 178 (the diagonal of a CU divides it into two equal A). The area of the O ABCH is equal to the product of its base, by its altitude. 365 Therefore the area of one-half the O, that is, the area of the A ABC, is equal to one-half the product of its base by its altitude. Hence, the area of the A ABC= %AB X DC. Q. E. D. 369, COR. 1 . Triangles having equal bases and equal alti tudes are equivalent. 370, COR. 2. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are fo each other as their bases ; any two triangles are to each other as the products of their bases by their altitudes. 186 PLANE GEOMETRY. BOOK IV. PROPOSITION VI. THEOREM. 371, The area of a trapezoid is equal to one-half the sum of the parallel sides multiplied by the alti tude. TT EJ) .A F b B Let ABCH be a trapezoid, and EF the altitude, To prove area of ABCH = (J3~C+ AB) EF Proof, Draw the diagonal AC. Then the area of the A ABC= % (AB X EF), 368 and the area of the A AHC= \ (HUX EF). By adding, area of ABCH= \ (AB + HO) EF. Q . E . D . 372, COR. The area of a trapezoid is equal to the product of the median by the altitude. For, by 191, OP is equal to l(HG+AB)\ and hence the area of ABCH= OP X EF. 373, SCHOLIUM. The area of an irregular polygon may be found by dividing the poly gon into triangles, and by finding the area of each of these triangles separately. But the method generally employed in practice is to draw the longest diagonal, and to let fall perpendiculars upon this diagonal from the other angular points of the polygon. The polygon is thus divided into right triangles and trape- zoids ; the sum of the areas of these figures will be the area of the polygon. D AEEAS OF POLYGONS. 187 PROPOSITION VII. THEOREM. 374, The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. Let the triangles ABC and ADE have the common angle A. A ABC ABxAC To prove Proof, Now A ADE ADxAE Draw BE. and A ABC _ A ABE A ABE AC ~AE AB A ADE AD (& having the same altitude are to each other as their bases). By multiplying these equalities, AABC_^ABxAO A ADE AD x AE 370 Q. E. D. Ex. 292. The areas of two triangles which have an angle of the one supplementary to an angle of the other are to each other as the products of the sides including the supplementary angles. 188 PLANE GEOMETRY. BOOK IV. COMPARISON OF POLYGONS. PROPOSITION VIII. THEOREM. 375, The areas of two similar triangles are to each other as the squares of any two homologous sides. A o A Let the two triangles be ACB and A O B . A ACB To prove A A C B 1 A B Draw the perpendiculars CO and C O 1 . T , A ACB _ ABx CO AB CO - o 7n A^L^ -^ XW = :Z^ X W (fo;o A are to each other as the products of their bases by their altitudes). But (the homologous altitudes of similar A have the same ratio as their homolo gous bases). Substitute, in the above equality, for its equal ; then A ACB AB X C O AB Al? AA C B A B A B Q. E. O. COMPARISON OF POLYGONS. 189 PROPOSITION IX. THEOREM. 376, The areas of two similar polygons are to each other as the squares of any two homologous sides. E C B C Let S and 8 denote the areas of the two similar polygons ABC etc., and A B C etc. To prove 8 : /S" = A$ : 1*1?. Proof, By drawing all the diagonals from the homologous vertices E and E\ the two similar polygons are divided into triangles similar and similarly placed. 332 AB* A ABE fBE*\ ABOE AA J? AJJ C E A CPE A C D E 1 375 (similar & are to each other as the squares of any two homologous sides). A ABE ABCE A ODE 1 AA B E* AB O E A O D E 1 A ABE+ BCE+ ODE _ A ABE = AB 2 E + B C E + C D E ~ AA B E 1 ^ B 1 * (in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent). .: 8:8 = 13*: AW 1 . Q . E . D . 377. COR. 1. The areas of two similar polygons are to each other as the squares of any two homologous lines. 378, Con. 2. The homologous sides of two similar polygons have the same ratio as the square roots of their areas. 190 PLANE GEOMETRY. BOOK IV. PROPOSITION X. THEOREM. 379. The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides. Let BE, CH, AF, be squares on the three sides of the right triangle ABC. To prove C* =c= AS -f AC*. Proof, Through A draw AL\\io CE, and draw AD and FO. Since A BAG, BAG, and CAH are rt. A, GAG and BAH are straight lines. Since BD = BG, being sides of the same square, and BA = BF, for the same reason, and since Z ABD = Z FBC, each being the sum of a rt. Z. and the Z. ABC, the A ABD = A FBC. 150 Now the rectangle BL is double the A ABD, (having the same base BD, and the same altitude, the distance between the Us AL and D), and the square-4-Fis double the A FBC, (having the same base FB, and the same altitude, the distance between the \\s FB and OG). Hence the rectangle BL is equivalent to the square AF. In like manner, by joining AE and BK, it may be proved that the rectangle CL is equivalent to the square CH. Therefore the square BE, which is the sum of the rectangles BL and CL, is equivalent to the sum of the squares CH and ^ 380, COR. The square on either leg of a right triangle is equivalent to the difference of the squares on the hypotenuse and the other leg. COMPAEISON OF POLYGONS. 191 Ex. 293. The square constructed upon the sum of two straight lines is equivalent to the sum of the squares constructed upon these two lines, increased by twice the rectangle of these lines. Let AB and BC be the two straight lines, and AC their sum. Con struct the squares ACGK and ABED upon AC and AB respectively. Prolong BE and DE until they meet KG and CG respectively. Then we have the square EFGH, with sides each equal to BC. Hence, the square ACGK is the sum of the squares ABED I) and EFGH, and the rectangles DEHK and BCFE, the dimensions of which are equal to AB and BC. ** B C H G H K Ex. 294. The square constructed upon the difference of two straight lines is equivalent to the sum of the squares constructed upon these two lines, diminished by twice the rectangle of these lines. Let A B and A C be the two straight lines, and BC their difference. Construct the square ABFG upon AB, the square ACKHnyon. AC, and the square BEDC upon BC (as shown in the figure). Prolong ED until it meets AG in L. The dimensions of the rectangles LEFG and HKDL are AB and AC, and the square BODE is evidently the difference between the whole figure and the sum of these rectangles ; that is, the square constructed G F upon BC is equivalent to the sum of the squares constructed upon AB and AC diminished by twice the rectangle of AB and AC. Ex. 295. The difference between the squares constructed upon two straight lines is equivalent to the rectangle of the sum and difference of these lines. Let ABDE and BCGF be the squares constructed upon the two D E straight lines AB and BC. The difference between these squares is the polygon ACGFDE, which poly gon, by prolonging CG to H, is seen to be composed of _, the rectangles ACHE and GFDH. Prolong AE and CHto Jand ^"respectively, making Eland HK each equal to BC, and draw IK. The rectangles GFDH and EHKI are equal. The difference between the squares ABDE and BCGF is then equivalent to the rectangle ACKI, which has for dimensions AI = AB- BC. K H F A C B AB + BC, and EH 192 PLANE GEOMETRY. BOOK IV. PROBLEMS OF CONSTRUCTION. PROPOSITION XL PROBLEM. 381, To construct a square equivalent to the sum of two given squares. Let R and R be two given squares. To construct a square equivalent to It -f- R. Construction, Construct the rt. Z A. Take AC equal to a side of R 1 , AB equal to a side of R ; and draw BC. Construct the square S, having each of its sides equal to BC. S is the square required. Proof, + (the square on the hypotenuse of a rt. A squares on the two . :.& + 379 equivalent to the sum of the s). Q.E.F. Ex. 296. If the perimeter of a rectangle is 72 feet, and the length is " to twice the width, find the area. Ex. 297. How many tiles 9 inches long and 4 inches wide will be required to pave a path 8 feet wide surrounding a rectangular court 120 feet long and 36 feet wide ? Ex. 298. The bases of a trapezoid are 16 feet and 10 feet; each leg is equal to 5 feet. Find the area of the trapezoid. PEOBLEMS OF CONSTRUCTION. 193 PROPOSITION XII. PROBLEM. 