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Full text of "Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson"

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IN MEMORIAM FLORIAN CAJO GEOMETRY: TEE ELEMENTS OF EUCLID AND LEGENDRE SIMPLIFIED AND ARRANGED TO EXCLUDE FROM GEOMETRICAL REASONING WITH THE ELEMEx\TS OF PLAXE AND SPHEPJCAL TRIGONOMETRY, ANT) EXERCISES Ii\ ELEMENTARY GEOMETRY AND TRIGONOMETRY. ADAPTED FOK SCHOOLS AND COLLEGES. • L-AWR^i^CE' S:' BENSON, n Author of "The Truth of the Bible Upheld,"— London, 18G4; "Geometrical Disquisi- tions," — London, 1864 ; " Scientific Disquisitions Concerning the Circle and Ellipse," 1862. Member of the New York Association for the Advancement of Science and Art ; Hon. Mem. Phi Eappa Society, University of Georgia ; Brothers' Society, Yale College, etc., etc. [all rights reserved.] NEW YORK: PUBLISHED FOR THE AUTHOR BY DAVTES & KENT, No. 183 WILLLAM STREET. 186V. 15f Rev. Thomas A. Boone, Professor in Carolina Female College, An- sonville, North Carolina, writes : " Your new work on the Elements op Geometry (Book First) haa been submitted to the President of Carolina Female College. He has examined it critically, and indorses it as an evident advancement of the science, in that it simplifies and meets the capacity of learners, retains all the essentials of the science, and is equally as competent for mental discipline as the old Eeductio Ad Abmrdum." STBREOTTPEli AND ElECTROTTPEIB,- .-, 183 Wjjaiaar ^treeln Nw Y. . • CAJORI Entered, according to Act of Congress, in the year 1867, by LAWRENCE S. BENSON, In the Cleric's Office of the District Court of the United States for the Southern District of New York. TO PROFESSOR GERARDUS BEEKMAN DOCHARTY, LL.D., COLLEGE OP THE CITY OF NEW YORK. Sm — In permitting me to inscribe to you this Treatise of Elementary Geometry, you do me great lionor. Your experience and success as a Teaclier and an Author will readily enable you to give a full scrutiny to the design and compass of this volume. Much originality can not be expected in a subject which has been, for more than two thousand yeare, enriched by a great number of eminent men ; but in these days of practi- cability, a modification of this science may be attempted, as you have yourself thought proper to do, with a view of utilizing the important principles of Geometry, and presenting them in such a manner that though " no royal road to Geometry" can be found, the path to a know- ledge of it may be rendered so clear that the impediments wiU be in tihe learner himself. And to remove much diflBculty in acquiring an easy acquaintance with its numerous theorems and problems, I have thought proper to exclude the inelegant Bedrictio ad absurdum from the methods of geometrical reasoning which you have expressed — " a consummation most devoutly to he wislied" and which accomplishment, resulting from my labors, I now present for the benefit and use of those whose education is in the future. I hav3 the honor to be, Very respectfully, yours, Lawrence S. Benson. Vvw York, AprU mh, 1867. TESTIMONIALS. The College op the City of New York, » Cor. Lexington Avenue and 23d Street. 1 New York, January Zd, 1867. I have had several interviews with Mr. Lawrence S. Benson on scientific subjects, and from his conversation, together with tlie Essays which he has published, I esteem him an excellent scholar and fine mathematician. He has a desire to establish the Elements of Euclid in all canes, independently of the demonstration known as the Beductio ad dbsurdum, " a consummation devoutly to be wished." Whatever aid or advice you can render him in the furtherance of this object will tend to the advancement of true science. Yours truly, G. B. DOCHARTY. Rooms op the New York Association for the Advancement ) of Science and Art, February )i8th, 18G7. I Extract from the transactions of the Association for the Advancement of Science and the Arts : " At a meeting of the New York Association held February 25, 1867, a paper on a new method of demonstrating the propositions of Geometry, denominated the Direct Method, in place of the one now in use,;ijid called the Indirect Method, was read by Lawrence S. Benson, Esq., which method the writer proposes to introduce into Schools and Academies. " After the reading of the paper, and the discussion of its merits, the subject was referred to Professor Fox, Principal of the Department of Free Schools of Cooper Union, and to Professor Cleveland Abbe, for ex- amination and report. It was also moved and carried that the Report when received be referred to the Section on Physical Science for final disposition. " The Section, after reading the Report of Professor Fox, the letter of Professor Abbe, and the opinion of Professor Docharty, who had been invited to examine the work, feel justified in commending this work as worthy of patronage. Professor Fox in his Report says : ' The design of arranging the Definitions, Axioms, and Propositions of Geometry, so as to use only the Direct Method of demonstration, is a good one, and when arranged in the form of a neat elementary text-book, will doubtless do much good, as the Direct Method is much more easily understood thaa the Lidirect Method, by beginners.' " L. D. Gale, M.D., Oen. Sec. of the New York Association for the Advancement of Science and tlie Art»." PREFACE. By way of preface, I will state what I have done in this edition, and explain the reason why I have done so. I have used such propositions only which are required to substantiate the principal theorems and problems by which the principles of Geometry have practical applica- tions in Trigonometry, Surveying, Mechanics, Engineering, Navigation, and Astronomy. I have generalized the various propositions found in the school editions of Geometry, and where particular cases arise under such general propositions, I have given the demonstrations for them. I have arranged the propositions to give the Direct Method of demonstra- tion in place of the Beductio ad ahsurdum or Indirect Method. My reasons for the foregoing changes are obvious to the experienced mind ; considering the extent and variety of modem education, the time devoted for the pupils to acquire rudimental knowledge becomes en- croached upon in order to make them acquainted with its numerous modifications ; and for the pupils to obtain such knowledge of the rudi- ments as will enable them to see the practical applications throughout all their extent and variety, which are very great in these days of ad- vancement and civilization, the rudiments which were taught centuries ago must be so abbreviated as to contain the essentials only. When materials for instructing the mind were scant, there were no opportunities to make close selection of them ; but now, when those materials are plentiful, a judicious selection of tlie best becomes imperatively necessary. And when geometrical principles have become extended by the Algebraic Analysis, and have been made practicable by Trigonometry, Surveying, Mechanics, Engineering, Navigation, and Astronomy, the mere mental exercises, which were regarded so beneiicent by the ancients, are unsuited for this practical age, which is continually bent on progress, while the intellect is sufficiently exercised by utilizing modern acquisitions; for this reason, I have reduced the number of propositions substantiating the principles of Geometiy, and I have classified them in such a manner that particular cases are enunciated by general propositions, a change which is likely to impress on the pupils the accuracy of geometrical principles, as they will be shown that geometrical principles are the same in all cases and under every circumstance. Many of the best geometers have objected to the Beductio ad dbsurdum in Geometry, while all geometers prefer the Direct Method of demonstra- VI PREFACE. tion. Any true proposition is susceptible of being directly demonstrated. And without entering into the merits or demerits of the Reductio ad ab' surdum, I have exchided it from geometrical reasoning, and have used the Direct Method only, a change which agrees with tlie spirit of the age, and fulfills the requirements of progress. I have omitted the various diagrams usually put among the definitions of Geometry, because when a magnitude is properly defined, the learner has a better conception of it from the definition than any diagram can give him ; and the omission of the diagrams will assist the mental exercise and cultivate the under- standing of the learner, which is the great object of geometrical study; and if the learner be made to draw the diagrams from the definitions, he will be better instructed than if they be given by the author. The time is not far distant when geometrical science may be attempted with- out using diagrams in the demonstrations. The diagrams are auxiliaries to the mind in the ascertainment of truth ; they are not necessary to the existence of truth, and " Geometry considers all bodies in a state of ab- straction, very different from that in which they actually exist, and the truths it discovers and demonstrates are pure abstractions, hypothetical truths." Hence, diagrams are like the pebbles used by Indians in count- ing, or other means of computing before the principle of numeration was discovered ; that when the intellect of man becomes more highly ex- panded and cultivated, diagrams will be regarded necessary to the first conceptions of geometrical knowledge, but altogether unsuited to a high development of geometrical science. I am greatly indebted to Hon. S. S. Randall, City Superintendent of the Board of Education of New York, for many valuable suggestions in the demonstrations and present arrangement of this work ; and also under many obligations to Professor Docharty, of the College of the City of New York ; and to L. D. Gale, M.D., General Secretary of the New York Asso- ciation for the Advancement of Science and Art, Cooper Union. Lawrence S. Benson, 61 MoETON Street, City or New Yobk, April mh, 1867. THE ELEMENTS OF EUCLID AND LEGENDRE. BOOK FIRST. ON THE STRAIGHT LINE AND TRIANGLE. DEFINITIOXS. 1. A definition is the precise term by which one thing is dis- tinguished from all other things. 2. Mathematics is that science which treats of those abstract quantities known as numbers, symbols, and magnitudes. 3. Geometry is that branch of Mathematics where the ex- tensions of magnitudes are considered without regard to the actual existence of those magnitudes. 4. A m^agnitude has one or more of three dimensions, viz., length, breadth, and thickness. 5. Geometers define a point, position without magnitude; but to give a point position, would entitle it to the three dimen- sions of magnitude, whereas a point in Geometry expresses no dimension. 6. A diagram, represents the abstractions of magnitudes, whereby their dimensions are determined, and geometrical reasoning conducted without regard to the actual properties of those magnitudes. '7. A line expresses length only, and is capable of two con- ditions — it can be straight or curved ; when its length is always in one direction, it is straight ; but when there is a continual variation in the direction of its length, it is curved, or in brevity called a curve. Scholiiim. A straight line can not be defined as having all its points in the same direction, because the points of a line are its extremities, and the extremities of a curved line can be 8 THE ELEMENTS OP [boOK I. placed on a straight line, and in this case the definition would not distinguish a straight line from a curve. And if a line be regarded composed of points, this would infer that a point has dimension ; but the intersection of lines is a point, which, how- ever, does not give position to the point, because a line is au abstraction, and position implies actual existence. 8. A surface expresses an inclosure by not less than three straight lines, or by one curved line, or by one straight line and one curved line ; consequently a surface has breadth and length, and the extremities of surfaces are lines, and the intersection of one surface with another is a line. Scho. A plane surface, or sometimes called a plane^ is one in which any line can be drawn wholly in the surface; and a c*rved surface is one in which a curve only can be drawn wholly in the surface in the direction of the curvature. 9. A volume or solid expresses an inclosure made by sur- faces, and has breadth, length, and thickness ; the extremities of a volume are surfaces, and the intersection of one volume with another is a surface. 10. An angle is formed by two straight lines meeting each other ; the point of intersection of the lines is called the vertex of the ano-le. When one straicjht line meets another straight line, so as to make two adjacent angles, these angles are right angles when they are equal ; and when one angle is greater than the other angle, the greater angle is an obtuse angle, and the less angle is an acute angle. The straight line which makes the two adjacent angles equal is the perpendicular to the other 6trai<;ht line. 11. When two straight lines on the same plane never meet each other on whichever side they be produced, they are called parallel lines. 12. Rectilinear surfaces are contained by straight lines, and are called polygons ; when a polygon has three sides, it is a triangle ; when it has four sides, it is a quadrilateral ; when it has five sides, it is z, pentagon ; when it has six sides, it is a hexagon / when it has seven sides, it is a heptagon / when it has eight sides, it is an octagon ; when it has nine sides, it is an enneagon / when it has ten sides, it is a decagon ; and so on, being distinguished by particular names derived from the Greek BOOK I.] EUCLID AND LEGENDRE. 9 language, denoting the number of angles formed by the sides. The straight line drawn through two remote angles of a poly- gon of four or more sides, is a diagonal. 13. When the triangle has its three sides equal, it is equi' lateral; when two of its sides are equal, it is isosceles; and when its sides are unequal, it is scalene. When its angles are equal, it is eqidangular y when one of its angles is a right angle, it is right-angled ; when one of its angles is an obtuse angle, it is obtuse-angled ; and when all its angles are acute, it is acute- angled. 14. When a quadrilateral has its opposite sides parallel, it is 2l parallelogram ; when it has two sides only parallel, or none of its sides parallel, it is a trapezium. 15. When a parallelogram has a right angle, it is a rectangle ; when it has two adjacent sides equal, but no right angle, it is a rhombus. When the rectangle has its sides equal, it is a square / and when its opposite sides only are equal, it is an ob- long. When the parallelogram has its opposite sides only equal, and no right angle, it is a rhomboid. 16. A plane surface contained by one line is a circle when every part of the line is equally distant from a point in the sur- face ; the point is the center of the circle, and the line is the circumference. 17. The straight line drawn from the center to the circum- ference is the radius ; the straight line drawn from one part of the circumference through the center to another part of the circumference is the diameter, which divides the circle and cir- cumference each into two equal parts. When the straight line does not pass through the center, it is a chord. 18. That portion of the circle contained by the semicircnm- ference and diameter is a semicircle; and that portion con- tained by the chord and a part of the circumference is a seg- ment ; a part of the circumference is an arc. 19. If the vertex of an angle be the center of a circle, that part of the circumference intercepted by the sides of the angle will give the value of the angle ; hence, the angle is measured by an arc when its vertex is the center of the circle, liut when the vertex is in the circumference, the angle is subtended by the arc intercepted by its sides ; hence, equal angles will be 10 THE ELEMENTS OF [BOOK I. measured by equal arcs, and subtended by equal arcs ; therefore equal arcs measure or subtend equal angles. 20. Two arcs are supplementary when both together are equivalent to the semicircumferenoe. And two angles are sup- plementary when both together are equivalent to two right angles, and complementary when equivalent to one right angle. 21. Things are equal when they have equal magnitudes and when they coincide in all respects; and are equivalent when they have equal magnitudes, but do not coincide in all respects. 22. The term, each to each, or sometimes respectively, is a limiting expression, and is used to denote the equality of lines or magnitudes taken in the same order; for without this quali- fication, two lines or magnitudes said to be equal to two other lines or magnitudes, would imply that their sums are equal, when it would be desirous of meaning that they are equal in the same order in which they are expressed — a difference very important in the demonstration of a proposition. 23. A proposition is demonstrated by superpositio7i when one figure is supposed applied to another, which is done in the first case of the third proposition of this book. 24. One proposition is the converse of another when, in the language of logic, the subject of the latter is the predicate of the former, and the predicate of the latter is the subject of the former. METHOD OF REASONING. 1. From the foregoing definitions, it is shown that the straight line and curve have certain relations, uses, and prop- erties which are important to be known. And in order that these relations, uses, and properties may be satisfactorily inter- preted, there are certain terms, expressive of certain facts or states of knowledge, by means of which the mind intuitively perceives a connection between the things known and those for elucidation, such as axioms^ hypotheses^ and postulates/ as demonstrations, theorems, problems, and lemmas,' as corollaries and scholiums. With the assistance of these, the mind is carried step by step in all its investigation of extension, and is able to discover by such investigation the properties, uses, and relations of geometrical magnitudes. They are the data by BOOK I.] EUCLID AND LEGENDRE. H which the hidden truths are revealed. Upon them a system of logic or argumentation is conducted, and by the conformity of the arguments and conclusions with the accepted truths, we have the science of Geometry. 2. Proposition in Geometry is a general term, expressing the subjects to be considered, and is either a problem or theorem. When it is the first, there is something required to be per- formed, such as drawing a line or constructing a figure ; and whatever points, lines, angles, or other magnitudes are given to efiect the purpose, they are the data of the problem ; and when it is the latter, a truth is proposed for demonstration, and whatever is assumed or admitted to be true, and from which the proof is to be derived, is the hypothesis. 3. Demonstration consists in evident deductions from clear premises, whereby the conclusion corroborates the premises and shows the argumentativeness of the deductions. In the course of demonstration, reference is often made to some previous proposition or definition. 4. Sometimes inferences arise involving another principle, but do not require any long process of reasoning to establish their truth — these are corollaries. Any remark made from the demonstration of a proposition is a scholium. A proposition which is preparatory to one or more propositions, and is of no other use, is a lemma. 5. And for the establishment of a proposition, there are four things required, viz. : the general enunciation, the particular enunciation, the construction, and the demonstration. 6. The hypotheses of demonstration are known as axiom, and postulate; the former is assumed to prove the truth of a theorem, and the latter is granted to pei'form the requisites of a problem. 7. An axiom is so evidently clear, that no process of reason- ing can make it more clear ; its truth is so easily recognized by the human mind, that so soon as the terms by which it is ex- pressed are understood, it is admitted; for instance, it is as- sumed as AXIOMS. 1. Things which are equal to the same, or to equals, are equal to one another. 12 THE ELEMENTS OF [bOOK I. 2. If equals or the same be added to equals, the wholes are equal, 3. If equals or the same be taken from equals, the remainders are equal. 4. If equals or the same be added to unequals, the wholes are unequal. 5. If equals or the same be taken from unequals, the re- mainders are unequal. 6. Things which are doubles of the same, or of equals, are equal to one another. I. Things which are halves of the same, or of equals, are equal to one another. 8. Magnitudes which exactly coincide with one another are equal. 9. The whole is greater than its part. 10. The whole is equal to all its parts taken together. II. All right angles are equal to one another. 12. If a straight line meet two other straight lines which are in the same plane, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines shall at length meet upon that side, if they be continually produced. These are the self-evident truths used by Euclid for geomet- rical demonstration ; but if the first eleven be considered for awhile, it will be seen that they can be reduced to two general axioms, viz., things which are equal to the same are equal, and things which are not equal to the same are unequal ; because when we add, subtract, multiply, or divide equals, the equality in each case is not destroyed; hence in each case equal to one another. And when we add unequals to or subtract un- equals from equals, the sums or remainders are not equal to the same, hence unequal to one another. And magnitudes which exactly coincide with one another are equal to the same, hence equal ; a whole and a part are not equal to the same, hence are unequal ; Avhile a whole and all its parts are equal to the same, hence are equal. From the definition of right angles, it is seen that when a straight line meets another straight line, so as to make the two adjacent angles formed by them equal to one BOOK I.] EUCLID AND LEGENDRE. 13 another, the two adjacent angles are right angles; then these two right angles are equal ; and since all right angles agree with the definition, they are equal to the same thing, hence equal to one another. But the twelfth axiom is not self-evident, be- cause the converse has been demonstrated, viz.: that two straisrht lines which meet one another make with any third line the interior angles less than two right angles. Geometers perceiv- ing this blemish in the Elements of Euclid, have endeavored in many ways to remove it, but without complete success. They employed three methods for this purpose : 1, By adopting a new definition of parallel lines. 2. By introducing a new axiom. 3. By reasoning from the definition of parallel lines, and the properties of lines already demonstrated.* The diffi- culty with parallel lines is, that geometers have confounded a definition with a proposition. Definition 11 is perfectly legiti- mate, as it simply defines what kind of lines are parallel ; but when it is inferred from it that these lines are equally distant from each other, this is no axiomic inference, because the curve and its asymptote are two lines which never meet, however far they be produced on the same plane, but they are not equally distant from each other ; hence the inference that parallel lines are equally distant, embodies a question which requires a dem- onstration to establish ; and to establish this question has given perplexity to geometers, for though they have proven the lines equally distant at particular points, they have not proven them so at every point; and here consists the incompleteness of their demonstrations, and here is required some general demonstra- tion which will embrace every part of the lines, however so far they be produced on the same plane.f 8. A postulate is a problem so easy to perform that it does not require any explanation of the manner of doing it, so that the geometer reasonably expects the method to be known ; for instance, it is granted as — * See notes to Playfair's Euclid, Legendre's Geometry, Leslie's Geom- etry, the ea-cursus to the tirst book of Camerer's Euclid, Berlin, 1825 ; Col. P. Thomson's Oeometry xoithout Axioim, Professors Thomson's and Simson's editions of Euclid — London, Glasgow, and Belfast. f See fifteenth and nineteenth propositions cf this book. 14 THE ELEMENTS OF [bOOK I. POSTULATES, 1. That a straight line can be drawn from any one point to any other point. 2. That a terminated straight line may be extended to any length in a straight line. 3. That a circle may be described from any center, at any distance from that center. EXPLAIS'ATION OF SIGSTS. In Algebra, the sign +, called Plus {more Jy), placed between the names of two magnitudes, is used to denote that these mag- nitudes are added together ; and the sign — , called Minus {less hy)^ placed between them, to signify that the latter is taken from the former. The sign z=, which is read equal to, signifies that the quantities between which it stands are equal to one another. The sign =o=, signifies that the quantities between which it stands are equivalent to one another. In the references, the Roman numerals denote the book, and the others, when no word is annexed to them, indicate the proposition ; otherwise the latter denote a definition, postulate, or axiom, as specified. Thus, III. 16 means the sixteenth proposition of the third book ; and I. ax. 2, the second axiom of the first book. So also hyp. denotes hypothesis, and const, construction. PROPOSITIONS. Prop. I. — Problem. — To describe an isosceles triangle on a finite straight line given in position. Let AX be the given straight line; it is required to describe an isosceles triangle having its base on AX. From a point C, without the line AX as a center, and a radius CA (I. post. 3), de- scribe a circle ABED, cutting the line AX in two points A and B; draw from these points the straight lines AE and BD (I. post. 1) passing through the center of the circle ; the triangle ACB is the one required. Because C is the center of the circle ABED (I. def 16), CA BOOK I.J EUCLID AND LEGENDRE. 15 is equal to CB, therefore the triangle ACB has two sides equal ; hence (I. def. 13) it is isosceles, and is described on AX, which was required to be done. Corollary 1. But the angle EAB is subtended by the arc EB (I. def. 19), and the angle DBA is subtended by the arc DA ; since AE and DB pass through the center of the circle (const,), they are both diameters of the circle (I. def 17) ; hence the arcs DEB and ADE are each a semicii-cumference, and (I. ax. 1) are equal ; therefore the sum of tlie arcs BE and ED is equivalent to the sum of the arcs ED and DA ; the arc ED is common ; hence (I. ax. 3) we have the arc EB equal to the arc DA ; therefore the angles EAB and DBA are subtended by equal arcs, consequently (I. def 19) the angles are equal. Hence, in an isosceles triangle, the angles opposite the equal sides are equal. Cor. 2. The line AE, which forms with AB the angle EAB, intercepts the line DB at C, which forms with AB the angle DBA, and the line DB intercepts AE at C also. C being the center of the circle ABED, CB, that portion of BD intercepted hy AE, is equal to CA, that portion of AE intercepted by BD (I. def 16) ; but CB and CA are the sides of the triangle ACB (I. 1) ; hence, when two angles of a triangle are equal, the opposite sides to them are also equal, and the triangle is isos- celes (I. def 13). Pkop. II. — Problem. — To describe an equilateral triangle on a finite straight line given in magnitude. Let AB be the given straight line ; it is required to describe an equilateral triangle having AB for its base. From A as a center, and a radius AB (I. post. 3), describe the circle FCD ; and from B as a center, and a radius BA, describe the circle HCE. The circles having equal radii (I. ax. 1) are equal ; draw from C through the center B, CH ; and from C through the center A, CF (I. post 1) ; the triangle ACB is the one required. Because the circles have equal radii (const.), AC is equal to 16 THE ELEMENTS OF [bOOK I. AB, and CB is equal to AB ; hence (I. ax. 1) the three sides of the triangle ACB are equal; the triangle (I. def. 13) is equi- lateral, and is described on AB, which was required to be done. Corollary 1. If AB be produced both ways (I. post. 2) to D and E, the angle DBC is subtended by the arc CD, and the angle FCB is subtended by the arc FB; the arcs OB and BF are together equivalent to the arcs BC and CD (I. ax. 1) ; hence (I. 1, cor. 1) the arc BF is equal to the arc CD, therefore (I. def 19) the angle FCB is equal to the angle DCB. Again: the angle ACH is subtended by the arc AH, and the angle CAE is subtended by the arc CE. But (I. ax. 1) the arcs AC and CE are equivalent to the arcs CA and AH, and (I. ax. 3) the arcs CE and AH are equal; therefore (I. def 19) the angles ACH and CAE are equal, but the angle ACH is the same as the angle FCB ; hence (I. ax. 1) the three angles of the triangle are equal ; therefore in an equilateral triangle the angles are equal. And in a manner similar to Cor. 2 of the first proi)osi- tion,it can be shown, conversely^ that when a triangle has three equal angles, the sides opposite them are also equal ; hence an equilateral triangle is also equiangular, and, conversely^ an equi- angular triangle is also equilateral. Prop. IH. — Theorem. — If two triangles have txoo sides of the one equal to two sides of the other, each to each, and have also an angle in one equal to an angle in the other simiiiarly situated with respect to those sides, the triangles have their bases or remaining sides equal / their other angles equal, each to each, viz., those to which the equal sides a.e opposite, and the triangles are equal. This general proposition has four cases, viz. : first, when the equal angles are contained by the respectively equal sides; Becond, when the equal angles are opposite to one pair of the respectively equal sides ; third, when the equal angles are op- posite to the other pair of the respectively equal sides ; and fourth, the limitation that when the least sides respectively of the triangles be equal, and the angles opposite the least sides "be equal, the angles opposite the greater of the respectively equal sides must be of the same kind, either both acute, or not acute. BOOK I.] EUCLID AND LEGENDRE. 17 First case. Let ABC and DEF be the two triangles having any two sides equal, each to each, viz., AC and CB equal to EF and DF, and the contained angles ACB and EFD equal ; the remaining sides AB and DE are equal, tlie angle CBA opposite AC equal to the angle FDE opposite FE, the angle CAB opposite CB equal to the angle FED opposite DF, and tlie triangles ABC and DEF are equal. If the triangle ABC be placed on the triangle DEF so that the vertex of the angle ACB will fall on the vertex of the angle DFE, the angle ACB being equal to the angle DFE (hyp.), the side CB will fall on FD, and the side CA will fall on FE ; CB and FD being equal (hyp.), the extremity B will fall on the extremity D. CA and FE being equal (hyp.), the ex- tremity A will fall on the extremity E ; and since AB is a straight line, it will coincide with DE (I. def. V), a straight line drawn from D to E. Therefore the triangle ABC has its three sides coinciding with the three sides of the triangle DEF ;; hence the angle CAB will fall on the angle FED, and be equal to it ; the angle CBA will fall on the angle FDE, and be equal to it ; consequently the two triangles have their three sides and. three angles equal, each to each, and (I. ax. 8) are equal. Second case. When the triangles ABC and DEF have the sides CA and CB respectively equal to FE and FD, and the angles ABC and EDF equal, respectively opposite to CA and FE, the remaining sides are equal ; the angle CAB opposite CB is equal to the angle FED opposite to FD, the angle ACB op- posite to AB is equal to the angle EFD opposite to DE, and the triangles are equal. Let the side DE b be put on AB so that D will fall on B, and the eqnal angles ABC and EDF will be on different sides of AB ; join CF (I. post. 1 ). BC and BF (hyp.) are 2 18 THE ELEMENTS OF [bOOK I. equal, the triangle CBF (I. def. 13) is isosceles, and (I. 1, cor. l) the angle BCF eqiial to the angle BFC; and hecause CA and, AF are equal (hyp.), the triangle CAF (I. def. 13) is isosceles, and (1, 1, cor. 1) the angle ACF is equal to the angle AFC ; then (I. ax. 2) the angles BCF and ACF are equal to the angles BFC and AFC, or the angle BCA equal to the angle BFA (I. ax. 1 and ax. 10). Hence we have in the triangles ABC and ABF, two sides, and the contained angle in each equal, .-«ach to each, therefore hj first case the triangles can be shown .-.equal in all respects. Third case. It can be proven in a similar manner as the second * case. When the angles CAB and FED of the second case are , obtuse, and the angles CBA and FDE of the third case are . obtuse, the proofs are given by the third axiom of the first book. Fourth case. When the triangles ABC and DEF have their least sides in each equal — viz., AB to DE — and another side in each, equal, the angles ACB and EFD being equal, the angles opposite the second pair of equal sides must be both acute or both not acute ; otherwise, two triangles can be formed having two sides .and an augl'-^ in each equal, each to each, and ? the triangles unequal. For, in the triano-lcs ABC and ACD, the side AC is common, the angle BAC equal to angle CAD, the sides BC and CD can be equal, and the tri- angles (I. ax. 9) unequal; hence in two triangles when the greatest and least sides are respectively equal, and the equal angles opposite to the least sides be given, the angles opposite the greatest sides must both be not acute to determine the tri- angles ; but when in two triangles the two less sides of each are respectively equal, and the equal angles opposite tlie least sides be given, the angles opposite the other equal sides must both be acute, to determine the triangles. The equality of the triangles can be proven by {ho second and third cases, using the second axiom when the angles are acute, and the third axiom when the angles are obtuse ; but when the angles are right-angled, the equality of the triangles is shown from tlie first corollary to the first proposition without those axioms. !BOOK I.] EUCLID AND LEGENDKE. 19 B Peop. IV. — Theok. — If the three, sides of one triangle be ■equal to the three sides of another, each to each: (1) the ajigles ■of one triayigle are equal to the angles of the other, each to each^ viz., those to which tJie equal sides are 02)posite, and (2) the triangles are equal. Let ABC and DEF be the two tri- c p angles having then- three sides equal, viz., AB to DE, CA to FE, and CB to FD, the angles are equal, viz., ACB to EFD, CAB to FED, and CBA to FDE ; and the triangles are equal. If the side DE be placed on the side AB so that the triangles Avill fall on different sides of AB, D will fall on B, and E on A, because DE is equal to AB, and the triangle DEF will take the position BFA, BF being the same as DF, and FA the same as FE. Join CF, a and because (hyp.) BC is equal to BF, the angles BCF and BFC are equal (I. 1, cor. 1). It would be shown in a similar manner that the angles FCA and CFA are equal, therefore (I. ax. 2) the angles BCA and BFA are equal — that is (I. ax. 1), the angles BCA and DFE are equal. But (hyp.) the sides CB and FD are equal, and the sides CA and FE, and it has been shown that the contained angles are equal, therefore (I. 3, first case) the other angles are equal — that is, CAB to FED and CBA to FDE, and the triangles are equal. Wherefore, if the three sides, etc. Prop. V. — Prob. — To bisect a given angle, that is, to divide it into tico equal angles. Let BAC be the given angle ; it is required to bisect it. From A as a center, and AD less than AB (I. post. 3), de- scribe the arc DE ; draw the chord DE, then upon DE, on the side remote from A, describe an equilateral triangle (I. 2), DFE, then join AF ; AF bisects the angle BAC. 20 THE ELEMENTS OF [book I. Because AD is equal to AE (L def. 16), and AF is comraoii to the two triangles DAF and EAF, the two sides DF and EF are equal (I. 2) ; therefore the two tri- angles DAF and EAF have their three sides equal, each to each, and the triangles are equal (I. 4) ; consequently the angle DAF opposite DF is equal to the angle EAF opposite EF, and the angle BAG is bisected by the line AF; which was to be done. OTHERWISE, Let BAG be the given angle ; in AB take any two points as B and D, and cut off AG and AE respectively equal to AB and AD, join BE and GD, and the straight line joining the inter- section of BE and CD with the vertex A bisects BAG. The proof is easy, and is omitted to exercise the ingenuity of the pupU. Prop. VI. — Prob. — To bisect a given finite straight line. Let AB be the given line ; it is required to bisect it. Describe (L 2) upon it an equilateral triangle ABG, and C bisect (I. 5) the angle AGB by the straight line GD ; AB is bisected in the point D. Because AG is equal to GB, and CD common to the two triangles AGD, BGD, the two sides AG, GD are equal to BG, B CD, each to each ; and the angle AGD is equal (const.) to the angle BGD; therefore the base AD is equal (L 3) to the base DB, and the line AB is bisected in the point D ; which was to be done. Sc/io. In practice, the construction is effected more easily by describing arcs on both sides of AB, from A as a center, and "with any radius greater than the half of AB ; and then, by de- gcribing arcs intersecting them, with an equal radius, from B as center, the line joining the two points of intersection will bi- sect AB. The proof is easy. Peop. VII. — Peob. — To draw a straight litie perpendieuiar BOOK I.] EUCLID AND LKGENDRE. 21 to a given straight line, from a giveyi point in that straight line. Let AB be the given straight line, and C a point given in it ; it is required to draw a perpendicular from the point C. From C as a center, and a radius CE (I. post 3), describe the semicircle EHF; then (I. def. 16) EC is equal to CF, and on EF (I. 2) describe the ^ equilateral triangle EDF ; then a line from C to the vez'tex D is the perpen- dicular required. Because EC is equal to CF (I. def. 16), ED is equal to DF (I. 2), and the angle DEC equal to angle DFC (I. 2, cor. 1) ; hence (I. 3) the triangles ECD and FCD are equal ; but the angle ECD is equal to the angle FCD, and therefore (I, def 1 0) DC is perpendicular to AB from C. Pkop. Vin. — Prob. — To draw a straight line perpendicular to a given straight line of an unlimited length, from a given •point without it. Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it. It is required to draw a straight line from C perpendicular to AB. Take any point D upon the other side of AB, and from the center C, at the dis- tance CD, describe (L post. 3) the circle ADB meeting AB in A and B ; bisect (I. 6) AB in G, and join CG ; the straight line CG is the perpendicular required. Join CA, CB. Then, because AG is equal to GB, and CG common to the triangles AGC, BGC, the two sides AG, GC are equal to the two, BG, GC, each to each ; and the base CA is equal (I. def 1 6) to the base CB ; therefore the angle CGA is equal (I. 4) to the angle CGB ; and they are adjacent angles ; therefore CG is perpendicular (I. def 10) to AB. Hence, from the given point C a perpendicular CG has been drawn to the •given line AB ; which was to be done. Scho. This proposition and the preceding contain the only .two distinct cases of drawing a perpendicular to a given straight 22 THE ELEMENTS OF [BOOK B- line through a given point ; the first, when the point is in the line ; the second, when it is zcithout it. In practice, the construction will be made rather more simple by describing from A and B, when found, arcs on the remote side of AB from C, with any radius greater than the half of AB, and joining their point of intersection with C. Prop. IX. — T^eor. — "When one straight line meets another straight line ayid forms two unequal angles on the same side of that line, the two angles will be equivalent to two right angles. ^ Let the straight line DC meet the straight line AB and form the two un- equal angles DCA and DCB on the same side of AB ; the two angles will be equiva- c lent to two right angles. At C, where the line DC meets AB, draw a perpendicular to AB from C (I. '7), then the angles ACE and ECB are two right angles (I. def. 10). But (I. ax. 10) the angle ECB is equivalent to the angles ECD and DCB both together ; likewise the angles ACE and ECB together are equivalent to the angles ACD and DCB both together; hence (I. ax. 1) the angles ACD and DCB together ai-e equivalent to two right angles. Wherefore, when one straight line meets, etc. Cor. Hence, if the straight line DC be pi-oduced on the other side of AB, the four angles made by DC produced and AB are together equivalent to four right angles. Hence, also, all the angles formed by any number of straight lines intersecting one another in a common point are together equivalent to four right angles. Prop. X. — Theor. — 7)^, at a point in a straight line, tico other straight lines on the ojyposite sides make the adjacent angles together equivalent to two right atigles, those two straight lines are in one and the same straight line. Let DC be the straight line which makes, at the point C, with AC and CB, two adjacent angles ACD and DCB togethei- equivalent to two right angles ; AC and CB are in one and the- same straight line. — B BOOK I.] EUCLID AND LEGENDEE. 23 From C draw a perpendicular to AC (I. ^ 7), then the angle ACE is a right angle (I. def. 10). But (hyp.) ACD and DCB are together equivalent to two right angles, so ACE and ECB are equivalent * to two right angles (I. ax, 1) ; hence ACE being a right angle (const, and I. def. 10), ECB must also be a right angle; then EC is perpendicular to CB (I. def. 10), and the angles ACE and ECB are equal (I. ax. 11) ; therefore (I. def 10) EC is a straight line which makes two equal angles with AB. But AC and CB make with EC the same equal angles ; hence AC and CB are the same straight line with AB. And DC makes with AC and CB (hyp.) two adjacent angles equivalent to two right angles, but DC makes with AB (I. 9) the same angles equivalent to two right angles ; hence AC and CB are the same straight line with AB (I. def 7). OTIIER^VISE, It is proven (I. 9) that the angles ACD and DCB are to- gether equivalent to two right angles ; then, as C is a point in AB, AC can be one line and CB another (I. def 7, scho.) ; hence AC and CB are in one and the same straight line, be- cause the two unequal angles ACD and DCB are equivalent to two right angles (I. 9). Wherefore if, at a point, etc. Prop. XI. — Theoe. — If two straight lines cut one another^ the vertical or opposite angles are equal. Let the two straight lines AB, CD cut one another in E ; the angle AEC is equal to the angle DEB, and CEB to AED. Because the straight line AE makes with CD the angles CEA, AED, these angles are together equivalent (I. 9) to two risrht angles. Again : because DE n makes with AB the angles AED, DEB these also are together equivalent to two right angles; and CEA, AED ^ have been demonstrated to be equivalent to two right angles ; wherefore (I. ax. 11 and 1) the angles CEA, AED are equal to the angles AED, DEB. Take away the common angle AED, and (I. ax. 3) the remaining angles CEA, DEB are equal 24 THE ELEMENTS OF [bOOK I. lu the same manner it can be demonstrated that the angles CEB, AED are equal. Therefore, if two straight lines, etc. Cor. If at a point in a straight line two other straight lines meet on the opposite sides of it, and make equal angles with the parts of it on opposite sides of the point, the two straight lines are in one and the same straight line. Let AEB be a straight line, and let the angles AEC, BED be equal, CE, ED are in the same straight line. For, by adding the angle CEB to the equal angles AEC, BED, we have BED, BEC together equal to AEC, CEB, that is (I. 9), to two right angles ; and therefore, by this proposition, CE, ED are in the same straight line. Scho. In the proof here given, the common angle is AED ; and CEB might with equal propriety be made the common angle. In like manner, in proving the equality of CEB and AED, either AEC or BED may be made the common angle. It is also evident, that when AEC and BED have been proved to be equal, the equality of AED and BEC might be inferred from the ninth proposition, and the third axiom. Prop. XIL — Pkob. — To describe a triangle of which the sides shall be equal to three given straight lines; but any txoo of these must be greater than the third. Let A, B, C be three given straight lines, of which any two are greater than the third ; it is required to make a tri- angle of which the sides shall be equal to A, B, C, each to each. Take an unlimited straight line DE, and let F be a point in it, and make FG equal to A, FH to B, and HK to C. From the center F, at the distance FG, describe (I. post. 3) the circle GLM, and from the center H, at -E the distance HK, describe the circle KLM. Now, because (hyp.) g FK is greater than FG, the cir- " ^ cumference of the circle GLM will cut FE between F and K, and therefore the circle KLM can not lie wholly within the circle GLM. In like manner, be- , cause (hyp.) GH is gi-onter than HK, the circle GLM can not lie wholly witluu the ciicle KLM. Neither can the circles be BOOK I.] ETTCLID AND LEGENDRE. 