382, To construct a square equivalent to the differ ence of two given squares. K i ._ %/_ \ -J- """ ~?7*,~~~-A i s ! L J Let R be the smaller square and R 1 the larger. To construct a square equivalent to R 1 JR. Construction, Construct the rt. Z A. Take AB equal to a side of R. From JB as a centre, with a radius equal to a side of R\ describe an arc cutting the line AX &\> O. Construct the square S, having each of its sides equal to A 0. S is the square required. Proof. AC 2 =0= W - A3 2 , 380 (the square on either leg of a rt. A is equivalent to the difference of the squares on the hypotenuse and the other leg). == R 1 - R. Q. E. F. Ex. 299. Construct a square equivalent to the sum of two squares whose sides are 3 inches and 4 inches. Ex. 300. Construct a square equivalent to the difference of two whose sides are 2 inches and 2 inches. Ex. 301. Find the side of a square equivalent to the sum of two squares whose sides are 24 feet and 32 feet. Ex. 302. Find the side of a square equivalent to the difference of two squares whose sides are 24 feet and 40 feet. Ex. 303. A rhombus contains 100 square feet, and the length of one diagonal is 10 feet. Find the length of the other diagonal. 194 PLANE GEOMETRY. BOOK TV. PROPOSITION XIII. PROBLEM. 383, To construct a square equivalent to the sum of any number of given squares. Let m, n, o, p, r be sides of the given squares. To construct a square =c= m 2 -f n* -f o 2 -f p 2 -f r 2 . Construction, Take AB = m. Draw AC = n and _L to AB at A, and draw BG. Draw GE = o and J_ to BG at O, and draw BE. Draw EF =p and _L to BE at E, and draw BF. Draw FH= r and J_ to BF at ^, and draw BH. The square constructed on BH\s the square required. Proof, ^ FIT + EF + =0= ^F 2 + EF* + ^C 2 + * FH* + EO -^r EF + CA* + AB\ 379 (the sum of the squares on the two leas of a rt. A is equivalent to the square on the hypotenuse). That is, BH* =c= m 2 + n* -f- o 2 -f p 2 + r. Q.E.F, PEOBLEMS OF CONSTRUCTION. 195 PROPOSITION XIV. PROBLEM. 384, To construct a polygon similar to two given similar polygons and equivalent to their sum. A! " B" P ....... "Li Let E and R be two similar polygons, and AB and A B two homologous sides. To construct a similar polygon equivalent to JR-\- IV. Construction. Construct the rt. Z P. Take PH= A , and PO = AB. Draw OH, and take A"" = OH. Upon A"H", homologous to AB, construct It" similar to E. Then J?" is the polygon required. Proof, PO* \r and fr=^ B76 &" A""* (similar polygons are to each other as the squares of their homologous sides). By addition, ^^ = ^ + ^ =1 . -B" A"" . .Sf-o-K + Sf. :** 19G PLANE GEOMETRY. BOOK IV. PROPOSITION XV. PROBLEM. 385, To construct a polygon similar to two given similar polygons and equivalent to their difference. /* A! B A 13 A" B" P O Let R and R f be two similar polygons, and AS and A B two homologous sides. To construct a similar polygon equivalent to R* H. Construction, Construct the rt. Z P, and take PO = AB. From as a centre, with, a radius equal to A B\ describe an arc cutting PX at IT, and join OH. Take A"" = PIT, and on A"J3", homologous to AB, construct P" similar to R. Then R" is the polygon required. Proof, E (similar polygons are to each other as the squares of their homologous sides). By subtraction, 72 - Q . E . F . PROBLEMS OF CONSTRUCTION. 197 PROPOSITION XVI. PROBLEM. 386, To construct a triangle equivalent to a given polygon. C I A E F Let ABGDHE be the given polygon. To construct a triangle equivalent to the given polygon. Construction, From D draw DE, and from J7draw EF II to DE. Produce AEto meet HF at F, and draw DF. Again, draw CF, and draw DK II to CF to meet AF pro duced at K, and draw CK. In like manner continue to reduce the number of sides of the polygon until we obtain the A CIK. Proof, The polygon ABCDF has one side less than the polygon ABGDHE, but the two are equivalent. For the part ABODE is common, and the A DEF^ A DEE, 369 (for the base DE is common, and their vertices F and H are in the line FH II to the base}. The polygon ABCK has one side less than the polygon ABCDF, but the two are equivalent. For the part ABCF is common, and the A CFK^ A CFD, 369 (for the base CF is common, and their vertices K and D are in the line KD II to the base}. In like manner the A CIK^= ABCK. Q. E. F, 198 PLANE GEOMETRY. BOOK IV. PROPOSITION XVII. PROBLEM. 387, To construct a square which shall have a given ratio to a given square. / / m IB n m "V.. R be the given square, and the given ratio. ___ ^ m To construct a square which shall be to R as n is to m. Construction, Take AB equal to a side of R, and draw Ay, making any acute angle with AB. On Ay take AE=m, EF n, and join EB. Draw FG \\ to EB to meet AB produced at C. On A C as a diameter describe a semicircle. At B erect the J. BD, meeting the semicircumference at D. Then BD is a side of the square required. Proof. Denote AB by a, BO by b, and BD by x. Now a : x x : b ; that is, x 1 = ab. 337 Hence, a 2 will have the same ratio to x* and to ab. Therefore 2 : x 2 = a 2 : ab = a : b. :b = m:n, 309 (a straight line drawn through two sides of a A, parallel to the third side, divides those sides proportionally). Therefore a 2 : x 2 = m : n. By inversion, x* : a 2 = n : m. Hence the square on BD will have the same ratio to R as n has to m. Q. E. F. PEOBLEMS OF CONSTRUCTION. 199 PROPOSITION XVIII. PROBLEM. 388, To construct a polygon similar to a given poly gon and having a given ratio to it. x ^x / \ m / ( / \ / V. _ ./ A n Let R he the given polygon and the given ratio. To construct a polygon similar to H, which shall be to R as n is to m. Construction, Find a line A B f , such that the square con structed upon it shall be to the square constructed upon AB as n is to m. 387 Upon A B as a side homologous to AB, construct the poly gon S similar to R. Then S is the polygon required. Proof, S : R = A B 2 : AB\ 376 (similar polygons are to each other as the squares of their homologous sides). But A B 1 : AB =n:m. Cons. Therefore S : R = n : m. Q. E. F. Ex. 304. Find the area of a right triangle if the length of the hypote nuse is 17 feet, and the length of one leg is 8 feet. Ex. 305. Compare the altitudes of two equivalent triangles, if the base of one is three times that of the other. Ex. 306. The bases of a trapezoid are 8 feet and 10 feet, and the alti tude is 6 feet. Find the base of an equivalent rectangle having an equal altitude. 200 PLANE GEOMETRY. BOOK IV. PROPOSITION XIX. PROBLEM. 389. To construct a square equivalent to a given ^parallelogram. f) 1 "x * : I \ 1 *" > r o Let ABCD be a parallelogram, b its base, and a its altitude. To construct a square equivalent to the O ABCD. Construction, Upon the line J/lTtake MN= a, and N0 = b. Upon MO as a diameter, describe a semicircle. At N erect NP -L to MO, to meet the circumference at P. Then the square R, constructed upon a line equal to NP, is equivalent to the O ABCD. Proof, MN : NP = NP : NO, 337 (a JL let fall from any point of a circumference to the diameter is a mean proportional between the segments of the diameter). That is, 390, COR. 1. given triangle, tween the base 391, COR. 2 <7z i>m polygon, triangle, and triangle ABCD. A square may be constructed equivalent to a by taking for its side a mean proportional be- and one-half the altitude of the triangle. A square may be constructed equivalent to a by first reducing the polygon to an equivalent then constructing a square equivalent to the PROBLEMS OF CONSTRUCTION. 201 PROPOSITION XX. PROBLEM. 392, To construct a parallelogram equivalent to a given square, and, having the sum of its base and altitude equal to a given line. R ip- ,1 \ 1 J JV o Let R be the given, square, and let the sum of the base and altitude of the required parallelogram be equal to the given line MN. To construct a O equivalent to R, with the sum of its base and altitude equal to MN. Construction, Upon MN&s a diameter, describe a semicircle. At M erect a J_ MP, equal to a side of the given square R. Draw PQ II to MN t cutting the circumference at 8. Draw SO A. to MN. Any.O having CM for its altitude and ON for its base is equivalent to R. Proof, 80= PM. 100, 180 But MC:SC=SC: ON, 337 (a J_ let fall from any point in the circumference lo the diameter is a mean proportional between ilic, segments of the diameter). Then ON. Q. E. F. NOTE. This problem may be stated : To construct two straight lines the sum and product of which are known. 202 PLANE GEOMETKY. BOOK IV. PROPOSITION XXI. PROBLEM. 393, To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line. s \ ~ 7 Afp- JN / R Let R be the given, square, and let the difference of the base and altitude of the required parallelogram be equal to the given line MN. To construct a O equivalent to R } with the difference of the base and altitude equal to MN. Construction. Upon the given line JOTas a diameter, describe a circle. From M draw MS, tangent to the O, and equal to a side of the given square R. Through the centre of the O draw SB intersecting the cir cumference at (7 and B. Then any O, as R 1 , having SB for its base and SC for its altitude, is equivalent to R. Proof, SB : SM= SM : SO, 348 (if from a point without aQa secant and a tangent are drawn, the tangent is a mean proportional between the whole secant and the part without the O). Then SM 1 ; - SB X SC, and the -difference between SB and SO is the diameter of the O, that is, MN. Q.E.F. NOTE. This problem may be stated : To construct two straight lines the difference and product of which are known. PKOBLEMS OF CONSTRUCTION. 203 PROPOSITION XXII. PROBLEM. 394, To construct a polygon similar to a given poly gon P, and equivalent to a given polygon Q. fr m- W A ~ D Let P and Q be two polygons, and AB a side of P. To construct a polygon similar to P and equivalent to Q. Construction, Find squares equivalent to P and Q, 391 and let m and n respectively denote their sides. Find A B , a fourth proportional to m, n, and AB. 351 Upon A B , homologous to AB, construct P similar to P. Then P 1 is the polygon required. Proof. m : n = AB : A B\ Cons. ,2 . ^2 _ But But . . m P=c=m 2 , and P: Q=--m 2 :tf--= Cons. 376 (similar polygons are to each other as the squares of their homologous sides). :.P:Q=P:P. Ax. 1 /. P is equivalent to Q, and is similar to P by construction. Q. E. F. 204 PLANE GEOMETRY. BOOK IV. PROBLEMS OF COMPUTATION. Ex. 307. To find the area of an equilateral triangle in terms of its side. Denote the side by a, the altitude by h, and the area by S. _5> O ~2 Then But 2 axh 2 \/3~ Ex. 308. To find the area of a triangle in terms of its sides. By Ex. 219, h = f V7(s - a) (s - b) (s - c). Hence, -X- Vs(s-a)(s-b)(s-c) 2 o Vs (s a) (s - b) (s c). Ex. 309. To find the area of a triangle in terms of the radius of the circumscribing circle. Tf R denote the radius of the circumscribing circle, a*nd h the altitude of the triangle, we have, by Ex. 222, A Multiply by a, and we have But a X h = 2 S. abc NOTE. The radius of the circumscribing circle is equal to 48 EXERCISES. 205 THEOREMS. 310. In a right triangle the product of the legs is equal to the product of the hypotenuse and the perpendicular drawn to the hypotenuse from the vertex of the right angle. 311. If ABC is a right triangle, C the_vertex_ of the right angle, BD a line cutting AC in D, then BD* + AC* = AB* + ~ rr ^ 312. Upon the sides of a right triangle as homologous sides three similar polygons are constructed. Prove that the polygon upon the hypotenuse is equivalent to the sum of the polygons upon the legs. 313. Two isosceles triangles are equivalent if their legs are equal each to each, and the altitude of one is equal to half the base of the other. .>\ 314. The area of a circumscribed polygon is equal to half the product of its perimeter by the radius of the inscribed circle. 315. Two parallelograms are equa^it-jfa^ adjacent sides of the one are equal respectively to two adjacent sidjfcf the other, and the included angles are supplementary. ^**^ ^m$\ Hf~ ^"^ \\ 316. Every straight line drawn through the, centre of a parallelogram divides it into two equal parts. 317. If the middle points of two adjacent sides of a parallelogram are joined, a triangle is formed which is equivalent to one-eighth of the entire parallelogram. 318. If any point within a parallelogram is joined to the four vertices, the sum of either pair of triangles having parallel bases is equivalent to one-half the parallelogram. 319. The line which joins the middle points of the bases of a trape- zoid divides the trapezoid into two equivalent parts. - 320. The area of a trapezoid is equal to the product of one of the legs and the distance from this leg to the middle point of the other leg. 321. The lines joining the middle point of the diagonal of a quadri lateral to the opposite vertices divide the quadrilateral into two equiva lent parts. \, 322. The figure whose vertices are the middle points of the sides of any quadrilateral is equivalent to one-half of the quadrilateral. 323. ABC is a triangle, M the middle point of AB, P any point in AB between A and M. If MD is drawn parallel to PC, and meeting BC at D, the triangle BPD is equivalent to one-half the triangle ABC. 206 * PLANE GEOMETRY. BOOK IV. NUMERICAL EXERCISES. 324. Find the area of a rhombus, if the sum of its diagonals is 12 feet, and their ratio is 3 : 5. 4, ^fl^T 325. Find the area of an isosceles right triangle if the hypotenuse is 20 feet I k * 326. In a right triangle, the hypotenuse is 13 feet, one leg is 5 feet. Find the area. : 327. Find the area of an isosceles triangle if the base = b, and leg = c. 328. Find the area of an equilateral triangle if one side = 8. /(rU 329. Find the area of an equilateral triangle if the altitude = h. Y> 330. A house is 40 feet long, 30 feet wide, 25 feet high to the eaves, and 35 feet high to the ridge-pole. Find the number of squarefeet in f its entire exterior surface. ^S 331. The sides of a right triangle are as 3 : 4 : 5. The altSrae upon the hypotenuse is 12 feet. Find the area. 332. Find the area of a right triangle if one leg = a, and the altitude upon the hypotenuse = h. *"^ "/ V-. 333. Find the area of a triangle if the lengths of the sides are 104 feet, 111 feet, and 175 feet. 7 334. The. area of a trapezoid is 700 square feet. The bases are 30 feet and 40 feet respectively. Find the distance between the bases. 335. ABCD is a trapezium; ^i=S7 feet, BC= 119 feet, CD = 41 feet, DA = 169 feet, AC=* 200 feet. Find the area. 336. What is the area of a quadrilateral circumscribed about a circle whose radius is 25 feet, if the perimeter of the quadrilateral is 400 feet? What is the area of a hexagon having an equal perimeter and circum scribed about the same circle ? 337. The base of a triangle is 15 feet, and its altitude is 8 feet. Find the perimeter of an equivalent rhombus if the altitude is 6 feet. / c 338. Upon the diagonal of a rectangle 24 feet by 10 feet a triangle equivalent to the rectangle is constructed. What is its altitude? 339. Find the side of a square equivalent to a trapezoid whose bases are 56 feet and 44 feet, and each leg is 10 feet. | $ 340. Through a point Pin the side AB of a triangle ABC, a line is drawn parallel to BC, and so as to divide the triangle into two equiva lent parts. Find the value of AP in terms of AB. EXERCISES. 207 341. What part of a parallelogram is the triangle cut off by a line drawn from one vertex to the middle point of one of the opposite sides ? 342. In two similar polygons, two homologous sides are 15 feet and 25 feet. The area of the first polygon is 450 square feet. Find the area of the other polygon. 343. The base of a triangle is 32 feet, its altitude 20 feet. What is the area of the triangle cut off by drawing a line parallel to the base and at a distance of 15 feet from the base ? 344. The sides of two equilateral triangles are 3 feet and 4 feet. Find the side of an equilateral triangle equivalent to their sum. 345. If the side of one equilateral triangle is equal to the altitude of another, what is the ratio of their areas ? 346. The sides of a triangle are 10 feet, 17 feet, and 21 feet. Find the al^K of the parts, into which the triangle is divided by bisecting the angle^^ed by the first two sides. 347. In a trapejaoid, one base is 10 feet, the altitude is 4 feet, the area is 32 square feet. v Find the length of a line drawn between the legs parallel to the base and distant 1 foot from it. 348. If the altitude A of a triangle is increased by a length m, how much must be taken from the base a in order that the area may remain the same ? 349. Find the area of a right triangle, having given the segments p, q, into which the hypotenuse is divided by a perpendicular drawn to the hypotenuse from the vertex of the right angle. PROBLEMS. 350. To construct a triangle equivalent to a given triangle, and having one side equal to a given. length I. 351. To transform a triangle into an equivalent right triangle. 352. To transform a triangle into an equivalent isosceles triangle. 353. To transform a triangle ABO into an equivalent triangle, hav ing one side equal to a given length I, and one angle equal to angle BAG. HINTS. Upon AB (produced if necessary), take AD = I, draw BE II to CD, and meeting AC (produced if necessary) at E\ A BED^&BEC. 354. To transform a given triangle into an equivalent right triangle, having one leg equal to a given length. 208 PLANE GEOMETRY. BOOK IV. 355. To transform a given triangle into an equivalent right triangle, having the hypotenuse equal to a given length. 356. To transform a given triangle into an equivalent isosceles tri angle, having the base equal to a given length. To construct a triangle equivalent to : 357. The sum of two given triangles. 