25 wholly without each other, since (hyp.) GF and HK are to- o-ether jrreater than FH, The circles must therefore intersect each other ; let them intersect in the point L, and join LF, LH ; the triangle LFII has its sides equal respectively to the three lines A, B, C. Because F is the center of the circle GLM, FL is equal (I. def. 16) to FG; but (const.) FG is equal to A; therefore (I. ax. 1) FL is equal to A, In like manner it may be shown that HL is equal to C, and (const.) FH is equal to B ; therefore the three straight lines LF, FH, HL are respectively equal to the three lines A, B, C ; and therefore the triangle LFH has been constructed, having its three sides equal to the three given lines, A, B, C ; which was to be done. Scho. It is evident that if MF, MH were joined, another tri- angle would be formed, having its sides equal to A, B, C, It is also obvious that in the practical construction of this problem, it is only necessary to take with the compasses FH equal to B, and then, the compasses being opened successively to the lenorths of A and C, to describe circles or arcs from F and H as centers, intersecting in L ; and lastly to join LF, LH. The construction in the proposition is made somewhat differ- ent from that given in Simson's Euclid, with a vicAv to obviate objections arising from the application of this proposition in the one that follows it. Prop. XIII. — Prob. — At a given point in a given straight line, to make a rectilineal angle equal to a given one. Let AB be the given straight line, A the given point in it, and C the given angle ; it is required to make an angle at A, in the straight line AB, that shall be equal to C. In the lines containing the angle C, take any points D, E, and join them, and make (L 12) the triangle AFG, the sides of which, AF, AG, FG, shall be equal to the three straight lines CD, CE, DE, each to each. Then, because FA, AG are equal to DC, CE, each to each, and the base FG to the base DE, the angle A is equal (I. 26 THE ELEMENTS OF [book 4) to the angle C. Therefore, at the given jsoint A in the given straight line AB, the angle A is made equal to the given angle C ; which was to be done. Scho. The construction is easy, by making the triangles isos- celes. In doing this, arcs are described with equal radii from C and A as centers, and their chords are mad« equal. It is evident that another angle might be made at A, on the other side of AB, equal to C. Prop. XIV. — Theor. — If two angles of one triangle he equal to two angles of another^ each to each, and if a side of the one he equal to a side of the other siniiliarly situated with respect to those angles ; (l) the remaining sides are equal, each to each ; (2) the remaining angles are equal ; and (3) the triangles are equcd. This proposition is the converse of the third proposition, and is susceptible of three cases, viz. : first, when the equal sides are between the equal angles ; secondly and thirdly, when the equal sides are opposite to the equal angles similarly situated. Q J, Let ABC and DEF be two triangles M-hich have the angles ACB and EFD equal; the angles BAG and DEF equal, and the sides CA and FE equal ; then the sides CB and FD are equal, also the sides AB and DE ; the angles CBA and FDE are equal, and the tri- angles are equal to one another. If the triangles be placed so as to have their sides CB and FD in the same straight line, but the triangles be on opposite sides of that line, and the vertex C on the vertex F ; then because the angles AFD and EFD are equal (hyp.), the angle AFE is bisected by FD ; and because AF and FE are equal (hyp.), the triangle AFE is isosceles (I. def 13), and the line AE (T. 6) is also bisected by FD; hence (I. 1, cor. 1) the angles FAE and FEA are equal ; therefore (I. 3) the triangles FAH and FEII are equal. But the angles FAD BOOK I.] EUCLID AND LEGENDKE. 27 and FED are eq^^al (byp.) ; tlien taking from each the equal angles FAII and FEH, there will remain (I. ax, 3) the angle HAD equal to the angle HED; hence (I. 1, cor. 2) the triangle AED is isosceles, and (I. def. 13 and I. cor. 2) the side AD is equal to the side ED, and AE being bisected by FD (I. 6), we have the triangles AHD and EHD (I. 3) equal ; therefore the angles HDA and HDE are equal; hence the triangles AFD and EFD have the sides AF and EF (hyp.) equal, the angles FAD and FED equal (hyp.), and the sides AD and DE equal (I. def 13 and I, cor. 2) ; therefore (I. 3) the angles FDA and FDE are equal ; the side FD is common, and the two triangles are equal. In a similar manner, the second and third cases can be dem- onstrated. Wherefore, if two angles of one triangle be, etc. Cor. Hence, from this proposition, it can also be shown that the second corollary to the first proposition is true. Let the triangle ABC have the angle CAB equal c to the angle CBA, then will AC be equal toCB. From the vertex C (I. S) draw CD per- pendicular to AB, then the angles CDA and CDB are both right angles (I. def 10), ^ the angles CAD and CBD are equal (hyp.), and the side CD coni- mon (const.); therefore (I. 14) the sides AC and CB are equal. 8cho. It will be seen from propositions third, fourth, and fourteenth of this book, that two triangles are in every respect equal when the three sides of the one are respectively equal to the three sides of the other, when two triangles have two angles and a side in each equal, each to each, and when one angle and two sides of one are equal to one angle and two sides of the other, each to each, and the equal angles in each triangle simi- larly situated with respect to those sides, but with this limita- tion, that when two equal sides respectively of the triangles are the least sides, that the angles opposite the greater sides respec- tively of the triangles must be of the same kind, either both acute, both right-angled, or both obtuse. And from these prop- ositions it is shown that of the sides and angles of a triangle, three must be given to determine the triangle, and these three can not all be angles. Were only three angles given, the sides, 28 THE ELEMENTS OF [bOOK I. as will appear from the twentieth prosposltion of this book, might be of any magnitude whatever. The pupil may occupy himself in proving these propositions by superposition or some other way, as by pursuing a course of demonstration different from what is given in the text, he will more readily familiarize himself with the process of geometrical reasoning. Prop. XV. — Theor. — Parallel straight liries are equally/ dis- tant from each other, however so far they be produced on the same plane. Let the straischt line AB be bisected (I. 6) at C, and let the perpendicular CD be drawn (I. 7) ; join AD and A c ' ^ BD, and if the triangle ADC be applied to the triangle BDC so that they will fall on difler- €nt sides of BD and have D and BD common, their sides DE and CB will be equally distant from each other, and so they will never meet, however so far they be produced on the same plane, and consequently (I. def 11) are parallel straight lines. Because DC is perpendicular to AB (const.), the angles ACD and BCD are both right angles (I, def 10), and are equal (I. ax. 11) ; hence the triangles ADC and BDC have the side DC common, the sides AC and CB equal (const.), and the angles ACD and BCD equal (I. def 10 and ax. 11); therefore (I. 3) the triangles are equal, having the sides AD and BD equal ; the angle ADC equal to the angle BDC, and the angle DAC equal to the angle DBC. Now, when ADC is applied to BDC so that they will fall on difterent sides of BD, and have D and BD common (hyp.), since AD is equal to BD, the point A will fall on B ; hence the angle EBD will be the same as ADC, and equal to BDC, the angle BED the same as ACD, and equal to DCB, the angle EDB the same as DAC, and equal to DBC, the side DE the same as AC, and equal to AC ; the side EB equal to DC; therefore the triangles BDC and BDE are equal (I. ax. 8), having their sides and angles equal, each to each. Since DE and CB are straight lines, they have no variation in the direction of their lengths from D to E or from C to B (I. def 7) ; and because DC is equal to EB, DE and CB are BOOK I.] EUCLID AND LEGENDEE. 29 equally distant from each other at their extremities, and having no variation in the direction of their lengths from D to E and from C to B (I. def. 7), they are also equally distant from each other at every part between D and E and C and B, each to each ; therefore DE and CB (I. post. 2) on being produced to any length are still the same straight lines, and will have no variation in the direction of their lengths (I. def. 7), conse- quently they will always be the distant DC or EB from each other, at every part, each to each ; and being always the same distant DC or EB from each other, will never meet, and are parallel straight lines (I. def 11). Wherefore, parallel straight lines, etc. Cor. 1. In like manner, it can be shown that DC and EB are parallel lines; hence DCBE is a parallelogram (I. def 14) ; and since the angles DCB and DEB are equal, and the angles EDB and BDC equal to the angles EBD and DBC (I. ax. 2), and the sides DE and CB equal, also the sides CD and BE equal, a parallelogram has its opposite sides and opposite angles equal. Cor. 2. Hence parallel lines, DE and CB, intercepted by a straight line, DB, make the alternate angles EDB and CBD equal; and, co?it»er5eZy, when the alternate angles EDB and CBD are equal, the lines DE and CB are parallel. Cor. 3. In the parallelogram, the opposite sides being equal, the straight lines which join the extremities of two equal and parallel straight lines toward the same parts — that is, the near- est extremities together — are themselves equal and parallel ; hence a quadrilateral which has two sides equal and parallel is (I. def. 14) a parallelogram. Cor. 4. Because the triangles DCB and DEB are equal, a diagonal, DB, bisects the parallelogram ; and if two parallelo- grams have an angle of the one equal to an angle of the other, and the sides containing those equal angles respectively equal, the parallelograms are equal, as the parallelograms can be bi- sected by diagonals subtended by the equal angles, and the tri- angles thus formed are equal (I. 3) ; hence (I. ax. 6) the par- allelograms are equal; hence, also, if a pai'allelograra and a tri- angle be upon the same or equal bases, and between the same parallels, the parallelogram is double the triangle. 30 THE ELEMENTS OF [bOOK I. Cor. 5. Hence, also, parallelograms upon the same or equal bases, and between the same parallels, are equal ; and triangles upon the same or equal bases and between the same parallels, are equal. Cor. 6. Hence, from the preceding corollary, it is plain that, triangles or parallelograms between the same parallels, but upon unequal bases, are unequal. Cor. T. And a straight line drawn from the vertex of a tri- angle to the point of bisection of the base, bisects the triangle ; and if two triangles have two sides of the one respectively equal to two sides of the other, and the contained angles supple- mental (I. def 20), the triangles are equivalent; the converse is also true. Cor. 8. If through any point in either diagonal of a parallelo- gram straight lines be drawn parallel to the sides of the four parallelograms thus formed, those through which the diagonal does not pass, and which are called the com2)lements of the other two, are equivalent. Peop. XVI. — Theoe. — If a straight linefallupon tioo parallel straight Ihies, (l) it makes the alternate angles equal to one an- other ; (2) the exterior angle equal to the interior and remote upon the same side, and (3) the tico interior angles upon the same side together, equivalent to two right angles. Let the straight lines AB, CD be parallel, and let EF fall upon them; then (l) the alternate angles AGH, GHD are equal to one another ; (2) the exterior angle EGB is equal to the interior and remote upon the same side, GHD; and (3) the two in- terior angles BGH, GHD ujDon the same side are together equivalent to two right angles. Since AB, CD are parallel (hyp.), theper- B pendiculars (I. 7) MG, LH make the angles MGH, GIIL equal to one another (I. 15, cor. 2), and AG3I, LHD are two riglit .jj angles (I. def 10); if Ave add AGM to MGH they will be equal to AGH (I. ax. ^' 10), and DHL added to LIIG are likewise equal to GIID ; hence (I. ax. 2) AGH and GHD are equal to one another. BOOK I.] EUCLID AND LEGENDRE. 31 Second. AGII is equal to EGB (I. 11), therefore (I. ax, 1) EGB is equal to GHD. Third. Add to EGB and GHD, each, the angle BGII ; there- fore (I. ax. 2 and ax. 10) EGB and BGH are equivalent to the angles GHD and BGH, but EGB and BGH are equivalent to two right angles (I. 9) ; therefore, also, BGH and GHD are to- gether equivalent to two right angles. Wherefore, if a straight line, etc. Cor. 1. Hence, conversely, tAvo straight lines are parallel to one another, if another straight line falling on them (1) makes the alternate angles equal ; (2) the exterior angle equal to the interior and remote upon the same side of that line ; and (3) the two interior angles upon the same side together equivalent to two right angles. Let EF fall on AB and CD, the perpendiculars (I. 1) GM and HL make two riglit angles (I. def. 10), AGM and DHL. But AGH and DHG are equal (hyp.) ; hence (L ax. 3) MGH is ■equal to LHG. But MGL and LHM are both right angles (const, and L def. 10) ; hence (I. ax. 3) HGL is equal to GHM; •therefore in the triangles HMG and GLH we have two angles in one equal to two angles in the other, each to each, and the side GH common, and (I. 14) the triangles are equal; hence GM and HL are equal, and (L 15) AB and CD are equally dis- tant from each other, and will never meet on being produced {L post. 2), and are parallel (L def' 11). Because EGB is equal to DHG (hyp.), and EGB equal to AGH (I. 11), the angle AGH is equal to DHG (L ax. 1), but they are alternate angles; therefore (L 16, cor. 1, part l) AB is parallel to CD. Again : because BGH and GHD are together equivalent to two right angles (hyp.), and AGH and BGH are also equivalent to two right angles (I. 9), the angles AGH and BGH are together equal to BGH and GHD (I. 7 and ax. 1) ; then (L ax. 3) AGH is equal to GHD, but they are alternate angles, therefore (I. 16, cor. 1, part l) AB and ED are parallel. Wherefore, two straight lines are parallel, etc. Cor. 2. When one angle of a parallelogram is a right angle, all the other angles are right angles; for since (I. 10) BGM and GMH are together equivalent to two right angles, if one of them be a right angle, the other must also be a right angle, and 32 THE ELEMENTS OF [boOK I. (I. 15, cor. 1) the opposite angles are equal. A rectangle, then (L def. 15), has all its angles right angles. Cor. 3. If two straight lines make an angle, two others par- allel to them contain an equal or supplemental angle ; thus LGM, and the vertical angle produced by AB and the continuance of MG through G, are each equal to LHM, while the angle AGM, and its vertical angle contained by GB and the continuance of MG, are each equal to the supplement of LHM ; hence we can divide a given straight line AB into any proposed number of equal parts. Prop. XVH. — Theor. — Two straight lines which are not in the same straight line, and which nre parallel to a third straight line, are parallel to one another. Let the straight lines AB, CD be each of them parallel to the straight line EF ; AB is also parallel to CD. Let the straight line LH cut AB, CD, EF ; and because LH L cuts the parallel straight lines AB, EF, the angle LGB is equal (L 16, part 2) to the . J} angle LHF. Again : because the straight line LH cuts the parallel straight lines CD, " ^ EF, the angle LKD is equal (L 1 6, part 2) to ^ ■ " — J. the angle LHF ; and it has been shown that the angle LGB is equal to LHF; wherefore, also, LGB is equal (L ax. 1) to LKD, the interior and remote angle on the same side of LH ; therefore AB is parallel (L 10, part 1) to CD. Wherefore, two straight lines, etc. Prop. XVIIL — Prob. — To draw a straight line parallel to a given straight line through a given poi^it without it. Let AB be the given straight line, and C the given point ; it is required to draw a straight line through C, parallel to AB. In AB take any point D, and join CD ; at the point C, in the c straight line CD, make (I. 13) the angle E — -^ F p^j, ^^^^^^ ^^ ^j^j, ^ ^^^ produce the A ^ B Straight line EC to any point F. Because (const.) the straight line CD, which meets the two straight lines AB, EF, makes the alternate BOOK I.] EUCLID AND LEGENDEE. 33 angles ECD, CDB equal to one another, EF is parallel (I. 15, eor, 2) to AB. Therefore the straight line EOF is drawn through the given point C parallel to the given straight linev AB ; which was to be done. • Prop. XIX, — Theok. — If a straight line meet two otho'^ straight lines xohich are in the same pkoie, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines shall at length meet upon that side, if they he continually i^i'oduced. Axiom twelfth. Elements of Euclid. Let EF be the straight line meeting AB, ^ CD on the same plane, so tliat BLO, DOL are less than two right angles, the lines AB, CD will meet if continually produced. At the point O, draw GH (I. 1 8) parallel g to AB, then BLO, HOL are equivalent to two right angles (L IG), but DOL is less c than HOL (I. ax. 9) ; therefore BLO, DOL are less than BLO, HOL, and less than two right angles. But GH, which fornij^ with EF the angle HOL, is parallel (const.) with AB, which! forms with EF the angle BLO, and (L def 11) GH and AB- never meet each other, because (L 15) they are equally distant from each other, and (L 16, cor. l) the interior angles HOL^ BLO together equivalent to two right angles ; therefore.^ then, CD, which forms Avith EF the angle DOL less than HOL, can not be parallel with AB (L ax. 9) ; and not being parallel wath AB, CD and AB can not preserve the condition of par- allel lines (L def 11), and will meet. And the line CD making with EF the angle DOL less than HOL, on the side of EF, where the interior angles BLO, DOL are less than two ricrht angles, the line OD must therefore be between GH and AB, and is consequently nearer to AB than GH is to AB (L ax. 9) ; but CD making with EF the angle EOC greater than the angle GOE, which GH makes with EF, therefore CO must be with- out AB and GH, and consequently is farther from AB than GH is from AB ; hence the straight line CD, Avhich is made of CO and OD, has parts unequally distant from AB ; therefore since CD approaches AB on the side of EF where OD is, CD must 3 S^ THE ELEMENTS OF [bOOK I. meet AB on that side wliea continuallj^ produced, but OD makes the angle DOL less than HOL, therefore CD will meet AB on the side of EF where the angles BLO, DOL are less than BLO, HOL. Wherefore, if a straight line, etc. Cor. 1. Hence a straight line which intercepts one of two or more parallel straight lines will intercept the others if continu- ally produced ; hence, also, two straight lines which intercept each other are not both parallel to the same straight line. Prop. XX. — Theor. — If a side of any triangle be produced, (1) the exterior angle is equivalent to the tico interior and re- mote angles ; and (2) the three interior angles of every triangle are together equivalent to two right angles. Let ABC be a triangle, and let one of its sides BC be pro- educed to D; (1) the exterior angle ACD is equivalent to the two interior and remote angles CAB, ABC; and (2) the three interior angles, ABC, BCA, CAB, are together equivalent to two right angles. Through tlae point C draw (L 18) CE parallel to the straight line AB. Then, because AB is parallel to EC, and AC falls '>iipon them, the alternate angles BAC, ACE are (L 16, part 1) equal Again: because AB is ijarallel to EC, and BD falls upon them, the exterior angle ECD is equivalent (L I G, part 2) to the interior and remote ano-le ^VBC : 'B^ ^ — D but the angle ACE has been sliown to F be equal to the angle BAC ; there- fore the whole exterior angle ACD is equivalent (I. ax. 2) to the two interior and remote angles CAB, ABC. To these equals add the angle ACB, and the angles ACD, ACB are equivalent (L ax. 2) to the three angles CBA, BAC, ACB ; but the angles ACD, ACB are equivalent (I. 9) to two right angles; therefore, also, the angles CBA, BAC, ACB are equivalent to two right angles. Wherefore, if a side, etc. Another proof of this important proposition can be given by producing the side AC through C to F. !N"ow (I. 11), the angle DCF is equal to the angle ACB ; EC being parallel (const.) to BA, the exterior angle ECD is equal to the intei'ior and remote angle ABC (L 10), and for the same reason the alternate angles r BOOK I.J EUCLID AND LEGENDRE. 35 EGA and CAB are equal; hence we have the three angles of the triangle, ACB, CBA, and CAB, equal to the three angles DCF, DCE, and EGA, each to each ; but (I. 9) tlie angles DCF, DCE, and ECA are equivalent to two right angles ; therefore (I. ax. 1) the three angles ACB, CBA, and CAB of the triangle are likewise equivalent to two right angles. And when a par- allelogram (I. def 14) is formed by drawing (I. 18) pai-allels to BA and AC respectively, it can be shown by the sixteenth proposition tliat two adjacent angles of the parallelogram are equivalent to two right angles, and the four angles together equivalent to four right angles; since (I. 15, cor. 4) a diagonal bisects the parallelogram and forms two equal triangles, the angles are also equally divided, hence each triangle has its three angles equivalent to two right angles. Cor. 1. All tlie interior angles of any rectilineal figure, to- gether Avith four right angles, are equivalent to twice as many right angles as the figure has sides. For any rectilineal figure can be divided into as many tri- angles as the figure has sides, by draAving straight lines from a point within the figure to each of its angles; and by the prop- osition, all the angles of these triangles are equivalent to twice as many right angles as there are triangles — that is, as there are rsides of the figure ; and the same angles are equivalent to the angles of the figure, together with the angles at the point Avhich is the common vertex of the triangles — that is (I. 9, cor.), together with four right angles. Tlierefore all the angles of the figure, together Avitli four i-ight angles, are equivalent (I. ax. 1) to twice as many riglit angles as the figure has sides. Scho. 1. Another proof of this corollary may be obtained by ■dividing the figure into triangles by lines drawn from any .ano-les to all the remote angles. Then each of the tAVO ex- treme triangles has tAVO sides of the polygon for tAvo of its sides, while each of the other triangles has only one side of the figure for one of its sides ; and hence the number of triangles is less by tAVO than the number of the sides of the figure. Biit the in- terior angles of the figure are evidently equivalent to all the interior angles of all the triangles — that is, to tAvice as many right angles as there are triangles, or tAvice as many right angles as the figure has sides, less the angles of two triangles — 36 THE ELEMENTS OP [bOOK I. tlfet is, four riglit angles. Hence, iu any equiangular figure, the number of the sides being known, the magnitude of each angle compared with a right angle can be determined. Thus, in a regular pentagon, the amount of all the angles being twice five *right angles less four— that is, six right angles, each angle will be one fifth j^art of six right angles, or one right angle and one fifth. In a similar manner it would appear that in the regular hexagon, each angle is a sixth part of eight right angles, or a right angle and a third ; that in the regular heptagon, each is a right angle and three sevenths ; in the regular octagon, a rio-ht ano;le and a half, etc. Cor. 2. All the exterior angles of any rectilineal figure are together equivalent to four right angles. Because each interior angle ABC, and the adjacent exterior ABD, are together equivalent (I. 9) to two right angles, therefore all the in- terior, together with all the exterior angles of the figure, are equivalent to twice as many right angles as there are B sides of the figure — that is, hj the fore- going corollary, they are equivalent to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equivalent (I. ax. .3) to four right angles. Sc/to. 2. It is to be observed, that if angles be taken in the ordinary meaning, as understood by Euclid, this corollary and the foregoing are not applicable when the figures have re-entrant angles — that is, such as open outward. The second corollary will hold, how- ever, if the difference between each re-entrant angle and tAvo riiiht ansrles be taken from the sum of the other exterior anccles: and the former will be applicable, if, instead of the angle which opens externally, the difterence between it and four right angles be used. Both corollaries, indeed, Avill hold without change, if the re-entrant angle be regarded as internal and greater than two right angles; and if, to find the exterior angles, the interior be taken, in the algeljraic sense, from two right angles, as in this case, the re-entering angles will give negative or subtractive results. BOOK I.] EUCLID AND LEGENDEE. 87 Cor. 3. If a triangle has a right angle, the remaining angles are together eqiiivaleut to a right angle ; and if one angle of a triangle be equivalent to the other two, it is a right angle. Cor. 4. The angles at the base of a right-angled isosctles tri- angle are each half a right angle. Cor. 5. If two angles of one triangle be equal to two angles of another, their remaining angles are equal. Cor. 6. Each angle of an equilateral triangle is one third of two right angles, or two thirds of one right angle. Cor. 1. Hence, a right angle may be trisected by describing an equilateral triangle on one of the lines containing the right angle. ScJio. 3. By this principle also, in connection with the fifth proposition, we may trisect any angle, Avhich is obtained by the successive bisection of a right angle, such as the half, the fourth, the eighth, of a right angle, and so on. Cor. 8. Any two angles of a triangle are less than two right angles. Cor. 9. Hence every triangle must have at least two acute angles. Peop. XXI. — Tiieor. — Iftxco sides of a triangle be unequal, (1) the greater side has the greater angle opposite to it ; and (2) conversely, if tioo angles of a triangle be unequal, the greater angle has the greater side opjiosite to it. D Let ABC be a triangle of which the side AB is greater than the side AC ; the angle ACB opposite AB is greater than the angle ABC opposite AC. Because AB is greater than AC, pro- duce AC (I. post. 2), and with A as a center, and a radius AB, describe a circle (I. post. 3) intercepting AC produced in D, and join BD ; the triangle ADB is isosceles (I. defs. 13 and 16) ; therefore the angle ADB is equal to the angle ABD (I. 1, cor. 1). But (I. 20) the exterior angle ACB is equivalent to the sum of the two remote interior angles CDB and DBC. And CDB is equal to DBA (I. 1, cor. 1), but ABC is less than ABD (I. ax. 9) ; therefore ACB is greater than CBA. The proof can also be given by laying oflf on the greater side 38 THE ELEMENTS OF [bOOK I. AB, tlie less side AC ; joining the vertex of the opposite angle with the point where AC terminates on AB ; and the demon- stration conducted similarly to the preceding. Again : by cutting oflf on AC a part equal to CB, hisect the angle BCA, and join the extremity of the part on AC equal to CB with the foot of the line bisecting the angle BCA ; this proof is given by means of (I. 3 and 20). Conversely: when the angle ACB is greater than the angle ABC, the side AB is greater than AC. At the vertex of BCA on AC make an angle ACD equal to the angle ABC (I. 13). Now, the angle ACD, the equal to the angle ABC, is subtended by AD, ~^ ^ ^ and the angle ACB is subtended by AB, but AB is greater than AD ; hence the greater angle is subtended by the greater line ; therefore, in the triangle ACB, the greater angle ACB is subtended by a greater side than the less angle ABC. And (L def. 19) if the angles ACD and ABC be subtended by arcs, the arc subtending ACB is greater than the arc subtending ABC ; but the side AB is the chord of the arc subtendins: ACB, and AC is the chord of the arc subtending ABC ; therefore AB is greater than AC. Wherefore, if two sides of a triangle, etc. Cor. Hence any two sides of a triangle are together greater than the remaining side. Scho. The truth of this corollary is so manifest, that it is given as a corollary to avoid increasing the number of axioms. Archimedes defined the straight line the shortest distance be- tween two points ; hence two straight lines connecting three points not in the same direction, are together greater than one straight line connecting any two of those jDoiuts. Prop. XXII, — Tiieor. — If two triangles have tico sides of the one equal to two sides of the other, each to each, hut the angles contained hg those sides loiequal, the base or remaining side of the one ichich has the greater angle is greater than the base or reraaining side of the other. Let DEG and DEF be two triangles which have the sides DE common, the sides DG and DF equal, but the angle EDG HOOK I.] EUCLID AND LEGENDEE. 39 greater tlian the angle EDF ; the side EG is also greater than the side EF. In the triang-les DEF and DEG Ave have (hyp.) DE commoii, DG equal to DF, and the an'He EDG crreater than the anHe EDF. Now, because DG and DF are equal, the angles DGF and DFG are equal (I. 1, cor. 1) ; but the angle DGF is greater than the angle EGF (I. ax. 9) ; therefore the angle DFG is greater than the angle EGF ; and much more is the angle EFG greater than the angle EGF. Then (I. 21), EG op- posite EFG is greater than EF opposite EGF. Wherefore, if two triangles have, etc. There are other cases of this proposition ; if a line equal to DE, the less side, be drawn through D, making with DF, on the same side of it with DE, an angle equal to EDG, the extremity of that line might fall on FE produced, or above, or helo^o it. Or the ano-le DFE could be in the triangle DEG, or on the other side of DE. And a very easy proof can be given by bisecting the angle FDG by a straight line cutting EG in a point, which call K, and joining DK and KF, for KG would be equal to KF (I. 3) ; adding EK, we would have EG equivalent to EK and KF; therefore (I. 21, cor.) EG greater than EF. Cor. Hence, conversely, if two triangles have two sides of the one equal to two sides of the other, each to each, but their bases unequal, the angles contained by the respectively equal sides of those triangles arc also unequal, the greater angle being in the triangle which has the greater base. For with a radius, DG, or its ecpial, DF, describe a circle (I. post. 3) from a center, D, and draw on the same side of DE with the angle EDF a line equal to EG from the other extremity of DE to the circumference, then in the triangles DEG and DEF we have (const.) DG equal to DF (I. def 16). But (hyp.) the side EF is less than EG ; hence the angle EDF is less than the angle EDG (I. ax. 9). Pkop. XXin. — Peob. — To describe a parallelogram upon a given straight line. Let DB be a given straight line ; it is required to describe a 40 THE ELEMENTS OF [book I. paralleloGjram. U])Ou it. From a point, D, on DB draw DF (I. F H B post. 1), then from the point F on DF draw FL parallel to DB (I. 18) ; and if through B, a point on DB, a parallel to DF (I. 18), he drawn, DBFG (I. def. 15) is a parallelogram. If from D a perpendicular, DII (I. 7), be drawn, then the angle HDB is a riglit angle (I. def 10) ; and from H a parallel to DB as HL be drawn (I. 18), and from B a parallel to HD as LB be drawn (I. 18), then DBIIL (I. def 15) is a rectangle. And if from D as a center, and a radius DB (I. post. 3), an arc be described intercepting DH, or DH produced, and a rectangle be described from that point where DH is intercepted, that rectangle will be a square (I. def 15). Co?: 1. Hence (const, and I. 15, cor. l) a square has all its sides equal, and (I. 15, cor. l) all its angles are right angles. Cor. 2. Hence the squares described on equal straight lines (I. 15, cor. 4) are equal. Cor. 3. If two squares be equal, their sides are equal (I. ax. 1). Cor. 4. If AB and AD, two adjacent sides of a rectangle BD, be divided into parts which are all equal, straight lines drawn through the points of section, parallel to the sides, divide the rectangle into squares which are all equal, and the number A B of which is equal to the product of the number of parts in AB, one of the sides, multiplied by the number of parts in AD, the other. For (const.) these figures are all parallelograms; and (I. c def. 15 and const.) the sides being equal, and the angles being (I. 16, part 2) equal to A, and there- fore right angles, hence (I. 15, cor. 4) they are all squares. Of these squares, also, there are evidently as many columns as there are parts in AB ; while in each column there are as many squares as there are parts in AD. The number of such squares con- tained in a figure is called, in the language of mensuration, the area of that figure. Cor. 5. Hence, since any parallelogram is equivalent (I. 15, cor. 5) to a rectangle on the same base and between the same BOOK I.] EUCLID AND LEGENDKE. 41 parallels, it follows that the area of any 2^'^^'>'(Melogram is equivale7it to the product of its base and its "perpendicidar height / and the area of a square is computed by multiplying a side by itself Cor. 6. Hence, also (I. 15, cor. 4), the area of a triangle is computed by m.idtiplying any of its sides by the perpendicidar draion to that side from the opposite angle., and taking half the product ; and the area of a trapezium is found by multi- P'lying either diagonal by the sum of the perpendicidar s draion to it from the ayigles which it subtends, and taking half the product. When, in consequence of one of the angles being re- entrant, the perpendiculars lie on the same side of the diagonal, the difference of the perpendiculars must evidently be used in- stead of their sum. Cor. 7. Every polygon may be divided into triangles or tra- peziums by drawing diagonals ; and therefore the area of any polygon whatever can be computed by finding the areas of those component figures by the last corollary, and adding them together. Scho. This corollary and the two foregoing contain the ele- mentary principles of the mensuration of rectilineal figures, and they form a connection between arithmetic or algebra and ge- ometry. They also explain the origin of the expressions, " the square of a number," " the rectangle of two numbers," and " the product of two lines." Prop. XXIV. — Theor. — If parallelograms be described on two sides of any triangle, and their sides which are parallel to the sides of the triangle be produced until they meet, the sum of the parallelograms will be equivalent to the parallelogram de- scribed on the base of the triangle having its adjacent sides to the base parallel to the straight line joining the vertex of the triangle with the point of intersection of the sides of the other parallelograms produced, and terminated by the latter sides or those sides produced. Let BAG be a triangle ; the parallelograms MB AD and CEFA, described on the two sides BA and CA, respectively, are to- gether equivalent to the parallelogram BCHK, described on the base BC ; the parallelograms MBAD, CEFA, and BCIIK 43 THE ELEMENTS OF [book I. beino; described ao;reeablv to the proposition. Describe on BA and CA (L 23) tlie parallelograms MB AD and CEFA ; and produce the M B L c E titles MD and EF until they meet in G ; draw GA, and produce it to L on the base BC (I. post. 2) ; describe the parallelogram BCIIK (I. 23) on BC. Then, since (const, and I. 15, cor. 1) BH and CK are parallel and equal to AG, they are parallel and equal to one another (I. ax. 1) ; also (1. 15, cor 3) HK is parallel and equal to BC ; hence (I. def. 15) BCHK is a parallelogram, and BLIIW and LCWK are also parallelograms. Xow, the parallelograms BLHW and HBAG (I. 15, cor. 5) are equal, and for similar reason the parallel- ograms HBAG and MBDA ; hence (I. ax. 1) MBDA is equal to BLHW. In similar manner, it can be shown tlnit CEFA is equiA^alent to LCKW; therefore the whole parallelogram BCHK (I, axs. 2 and 10) is equivalent to the sum of the two parallelo- grams MB AD and CEFA. Wherefore, if on any two sides of a triangle, etc. Cor. 1. A paiticular case of this proposition is, lohen the tri- angle is right-angled., then the sqiuires described on the legs — that is, the sides containing the right angle, are together equiva- lent to the square on the liypotlieniise — that is, the side opposite the right angle. Let ABC be a rio-ht-anoled triangle, having the right angle BAC ; the square described on tlie hypothenuse BC is equiva- lent to the sum of the squares on BA and AC. On BC, BA, Q and AC (L 23) desciibe the squares BCHK, BADE, and ACFG ; through A draw AL parallel to BH (L 18), and draw AH and DC (L post. 1). Tlien, because the angles BAC and BAE are both right angles (L def 10), the two BOOK I.] EUCLID AND LEGENDEE. 43 straight lines CA and AE are in the same straight line (I, 10). For like reason, BA and AF are in the same straight line. Again : because the angle IIBC is equal to the angle DBA (I, ax. 1), if the angle ABC be added to each, we have (I. ax. 2) HBA equal to DBC ; and because AB is equal to DB (const.), and BIT equal to BC, therefore (I. 3, case 1) the triangle ABH is equal to the triangle DBC. But (I. 15, cor. 4) the parallelo- gram BPIIL is double the ti-iangle ABH ; for like reason the square BADE is double the triangle DBC ; hence (I. ax. 6) BPIIL is equivalent to BADE. In like manner, PCLK can be shown equivalent to ACFE. Xow (I. ax. 10), BCHK is equiva- lent to BPHL and PCLK together; hence (L ax. 1) BCHK is equivalent to BADE and ACFG together. Wherefore, if the triangle is rio-ht-angled, etc. OTHERWISE, Let the squares on AB and BC fall on the same side of BC. Describe the square BAED on the side BA (L 23), and th.e square BCMN" on the side BC (L 23), and produce AE to F (I. post. 2) ; then tlirough D draw PF parallel to BC (I. 18). Because AF is parallel to BD (L def. 14, and 15, cor. 1), BC is equal to DF, and BA is equal to '^^ DE, and the angles BAC and DEF are both right angles (1. def 10), and equal (I. ax, 11); therefore the triangle BAC is equal to the triangle DEF (I. 3) ; and because BCED is com- mon to the square BAED and the parallelogram (I. def 14) BCDF, and the triangles BAC and DEF equal, the square BAED is equivalent to the parallelogram BCDF. And PF being parallel to BC (const.), the parallelograms (I. def 14) BCDF and BCPL have a common base, BC, and equal altitudes; hence (I. 15, cor. 5) they are equivalent, and (I. ax, 1) the square BAED is equivalent to the parallelogram BCPL. From A draw (I. 18) AK parallel to BM, and produce DE (I. post. 2) to BM; then AK and BM being parallel (const.), ED and 44 THE ELEMENTS OF [book I. BA, being opposite sides of the same square, are also parallel (I. def. 15) ; hence MR is equal to BA, and equal also to DE (I. ax. 1). But BM is equal to BC (const.) ; therefore the par- allelogram BAjMR is equal to the parallelogram BCDF ; and BAMR having the same base and equal altitude with the par- allelogram BGMK, is equivalent to it (I. 15, cor. 5); hence BGMK is equivalent to BCDF, equivalent to BCPL, aud equivalent to the square BADE (I. ax. 1). Or, the square described upon BA is equivalent to the rect- angle of the hypothenuse B C and the part BG of the hypjoth- enuse nearest to BA intercepted by the p)erpe7idicular drawn from the vertex of the right angle to the liypothenuse. In a similar manner, it can be shown that the square described on A G is equivalent to the rectangle of the hypothenuse B (7, and the remaining part G G of the hypothenuse intercepted by the same p)erpendicxdar. But the two rectangles are equivalent to the square of the hyi^othenuse (I. ax. 10) ; hence the tico squares described on the sides AB and AG (I. ax. 1) are equivalent to the square described on the hypothenuse. E F And if we make the tiiangle an isosceles rio;ht-an(2;led tri- angle as ABC, the square de- sciibed on AB will contain four equal triangles, ACB,BCF, FCE, and ECA, while each of the squares described on AC and CB will contain two such triangles, and both together will be equivalent to the four equal triangles, or equivalent to the square on AB. The demonstration is very simple, and it would be well for the pupil to undertake it. Hence, conversely, if the square described upon one side of a triangle be equivalent to the sum of the square described upon the two other sides of the triangle, the angle contained by those two sides is a right angle ; and when those two sides form two equal squares, the triangle is a right-angled isosceles triangle. Scho. 1. The proof of the corollary can be shown, also, either by describing the square of the hypothenuse on the other side of BC; and the other squares sometimes on one side and some- times on the other; and since drawing a perpendicular from the BOOK I.] EUCLID AND LEGEXDKE. 45 vertex of tlie riglit angle to the hypothenuse makes two ri(iht angles, so a line can be drawn from the same vertex to the point of bisection of the hypothenuse and make two supple- niental avgles and two equivalent triangles, and the demonsti'a- tion conducted by supplemental angles (I. 15, cor. V) instead of right angles. Proportion also gives neat and easy solutions to this corollary. (See V. 8, scho.) Cor. 2. If two right-angled triangles have their hypothenuses equal, and a side similarly situated in each also equal, the two triangles are equal by the third proposition of this book ; and, conversely, if the legs of a right-angled triangle be equal to the legs of another right-angled triangle, each to each, their hypoth- enuses can be in a similar manner shown equal. Cor. 3. Hence, also, we can find a square equivalent to the sum of more than two squares ; thus, let AB be the side of one square, and AC, perpendicular to it, the side of another squai'e ; join CB ; the square on CB (I, 24, cor. l) is equivalent to the sum of the squares on CA and AB. In like manner, if CD be drawn perpendicular to CB, and DB be drawn, the square on DB is equivalent to the squares on DC, CA, and AB, and by drawing a perpendicular to DB, a square can be found equiva- lent to the sum of four squares ; hence a square can be found equivalent to the sum of any number of squares. Cur. 4. Since (CB)- o (AB)= + (CA)=, we have (AB)=o (CB)" — (CA)-; hence a square can be found equivalent to the difference of two squares. Cor. 5. If a perpendicular be drawn from the vertex of the angle A, in the triangle BAC (diagram to cor. l), to P on the hypothenuse BC, cutting BC into two segments, BP and PC, the difference of the squares on the sides AB and AC is equiva- lent to the difference of the squares on the segments BP and PC. For the square on AB is equivalent to the squares on BP and PA, and the square on AC is equivalent to the squares on PC and PA; therefore (AC)^ — (AB)^ =o- (PC)' — (BP)^ Cor. 6. Hence the squares on txoo sides of a triangle are to- gether equivalent to txoice the square of half the remaining side^ and twice the square of the straight line from its point of M- 46 THE ELEMENTS OF [book I. section to the opposite anrjle. Suppose P in tlic triangle to te the point of bisection of the side BC ; tlion, when AP is per- pendicuhir to BC, we have (AB)- + (AC)= o (BP)- + (AP)' + (PC)' + (AP)^ .0= 2 (BP)= + 2 (AP)\ And when AP is not perpendicular to BC, the equivalence of the squares on two sides to the same will be shown in tlic next book. Scho. 2. In proof of coj-. 