358. The difference of two given triangles. 359. To transform a given triangle into an equivalent equilateral triangle. To transform a parallelogram into : 360. A parallelogram having one side equal to a given length. 361. A parallelogram having one angle equal to a given angle. 362. A rectangle having a given altitude. To transform a square into : y^ 363. An equilateral triangle. 364. A right triangle having one leg equal to a given length. 365. A rectangle having one side equal to a given length. To construct a square equivalent to : 366. Five-eighths of a given square. >867. Three-fifths of a given pentagon. ""368. To draw a line through the vertex of a given triangle so. as to divide the triangle into two triangles which shall be to each other as 2:3. 5* 369. To divide a given triangle into two equivalent parts hy drawing a line through a given point P in one of the sides. 370. To find a point within a triangle, such that the lines joining this point to the vertices shall divide the triangle into three equivalent parts. """371. To divide a given triangle into two equivalent parts by drawing a line parallel to one of the sides. 372. To divide a given triangle into two equivalent parts by drawing a line perpendicular to one of the sides. ^>373. To divide a given parallelogram into two equivalent parts by drawing a line through a given point in one of the sides. 374. To divide a given trapezoid into two equivalent parts by draw ing a line parallel to the bases. "^375. To divide a given tranezoid into two equivalent parts by draw ing a line through a given point iu one of the bases. BOOK V. REGULAR POLYGONS AND CIRCLES. 395, A regular polygon is a polygon which is equilateral and equiangular ; as, for example, the equilateral triangle, and the square. PROPOSITION I. THEOREM. 396, An equilateral polygon inscribed in a circle is a regular polygon. c Let ABC etc., be an equilateral polygon inscribed in a circle. To prove t fie polygon ABC etc., regular. Proof, The arcs AE, EC, CD, etc., are equal, 230 (in the same O, equal chords subtend equal arcs). Hence arcs ABO, BCD, etc., are equal, Ax. 6 and the A A, B, C, etc., are equal, (being inscribed in equal segments). Therefore the polygon ABC, etc., is a regular polygon, being equilateral and equiangular. a E . Dt 210 PLANE GEOMETRY. BOOK V. PROPOSITION II. THEOREM. 397, A circle may be circumscribed about, and a circle may be inscribed in, any regular polygon. D Let ABODE be a regular polygon. I. To prove that a circle may be circumscribed about ABODE. Proof, Let be the centre of the circle passing through A, B, a Join OA, OB, 00, and OD. Since the polygon is equiangular, and the A OBCis isosceles, and By subtraction, 154 Z OB A = Z. OCD. Hence in the A OB A and OCD the Z OB A = Z OCD, the radius OB = the radius OC, and AB=OD. 395 .-.A OAB = A OCD, 150 (having two sides and the included Z of the one equal to two sides and the included Z of the other). , .OA = OD. Therefore the circle passing through A, B, and C, also passes through D. REGULAR POLYGONS AND CIRCLES. 211 In like manner it may be proved that the circle passing through B t C, and D, also passes through E\ and so on through all the vertices in succession. Therefore a circle described from as a centre, and with a radius OA, will be circumscribed about the polygon. II. To prove that a circle may be inscribed in ABODE. Proof, Since the sides of the regular polygon are equal chords of the circumscribed circle, they are equally distant from the centre. 236 Therefore a circle described from as a centre, and with the distance from to a side of the polygon as a radius, will be inscribed in the polygon. Q. E.D. 398, The radius of the circumscribed circle, OA, is called the radius of the polygon. 399, The radius of the inscribed circle, OF, is called the apothem of the poly yon. 400, The common centre of the circumscribed and in scribed circles is called the centre of the polygon. 401, The angle between radii drawn to the extremities of any side, as angle AOB, is called the angle at the centre of the polygon. By joining the centre to the vertices of a regular polygon, the polygon can be decomposed into as many equal isosceles triangles as it has sides. Therefore, 402, COR. 1. The angle at the centre of a regular polygon is equal to four right angles divided by the number of sides of the polygon. 403, Con. 2. The radius drawn to any vertex of a regular polygon bisects the angle at the vertex. 404, COR. 3. The interior angle of a regular polygon is the supplement of the angle at the centre. For the Z ABC = 2 Z ABO = Z ABO + /.BAO. Hence the Z ABC is the supplement of the Z AOB. 212 PLANE GEOMETRY. BOOK V. PROPOSITION III. THEOREM. 405, If the circumference of a circle is divided into any number of equal parts, the chords joining the successive points of division form a regular inscribed polygon, and the tangents drawn at the points of division form a regular circumscribed polygon. I D H F Let the circumference be divided into equal arcs, AJ3, BC, CD, etc., be chords, FBG, GCH, etc., be tangents. I. To prove that ABODE is a regular polygon. Proof, The sides AB, EC, CD, etc., are equal, 230 (in the same O equal arcs are subtended by equal chords). Therefore the polygon is regular, 396 (an equilateral polygon inscribed in a O is regular). II. To prove that the polygon FGIIIKis a regular polygon. Proof, In the A AFB, BGC, CUD, etc. AB - BC= CD, etc. 395 Also, Z BAF= Z ABF= Z CBG = Z BOO, etc., 269 (being measured by halves of equal arcs). Therefore the triangles are all equal isosceles triangles. Hence Z.F=/.G = Z.H, etc. Also, FB = BG=GC = CH, etc. Therefore FG = GH, etc. /. FGH1K is a regular polygon. 395 Q. E. D. 406. COR. 1. Tangents to a circumference at the vertices of a regular inscribed polygon form a regular circumscribed poly gon of the same number oj REGULAR POLYGONS AND CIRCLES. 213 407, COR. 2. If a regular polygon is inscribed in a circle, the tangents drawn at the middle points of the arcs subtended by the sides of the polygon form a circumscribed regular polygon, whose sides are parallel to the sides of the inscribed polygon and whose vertices lie on the radii (prolonged) of the inscribed polygon. For any two cor responding sides, as AB and A B , perpendicular to OM, are parallel, and the tangents MB and NB , intersecting at a point equidistant from OM &nd 0_ZV( 246), intersect upon the bisector of the Z. MON( 163) ; that is, upon the radius OB. 408, COR. 3. If the vertices of a regular inscribed polygon are joined to the middle points of the arcs sub tended by the sides of the polygon, the joining lines form a regular inscribed polygon of double the number of sides. 409, COR. 4. If tangents are drawn at the middle points of the arcs between adjacent points of contact of the sides of a regular cir cumscribed polygon, a regular circumscribed polygon of double the number of sides is formed. D K 410, SCHOLIUM. The perimeter of an inscribed polygon is less than the perimeter of the inscribed polygon of double the number of sides; for each pair of sides of the second polygon is greater than the side of the first polygon which they replace ( 137). The perimeter of a circumscribed polygon is greater than the perimeter of the circumscribed polygon of double the num ber of sides ; for every alternate side FG, HI, etc., of the poly gon FGJTI, etc., replaces portions of two sides of the circum scribed polygon ABCD, and forms with them a triangle, and one side of a triangle is less than the sum of the other two sides. 214 PLANE GEOMETRY. BOOK V. PROPOSITION IV. THEOREM. 411, Two regular polygons of the same number of sides are similar. Let Q and Q 1 be two regular polygons, each having n sides. To prove Q and Q similar polygons. Proof, The sum of the interior A of each polygon is equal to O-2)2rt,Zs, 205 (the sum of the interior A of a polygon is equal to 2 rt. A taken as many times less 2 as the polygon has sides). (n 2) 2 rt. A R nA Each angle of either polygon = * ^ S AJO ( for the A of a regular polygon are all equal and hence each Z is equal to the sum of the A divided by their number). Hence the two polygons Q and Q are mutually equiangular. Since AB = BO, etc., and A B = , etc., 395 AE\ A = C: B C , etc. Hence the two polygons have their homologous sides proportional. Therefore the two polygons are similar. 319 a E. D. 412, COR. The areas of two regular polygons of the same number of sides are to each other as the squares of any two homokgous. sides. REGULAR POLYGONS AND CIECLES. 