5, the obvious principle is employed, that the difference of two magnitudes is the same as the differ- ence obtained after adding to each the same third magnitude. Thus the difference of the squares on BP and PC is the same as the difference between the sum of the squares BP and PA and of PC and PA. Pkop. XXy. — Theok. — Tlie side and diagonal of a square are incommeiisuraUe to one another — that is, there is no Ivie which is a measure of both. Let ABCD be a square, and BD one of its diagonals ; AB, BD are incommensurable. Cut off DE equal to DA, and join AE. Then, since (I. 1, cor. 1) the angle DEA is equal to the acute angle DAE, AEB A B i^ obtuse, and therefore (I. 22, cor.) in the triangle ABE, BE is less than AB, or than AD ; wherefore AD is not a measure of BD. Draw EF ])erpendicular to BD. Then tiie angles FAE, FEA, being the complements of the equal angles DAE, DEA, are equal, and therefore AF, FE are equal. But (I. 20, cor. 4) ABD is half a riglit angle ; as is also BFE, since BEF is a right angle ; wherefore BE is equal to FE, and therefore to AF. From FB, which is evi- dently the diagonal of a square of which FE or EB is the side, cut off FG equal to FE, and join GE. Then it would be shown, as before, that BG is less than BE ; and therefore BE, the dif- ference between the side and diagonal of tlie square .VC, is con- tained twice in the side AB, with the remainder GB, wliich is itself the difference between the side FE or EB, and the diag- onal FB of another squai*e. By repeating the process, we Bhould find, in exactly the same manner, that BG would be BOOK I.] EUCLID A:SD LEGENDEE. 47 contained twice in BE, with a remainder, which would be the difierence between the side and diagonal of a square described on I3G ; and it is evident that a like process might be repeated continually, as no excess of a diagonal above a side would be contained in the side witliout remainder; and as this pro- cess has no termination, theiv is no line, however small, which will be contained without remainder in both AB and BD ; they are, therefore, incommensurable. Scho. Tills proposition can be illustrated by numbers. Let 10 be the side of tlie square ; then (I. 24, cor. 1) the diagonal will be expressed by the square root of 200, or 14'142 + ; there being no common multiple of 10 and 14-142 +, these numbers are incommensurable with each otlier. Or, wlicn any two lines are taken which by division and subdivision no common meas- ure can be found which can be contained in each v.'ithout a re- mainder, the two lines are said to be ijicommensicrable with each other, and such lines are tlie side and diagonal of a square ; and any two magnitudes wiiatever which have no common unit of measure are incommensurable with one another. E^TD or BOOK FIRST. BOOK SECOND.* ON" THE llECTANGLE AND SQUARE. DEFINITIONS. 1. A rectangle is said to be contained by the two straight lines which are about any of the right angles. For the sake of brevity, the rectangle contained, by AB and CD is often ex- pressed simply by AB. CD, a jooint being placed between the letters denoting the sides of tlie rectangle ; and the square of a line AB is often written simply AB". 2. A gno7non is the part of a parallelogram which remains when either of the parallelograms about one of the diagonals is taken away. PROPOSITIONS. Pkop. I. — TiiEOR. — If there be tioo straight lines, one of which is divided into any number of parts, the rectangle con- tained by the two lines is equivalent to the rectangles contained by the undivided line, and the several j^arts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in tlie points D, E ; tlie rectangle contained by A and BC is equivalent to the rectangles contained by A and BD, A and DE, and A and EC. From B draAV (I. V) BG perpendicular to BC, and make it equal to A ; through G draw (I. 1 8) GH par- ^ DEC ^^^^^ ^^ J3Q . ^^^^ through D, E, C draw DK, EL, CH parallel to BG. Then BH, BK, DL, and EH are evidently rectangles ; and BH is equivalent (I. ax. 10) to BK, DL, EH. But BH is contained by A and BC, for (H. dcf. 1) ■^ it is contained by GB and BC, and GB is * The second and third books are arrantred very similarly to those books in the edition of Euclid i)y Professor Thomson of the University of Glasgow, Scotland. G K L n BOOK 11.^ EUCLID AND LEGENDEE. 49 equal (const.) to A; and BK is contained by A and BD, for it is contained by GB and BD, of whicli GB is equal to A. Also DL is contained by A and DE, because DK, that is (I. 15, con 4) BG, is equal to A ; and in like manner it is shown that EH is contained by A and EC. Therefore the rectangle contained by A and BC is eqidvalent to the several rectangles contained by A and BD, by A and DE, and by A and EC. Wherefore, if there be two straight lines, etc. Prop. II. — Theoe. — If a straight line he divided into any ttoo parts, the rectangles contained by thexchole and each of the parts are together equivalent to the square of the whole line. Let the straight line AB be divided into any two parts in the point C ; the rectangles AB.AC and AB.BC are equivalent tc» the square of AB, Upon AB describe (I. 23) the square AE, and through C draw (I. 18) CF parallel to AD or BE. Then AE is equal (I. ax. 10) to the rectangles AF and CE. But iVE is the square of AB, and AF is the rectangle contained by BA, AC ; for (II, def 1) it is contained by DA, AC, of which DA is equal to AB ; and CE is contained by AB, BC, for BE is equivalent to AB ; therefore the rectangles under AB, AC, and AB, BC are equivalent to the- square of AB. If, therefore, etc. 8cho. This proposition may also be demonstrated in the fol- lowing manner : Take a straight line D equal to AB, Then (II, 1) the rect- angles AC.D and BCD are together equiva- lent to AB.D. But since D is equal to AB, ^ ^ ^ the rectangle AB.D is equivalent (I. def 15) to the square of AB, and the rectangles AC.D and BCD are respectively equivalent ° to ACAB and BCAB ; wherefoi-e the rect- angles ACAB and BCAB are together equivalent to the square of AB. In a manner similar to this, several of the following proposi- tions may be demonstrated. Such proofs, thougli perhaps not 60 easily understood at first by the learner, are shorter than 4 so THE ELEMENTS OF [bOOK H. those given by Euclid; aud they have tlie advantage of being derived from those preceding them, instead of being estab- lished by continual appeals to original principles. Pkop. III. — Theor. — If a straight line he divided into any two 2ici7'ts, the rectangle contained hy the lohole and one of. the parts is equivalent to the square of that part^ together loith the rectangle contained hy the tioo parts. Let the straight line AB be divided into two parts in the point C ; the rectangle AB.BC is equivalent to the square of -BC, together with the rectangle AC.CB Upon BC describe (I. 23) the square CE ; produce ED toF; and through A draw (I. 18) AF parallel to CD or BE. Then the rectangle AE is equivalent (I. ax. 10) to the rectangles CE, AD. But AE is the rectangle con- tained by AB, BC, for it is contained by x^B, BE, of which BE is equal to BC ; and AD is contained by AC, CB, for CD is equal to CB ; also DB is the square of BC. Thei-e- fore the rectangle AB.BC is equivalent to the square of BC, together with the rectangle AC.CB. If, therefore, etc. 8cho. Otherwise : Take a line D equal to CB. Then (II. 1) the rectangle AB.D is equivalent to the -^ ^ ^ rectangles BCD and AC.D ; that is (const. and L def 15), the rectangle AB.BC is ^ equivalent to the square of BC togctlier with the rectangle AC.CB. Prop. IV.— Tiieor.— 7/" a straight line he divided into any tico j^^-^^ts, the square of the v-hole line is equivalent to the squares of the tioo i)arts\ together with twice their red angle. Let the straight line AB be divided into any two parts in C ; the square of AB is equivalent to the squares of AC and CB, together with twice the rectangle under AC and CB. On AB describe (L 23) the square of AE, and join BD; through C draw (I. 18) CGF parallel to AD or BE; and through G draw UK parallel to AB or DE. Then, l)ecause G / BOOK II.] EUCLID AND LEGENDEE. 51 CF is parallel to AD, aucl BD falls upon them, the exterior angle CGB is equivalent (I. 16) to the inte- rior and remote angle ADB ; but ADB is ^ ^ ? •equal (I. l) to ABD, because BA and AD are equal, being sides of a square; wherefore (I. ax. 1) the angle CGB is equal to GBC; iiad therefore the side BC is equal (I. 1, cor.) to the side CG. But (const.) the figure CK » ^ B is a parallelogram ; and since CBK is a right angle, and BC equal to CG, CK (I. def 15) is a square, and it is upon the side CB. For the like reason HF also is a square, and it is upon the side HG, which is equal (I. 15, cor. 1) to AC; therefore HF, CK are the squares of AC, CB. And because (I. 15, cor. 8) the complements AG, GEare equivalent, and that AG is the rectangle contained by AC, CB, for CG has been proved to be equal to CB ; therefore GE is also equivalent to the rectangle AC.CB; wherefore AG, GE are equivalent to twice the rectangle AC.CB. The four figures, therefore, HF, -CK, AG, GE are equivalent to the squares of AC, CB, and twice the rectangle AC.CB. But HF, CK, AG, GE make up the Avhole figure AE, which is the square of AB ; therefore the square of AB is equivalent to the squares of AC and CB, and twice the rectangle AC.CB. Wherefore, etc. Otherwise: AB^ = AB.AC + AB.CB (H. 2). But (H. 3) A.B. AC =AC=+ AC.CB, and AB.CB=zCB-+AC.BC. Hence <I. ax. 2) AB.AC + AB.BC, or AB-=AC^+CB-+2AC.CB. Cor. 1. It follows from this demonstration, that the parallel- ograms about the diagonal of a square are likewise squares. Cor. 2. Hence the square of a straight line • is equivalent to four times the square of its half; for the straight line being iDisected, the rectangle of the parts is equivalent to the square of one of them. Pkop. V. — Theor. — If a straight line be divided into two ' equal parts, and also into tioo unequal parts ; the rectangle con- tained by the unequal parts., together with the square of the line hettoeen the points of section, is equivalent to the square of half the line. Let the straio;ht line AB be bisected in C, and divided un- H / K L / 52 THE ELEMENTS OF [bOOK II. equally in D ; the rectangle AD.DB, together with the square of CD, is equivalent to the square of CB. On CB describe (I. 23) the square CF, join BE, and through D draw (I. 18) DHG parallel to CE or BF ; also through H draw KLM parallel to CB or EF ; and ■^ c D B through A draw AK parallel to CL or BM. Then (I. 15, cor. 5) AL and CM ^^ are equal, because AC is equal to CB ; and (I. 15, cor. 8) the complements CH E G F and HF are equivalent. Therefore (I. ax. 2) AL and CH together are equal to CM and HF together ; that is, AH is equivalent to the gnomon CMC To each of those add LG, and (I. ax. 2) the gnomon CMG, together with LG, is equivalent to AH together with LG. But the gnomon CMG, and LG make up the figure CEFB, which is the square of CB ; also AH is the rectangle imder AD and DB, because DB is equal (IL 4, cor. 1) to DH ; and LG is the square of CD. Thei-efore the rectangle AD.DB and the square of CD are equivalent to the square of CB. Wherefore, if a straight line, etc. Otherwise : Since, as is easily seen from the proof in the above, DF is equal to AL, take these separately from the entire figure, and there remain AH and LG equivalent to the square CF, as before. The proof may also be as follows: AD.DB = CD.DB + AC.DB (H. 1) or AD.DB =CD.DB+ CB.DB, because CB=:AC. Hence (H. 3) AD.DB =CD.DB + CD.DB+DB^=i:2CD.DB + DBl To each of these add CD=; then AD.DB + CD^=2CD.DB+DB^+CD^ or (H. 4) AD.DB + CD'=CB\ Cor. 1. Hence the ditference of the squares of CB and CD is equivalent to the rectangle under AD and DB. But since AC is equal to CB, AD is equivalent to the sum of CB and CD, and DB is the difterence of these lines. Hence the difference of the squares of two straight lines is equivalent to the rectangle under their sum and difference. Cor. 2. Since the square of CB, or, which is the same, the rectangle AC.CB, is greater than the rectangle AD.DB by the equare of CD, it follows, that to divide a straight line into two BOOK n.] EUCLID AND LEGENDKE. 53 parts, the rectangle of wliich may be the greatest possible, or, as is termed, a tnaxlmiim, tlie line is to be bisected. Cor. 3. Hence also tlie sum of the squares of tlie two parts into which a straight line is divided, is the least possible, or is, as it is termed, a ininimuni, Avlien the line is bisected. For (II. 4) the square of tlie line is equivalent to the squares of the parts and twice their rectangle ; and therefore the greater the rectangle is, the less are the squares of the j^arts; but, by the foregoins; corollarv, the rectangle is a maximum when the line is bisected. C<yr. 4. Since (I. 24, cor. 5) the difference of the squares of the sides of a triangle is equivalent to the difference of the squares of the segments of the base, it follows, from the first corollary above, that the rectangle under the sum and differ- ence of the sides of a triangle is equivalent to the rectano-le under the sum and difference of the segments, intercepted be- tween the extremities of the base and the jjoint in which the perpendicular cuts the base, or tlie base produced. Cor. 5. Hence, also, if a straight line be drawn from the ver- tex of an isosceles triangle to any point in the base, or its con- tinuation, the difference of the squares of that line and either of the equal sides is equivalent to the rectangle under the seoments intercepted between the extremities of the base and the jDoint. Cor. 6. Since (I. 24, cor. 4) the square of one of the legs of a right-angled triangle is equivalent to the difference of the squares of the hypothenuse and the other leg, it follows (IT. 5, cor. 1) that the square of one leg of a right-angled triangle is equivalent to the rectangle under the sum and difference of the hypothenuse and the other. ^cho. And a parallelogram can be constructed equivalent to a ffiven triangle, or anv given rectilinear fiorure havinor an angle equal to a given angle, or applied to a given straight line — that is, having that straight line for one of its sides, when the jjarallelogram shall be equivalent to a given triangle or given rectilinear figure, and have one of its angles equal to a given angle, by applying a parallelogram equivalent to the given triangle with an equal angle, to a given straight line, and then constructing an equal triangle to the given triangle (I. 15 and 15, cor. 4). A C B D K H / L / 54 THE ELEMENTS OF [bOOK II, Prop. VI. — Theok. — If a straight line he bisected, and be produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equivalent to the square of the straight line which is made %ip of the half and the p>art produced. Let the straight line AB be bisected in C, and produced to D ; the rectangle AD.DB, and the square of CB, are equiva- lent to the square of CD. Upon CD describe (I. 23) the square CF, and join DE ; through B draw (I. 18) BHG parallel to CE or DF ; through H draw KLM parallel to AD or EF, and through A draw AK parallel to CL or jj DM. Then because AC is equal to CB, the rectangle AL is equal (I. 15, cor. 5) to CH ; but (1.15, cor. 8) the complements CH, J, ^j J, HF are equivalent ; therefore, also, AL is equal to HF, To each of these add CM and LG ; therefore AM and LG are equivalent to the whole square CEFD, But AM is the rectangle under AD and DB, because (11. 4, cor. 1) DB is equal to DM; also, LG is the square of CB, and CEFD the square of CD. Therefore the rectangle AD.DB and the square of CB are equivalent to the square of CD. Wherefore, if a straight line, etc. Otherwise : Produce CA to N, and make CN equal to CD. To these add the equals CB and CA ; ^ ^ ^ ^ ° therefore NB is equal (L ax. 2) to AD. But (IL 5) the rectangle NB.BD, oi- AD.BD, together with the square of CB, is equivalent to the square of CD ; which was to be proved. Prop. VIL — Theor. — If a straight line be divided into a?2y two parts, the squares of the whole line and of one of the p>arts are equivalent to ticice the rectangle contained by the whole and that part, together tcith the square of the other part. Let the straight line AB be divided into any two parts in the point C ; the squares of AB, BC are equivalent to twice the rectangle AB.BC, together with the square of AC. BOOK II.] EUCLID AND LEGENDRE. 65 Upon AB describe (I. 23) the square AE, and construct the figure as in the preceding propositions. Then, because (I. 15, cor. 8) the complements CII, ^ c B FK are equivalent, add to each of thorn CK; the whole AK is therefore equal to the whole ^ CE ; therefore AK, CE are together double of AK. But AK, CE are the gnomon AKF, K together with the square CK ; therefore the i> F E gnomon AKF and the square CK are double of AK, or double of the rectangle AB.BC, because BC is equal (II. 4, cor. 1) to BK. To each of these equals add HF, which is equal (II. 4, cor. 1 ) to the square of AC ; therefore the gno- mon AKF, and the squares CK, HF are equivalent to twice the rectangle AB.BC and the square of AC. But the gnomon AKF, and the squares CK, HF make up the whole figure AE, together with CK; and these are the squares of AB and BC ; therefore the squares of AB and BC are equivalent to twice the rectangle AB.BC, together with the square of AC ; wherefore, if a straight line, etc. Otherwise: Because (II. 4) AB^ = AC=+BC-+2AC.BC, add BC= to both; then AB=+BC==AC= + 2BC=+2AC.BC. But (II. 3) BC=+AC.BC= AB.BC, and therefore 2BC=+2AC.BC =2AB.BC; wherefore AB-+BC^=AC=+2AB.BC. Cor. 1. Since AC is the diflierence of AB and BC, it follows that the square of the difterence of two straight lines is equiva- lent to the sum of their squares, wanting twice their rectangle. Cor. 2. Since (II. 4) the square of the sum of two lines ex- ceeds the sum of their squares by twice their rectangle, and since, by the foregoing corollary, the square of their difference is less than the sum of their squares by twice their rectangle, it follows that the square of the sum of two lines is equivalent to the square of their difference, together with four times their rectangle. Pkop. VIII. — TiiEOR. — If a straight line he divided into two equal, and also into tvw imequal parts, the squares of the un- equal parts are together double of the square of half the linCj and of the square of the line hetioeen the points of section. Let the straight line AB be divided equally in C, and un- 56 THE ELICMENTS OF [book n. E equally in D ; the square? of AD, DB are together double of the squares of AC, CD. From C draAV (I. 7) CE j)ei-|ie!Klicular to AB, and make it equal to AC or CB, and join EA, EB ; through D draw (I. 18) DF jjarallel to CE, and tlirougli V draw FG parallel to AB ; and join AF. Then because (const.) the triangles ACE, BCE are right-angled and isosceles, tlie angles CAE, AEC, CEB, O O 7 O 7 3 7 EBC are (I. 20, cor. 4) each half a riglit angle. So also (I. 16, part 2) are EEC, BED, because FG is parallel to AB, and ED to EC ; and for the same reason EGF, ADF are right angles. The angle AEB is also a right angle, its parts being each half a right angle. Hence (I. 1) EG is equal to GF or CD, and ED to DB. Again (I. 24, cor. 1) : the square of AE is equivalent to the squares of AC, CE, or to tAvice the square of AC, since AC and CE are equal. In like manner, the square of EF is equivalent to twice the square of GF or CD. Now (I. 24, ^ cor. 1), the squares of AD and DF, or of AD and DB, are equivalent to the square of AF ; and the squares of AE, EF, that is, twice the square of AC and twice the square of CD, are also equivalent to the square of AF; therefore (I. ax. 1) the squares of AD, DB are equivalent to twice the square of AC and twice the square of CD. If, there- fore, a straight line, etc. Otherwise: DB^+2BC.CD=BC=+CD^ (II. 7), or DB=+ 2AC.CD=:AC= + CD^-; also (11. 4) AD==:AC^+CD=+2AC. CD. Add these equals togethei-, and from the sums take 2AC.CD; then AD-+DB^=2AC^+2CD^ Prop. IX. — ^Theor. — If a straight line he bisected, and pro- duced to any point, the squares of the whole line thus produced, and of the part of it produced, are together double of the square of half the lifie bisected, and of the square of the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D ; the squares of AD, DB arc double of the squares of AC, CD. From C draw (I. 7) CE perpendicular to AB ; and make it BOOK II.] EUCLID AND LEGENDRE. 57 A equal to AC or CB ; join AE, EB, and tln-ough E and D draw (I. 1 8) EF parallel to AB, and DF parallel to CE. Then be- cause the straiu'lit line EF meets the parallels EC, FD, the ano-les CEF, EFD are equivalent (I. 9) to two right anixles : and therefore the an- gles BEF, EFD are less than two right angles; therefore (I. 19) EB, FD will meet, if pro- duced toward B, D ; let them meet in G, and join AG. Then it would be proved, as in the last proposition, that the angles CAE, xVEC, CEB, EBC are each half a right angle, and AEB a right angle. BDG is also a right angle, being equal (I. 16) to ECB, since (const.) EC, EG are parallel ; DGB, DBG are each half a right angle, being equal (I. 10 and I. 11) to CEB, CBE, each to each ; and FEG is half a right angle, being (I. 16) equal to CBE. It would also be proved, as in the last proposition, that the squai-e of AE is twice the square of AC, and the square of EG twice the square of EF or CD. Now (I. 24, cor. l),the squares of AD, DG, or of AD, DB, are equiv- alent to tlie sipiare of AG; and the squares of AE, EG, or twice the square of AC and twice the square of CD, are also equiva- lent to the square of AG. Therefore (I. ax. 1) the squares of AD, DB are equivalent to twice the square of AC and twice the square of CD. If, therefore, a straight line, etc. Otherwise: Produce CA, making CH equal to CD. To these add CB, CA ; therefore IIB, AD are equal. Then (II. 8) IIB--+BD^ or AD=+ BD^ = 2CD=-f2AC-. &eho. The nine foregoing propositions may all be proved very easily by means of algebra, in connection with the princi- ples of mensuration, already established in the corollaries to the 23d proposition of the first book. Thus, to pi'ove the fourth proposition, let AC=:a, CB = 5, and, consequently, AB=a+5. Now, the area of the square described on AB will be found (I. 23, cor. 4) by multiplying a-\-h by itself. This product is found, by performing tlie actual operation, to be «'+2aJ+^'; an expression, the first and third parts of which are, by the n A B D 58 THE ELEMENTS OF [bOOK H. same corollary, the areas of the squares of AC and CB, and the second is twice the rectangle of those lines. In like manner, to prove the eighth, adopting the same nota- tion, Ave have the line which is made up of the whole and CB = a + 2$y and, multiplying this by itself, we get for the area ot the square of that line, cr+iab+ib", or a'-\-4{a-]-b)b, the first part of which is the area of the square of AC, and the second four times the area of the rectangle under AB and CB. It will be a useful exercise for the student to prove the other propositions in a similar manner. He will also find it easy to investigate various other relations of lines and their parts by means of algebra. All the properties delivered in these propositions hold also respecting numbers, if products be substituted for rectangles. Thus, 7 being equal to the sum of 5 and 2, the square, or sec- ond power of 7, is equal to the squares of 5 and 2 and twice their product ; that is, 49 = 25 + 4+20. Pkop. X. — Peob. — To divide a given straight line into txoo parts, so that the rectangle contained by the \chole and one of the parts may be equivalent to the square of the other part. Let AB be the given straight line ; it is required to divide it into two parts, so that the rectangle under the whole and one of the parts may be equivalent to the square of the other. Upon AB describe (I. 23) the square AD ; bisect (I. 6) AC in E, and join E with B, the remote extremity of AB ; produce CA to F, making EF equal to EB, and cut off AH equal to AF ; AB is divided in H, so that the rectangle AB.BH is equivalent to the square of AH. Complete the parallelogram AG, and produce GH to K. Then, since BAC is a right angle, FAH is also (I. 9) a right angle; and (I. def 15) AG is a square, AF, AH being equal by construction. Because the straight line AC is bisected in E, and produced to F, the rectangle CF.FA and the square of AE are together equivalent (II. 6) to the square of EF or of EB, since (const.) EB, EF are equal. But the squares of BA, AE are equivalent (I. 24, cor. 1) to the square of EB, because the angle EAB is a right angle; therefore the rectangle CF.FA and the square of AE are equivalent (I. ax. 1) to the squares of BOOK n.] EUCLID AND LEGENDRE. 59 BA, AE. Take away the square of AE, -u-hich is common to both; therefore the remainino- rectano-le CF.FA F G E H -" B D is equivalent (I. ax. 3) to the square of AB. But the figure FK is the rectangle contained by CF, FA, for AF is equal to FG ; and AD is the square of AB ; therefore FK is equal to AD. Take away the common part AK, and (I. ax. 3) the remainders FH and HD are equivalent. But HD is the rectangle AB.BH, for AB is equal to BD ; and FH is the square of AH. Therefore the rectangle AB.BH is equivalent to the square of AH ; Avherefore the straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH ;, which was to be done. Sc/w. 1. In the practical construction in this proposition, and in the 2d cor. to 9 of the fifth book, which is virtually the same, it is sufficient to draw AE perpendicular to AB, making it equal to the half of AB, and producing it through A ; then, to make EF equal to the distance from E to B, and AH equal to AF. It is plain that BD might be bisected instead of AC, and that in this way another point of section would be obtained. While the enunciation in the text serves for ordinary pur- poses, it is too limited in a geometrical sense, as it comprehends only one case, excluding another. The following include& both : In a given straight line^ or its continuatio7i, to find ai^ointy such that the rectangle contained hy the given line^ and the seg- ment between one of its extremities and the required point, may be equal to the square of the segment between its other extremity and the same point. The point in the continuation of BA will be found by cutting ofiT a line on EC and its continuation, equal to EB, and describ- ing on the line composed of that line and AE a square lying on the opposite side of AC from AD ; as the angular j^oint of that square in the continuation of BA is the point required. The proof is the same as that given above, except that a rectangle corresponding to AK is to be added instead of being sub- tracted. Scho. 2. The line CF is equivalent to BA and AH ; and since 60 THE ELEMENTS OF [bOOK H, it has been shown that the rectangle CF.FA is eqiiivalent to the square of BA or CA, it follows that if any straight line AB (see the next diagram) be divided according to this proposition in C, AC being the greater part, and if AD be made equal to AB, DC is similarly divided in A. So also if DE be made F E D A C B equal to DC, and EF to EA, EA is divided similarly in D, and FD in E ; and the like additions may be continued as far as we please. Conversely, if any straight line FD be divided according to this proposition in E, and if EA be made equal to EF, DC to DE, etc., EA is similarly divided in D, DC in A, etc. It fol- lows also, that the greater segment of a line so divided will be itself similarly divided, if a part be cut oif from it equal to the less ; and that by adding to the whole line its greater segment, another line will be obtained, which is similarly divided. Prop. XI. — Theok. — In an oUuse-angkd triangle^ the square of the greatest side exceeds the squares of the other tioo^ hy tioice the rectangle contained by either of the last-mentioned sides, and its continuation to meet a perpendicular drawn to it from the opposite angle. Let ABC be a triangle, having the angle ACB obtuse ; and let AD be perpendicular to BC produced ; the square of AB is equivalent to the squares of AC and CB, and twice the rectan- gle BC.CD. Because the straight line BD is divided into two parts in the point C, the square of BD is equivalent (II. 4) to the squares of BC and CD, and tAvice the rectangle BC.CD. To each of these equivalents add the square of DA ; and the squares of DB, DA are equivalent to the squares of BC, CD, DA, and twice the rectangle BC.CD. But, because the angle D is a right angle, the square of BA is equivalent (I. 24, cor. 1) to the squares of BD, DA, and tlie square of CA is equivalent to the squares of BOOK II.] EUCLID AND LEGENDRE. 61 CD, DA ; therefore the square of BA is equivalent to the squares of BC, CA, and twice the rectangle BC.CD. Therefore, in an obtuse-angled triangle, etc. Pkop. XII. — TiiEOR. — I/i any triangle^ the square of a side subtending an acute a?igle is less than the squares of the other sides, by twice the rectangle contained by either of those sides, and the straight line hitercej^ted between the acute angle and the perpendicular drawn to that side from tJie opposite angle. Let ABC (see this figure and that of the foregoing proposi- tion) be any triangle, having the angle B acute ; and let AD be perpendicular to BC, one of the sides containing that angle; the square of AC is less than the squares of AB, BC, by twice the rectangle CB.BD. The squares of CB, BD are equivalent (II. 1) to twice the rectangle contained by CB, BD, and the square of DC. To each of these equals add the square of AD ; therefore the squares of CB, BD, DA are equivalent to twice the rect- angle CB.BD, and the squares of AD, DC. But, because AD is perpendicu- lar to BC, the square of AB is equiva- lent (I. 24, cor. 1) to the squares of BD, DA, and the square of x\.C to the squares of .VD, DC ; therefore the squares of CB, BA are equivalent to the square of AC, and twice the rectangle CB.BD ; tliat is, tlie square of AC alone is less than the squares of CB, BA, by twice the rectangle CB.BD. If the side AC be perpendicular to BC, then BC is the straight line between the perpendicular and the acute angle at B ; and it is manifest that the squares of AB, BC are equiva- lent (I. 24, cor. 1) to the square of AC and twice the square of BC. Therefore, in any triangle, etc. Scho. By means of this or the foregoing proposition, the area of a triangle may be computed, if the sides be given in num- bers. Thus, let AB = 17, BC = 28, and AC = 25. From AB=+ BC take AC; that is, from l7'4-2SUake 25=; the remainder 448 is twice CB.BD. Dividing this by 56, twice BC, the quo- tient 8 is BD, Hence, from either of the triangles ABD, 62 THE ELEMENTS OF [bOOK II. ACD, we find the perpendicular AD to be 15 ; and thence the area is found, by taking half tlie product of BC and AD, to be 210. The segments of the base are more easily found by means -of tJie 4th corollary to the fifth proposition of this book, in connec- tion with the pi-inciple, that if half the difference of two mag- nitudes le added to half their sum ^ the resvlt is the greater; and if half the difference he taken from half the sum, the remainder is the less. Thus, if 42, the sum of AB and AC, be multiplied Tjy 8, their difference, and if the product, 336, be divided by 28, the sum of the segments of the base, the quotient 12 is their difference. The half of this being added to the half of 28, the sum 20 is the greater segment CD ; and being subtracted from it, the remainder 8 is BD. To prove the principle mentioned alxjve, let AB be the greater, and BC the less of two ^ ^ ^ ^ ^ magnitudes. Bisect AC in D, and ' make AE equal to BC. Then AD or DC is half the sum, and ED or DB half the difference of AB and BC ; and AB the greater is equivalent to the sum of AD and DB, and BC the less is equiv- alent to the difference of DC and DB. Cor. Hence, when AP (I. 24, cor. 6) is not perpendicular to BC — the truth of the corollary can be shown by this and pre- -ceding propositions. Hence, if the sides of a ti-iangle be given in numbers, the line AD can be computed. Thus, if AC^ll, AB = 14, and BC = 7, we have AC-f BC= = 121 + 49 = 170, and 2AD-=:98. Then 170 — 98 = 72, the half of Avhich is 36 ; and 6, the square root of this, is CD. Prop. XHI. — Vrov,.— To describe a square that shall he •equivalent to a given rectilineal figure. Let A be the given rectilineal figure ; it is required to de- scribe a square that shall be equal to it. Describe (H. 5 scho.) the rectangle BD equal to A. Then, if the sides of it, BE, ED, be equal to one another, it is a square, and what was required is done. But if they be not equal, produce one of them BE to F, and make EF equal to ED; bisect BF ia BOOK II.] EUCLID AND LEGENDEE. 63 O, and from the center G, at the distance GB, or GF, describe (I. post. 3) the semicircle BIIF; produce DE to H, and join GH. Therefore, because the straight line BF is divided equally in G, and unequally in E, the rectangle BE.EF, and the square of EG, are equivalent (II. 5) to the square of GF, or of GH, because GH is equal to GF. But the squares of HE, EG are equal (I. 24, cor, l) to the square of GH ; therefore the rectangle BE.EF and the square of EG are equiv- alent to the squares of HE, EG. Take away the square of EG, which is common, nvA the remaining rectangle BE.EF is equiv- alent to the square of EH. But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED ; therefore BD is equivalent to the square of EH ; but BD is equivalent to the figure A; therefore the square of EH is <^quivalent to A. The square described on EH, therefore, is the required square. Scho. Tliis is a particular case of the sixteenth proposition of the fifth book, and an easy solution can be effected by means of Proportion. Pkop. XIY. — TiiEOR. — T/ie siwi of the squares of the sides of a trapezium is equivalent to the sum of the squares of the diagonals, tor/ether icith four times the square of the straight line joining the points of bisection of the diagonals. Let ABCD be a trapezium, having its diagonals AC, BD bisected in E and F, aud let EF be joined ; the squares of AB, BC, CD, DA are together equivalent to the squares of AC, BD, together with four times the squai-e of EF. Join AF, EC. The squares of AB, AD are together equivalent (II. 12, cor.) to twice the sum of the squares of DF and AF; and the squares of BC, CD are equiv- alent to twice the sum of the squares of DF and CF. Add these equivalents together, and the sum of the squares of AB, BC, CD, DA is equivalent to four times the square of DF, to- B 64 THE ELEMEXT3 OF [BOOK H. getlier with twice the sum of the squares of AF and CF. But tv/ice the squares of AF and CF are equivalent (11. 12, cor.) to four times the squares of AE and EF; and (II. 4, cor. 2) four times the square of DF is equivalent to the square of BD, and four times the square of AE to the square of AC. Hence the squares of AB, BC, CD, DA are equivalent to the squares of AC and BD, together with four times the square of EF. Therefore, the sum, etc. Cor. Hence the squares on the diagonals of a parallelogram are together equivalent to the sum of the squares on its sides — for in the case of a parallelogram the line EF vanishes, as the diagonals of a parallelogram bisect each other. /S'cAo. Hence, if we have the sides and one of the diagonals of a parallelogram in numbers, we can compute the remaining diagonal. Thus, if AB, DC be each =9, AD, BC each = 7, and AC =8, we have AB^+BC'+CD= + DA= = 81 + 49+81 + 49 = 260, and AC-r=64. Taking the latter from the fonner, and ex- tracting the square root, we find BD = 14. END OF BOOK SECOND. BOOK THIRD. ON THE CIRCLE, AND LINES AND ANGLES DE- PENDING ON IT, AND RECTILINEAL FIGURES DESCRIBED ABOUT THE CIRCLE. DEFIXinOIfS. 1. A STRAIGHT line is said to touch a circle, or to be a tangent to it, when it meets the circle, and being produced does not cut it. 2. Circles are said to touch one another, which meet, but do not cut one another. 3. In a circle, chords are said to be equally distant from the center, when the perpendiculars drawn to them from the center are equal. 4. And the chord on which the greater perpendicular falls, is said to \)e farther from the center. 5. An angle in a segment of a circle is an angle contained by two straight lines drawn from any point in the arc of the segment to the extremities of its chord ; 6. And an angle is said to stand upoti the arc intercepted between the straight lines that contain the angle. 7. A sector of a circle is a figure contained by any arc of the circle, and two radii drawn through its extremities. 8. A quadrant is a sector whose radii are perpendicular to each other. It is easy to show by superposition, that a quad- rant is half of a semicircle, and therefore a fourth part of the entire circle. 9. Similar segments of circles are those which contain equal angles. 10. Concentric circles are those which have the same center. 11. A regular polygon is equilateral. 12. When the sides of one rectilineal figure pass through the 5 66 THE ELEMENTS OF [bOOK HI. angular points of another, the figures not coinciding with one another, the interior figure is said to be inscribed in the exte- rior, and the exterior to be circutnscribed, or described, about the interior one. 13, When all the angular points of a rectilineal figure are upon the circumference of a circle, the rectilineal figure is said to be inscribed in the circle, and the circle to be circumscribed, or described, about the rectilineal figure. 14. When each side of a rectilineal figure touches a circle, the rectilineal figure is said to be circumscribed, or described, about the circle, and the circle to be inscribed in the rectilineal figure. PROPOSITIONS. Prop. I. — Prob. — To find the center of a given circle,. Let ABC be the given circle, and draw any chord AB. Bisect AB (I. 6) by the perpendicular EDO drawn to the circumference on both sides AB. Since EDC bisects AB (const.) and is perpendicular to AB (L 6), the angles AEB and ACB are also bisected by EDC, and the subtended arcs ACB and AEB (I. 5) are likewise bisected by EDC; therefore the arcs CA+AEo the arcs CB+BE; hence CA + AE or CE is the semicircumference, and the perpendicular EDC is a diameter — consequently it passes through the center of the circle. Then taking any other chord FIT, and in same manner it can be shown that the perpendicular MN which bisects the chord is also a diameter, and since all diameters pass through the center of the circle, the point at which they intersect each other, being the only point they have in common, that point is the center of the circle. Cor. 1. Hence, to find center of any regular polygon (L 6), bisect the sides by perpendiculars drawn from the points of bisection (I. 7), and the point where the perpendiculars inter- eect each other is the center of the polygon. BOOK m,] EUCLID AND LEGENDRB. 67 Cor. 2. In a triangle, straight lines drawn from the points of bisection of the three sides to the opposite angles all pass throue'h the same point. 8cho. From the preceding coroUarj'^ it can be shown that each of the straight lines is divided into two segments at the common point of bisection, of which the segment nearest the angle is double the other. Prop. II. — Theor. — If a straight line drav^nfrom the center of a circle bisect a chord which does not pass through the ce?i- ter, it cicts it at right angles/ and {2) if it cut it at right angles, it bisects it. Let ABC be a circle ; and let ED, a straight line drawn from the center E, bisect any chord AB, which does not pass through the center, in the point D ; ED cuts AB at right angles. Join EA, EB. Then in the triangles ADE BDE, AD is equal to DB, DE common, and (I. def 16) the base EA is equal to the base EB ; therefore (I. 4) the angles ADE, BDE are equal; and consequently (I. def 10) each of them is a right angle; wherefore ED cuts AB at right angles. Next, let ED cut AB at right angles ; ED also bisects it. The same construction being made, because the radii EA, EB are (I. def 16) equal, the angle EAD is equal (I. 1, cor.) to EBD ; and the right angles ADE, BDE are equal ; therefore in the two triangles EAD, EBD there are two angles in one equal to two angles in the other, each to each, and the side ED, which is opposite to one of the equal angles in each, is common; therefore (L 14) AD is equal to DB. If a straight line, therefore, etc. > Cor. 1. Hence, in an isosceles triangle, a straight line drawn from the vertex bisecting the base is perpendicular to it ; and a straight line drawn from the vertex perpendicular to the base, bisects it. Cor. 2. Let the straight line AB cut the concentric circles ABC, DEF in the points A, D, E, B ; AD is equal to EB, and AE to DB. From the common center G, draw GH perpen- 68 THE ELEMEi^TS OF [book ni. dicular to AB. Then (III. 2) AH is equal to HB, and DH to HE. From AH take DH, and from HB take HE, and the remainders, AD, EB, are equal. To these equals add DE, and the sums AE, DB are equal. Cor. 3. Any numljer of parallel chords in a circle are all bisected by a diameter perpendicular to them. Pkop. hi. — Theok. — Tico chords of a circle which are not both diameters, can not bisect each other. From the definition of the circle, the center is the only point in the circle which is equally distant D from all parts of the circumference ; then any chord which passes through the center is bisected at the center ; the diameter is the only chord (HI, 1) which passes through the center, there- fore any other chord AB can not bisect a diameter CD. And when neither chord AB nor EF is a diameter, their point of intersection not being- the center of the circle, is unequally distant from the circumference ; therefore the chords AB and EF are not bisected by each other. Prop. IV. — Theor. — Iftico circles cut 09ie another, they have not the same center. Let ABC and DBE be two circles which cut one another in B; they will not have the same center. For AC is the diameter of ABC (III. 1), and DE is the diameter of DBE. But the intersection of the two diameters AC and FH is the center of the circle ABC (in. 1), and the inter- section of the two diameters LM and DE is the center of the circle DBE. Now the points BOOK m.] EUCLID AND LEGENDRE. 