215 PROPOSITION V. THEOREM. 413, The perimeters of two regular polygons of the same number of sides are to each other as the radii of their circumscribed circles, and also as the radii of their inscribed circles. A M B A M B Let P and P denote the perimeters, and O> the centres, of the two regular polygons. From 0, draw OA, O A , OB, O B , and Ja OM, O M . To prove P : P = OA : O A = OM: O M . Proof, Since the polygons are similar, 411 333 In the isosceles A OAB and O A B and OA : OB = O A : O B . . . the A OAB and O A B are similar. 326 .:A:A =OA:0 A I . 319 Also AB : A B = OM: O M , 328 (the homologous altitudes of similar A have the same ratio as their bases). /. P:P =OA: O A = OM: OM . Q. E. D. 414, COR. The areas of two regular polygons of the same number of sides are to each other as the squares of the radii of their circumscribed circles, and also as the squares of the radii of their inscribed circles. 376 216 PLANE GEOMETRY. BOOK V. PROPOSITION VI. THEOREM. 415. The difference between the lengths of the perim- eters of a regular inscribed polygon and of a similar circumscribed polygon is indefinitely diminished as the number of the sides of the polygons is indefinitely increased. Let P and P r denote the lengths of the perimeters, AB and A B two corresponding sides, OA and OA 1 the radii, of the polygons. To prove that, as the number of the sides of the polygons is indefinitely increased, f P is indefinitely diminished. Proof, Since the polygons are similar, P:P=OA :OA. 333 Therefore P P : P : : OA OA : OA. 301 Whence OA(P P) = P(OA - OA). 295 Now OA is the radius of the circle, and P, though an increasing variable, always remains less than the circumference of the circle. Therefore P Pis indefinitely diminished, if OA OA is indefinitely diminished. Draw the radius 00 to the point of contact of A JB . In the A OA C, OA OC< AC. 137 Substituting OA for its equal 00, we have OA OA<A O. REGULAR POLYGONS AND CIRCLES. 217 But as the number of sides of the polygon is indefinitely increased, the length of each side is indefinitely diminished ; that is, A E , and consequently A C, is indefinitely diminished. Therefore OA OA, which is less than A C, is indefinitely diminished. Therefore P 1 P is indefinitely diminished. Q E D 416, COR. The difference between the areas of a regular inscribed polygon and of a similar circumscribed polygon is indefinitely diminished as the number of the sides of the poly gons is indefinitely increased. For, if 8 and S 1 denote the areas of the polygons, S 1 : 8= OZ 2 : OA 2 = OA * : 00\ 414 By division, S - S: S= OA 1 - 00* : 00*. Whence S -8= 8x OA *IL OC * = 8x 4 00 OO Since A C can be indefinitely diminished by increasing the number of the sides, /S 8 can be indefinitely diminished. 417, SCHOLIUM. The perimeter P 1 is constantly greater than P } and the area S 1 is constantly greater than 8\ for the radius OA is constantly greater than OA. But P 1 constantly decreases and P constantly increases ( 410), and the area S constantly decreases, and the area S constantly increases, as the number of sides of the polygons is indefinitely increased. Since the difference between P and P can be made as small as we please, but cannot be made absolutely zero, and since P is decreasing while P is increasing, it is evident that P 1 and P tend towards a common limit. This common limit is the length of the circumference. 259 Also, since the difference between the areas S and S can be made as small as we please, but cannot be made absolutely zero, and since /S" is decreasing, while S is increasing, it is evident that S and S tend towards a common limit. This common limit is the area of the circle. 218 PLANE GEOMETRY. BOOK V. PROPOSITION VII. THEOREM. 418. Two circumferences have the same ratio as their radii. Let C and C be the circumferences, R and R the radii, of the two circles Q and Q f . To prove C\C = E\ Iff. Proof, Inscribe in the two similar regular polygons, and denote their perimeters by P and P. Then P-P = R:R ] (l 413) ; that is, Iff X P= R X P. Conceive the number of the sides of these similar regular polygons to be indefinitely increased, the polygons continuing to have an equal number of sides. Then R X P will continue equal to Rx P, and P and P will approach indefinitely C and C 1 as their respective limits. /. R XC= RxC ( 260) ; that is, C: C = R : Iff. Q. E. D. 419. COR. The ratio of the circumference of a circle to its diameter is constant. For, in the above proportion, by doubling both terms of the ratio R : J3 f , we have C:C =2:2 . By alternation, C:2R=C : 2 J2 . This constant ratio is denoted by IT, so that for any circle whose diameter is 2 R and circumference C, we have -- = ir, or C=2>rrR. 420, SCHOLIUM. The ratio TT is incommensurable, and there fore can be expressed in figures only approximately. REGULAR POLYGONS AND CIRCLES. 219 PROPOSITION VIII. THEOREM. 421, The area of a regular polygon is equal to one- half the product of its apothem by its perimeter. A M B Let P represent the perimeter, R the apothem, and 8 the area of the regular polygon ABC etc. To prove 8= % E X P. Proof. Draw OA, OB, OC, etc. The polygon is divided into as many A as it has sides. The apothem is the common altitude of these A, and the area of each A is equal to -J R multiplied by the base. 368 Hence the area of all the A is equal to ^ It multiplied by the sum of all the bases. But the sum of the areas of all the A is equal to the area of the polygon. and the sum of all the bases of the A is equal to the perim eter of the polygon. Therefore S = RX P. Q. E. D. 422, In different circles similar arcs, similar sectors, and similar segments are such as correspond to equal angles at the centre. 220 PLANE GEOMETRY. BOOK V. PROPOSITION IX. THEOREM. 423, The area of a circle is equal to one-half the product of its radius by its circumference. BMC Let E represent the radius, C the circumference, and S the area, of the circle. To prove S=%RxC. Proof, Circumscribe any regular polygon about the circle, and denote its perimeter by P. Then the area of this polygon = | E X P, 421 Conceive the number of sides of the polygon to be indefi nitely increased ; then the perimeter of the polygon approaches the circumference of the circle as its limit, and the area of the polygon approaches the circle as its limit. But the area of the polygon continues to be equal to one- half the product of the radius by the perimeter, however great the number of sides of the polygon. Therefore S - %R X & 260 Q. E. D. 424. COR. 1. The area of a sector equals one-half the product of its radius by its arc. For the sector is such a part of the circle as its arc is of the circumference. 425, COR. 2. The area of a circle equals IT times the square of its radius. For the area of the O = | R X O = $ E X REGULAR POLYGONS AND CIRCLES. 221 426, COE. 3. The areas of two circles are to each other as the squares of their radii. For, if 8 and S denote the areas, and R and JR 1 the radii, 427, Con. 4. Similar arcs, being like parts of their respective circumferences, are to each other as their radii ; similar sectors, being like parts of their respective circles, are to each other as the squares of their radii. PROPOSITION X. THEOREM. 428. The areas of two similar segments are to each other as the squares of their radii. C ^.4^ JB JL~ P Let AC and A C 1 be the radii of the two similar seg ments ABP and A B P . To prove ABP : A B P = AC* : A C 1 *. Proof. The sectors ACB and A* C B are similar, 422 (having the A at the centre, C and C , equal). In the A ACBsuiAA C B Z.0=/. C , AC= CB, and A C = C B . Therefore the A ACB and A C B are similar. 326 Now sector ACB : sector A C B = AC* : A C *, 427 and AACJ3:AA C B = AC: A^ . 375 TT n sector ACS- A AC3 _ AC* , qm sector A C - A ~~ That is, ABP : A B P 1 = AC ; AW Q.E.O. 222 PLANE GEOMETKY. BOOK V. PEOBLEMS OF CONSTRUCTION. PROPOSITION XI. PROBLEM. 429. To inscribe a square in a given circle. Let be the centre of the given circle. To inscribe a square in the circle. Construction, Draw the two diameters AC and ED J_ to each other. Join AE, EC, CD, and DA. Then AECD is the square required. Proof. The A ABC, BCD, etc., are rt, A, 264 (being inscribed in a semicircle), and the sides AE, EC, etc., are equal, 230 (in the same O equal arcs are subtended by equal chords ). Hence the figure A BCD is a square. 