69 of intersection of these diameters are different, therefore ABC has not the same center with DBE. Wherefore, if two cir- cles, etc. Cor. 1. Hence, if one circle touches another internally, they have not the same center. Cor. 2. One circle can not cut another in more than two points, nor touch another in more than one point. Prop. V. — Theoe. — If from any jjoint icithin a circle, which is not the center, straight lines be drawn to the circum- ference ; (1) the greatest is that tchich passes through the cen- ter, and (2) the continuation of that line to the circumference, in the opposite direction, is the least y (3) of others, one nearer to the line passing through the center is greater than one more remote ; and (4) from, the same point there can be drawn ordy two equal straight lines, one upon each side of either the longest or shortest line, and making equal angles vnth that line. Let ABCD be a circle, E its center, and AD a diameter, in which let any point F be taken, which is not the center ; of all the straight lines FA, FB, FC, etc., that can be drawn from F to the circumference, FA is the greatest, and FD the least ; and of the others, FB is greater than FC. 1. Join BE,CE. Then (I. 21, cor.) BE, EF are greater than BF ; but AE is equal to EB ; therefore AF, that is, AE, EF, is greater than BF. 2. Because CF, FE are greater (I. 21, cor.) than EC, and EC is equal to ED ; CF, FE are greater than ED. Take away the common part FE, and (I. ax. 5) the re- mainder CF is greater than the remain- der FD. 3. Again : because BE is equal to CE, and FE common to the triangles BEF, CEF ; but the angle BEF is greater than CEF ; therefore (L 22) the base BF is greater than the base CF. 4. Make (I. 13) the angle FEH equal to FEC, and join FH. Then, because CE is equal to HE, EF common to the two triangles CEF, HEF, and the angle CEF equal to the angle HEF ; therefore (I. 3) the base FC is equal TO THE ELEMENTS OF [bOOK III. to the base FH, and the angle EFC to the angle EFH. But, besides FH, no other straight line can be drawn from F to the circumference equal to FC (L ax, 9). Prop, VI, — Theok, — If from any point without a circle straight lines he drawn to the circrimference; (l) of those which fall npon the concave part of the circumference, the greatest is that which passes through the center ; unci (2) of the rest, one nearer to the greatest is greater than one more remote. (3) Hut of those which fall upon the convex part, the least is that which when produced passes through the center' and (4) of the rest, one nearer to the least is less than one more remote. And (5) only two equal straight lines can be drawn from the point to either part of the circumference, one upon each side of the line passing through the center, and making equal angles with it. Let ABF be a circle, M its center, and D any point without it, from which let the straight lines DA, DE, DF be drawn to the circumference. Of those which fall upon the concave pait of the circumference AEF, the greatest is DlNLiV, which passes through the center ; and a line DE nearer to it is greater than DF, one more remote. But of those which fall upon the con- vex circumference LKG, the least is DG, the external part of DMA ; and a line DK nearer to it is less than DL, one more remote, 1. Join ME, MF, ML, MK ; and because MA is equal to ME, add MD to each ; therefore AD is equal to EM, MD ; but (I. 21, cor,) EM, MD are together greater than ED; therefore, also, AD is greater than ED. 2. Because ME is equal to MF, and MD common to the triangles EMD, FMD, but the angle EMD is greater than FMD ; therefore (L 22) the base ED is greater than the base FD. 3, Because (I, 21, cor.) MK, KD are greater than MD, and MK is equal to MG, the remainder KG is greater (I, ax. 5) than the remainder GD ; that ie, GD L* less than KD. BOOK m.] EUCLID AND LEGENDKE. 71 4, Because MK is equal to ML, and MD common to the triangles KMD, LMD, but the angle DMK less than DML; therefore the base DK is less (I. 22) than the base DL. 5. Make (I. 13) the angle DMB equal to DMK, and join DB. Then, because MK is equal to MB, MD common to the triangles KMD, BMD, and the angle KMD equal to BMD ; therefore (I. 3) the base DK is equal to the base DB, and the angle MDK to the angle MDB, But, besides DB, there can be no straight line drawn from D to the circumference equal to DK (I. ax. 9). Prop. VII. — Theoe. — Equal chords in a circle are equally distant from the center ; and (2) chords which are equally dis- tant from the center are equal to one another. Let the chords AB, CD, in the circle ABDC, be equal to one another; they are equally distant from the center. Take (TIL 1) E the center of the circle, and draw (L 7) EF, EG perpendiculars to AB, CD ; join also EA, EC. Then, be- cause the straight line EF, passing through the center, cuts the chord AB, which does not pass through the center, at right angles, it also (III. 2) bisects it ; where- fore AF is equal to FB, and AB is dou- ble of AF. For the same reason, CD is double of CG ; but AB is equal to CD ; therefore AF is equal (I. ax. V) to CG. Then, in the right-angled triangles EF A, EGC, the sides EA, AF are equal to the sides EC, CG, each to each, therefore (L 3) the sides EF, EG are equal. But chords in a circle are said (IIL def, 3) to be equally distant from the center, when the perpendiculars drawn to them from the center are equal; therefore AB, CD are equally distant from the center. Next, if the chords AB, CD be equally distant from the cen- ter, that is, if FE be equal to EG, AB is equal to CD. For, the same construction being made, it may, as before, be demon- strated that AB is double of AF, and CD of CG ; and because tlie right-angled triangles EFA, EGC have the sides AE, EF equal to CE, EG, each to each, the sides AF, CG are also (L 3) 72 THE ELEMENTS OF [bOOK m. equal to one another. But AB is double of AF, and CD of CG ; wherefore AB is equal (I. ax. 6) to CD. Therefore equal chords, etc. Cor. Hence, the diameter of a circle is the greatest chord ; (2) of others, one nearer to the center is greater than one more remote ; and (3) the greater is nearer to the center than the less. Prop. VIII. — Thkoe. — The straight line drawn perpendicu- lar to a diameter of a circle^ through its extremity.^ falls icith- out the circle ; hut any other straight line drawn through that point cuts the circle. Let ABC be a circle, of which D is the center, and AB a diameter ; if AE be drawn through A perpendicular to AB, it falls without the circle. In AE take any i:)oint F ; and draw DF, meeting the circumference in C. Because DAF is a right angle, it is greater (I. 20) than DFA; and there- fore (I. 21) DF is greater than DA. But (I. def. 16) DA is equal to DC; therefore DF is greater than DC, and the point F is therefore without the circle ; and in the same manner it may be shown, that any other point in AF, except the point A, is without the circle. Again : any other straight line drawn through A cuts the circle. Let AG be drawn in the angle DAF, and draw (I. 8) DH perpendicular to AG, and meeting the circumference in C. Then, because DHA is a right angle, and DAH less than a right angle, being a part of DAE, the side DH is less (I. 20) than the side DA, But (L def 16) DK is equal to DA; there- fore DH is less than DK ; the point H is therefore within the circle ; and AG cuts the circle, since its continuation through A must fall on the opposite side of EAL, and must therefore be without the circle. Therefore the straight line, etc. Cor. From this it is manifest that the straight line which is drawn at right angles to a diameter of a circle from its extrem- BOOK in.] EUCLID AND LEGENDRE. 73 ity touches (III. def. 1) the ch'cle; and that it touches it only in one poiut, because at every point except A, it falls without the circle. It is also evident, that there can be but one tan- gent at the same point of a circle. Pkop. IX. — Pbob. — From a given pointy either -without a given circle^ or in its circumference^ to draw a straight line touchi^ig the circle. First : let A be a given point without the given circle BCD ; it is required to draw from A a straight line touching the circle. Find (III. 1) E the center of the circle, and draw AE cutting the circumference in D ; from the center E, at the distance EA, describe (I. post. 3) the circle AFG; from D draw (I. 7) DF at right an- gles to EA; and draw EBF, and join AB. AB touches the circle BCD. Because E is the center of the cir- q cles, EA is equal to EF, and ED to EB ; therefore the two sides AE, EB are equal to the two FE, ED, each to each, and they contain the angle AEF common to the two tri- angles AEB, FED ; therefore the angle EBA is equal (I. 3) to EDF, and is, therefore, a right angle, because (const.) EDF is a right angle. Now, since EB is drawn from the cen- ter, it is part of a diameter of which B is one extremity ; but a straight line drawn from the extremity of a diameter at right angles to it touches (III. 8, cor.) the circle; therefore AB touches the circle ; and it is drawn from the given point A ; which was to be done. Secondlj : if the given point be in the circumference of the circle, as the point D, draw DE to the center E, and DF at right angles to DE ; DF touches (III 8, cor.) the circle. Cor, 1. If AB be produced to H, AH is bisected (III. 2) in B. Hence a chord in a circle touching a concentric one is bisected at the point of contact. Scho. 1. It is evident that from any point A without the cir- 74 THE ELEMENTS OF [BOOK HI. cle, two tangents may be drawn to the circle, and that these are equal to one another, being equal respectively to the equal lines DF and DF'. Scho. 2. The construction of the first case of this problem is as easily efiected in practice, by describing a circle on AE as diameter, as its circumference will cut that of the given circle in the points B and B^ The reason of this will be evident from the twelfth proposition of this book. Cor. 2. Hence in a given straight line AB, a point may be found such that the difference of its distances from two given points C, D, may be equal to a given straight line. Join CD, and from D as center, with a radius, DE equal to the given difference, describe a circle ; draw CP"G perpendicular to AB, and make FG D equal to FC ; through C, G describe a cir- cle touching the other circle ; join B, D, the centers of the two circles, and draw BC ; BC, BD are evi- dently the required lines. Cor. 3. In the same manner, if a circle were described fi-om D as center, with the sum of two given lines as radius, and a circle were described through C and G touching that circle, straight lines drawn from the center of that circle to C and D would be equal to the given sum. Pkop. X. — Theor. — The angle at the center of a circle is double of the angle at the circuntference., xipon the same base^ that is, upon the same part of the circumference. In the circle ABC, let BEC be an angle at the center, and BAC an angle at the circumference, which have the same arc BC for their base ; BEC is double of BAC. Draw AE, and produce it to F ; and first, let E, the center of the circle, be wiiliin the angle BAC. Because EA is equal to EB, the angle EAB is equal (I. 1, "p ^ cor.) to EI>A ; therefore the angles EAB, EBA are together double of EAB ; but (I. 20) the angle BEP is equal to EAB, EBA; therefore also BEF is double of EAB. BOOK in.] EUCLID AND LKGENDRE. 75 For the same reason, the angle FEC is double of the angle EAC ; therefore the whole angle BEC is double of the whole BAG. Again : let E the center of the circle be without the an^le CD O BAG ; it may be demonstrated, as in the first case, that the angle FEG is double of FAG, and that FEB, a part of the first, is double of FAB a part of the other; there- fore the remaining angle BEC is double of the remaininsj ansjcle BAG. The angle at the center, therefore, etc. Scho. That if two magnitudes be double of two others, each of each, the sum and diff'erence of the first two are respectively double of the sum and difference of the other two. It is thus proved by Play fair: "Let A and B, G and D be four magnitudes, such that A = 2G, and B = 2D; then A+B = 2(G + D). For since A = G + G, and B=D-f-D, adding equals to equals, A4-B = (G + D) + G + D = 2(G + D). So, also, if A be greater than B, and therefore G greater than D, since A=:G4-G, and B=D4-D, taking equals from equals, A— B = (G— D) + (G— D), that is A— B = 2(G— D)." The following is an outline of another proof of the second case: from the triangle BGE we have (I. 20) BGG=BEG-!-B =:BEG4-EAG (I. 1). We have also from the triangle AGG, in a similar manner, BGG = BAG + G = BAG + EAG = 2BAG + EAG. Hence (I. ax. 1) BEG + EAG = 2BAG + EAG. Take away EAG, etc. If in the first diagram, GE were produced to meet AB, the first case might be proved in a similar manner. The second case might also be proved by drawing from G to AB a straight line meeting it in a point H, and making Avith AC an angle equal to BAG. Then, by taking the difference between the equal angles EAC, EGA, and the equal ones HAG, lie A, we have EAB and EGH equal, and therefore EBA= EGH. But EGB=HGG; and therefore (I. 20, cor. 5) BEG = BHG=:2BAG. Tl e same proof, with some obvious variations, would be applicable in the first case. There is evidently a third case, viz., when AB or AG passes through the center; but though this case is not given in a sep- arate form, its proof is contained in that of either of the others. 76 THE ELEMENTS OF [book nic PpwOp. XL — Theor. — In a circle, (1) the angle in a semicir- cle is a right angle ; (2) the angle in a seginent greater than a semicircle is acute ; and (3) the angle in a segment less than a semicircle is obtuse. Let ABC be a circle, of which F is the center, BC a diame- ter, and consequently BAG a semicircle ; and let the segment BAD be greater, and BAE less than a semicircle ; the angle BAG in the semicircle is a right angle ; but the angle BAD in the segment greater than a semicircle is acute ; and the angle BAE in the seg- ment less than a semicircle is obtuse. Draw AF and produce it to G. Then (III. 10) the angle BAG at the circum- ference, is half of BFG at the center, both standing on the same arc BG ; and for the same reason, GAG is half of GFC. Therefore the whole angle BAG is half of the angles BFG, GFG ; and (I. 9) these are together equivalent to two right angles ; therefore DAG is a riglit angle, and it is an angle in a semicircle. But (I. ax. 9) the angle BAD is less, and BAE greater than the right angle BAG ; therefore an angle in a seg- ment greater than a semicircle is acute, and an angle in a seg- ment less than a semicircle is obtuse. Pbop. XII. — Theoe. — If a straight line touch a circle, the straight line drawn from the center to the point of contact is 'perpendicular to the line touching the circle. Let the straight line FH touch the circle ABGD at the point C ; the straight line GA drawn from that point to the center of the circle is perpendicular to the line touching the circle. Draw a diameter BD parallel to FH (I. 18). and with the diameter BD as a base, and the point C as a ver- tex, make the triangle BGD. But BGD is a right angle (III. 11). FGB and HCD are equivalent to a right angle (L 9), and because BD is parallel to FH (const.), FGB is equal to GBD, and HGD is equal to CDB (L 15, cor. 2) ; and BOOK in.] EUCLID AND LEGENDKE. 77 at the point C draw a perpendicular CA to FH (I. 7), it will also be perpendicular to BD (I. 17), then FCB and BCE are equivalent to BCD (I. ax. 1) ; taking a^vay the common angle BCE, we have FCB equal to ECD ; but FCB is equal to CBD, hence ECD is equal to CBD ; and in same manner it can be shown that ECB is equal to CDB. The triangles EBC and EDC having EC common, the angle EBC equal to the angle ECD, and the angles BEC and CED both right angles, are equal (I. 3), therefore EB is equal to ED, and the diameter BD is bisected by CA, then CA passes through the center of the circle (III. 1). Wherefore, if a straight line, etc. Cor. Hence, conversely, if a straight line touch a circle, a straight line drawn from the point of contact, perpendicular to the tangent, passes through the center. Prop. XIII. — Theoe. — If one circle touch another inter- nally or externally in any point, the straiyht line rohich joins their centers being jyroducecl, passes through that point. Let al)Q and DEC be two circles which touch one another internally, their centers will be in the same straight line with their point of contact. At the point of contact C draw the tangent FH, and at this point erect a perpen- dicular CaE (I. 7). B p D The diameter of aJC is perpendicular to the tangent at the point of contact (HI. 12), hence Ca is the diameter of a6C, and passes through the center of a5C(IH. 1). NowFH is also tangent to the circle DEC (const.) at the point C ; hence, the perpendicular CE (HI. 12) is the diameter of DEC, and passes (HI. 1) through the center of DEC. But Ca and CE are in the same straight line, hence the centers of the cir- cles ahQi and DEC, and the point of contact C, are in the straight line CaE. Wherefore, if one circle touch, etc. Or, let ABC and CDE be two circles which touch one an- 78 THE ELEMENTS OF [bOOK IH. Other externally in the point C ; their centers will be in the same straight line with the point of contact. At the point C draw a tangent FH, which will be perpen- dicular to the diameter of ABC (III. 12). FH being tangent at the point of contact of the circles, is also pei-pendicular to the diameter of CDE (IIL 12). ACF is a right angle (I. def. 10), and FCE for the same reason is a right angle, and both equal to one another (I. def, 10, and ax. 11), hence are two right angles ; then (I. 10) AC and CE form the same straight line; but AC passes through the center of ABC, and CE passes through the center of CDE, being their respective diameters (IIL 1), therefore the centers of the circles are in the same straight line with the point C. Wherefore, if two circles touch each other, etc. Prop. XIV. — Tueor. — Similar segmevU of circles upon equal bases are equal to one another, and have equal arcs. Let AEB, CFD be similar segments of circles upon the equal straight lines or bases, AB, CD ; the segments are equal ; and likewise the arcs AEB, CFD are equal. For if the segment AEB be applied to the segment CFD, so that the point A may E y_ be on C, and the straight line AB on CD, the point B will D coincide with D, be- cause AB is equal to CD ; therefore the straight line AB coinciding with CD, the segment AEB must coincide (I. def 16) with the segment CFD, and is therefore equal (I. ax. 8) to it ; and the arcs AEB, CFD are equal, because they coincide. Therefore similar seg- ments, etc. Prop. XV. — Prob, — A segment of a circle being given, to complete the circle of lohich it is a segment. Assume three points in the arc of the segment, and find (IIL 1) the center of the circle. From that center, at the distance between it and any point in the arc describe a circle, and it will evidently be the one required- BOOK ni.] EUCLID AND LEGENDEE. 79 Prop. XV L — Treor. — Ifi equal circles, or in the same circle, equal a?igles stand upon equal arcs, whether they are at the centers or the circumferences. Let ARC, DEF be equal circles, having the equal angles BGC, EHF at their centers, and BAG, EDF at their circum- ferences ; the arc BKC is equal to the arc ELF. Join BG, EF ; and because the circles ABG, DEF are equal, their radii are equal ; therefore the two sides BG, GG are equal to tlic two, EH, HF ; and (hyp.) the angles G and H are equal ; therefore (I. 3) the base BG is equal to the base EF. Then, because the angles A and D are equal, the segment BAG is similar (III, def 8) to the segment EDF ; and they are upon equal straight lines BG, EF ; but (III. 14) similar segments of circles upon equal straight lines have equal arcs ; therefore the arc BAG is equal to the arc EDF. But the whole circumfer- ence ABG is equal to the whole DEF, because the circles are equal ; therefore the remaining arc BKG is equal (I. ax. 3) to the remaining arc ELF. Wherefore, in equal circles, etc. Cor. 1. Gonversely, in equal circles, or in the same circle, the angles which stand upon equal arcs are equal to one another, •whether they are at the centers or the circumferences. (I. def 19.) Cor. 2. Hence, in a circle, the arcs intercepted between par- allel chords are equal. For if a straight line be drawn trans- versely, joining two extremities of the chords, it will (I. 16) make equal angles with the chords ; and therefore the arcs on "which these stand are eqiial. Cor. 3. Hence, in equal circles, equal chords divide the cir- cumferences into parts which are equal, each to each. 80 THE ELEMENTS OF [book nr. D Pkop. XVII. — Peob. — To bisect a given arc of a circle. Let ADB be a giA-en arc ; it is required to bisect it. Draw AB, and (I. 6 and 7) bisect it in C, by the perpendicu- lar CD ; the arc ABD is bisected in the point D. Join AD, DB. Then, because AC is equal to CB, CD com- mon to the triangles ACD, BCD, and the angle ACD equal to BCD, each of them being a right angle; therefore (I. 3) AD is equal to BD. But (III. 16, cor. 3), in the same circle, equal lines cut off equal arcs, the greater equal AC B to the greater, and the less to the less; and AD, DB are each of them less than a semicircle, because (III. 1) DC^ or DC produced, passes througli the center; wherefore the arc AD is equal to the are DB; therefore the given arc is bisected in D ; which was to be done. Prop. X'N'III. — Theoe. — If a straight line touch a circle, and from thepoifit of contact a straight line be drawn dividing the circle into tico segments ^ the angUs made by this line with the tangent are equivaletit to the angles which are in the alter- nate segments. Let the straight line DE touch the circle BAG in the point B, and let the straight line BA be drawn dividing the circle into the segments AGB, AB ; the angle ABE is equal to any an- gle in the segment AGB, and the angle ABD to any angle in AB. If AB (fig. 1 ) be perpendicular to DE, it passes (III. 1 2, cor.) through the center, and the segments being therefore semicir- cles, the angles in them are (HI. 11) right angles, and conse- quently equal to those which AB makes with DE. But if BA (fig. 2) be not perpendicular to DE, draw BF per- ■ A F D B E D B E pendicular to it ; join FA and produce it to E; join also CA, BOOK in.] EUCLID AND LFGKNDRE. 81 CB, C being any point in the arc ACB. Then (TIT. 12, cor.) BF is a diameter, and (III. 11, and I. 9) the angles BAF, BAE are riglit angles. Therefore, in the triangles J^AE, FBE, the angle E is common, and the angles BAE, FBE equal, being right angles; wherefore (I. 20, cor. 5) the remaining angle ABE is equal to the remaining angle F, which is an angle in the remote or alternate segment BFA. Again : the two angles ABD, ABE are equal (I. 9) to two right angles ; and because ACBF is a quadrilateral in the circle, the opposite angles C and F are also equivalent (I. 20, cor. 1) to two right angles ; therefore (I. ax. 1) the angles ABD, ABE are together equal to C and F*. From these equals take away the angles ABE and F, which have been proved to be equal ; then (I. ax. 3) the remaining angle ABD is equal to the re- maining angle C, which is an angle iu the remote segment ACB. If, therefore, a straight line, etc.. ScliO. 1. The first case is wanting in most editions of Euclid. In the second diagram, FA and DB will meet (I. 19) if pro- duced, the angles FBE, BFA together being evidently less than two right angles. Scho. 2. Let AB be a fixed chord, and ^ through B draw any other line DE cut- ting the circle in C ; and join AC. Now (III. 16, cor. 1), wherever C is taken in the arc ACB, the angle ACE is con- stantly of the same magnitude ; and so also (I. 9) is the exterior angle ACD. If C be now taken as coinciding with B, the straight line DE will become the tangent D'BE', AC will coincide with AB, axid the angles 6 82 THE ELEMENTS OF [bOOK HI. AGE, ACD will become ABE', ABD'. If again C take the position C^ the angles ACE, ACD will become AC'^'', AC"T>'\ Now, the equality of ABE', ACE, and of ABD', ACD'' is what is provecl in the eighteenth proposition, and from the equality of ACD and AC'D" cor. 2 follows by the addition of ACB. Cor. 1. Angles in the same segment of a circle are equal to one another. Cor. 2. The opposite angles of any quadrilateral figure de- scribed in a circle are together equivalent to two right angles; and conversely^ if two opposite angles of a quadrilateral be to- gether equal to two right angles, a circle may be described about it. * Cor. 3. If the circumference of a circle be cut by two straight lines which are perpendicular to one another, the squares of the four segments between the point of intersection of the two lines and the points in which they meet the circum- ference, are together equal to the square of the diameter. Prop. XIX. — Prob. — TJpon a given straight line, to describe a segment of a circle containing an angle equal to a given angle. Let AB be the given straight line, and C the given angle ; it is required to describe on AB a segment of a circle containing an angle equal to C. First : if C be a right angle, bisect (I. 6) AB in F, and from the center F, at the distance FB, de- scribe the semicircle AHB ; therefore (III. 11) any angle AHB in the semicir- cle is equal to the right angle C. But if C be not a right angle, make (I. 13) the angle BAD equal to C, and (I. 7) from A draw AE pei-pendicular to AD ; bisect (I. 5 and 6) AB by the perpendicular FG, and join GB. Then, because AF is equal to FB, FG common to the triangles AFG, BFG, and the angle AFG equal to BFG, therefore (I. 3) AG is equal to GB ; and the circle described from the center G, at the distance GA, will pass through the point B ; let this be the circle AIIB, Then, because from the poiut A, the extremity [book III. EUCLID AND LEGENDRE. 83 of the diameter AE, AD is drawn at right angles to AE, AD (III. 8, cor.) touches the circle; and (III. 18) because AB, drawn from the point of contact A, cuts the circle, the angle DAB is equal to any angle in the alternate segment AHB ; but DAB is equal to C ; therefore also C is equal to any angle in the segment AHB ; wherefore upon the given straight line AB the segment AHB is described, which contains an angle equal to C j which was to be done. Scho. It is evident there may be two segments answering the conditions of the problem, one on each side of the given line. It is also plain, that when C is an acute angle, and the segment is to be above AB, G is above AB; but when obtuse, it is below it. It is likewise plain, that the angle BAE is the complement of the given angle C ; that is, the difference between it and a right angle. Cor. Hence from a given circle can be cut off a segment which shall contain an angle equal to a given angle. Prop. XX. — Theor. — If two chords of a circle cut one an- other, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. In the circle LBM, let the two chords LM, BD cut one an- other in the point F ; the rectangle ].<F.FM ^ is equal to the rectangle BF.FD. If LM, BD both pass through the center, 80 that F is the center, it is evident that LF, FM, BF, FD, being (I. def 16) all equal, the rectangle LF.FM is equal to the rectangle BF.FD. But let one of them, BD, pass through the center and cut the other, AC, which does not pass through the center, at right angles, in the point E. Then, if BD be bisected in F, F is the center. Join AF; and because BD which passes through the center, is perpendicular to AC, AE, EC are (III. 2) equal to one another. Now, because BD is 84 THE ELEMENTS OF [bOOK HI. divided equally in F, and unequally in E, the rectangle BE.ED, and the square of EF are equivalent (II. 5) to the square of FB ; that is (I. 23, cor. 2), to the square of FA. But (I. 24, cor. I) the squares of AE, EF are equivalent to the square of FA ; therefore the rectangle BE.ED and the square of EF are equivalent to the squares of AE, EF. Take away the common square of EF, and the remaining rectangle BE.ED is equivalent to the remaining square of AE; that is, to the rectangle AE.EC. Next, let BD pass through the center, and cut AC, which does not pass through the center, in E, but not at right angles. Then, as before, if BD be bisected in F, F is the center of the circle. Join AF, and (I. 8) draw FG per- pendicular to AC ; therefore (III. 2) AG is equal to GC ; wherefore (II. 5) the rectan- gle AE.EC and the square of EG are equiv- alent to tlie square of AG. To each of these equivaleuts add the square of GF; therefoi'e the rectangle AE.EC and the squares of EG, GF are equivalent to the squares of AG, GF ; but (I. 24, coi-. l) the squares of EG, GF are equivalent to the square of EF; and the squares of AG, GF are equivalent to the square of AF ; therefore the rectangle AE.EC and the square of EF are equivalent to the square of AF ; that is (I. 28, cor. 2), to the square of FB. But (II. 5) the square of FB is equivalent to the rectangle BE.ED, together with the square of EF ; therefore the rectangle AE.EC and the square of EF are equivalent to the rectangle BE.ED and the square of EF; take away the common square of EF, and the remain- ing rectangle AE.EC is equivalent to the remaining rectangle BE.ED. Lastly : let neither of the lines pass through the center, and through E, the point of intersection, draw a diameter. Then, the rectangle AE.EC is equivalent, as has been shown, to the rectangle DE.EB ; and, for the same reason, the rectangle of the other chord is equivalent to the same rectangle DPIEB ; therefore (I. ax. 1) the rectangle AE.EC is equivalent to the rectangle of the other chord. If, therefore, two chords of a circle, etc. BOOK III.] EUCLID AND LEGENDI E. 85 Scho. The second and third cases may he thus demonstrated in one: Join FC. Then the rectangle AE.EC is equivalent (IT. 5, cor. 5) to the difference of the squares of AF and FE, or ot DF and FE,or (II. 5, cor. l)to the rectangle DE.EB. Proportion, however, affords much the easiest method of demonstrating both this proposition and the following. Peop. XXI. — TiiEOR. — If from any point without a c'rcle two straight li?ies be draim, one of which cuts the'circle, and the other touches it ; the rectangle contained hy the whole line which cuts the circle^ and the />«?•« of it without the circle, is equivalent to the square of (he line which touches it. Let D he any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle in C and A, and DB touches it in B ; the rectangle AD.DC is equivalent to the square of DB. There are two cases — first : let DCA pass through the center E, and join EB. Therefore (III. 12) the angle EBD is a right angle; and (II. 6) "because the straight line AC is bisected in E, and produced to D, the rectangle AD.DC and the square of EC are together equiva- lent to the square of ED ; and CE is equal to EB ; therefore the rectangle AD.DC and the square of EB are equivalent to the square of ED. But (I. 24, cor. l) the square of ED is equivalent to the squares a of EB, BD, because EBD is a right angle ; therefore the rectangle AD.DC and the square of EB are equivalent to the squares of EB, BD ; take away the common square of EB, and the remaining rectangle AD.DC is equiva- lent to the square of the tangent DB. Second : if DCA do not pass through the center, take (III. 1) the center E, and draw (I. 8) EF perpendicular to AC, and join EB, EC, ED. Then, because the straight line EF, which passes through the center, is perpendicular to the chord AC, AF is equal (III. 2) to FC. And (II. 6) because AC is bisected iu F, and produced to D, the rectangle AD.DC and the square 86 THE ELEMENTS OF [book ni. of FC are equivalent to the square of FD. To each of these equals add the square of FE ; therefore the rectangle AD.DC, and the squares of CF, FE are equivalent to the squares of DF, FE ; but (I. 24, cor. 1) the square of ED is equivalent to the squares of DF, FE, because EFD is a right angle ; and the square of EC, or (I, 23, cor. 2) of EB, is equivalent to the squares of CF, FE ; therefore the rectangle AD.DC and the square of EB are equivalent to the square of ED. But (I. 24, cor. 1) the squares of EB, BD are equivalent to the square of ED because EBD is a right angle; therefore the rectangle AD.DC and the square of EB are equivalent to the squares of EB.BD. Take away the common square of EB ; therefore the remaining rectangle AD.DC is equivalent to the square of DB ; wherefore, if from any point, etc. Scho. The second case may be demonstrated more briefly thus: Join AE. Then the rectangle AD.DC is equal (II. 5, cor. 5) to the difference of the squares of ED and EC, or of ED and EB, or (I. 24, cor. 1) to the square of DB. Cor. 1. If from any point without a circle there be drawn two straight lines cutting it, as AB, AC, the rectangles contained by the whole lines and the parts of them without the circle are equivalent to one another, viz., the rectangle BA.AE to the rectangle CA.AF; for each rectangle is equivalent to the square of the tangent AD. Cor. 2. If two straight lines intersect each other, so that the rectangle under the seg- ments of one of them is equal to the rectangle under the segments of the other, their ex- tremities lie in the circumference of a circle. Scho. By means of the first corollary, it would be shown in a similar manner, that if two straight lines meet in a point, and if they be so divided that the rectangle under one of them and its segment next the common point is equal to the rectangle BOOK. III.] EUCLID AND LEGENDRE. 87 under the otlier and its corresponding segment, the points of section and the extremities remote from the common point lie in the circumference of the same circle. D Pkop. XXII. — TiiEOR. — If frotn a point without a circle there be draw:i two straight lines, one of ichich cuts the ctrcle^ and the other meets it ; and if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equivalent to the square of the line which meets it, the line which meets the circle touches it. If from a point without the circle ABC two straight lines DCA and DB be drawn, of which DCA cuts the circle, and DB meets it ; and if tlie rectangle AD.DC be equivalent to the square of DB ; DB touches the circle. Draw (III. 9) the straight line DE touching the circle ABC ; find (III. 1) the center F; and join FE, FB, FD. Then (III. 12) FED is a right angle; and (fll. 21) because DE touches the circle ABC, and DCA cuts it, the rectangle AD.DC is equivalent to the square of DE. But (hyp.) the rectangle AD.DC is equivalent to the square of DB ; therefore the square of DE is equivalent to the square of DB ; and the straight line DE equal (I. 23, cor. 3) to the straight line DB. But FE is equal to FB, and the base FD is common to the two tri- angles DEF, DBF ; therefore (I. 4) the angle DEF is equal to DBF; but DEF is a right angle; therefore, also, DBF is a right angle ; and FB is a part of a diameter, and the straight line which is drawn at riiiht angles to a diameter, from its ex- tremity, touches (III. 8, cor.) the circle ; therefore DB touches the circle ABC. Wherefore, if from a point, etc. Prop. XXIII. — Prob, — To divide a given straight line into two j)arts, such that the sjuare of one of them may be equivalent to the rectangle contained by the other, and a given straight line. Let AB, AC be two given straight lines ; it is required to divide AB into two parts, such that the square of one of them may be equivalent to the rectangle under AC and the other. 88 THE ELEMENTS OF [book m. On CB, the sum of the given lines, describe the semicircle CDB, and draw AD jterpendicular to CB ; bisect CA in E, and join DE ; and make EF equal to ED ; then the square of AF is equivalent to the rectangle AC.FB. Describe the semicircle CGA, cutting ED in G, and join GF. Then, because FE, EG are respectively equal to DE, EA, and the angle FED common, GF is equal to AD, and the an- gle EGF to EAD, whicli is a right angle, and therefore GF touches the circle CGA. Hence (III. 20) the rectangle CA.AB is equivalent to the square of AD, or of FG, or (HI. 21) to the rectangle CF.FA. But (11. 1) the rectanglo CA.AB is equivalent to the two CA.AF, CA.FB, and (II. 3) the rectangle CF.FA is equivalent to CA.AF and the square of AF. From these equivalents take away the rectangle CA.AF, and there remains the rectangle CA.FB equivalent to the square of AF. • Scho. The tenth proposition of the second book is the par- ticular case of this problem in which the given lines are equal. Prop. XXIV. — Prob. — To draw a common tangent to two given circles. Let BDC, FUG be given circles ; it is required to di-aw a common tangent to them. Join their centers A, E, and make BK equal to FE ; from A as center, with AK as radius, describe a circle cutting another, described on AE as diameter, in M ; draw AM, meeting tlie circle BCD in D ; and draw Eli parallel to AM; join Dil; it touches both the given circles. BOOK III.] EUCLID AND LEGENDRE. 89 For MD, EII, which (const.) are parallel, are equal to one another, because each of them is equal to KB; therefore (I. 15, cor. B) Dll is parallel to ]\1E. Now (III. 11), the angle AME in a semicircle is a right angle; and therefore (I. 16) the angles ADII, DUE are right angles, and (III. 8, cor.) DH touches both the ciicles, since it is perpendicular to the radii AD, EH. ScJio. In the figure the circles lie on the same side of the tangent, which is therefore exterior to them ; but the tangent can be transverse, or lie between the circles. It is plain also, tliat in these figures, by using the point L instead of M, an- other exterior and another transverse tangent would be ob- tained ; and this will always be so, when each of the circles lies wholly without the other, and does not touch it. But if the circles touch one another externally, the two transverse tangents coalesce into a single line passing through their point of contact ; if they cut one another, there will be two exterior tangents, but no transverse one ; if one of the circles touch the other internally, they carf have only one common tangent, and this passes through their point of contact ; and, lastly, if one of them lie wholly within the other without touching it, they can have no common tangent. If the circles be equal, the points of contact of the exterior common tangents are the extremities of the diameters perpen- dicular to the line joining the centers ; for (I. 15, cor. 3, and 1 6) the lines connecting the extremities of these toward the same parts are perpendicular to the diameters, and therefore (III. 8, cor.) they touch the circles. Prop. XXV.— Prob.— 7b insa-ibe a7iy regular polygon in a given circle. Let ABDC be the given circle ; it is required to inscribe any polyg<m in it. Since (I. 9, cor.) all the angles formed by any number of straight lines inter- secting each other in a common point are equivalent to four right angles, and (I, 20, cor. 1) any rectilinear figure can "be divided into as many triangles as the figure has sides, by straight lines joining the extremities of the 90 THE ELEMENTS OF [BOOK m. sides with a poiut within the figure — a regular polygon being equilateral — the straight lines connecting the extremities of those sides with the center of the regular polygon will divide the regular polygon into equal triangles ; for if ABC be a regu- lar polygon of three sides, AB is equal BC, and also equal to CA, and draw from E (III. 1, cor. 1), the center of the polygon, EA, EB, and EC. Now the triangles AEB, BEC, and CEA, having their bases equal (hyp.), their other sides common, (const.) are equal; therefore (I. 4) the angles AEB, BEC, and CEA are equal ; but those angles are equivalent to form right angles (I. 20, cor. 1) ; hence, each is one third of four right angles, or two thirds of two right angles ; consequently, a reg- ular polygon can be inscribed in a circle by making an angle on the diameter (I. 13) with the center of the circle the vertex of the angle, equal to that part of four right angles that the regular polygon has sides, viz. : if the regular polygon has four sides, each angle at the center will be one fourth of four right angles ; if five sides, one fifth of fourS-ight angles; and so on. Or, by bisecting (I. 5) the angle or the arc (III. 11), a regular polygon of double the number of sides can be inscribed in the circle. Having by these means the angle at the center of a regular polygon, the chord of the arc intercepted by the sides of the angles will be the side of the regular polygon required, from which (I. 12 and 13) the regular polygon can be inscribed in the given circle, which was to be done. Cor. 1. If the sides of the angles be produced beyond the circumference (I. post. 2), and parallels to the exterior sides of the polygon (I. 18) be drawn touching the circle (III. 9), then a similar and regular polygon can be described about the given circle. Cor, 2. Find the center of a given regular polygon (III. 1, cor. 1), and a circle can be inscribed in the polygon, or circum- scribed about it, by taking the straight line drawn from the center of the polygon to an extremity of the side of the poly- gon i'or a radius lor the circumscribed circle, and a line drawn from the center to the point of bisection of the side for a radius of the inscribed circle. Cor. 3. Since the triangles AEB, BEC, and AEC are equiv- alent respectively to the rectangles (I. 23, cor. 6) under the ra- BOOK in.] EUCLID AND LEGENDRE. 01 dius of an inscribed circle and the halves of AB, BC, and CA, it follows (IL 1) that the area of ABC is equivalent to tlie rect- angle under the radius and half the perimeter. Hence, if the sides be given in numbers, the length of the radius may be computed by calculating (II. 12, scho.) the area, and dividing it by half the sum of the sides, or its double by the sum of the sides. Cor. 4. Since the angles formed about the center of a circle are together equivalent to four right angles (I. 9, cor.), and since the angles of a triangle are together equivalent to two right angles (I. 20), it follows that when an equiangular trian- gle is formed (I. 2) having a vertex at the center and the radius for a side, each angle of the triangle is one tliird of two riglit angles, or one sixth of four right angles ; hence, six equal angles can be formed (I. 1.3) about the center of a circle, and since the sides about tliese angles are intercepted by the cir- cumference (const.), they are equal (I. def 16) ; and (I. 2, cor.) an equiangular triangle being also equilateral, the side of the triangle opposite the angle at the center of the circle is equal to the radius of the circle (T. ax. 1) ; therefore the radius can be made to subtend six equal arcs of the circumference. Cor. 5. When in the case of a general proposition to descrV^e a circle touching three given straight Imes ichich do vot pass through the same point, and which are not all parallel to one another. If two of the lines be parallel, there may evideiidy be two equal circles, one on each side of the line falling on the parallels, each of which will touch the three given lines, and their centers will be the intersections of the lines bisectimr the angles made by the parallels with the third line. But if the lines form a triangle by their intersections, there will be four circles touching them; one, inscribed, and the others each touching one side externally and the other two produced. The centers of the external circles will be the intersections of the lines bisecting two exterior angles; and the line bisecting the remote interior angle will pass through the same point. The method of proof is tl.e same as that given in the proposition. Cor. 6. If straight lines be drawn from the center of one of the external circles to the vertices of the triangle, the three tri- angles formed by the sides of the triangle and the straight lines 92 THE ELEMENTS OF [bOOK HI. are respectively equivalent to the rectangles (I. 15, cor. 4) under the radius of that circle and halves of tlie sides of the triangle. And if the triangle formed by the b.ide of the original triangle nearest the center of the circle, and the lines drawn to the vertices at the extremities of that side, be taken from the sum of the two other triangles, there remains the original tri- angle equivalent to the rectangle under the radius, and the ex- cess of half the sum of the two other sides of the orin-inal trian- gle above half the side nearest the center of the circle, or, which is the same thing, to the rectangle under the radius, and the excess of half the perimeter of the original triangle above the Bide nearest the center of the circle. The radius, therefore, of any of the external circles may be computed by dividing the area of the original triangle by the excess of half its perimeter above the side which the circle touches externally. Scho. The polygons considered in this proposition, and those which may be derived from them by the process of bisecting the angles or arcs subtending the sides of the polygons, are the only ones till lately which geometers have been able to de- scribe by elementary geometry, that is, by means of the straight line and circle. M. Gauss, of Gottingen, in \n?, Disqui- sitioiies Arithmeticoe, has shown that by elementary geometry, every regular poly. on may be inscribed in a circle, the number of whose sides is a power of 2 increased by unity, and is a prime number, or a number which can not be produced by the multiplication of two whole numbers, such as 17 — the fourth power of 2 increased by unity — and polygons of 257 and C5537 sides. But the investigation is too complex and difficult for an ordinary school text-book. END OP b60K third. BOOK FOURTH.* THE GENERAL THEORY OF PROPORTION". DEFINITIONS. 1. A LESS number or magnitude is said to measure a greater, or to be a measure^ a part^ or a suhmuUxple of tlie greater, when the less is contained a certain number of times, exactly, in the greater; and 2. The greater is said to be a multiple of the less. 3. Magnitudes which can be compared in respect of quantity, that is, which are either equal to one another, or unequal, are Baid to be of the sam,e kind, or homogeneous. Scho. 1. Thus, lines, whether straight or curved, are magni- tudes of the same kind, or are homogeneous, since they may be equal or unequal. In like manner, surfaces, solids, and angles form three other classes of homogeneous magnitudes. On the contrary, lines and surfaces, lines and angles, surfaces and solids, etc., are heterogeneous. Thus, it is obviously improper to say, that the side and area of a square are equal to one an- other, or are unequal. So likewise we cannot say that the area and one of the angles of a triangle are equal to each other, or are unequal ; and they are therefore heterogeneous. 4. If there be two magnitudes of the same kind, the relation which one of them bears to the other in respect of quantity, is called its ratio to the other. The first term, or magnitude, is called the antecedent of the ratio, and the second the consequent. 5. If there be four magnitudes, and if any like multiples whatever be taken of the first and third, and any whatever of * The accurate but prolix method of Euclid is substituted by the fol- lowing more concise method, by employing the notations and simple principles of Algebra. See Sup. to Book V. Thombon's Euclid. ^^ THE ELEMENTS OF [bOOK IV. the soconrl anfl fourth ; and if, according as the multiple of the first is gi-eater than the multiple of the second, equal to it, or less, the multiple of the third is also greater than the multiple of the fourth, equal to it, or less ; then the first of the magni- tudes is said to have to the second the same ratio that the third has to the fourth. 6. Magnitudes which have the same ratio are called propor- tionals ; and equality or identity of ratios constitutes /irojoor- tion or analogy. When magnitudes are proportionals, the relation is expressed briefly by saying, that the first is to the second, a« the third to the fourth, the fifth to the sixth, and so on. 1. When of the multiples of four magnitudes, taken as in the fifth definition, the multiple of the first is greater than that of the second, but the multiple of the third is not greater than that of the fourth ; then the first is said to have to the second a greater ratio than the third has to the fourth ; and, on the conti-ary, the third is said to have to the iomth a. less ratio than the first has to the second. 8. When there are three or more magnitudes of the same kind, such that the ratios of the first to the second, of the sec- ond to the third, and so on, whatever may be their number, are all equal; the magnitudes are said to be continual propor- tionals. 9. The second of three continual proportionals is said to be a mean proportional between the other two. 10. When there is any number of magnitudes of the same kind, greater than two, the first is said to have to the last the ratio compounded of the ratio which the first has to the second, and of the latio which the second has to the third, and of the ratio which the third has to the fourth, and so on to the last magnitude. For exam].]e, if A, B, C, D be four magnitudes of the same kind, the first, .>, is said to have to the last, D, the ratio com- pounded of tlie ratios of A to B, B to C, and C to D. 11. When iln-ee magnitudes are continual proportionals, the ratio of tlie first to the third is said to be duplicate of the ratio of the first Xo the second, or of the second to the third. 12. When iour magnitudes are continual proportionals, the BOOK rv.] EUCLID AND LEGENDEE. 95 ratio of the first to the fourth is said to be triplicate of the ratio of the first to the second, of the ratio of the second to the third, or of tlie ratio of the third to the fourth. Scho. 2. In continual proportionals, by their own nature, and that of compound ratio, the ratio of the first to the third is compounded of two equal ratios ; and the ratio of the first to the fourth, of three equal ratios ; and hence we see the reason and the propriety of calling the first duplicate ratio, and the second triplicate. It is plain, that on similar principles, the ratio of the first to the fifth would be said to be quadruplicate of the ratio of the first to the second, of the second to the third, etc., and thus we might form other similar terms at pleasure. The terras subduplicate, subtriplicate^ and sesquiplicate^ which are sometimes employed by mathematical wiiters, are easily understood after the explanations given above. In con- tinual proportionals, the ratio of the first terra to the second is said to be subcJuplicate of the ratio of the first to the third, and subtriplicate of that of the first to the fourth. Again : if there be four continual proportionals, the ratio of the first to the fourth is said to be sesquiplicate of the ratio of the first to the third ; or, which amounts to the same, the ratio which is com- pounded of another ratio and its subduplicate, is sesquiplicate of that ratio. 1 3. In proportionals, the antecedent terms are called homolo- gous to one another, as also the consequents to one another. Geometers make iise of the following technical words to denote diiferent modes of deriving one proportion from an- other, by changing either the order or the magnitudes of the terms. 14. Alternately : this word is used when there are four pro- portionals of the same kind ; and it is inferred that the first has the same ratio to the third which the second has to the fourth ; or that the first is to the third as the second to the fourth ; as is shown in the fourth proposition of this book. 15. By inversion: when there are four proportionals, and it is inferred that the second is to the first as the fourth to the third. Prop. 3, Book IV. 16. By composition: when there are four proportionals, and it is inferred, that the first, together with the second, is to the 96 THE ELEMENTS OF [BOOK IV. second as the tliii-fl, together with the fourth is to the fourth. Tenth Prop., Book IV. 17. By division : when there are four proportionals, avd it is inferred that the exjcess of the first above the second is to the second, as the excess of the third above the fourth is to the foui-th. Tenth Prop., Book IV. 18. My coyiveraion : when there are four proportionals, and it is inferred that the first is to its excess above tlie second, as the third to its excess above the fourth. Eleventh Prop., Book IV. Sc'ho. The substance of the five preceding definitions may be exhibited briefly in the following manner, the signs + and — denoting addition and subtraction, as has been explained already at the beginning of the second book: Let A: B:: C:D; Alternately, A : C : : B : D ; By inversion, B : A : : D : C ^ By composition, A + B : B : : C + D : D ; By division. A— B : B : : C— D : D ; By conversion, A : A — B : : C : C — D. 19. Ex mquo^ or ex equali (scil. distantid)^ from equality of distance: when there is any number of magnitudes more than two, and as many others, which, taken two in the one rank, and two in the other, in direct order, have the same ratio; and it is inferred that the first has to the last of the first rank the same ratio which the first of the other rank has to the last. This is demonstrated in the thirteenth proposition of this book. 20. Ex cequo^ inversely : when there are three or more mag- nitudes, and as many others, which taken two and two in a cross order, have the same ratio ; that is, when the first magni- tude is to the second in the first rank, as tlie last but one is to the last in the second rank; and the second to the third of the first rank, as the last but two is to the last but one of the second rank, and so on ; and it is inferred, as in the preceding defini- tion, that the first is to the last of the first rank, as the first to the last of the other rank. This is proved in the fourteenth proposition of this book. BOOK IV.] EUCLID AND LEGENDRE. 97 AXIOMS. 1. Like multiples of the same, or of equal magnitudes, are equal to one another. 2. Those magnitudes of which the same, or equal magni- tudes, are like multiples, are equal to one another. 3. A multiple of a greater magnitude is greater than the same multiple of a less. 4. That magnitude of which a multiple is greater than the same multiple of another, is greater than that other magnitude. EXPLANATION OF SIGNS. 1. The product arising from multiplying one number by an- other is expressed by writing the letters representing them, one after the other, without any sign between them ; and some- times by placing between them a point, or the sign x. 2. A product is called a power ^ when tlie factors are all the same. Thus, AA, or as it is generally written, A^, is called the second power, or the square of A ; AAA, or A', the third power, or cube of A; AAA A, or K\ its fourth power, etc. In relation to these powers, A is called their root. Thus, A is the second or square root of A'^, the third or cube root of A% the fourth root of A^, etc. In like manner, the second or square root of A is a number Avhich, when multiplied by itself, pro- duces A; the th rd or cube ro ^ of A is such a number, that if it be multiplied by itself, and the product by the same root again, the final product will be A. The square root of A is denoted by -/A or A*, its cube root by -^A, or A*, its fourth root by Ai, etc. 3. The quotient arising from dividing one number by an- other is denoted by writing the dividend as the numerator of a fraction, and the divisor as its denominator. 4. The signs =, =0, >, <, signify respectively equal to^ equivalent to, greater than, less than. PROPOSITIONS. Prop. I. — Treor, — If there be four numbers such that the quotients obtained by dividing the first by the second, and the 1 98 THE ELEMENTS OF [bOOK IV. third hy the fourth, are equal ; the first has to the second the same ratio that the third has to the fourth. Let A, B, C, D be four magnitudes, such that o— fj > ^^^^ A: B:: C :D. For, let m and n be any whole numbers, and multiply the AC fractions — and ^ by m, and divide the product by n y then — =- = — ^-. Now, if mA be srreater than riB, mC will also be nB riD greater than nD ; for, if this were not so, — rr would not be equal to -y:- In like manner it might be shown, that if mA be equal to 7iB, mC will be equal to iiQ ; and that if raK be less than nB, rriG will be less than wD. But wA, raQ are any like multiples whatever of A, C ; and ?iB, 7^D any whatever of B, D ; and therefore (IV. def 5) A : B : : C : D. Therefore, if there be four numbers, etc. AC 1 Scho. This proposition is the same, when ^ or Y\—P ^^ ^iP being a whole number. A C Cor, If AD=BC, by dividing by B and D, we get ^=t^> and therefore, by this proposition, A : B : : C : D. Hence, if the product of two numbers be equal to that of two others, the one pair may be taken as the extremes and the other as the means of an analogy. Prop. II. — Theor. — If any four numbers he proportional^ and if the first be divided by the second, and the third by the fourth, the quotients are equal. A C Let A : B : : C : D ; then ^= jt If A and B be whole numbers, let the first and third terms be multiplied by B, and the second and fourth by A, and the products are AB, AB, BC, AD. Now, since the first and sec- ond of these are the same, the third and fourth are (IV. def. 6) equal; that is, AD=BC; and by dividing these by B and D, A C we find (IV. ax, 2) =t = ^. BOOK IV.] EUCLID AND LEGENDEE. 99 If A and B be fractions, let A=— , and B=-, so that niA— ' in n E, and nB=F; the numerators and denominators E, F, m, n E F beinii whole numbers. Then (hyp.) — : - : : C : D. Multiply the first and third of these by mF, and the second and fourth l)y wE, and the products are EF, EF, mYQ, and nED. Now, the first and second of these being the same, the third and fourth (IV. def 5) are equal ; that is, nED=mFC, or mnP^ — wiftBC, since E=imA, and F=wB. Hence, by dividing these by m and w, we get (IV. ax. 2) ADi=BC ; and the rest of the proof is the same as in the first case. Therefore, if any four numbers, etc. Scho. 1. If either A or B be a whole number, the proof is in- cluded in the second part of the demonstration given above. Thus, if A be a whole number, we have simply E = A and m=: 1, and everything will proceed as above. The proof would also be readily obtained by substituting for B as before, but retain- inc: A unchanged. If A and B be incommensurable, such as the numbers ex- pressing the lengths of the diagonal and side of a square, the lengths of the diameter and circumference of a circle, etc., their ratios may be approximated as nearly as we please. Thus the diagonal of a square is to its side, as |f : 1, nearly ; as \%\ : 1, more nearly; as \%\\ : 1, still more nearly, etc. Hence, in such cases we can have no hesitation in admitting the truth of the proposition, as we see that it holds with respect to numbers the ratio of which differs from that of the proposed numbers by a quantity which may be rendered as small as we please — smaller, in fact, than anything that can be assigned. A 1 Scho. 2. This proposition is the same, when T>=i? or -, p being a whole number. Scho. 3. From this proposition and the foregoing, it appears, that if two fractions be equal, the numerator of the one is to its denominator as the numerator of the other to its denominator ; and that if the first and second of four proportional numbers be made the numerator and denominator of one fraction, and the third and fourth those of another, the two fractions are equal. 100 THE ELEMENTS OF [bOOK TV. This is the same in substance as that the two expressions, A : A C B : : C : D, and |T=t-o are equivalent, and may be used for one another. Cor. 1. It appears in the demonstration of this proposition, that AD = BC; that is, if four numbers be pT'oportionals, the product of the extremes is equal to the product of the means. Hence, if the product of the means be di\ided by one of the ex- tremes, the quotient is the other; and thus we have a proof of the ordinary ai'ithmetical rule for finding a fourth proportional to three uiven numV)ers. Cor. 2. It is evident, that if A be greater than B, C must be greater than D ; if equal, equal ; and if less, less ; as otherwise =5" and =^ could not be equal. A C Cor. 3. If A : B : : C : D, and consequently —=y^, by multi- plying these fractions by—, we get — 7^= — :Fr, or mA : wB : : mC : nT>. A Cor. 4. If A be greater than B, the fraction — is evidently T> C C greater than -^, and the fraction -^ less than ry ; that is, of two unequal numbers, the greater has a greater ratio to a third than the less has ; and a thii'd number has a greater ratio to the less than it has to the greater. A B Cor. 5. Conversely, if -^ be greater than -r^, A is greater than — be less than =, A B' B ; and, if — be less than z^, A is also greater than B. Prop. III. — Theor. — If four numbers he proportionals, they are proportionals also when taken inversely. If A : B : : C : D ; then, inversely, B : A : : D : C. For (IV. 2, cor. 1) BC = AD ; and hence by dividing by A and C, we obtain -r=p5 or (I^- 2, scho. 2) B : A : : D : C. Therefore, if four numbers, etc. / BOOK IV,] EUCLID AND LEGENDRE. 101 Prop. IY. — Theor. — If four numbers be proportionals, they are also proportionals when taken alternately. If A : B : : C : D ; tlien, alternately, A : C : : B : D. For (IV, 2, cor. 1) AD=;:BC ; whence, by dividing by C and A B D we get 7s =yt; or (IV. 2, scho. 2) A : C : : B : D, There- fore, if four numbers, etc, Scho. When the first and second terms are not of the same kind as the third and fourth, the terms can not be taken altern- ately, as ratios would thus be instituted between heterogene- ous macrnitudes. o Prop, V. — Theor. — Ejual numbers have the same ratio to the same number ^ and the same has the same ratio to equal numbers. Let A and B be equal numbers, and C a third ; then A : C : : B : C, and C : A : : C : B. A B For, A and B being equal, the fractions -^ and y, are also equal, or, which is the same, A : C : : B : C ; and, by inversion, (IV. 3) C : A : : C : B, Therefore, equal numbers, etc. Prop. VI. — Theor. — JVmnbers which have the same ratio to the same nwnber are equal ; and those to which the same has the same ratio are equal. If A : C : : B : C, or if C : A : : C : B, A is equal to B. A B For, since —7=., ,by multiplying by C we get A = B. The proof of the second part is the same as this, since, by in- version (IV. 3), the second analogy becomes the same as the first. Therefore, numbers, etc. Prop. VII. — Theor. — Ratios that are equal to the sameratio are equal to one another. If A : B : : C : D, and E : F : : C : D ; then A : B : : E : F. -. . AC , E C . „ ,^ . A E . ^ ror, smce p=-r., and = - ,therciore (1. ax. 1) p=p, ; that , is, (IV. 2, scho. 2) A : B : : E : F. Therefore, ratios, etc. 102 THE ELEMENTS OF [bOOK 17. Pkop. Vin. — Theok. — Ofnumhers which are proportionals^ as any one of the antecedents is to its consequent^ so are all the antecedents taken together to all the conseqzients. If A : B : : C : D : : E : F ; then A : B : : A+C+E : B+D +F. ACE Since :^==: ==-7, put each fraction equal to q, and multiply hj the denominators ; then A=zBq, C=:Dq, and E^Fg-. Hence, by addition, A+C-}-E = (B + D-f F)^'/ and by dividing by B+D+F, we get ^=WTWZw' ^^^ ?— g 5 ^^^ there- foi'e 4"= t"!"^tS > or A : B : : A+C + E : B+D + F. There- a ±>+jj+i^ fore, etc. Prop. IX. — Theor. — 3fagmtudes have the same ratio to one another that their like multiples have. Let A and B be two magnitudes ; then, n being a whole number, A : B : : nA : nB. For :jj=— ^, or A : B : : nA : wB. Therefore, magnitudes, etc. Prop. X. — ^Theor. — If four mimbers be prop>ortionals ; then (1) hy composition, the sum of the first and seco7\d is to the second, as the sum of the third and fourth to the fourth ; and (2), 5y division, the excess of the first above the second is to the second, as the excess of the third above the fourth is to the fourth. If A : B : : C : D ; then, by composition, A+B : B : : C+D: D; and by division, A— B : B : : C — D : D. 1. Since (hyp.) vi = i-:i and since -rr = ^: add the latter frac- ^ "^ ^ ' B D B D tions to the former, each to each, and there results — rr~ = 5^, or A+B : B : : C +D : D. 2. By subtracting the latter pair of the same fractions from the former, each from each, we obtain — =—- = —=-— ; or A — B i) B : B : : C— D : D. If, therefore, etc. BOOK rV.] EUCLID AND LEGENDRE. 103 Cor. By dividing the fractions which were found above by- addition, by those which were found by subtraction, we get :^^=^^; or (IV. 2, scho. 2) A+B : A-B : : C+D : C- D; that is, if four numbers be proportional, the sura of the fii-st and second terms is to their difference, as the sum of the third and fourth terms is to their difference. It is evident, that if B be greater than A, the analogy would become B+A : B— A : : D + C : D— C. Prop. XI. — Theoe. — If foumumhers he proportional ; then,, by conversion, the first is to its excess abuve the second^ as the third to its excess above the fourth. If A : B : : C : D ; then, by conversion, A : A — B : : C : C— D. x^ • n ;,- .BD -. AC. Jbor, smce (hyp. and mver.) —=^^ and smce — = -; take the former fractions from the latter, each from each, and there remains — -r — = — ^ — , or (by inver.) A:A — B:: C:C — D. Therefore, if four numbers, etc. Prop. XII. — Theor. — If there be members forming two or more analogies which have common consequents^ the sum of all the first antecedents is to their common consequent, as the sum of all the other antecedents is to their common consequent. If A : B : : C : D, and E : B : : F : D ; then A + E : .B : : C+F : D. For (hyp.) -j5=— ,and p = Tx; and hence, by addition, — -^— =5^, or A+E : B : : C + F : D. If, therefore, etc. Prop. XIII. — Theor. — If there be three or more niiyn^ers, and OS many others, which, taken two and two in order, have the sa7ne ratio ; then, ex aequo, the first has to the last of the first rank the sayne ratio that the first has to the last of the second rank. If the two ranks of numbers. A, B, C, D, and E, F, G, II, be 104 THE ELKMENT8 OF [bOOK IV. Buch that A : B : : E : F, B : C : : F : G, and C : D : : G : 11 ; then A : D : : E : H. A E B F C C For, since (hyp.) -- =-, - = - and ^ = - ; by multiplying together the first, third, and fifth fractions, and the second, fourth, and sixth, we obtain jTnn^KnxT 5 ^r, by dividing the terras of the first of these fractions by BC, and those of the A E second by FG, j)=g, or A : D : : E : II. Therefore, if there be three, etc. This proposition might also be enunciated tlius : If there be numbers forming two or more analogies, such that the conse- quents in each are the antecedents in the one immediately fol- lowing it, an analogy will be obtained by taking the antece- dents of the first analogy and the consequents to the last for its antecedents and consequents. Prop. XIV. — Theok. — If there be three or more numbers, and as many others, which, taken two and two in a cross order, have the same ratio ; then, ex tequo inversely, the first has to the last of the first rank the same ratio uhlch the first has to the last of the second ranTc. If the two ranks of numbers. A, B, C, D, and E, F, G, H,be such that A : B : : G : H, B : C : : F : G, and C : D : : E : F ; then, ex mqrco inversely, A : D : : E : H. A C B F P F For, since (hyp.) -^=^, ^=^,and ^ =^, by multiplying to- gether the fractions as in the preceding proposition, we get ABC GFE , ,,..,., . , . ^ ^pY)=TT7TT^j whence, by dividing the terms of the first of these fractions by BC, and those of the second by GF, we ob- A E tain jx = fT, or A : D : : E : H. If, therefore, etc. This proposition may also be enunciated thus: If there be numbers forming two or more analogies, such that the means of each are the extremes of the one immediatelj' following it, another analogy may be obtained by taking the extremes of the first analogy and the means of the last for its extremes and means. BOOK IV.] EUCLID AND LEGENDKE. 105 Prop. XV. — Theor. — If there he numbers forming tico or more analogies^ the products of their corresponding terms are proportionals. If A : B : : C : D, E : F : : G : II, and K : L : : M : N ; then AEK : BFL : : CGM : DUN. ^ ,, - A C E G TC M ^ . . For (hyp.) j^ =0, ^=|j>' aucl jy = ^; and taking the pro- ducts of the corresponduig terms of these fractions, we obtain A FTC OC \f ^^=^^., or AEK : BFL :: CGM : DHN. Therefore, if Bi'L DHIS' there be numbers, etc. Cor. 1. Hence, if there be two analoj^ies consistinsr of the sane terms, A, B, C, D, we have A" : B' : : C^ ; D^ ; if there be three, we have A' : B' : : C* : D', etc. ; and it thus appears, that like powers of proportional numbers are themselves pro- portional. Cor. 2. Like roots of proportional numbers are proportional. Thus, if A : B : : C : D, let 4/ A : VB : : VC : VE. Then, by the preceding corollary, A : B : : C : E. But (hyp.) A : B : : C : D ; and therefore (IV. 7) C : E : : C : D, and (IV. 6) E = D, and consequently V : A -/ B : : VC : VE, or VD. Prob. XVI. — Theor. — The sum of the greatest and least of four p. oportional members is greater than the sum of the other two. If A : B : : C : D, and if A be the greatest, and therefore (IV. 2, cor. 2) D the least ; A and D are together greater than B and C. For (by conversion) A : A — B :: C : C — D, and, altern- ately, A : C : : A— B : C— D. But (hyp.) A>C, and there- fore' (IV. 2, cor. 2) A— B > C — D. To each of these add B ; then A>B + C— D. Add again, D; then, A-hD>B4-C. Therefore, etc. Cor. Hence the mean of three proportional numbers is less than half the sum of the extremes. Prob. XVII. — Theor. — In numbers which are continual proportiojials, the first is to the third as the second power of the first to the second power of the second y the first to the 106 THE ELEMENTS OF [bOOK IV. fourth as the third power of the first to the third power of the second ; the first to the fifth as the fourth power of the first to the fourth power of the second ; and so on. 1 If A, Vy, C, D, E, etc., be continual proportionals; A : C : : A^ B^ ; A : D : : A^ : B'; A : E : : A^ : B', etc. For, since (IV. def. 8) A : B : : B : C, and since A : B : : A : B, we have (IV. 15) A^ : B^ : : AB : BC, or, dividing the third and fourth terms by B, A* : B' : : A : C. Again : since A* : B' : : A : C, and A : B : : C : D we have (IV. 15) A' : B' : : AC : CD, or dividing the third and fourth terms by C, A^ : B^ : : A : D ; and so on, as far as we please. There- foi'e, etc. Cor. Hence (IV. defs. 11 and 12) the ratio which is duplicate of that of any two numbers, is the same as the ratio of their squares ; that which is triplicate of their ratio, the same as the ratio of their cubes, etc. Prop, XVIII. — ^Theor. — A ratio xcMch is compounded of other ratios, is the same as the ratio of the products of their homologous terms. Let the ratio of A to D be compounded of the ratios of A to B, B to C, and C to D ; the ratio of A to D is tlie same as that of ABC, the product of the antecedents, to BCD, the product of the consequents. For, since A : D : : A : D, multiply the terms of the sec- ond ratio by BC ; then (IV. 9) A : D : : ABC : BCD. There- fore, etc. Prop. XIX. — Theor. — In numbers which are continual pro2)ortionalsj the difference of the first and second is to the first, as the difference of the first and last is to the sum of all t/ie terms excejjt the last. If A, B, C, D, E be continual proportionals, A — B : A : : A— E: A + B + C + D. For, since (hyp.) A : B : : B : C : : C : D : : D : E, we have (IV. 8) A : B (conv.) A : A — B : A— B : A :: A— E A+B+C+D : B + C + D + E. Hence A + B + C + D : A— E; and (inver.) A+B + C + I). BOOK IV.] EUCLID AND LEGENDRE. 107 It is evident that if A were the least term, and E the great- est, we should get in a similar manner, B^ — A : A : : E — A : A-f-B-f-C+D. Therefore, in numbers, etc. Cot. If the series be an infinite decreasing one, the last term will vanish, and if S be put to denote the sum of the series, the analogy will become A — B : A : : A : S ; and this, if rA be put instead of B, and the first and second terms be divided by A, will be changed into 1 — r : 1 : : A : S. The number r is called the common ratio, or common m,ultiplier, of the series, as by multiplying any term by it, the succeeding one is ob- tained. END OP BOOK FOURTH. BOOK FIFTH. DEFINITIONS. 1. Similar rectilineal figures are those which have their several angles equal, each to each, and the sides about the equal angles proportionals. 2. Two magnitudes are said to be reciprocally proportional to two others, when one of the first pair is to one of the second, as the remainincc one of the second is to the remaining one of the first. 3. A straight line is said to be cut in extreme andmean ratio^ ■when the whole is to one of the segments as that segment is to the other. 4. The altitude of any figure is the straight line drawn from its vertex perpendicular to its base. 5. A sti-aight line is said to be cut harmonically, Avhen it is divided into three segments, such that the whole line is to one of the extreme segments as the other extreme segment is to the middle one. PROPOSITIONS. Prop. I. — Theor. — Triangles and parallelograms of the same altitude are one to another as their bases. ^ Let the triangles ABC, ACD, and the parallelograms EC, CF have the same altitude, viz., the perpendicular drawn from the point A to BD ; then, as the base BC is to the base CD, so is the triangle ARC to the triangle ACD, and the parallelo- gram EC to the parallelogram CF. Produce BD both ways, and take any number of straight lines BG, Gil, each equal to BC ; and any number DK, KL, each equal to CD; and join AG, AH, AK, AL. Then, because CB, BG, Gil are all equal, the triangles ABC, AGB, AUG are (L 15, cor.) all equal. Therefore, whatever multiple the base BOOK v.] EUCLID AND LEGENDRE. loa HC is of BC, the same multiple is the triangle AUG of ABC. For the same reason, wliatver multiple LC is of CD, the same multiple is the triangle ALC of ADC. Also, if the'base HC - be equal to CL, the triangle AHC is equal (I, 15, cor.) to ALC ; and if the base HC be greater than CL, likewise (L 15, cor. 6) the ti-iangle AHC is greater than ALC; and if less, less. Therefore, since there are four magnitudes, viz., the two bases, BC, CD, and the two tri- angles ABC, ACD; and of the base BC, and the triangle ABC, the first and third, any like multiples whatever have been taken, viz., the base HC, and the triangle AHC ; and of the base CD, and the triangle ACD, the second and fourth, have been taken any like multiples whatever, viz., the base CL, and the triangle ALC ; and that it has been shown that, if the base HC be greater than CL, the triangle AHC is greater than ALC ; if equal, equal; and if less, less; therefore (IV. def. 5) as the base BC is to the base CD, so is the triangle ABC to the trian- gle ACD. Again : because (L 15, cor.) the parallelogram CE is double of the triangle ABC, and the parallelooi-ani CP"" of the triangle ACD, and that (IV. 9) magnitudes have the same ratio which their like multiples have ; as the triangle ABC is to the trian- gle ACD, so is the parallelogram EC to the parallelogram CF. But it has been shovvn, that BC is to CD, as the triangle ABC to the triangle ACD ; and as the tiiansjfle ABC is to the trian- gle ACD, so is the parallelogram EC to the parallelogram CF; therefore (TV. 7) as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF. Wheiefore, trian- gles, etc. Scho. This proposition may be briefly demonstiated thus: Let a perpendicular drawn from A to BD be called P. Then, ■J^P.BC will be equivalent to the area of the triangle ABC, and Ap.CD that of ACD. Dividing, therefore, the former of these , t, , 1 il'-KC BC ABC ,„, „ equals by the latter, we get yp-^ or, qy)~A<Jd' °^" ^ ' scbo. 2) BC : CD : : ABC : ACD. In extending this method 110 THE ELEMENTS OF [BOOK V. of proof to the parallelograms, we have merely to hbc P instead ofiP. Cor. 1. From this it is plain, that triangles and parallelo- grams which have equal altitudes, are one to another as their bases. Let the figures be placed so as to have their bases in the same straight line ; and perpendiculars being drawn from the vertices of the triangles to the bases, the straight line Avhich joins the vertices is parallel (I, 15, cor.) to that in Avhich their bases are, because the perpendiculars ai-e both equal and paral- lel to one another. Then, if the same construction be made as in the proposition, the demonstration will be the same. Cor. 2. Hence, if A, B, C be any three straight lines, we have A : B : : A.C : B.C. Cor. 3. So, likewise, if the straight lines A, B, C, D be pro- portional, and E and F be any other straight lines, we shall have, according to the preceding corollary, and the seventh proposition of the fourth book, A.E : BE : : C.F ; D.F. Prop. II. — Theor. — If a straight line he parallel to the base of a triangle, it cuts the other sides, or those produced, propor- tionally, and the segments between the parallel and the base are homol'jgous to one another ; and (2) if the sides of a triangle, or the sides produced, be cut proportionally, so that the seg- ments between the points of section and the base are homologous to one another, the straight line which jo. ns the points of sec- tion is parallel to the base. The enunciation of this proposition which is given by Dr. Simson and others, is defective, and might lead to error in its application, as it does not point out what lines are homolo- gous to one another in the analogies. It is plain that, instead of one proposition, this is in reality two, which are converses of one another. 1. Let DE be parallel to BC, one of the sides of the triangle ABC ; BD : DA : : CE : EA. Join BE, CD. Then (L 15, cor.) the triangles BDE, CDE are equivalent, because they are on the same base DE, and be- tween the same parallels. DE, BC. Now ADE is another tri- BOOK v.] EUCLID AND LEGENDRK 111 angle, and (IV. 5) equal magnitudes have to the same the same ratio; therefore, as the triangle BDE to ADE, so is the triangle ^ ^ » CDE to ADE. But, (V. ]) as the triangle BDE to ADE, so is BD to DA ; Because, having the same alti- tude, viz., the perpendicular drawn from E to AB, they are to one another as their bases; ^ c B O and for the same reason, as the triangle CDE to ADE, so is CE to EA. Therefore (IV. 1) as BD : DA : : CE : E A. 2. Next, let the sides AB, AC of the triangle ABC, or those produced, be cut proportionally in the points D, E ; that is, so that BD : DA : : CE : EA, and join DE ; DE is parallel to BC. The same construction being made, because (hyp.) as BD : DA : : CE : EA ; and (V. 1) as BD to DA, so is the triangle BDE to the triangle ADE; and as CE to EA, so is the triangle CDE to ADE ; therefore (IV. 7) the triangle BDE is to ADE, as the triangle CDE to ADE ; that is, the triangles BDE, CDE have the same ratio to ADE ; and therefore (IV. 6) the triangles BDE, CDE are equal ; and they are on the same base DE, and on the same side of it ; therefore (I. 15, cor.) DE is parallel to BC. Wherefore, if a straight line, etc. Cor. The triangles which two intersecting straight lines form with two parallel ones, have their sides which are on the intersecting lines proportional ; and those sides are homologous which are in the same straight line; and (2), conversely, if two straight lines form with two intersecting ones triangles which have their sides that are on the intersecting lines proportional, the sides which are in the same Rtrai<irht line with one another being homologous, those straight lines ai-e parallel. 1. Let DE and BC (first and second figures) be the parallels, and let them be cut by the straight lines BD, CE, which inter- sect each other in A; then BA : AC : : DA : AE. For, since BD : DA : : CE : EA, we have, by composition in the first fig- 112 THE ELEMENTS OF [BOOK V. tire, and by division in the second, BA : DA : : CA : EA, and, alternately, BA : AC : : DA : AE. 2. But if BA : AC : : DA : AE, DE and BC are parallel. For, alternately, BA : DA :: CA : EA ; then, in the first figure by division, and in the second by composition, we have BD : DA : : CE : EA ; and therefore, by the second part of this proposition, DE is pai-allel to BC. Prop. III. — TheoPw — Tlie sides about the equal aiigles of equiangular triangles are proportionals ; and those ichich are ojyposite to the equal angles are homologous sides^ that is, are the antecedents or consequents of the nitios. Let ABC, DCE be equiangidar triangles, having the angle ABC equal to DCE, and ACB to DEC, and consequently (I. 20, cor. 5) BAC equal to CDE ; the sides about the equal angles are proportionals ; and those are the homologous sides which are opposite to the equal angles. Let the triangles be placed on the same side of a straight line BE, so that sides BC, CE, which are opposite to equal ano-les, may be in that straiu^ht line and contii^nous to one an- other; and so that neither the equal angles ABC, DCE, nor ACB, DEC at the extremities of those sides may be adjacent. Then, because (I. 20) the angles ABC, ACB are together less than two right angles, ABC and DEC, which (hyp.) is equal to ACB, are also less than two right angles ; wherefore (I. 19) BA, ED will meet, if produced ; let them be produced and meet in F. Again : because the angle ABC is equal to DCE, BF is parallel (L 16, cor.) to CD ; and, because the angle ACB is equal to DEC, AC is par- allel to FE. Therefore, FACD is (I. def 15) a parallelogram; and consequently (L 15, cor.) AF is equal to CD, and AC to FD. Now (V. 2) because AC is parallel to FE, one of the sides of the triangle FBE, BA : AF : : BC : CE. But AF is equal to CD ; therefore (IV. 5) as BA : CD : : BC : CE, BOOK v.] EUCLID AND LI GKNDRE. 113 and alternately (IV. 4) as AB : BC : : DC : CE. Ai^ain : (V. 2) because CD is parallel to 15F, as BC : CE : : FD : DE; but FD is equal to AC; therefore, as BC : CE : : AC : DE ; and, alternately, as BC : CA : : CE : ED. Therefore, because it has been proved that AB : BC : : DC : CE, and as BC : CA : : CE : ED; ex mquo (IV. 13), BA : AC : : CD : DE. Therefore, the Bides, etc. Scho. 1. Hence (V, def l) equianirnlar triangles are similar. Cor. If two angles of one triangle be respectively equal to two angles of another, their sides are proportional, and the sides opposite to equal angles are homologous. For (I. 20, cor. 5) the remaining angles are equal, and therefore the triangles are equiangular. Scho. 2. In a similar manner we may produce a given straight line, so that the whole line so produced may have to the part produced the ratio of two given straight lines. Thus, if BA be the line to be produced, make at B an angle of any magnitude, and take BE and CE equal to the other given lines; join AC, and draw FE parallel to it. Then, since FE is paral- lel to AC, a side of the triangle ABC, we have (V. 2) BF : AF : : BE : CE, so that BF has to AF the given ratio. Prop. IV. — Theor. — The straight line which bisects an aru- gle of a triangle, divides the opposite side into segments ichich have the same ratio to one another as the adjacent sides of the triangle have ; and (2) if the segm.ents of the base have the same ratio as the adjacent sides, the straight line draicn from the vertex to the point of section, bisects the vertical angle. 1. Let the angle BAC of the triangle ABC be bisected by the straight line AD ; then BD : DC : : BA : AC. Through C draw (I. 18) CE par- allel to DA; then (I. 16, cor. l) BA produced will meet CE; let them meet in E. Because AC meets the parallels AD, EC, the g D c angle ACE is equal (I. 16) to the alternate angle CAD ; and because BAE meets the same paral- 8 114 THE ELEMENTS OF [book lels, the angle E is equal (I. ] 6, part 2) to BAD ; therefore (I. ax. 1) the angles ACE, AEC are equal, because they are re- spectively equal to the equal angles, DAC, DAB; and conse- quently AE is equal (I. 1, cor.) to AC. Now (V. 2) because AD is parallel to EC, one of the sides of the triangle BCE, BD : DC : : BA : AE ; but AE is equal to AC; therefore (IV. 5) BD : DC : : BA : AC. 2. Let now BD : DC : : BA : AC, and join AD ; the angle BAC is bisected by AD. The same construction being made, because (hyp.) BD : DC : : BA : AC; and (V. 2) BD : DC : : BA : AE, since AD is parallel to EC ; therefore (IV. 1) BA : AC : : BA : AE; consequently (IV. 6) AC is equal to AE ; and (I. 1) the angles AEC, ACE are therefore equal. But (I. 16) the angle BAD is equal to E, and DAC to ACE ; wherefore, also, BAD is equal (I. ax. 1) to DAC; and therefore the angle BAC is bisected by AD. The straight line, therefore, etc. And if an exterior angle of a triangle be bisected by a straight line which also cuts the base produced, the segments between the bisecting line and the extremities of the base have the same ratio to one another as the other sides of the triangle have ; and (2) if the segments of the base produced have the same ratio which the other sides of the triangle have, the straight line drawn fiom the vertex to the point of section bisects the exterior angle of the triangle. 1. Let an exterior angle CAE of any triangle ABC be bisected by AD which meets the opposite side produced in D ; then BD: DC:: BA : AC. Through C draw (T. 18) CF parallel to AD; and because AC meets the parallels AD, FC, the angle ACF is equal (I. 16) to CAD ; and because the straight line FAE meets the parallels AD, FC, the angle CFA is equal to DAE ; therefore, also, ACF", CFA are (I. ax. 1 ) equal to one another, because they are respect- ively equal to the equal angles DAC, DAE ; and consequently (I. cor.) AF is equal to AC. Then (V. 2) because AD is parallel to FC, a side BOOK v.] EUCLID AND LEGENDEE. 115 of the triangle BCF, BD : DC : : BA : AF ; but AF is equal to AC ; as therefore BD : DC : : BA : AC. 2. Let now BD : DC : : BA : AC, and join AD ; the angle CAD is equal to DAE. The same construction being made, because BD : DC : : BA : AC; and that (V. 2) BD : DC : : BA : AF; therefore (IV. V) BA : AC : : BA : AF ; wherefore (IV. 6) AC is equal to AF, and (I. 1) the angle AFC to ACF. But (I. 16) the angle AFC is equal to EAD, and ACF to CAD ; therefore, also (I. ax. 1), EAD is equal to CAD. Wherefore, etc. Cor. If G be the point in which BC is cut by the straight line bisecting the angle BAC, we have (V. 4) BG : GC : : BA : AC ; and by this proposition, BD : DC : : BA : AC ; whence (IV. 7) BD : DC :: BG : GC, and therefore (V. def 5) BD is divided harmonically in G and C. 8cho. If the triangles be isosceles, the line bisecting the ex- terior angle at the vertex is parallel to the base. In this case, the segments may be regarded as infinite, and therefore equal, their difference, the base, being infinitely small in comparison of them. Prop. V. — Theor. — If the sides of two triangles^ about each of their angles^ he proportionals, the triangles are equiangular, and have their equal angles opposite to the homologous sides. Let the triangles ABC, DEF have their sides proportionals, that AB : BC : I DE : EF ; and BG : CA : : EF : FD ; and consequently, ex aequo, BA : AC : : ED : DF; the triangles are equiangular, and the equal angles are opposite to the homolo- gous sides, viz., the angle ABC equal to DEF, BCA to EFD, and BAC to EDF. At the points E, F, in the straight line EF, make (L 13) the angle FEG equal to B, and EFG equal to C. Then (V. 3, cor.) the triangles ABC, GEF have their sides opposite to the equal angles proportionals ; wherefore, AB:BC::GF:EF; but (hyp.) AB : BC : : DE ; EF. 116 THE ELEMENTS OF [bOOK V. Therefore (TV. 7) DE : EF : : GF : EF ; whence, since DE and GF have the same ratio to EF, they are (IV. 6) equal. It may be shown in a similar manner that DF is equal to EG; ani because, in the triangles DEF, GEF, DE is equal to FG, EF common, and DF equal to GE ; therefore (I. 4) the angle DEF is equal to GFE, DFE to GEF, and EDF to EGF. Then, because the angle DEF is equal to GFE, and (const.) GFE to ABC ; therefore the angle ABC is equal to DEF. For the same reason, ACB is equal to DFE, and A to D. Therefore the ti'iangles ABC, DEF are equiangular. Wherefore, if the sides, etc. Prop. VI.