5m Q. E. F. 430. COR. By bisecting .the arcs AE, EC, etc., a, regular polygon of eight sides may be inscribed in the circle ; and, by continuing the process, regular polygons of sixteen, thirty-two, sixty-four, etc., sides may be inscribed. Ex. 376. The area of a circumscribed square is equal to twice the area of the inscribed square. Ex. 377. If the length of the side of an inscribed square is 2 inches, what is the length of the circumscribed square ? PROBLEMS OF CONSTRUCTION. 223 PROPOSITION XII. PROBLEM. 431, To inscribe a regular hexagon in a given circle. Let be the centre of the given circle. To inscribe in the given circle a regular hexagon. Construction, From draw any radius, as OC. From (7 as a centre, with a radius equal to 00, describe an arc intersecting the circumference at F. Draw O^and OF. Then CF is a side of the regular hexagon required. Proof, The A OFC is equilateral and equiangular. Hence the Z FOO is of 2 rt. A, or \ of 4 rt, A 138 And the arc FO is -jt of the circumference ABCF. Therefore the chord FC, which subtends the arc FC, is a side of a regular hexagon ; and the figure CFD etc., formed by applying the radius six times as a chord, is a regular hexagon. Q . E . F . 432, COR. 1. By joining the alternate vertices A, O, D, an equilateral triangle is inscribed in the circle. 433, COR. 2. By bisecting the arcs AB, BO, etc., a regular polygon of twelve sides may be inscribed in the circle ; and, by continuing the process, regular polygons of twenty-four, forty- eight, etc., sides may be inscribed. 224 PLANE GEOMETRY. BOOK V. PROPOSITION XIII. PROBLEM. 434, To inscribe a regular decagon in a given circle. B Let be the centre of the given circle. To inscribe a regular decagon in the given circle. Construction, Draw the radius OC, and divide it in extreme and mean ratio, so that 00 shall be to OS. as OS is to SO. 356 From C as a centre, with a radius equal to OS, describe an arc intersecting the circumference at B, and draw EG. Then BC\$ a side of the regular decagon required. Proof, Draw BS and BO. By construction OC:OS=OS: SO, and BC=OS. . . OC\BC--=BC:SC. Moreover, the Z 0GB = Z SCB. Iden. Hence the A 0GB and BC8 are similar, 326 (having an Z of the one equal to an Z of the other, and the including sides proportional). But the A 0GB is isosceles, (its sides 00 and OB being radii of the same circle). . . A BCS, which is similar to the A OCB, is isosceles, and CB^BS^OS. PEOBLEMS OF CONSTRUCTION. 225 . . the A SOB is isosceles, and the Z = Z SBO. But the ext. Z CSB = Z + Z 50 = 2 Z 0. 145 Hence Z #<?.(= Z OS5) - 2Z 0, 154 and Z OBC(= Z flCS) = 2 Z 0. 154 /. the sum of the A of the A OCB = 5 Z = 2 rt. A, and Z = of 2 rt. A, or ^ of 4 rt. A Therefore the arc BG is -fa of the circumference, and the chord BCia a side of a regular inscribed decagon. Hence, to inscribe a regular decagon, divide the radius in extreme and mean ratio, and apply the greater segment ten times as a chord. Q.E. F. 435, COR. 1. By joining the alternate vertices of a regular inscribed decagon, a regular pentagon is inscribed. 436, COR. 2. By bisecting the arcs BG, OF, etc., a regular polygon of twenty sides may be inscribed; and, by continuing the process, regular polygons of forty, eighty, etc., sides may be inscribed. Let R denote the radius of a regular inscribed polygon, r the apothem, a one side, A an interior angle, and the angle at the centre ; show that Ex. 378. In a regular inscribed triangle a = R \/3, r = R, A = 60, C= 120. Ex.379. In an inscribed square a = R V2, r = %ftV2, ^1 = 90, Ex. 380. In a regular inscribed hexagon a = , r = %R V3, A = 120, 0=60. Ex. 381. In a regular inscribed decagon , A = 144, (7=36. 226 PLANE GEOMETRY. BOOK V. PROPOSITION XIV. PROBLEM. 437, To inscribe in a given circle a regular pente- decagon, or polygon of fifteen sides. E F Let Q be the given circle. To inscribe in Q a regular pentedecagon. Construction, Draw EH equal to a side of a regular inscribed hexagon, 431 and EF equal to a side of a regular inscribed decagon. 434 Join FH. Then FH will be a side of a regular inscribed pentedecagon. Proof, The arc EH\& \ of the circumference-, and the arc EF is -^ of the circumference. Hence the arc FJIis J- y^, or -j^, of the circumference, and the chord FIT is a side of a regular inscribed pente decagon. By applying FH fifteen times as a chord, we have the polygon required. 438, COR. By bisecting the arcs FH, HA, etc., a regular polygon of thirty sides may be inscribed; and, by continuing the process, regular polygons of sixty, one hundred twenty, etc.> sides, may be inscribed. PROBLEMS OF CONSTRUCTION. 227 PROPOSITION XV. PROBLEM. 439, To inscribe in a given circle a regular polygon similar to a given regular polygon. Let ABCD etc., be the given regular polygon, and C D E the given circle. To inscribe in the circle a regular polygon similar to ABCD, etc. Construction, From 0, the centre of the given polygon, draw OD and 00. From , the centre of the given circle, draw aO r and O D , making the Z =Z 0. Draw C D . Then C D will be a side of the regular poly^ Proof, Each polygon will have as many (= Z ) is contained times in 4 rt. A. Therefore the polygon C D E etc., is similar to the poly gon CDE etc., 411 (two regular polygons of the same number of sides are similar). 228 PLANE GEOMETRY. BOOK V. PROPOSITION XVI. PROBLEM. 440, Given the radius and the side of a regular inscribed polygon, to find the side of the regular inscribed polygon of double the number of sides. D LetAB be a side of the regular inscribed polygon. To find the value of AD, a side of a regular inscribed poly gon of double the number of sides. From D draw DH through the centre 0, and draw OA, AH. DITis _L to AB at its middle point C. 123 Inthert.A(L4<?, OC*= OA* - A0\ 339 That is, OC= hence Therefore, 00= In the rt. A DAH, 264 334 and = 20A(OA-OC), AD = V2 OA (OA-00). If we denote the radius by R, and substitute Vj? 2 _ for 00, then AD =- Q.E.F. PROBLEMS OF COMPUTATION. 229 PROPOSITION XVII. PROBLEM. 441, To compute the ratio of the circumference of a circle to its diameter approximately. Let C be the circumference, and R the radius. To find the numerical value of IT. No. Bides. 12 =2-V4- 419 Therefore when E = I, ir = %C. We make the following computations by the use of the formula obtained in the last proposition, when R = 1, and AB = 1 (a side of a regular hexagon). Form of Computation. Length of Side. Length of Perimeter. 0.51763809, 6.21165708 24 c a =V2-V4-(0.5176380l^ 0.26105238 6.26525722 48 c, = V2 - V4~- (0261052^7 0.13080626 6.27870041 0.06543817 6.28206396 0.03272346 6.28290510 384 c 6 = V2 - V4 - (0.03272346) 2 0.01636228 6.28311544 768 c T =V2-Vi- (0.016362217 0.00818121 6.28316941 Hence we may consider 6.28317 as approximately the cir cumference of a O whose radius is unity. Therefore TT = -J(6.28317) = 3.14159 nearly. aE . F . 442, SCHOLIUM. In practice, we generally take <* = 3.1416, 96 4 =2-4- 192 Cft = V2- V4^(0706543817f 1 = 0.31831. 7T 230 PLANE GEOMETRY. BOOK V. MAXIMA AND MINIMA. SUPPLEMENTARY. 443, Among magnitudes of the same kind, that which is greatest is the maximum, and that which is smallest is the minimum. Thus the diameter of a circle is the maximum among all inscribed straight lines ; and a perpendicular is the minimum among all straight lines drawn from a point to a given line. 444, Isoperimetric figures are figures which have equal perimeters. PROPOSITION XVIII. THEOREM. 445, Of all triangles having two given sides, that in which these sides include a right angle is the maximum. A E Let the triangles ABC and EEC have the sides AB and EG equal respectively to EB and EC ; and let the angle ABC be a right angle. To prove A ABC > A EEC. Proof, From E let fall the -L ED. The A ABC and EEC, having the same base BO, are to each other as their altitudes AB and ED. 370 Now EB > ED. 114 By hypothesis, EB = AB. .-. AB > ED. Q.E.O MAXIMA AND MINIMA. 231 PROPOSITION XTX. THEOREM. 446, Of all triangles having the same base and equal perimeters, the isosceles triangle is the maximum. Let the A ACB and ADB have equal perimeters, and let the &ACB be isosceles. To prove A A CB > A ADB. Proof, Produce AC to H, making CH= AC, and join HB. ABH\& a right angle, for it will be inscribed in the semi circle whose centre is C, and radius CA. Produce HB, and take DP= DB. Draw CK and DM\\ to AB, and join AP. Now AH= AC+ CB = AD+DB = AD + DP. Rut AD + Z>P>AP, hence AH> AP. Therefore HB > BP. 120 But KB = i HB and MB =IBP. 121 Hence JT.>Jf.. By 180, KB = CE and MB = DF, the altitudes of the & ACB and ADB. Therefore A ABC> A ADB. 370 Q. E. D. 232 PLANE GEOMETRY. BOOK V. PROPOSITION XX. THEOREM. 