— Theor. — If two triangles have one angle of the one equal to one angle of the other^ and the sides about the equal angles proportionals ; the remaining angles are equal^ each to each^ viz.y those which are opposite to the homologous sides. Let the triangles ABC, GEF, of the previous diagrams, have the angles ABC, EFG equal, and the sides about those angles proportionals ; that is, BA : BC : : GF : EF ; the angle BAG is equal to EGF, and ACB to EFG. Make (I. 13) the angle FED equal to either of the angles ABC, GFE ; and the angle EFD equal to ACB. Then (V. 3, cor.), BA : BC : : DE : EF. But (hyp.) BA : BC : : GF : EF ; and therefore (IV. 7) GE : EF : : DE : EF; wherefore ED is equal (IV. 6) to FG. Now EF is common to the two trian- gles GEF, DEF; and the angle GFE is equal (const.) to DEF ; therefore the angle EFD is equal (I. 3) to FEG, and D to G. But (const.) the angle EFD is equal to ACB ; there- fore ACB is equal to FEG ; and (hyp.) the angle ABC is equal to GEF ; wherefore, also (I. 20, cor 5), the remaining angles A and G are equal. Therefore, if two triangles, etc. Prop. VII. — Theor. — If two triangles have two sides of the one proportional to tioo sides of the other, and if the angles opposite to one pair of the homologous sides be equal, and those opposite to the other pair be either both acute, or not BOOK v.] EUCLID AND LEGENDRE. lit G acute, the angles contained by the proportional sides are equal. Let ABC and EFG be two triangles which have tlie sides CB, CA proportional to GF, GE ; the angles CBA, GFE equal, and the ether angles CAB, GEF acute ; then the angles ACB, EGF, contained by the proportional sides, are equal. If the triangle ABC be applied to EFG, so that^ CB will foil on GF, and the vertex C on the vertex G, and make GP equal to CB ; then, because the angle CBA is equal to the angle GFE, AB will take the direction OP parallel to EF (I. 16). Since OP is parallel to EF, GOP is equal to GEF (I. 16, cor.), and we have GF : GP :: GE : GO (V. 3) ; but by hypothesis GF : CB : : GE : CA, and GP is equal to CB. Hence, CA is equal to GO ; therefore OP, which joins the extremities of GP and GO, is equal to AB, which joins the extremities of CB and CA (I. ax. 1), and the triangles ABC, OPG are equal. Hence, the angles GOP, CAB are equal (I. 16, cor, 1), but the angle GOP is equal to the angle GEF (I. 16, cor. 1); consequently (I. 14), the re- maining angle of GOP is equal to the remaining angle of EFG, and the angles ACB, EGF are equal. In the same manner it can be shown that the angles ACB, EGF are equal when CAB, GEF are not acute. Wherefore, if two triangles have, etc. Prop. VIII. — Theor. — Tn a right-angled triangle, if a per- pendicular be drawn from the right angle to the hypothenuse, the triangles on each side of it are simdar to the whole triangle, and to one another. ' Let ABC be a right-angled triangle, having the riijht angle BAC; and from the point A let AD be drawn perpendicular to the hypothenuse BC ; the triangles ADB, ADC are sitnilai- to the whole triangle ABC, and to one another. Because the angle BAC is equal (I. ax. 11) to ADB, each of 118 THE ELEMENTS OF [bOOK V. them beiiiff a right anejle, and that the ansjle B is common to the two triangles ABC, ABD ; the remaining angle C is equal (I. 20, cor. 5) to the remaining angle BAD. Therefore the triangles ABC, ABD are equiangular, and (V. 3) the sides about their equal angles are proportionals ; wherefore (V. def, l) the triancfles are similar. In the same manner it mioht be demon- strated, that the triangle ADC is equiangular and similar to ABC ; and the triangles ADB, ADC, being each equiangular to ABC, are (I. ax. 1) equiangular, and therefore (V. 3 and def. 1) similar to each other. Therefore, etc. Cor. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the hypothe- nuse, is a mean proportional (IV. def 9) between the segments of the hypothenuse; and also that each of the sides is a mean proportional between the hypothenuse and its segment adjacent to that side. For (V. 3) in the triangles BDA, ADC, as BD : DA : : DA : DC ; in the triangles ABC, DBA, as BC : BA : : BA : BD ; and, in the triangles ABC, ACD, as BC : CA : : CA : CD. Scho. This proposition affords an easy way of solving the first corollary to the twenty-fourth proposition of the first book, as follows: Let ABC be a triangle, right-angled at A ; the square of the hypothenuse BC is equivalent to the squares of the legs AB, AC. Draw AD perpendicular to BC. Then (V. 8, cor.) BC : BA : : BA : BD, and BC : CA : : CA : CD. Hence (IV. 2, cor. 1) the rectangle BC.BD is equivalent to the square of AB, and the rectangle BC.CD to the square of AC. Hence (I. ax. 2) BC.BD + BC.CD, or (II. 2) BC==^AB-+ACl Prop. IX. — Puob. — To find a third proportioyial to two given straight lines. Let A and B be two given straight lines ; it is required to find a third proportional to them. EUCLID AND LEGENDKE. 119 BA BOOK v.] Take two straight lines CF, CG, containing any angle C ; and upon these make CD equal to A, and DF, CE each equal to B. Join DE, and (I. 18) draw P'G parallel to it. EG is the third proportional required. For (V. 2) since DE is parallel to FG, CD : DF : : CE : EG. But (const.) CD is equal to A, and DF, CE each equal to B; therefore A : B :: B : EG; wherefore to A and B the third proportional EG is found, which was to be done. aS'c'/^o, Other modes of solving this problem may sometimes be employed with advantage. The following are among the most useful : 1. Draw AD (fig. to prop. 8) perpendicular to the indefinite straight line BC, and make DB, DA equal to the given lines ; join AB, and draw AC perpendicular to it ; DC is the third proportional required. For (V. 8, cor.) BD : DA : : DA : DC. 2. Draw BC perpendicular to AB, and having made BAand AC equal to the less and greater of the given lines, draw CD and BE perpendicular to AC ; AD will be a third proportional to AB and AC, and AE to AC and AB. This follows from the third proposition of this book, since the triangles ABC, ACD are equiangular, as are also ABC, AEB. The angles ABC, ACD, etc., are hex-e made right angles. They may be of any magnitude, however, provided they be equal. It is sufficient, therefore, to draw two straight lines, AB, AC, making any angle ; to cut off AB, AC equal to the given ))roportionals ; and then, BC being joined, to make the angle ACD equal to ABC, and to draw BE parallel to CD; or to make the angle ABE equal to AL'B, and to draw CD paral- lel to BE. This method affords an easy means of continuing a vseries of lines in contiimal proportion, both ways, when any two succes- sive terms are given. Thus, after CD and 1)E are drawn, it is only necessary to draw DF, EG, etc., parallel to BC, and F'H, GK, etc., parallel to CD ; as AD, x\F, iVH, etc., will be the siic- G B D H 120 THE ELEMENTS OF [cook V. A C B ceedinpj terms of the ascending scries, and AE, AG, AK, etc., those of the desceiidiiify one. Prop. X. — Prob. — To find a fourth proportional to three given straight lines. Let A, B, C be three given straight lines; it is required to find a fourth proportional to them. Take two straight lines DE, DF, contain- ing any angle EDF, and make DG equal to ( A, GE equal to B, and DH equal to C ; and having joined GH, draw (I. 18) EF parallel to it through the point E; HF is the fourth proportional required. For (V. 2) since Gil is parallel to EF, as DG : GE : : DH : HF; but DG is equal to A, GK to B, and DH to C ; therefore as A : B : : C : HF; wherefore, to the three given straight lines. A, B, C, the fourth proportional HF is found ; which was to be done. ^cho. 1. The solution of this problem may also be effected in several different ways; some of which may be meiely indicated to the student, as the proofs present no difficulty ; and it is evi- dent that, with slight modification, they are applicable in solv- ing the ninth proposition, which is only a particular case of the tenth. 1. Let AE,EC (last fig. to HL 20) be the second and third terms, placed contiguous, and in the same straight line, and draw BE, making any angle with AC, and equal to the first term ; through the three jioints, A, B, C, describe a circle cut- ting BE produced in D; ED is the fourth proportional. If AB and CD be joined, the proof will be obtained by means of the triangles ABE, CDE, which are similar. 2. Draw AB, AC (fig. to IH. 21, cor.) making any angle, and make AB, AE equal to the second and third tei'uis; then if AC be taken equal to the first term, and a circle be described pass- ing thiough B, C, and E, and meeting AC in F, AF will be the required line. If BF and EC be joined, the triangles ABF and ACE are similar, and hence the proof is immediately ob- tained. BOOK v.] EUCLID AND LEGENDE"S. 121 3, Make BD and DA (fig. to V. 8) perpendicular to each otlicT, and equal to tlie first and second terms; join AB, and draw AC perpendicular to it; in DA, produced through A, if necessary, take a line equal to the third term, through the upper extremity of which draw a line parallel to AC ; the line intercepted on DC, produced if necessary, hetween D and tliis parallel, is the fourth proportional required. ^ Cor. 1. If four straight lines be proportionals, the rectangle contained by the extremes is equivalent to that conta,ined by the means; and (2) if the rectangle contained by the extremes be equivalent to that contained by the means, the four straight lines are proportionals (V. 2, cor.). iScko. 2. This corollary, of which the corollary immediately following is a case, affords the means of deriving the equality of rectangles, and the proportion of straight lines containing them, from one another. It evidently corresponds to IV. prop. 1, cor., and prop. 2, cor. 1, and it might be regarded as an immediate result of those corollaries, without any distinct proof, if the lines were expressed (T-. 2-3, cor. 4) by lineal units, and the rectangles by superficial ones. This corollary and the third proposition of this book, when employed in connection with one another, form one of the most powerful instruments in geometrical investigations, and they facilitate in a peculiar degree .the application of algebra to such inquiries. Cor. 2. If three straight lines be proportionals, the rectangle contained by the extremes is equivalent to the square of the mean ; and (2) if the rectangle contained by the extremes be equivalent to the square of the mean, the three straight lines are proportionals. Prop. XI. — Prob. — To find a mean proportional hetween two given straight lines. Let AB, BC be two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC as diameter describe the semicircle ADC ; fi-om B (T. 7) draw BD at right angles to AC ; BD is the mean proportional between AB and BC. Join AD, DC. Then, because the angle ADC in a semicircle 122 THE ELEMENTS OF [bOOK V. is (III. 11) a right angle, and because in the right-angled tri- angle ADC, DB is drawn from the right angle perpendicular to AC, DB is a mean proportional (V. 8, cor.) between AB, BC, the segments of the base. Therefore be- tween AB, BC, the mean proportional DB is found ; Avhich was to be done. Scho. Out of several additional ways of solving this prob- lem, the following may be mentioned : 1. On the greater extreme as diameter describe a semicircle ; from the diameter cut oif a segment equal to the less extreme, through the extremity of which draw a perpendicular cutting the circumference ; and the chord drawn from that intersection to the extremity of the diameter common to the two extremes is the required mean. The proof is manifest from III. 11, and V. 8, cor. 2. Make AD, DC (2d fig. to III. 21) equal to the given ex- tremes ; on AC as chord describe any circle, and a tangent drawn from D will be the required mean. The proof of this is obtained by joining AB and BC, as the triangles ADB, BDC are similar. When one mean is determined, others may be found between it and the given extremes, and thus three means will be insert- ed between the given lines ; and by finding means between each successive pair of the five terms of which the series then consists, the number of means will be increased to seven. By continuing the process Ave may find fifteen nieans, thirty-one means, or any number which is less by one than a power of 2. We cannot find, however, by elementary geometry, any other number of means, such as two, four, or five. Cor. A given straight line can be divided in extreme and mean ratio. Prop. XTI. — Theor. — J^quivalent pnrnllelograms inhich Imve an an.fjle of the one equal to an angle of the other, have their sides about those angles reciprocally i/roportioiHil ; atid (2) parallelograms lohich have an angle of the one equal to an an^ gle of the < titer, and the sides a,bout those angles reciprocally proportional., are equioale^it to one another. BOOK v.] EUCLID AND LEGENDKE. 123 1. Let AB, BC be equivalent parallelograms, which have the angles at B equal ; the sides about those angles are recip- rocally proportional; that is, DB : BE : : GB : BF. Let the sides DB, BE be placed in the same straight line, and contiTUOus, and let the parallelograms be on opposite sides of DE ; then (I. 10, cor.) because the angles at B are equal, FB, BG are in one straight line. Complete the parallelogram FE, and (IV. 5) because AB is equal to BC, and that FE is another parallelogram, AB: FE::BC:FE. But (V. 1) as AB to FE, so is the base DB to BE ; and as BC to FE, so is the base GB to BF ; therefore (IV. 1) as DB : BE : : GB : BF. The sides, therefore, of the parallelograms AB, BC, about their equal angles, are (V. def 2) reciprocally proportional. 2. But let the sides about the equal an- gles be reciprocally proportional, viz., DB : BE : : GB : BF ; the parallelograms AB, BC are equivalent. The same construction being made, because, as DB : BE : : GB : BF; and (V. 1) as DB : BE : : AB : FE ; and as GB : BF : : BC : FE ; therefore (IV. 1) as AB : FE : : BC : FE ; wherefore (IV. 6) the parallelogram AB is equivalent to the parallelogram BC. Therefore equivalent parallelograms, etc. Scho. 1. If AD, CG were produced to meet, it would be easy to show, that AB and BC would be the complements of the parallelograms about the diagonal of the whole parallelogram AC. In the demonstration, it should in strictness be proved that AF and CE meet when produced. This follows from I. 19. Cor. Hence, equivalent triangles which have an angle of the one equal to an angle of the other, have their sides about those angles reciprocally proportional ; and (2) triangles which have an angle of the one equal to an angle of the other, and the sides about those angles reciprocally proportional, are equivalent to one another. JScho. 2. It is evident from the seventh corollary to the fif- 124 THE ELEMENTS OF [book V. teenth proposition of the first book, that this proposition is true as well as when the angles are supplemental as when they are equal. Prop. XIIT. — Prob. — TIpon a given straight line to describe a figure similar to a. given rectilineal fi ure, and such that the given line shall be homologous to an assigned side of the given figure. Let AB be the given straight line, on which it is reqnirefi to describe a rectilineal figure similar to a given rectilineal figure, and such that AB may be homologous to CD, a side of the given figure. First, let the given rectilineal figure be the triangle CDE. Make the angles BAF, ABF respectively equal to DCP], CDE; and (V, 3, cor.) the triangle ABF is similar to CDE, and has AB homologous to CD. Again: let the given figure be the quadrilateral CDOE. Join DE, and, as in the first part, describe the triangle ABF having the angles BAF, ABF respectively equal to DCE, CDE, and also the triangle BFH having the angles FBH, BFH respectively equal to^EDG, DEC. Then (I. 20, cor. 5) the angles BHF, DGE are equal, and (const.) A and C are equal. Also, since (conet.) ABF, FBH are respectively equal to CDE, EDG, the whole ABH is equal to the whole CDG. For the same reason AFH is equal to CEG ; and therefore the quadri- lateral figui-es A BHF, CDGE are equiangular. But likewise these figures have their sides about the equal angles propor- tional. For the triangles ABF, CDE being equiangular, and also BFH, DEG; as BA : AF : : DC : CE; and^as FH. : nB::EG: GD. Also, in the same triangles, AF : FB :: CE : ED ; and as FB : FH : : ED : EG ; therefore, ex mquo, AF : FH : : CE : EG. In the same maimer it may be pi'oved, that AB : BH : : CD : DG; whcrefoi-e (V. def 1) the figures ABIIF, CDGE are similar to one another. Next : let the given figure be CDKGE. Join DG ; and, as in the second case, deiscribe the figure ABHF, similar to CDGE, BOOK v.] EUCLID AND LKGENDRE. 125 and similarly situated; also, describe the triangle BHL having the angles BHL, IIBL respectively equal to DGK, GDK. Then (I. ax. 2) the whole angles FIIL, ABL are respectively equal to the whole angles EGK, CDK ; and (I. 20, cor. 5) the angles L and K are equal. Therefore the figures ABLHF, CUKGE are equiangular. Again : because the quadrilaterals ABHF, CDGE, and the triangles BLH, DKG are similar; as FH : HB :: EG : GD, and HB : HL : : GD : GK ; therefore, ex cequo, FH : HL : : EG : GK. In like manner it may be shown, that AB : BL : : CD : DK ; and because the quadrilateials ABHF, CDGE, and the triangles BLH, DKG are similar, the sides ?,bout the angles A and C, L and K, AFH and CEG are proportional. Therefore (V. def l) the five-sided figures ABLHF, CDKGE are similar, and the sides AB and CD are homologous. In the same manner, a rectilineal figure of six or more sides may be described on a given straight line, similar to one given ; which was to be done. Scho. In practice, if AB be parallel to CD, the cojistruction is most easily efll-cted by drawing AF and BF parallel to CE and DE ; then FH and BH parallel to EG and DG ; and lastly, HL and BL parallel to GK and DK. The doing of this is much facilitated by employing tlie useful instrument, the par- allel ruler. For the easiest methods, however, of performing this and many other problems, the student must have recourse to works that treat expressly on such subjects, particulai'ly treatises on practical geometry, surveying, and the use of mathematical instruments. Prop. XIV. — Theor. — Similar plane figures are to one an- other in the duplicate ratio of their homologous sides. Let ABC, DEF be sitnilar triangles, having the angles B and E equal, and AB : BC :: DE : EF, so that (IV. def. 13) the side BC is homologous to EF. A The ti'iangle ABC has to the trian- gle DEF the duplicate ratio of that which BC has to EF. Take (V. 9) BG a third propor- tional to BC, EF, so that BC : EF : : EF : BG, and join GA. Then, because 126 THE ELEITENTS OF [bOOK V. AB : BC : : DE : EF ; alternately, AB : DE : : BC : EF ; but (const.) as BC : EF : : EF : BG ; therefore (IV. V) as AB : DE : : EF : BG. The sides, therefore, of the triangles ABG, DEF -which are about the equal angles, are reciprocally proportional, and therefore (V. 12, cor.) the triangles ABG, DEF' are equal. Again : because BC : EF : : EF : BG ; and that if three straight lines be proportionals, the first is said (IV, def 11) to have to the third the duplicate ratio of that which it has to the second ; BC therefore has to BG the duplicate ratio of that which BC has to EF. But (V. 1) asBC to BG, so is the triangle ABC to ABG. Therefore (IV. 7) the triangle ABC has to ABG the duplicate ratio of that which BC has to EF. But the triangle ABG is equal to DEF ; wherefore, also, the triangle ABC has to DEF the duplicate ratio of that which BC has to EF. Therefore, etc. Scho. 1. From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar triangle similarly described on the second. The third proportional might be taken to EF and BC, and placed from E along EF produced ; and then a triangle equal to ABC would be formed by joining D with the extremity of the produced line. Scho. 2. The above case might also be proved by making BH equal to EF, joining AH, and drawing through H a par- allel to AC. The triangle cut off by the parallel is equal (1. 14) to DEF. But (V. 1) the triangle ABC is to the triangle ABH as BC to BH; and, for the same reason, the triangle ABH is to the triangle cut off by the parallel, or to DEF, as BC to BH, or (const.) as BH to BG. Therefore, ex cequo^ the triangle ABC Las to the triangle DEF the same ratio that BC has to BG, or (IV. def. 11) the duplicate ratio of that which BC has to EF. Again : let ABCD and AEFG be two squares, then will they be to one another in the duplicate ratio of their sides — as it has been previously demonstrated that similar triangles are to one another in the duplicate ratio of their liomologous sides. Therefore : ABD : AEF in the duplicate ratio of AB to AE ; and ADC : BOOK v.] EUCLID AND LEGENDEE. 127 D / / F AFG in the duplicate ratio of AB to AE ; hence, by compo- pition, ABD+ADC : AEF+ AFG in the duplicate ratio of AB to AE ; but ABD + ADC=ABDC, and AEF+AFG = AEFG ; wherefore the squares are to one another in the duplicate ratio of their sides. For like reason, the, triangles AIID, AFC are to one another in the duplicate ratio that PH is to DF, or that AD is to AC ; hence, similar triangles are one to another in the duplicate ratio of their alti- tudes or bases ; since the segments AHD, AFC have the same bases and altitudes as the triangles AHD and AFC, and being segments of quadrants, are similar ; hence, they are to one another in the du- plicate ratio of their bases and altitudes ; therefore, by composition, AED + seg. AHD : ABC + seg. AFC in the duplicate ratio of their homologous sides, or simi- lar polygons are to one another in the duplicate ratio of their homologous sides, and quadrants of circles are one to an- other in the duplicate ratio of their radii ; hence, semicircles are one to another in the duplicate ratio of their diameters, and. circles are one to another in the duplicate ratio of their diame- ters. Wherefore, similar surfaces are, etc. Cor. 1. Therefore, universally, if three lines be proportionals, the first is to the third as any plane figure upon the first to a similar and similarly described figure upon the second. Cor. 2. Because all squares are similar figures, and all circles are similar figures, the ratio of any two squares to one another is the same as the duplicate ratio of their sides ; and the ratio of any two circles to one another is the same as the duplicate ratio of their diameters ; hence, any two similar plane figures are to one another as the squares or circles (IV. 7) described on their homolocrous sides. Cor. 3. Because the sides of similar plane figures are propor- tional, therefore (IV. 8) their perimeters or peripheries are propoi'tional to the homologous sides ; hence, the perimeters of 128 THE ELKMENTS OF [bOOK V. similar polyo^ons are proportional to their apotliems ; the cir- cumfei-enees of circles are proportional to their diameters. Cor. 4. Hence plane figures which are similar, to the same figure are similar to one another (I. ax. l). Cor. 5. If four straight lines be proportionals, the similar plane figures, similarly desci-ibed upon them, are also propor- tionals; and, conversely., if the similar plane figures, similarly- described upon four straight lines, be proportionals, those lines are proportionals. Cor. 6. Similar polygons inscribed in circles are to one an- other as the squares of the diameters. Cor. 1. When there are three parallelograms, AC, CH, CF, the first, AC (IV. def 10), has to the third, CF, the ratio which is compounded of the ratio of the first, AC, to the second, CIT, and of the ratio of CH to the third, CF; but AC is to CH as their bases ; and CH is to CF as their bases ; therefore AC has to CF the ratio which is compounded of ratios that are the same with the ratios of the sides. Scho. 3. Dr. Simson remarks in his Note on this corollaiy, that "nothing is usually reckoned more difficult in the elements of geometry by learners, than the doctrine of compound ratio." This distinguished geometer, however, has both freed the text of Euclid from the errors introduced by Theon or others, and has explained the subject in such a manner as to remove the difficulties that were formeily felt. According to liim, " every proposition in which compound ratio is made use of, may with- out it be both enunciated and demonstrated ;" and " the use of compound ratio consists wholly in this, that by means of it, cir- cumlocutions may be avoided, and thereby propositions may- be more briefly either enunciated or demonstrated, or both may be done. For instance, if this corollary were to be enun- ciated, without mentioning compound ratio, it might be done as follows: If two pai-allelograms be equiangular, and if as a side of the first to a side of the second, so any assumed straight line be made to a second straight line ; and as the other side of the first to the other side of the second, so the second straight line be made to a third ; the first parallelogram is to the second as the first straight line to the third ; and the demonstration would be exactly the same as we now have it. BOOK v.] EUCLTD AND LEGENDKE. 129 But the ancient cceometers, when they observed this ennncia- ti(ni could l)e made nhortev, l)y giving a name to the ratio which the first straiglit line lias to the last, by which name the intermediate ratios miglA likewise be signified, of the first to the second, and of the second to the third, and so on, if there ■were more of them, they called this ratio of the fiist to the last, the ratio compounded of the ratios of the first to the second, and of the second to the third straight line; that is, in the present example, of the ratios which are the same with the ratios of the sides." Scho. 4. The seventh corollary will be illustrated by the fol- lowing proposition, which exhibits the subject in a different, and, in some respects, a preferable light : Triangles which have an angle of the one equal to an angle of the othei\ are proportional to the rectangles covitcined by the sides about those angles ; and (2) equiangular parallelO' grams are proportional to the rectangles co7itained by their ad- jacent sides. 1. Let ABC, DBE be two triangles, having the angles ABC, DBE equal; the first triangle is to the second as AB.BC is to DB.BE. Let the triangles be placed with their equal angles coinciding, and join CD. Then (V. 1) AB is to DB as the triangle ABC to DBC. But (V. 1, cor. 2) AB : DB :: AB.BC :DB.BC; theiefore (IV. 7) AB.BC is to DB.BC as the triangle ABC to DBC. Li the same maimer it would be shown that DB.BC is to DB.BE as the triangle DBC to DBE ; and, there- fore, ex ceq^io^ AB.BC is to DB.BE as the triangle ABC to DBE. 2. If parallels to BC through A and D, and to AB through C and E were drawn, parallelograms would be formed which would be re- spectively double of the triangles ABC and DBE, and which (IV. 9) would have the same ratio as the triangles; that is, the ratio of AB.BC to DB.BE; and this proves the second part of the proposition. Comparing this proposition and the corollary, we see that the ratio which is compounded of ihe ratio of the sides, is the same 9 130 THE ELEMENTS OF [book V. as the ratio of their rectangles, or the same (T. 23, cor. 5) as the ratio of their products, if they he e:q)ressed in numbers. This conclusion might also be derived from the proof given in the text. For (const.) DC : CE : : CG : K ; whence (V. 10, cor.) K.DC = CE.CG. But it was proved that BC : K : : AC : CF; or (V. 1) BC.DC : K.DC : : AC : CF; or BC.DC : CE.CG :: AC : CF; because K.DC=CE.CG. The twelfth proposition of this book is evidently a case of this proposition; and the fourteenth is also easily derived from it. Pijop. XV. — TriEOPw — The parallelograms ahout the diago- nal of any parallelogram are similar to the whole, and to one another. Let ABCD be a parallelogram, and EG, HK the parallelo- grams about the diagonal AC ; the parallelograms EG, HK are similar to the whole parallelogram, and to one another. Because DC, GF are parallels, the angles ADC, AGF are (I. 16) equal. For the same reason, be- cause BC, EF are parallels, the angles ABC, AEF are equal ; and (T. 15, cor. 1) each of the angles BCD, EFG is equal to the opposite angle DAB, and there- fore they are equal to one another ; Avherefore, in the parallelograms, the angle ABC is equal to AEF, and BAC common to the two tri- angles BAC, EAF ; therefore (V. 3, cor.) as AB ; BC : : AE : EF. And (IV. 5), because the opposite sides of parallelograms are equal to one another, AB : AD :: AE : AG; and DC : BC : : GF : EF; and also CD : DA : : FG : GA. Therefore the sides of the parallelograms BD, EG about the equal angles are proportionals; the parallelograms are, therefore (V. def l), similar to one another. In the same manner it would be shown that the parallelogram BD is similar to HK. Therefore each of the parallelograms EG, HK is similar to BD. But (V. 14, cor.) rectilineal figures which are similar to the same figure, are similar to one another; therefore the parallelogram EG is simi- lar to HK. Wherefore, etc. Scho. Hence, GF : FE : : FH : FK. Therefore the sides of BOOK v.] EUCLID AND LEGENDRE. 131 the paralleloojrains GK and EH, about the equal angles at F, are reciprocally proportional ; and (V. 12) these parallelograms are equivalent ; a conclusion which agrees with the eighth corollary to the fifteenth proposition of the first book. Prop. XVI. — Prob. — To describe a rectllmeal figure which shall be similar to one given rectilineal figure., and equivalent to one another. Let ABC and D be given rectilineal figures. It is required to describe a figure similar to ABC, and equal to D. Upon the straight line BC describe (II. 5, scho.) the parallelo- gram BE equivalent to ABC ; also upon CE describe the parallel- ogram CM equivalent to D, having the angle FCE equal to CBL. Therefore (I. 16 and 10) BC and CF are in a straight line, as also LE and EM. Between BC and CF find (V. 11) a mean proportional GH, and ^ on it describe (V. 13) the figure GHK similar, and similarly situated, to ABC; GHK is the figure required. Because BC : GH : : GH : CF, and if three straight lines be proportionals, as the first is to the third, so is (V. 14, cor. 2) the figure upon the first to the similar and similarly described figure upon the second ; therefore, as BC to CF, so is ABC to KGH ; but (V. 1) as BC to CF, so is BE to EF ; therefore (IV. 7) as ABC is to KGH, so is BE to EF. But (const.) ABC is equivalent to BE ; therefore KGH is equivalent (IV. ax. 4) to EF; and (const.) EF is equivalent to D; where- fore, also, KGH is equivalent to D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described, simi- lar to ABC and equivalent to D ; which was to be done. Prop. XVII. — Theor. — If two similar parallelograms have a comm,on angle, and be similarly situated, they are about the same diagonal. Let ABCD, AGEF be two similar parallelograms having a 132 THE ELEMENTS OF [book V. C E F B common angle CAB, tlicy will be about the same diagonal AD. Similar parallelograms have their sides about equal angles pro- portional (V. def. l). Draw the diagonals EG and CB ; hence, AB : AG : : AC : AE ; therefore EG ia parallel to CB (V. 3), and the angles AEG, ACB are equal (I. 16); like- wise the angles EGA, CBA. The triangles E AG, FG A are equal (I. 15, cor. 5); likewise the tri- angles CAB, DBA ; therefore AF is equal to EG, and AD is equal to CB ; but AF is the diagonal also of AGEF, and is in the same straight line with AD, the diagonal of ABCD. Wherefore, if two similar parallelograms, etc. Cor. Hence, equiangular parallelograms have to one another the ratio which is compounded of the ratio of their sides; hence, triangles which have one angle of the one equal, or sup- plemental, to one angle of the other, have to one another the ratio which is compounded of the ratio of the sides containing those angles. Pkop. XVIII. — ^Theok. — Of all the parallelof/rmns that can he inscribed in any triangle^ that which is described on the half of one of the sides as base is the greatest. Let ABC be a triangle, having BC, one of its sides, bisected in D; draw (I. 18) DE parallel to BA, and EF to BC ; let also G be any other point in BC, and describe the parallelogram GK; FD is greater than KG. If G be in DC, through C draw CL parallel to BA, and produce FE, KH, Gil as in the figure. Then (I. 15, cor. 8) the comj>le- ments LTI and IID are equivalent ; and since the bases CD, DB are equal, the parallelograms ND, DK (I. 15, cor. 5) are equivalent. To LH add ND, and to HD add DK ; then (I. ax. 2) the gnomort MND is equivalent to the parallelogram KG. But (L ax. 9) DL is greater than MXD ; and therefore BOOK V.J EUCLID AND LFGENDEE. 133 FD, which (T. 15, cor. 5) is equal to DL, is greater than KG, which is equivalent to the gnomon JNIND. - If G were in BD, since BD is equal to DC, AE is equal (V. 2) to EC, and AF to FB; and by drawing through A a jiaral- lel to BC, meeting DE produced, it would he proved in the same manner that FD is greater than the inscribed ])aralk'lo- gram applied to BG. Therefore, of all the parallelogi-ams, etc. Cor. Since (V. lo) all parallelograms having one angle coin- ciding with BCL, and their diagonals with CA, arc similar, it follows from this proposition that if, on the segments of a given straight line, BC, two parallelograms of the same altitude be described, one of them, DL, similar to a given parallelogram, the other, DF, will be the greatest possible when the segments of the line are equal. Scho. The parallelogram FD exceeds KG by the parallelo- gram OM similar to DL or DF, and described on OH, which is equal to DG, the difference of the bases BD and BG. Hence we can describe parallelograms equivalent and similar to given rectilineal figures. The enunciation of this proposition here given is much more simple and intelligible than that of Euclid, and the proof is considerably shortened, Euclid's enunciation, as given by Dr. Simson, is as follows: "Of all pai'allelograms applied to the same straight line, and deficient by parallelograms, similar and similarly situated to that which is described upon the half of the line ; that which is applied to the half, and is similar to its defect, is the greatest." It may be remaiked, that this piopo- sition, in its simplest case, is the same as the second corollary to the fifth proposition of the second book. Prop. XIX. — ffHEOR. — In equal circles^ or in the same cir- cle^ angles, whether at the centers or circumferences^ have the same ratio as the arcs on which they stand have to one another^ so also have the sectors. Let, in the equal circles ALB, DNE, the angles LGK, KGC, CGB, DHN, NHM, and IMITF be at the centers, and the angles BAG and EDF be at the circumferences, then will those angles have to each other the same ratio as the arcs KL, KC, CB, DN, NM, MF, and FE have to one another. 134 THE ELEMENTS OF [book V. Since (T. def. 19) all angles at the center of a circle are measured by the arcs intercepted by the sides of the angles, and all ancjles at the circum- ference are subtended by the arcs intercepted by the sides of the angles, and (III. 16) equal angles will have equal arcs whether they be at the center or the circumference; hence, the same ratio which the arcs have to each other, will the angles also have to one another — that is, when the arcs be greater, the angles will be greater; less, less ; and equal, equal. And since the sectors are contained (III. def V) by the sides of the anirles and the arcs, the same ratio between the sectors will evidently exist as there is between the arcs. Wherefore, in equal circles, etc. Cor. Hence, conversely^ arcs of the same or equal circles will have the same ratio as the angles or sectors which they measure or subtend — when greater, greater ; less, less ; or equal, equal. Pkop. XX. — ^Peob. — Tlie area of a regular inscribed poly ' gon^ and that of a regular circumscribed one of the same nvirv- her of sides being given y to find the areas of the regidar in- scribed and circumscribed polygons having double the number of sides. ].et A be the center of the circle, BC a side of the inscribed polygon, and DE parallel to BC, a side of the circumscribed one. Draw the perpendicular AFG, and the tangents BH, CK, and join BG; then BG will be a side of the inscribed polygon of double the number of sides; and (111.^6, cor. l) IIK is a side of the similar cii-cumscribed one. / \\ // \ Now, as a like construction would be / x/^ \ Tinule at each of the remaining angles of MAN ^^Ijp iwlygon, it will be sufficient to con- sider the i)art here represented, as the tri- angles connected with it are evidently to each other as the poly- gons of which they are parts. For the sake of brevity, then. n G K BOOK v.] EUCLID AND LEGENDKS. 135 let P denote the polygon whose side is BC, and P' that whose side is DE; and, in like manner, let Q and Q' represent those whose sides are BG and HK ; P and P/ therefore, are given ; Q and Q' required. Now (V. ]) the triangles ABF, ABG are proportional to their bases AF", AG, as they are also to the polygons P, Q ; therefore AF : AG : : P : Q. The triangles ABG, ADG are likewise as their bases AB, AD, or (V. 3) as AF, AG; and they are also as the polygons Q and P' ; therefore AF : AG : : Q : P' ; wherefore (IV. 7)" P : Q : : Q : P' ; so that Q is a mean proportional between the given polygons P, P'; and, re))re- senting them by numbers, we have Q-=PP'', so that the area of Q will be comjyuted by -tnaltlphjing P by P', and extracting the square root of the product. Again : because AH bisects the angle GAD, and because the triangles AHD, AHG are as their bases, Me have Gil : HD : : AG : AD, or AF : AB : : AHG : AHD. But we have already seen that AF : AG :: P : Q; and therefore (IV. 7) AHG : AHD : : P : Q. Hence (IV. 11) AUG : ADG : : P : P-|-Q; whence, by doubling the antece- dents, 2 AHG : ADG : : 2P : P+Q. But D IT P K" "R it is evident, that whatever part the tri- angle ADG is of P^, the same part of the polyg )n Q' is the triangle AHK, which is double of the trianojle AHG, Hence the last analogy becomes Q' : P^ : : 2P : P+Q. Now (IV. 2, cor. 1) the product of the ex- tremes is equal to the product of the means; and therefore Q' will be computed by dividing twice the product o/'P and V by P+Q; and the mode of finding Q has been pointed out already. Prop. XXI. — ^Theor. — Of regular polygons which have eqiial perimeters., that which has the greater mimber of sides is the greater. Let AB be half the sides of the polygon which has the less number of sides, and BC a perpendicular to it, which will evi- dently pass through the center of its inscribed or circumscribed circle; let C be that center, and join AC. Then, ACB will be 136 THE ELEMENTS OF [book V. the angle at tlie center subtended by the half side AB. ]\I;ike BCD equal to the angle subtended at the center of the other polygon by half its side, and from C as center, with CD as ra- dius, describe an arc cutting AC in E, and CB i)roduced in F. Then, it is plain, that the angle ACB i'* to four right angles as AB to the common peiimeter; and four right angles are to DCB, as the common pei-imeter to the half of a side of the other polygon, which, for brevi- ty, call S; then, ex mquo^ the angle ACB is to DC]} as AB to S. But (V. 19) the angle ACB is to DCB as the sector ECF to the sector DCF; and consequently (IV. 1) the sector ECF is to DCF as AB to S, and, by division, the sector ECD is to DCF as AB — S to S. Now the triangle ACD is greater than the sector CED, and DCB is less than DCF. But (V. 1) these triangles are as their bases AD, DB ; therefore AD has to DB a greater i-atio than AB — S to S. Hence AB, the sum of the first and second, has to DB, the second, a greater ratio than AB, the sum of the third and fourth, has to S, the fourth ; and therefore (IV. 2, cor. 5) 8 is greater than DB. Let then BG be equal to S, and draw GH parallel to DC, meeting FC pro- duced in H. Then, since the angles GHB, DCB are equal, BH is the perpendicular drawn from the center of the polygon hav- ing the greater number of sides to one of the sides ; and since this is greater than BC, the like perpendicular in the other polygon, while the perimeters are equal, it Ibllows that the area of that which has the gi-eater number of sides is greater than that of the other. Prop. XXIT. — Theor. — If the diameter of a circle 1 e divided into any two parts, AB, BC, and if semicircles, ADBjBECy be described on opposite sides of these, the circle is divided by tJieir arcs into ttoo figures, GDE, FED, the boundary of each <f tchich is equal to the circumference q/'FG ; and which are such that AC : BC : : FG : FED, and AC : AB :: ¥G : GDE. For (V. 14, cor. 3) the circumferences of circles, and conse- quently the halves of their circumferences, are to one another as their diameters ; thereiore AB is to AC as the arc ADB to BOOK v.] EUCLID AND LEGENDRE. 137 AFC, and BC is to AC as the arc BEC to AFC. Hence (IV. 1) AC is to AC as the compound arc ADEC to AFC; therefore ADEC is p equal to half the circumference ;and the entire boundaries of the fiirures GDE, FED are each equal to the circumfer- ence of FG. Again (V. 14, cor, 2): circles, and consequently semicircles, are to one an- other as the squares of their diameters ; therefore AC' is to AB^ as the semicircle AFC to the semicircle ADB. Hence, since (H. 4) AC' = AB» -f2AB.BC + BC^ we find by conversion that AC is to 2AB. BC + BC% as the semicircle AFC to the remaining space BDAFC; whence, by inversion, 2AB.BC + BC' is to AC' as BDAFC to the semicircle AFC. But BC' is to AC= as the Bemicircle BEC to the semicircle AFC; and therefore (IV. 1) 2AB.BC + 2BC' is to AC as the compound figure FDE to the semicircle AFC. But (II. 3) 2AB.BC+2BC = 2AC.BC, and (V. 1) 2AC.BC : AC^ : : BC : ^AC. Hence the preceding analogy becomes BC to ^AC, as FED to the semicircle, or by doubling the consequents, and by inversion, AC to BC, as FG to FED ; and it would be proved, in the same manner, that AC : AB : : FG : GDE. Cor. Hence we can solve the curious problem, in which it is required to divide a circle into any proposed number of parts, equal in area and boundary; as it is only necessary to divide the diameter into the proposed number of equal parts, and to desci'ibe semicircles on opposite sides. Then, whatever part AB is of AC, the same part is AEG of the circle. Their bound- aries are also equal, the boundary of each being equal to the circumference of the circle. Scho. Another solution would be obtained, if the circumfer- ence were divided into the proposed number of equal parts, and radii drawn to the points of division. This division, hoM'ever, can be eftected only in some particular cases by means of ele- mentary geometry. Pkop. XXIII. — Prob» — To divide a given circle ABC into 138 THE ELEMENTS OF [book V. any proposed number of equal parts by means of concentric circles. Divide the radius AD into the proposed number of equal parts, suppose three, in the points E, F, and through these points draw perpendiculars to AD, meeting a semicircle de- scribed on it as diameter in G, H; from D as center, at the distances DG, DH, describe the circles GL, HK ; their circumferences divide the circle into equal parts. Join All, DH. Then (V. 17, cor. 2) AD, DH, DF being continual propor- tionals, AD is to DF as a square de- scribed on AD is to one described on DH. But (V. 14, cor. 2) circles are proportional to the squares of their diameters, and consequently to the squares of their radii. Hence (IV. 7) AD is to FD as the circle ABC to the circle HK ; and therefore, since FD is a third of AD, HK is a third of ABC. It would be proved in a similar manner that AD is to ED as ABC to GL. But ED is two thirds of AD, and therefore GL is two thirds of ABC ; wherefore the space between the circumferences of GL, HK is one third of ABC, as is also the remaining space between the circumferences of ABC and GL. Cor. Hence it is plain that the area of any annulus^ or nng, between the circumferences of two concentric circles, such as that between the circumferences of ABC and GL is to the cir- cle ABC as the difference of the squares of the radii DM, DL to the square of DM ; or (II. 5, cor. 1, and HI. 20) as the rect- angle AL.LM, or the square of the perpendicular LB to the square of DM; and it therefore follows (V. 14) that the ring is equivalent to a circle described with a radius equal to LB. Prop. XXIV. — Theor. — If on BC the hypothenuse of a ri^/it-angled triangle ABC, a semicircle, BAC, be described on the same side as the triangle, and if semicircles, ADB, AEC, ba described on the legs, falling without the triangle, the lunes or crescents ADB, AEC, bounded by the arcs of the semicircles^ are together equal to the right-angled triangle ABC. BOOK v.] EUCLID AND LEGENDRE. 139 For (V, 14 and lY. 9) the semicircle ADB is to the semicircle BAG as the square of AB to the square of BC, and AEC to BAG as the square of AC to the square of BC ; whence (IV. 1) the two semicircles ADB, AEG taken together, have to BAG the same ratio as the sum of the squares of AB, AG to the square of BC, that is, the ratio of equality. From these equals take the segments AFB, AGG, and there remain the luncs DF, EG equal to the triangle ABC. Cor. If the legs AB, AC be equal, the arcs AB'B, AGG are equal, and each of them an arc of a quadrant ; also the radius drawn from A is perpendicular to BC ; and since the halves of equals are equal, each lune is equal to half of the triangle ABC. -If, therefore, ABC be a qnadi'ant, and on its choi'd a semicircle be described, the lune comprehended between the circum- ferences is equal to the triangle ABC ; and since (II. 13) a square can be found equal to ABC, we can thus effect the quadrature of a space, ADC, bounded by arcs of circles. Prop. XXV. — Prob. — To find the area of a circle. Sc/io. 1. The approximate area of a circle can be found by means of the twentieth proposition of this book, by what is called the method of exha'ustions, giving an error in excess ; viz., the approximate area thus obtained is square of radius multiplied by .3.1415926, etc. Geometry being an exact science, and its conclusions being derived from accurate principles, the apj)roximate area for the circle is not consistent with the strictness of geometrical rea- soning, and the area of the circle must be established exactly before it can be regarded a geometrical truth. The reason why the method of e haustions gives the a]yjyroxim,ate result, is be« cause — by the twentieth proposition of this book — the circum- scribed and inscribed regular and similar polygons about the circle are supposed^ by continually doubling the number of their 140 THE ELEMENTS OF [BOOK V. sides, to he made equivalent to the circle; but Carnot, in his Reflexions sur la Metaphysique da Calcul Infinitesimal^ states, "That the ancient geometers did 'not consider it con- sistent with the strictness of o-eometrical reasnTiing to re>2:ard curve lines as polygons of a great number of sides." Now, the area of any regular ])olygon is the rectangle of its a])othem and semi-perimeter; but this area is derived from the sixth cor- ollary of the twenty-third proposition of the first book — since it has been shown in the first corollary of the twentieth propo- sition of the same book, that any rectilineal figure can be divided into as many triangles as the figure has sides ; there- fore, in case of a regular polygon, when triangles are formed ia it by straight lines drawn from the center to the extremities of the several sides of the polygon, the area of the polygon be- comes by the tenth axiom of the firj^t book equivalent to the sum of these triangles ; hence (I. 23, cor. 6) each triangle is the rectangle of the apothem of the polygon and a semi-side of the polygon; therefore the area of the polygon is (I. ax. 10) the rectangle of its apothem and its semi-perimeter. Since (I. 23, C01-. 4) the aiea of a triangle is derived from the properties of parallel straight lives, and any polygon has its sides straight lines (I. def 12), the pre perties o^ parallel straight lines are applicable to all polygons; but the circle being formed by a curve li7ie, the properties of parallel straight lines are not ap- plicable to it ; hence the reason is evident why the ancient geometers objected to the curve line being regarded a polygon of a great number of sides. Euclid, in his ^/e??<e/i^<f, endeav- ored to sustain the proposition, that the circle is the I'cctangle of its radius and semicircumfereiice, by what is called the indirect^ apogogic, or Meductio ad ahsurdwm, method. Now, every true pro])osition can be directly demonstrated, and a fair test of the truth or falsity of this proposition can be in the success or failure of it being directly demonstrated. I have given the d reel demonstrations for every other ])roposition in geometry; but I can not do so in this case — therefore I believe the proj)0- sition fallacious. Archimedes has shown that the relation of diameter to the circumference of a circle expressed in numbers, to be as 7 to 22 — which is practically correct. Among isoperi- metrical figures, the circle contains the greatest area ; there- BOOK v.] EUCLID AND LEGENDRE. 141 fore when 22 expresses the circumference of a circle, the perim- eter of its equivalent square must he greater than 22; and if a cube be inechauically constructed upon a base whose perimeter is 24.2487 + , it will be equivalent to a cylinder of same height, the diameter of whose base is 7. Now, when 24.2487+ expresses the perimeter of a square, each of its sides (I. 23, cor. 1) will be 6.0621 + ; and its area ■will be 36.75, or three times square of the radius of the circle. Hence we get by mechanical construction less than what is ob- tained by the method of exhaustions. The geometrical con- firmation of the mechanical construction is given in the second corollary to the seventeenth proposition of the sixth book. Scho. 2. Euclid has endeavored to demonstrate that the cir- cle is the rectangle of circumference ajid semi-i'adius. Now, the square equal to circle is somewhere between the inscribed and circumscribed squares, and its area is equal to its perime- ter multiplied by less than semi-radius; consequently the rect- angle of circumference and semi-radius will produce more than area of circle. {^Or Thomson^ s Eficlid, Appendix, DooJc I., Prop. JTJTXZX) " The area of a circle is equal to the rectangle under its ra- dius, and a straight line equal to half its circumference. Let AB be the radius of the circle BC ; the area of BC is equal to the rectangle under AB and a straight line D equal to half the circumference. " For if the rectangle AB. D be not equal to the circle BC, it is equal to a circle either greater or less than BC. First, sup- pose, if possible, the rectangle AB. D to be the area of a circle EF, of which the radius AE is greater than AB." Here Euclid is inconsistent with his own proposition : at the very stait he bases his argument upon a contradiction. He premises that the area of a circle is equal to the rectangle under its radius, and a straight line equal to half its circumference ; then sujy- pose, i. 6., asks to be granted for the sake of argument, that that same rectangle is equal to a larger circle. Why does he resort to this subterfuge? It will be said to show the Heductio ad absurduni ; very well, let us follow his argument : "and let GHK be a regular polygon described about the circle BC, such 142 THE ELKMENT8 OF [bOOK V. that its sides flo not meet the circumference of EF. Then, by dividing this polygon into triangles by radii drawn to G, H, K, etc., it would be seen that its area is equal to the rectangle under AB and half its perimeter. But the perimeter of the polygon is greater than the ciixiumference of BC, and therefore the area of the polygon is greater than the rectangle AB. D ; that is, by hypothesis, than the area of the circle EF, which is absurd." What is absurd ? That the circle EF is greater than the polygon GHK, etc., or Euclid's argument? The absurdity is in considering the area of a circle equal to a larger circle. An argument based upon absurdity must necessarily lead to absurdity, which in fact has been the case. When Euclid sup- posed, i. e., asked to be granted for the cake of argument, AB. D= circle EF, it does not prove the area of polygon greater than the area of circle EF, because he at the start supposed AB. D = circle EF, and consistently with his hypothesis and his argument, it must be so to the end ; therefore, consistently with his argument and his hypothesis, AB. D is greater than the area of polygon GHK, etc. The first part of Euclid's prop- osition is nothing more than a demonstration to prove the area of a circle is greater than the area of a polygon drawn within the circle. And the second part of Euclid's proposition is nothing more than a demonstration to prove a circle less than the circumscribing polygon. This proposition of Euclid is very sophistical, and consequently its fallacy has been imde- tected, owing no doubt to the repute of Euclid, and to the sup- position that Euclid argued from axioms, and consistently with the principles of geometry, which he did ; but in this instance he deceived himself, and consequently all those who believe him the oracle of geometry. When he attempted to prove AB. D=area of circle BC, it was contradictory to his argument to suppose AB.'D = area of circle EI*'; because when he based his argument upon the premies that AB. D=:area of circle EF, consistency demanded that he should stand by his premise, and not forsake it as soon as it led to an absurdity, and judge a cir- cle less than a polygon within a cii'cle. Tlie absurdity is in his own argument, to base it upon a supposition which he knew was inconsistent with his proposition, and the inconsistence to drop his premiss when he perceived it led to an absurdity ; as BOOK v.] EUCLID AND LEGENDRE. 143 AB. D is less than the circle EF, it is a very fallacious argu- ment, when based on the supposition that they are equal, and it leads to an absurdity ; and very inconsistent with geometri- cal reasoning for Euclid to drop at the conclusion of his argu- ment the very premiss upon which he based his argument. Every method of demonstration, as well as that method termed Heductio adalsurdttm, require that the premiss which is adopt- ed at the start be retained to the conclusion. And when Euclid adopted AB. D=circle EF at the commencement of his demon- stration, consistence of reason and science demanded that he fihould have kept it to the conclusion, and then there would have been no absurdity, but a demonstration to prove that the polygon GHK, etc., is less than circle EF. But Euclid had in his mind AB. D = circle BC ; forgetting that he had adopt- ed AB. D= circle EF, and had stiU to prove AB. D= circle BC. END OF BOOK FIFTH. BOOK SIXTH. ON THE PLANE AND S0LID3. DEFINITIONS. 1. A STRAIGHT line is said to be perpendicular to a plane when it makes ri'jrht ansjles with all straiiiht lines meeting' it iu that plane. 2. The inclination of two planes which meet one another is the angle contained by two straight lines drawn from any point of their common section at right angles to it, one upon each plane. The angle which one plane makes >.ith another is Boraetimes called a dihedral angle. 3. If that angle be a right angle, the planes arc perpendicu- lar to one another. 4. Parallel planes are such as do not meet one another, though produced ever so far in every direction. 5. A solid angle is that which is made by more than two plane angles meeting in one point, and not lying in the same plane. If the number of plane angles be three, the solid angle is tri- hedral ; if four, tetrahedral ; if more than io\x\\ polyhedral. 6. Kpolyfiedron is a solid figure contained by plane figures. If it be contained by four plane figures, it is called a tetrahe- dron ; if by six, a hexahedron ; if by eight, an octahedron ; if by twelve, a dodecahedron ; if by twenty, an icosahedron^ etc. 1. A regular body^ or regular polyhedron^ is a solid con- tained by plane figures, which are all equal and similar. 8. Of solid figures contained by planes^ those are similar which have all their solid angles equal, each to each, and which are contained by the same number of similar plane figures, simi- larly situated. 9. A pyra^nid is a solid figure contained by one plane figure called its base, and by three or more triangles meeting in a point without the plane, called the vertex of the pyramid. BOOK VI.] EUCLID AND LEGENDEE. 145 10. A ]jrism is a solid fijjiire, the ends or hases of \vliich are parallel, and are equal and similar plane figures, and its otlier boundaries are parallelograms. One of tliese parallelograms also is sometimes regarded as the base of the j)rism. 11. Pyramids and prisms are said to be triangular wlien their bases are triangles ; quadrangular, when their bases are quadrilaterals ; pentagonal, when ihey are pentagons, etc. 12. The altitude of a pyramid is the perpendicular drawn from its vertex to its base; and the altitude of a prism is either a perpendicular drawn from any point in one of its ends or bases, to the other; or a perpendicular to one of its bounding parallelograms from a point in the line opposite. The first of these altitudes is sometimes called the length of the prism. 13. A prism, of which the ends or bases are perpendicular to the other sides, is called a right pris7n y any other is an ohluiue pristn. 14. A parallelopiped is a prism of which the bases are par- allelograms. 15. A parallelopiped of which the bases and the other sides are rectangles, is said to be rectangular. 16. A cube is a rectangular parallelopiped, which has all its six sides squares. 17. A sphere is a solid figure described by the revolution of a semicircle about its diameter, which remains unmoved. 18. The axis of a sphere is the fixed straight line about which the semicircle revolves. 19. The center of a sphere is the same as that of the generat- ing semicircle. 20. A diameter of a sphere is any straight line which passes through the center, and is terminated both ways by its surface. 21. A cone is a solid figure described by the revolution of a riirht-aniiled triangle about one of the legs, which remains fixed. If the fixed leg be equal to the other leg, the cone is called a right-angled cone ; if it be less than the other leg, an obtuse- angled, and if greater, an acute-angled cone. 22. The axis of a cone is the fixed straight line about which the triangle revolves. 23. The base of a cone is the circle described by the leg ■which revolves. 10 146 THE ELEMENTS OF [book VI. 24. A cylinder is a solid figure described by the revolution of a rectangle about one of its sides, which remains fixed. 25. The axis of a cylinder is the fixed straight line about which tlie rectano-le revolves. 26. The bases or ends of a cylinder are the circles described by the two revolving opposite sides of the rectangle. 27. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals. PROPOSITIONS. Prop. I. — Theor. — One part of a straight line can not be in a plane and another part above it. J^et EFGH be a plane, then the straight line AB will be wholly in the plane. By def 1, Book VI., and def 7, Book L, AB, being a straight line in the plane EFGH, is wholly in that plane, and can not have one part in the plane and another part above it. Cor. 1. Hence two straight lines which cut one another are in the same plane ; so also are three straight lines which meet one another, not in the same point. Cor. 2. Hence, if two planes cut one another, their common section is a straight line. Prop. H. — Theor, — If a straight li?7£ be perpendicular to each of two straight lines at their point of intersection^ it is also perpendicvlar to the plane in which they are. Let the straight line EF be perpendicular to each of the straight lines AB, CD at their intersection E ; EF is also per- pendicular to the plane passing through AB, CD. T.ake the straight lines EB, EC equal to one another, and join BC ; in BC and EP" take any points G and F, and join EG, FB, FG, FC. Then, in the triangles BEF, CEF, BE is equal to CE; EF common; and the angles BEF, CEF are equal, being (hyp.) right angles; therefore (I. 3) BF is equal to CF. The triangle BFC is therefore isosceles ; and (IL 5, BOOK VI.] EUCLID AND LKGENDRE. IIT cor, 5) the square of BF is equivalent to the square of FG and the rectangle BG.GC. Foi" the same reason, because (const.) the triangle BEG is isosceles, the square of BE is equivalent to the square GE and the rect- angle BG.GC. To each of these add the square of EF ; then tlie squaies of BE, EF are equivalent to the squares of GE, EF, and the rectangle BG.GC, But (T. 24, cor, 1) the squares of BE, EFare equivalent to the square of V>F, because BEF is a right angle; and it has been shown that the square of BF is equiva- lent to the .square of FG and the rectangle BG.GC; therefore the square of FG and the rectangle BG.GC are equivalent to the squares of GE, EF, and the rectangle BG.GC. Take the rectangle BG.GC from each, and there remains the square of FG, equivalent to the squares of GE, EF ; wherefore (I. 24, cor.) FEG is a right angle. In the same manner it would be proved that EF is perpendicular to any other straight line drawn through E in ihe plane passing through AB, CD. But (VI. def 1) a straight line is perpendicular to a plane when it makes right angles with all straight lines meeting it in that plane; therefore EF is perpendicular to the plane of AB, CD. Where- fore, if a straight line, etc. Cor. Hence (VI. def. l) if three straight lines meet all in one point, and a straight line be perpendicular to each of them at that point, the three straight lines are in the same plane, Pkop, III. — Theor, — Tf tioo straight lines he perpendicular to the same plane, they are parallel to one another. Let the straight lines AB, CD be at right angles to the same plane BDE; AB is parallel to CD. Let them meet the plane in the points B, D ; join BD, and draw DE perpendicular to BD in the plane BDE ; make DE equal to AB, and join BE, AE, AD. Then, because AB is perpendicular to the plane, each of the angles ABD, ABE is (VL def 1) a right an- gle. For the same reason, CDB, CDE are right angles. And because AB is equal to 148 THE ELKMENTS OF [lIOOK VI. DE, BD common, and the angle ABD equal to BDE, AD is equal (I. 3) to DE. A-^ain : in the triangles ABE, ADE, AB is equal to DE, BE to AD, and AE common ; therefore (I. 4) the angle ABE is equal to EDA ; but ABE is a right angle ; therefore EDA is also a right angle, and ED perpendicular to DA ; it is also perpendicular to each of the two BD, DC; therefore (VI. 2, cor.) these three straight lines DA, DB, DC are all in the same plane. But (VI. 1, cor. 1) AB is in the plane in which are BD, DA ; therefore AB, BD, DC are in one plane. Now (hyp.) each of the angles ABD, BDC is a right angle ; theiefore (L 16, cor. 1) AB is parallel to CD. Wherefore, etc. Cor. 1. Hence (I. def. 11) if two straight lin^s be parallel, the straight line drawn from any point in the one to any point in the other is in the same plane with the parallels. Cor. 2. Hence, also, if one of two parallel straight lines be perpendicular to a plane, the other is also perpendicular to it. Also, two straight lines which are each of them T)arall(l to the same straight line, and are not both in the same plane with it, are parallel to one another. Scho. The same has been proved (I. 1 7) respecting straight lines in the same plane; therefore, universally, straight hues •which are parallel to the same straight line, are parallel to one another. Prop. TV. — Tfieor. — If two^ straight lines meeting one an- other be parallel to two others that meet one another^ and are not in the same plane with the first two ; tlie first two and the other two contain equal angles. Let the straight lines AB, BC, which meet one another, be parallel to DE, EF, which also meet one another, but are not in tlie same ])lane with AB, BC ; the angle ABC is equal to DEF. Take BA, BC, ED, EF, all equal to one another, and join AD, CF, BE, AC, DF. Because BA is equal and parallel to ED, thereft)re AD is (I. 15, cor. l) both equal and parallel to BE. For the same reason, CF is equal and ^ N E \ D f BOOK VI.] EUCLID AND LEGKNDKK. 149 parallel to BE. Therefore AD and CF Ix'ing each of them parallel to BE, are (VI. 3, cor, 2) parallel to one another. They are also (I. ax, 1) equal; and AC, DF join them toward the same parts; and therefore (I. 15, cor. 1) AC is equal and parallel to DF. And because AB, BC are equal to DE, EF, and AC to DF, the angle ABC is equal (I. 4) to DEF. There- fore, if two stiMight line?, etc. Schu. Or supplemental ones, as will be plain after the de- monstration here given, if AB be produced through 15. This generalizes the third corollary to the sixteenth proposition of the first book. Prop. V. — Prob. — To clrmo a straight line perpendicular to a plane, from a given point above it. Let A be the given point above the plane BII ; it is required to draw from A a perpendicular to BH. In the plane draw any straight line BC, and (I. 8) from A draw AD perpendicular to BC. Then, if AD be also perpen- dicular to the plane BH, the thing required is done. But if it be not, from D (I. 7) draw DE, in the i)lane BII, at right angles to BC ; from A draw AF perpen- dicular to DE; and through F draw (I. 18) Gil parallel to^ BC. Then, because BC is at right an- gles to ED and DA, BC is at right angles (VI. 2) to the plane passing through ED, DA ; and GH being parallel to BC, is also (VT. 3, cor. 2) at right angles to the plane through ED, DA ; and it is therefore perjjendicular (VI. def 1) to every straight line meeting it in that ])lane; GM is consequently perpendicular to AF. Therefore AF is per- pendicular to each of the straight lines Gil, DE ; and conse- quently (VI. 2) to the plane BHj wherefore AF is the perpen- dicular required. Prop. VI. — ^Prob. — To draw a straight line perpendicular to a given plane from a point given in the plane. 150 THE ELEMENTS OF [book VI. D B Let A be the point given in the plane ; it is required to draw a perpendicular from A to the plane. From any point B, above the plane, draw (VI. 5) BC per- pendicular to it ; if this pass through A, it is the perpendicular required. If not, from A draw (I. 18) AD parallel to BC. Then, be- cause AD, CB are parallel, and one of them, BC, is at right angles to tlie given plane, the other, AD, is also (VI. 3, cor. 2) at right angles to it. Scho. From the same point in a given plane there can not be two straight lines drawn perpendicular to the plane upon the same side of it ; and there can be but one perpendicular to a plane from a point above it. Cor. Hence planes to which the same straight line is perpen- dicular, are i^arallel to one another. E Prop. VII. — Theor. — Two planes are parallel^ if tico straiglit lines which meet one another on one of them be parallel to two which meet on the other. Let the straight lines AB, BC meet on the plane AC, and DE, EF on the plane DF ; if AB, BC be parallel to DE, EF, the plane AC is parallel to DF. From B draw (VI. 3, cor. 2) BG perpendicular to the plane DF, and let it meet that plane in G ; and through G draw (L 18) Gil parallel to ED, and GK to EF. Then, because BG is perpendicu- lar to the plane DF, each of the angles BGII, BGK is (VI. def 1) a right an- gle ; and because (VI. 3, cor. 2) BA is parallel to Gil, each of thom being par- allel to DE, the angles GBA, BGII are together equal (1. 1 6, cor. 1 ) to two right angles. But BGII is a risjht aniile ; therefore, also, GBA is a right angle, and GB perpendicular to BA. For the same reason, (iB is pei-pendicular to BC. Since, therefore, GB is perpendicular to BA, BC, it is perpendicular (VL 2) to the plane AC; and (const.) it is perpendicular to the plane DF. But (VI. 6, cor.) planes to which the same straight BOOK VI.] EUCLID AND LEGENDRE. 151 line is perpendicular are parallel to one another; therefore the planes AC, DF are parallel. Wherefore, two planes, etc. Cor. 1. Hence, if two parallel planes be cut by another plane, their common sections with it are parallels. Cor. 2. If a straight line be perpendicular to a plane, every plane which passes through it is perpendicular to that plane. C-yr. 3. Hence, if two planes cutting one another be each perpendicular to a third plane, their common sectiiBU is perpen- dicular to the same plane. Paop. VIH. — Theor, — If two straight lines be cut by parallel planes, they are cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B ; C, F, D; as AE : EB : : CF : FD. Join AC, BD, AD, and let AD meet KL in X ; join also EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are (VL 7, cor.) parallel. For the same reason, because GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel. Then (V. 2) because EX is parallel to BD, a side of the triangle ABD, AE : EB : : AX : XD ; and be- jf cause XF is parallel to AC, a side of the triangle ADC, AX : XD : : CF : FD ; and it was proved that AX : XD : : AE : EB ; therefore (IV. 7) AE : EB : : CF : FD. Wherefore, if two straight lines, etc. Prop. IX — Theor. — If a solid angle be contained by three plane angles, any two of them, are greater than the tliird. Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB ; any two of these are greater than the third. If the angles be all equal, it is evident that any two of them are greater than the third. But if they be not, let BAC be that angle which is not less than either of the other two, and 153 THE ELEMENTS OF [book VI. is rrreator than one of thorn, DAB; aiul make in tlie plane of BA, AC the angle BAE equal (I. 13) to DAB; make AE equal to AD; through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. Then, in tlie triangles BAD, BAE, because DA is equal to AE, AB common, and the angle DAB is equal to EAB, DB is equal (I. 3) to BE. Again : because (I. 2], cor.) BD, I C are greater than CB, and one of them, BD, has been pioved equal to BE, a part of CB, there- fore the other, DC, is greater (I. ax. 5) than the remaining part, EC. Then, because DA is equal (const.) to AE, and AC common, but the base DC greater than the base EC, therefore (T. 21) the angle DAC is greater than EAC, and (const.) the angles DAB, BAE are equal ; Avherefore (I. ax. 4) the angles DAB, DAC are together greater than BAE, EAC, that is, than BAC. But BAC is not less than either of the angles DAB, DAC ; therefore BAC, with either of them, is greater than the other. Wherefore, if a solid angle, etc. Cor. 1. If every two of three plane angles be greater than the third, and if the straight lines which contain them be all equal, a ti-iangle may be made, having its sides equal, each to each, to the straight lines that join the extremities of those equal straight lines. Cor. 2. If two solid angles be each contained by three plane angles, equal to one another, each to each ; the planes in which the equal angles are, have the same inclination. Cor. 3. Two solid angles, contained each by three plane an- gles which are equal to one another, each to each, and alike situated, are equal to one another. Cor. 4. Solid figures contained by the same number of equal and similar planes, alike situated, and having none of their solid angles contained by more than three plane angles, are equal and similar to one another. Cor. 5. If a solid be contained by six planes, two and two of which are parallel, the opposite planes are similar and equal parallelograms. BOOK VI.] EUCLID AND LEGENDKE. 153 Prop. X. — Theor. — Ecery solid angle is contained hy plane angles^ which are together less than four right angles. Let the solid angle <it A be contained hy any nnmher of plane angles, BAC,^ CAD, DAE, EAF, FAB ; these together are less than four right angles. Let the planes in which the angles are be cut by a plane, and let the connnon sections of it with those planes be BC, CD, DE, EF, FB. Then, because the solid angle at W is contained by three plane angles, CBA, ABF, FBC, of which (VL 9) any two are greater than the third, CBA, ABF are greater than FBC. For the same reason, the two plane angles at each of the ])oints C, D, E, F, viz., the angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the ansfles of the figure BCDEF. Therefore all the angles at the bases of the triangles are together greater than all the angles of that figure; and because (L 20) all the anoles of the triangles are together equal to twice as many right angles as there are triangles — that is, as there are sides in the figure BCDEF; and that (L 20, cor. 1) all the angles of the figure, together with four right angles, are likewise equal to twice as many riuht angles as there are sides in thefi<»-ure; therefore all the angles of the triangle are equal to all the angles of the figure, together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the figure, as has been proved ; wherefore the re- maining angles of the triangles, viz., those at the vertex, wiiich contain the solid angle at A, are less than four ri^-ht angles. Therefore, every solid angle, etc. Scho. This proposition does not necessarily hold, if any of the angles of the rectilineal figure BCDEF be re-entrant; or, ■which is the same, if any of the planes foi-ming the solid angle at A, being produced, pass through that angle. Prop. XI. — Prob. — To make a solid angle having the angles containing it equal to three given jL>la?ie angles^ any two of 154 THE ELEMENTS OF [book VI. which are greater than the third^ and all three together less than four right angles. Let B, E, H be given plane angles, any two of which are greater than the third, and all of them together less than four right angles; it is required to make a solid angle contained by plane angles equal to B, E, H, each to each. Fjom the lines containing the angles, cut oif BA, BC, ED, EF, IIG, HK, all equal to one another, and join AC, DF, GK ; then (VI. 9, cor. 1) a triangle may be made of three straight lines equal to AC, DF, GK. Let this (L 12) be the triangle LMN, AC being equal to LM, DF to MN", and GK to LN. About LiMN describe (IIL 25, cor. 2) a circle, and draw the radii, LO, JMO, NO ; draw also OP (VI. 6, cor.) perpendicu- lar to the plane LMN. Then, any of the radii LO, MO, NO is less than AB. Find (I. 24, cor. 3) the side of a square equiva- lent to the difference of the squares of AB and LO ; make OP equal to that side, and join PL, PM, PN ; the plane angles LPM, MPN, and NPL form the solid angle required. For, since OP is (const.) perpendicndar to the ])lane LIVIN, the angles LOP, MOP, NOP are (VI. def 1) right angles; and therefore, since in the triangles OLP, OMP, ONP the sides OL, OM, ON are equal, OP common, and the contained angles equal, the bases LP, MP, NP are (I. 3) all equal. Also (const ) the square of AB is equivalent to the squares of LO, OP; and (I. 24, cor. 1) the vsquare of LP is also equivalent to the squares of LO, OP, because LOP is a right angle. There- fore (I. ax. 1) the square of AB is equal to the square of PL, and (I. 23, cor. 3) AB to PL ; and hence all the straight lines LP, IMP, NP, liA, BC, ED, etc., are equal. Then, in^lhe tri- angles LPM, ABC, the sides AB, BC are equal to LP, PM, BOOK VI.] EUCLID AND LEGENDRE. 155 eacli to each ; and (const.) AC is equal to LM ; therefore (I. 4) the angles AI>C, LPM are equal ; and it would be shown in a similar manner, that the angle E is equal to ]\IPN, and H to NPL. The solid angle at P, therefore, being contained by- three plane angles, which are equal to the three given angles, B, E, H, each to each, is such as was required. Piiop. XTI. — Theor. — A plane cutting a solid., and parallel to two of its opposite planes., divides the whole into two solids^ the base of one of which is to the base of the other as the one solid is to the other. Let the solid BC be cut by the plane GF which is parallel to the opposite planes BY and IIC, and divides the whole into two s( lids, BFand GC; as the base of the first is to the base of the second, so is BF to GC. N M H G B ./LTP /■ V LJ-L V t Q K Produce YC both ways, and take on one side any number of straight lines, YK and KL, each equal to YF, and complete parallelograms similar and equal to BY. Then, because BY, YK, and KL are all equal, the parallelograms on them are also equal (I. 15, cor. 5); and for same reason the parallelograms on the other side of YC, on the straight lines CQ, QS, each equal to FC, are also equal ; therefore three ])lanes of the solid XK are equal and similar to three planes of KB, as also to three planes of YG. But (VL 9, cor. 5) the three planes oppo- site to these three are equal and similar to them hi the several solids ; and none of their solid angles are contained bv more than three plane angles; therefore (VI. 9, cor. 4) the solids XK, KB, and YC are equal. For the same reason, FH, CM and QN are also equal. Therefore whatever multiple the base LF is of YF, the same multiple is the solid LG of YG. For same reason, whatever multiple the base FS is of FC, the same 156 THE ELEMENTS OF [BOOK TI. multiple is the solid FT of FH. And if the base LF be equal to SF, the solid LG is equal (VI. 9, cor. 4) to FT ; if greater, greater; and if less, less. Therefore (IV. def 5) as the base YP^ is to the base FC, so is the solid BF to the solid GC ; wherefore, a plane, etc. Prop. XTII. — Prob. — At a given point in a given straight line, to make a solid angle equal to a given solid angle conr tained by three plane angles. Let A be a given point in a given straight line AB, and D a given solid angle contained by the three plane angles EDO, EDF, F'DC; it is required to make at A in the straight line AB a solid an le equal to the solid angle D. In DF take any point F, from which draw (VI. 5) FG per- pendicular to the plane EDO, meeting that plane in G ; join DG, and (I. 13) make the angle BAL equal to EDO, and in the plane BAL make the angle BAK equal to EDG ; then make AK equal to DG, and (VI. 6) draw KH perpendicular to the plane BAL, and equal to GF, and join AH. Then the solid angle at A, which is contained by the plane angles BAL, BAH, HAL, is equal to the given solid angle at D. Take AB DE equal to one another; and join IIB, KB, FE, GE; and (VI. def 1) because FG is perpendicular to the plane EDO, FGD, F'GE are right angles. For the same reason, HKA, HKB are right angles; and because KA, AB are equal to GD, DE, each to each, and contain equal angles, BK is equal (I. 3) to EG ; also KH is equal to GF, and HKB, FGE are right angles ; therefore HB is equal to FE. Again : be- cause AK, KH are equal to DG, GF, and contain right angles. AH 19, equal to DF; also AB is equal to DE, and IIB to FE; therefore (L 4) the angles BAH, EDF are equal. Again : since BOOK VI.] EUCLID AND LEGENDRE. 157 (const.) the angle BAL is equal to EDC, and BAK to EDG, tlie remaining angles KAL, GDC are (I. ax. 3) eqnal to one another; and, by taking AL and DC eqnal, and joining LH, LK, CF, CG, it would be proved, as in the foregoing part, that the angle HAL is equal (I. 4) to FDC. Therefore, because tlie three plane angles BAL, BAII, HAL, which contain the solid angle at A, are equal to the three EDC, EDF, FDC, which contain the solid angle at D, each to each, and are situated in the same order, the solid angle at A is equal (VL 9, cor. 3) to the solid angle at D. Therefore, what was required has beeu done. Prop. XIV. — Theor. — If a parallclopiped he cut by a plane passing through the diagonals of two of the opposite planes^ it is bisected by that plane. Let AB be a parallclopiped, and DE, CF the diagonals of the opposite parallelograms AH, GB, viz., those which join the equal angles in each. Then (VI. 3, cor. 2) CD, P^E are paral- lels, because each of them is parallel to GA ; wherefore (VL 3, cor, 1) the diagonals CF, DE are in the plane in which the par- allels are, and (VL 8) are themselves parallels. Again: be- cause (L 15, cor. 1) the triangle CGF is equal to CBF, and DAE to DHE; and that (VL 12) the parallelogram CA is equal and similar to the opposite one BE; and GE to CH ; therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal (VL 9, cor. 4) to the prism contained by the two tri- angles CBF, DHE, and the three parallelo- grams BE, CH, EC ; because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane an- gles. Therefore, if a parallclopiped, etc. Seho. The insisting litres of a parallclopiped are the sides of the parallelograms between the base and the opposite plane. Cor. In a parallclopiped, if the sides of two of the opposite planes be each bisected, the common section of the planes pass- ing through the points of division, and any diagonal of the par- allelepiped bisect each other. 158 THE ELEMENTS OF [book VI. Pkop. XV. — Theor. — ParaUelopipeds upon the same base^ and of the same altitude, the insisting lines of which are ter- minated in the same straight lines of the />^t</ie opposite to the base, are equal to one another. Let the parallelopipeds All, AK (2fl fig.) be upon tlie same base AB, and of tlie same altitude; and let their insisting lines AF, AG, LM, LN be terminated in the same straight line F'N, and CD, CE, BH, BK in the same DK ; AH is eqlial to AK. First, let the parallelograms DG, HN, whicli are opposite to the base AB, have a comtnon side HG. Then because AH is cut by the plane AGHC passing through the di- agonals AG, CH of the opposite planes ALGF, CBHD, All is bisected (VI. 14) by the plane AGHC. For the A L same reason, AK is bisected by tlie plane LGHB through the diagonals LG, BH. Therefore the solids AH, AK are equal, each of them being double of the prism contained between the trian- gles ALG^ CBH. But let the parallelograms DM, EX, opposite to the base, have no common side. Then, because CH, CK are parallelo- grams, CB is equal (I. 15, cor. 1) to each of the ojjposite sides DH, EK; wherefore DH is equal to EK. P^'rom 1)K take sep- arately EK, DH; then DE is equal to HK ; wherefore, also (I. 15, cor. 5), the ti'iangles CDE, BHK are equal ; and (I. 15, cor. 5) the parallelogram DG is equal to HN. For the same reason, the triangle AFG is equal to LMN; and (VI. 12) the parallelo- gram CF is equal to BM, and CG to BN ; for they are oppo- site. Therefore (VI. 9, cor. 4) the prism which is contained by the two triangles AFG, CDE, and the three jiarallelograms AD, DG, GC, is equal to the prism contained by the two tri- angles LMN, BHK, and the three parallelograms BM, MK, KL. If, therefore, the prism LMNBHK be taken from the solid of which the base is the parallelogram AB, and in which FDKN is the one opposite to it ; and if from the same solid there be BOOK VI.] EUCLID AND LEGENDRE. 150 taken the prism AFGCDE, the remaining solids AH, AK are equal. Therefore, parallelopipeds, etc. Cor. 1. Also parallelopipeds upon the same base and of the same altitude, the insisting lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another. Cor. 2. Hence parallelopipeds which are upon equal bases, and of the same altitude, are equal to one another (I. 15, cor. 5). Co:: 3. Parallelopipeds which have the same altitude, are to one another as their bases (VI. 12). Cor. 4. If thei-e be two triangular prisms of the same alti- tude, the base of one of which is a parallelogram, and that of the other a triangle ; if the parallelogram be double of the tri- angle, the prisms are equal (I. 15, cor. 4). o M -o4 V. y\ jy Vti N Prop. XVI. — Theor. — Similar solids are one to another in the tnplicate ratio of their homologous sides. Let AB, CD be similar parallelopipeds, and the side AE homologous to CF ; AB has to CD the triplicate ratio of that which AE has to CF. Produce AE, GE, HE, and in these produced take EK equal to CF, EL equal to FN, and EM equal to FR ; and com- plete the parallelogram KL and the solid KO. Because IvE, EL are equal to CF, FN, and the angle KEL equal to the angle CFN, since it is equal to the angle AEG, which is equal to CFN, be- cause the solids AB, CD are similar; therefore the parallelogram KL is similar and equal to CN. For the same reason, the parallelogram MK is similar aii<l equal to CR, and also OE to FD. Therefore three paral- lelograms of the solid KO are equal and similar to three paral- lelograms of the solid CD; and (VI. 12) the three opposite ones in each solid are equal and similar to these. Therefore (VI. 9, cor. 4) the solid KO is equal and similar to CD. Com- 100 THE FXKMENT8 OF [book VI. N "sC Jj R plete the paralIeloG:rani GK, and the solids EX, LP upon the bases GK, KL. so that EH may be an insisting line in each of them ; and thus they are of the same altitude with the solid >^B. Then, because the sol- ids AB, CD are similar (VI. def. 8, and alternately), as AE is to CF, so is EG to FN, and so is EH to FR ; and FG is equal to EK, and FN to EL, and FR to EM ; there- fore, as AE to EK, so is p]G to EL ; and so is HE to EM. But (V. 1) as AE to EK, so is the parallelogram AG to GK ; and as GE to EL, so is GK to KL ; and as HE to EM, so is BE to KM ; therefore (IV. 7) as the parallelogram AG to GK, so is GK to KL, and PE to KM. But (VI. 13) as AG to GK, so is the solid AB to EX ; and as GK to KL, so is the solid EX to PL ; and as PE to KM, so is the solid PL to KG; and therefore (IV, 7) as the solid AB to EX, so is EX to PL, and PL to KG. But if four magnitudes be continual proportionals, the first is said to have to the fourth the trijtlicate ratio of that which it has to the sec- ond; therefore the solid AB has to KG the triplicate ratio of that which AB has to EX. But as AB is to EX, so is the par- allelogram AG to GK, and the straight line AE to EK. Wherefore the solid AB has to the solid KG the triplicate ratio of that which AE has to EK; and the solid KG is equal to CD, and the straight line EK to CF. Tliei-efore the solid AB has to CD the ti-i]tlicate ratio of that which the side AE has to the homologous side CF. Since (VI- 15) CNF is half of the base of CD, CRF half of CR, and NRF half of NR, which planes form the triangular pyramid CNF, K ; and since the triangular pyramid AGE, II is formed ia a similai" manner, and the parallelo])iped AB has to the paral- lcl()pi|)C(l CI) the triplicate ratio of that which the side AE has to tlie side CF ; hence (IV, ax. 1) the triangular ])yramid CNF, R has to the triangular j)yrainid AGE, II the triplicate ratio of tliat which AE has to CF. And (L 20, cor. 1) all polygons can be divided into triangles; therefore (V. 14) solids BOOK VI.] EUCLID AND LEGENDRE. 131 on similar bases liave to one another tlie triplicate ratio of lioni<)'o<r()iis >i(les. Wherefore, siniihir solids aie, etc. Cor. 1. Hence, similar solids of the same or equal bases are to one another as their altitudes — and, conversch', those of the same ov equal altitudes are to one another as their bases. Cor. 2. Hence, also, the bases and altitudes of equivalent polids are reciprocally proportional; and conversely, solids lia villi? their bases and altitudes reciprocally proportional, are equivalent. Cor. 3. Hence, cones and cylinders upon equal bases are a3 their altitudes, and their bases and altitudes are reciprocally proportional when the cones and cylinders are equivalent. And similar cones and cylinders have to one another the tripli- cate ratio of that which the diameters of their bases have ; and gpheres have the triplicate ratio of their diameters (V. 14 and VI. 1(3). Cor. 4. From this it is manifest, that if four straight lines be continual proportionals, as the first is to the fourth, so is the parallelopiped described from the first to the similar solid simi- larly described from the second ; because the first strais^ht line has to the fourth the triplicate ratio of that which it has to the second. Cor. 5. Parallelopipeds contained by parallelograms equi- angular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio which is the same with the ratio compounded of the ratios of their sides (V. 14, cor. 7). Cor. 6. The bases and altitudes of equivalent parallelopipeds are reciprocally proportional; and (2) if the bases and altitudes be reciprocally proportional, the pai'alleloj)ipeds are equivalent. Prop. XVH. — Tiieor. — Every pyramid is one third the prism of the same b-fse and altitude, and every co?ie is one third of the cylinder with the same base and altitude., or every pyramidal solid is one third the solid of the same base and altitude. Let there be a prism of which the bases are the triangles ABC, DEF ; the prism may be divided into three equal tri- angular pyrajnids. 11 162 THL ELEMENTS OF [book VL- Join BD, EC, CD; and because ABED is a parallelogram, and BD its diagonal, the triangles ABD, EBD are (I. 15, cor. 1) equal; therefore (VI. IC, cor. 3) the pyramid of which the base is ABD, and vertex C, is equal to tho pyramid of which the base is EBD and vertex C. But EBC may be taken as the base of this pyramid, and D as its veitex. It is there- fore equal (VI. 16, cor. 3) to the pyramid of which ECF is the base, and D the vertex; for they l)ave the same altitude, and (I. 15, cor. 1) equal bases ECF, ECB; and it has been al- ready proved to be equal to the pyramid ABDC. Therefore the prism A13CDEF is divided into three equal pyramids having tri- angular bases, viz., into the pyramids ABDC, EBDC, ECFD. Therefore, every ti'iangular piism, etc. Now every polygon can be divided into triangles (I. 20, cor. 1) ; hence every j)risni with a polygonal base can be divided into i)risms havinij trianirular bases : and as each of these tii- angu.lar prisms can be divided into three equivalent triangular pyramids, therefjre every pyramid is one third the prism of Bame base and altitude. As it has been shown (V. 14) that nil surfaces are to one another in the duplicate ratio of their liomologous sides, and (VI. IC) all solids are to one another in the trij)licate ratio of their homologous sides, it follows that all Bolids of similar bases and altitudes have the same propoition to one another (VI. 16, cor. 3); hence, cones having similar bases and altitudes to cylinders, have the same proportion to those cylinders which pyramids have to prisms of similar bases and altitudes; therefore the cones are one thiid the cylinders, as it is evident that the section of the cone and cylinder is similar to the section of the pyramid and prism of whatever regular and similar base. For ABC can be the section of a cone and of a pyramid of any regular base, and ABED can be the section of a prism of any similar base, and pcction of a cylinder of similar base with cone ; therefore the same proportion which regulates the respective magnitudes BOOK VI,] EUCLID AND LEGENDRE. 