447, Of all polygons with sides all .given "but one, the maximum can be inscribed in a semicircle which has the undetermined side for its diameter. - Let ABODE be the maximum of polygons with sides AB, BC, CD, DE, and the extremities A and E on the straight line MN. To prove ABODE can be inscribed in a semicircle. Proof, From any vertex, as C, draw CA and CE. The A AGE must be the maximum of all A having the given sides CA and CE; otherwise, by increasing or diminish ing the Z ACE, keeping the sides CA and CE unchanged, but sliding the extremities A and E along the line MN, we can increase the A ACE, while the rest of the polygon will remain unchanged, and therefore increase the polygon. But this is contrary to the hypothesis that the polygon is the maximum polygon. Hence the A ACE with the given sides CA and CE is the maximum. Therefore the Z ACE is a right angle, 445 (the maximum of A having two given sides is the A with the two given sides including a rt. Z). Therefore C lies on the semi-circumference. 264 Hence every vertex lies on the circumference ; that is, the maximum polygon can be inscribed in a semicircle having the undetermined side for a diameter. % o. E. D. MAXIMA AND MINIMA. 233 PROPOSITION XXI. THEOREM. 448, Of all polygons ivith given sides, that which can be inscribed in a circle is the maximum. Let ABODE be a polygon inscribed in a circle, and A B C D E be a polygon, equilateral with respect to ABCDE, which cannot be inscribed in a circle. To prove ABCDE greater than A B Proof, Draw the diameter AH. Join OS" and DIL Upon O D (= CD) construct the A C IT D = A CHD, and draw A H . Now ABCH> A B C H , 447 and AEDH>A E D IF, (of all polygons with sides all given but one, the maximum can be inscribed in a semicircle having the undetermined side for its diameter). Add these two inequalities, then ABCHDE> A B C IT D E . Take away from the two figures the equal A CHD and C H D . Then ABCDE > A B C D E 1 . Q . E . D . 234 PLANE GEOMETRY. BOOK V. PROPOSITION XXII. THEOREM. 449, Of isoperimetric polygons of the same number of sides, the maximum is equilateral. K Let ABCD etc., be the maximum of isoperimetric polygons of any given number of sides. To prove AB, EC, CD, etc., equal. Proof, Draw^(7. The A ABC must be the maximum of all the A which are formed upon AC with a perimeter equal to that of A AEG. Otherwise, a greater A AKC could be substituted for A ABC, without changing the perimeter of the polygon. But this is inconsistent with the hypothesis that the poly gon ABCD etc., is the maximum polygon. /. the A AEG is isosceles, 446 (of all A having the same base and equal perimeters, the isosceles A is the maximum). In like manner it may be proved that EG= CD, etc. Q . E . D . 450, COR. The maximum of isoperimetric polygons of the same number of sides is a regular polygon. For, it is equilateral, 449 (the maximum of isoperimetric polygons of the same number of sides is equilateral). Also it can be inscribed in a circle, 448 (the maximum of all polygons formed of given sides can be inscribed in a O). That is, it is equilateral and equiangular, and therefore regular. . 395 Q. E. D. MAXIMA AND MINIMA. 235 PROPOSITION XXIII. THEOREM. 451, Of isoperimetric regular polygons, that which has the greatest number of sides is the maximum. o A D Let Q be a regular polygon of three sides, and Q f a regular polygon of four sides, and let the two poly- gons have equal perimeters, To prove Q greater than Q. Proof, Draw CD from C"to any point in AB. Invert the A CD A and place it in the position DCE t let ting D fall at C, C at D, and A at E. The polygon DBCE is an irregular polygon of four sides, which by construction has the same perimeter as Q , and the same area as Q. Then the irregular polygon DBCE of four sides is less than the regular isoperimetric polygon Q of four sides. 450 In like manner it may be shown that Q is less than a regular isoperimetric polygon of five sides, and so on. Q . E- Dp 452, Con. The area of a circle is greater than the area of any polygon of equal perimeter. -" 382. Of all equivalent parallelograms having equal bases, the rec tangle has the least perimeter. 383. Of all rectangles of a given area, the square has the least perimeter. 384. Of all triangles upon the same base, and having the same alti tude, the isosceles has the least perimeter. 385. To divide a straight line into two parts such that their product shall be a maximum. 236 PLANE GEOMETRY. BOOK V. PROPOSITION XXIV. THEOREM. 453, Of regular polygons having a given area, that which has the greatest number of sides has the least perimeter. Let Q and Q be regular polygons having the same area, and let Q 1 have the greater number of sides. To prove the perimeter of Q greater than the perimeter of Q 1 . Proof, Let Q 1 be a regular polygon having the same perim eter as Q , and the same number of sides as Q. Then Q > Q", 451 (of isoperimetric regular polygons, that which has the greatest number oj sides is the maximum). But Q = Q . .: Q > Q". . . the perimeter of Q > the perimeter of Q". But the perimeter of Q = the perimeter of Q". /. the perimeter of Q > that of Q . Cons. a E. D. 454, COR. The circumference of a circle is less than the perimeter of any polygon of equal area. 386. To inscribe in a semicircle a rectangle having a given area; a rectangle having the maximum area. 387. To find a point in a semi-circumference such that the sum of its distances from the extremities of the diameter shall be a maximum. EXERCISES. 237 THEOREMS. 388. The side of a circumscribed equilateral triangle is equal to twice the side of the similar inscribed triangle. Find the ratio of their areas. 389. The apothem of an inscribed equilateral triangle is equal to half the radius of the circle. 390. The apothem of an inscribed regular hexagon is equal to half the side of the inscribed equilateral triangle. 391. The area of an inscribed regular hexagon is equal to three- fourths that of the circumscribed regular hexagon. 392. The area of an inscribed regular hexagon is a mean proportional between the areas of the inscribed and the circumscribed equilateral triangles. 393. The area of an inscribed regular octagon is equal to that of a rectangle whose sides are equal to the sides of the inscribed and the cir cumscribed squares. 394. The area of an inscribed regular dodecagon is equal to three times the square of the radius. ^ 395. Every equilateral polygon circumscribed about a circle is regu lar if it has an odd number of sides. 396. Every equiangular polygon inscribed in a circle is regular if it has an odd number of sides. 397. Every equiangular polygon circumscribed about a circle is regular. 398. Upon the six sides of a regular hexagon squares are constructed outwardly. Prove that the exterior vertices of these squares are the ver tices of a regular dodecagon. 399. The alternate vertices of a regular hexagon are joined by straight lines. Prove that another regular hexagon is thereby formed. Find the ratio of the areas of the two hexagons. 400. The radius of an inscribed regular polygon is the mean propor tional between its apothem and the radius of the similar circumscribed regular polygon. 401. The area of a circular ring is equal to that of a circle whose diameter is a chord of the outer circle and a tangent to the inner circle. 402. The square of the side of an inscribed regular pentagon is equal to the sum of the squares of the radius of the circle and the side of the inscribed regular decagon. 238 PLANE GEOMETRY. BOOK V. If R denotes the radius of a circle, and a one side of a regular inscribed polygon, show that : R 403. In a regular pentagon, a = ~ VlO 2\/5. 404. In a regular octagon, a = 405. In a regular dodecagon, a = 406. If on the legs of a right triangle, as diameters, semicircles are described external to the triangle, and from the whole figure a semicircle on the hypotenuse is subtracted, the remainder is equivalent to the given triangle. NUMERICAL EXERCISES. 407. The radius of a circle = r. Find one side of the circumscribed equilateral triangle. -^ y-5 408. The radius of a circle = r. Find one side of the circumscribed regular hexagon. ^Vv^ 409. If the radius of a circle is r, and the side of an inscribed regular polygon is a, show that the side of the similar circumscribed regular polygon is equal to 2ar ^ V4r 2 -a 2 410. The radius of a circle = r\ Prove that the area of the inscribed regular octagon is equal to 2r 2 \/2.L- 411. The sides of three regular octagons are 3 feet, 4 feet, and 5 feet, respectively. Find the side of a regular octagon equal in area to the sum of the areas of the three given octagons. 412. What is the width of the ring between two concentric circum ferences whose lengths are 440 feet and 330 feet? 413. Find the angle subtended at the centre by an arc 6 feet 5 inches long, if the radius of the circle is 8 feet 2 inches. 414. Find the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle. 415. What is the length of the arc subtended by one side of a regular dodecagon inscribed in a circle whose radius is 14 feet? 416. Find the side of a square equivalent to a circle whose radius is 56 feet. EXERCISES. 239 417. Find the area of a circle inscribed in a square containing 196 square feet. 418. The diameter of a circular grass plot is 28 feet. Find the diam eter of a circular plot just twice as large. 419. Find the side of the largest square that can be cut out of a cir cular piece of wood whose radius is 1 foot 8 inches. 420. The radius of a circle is 3 feet. What is the radius of a circle 25 times as large ? ^ as large ? -^ as large ? 421. The radius of a circle is 9 feet. What are the radii of the con centric circumferences that will divide the circle into three equivalent parts? ^ 22. The chord of half an arc is 12 feet, and the radius of the circle is feet. Find the height of the arc. 423. The chord of an arc is 24 inches, and the height of the arc is 9 inches. Find the diameter of the circle. 424. Find the area of a sector, if the radius of the circle is 28 feet, and the angle at the centre 22J. 425. The radius of a circle = r. Find the area of the segment sub tended by one side of the inscribed regular hexagon. 426. Three equal circles are described, each touching the other two. If the common radius is r, find the area contained between the circles. PROBLEMS. To circumscribe about a given circle : 427. An equilateral triangle. 429. A regular hexagon. 428.. A square. 430. A regular octagon. 431. To draw through a given point a line so that it shall divide a given circumference into two parts having the ratio 3 : 7. % 132. To construct a circumference equal to the sum of two given (^jumferences. 433. To construct a circle equivalent to the sum of two given circles. 434. To construct a circle equivalent to three times a given circle. 435. To construct a circle equivalent to three-fourths of a given circle. To divide a given circle by a concentric circumference : 436. Into two equivalent parts. 437. Into five equivalent parts, 240 PLANE GEOMETRY. BOOK V. MISCELLANEOUS EXEKCISES. THEOREMS. 438. The line joining the feet of the perpendiculars dropped from the extremities of the base of an isosceles triangle to the opposite sides is parallel to the base. 439. If AD bisect the angle A of a, triangle ABC, and BD bisect the exterior angle CBF, then angle ADB equals one-half angle ACB. 440. The sum of the acute angles at the vertices of a pentagram (five- pointed star) is equal to two right angles. 441. The bisectors of the angles of a parallelogram form a rectangle. 442. The altitudes AD, BE, CF of the triangle ABC bisect the angles of the triangle DEF. HINT. Circles with AB, BO, AC as diameters will pass through E and D, E an^ F t D and F, respectively. 443. the portions of any straight line intercepted between the cir cumferences of two concentric circles are equal. 444. Two circles are tangent internally at P, and a chord AB of the larger circle touches the smaller circle at C. Prove that PC bisects the angle APB. HINT. Draw a common tangent at P, and apply \\ 263, 269, 145. 445. The diagonals of a trapezoid divide each other into segments which are proportional. 446. The perpendiculars from two vertices of a triangle upon the opposite sides divide each other into segments reciprocally proportional. 447. If through a point P in the circumference of a circle two chords are drawn, the chords and the segments between P and a chord parallel to the tangent at Pare reciprocally proportional. 448. The perpendicular from any point of a circumference upon a chord is a mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord. 449. In an isosceles right triangle either leg is a mean proportional between the hypotenuse and the perpendicular upon it from the vertex of the right angle. 450. The area of a triangle is equal to half the product of its perim eter by the radius of the inscribed circle. MISCELLANEOUS EXERCISES. 241 451. The perimeter of a triangle is to one side as the perpendicular from the opposite vertex is to the radius of the inscribed circle. 452. The sum of the perpendiculars from any point within a convex equilateral polygon upon the sides is constant. 453. A diameter of a circle is divided into any two parts, and upon these parts as diameters semi-circumferences are described on opposite sides of the given diameter. Prove that the sum of their lengths is equal to the semi-circumference of the given circle, and that they divide the circle into two parts whose areas have the same ratio as the two parts into which the diameter is divided. 454. Lines drawn from one vertex of a parallelogram to the middle points of the opposite sides trisect one of the diagonals. 455. If two circles intersect in the points A and B, and through A any secant CAD is drawn limited by the circumferences at C and D, the straight lines BC, BD, are to each other as the diameters of the circles. 456. If three straight lines A A , BB , CC f , drawn from the vertices of a triangle ABC to the opposite sides, pass through a common point within the triangle, then OA OB PC = l AA BE CC* 457. Two diagonals of a regular pentagon, not drawn from a common vertex, divide each other in extreme and mean ratio. | !? Loci. 458. Find the locus of a point P whose distances from two given points A and B are in a given ratio (ra : ri). 459. OP is any straight line drawn from a fixed point to the cir cumference of a fixed circle ; in OP a, point Q is taken such that OQ: OP is constant. Find the locus of Q. 460. From a fixed point A a straight line AB is drawn to any point in a given straight line CD, and then divided at P in a given ratio (m : n). Find the locus of the point P. 461. Find the locus of a point whose distances from two given straight lines are in a given ratio. (The locus consists of two straight lines.) 462. Find the locus of a point the sum of whose distances from two given straight lines is equal to a given length k. (See Ex. 73.) 242 PLANE GEOMETRY. BOOK Y. PROBLEMS. 463. Given the perimeters of a regular inscribed and a similar circum scribed polygon, to compute the perimeters of the regular inscribed and circumscribed polygons of double the number of sides. 464. To draw a tangent to a given circle such that the segment inter cepted between the point of contact and a given straight line shall have a given length. 465. To draw a straight line equidistant from three given points. 466. To inscribe a straight line of given length between two given circumferences and parallel to a given straight line. (See Ex. 137.) 467. To draw through a given point a straight line so that its dis tances from two other given points shall be in a given ratio (ra : n). HINT. Divide the line joining the two other points in the given ratio. 468. Construct a square equivalent to the sum of a given triangle and a given parallelogram. 469. Construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon. 470. Construct a pentagon similar to a given pentagon and equiva lent to a given trapezoid. 471. To find a point whose distances from three given straight lines shall be as the numbers ra, n, and p. . (See Ex. 461.) 472. Given two circles intersecting at the point A. To draw through A a secant B AC such that AB shall be to AC in a given ratio (ra : n). HINT. Divide the line of centres in the given ratio. 473. To construct a triangle, given its angles and its area. 474. To construct an equilateral triangle having a given area. 475. To divide a given triangle into two equal parts by a line drawn parallel to one of the sides. 7 476. Given three points A, B, C. To find a fourth point Psuch that the areas of the triangles AP, APC, PC, shall be equal. 477. To construct a triangle, given its base, the ratio of the other sides, and the angle included by them. 478. To divide a given circle into any number of equivalent parts by concentric circumferences. 479. In a given equilateral triangle, to inscribe three equal circles tangent to each other and to the sides of the triangle. 14 DAY USE RN TO DESK FROM WHICH BORROWED This bdolos due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. LD 21-50m-6, 60 (B1321slO)476 General Library University of California Berkeley p YB 72971 UNIVERSITY OF CALIFORNIA LIBRARY