163 of tlie pyramid and pi-ism, also regulates the respective mag- nitudes of cone and cylinder — as all surfaces are (V. 14) to one another in the duplicate ratio of their homologous sides, and (VI, 16) all solids are to one another in the triplicate ratio of their homologous sides. Cor. 1, p]very sphere is two thirds of its circumscribing cyl- inder. Let ABCD be a cylinder circumscribing the sphere EFGH ; then will the sphere EFGH be two thirds of the cylin- der ABCD, For, let the plane AC be a section of the sphere and cylinder through the center J, Join AJ, B.T. Also, let FJII be paiallel to AD or BC, and OKL be parallel to AB or DC, the base of the cylinder, the latter line, KL, meeting BJ in M, and the circular section of the sphere in N. Then, if the whole plane HFBC be conceived to revolve about the line IIF as an axis, the squai'e FG will describe a cylinder, AG ; the quadrant JFG will describe a hemis- phere, EFG; and the triangle JFB will describe a cone, JAB. Also, in the rota- El ^^ |o tion, the three lines or parts KL, KN, KM, as radii, will describe correspond- ing circular sections of those solids, viz., D u c KI^, a section of the cylinder; KN, a «;ection of the sphere; and KM, a section of the cone. Now, FB being equal to FJ or JG, and KL parallel to FB, then by eimilar triangles JK is equal to KM. And since in the right- angled triangle JKN, JN-OJK--|-KN- (L 24, cor. 1), and be- cause KL is equal to the radius JG or JN, and KM is equal to JK, therefore KL-<;>KM"-|-KN-; and because circles areas the squares of their diameters or the squares of their radii, therefore (V. 14, cor. 2) circle of KL is equivalent to circles of KM and KN, or section of cylinder is equivalent to both cor- responding sections of sphere and cone. And as this will always be, it follows that the cylinder EB, which is all the fonner sections, is equivalent to the hemisphere EFG and cone JAB, which are all the latter sections. But JAB is one third of the cylinder EB (VI. 1 7) ; therefore (I. ax. 3) the hemisphere EFG is two thirds of the cylinder EB. Cor. 2. If the parallelogram BEGC be revolved around the K^ mX ' \ /^ J V y 16i THE ELEMENTS OF [book VI. fixed axis BC, it will generate a cylincler (VI. def. 24) ; the semicircle BNC will generate a sphere (VI. def. 17) ; and the triangle BGC will generate a cone (VI. def. 21). The cone Avill be one third the cylinder (VI. 17), and the sphere will bo two thirds the same cylinder (VI. 17, cor 1). The triangle BOP having one half altitnde and one half base of the triangle BGC, will generate a cone one eighvh of the cone generated by the triangle BGC (VI. 16, cor. 3) ; hence, one twelfth of the cylinder generated by the square BENP;and the cone generated by the triangle BNP is one half cone gener- ated by the triangle BGC (VI. 16, cor. 1) ; hence, four times cone generated by the triangle BOP. And the hemisphere gen- erated by the quadrant BNP is two thirds cylinder generated by the square BENP VI. 17, cor. 1), or eight times cone gen- erated by the triangle BOP. Let the triangle BSN be described on BN, equal to the tri- angle BON (I. 23,and 15, cor. 4). Then the trapezium BSNP will generate a solid equivalent to the sum of a cylinder one half cylinder generated by the square BENP, and a cone one sixth of the same cylinder, or eight times the cone generated by the triangle BOP, making a solid equivalent to the hemis- phere generated by the quadrant BNP on the same radius PN" and same altitude BP. But the triangle BNP is common to both the trapezium B>NP and the quadiant BNP, and gener- ates in each case the solid equivalent to four times cone gener- ated by the tiiangle BOP ; therefore the segment BN and the triangle BSN generate an equivalence of solid, or four times cone generated by the triangle BOP; consequently the seg- ment BN and the triangle BSN are equivalent (I. ax. 1). Again : the triangle BNP generates a cone one third the cylinder generated by the square BENP (VI. 17), and the quadrant BNP generates a heniisphere two thiids of the same cylinder (VI. 17, cor. 1). The triangle BNP is one half the square BlilNP. Now, the trapezium BSNP, equivalent to three fourths of the BOOK VI.] EUCLID AND LEGENDRE. 165 gquiire, on same raflius aiul altitude as the square, generates a solid two thirds of the solid generated by the sqnuiv, and the quadrant BNP witli same radius and altitude as the square BENP generates an equivalent solid with the trapezium BSNP. That an equivalence of surfaces ujion the same radius •will generate an equivalence of solids can be illustrated by- taking a trapezium greater than the trapezium BSNP, having tlie same radius. It can easily be shown that the greater trape- zium generates a greater solid than the less trapezium, and in a similar manner it can be shown that a less trapezium than the trapezium BSNP generates a less solid ; hence, very evidently, when a greater surface upon same radius generates a greater solid, and a less surface generates a less solid, equivalent surfaces must generate equivalent solids on the same radius; and, conversely, when we have equivalent solids generated npon the same radius, the generatirig surfaces are equivalent ; therefore (I. ax. l) the quadrant BNP is three fourths of the square BENP, or the semicircle BNC is three fourths of the parallelogram BEGC, or any circle is three fourths of the cir- cumscribing square, or Three Times Square of Badius. Hence, we have a geometrical confirmation of the mechanical construction in scholium to twenty-fifth proposition of book fifth. < TIIER'WISR : Tlie triangle BGC generates a cone one third (VI. 17) of the cylinder generated by tlie rectangle BEGC, and the semi- cii'cle BXC genei'ates a sphere two thirds ( VI. 1 7, cor. 1 ) of the same cylinder ; the s] here is the mean between the cone and cylinder; therefore the semicircle is evidently the meanhe- tween the triangle BGC and the rectangle BEGC, or three fourths of the rectangle BEGC ; or any circle is three fourths square of its diameter, or three times square of its radius. OTHERWISE : Circles are to one another as the squares described on their diameters (V. 14) ; consequently squares are to one another as the circles described on their sides; therefore there is an equality of proportion (V. 24) ; hence, we derived the arith* metical proportion : 166 THE ELEMENTS OF [boOK YI. Rectangle BEGC, semicircle BNC, triangle BGC. The sum of extremes is equivalent to twice the mean ; there- fore we have — Rectangle BEGC -f- triangle BGCo2 semicircle BNC; or, semicircle BNCOi rectangle BEGC + i triangle BGC. Also, the difference between first and second terms of an arithmetical proportion is the same as the difference between the second and third terms, as the diffei'ence between third and fourth terms, and so on ; hence we have — Rectangle BEGC— semicircle BNCOsemicircle BNC -tri- angle BGC ; therefore we get segment BXOtriangle BSN, or one fourth square BENP; consequently, eircle =0 three fourth square of diameter, or tliree times square of radius. Cor. 3. Archimedes discovered the proi)ortion 1, 2, 3 be- tween the cone, sjihere, and cylinder of similar dimensions; but from the previous corollary we obtain the proportion 1, 2, 3, 4 for the cone, sphere, cylinder, and cube of similar diuiensions; because the cube is eight times cube of radius of th^ sjyhere (Vr. 10) ; the cylinder is six times cube of radius of the sphere (VI. 17, cor, 2) ; the sphere infour times cube of radius (f the sphere (VI. 17, cors. 1 and 2) ; and the cone is twice ctibe of radius of the sphere (VI, 17, cors, 1 and 2). Hence the sphere is the mean proportional of the cone, and the cube circum- scribing the si)here, or one half the circumscribiug cube; therefore the surface of the sphere is fur times the area of one of its great circles, or two thirds the surface of the circum- scribing cylinder. Hence, there is the identical proportion be- tween the surfaces of the sphere and cylinder as there is between \}\q solidities of the s[ihere and cylinder. Scho. 1. Therefore the second corollary gives the solution to the long mooted and much vexed question of the Quadra- ture of the Circle, showing that the perplexity of it arose from the uii geometrical sitjyposition (V. 25, scho.) that "the circle is a regular polygon of an infinite inind)er of sides." Hence it is evident that all conclusions derived fi'oni a fdlacious su])])0- sition will give ])erplexity so long as the snpposilion is main- tained, and must necessarily involve conti'adiction*^ to the rigor of geometrical reasoning. And wlicii demonstrations are con- ducted consistently with established definitions, axioms, and BOOK VI.] EUCLID AND LKGKNDRE. 16T propositions, all conclusions derived from tlieni are unimpeach- able, and arc valuable to a system of scientific truths. iicho. 2. Geometry, like all other sciences, is based upon cvviiuuj'undamencal i)rinciples, and a close examination of tliia science reveals the fact that, throughout its whole extent and its vai'ious applications, tlie principle that the siraljht line is the shortest line bcttoeen tv:o fjiven points is the fundamental ])rinciple of the science ; by this principle the dimensions of magnitudes are determined, distances of objects made known, and other useful and practical results ascertained. iSince this jjrinciple is so important, it would be interesting to inquire the reason why it has such manifest usefulness. The angle is a magnitude contained l)y the intersection of two straifjht lii.es^ and the polygon is another magnitude bounded by three or moYG straight lines ^ lience we see how intimate the connec- tion between the straight line and the angle and polygon ; therefore we find that the functions of the angle are straight lines, such as sines, co-sines, tangents, etc. ; and the properties of the polygon ai"e defined by straigld lines, such as its jjerime- ter and apothem ; therefore in all rectilineal magnitudes we discover a use for the straight line above all other lines, and evidently the principle of the straight line has a i)eculiar force to all rectilineal figures; consequently we adopt x}w straight line as a means of measure for all rectilineal magnitudes. The adoption of this m^eans of measure constitutes the straight line a standard by which all measurements of rectilineal magni- tudes are compared. Hence very naturally there is a consist- ency between the measurements and other properties of recti- lineal magnitudes. Now, when w(^ examine the circle or portions of the circle, as the segments, sectors, arcs, etc., we at once discover a nou' coincidence between the curve which bounds them and the straight line which bounds rectilineal magnitudes ; hence, very evidently, the superficies of curvilmear spaces require a pecu- liar coni-ection between tiiem and the bounding curve, as there is a peculiar connection between the superficies of rectilineal spaces and the bounding straight lines. In other words, since "vve derive the measurements and other properties of rectilineal magnitudes from the principle of the straight line^ so we nmst V 168 THE ELEMKNTS OF [book VI. determine tlie measurements and other properties of curvilinear magnitudes from the prini-iple of tlie curve, and thus we per- ceive wh^, when we endeavor to obtain the area of circle by the method of exhaustions, using the straight line as a mean$ of measure, we can get the a2:>proxirnate area only, and ahy it is necessai-y to obtain accurate results to use the [trniciple of the curm. Geometry, in its pi'esent state, is the science of iho straight line, and the introduction of the principle of t!ie curve into geometrical consideration would usher in a distinct science^ but eminently useful in solving problems of cui'vilincar spaces and boundaries which were before unsolved, inasmuch as the approximate results only were given for them. The method of exhaustions is applicable to rectilineal magni- tudes, and its results are consistent with the ])rinciple of the Btraight line, because the straight line is adopted as a means and standard oi' measurement; but since the straight line and curve do not coincide, the principle and propeities of the straight line are not applicable to curvilinear spaces or bound aries ; hence, what is true in one case, becomes absurd in the other. Prop. XVIIT. — Tiieou. — 77te sections of a solid I y parallel planes are similar figures. Let the prism JMN be cut by the two parallel planes AD, FK ; their sections with it are similar figures. For (VI. 7, cor.) the sections have parallel sides (T. 15, cor. 2). The figures AD, FK, thei-efore, have their sides similar, each to each. Their several an- gles are also (VI. 4) equal ; for they ai"e con- tained by straight lines wliich are parallel ; and therefore the figures are similar. Cor. 1. A section of a prism by a plane par- allel to the base is equal and bimilar to the base. Scho. 1. Since (VI. def. 24) a cylinder is de- scribed by the revolution of a rectangle about one of its sides, it is plain that any straight line in the rectangle perpendicular to the fixed line will describe a circle parallel to the base ; and hence BOOK VI.] EUCLID AND LEGENDEE. 169 every section of a cylinder by a plane parallel to the base is a circle equal to the base. Cor. 2. The section of a pyramid by a plane parallel to its base is a fiijure similar to the base. Scho. 2. Since (VI. def. 21) a cone is described by the revo- lution of a right-angled triangle about one of its legs, it is plain that any straight line in the triangle perpendicular to the fixed leg will describe a circle parallel to the base; and the radius of that circle will be to the radius of the base as the altitude of the cone cut off to that of the whole cone. Cor. 3. A section of a sphere by a plane is a circle. Since the radii of the sphere are all equal, each of them being equal to the radius of the describing semicircle, it is plain that if the section pass through the center, it is a circle of the same radius as the sphere. But if the plane do not pass through the center, draw (VI. 5) a perpendicular to it from the center, and draw any number of radii of the sphere to the intersection of its surface with the plane. These radii, which are equal, are the hypothenuses of right-angled triangles, which have the per- pendicular from the center as a common leg; and therefore (I. 24, cor. 2) their other legs are all equal ; wherefore the section of the sphere by the plane is a circle, the center of which is the point in which the perpendicular cuts the plane. 8cho. 3. All the sections through the center are equal to one another, and are greater than the others. The former are therefore called great circles, the latter small or less circles. tScho. 4. A straight line drawn through the center of a circle of the sphere perpendicular to its plane is a diameter of the sphere. The extremities of this diameter are called the poles of the circle. It is plain (I. 24, cor. 2) that chords drawn in the sphere from either pole of a circle to the circumfeience are all equal ; and therefore (III. 16, cor. 3) that arcs of great cir- cles between the pole and circumference are likewise equal. Scho. 5. The pyramid or cone cut off from another pyramid or cone by a plane parallel to the base is similar (VI. defs. 8 and 27) to the whole pyi-amid or cone. Prop. XIX. — Theor. — If the altitixde of a parallelopipedy and the length and p rpendicular breadth of its base be all 170 THE ELKMKNTS OF [bOOK VI. d'vided into parts equal to one another^ the continued p oduct of the numler of parts in the three lines is the nvmhtr ofctiba contained in the parcdldopiped, each cube havinrj the side of its base equal to one (f the parts. P'irst, suppose the ])nralk'loi)ipefl to 1)8 rcctanccular, Tlien ])laiies panvlU'l to the base i)assiiii; tliroiiu'h the points ot'seetion of the nh.itude will evidently divide the so'id into as many- equal solids as there ave parts in the altitude; and each of tliese partial solids will l)e composed of as many cuius as the base contains squares, each equal to a base of one of the cubes. 13ut (I. 23, cor. 4) the number of these squares is the product cf the length and breadth of the base; and hence the entire number of cubes will be equal to the product of the three dimensions, the length, breadth, and altitvide. If the base be not rectanj,ndar, its area (I. 23, cor. 5) will be the product of its lenjxth and j.erpcndicular breadth; and it is evident that the product of this by the altitude will be the number of cubes as before. Lastly, if the iusistiuii lines be not perpendicular to tlie base, still the oblique parallelopi])ed is equal (VI. 15, and cor. 1) to a rectano-ularone of the same altitude; and therefore the num- ber of cubes will be found as before, by multiplying the area of the base by the altitude. Cor. 1. Hence it is evident that the volume, or numerical solid C07itent, or, as it is also called, the solidify, of a parallelo- piped is the product of its altittule and the area of its base both expressed in numbers; and it is ])lain that the same holds in i-egard to any prism whatever, and also in regard to cylinders. Cor. 2. The content of a pyramid or cone is found by multi- plying the area of the base by the altitude, and tak'ug a third of the product. For (VI. 17) a pyramid is a third part of a prism, and (VI. 17) a cone a third jtart of a cylinder, of the same base and altitude. An easy method of comjmting the content of a truncated pyramid or cone, that is, the frustum which remains when a part is cut from the top by a plane parallel to the base, may be thus investigated by the help of algebra. The solid cut off is (VI. 18, scho. 3) similar to the whole; and tlierefore the areas BOOK VI.] EUCLID AND LEOENDRE. 171 of their bases will be proportional to the squares of their cor- respondino^ dimensions, and consequently to the squares of tlieir altitudes. Hence puttins^ V to denote the volume or con- tent of the frustum, 11 and B the altitudes and base of the whole solid, and h and h those of the solid cut oif, if we put ^IP to denote B, since B : 6 : : H* : A", or B : 6 : : qW \ qh^^ we shall have (IV. 2, cor. 1) b = qh'^-^ and therefore (VI. 19, cor. 2) the contents of the whole cone and the part cut off are equal respectively to ^^■IP and ^qlf \ wherelbre \ = ^q(\¥—h')^ ov^ by resolvinnf the second member into factors, V=:i'/(H-+ HA + /r-) {\l-h)=^{qW+q\lh + qh') (H-A). Now ^H^ is equal to B, qli' to 6, qWh to a mean proportional between them, and II — h to the height of the fi'ustum. Hence, to findthe con- tent of a truncated pyramid or cone, add to<jet/ter the areas of its two bases aud a mean proportional between theni^ mxdtiply the sum by the height of the frustum, anddivide the productby 3, This admits of convenient modifications in particular cases. Thus, if the bases be squares of which S and s are sides, and if a be, the altitude of the frustum, we shall have V=ia(S'+Ss-f-0=^»(3Ss+S-— 2Ss+5'); or, Vz^ia{3S5+(S— s)^}=a{S5+i(S— s)'}. Hence, to find the content (f the frustum of a square pyramid^ to the rectangle under the sides of its bases add a third of the square of their difference, and mxdtiply the sum. by the height. It would be shown in like manner (V. 25, scho. and VI. IV, cor. 2), that if II and r be the radii of the bases of the frustum of a cone, and a its altitude, N '^Za{Vxr\-\^—rY\. Solidity of cylinder, 3 x 11" x a. Solidity of cone, R^Xflf. Solidity of sphere, is 4R'. Solidity of spherical sector, is 211' X or. Solidity of spherical segment, when it has two bases, is f R'4-r)xa+ia'; and when it has but one base, ^R-'x« + ^a'. Cor. 3. The content of a polyhedron may be found by divid- 172 THE ELEMENTS OF [book VI. ing it into pyramids, and adding togetlicr tlieir contents. The division into pyramids may be matle either by phuies passini^ throng!) tlie vertex of one of the soliil angles, or by planes passing throngh a point within tlie body. Piiop. XX. — Theor. — The surfaces of two similar polyhe- drons may be divLdedinto the same number cf similar triangles similarly situated. This fullows immediately from the definition (VT. def. 8) of similar bodies bounded by planes, if the sides or faces of the polyhedron be triangles; and any face in the one, and the cor- responding face in the other, which are not triangles, are yet similar, and may be divided (I. 20) into the same number of similai- tiiangles similarlj' situated. Cor. Hence it would be shown, as in tlie fourteenth proposi- tion of the fifth bo()k, that the surfaces of the polyhedions are proportional to any two of their ^similar triangles ; and there- fore they are to one another in the duplicate ratio of the homologous sides of those triangles, that is, of the edges or in- tersections of the similar planes. Hence also the surfaces are proportional (V. 14, coi'. 2) to the squares of the edges. Prop. XXI. — Theor. — Triangular pyramids are similar, if two faces in one of them be similar to twj faces in the other^ each to each, and their inclinations equ<d. Let ABC, abc be the bases, and D, d the vertices of two tn- angular pyramids, in which ABC, DBC are respectively similar to abc, dbc, and the inclination of ABC, DBC equal to that of abc, dbc ^ the pyi'amids are similar. To demonstrate this, it is sufficient to show that the triangles j^ AI>D, ACD are similar to ahd^ acd, i\)Y then the solid angles (Vi. 9, cor. 3) will be equal, each to each, and (VI. def 8) the pyr- amids similar. Since the plane angles at B and b are equal, the inclinations of ABC, DBC, and of vhc, dbc, are (VI. 9, cor, 2) equal ; therefore ABl), abd are equal. Then (hyj..) DB : BC : ; c/i : ic, and BC ; BA : : dc : BOOK VI.] EUCLID AND LEGENDRE. 173 ba/ whence, ex ceqnn, DB : BA : : clb : ha ; and therefore (V. 6) the trianirles ABD, al>d are equiangular, and consequently Bimilar ; and it would be proved in the same maimer that ACD, acd are similar. Therelure (VI. def. 8) the pyramids are similar. Cor. Hence triangular pyramids are similar, if three faces of one of them be respectively similar to three faces of the other. In the triangular pyramids ABCD, ahcd (see the preceding figure), let the faces ABC, ABD, DBC be similar to ahc^ abd, doc, each to each ; the pyramids are similar. For (V. def 1) AD : DB ■.-. ad : dh, and DB : DC : : cf5 : dc ; whence, ex aequo, AD : DC : : ad '. dc. Also DC : CB : : dc : cb, and CB : CA : : cb : ca ; whence, ex ceqi/o, DC : CA : : dc : ca ; and therefore (V. 5) the triangles ADC, adc are equiangular, and (VI. 9, cor. 3, and def. 8) the pyramids are similar. Prop. XXII. — Tiieor. — Similar polyhedrons may be divid- ed nfo the same riumfer of triangular pyratnids, similar, each to each, and similarly situated. Let ABCDEFG and ahcdefg be similar polyhedrons, having the solid angles equal which are marked with the corresponding large and small letters ; they may be divided into the same number of similar triangular jjyramids similarly situated. The surfaces of the polygons may be divided (VI. 20) into the same number of sim lar triangles, similarly situated; then planes passing through any two corresponding solid angles, A, a, and through the sides of all these triangles, except those forming the solid angles. A, a, will divide the polyhedrons into triangular pyramids, similar to one another, and similarly sit- uated. 174 TnE ELEMENTS OF [book VT. The pyramids thus formocl have each one solid anole at the common vertex A or a ; and these solid angles may be of three classes: Is^ those which have two of their faces coincidins with faces of one of the polyhedi'ons; 2f/, those which have only one face coinciding ; and 3o?, those which lie wholly with- in the solid angle A or a. Now those of the first kind in one of the polyhedrons are similar to the corresponding ones in the other, by the corollary to the twenty-first proposition of this Look ; and those of the second kind by the twenty-first From the polyhedrons take two of these similar ])yramids, and the remaining bodies will be similar, as the boundaries common to them and the pyramids are (VI. 21, and cor.) similar trian- gles ; and their other boun<laries are similar, being faces of the proposed polyhedrons. Also the solid angles of the remaining bodies are equal, as some of them are angles of the primitive polyhedrons, and the rest are either trihedral angles which are contained by equal plane angles, or may be divided into such. From these remaining bodies other similar triangular pyramids may be taken in a similar manner, and the process may be con- tinued till only two similar triangular pyramids remain ; and thus the polyhedrons are resolved into the same number of eimilar triangular pyramids. Prop. XXIII. — Pitoc. — To find the diameter of a given sphere. Let A he any point in the surface of the civen sphere, and take any three jjoints B, C, I) at equal distances from A. De- ecribe the triangle hcd having he equal to the distance or chord BC, cd equal to CD, and hd to BD. Find e the center of the circle described about hcd^ and join he; draw a^perpendicular to he^ and make ha eqtial to BA ; draw ^perpendicular to 6a, and af is equal to the diameter of the sphere. BOOK VI."] EUCLID AND LEGENDRE. 175 Conceive a circle to be described through BCD, and E to be its center; that circle will evidently be the section of the sphere by a plane throui;h B, C, D ; and it will be equal to the circle described about bed. Conceive the diameter AEF to be drawn, and liA, BE, BF to be joined. Then, in the right-angled triangles ABE, abe^ the sides AB, BE are respect- ively equal to ab^ be, and therefore (I. 24, cor. 2) the angles A, a are equal. Again : in the right-angled triangles ABF, abf, the angles A, a are equal, and also the sides AB, ab/ hence (I. 14) the Bides AF, a/' are equal; that is, a/ is equal to the diameter of the sphere. Pkop. XXIV. — TiiEOR. — T/ie angle of a spherical triangle is the angle formed by the tangents of the arcs forming the fpherical angle, and is measured by the arc of a great circle described from the vertex as apcle, and intercepted by the sides, produced if necessary. Let BAC be a spherical angle formed by the arcs AB and AC, then it is the same as the angle EAD fonncd by the tan- gents EA ajid DA, and is measured by the arc of a great circle intercepted by the arcs AB and AC, produced if necessary. The tangents AE and AD are both jierpendicular to the common diameter AH (HI. 12), and being in the same planes as the arcs AB and AC, form an angle EAD equal to the spherical angle BAC. ^ Again : the radii FB and FC of the great circle described from the vertex as a pole, being in the same planes as the arcs AB and AC, and perpendicular to AH, are parallel with AE and AD respectively, hence the angle EAD is equal to the angle BFC. But the angle BFC is measured by the arc BC (I. def. 19) ; therefore, also (I. ax. 1), the spherical angle BAC is measured by the aic BC. ^ho. The angles of spli( rical triangles may be compared to- gether by means of the arcs of great circles described from their vertices as poles aid included between the arcs forming 176 THE ELEMENTS OP [bOOK TI. the angles, and it is easy to make a spherical angle equal to a given angle. Cor. 1. If from the vertices of the three angles of a spherical triangle as poles, arcs be described forming a spherical tiiangle, then the veitices of the angles of this second triangle will be respectively poles of the sides of the first, and each angle will be measured reciprocally by a semicircumference less the side of the other triangle opposite to the angle. Because A, B, and C are poles respectively of the arcs FE, ED, and DF, the distances of the poles from the extremities of their respective arcs are, in each case, a quadrant ; hence the extremities of the ai'cs FE, ED, and DF are respectively re- moved the length of a quadrant from the extremities of the arcs AB, BC, and AC ; therefore the extremities of the former arcs are the poles of the latter arcs, each to each. Since A is the pole of the arc GH, the angle BAG is meas- ured by that arc (VI. 24), and F being the pole of AH, FH is a quadrant, and E being the pole of AG, GE is a quadrant; hence Fllf GEO semicircumference; but FH + GEoFE-f GIl, or the arc GH, which measures the angle BAG, is equiva- lent to a semicircumference less the arc FE. In like manner, the angle ABC can be shown to be measured by a semicircum- ference less the arc DE, and the angle ACB to be measured by a semicircumference less the arc DF. And, reciprocally, the angle FDE is measured by the arc LO. But LO + BCoLC 4-B0=O semiciivumfcrence; hence L0=O semicircumference minus BC ; and a similar condition can be shown for the other ambles of the triauijle FED. BOOK VI,] EUCLID AND LEGENDRE. 177 Cor. 2. As each angle of a spherical triangle is less than two right angles, the three angles are less than six right angles. And as the sum of the sides of a spherical triangle is less than the cij'cumference of a great circle, and the angles being meas- ured (VI. 24, cor. 1) by three semicircumferences less the three sides of the polar triangle, taking away the sides, we have the remainder greater than one semicircumference — or the three angles greater than two right angles. Hence the angles of a spherical triangle vary between two right angles and six right angles, without reaching either limit ; therefore two an- gles given can not determine the third. 12 END OF BOOK SIXTH, THE ELEMENTS OF PLANE TRIGONOMETRY. DEFIXITIOXS. 1. TRiGOJfOMETRY is the practical application of geometrical principles for the investigation of ratios of the sides of triangles in connection with the magnitudes of their angles. Yor perspi- cuity, the vertex of the angle is placed in the center of a circle, and the arc of the circumference intercepted by the sides con- taining the angle is used as 2i measure of the angle (L def 19). Let a straight line be supposed to move around a fixed point ; it will make with a stationary line angles which will vary as the line is moved, and when it has passed around until it coincides with the stationary line from which it is supposed to have started, it will have gone over the magnitude of four right angles (I. 9, cor.), the extremity of the movable line will trace the circumference of a circle, and the successive arcs intercept- ed between the movable and stationary lines will give the magnitude of the angles (I. def 19). 2. For the purposes of calculation, a right angle is divided into an arbitrary number of equal parts; each one of these parts is subdivided into other equal parts, and each part of this subdivision undergoes a second subdivision of equal parts, and when particular nicety and precision are desired, there is a third subdivision of equal parts. Thus, the first division of a right angle is into degrees. Among the English mathematicians, a right angle has ninety degrees, which divi- sion is derived from Greek works, and has great antiquity, being used by the remotest ancient mathematicians and astron- omers of whom we have any account. Among some modem French mathematicians, a right angle is divided into one hun- dred degrees, which centesimal division is continued through- out all the various subdivisions. But in the Greek division, which is more generally used on account of the great facility with which 360 can be subdivided, each degree has sixty equal PLANE TKIGONOMETET, 179 parts called minntes^ each minute has sixty equal parts called seconds, and each second is sometimes subdivided in deci- mal parts, and thus the most extreme minuteness can be ob- tained. 3. The symbols for abbreviation in the expression of the value of angles are as follows: °, \ "\ thus, 60°, 15', "lb" are read, sixty degrees, fifteen minutes, and twenty-five seconds. 4. The reason why the right angle is assumed for division is because that angle preserves an inversion between every angle less than a right angle and its complement (L def 20), and a similarity between every angle greater than a right angle and its supplement (I. def 20) ; thus, in the first case, the functions of the angle are inverted in respect to its complement, and in the latter case the functions are the same in respect to its supplement, as will more readily be seen by the seventh defi- nition and following corollaries. 5. The straight line drawn from one extremity of an arc, perpendicular to the diameter passing through the other ex- tremity, is the sine of the angle measured by it ; and the part of that diameter intercepted between the sine and the arc is the versed sine of the angle which it measures. 6. If a straight line touch a circle at one extremity of an arc, the part of it intercepted between that extremity and the diameter produced, which passes through the other, is the tati- gent of the angle which it measures; and the straight line drawn from the center to the remote extremity of the tangent is the secant of the angle. Y. The cosine of an angle is the sine of its complement. In like manner, the coversed sine, cotangent, and cosecant of an angle are respectively the versed sine, tangent, and secant of its complement. The sine, versed sine, tangent, and secant may be denoted by the abbreviated expressions, sin, versin (or vs.), tan, and sec/ and the cosine, coversed sine, cotangent, and cosecant, by cos, coversin (or covs), cotan, and cosec. For the sake of eimplicity, the radius of the circle employed for comparing dif- ferent angles is generally taken in investigations as unity; when this is not done, it is denoted by its initial letter R. The sides of a triangle are often conveniently denoted by tha 180 THE ELEarENTS OF Bmall letters corresponding to the capital ones placed at the opposite angles. Thus, a denotes the side opposite to the an- gle A, etc. To prevent ambiguity, we may read A, B, C ; angle A, angle B, angle C ; while a, 6, c may be called side a, side b, side c. To illustrate the foregoing definitions, let C be the center of a circle, and AB, DE two diameters perpendicular to each other. Through any point F in the circumference draw the diameter FL; draw FG perpendicular to AB, and FI to DE; through A draw AH perpendicular to AB, and therefore (III. 8, cor.) touching the circle in A ; and let it meet LF produced in H ; and, lastly, draw DK perpendicular to DE, meeting FL H D /^ ~^^ ~y !^ T^N. ( / r ^ \/ Q C \ ; k J ^ N \. V E produced in K. Then the arc AF contains the same number of degrees, etc., as the angle ACF ; and FG is the sine of this angle ; FI, or its equal CG, the cosine ; AG the versed sine, and DI the coversed sine ; AH the tangent, and CH the secant j DK the cotangent, and CK the cosecant. From these definitions we derive the following corollaries : Cor. 1. The sine, of an angle ACF is half the chord of double the arc measuring it. For if FG be produced to meet the cir- cumference in N, FN is bisected (III. 2) in G, and (III. 17) the arc FAN in A. Cor. 2. The sine of the right angle ACD is the radius CD. Cor. 3. If AF bo half of AD, and consequently ACF half a right angle, the tangent AH is equal to the radius. For A be- ing a right angle, II must be half a right angle, and (I. 1, cor. 2) AH equal to AC. Cor. 4. Put the angle = A, and the radius=:l. Then (I. 24, cor. 1) FG-+CG'=CF^; that is, siu=A+cos'A = l. In like PLANE TRIGONOMETRY. 181 manner, we find from the right-angled triangles CAH, CDK, that Cir=CA^+AH^ and CK^ = CD'+DK^; that is, sec^A= l-ftan^A, and cosec^'A^l+cot'^A. Cor. 5. In the similar triangles CGF, CAH, CG : CF, or CA : : CA : CH ; that is, the cosine of an angle is to the ra- dius as the radius to its secant. Hence also (V. 9, cor.) CG.CH =:CA^; that is, cosAsecA = l. It would be found in like manner from the triangles GIF, CDK, that sin A cosecA=l, CI being equal to the sine FG. Cor. 6. In the same triangles CGF, CAH, the cosine CG is to the sine GF as the radius CA to the tangent AH ; whence (V. 8) cos A tanA=sin A. The triangles CIF, CDK give in like manner sin A cot A = cos A. Cor. 7. The radius is a mean proportional between the tan- gent of an angle and its cotangent. For the triangles CAH, CDK are similar; and therefore HA : AC : : CD, or CA : DK. Hence (V. 9, cor.) tan A cot A=:l. Cor. 8. The sine of an angle, and the sine of its supplement are equal. So likewise are their cosines, tangents, cotangents, secants, and cosecants. Let ACF be an angle, FG, AH its sine and tangent, and CG, DK its cosine and cotangent. Make the angle BCM equal to ACF ; draw the perpendicular MO ; and produce MC both ways to meet HA, KD produced in P and Q. Then (I. def. 20, and I. 9) the angles BCM, ACM, or ACF, ACM are sup- plements of each otlier ; as are also the arcs BM, AM, or AF, AM, since (HI. 16) BM, AF are equal. Now the triangles CGF, COM are equiangular, and have the sides CF, CM equal ; 182 THE ELEMENTS OF Q T) / ! \ / ^ \ V C \ J 4 H therefore (I. 14) MO is equal to FG, and CO to CG; and MO, FG are the sines of ACM, ACF, and CO, CG their cosines. Again : the triangles ACP, ACH are equiangular, and have AC common; therefore (I. 14) AP is equal to AH, and CP to CH ; and AP, AH are the tangents, and CP, CH the secants of ACM, ACF. In like manner it would be proved, by means of tlie trian- gles CDQ, CDK, that DQ, the cotangent of ACM, is equal to DK, the cotangent of ACF, and that their cosecants CQ, CK are equal. PROPOSITIOXS. » Pkop. I. — TnEOR. — T/i a right-ancjled triangle the hypothe- 7iuse is to either of the legs as the radius to the sine of the an^ gle opposite to that leg, or to the cosine of the adjacent angle ; (2) either of the legs is to the other as the radius to the tangent of the angle opposite to the latter ; and (3) either of the legs is to the hypothenuse as the radius to the secant of the contained angle. Let ABC be a triangle, right-angled at C ; then (1) c : h :: R : sin B, or cos A ; (2) a : 6 : : R : tau B ; and (3) a : c : : R : sec B. From B as center, with any radius, describe an arc cutting AB, BC in D, E; and through D, E draw (I. 8 and V) DF, ^ EG perpendicular to BC. Then (Trig. defs. 5 and 6) FD, EG, and BG are re- spectively the sine, tangent, and secant of the angle B ; and, since C is a right angle, A and B (Trig, def 4) are complements of each other; and therefore (Trig, def 7) sinB=cosA. Ao-ain : since the an^le B is common to the triangles ABC, DBF, GBE, and the angles at C, F, E right angles, these triangles (I. 20, cor. 5) are equiangular. Hence (V. 3) in the triangles ABC, DBF, BA : AC : : BD : DF ; that is, c : ^< : : R : sin B, or cos A. Again : (V. 3) in the triangles AP.C, GBE, PLANE TRIGONOMETRY. 183 BC : CA : : BE : EG ; that is, a : 5 : : R : tan B ; and BC : BA : : BE : BG ; that is, a : c : : R : sec B. Cor. Hence (V. 10, cor.) RJ=c sinB=c cos A; that is, the 'product of either leg and the radius is equal to the product of the hypothenuse and the sine of the angle opposite to that leg^ or of the hypothenuse and the cosine of the adjacent angle. When R=:l, this becomes simply b=c siu B=c cos A. Again : J=a tan B and c=:a sec B. Prop. II. — Theor. — The sides of a plane triangle are pro- portional to the sines of the opposite angles. Let ABC be any triangle ; then a : b : : sin A : sin B ; a : c : : sin A : sin C ; and b : c : : sin B : sin C. Draw AD perpendicular to BC; then AD is a leg of each of the right-angled triangles ADB, ADC; and therefore (Trig, 1, cor.) R.AD=AB sin B, and R.AD^AG sin C. Hence (I. ax. 1) AB sin B = AC sin C, or c sin B = 5 sin C; whence (V. 10, cor.) b : c : : sin B : sin C ; and, by drawing perpendic- ulars from B and C to the opposite sides, it would be proved in a similar manner that a : c : : sin A : sin C, and a : b :: sin A : fin B. Cor. From B as center M'ith BA as radius, describe an arc AD ; and from C as center, with an equal radius, describe an arc EF. Draw AG, EH perpendicular to BC ; these (Trig. def. 5) are respectively the sines of B and C to equal radii. Then the triangles AGC, EHC are equiangular, the angle at C being common, and the angles at G and H right angles. Hence (V. 3) CA : AG : : CE, or (const.) AB : EH ; and, alternately, CA : AB : : AG : EH ; that is, 5 : c : : sin B : sin C. The demonstration is simplified by taking, as here, one of the sides, AB or AC, as radius. This, however, is not essential, as arcs may be described from B and C as centers, with equal radii of any magnitude, and their sines, and a perpendi-cular from A to BC being drawn, the proof will be readily obtained. Scho. From one of the foregoing analogies we have, by in- version, c : J : : sin C : sin B. If C be a right angle, this 184 THE ELEMENTS OF (Trig. def. cor. 2) becomes c : Z> : : R : sin B, as in Prop. L The first part, therefore, of that proposition is a particular case of this one. Prop. ITT. — Theor. — The sum of any Uco sides of a trian- gle is to their difference as the tangent of half the snm of the angle opposite to those sides is to the tangent of half their dif- ference. ].et ABC be a triangle, a, h any two of its sides, of which a is the greater, and A, B the angles opposite to them ; then a+b : a-h :: tan i(A + B) : tan i(A— B). From C as center, with the greater side a as radius, describe the circle DBE, cutting AC produced in D and E, and BA produced in F ; join BD, BE, CF ; and draw EG par- allel to AB, meeting DB produced in G. Then because DC and CE are each equal to a, DA is equal to a+5, and AE to a—h. Also (I. 20) the exterior angle DCB is equal to A + B ; and DEB, which is at the circumference, is (III. 10) half of DCB, •which is at the center ; therefore DlEB=i(A + B). Again (I. 1, cor. l) : the angle F is equal to B ; and (I. 20) in the triangle ACF, the exterior angle A = ACF+F=ACF+ B; and consequently, ACF= A— B ; and (III. 10) ABE, or (I. 16) its equal, BEG = |(A— B). Now, since (III. 11) EBD, being in a semicircle, is a right angle, as also (I. 9) EBG ; if a circle were described from E as center, with EB as radius, DBG (III. 8, cor.) would touch it, and (Trig, def 6) DB would be the tangent of DEB, and BG of BEG; and therefore DB, BG will evidently be proportional to the tangents of those angles to any other radius. Or strictly, EB : BD : : 1 : tan DEB (Trig. 1) and (inver.) BD : EB :: tan DEB : 1. Also (Trig. 1) EB : BG :: 1 : tan BEG. Hence, ex mquo, BD : BG : : tan DEB : tan BEG. Lastly, since BA (const.) is parallel to GE, we have (V. 2) DA : AE : : DB : BG ; that is, a-\-b : a—b : : tan | (A+B) : tan i(A— B). PLANE TRIGONOMETKT. 185 Prop. IV — Tiieor. — hi a plane triangle^ the cosine of half the difference of any two angles is to the cosine of half their sum, as the sum. of the opposite sides to the third side ; and (2) the sine of half the difference of any two angles is to the sine of half their sum, as the difference of the opposite sides to the third side. Let ABC (see the last proposition) be any plane triangle ; then cosi(A — B) : cos^(A + B) : : a+b : c; and sin \{A — B) : sin -KA+B) : : a — b : c. For it was shown in the preceding proposition, that BED= i(A+B), and ABE=i(A-B) ; and since DBE is a right an- gle, DBA is the complement of ABE, and D of BED. But (Trig. 2) in the triangle ABD, sin ABD : sin D : : AD : AB ; that is, (Trig. def. 7) cosi(A— B) : cosi(A+B) :: a+b : c. Again (Trig, 2) : in the triangle ABE, sin ABE : sin AEB : : AE : AB J that is, sin ^(A — B) : sin i(A+B) : : a — b : c. Prop. V. — Theor. — In any plane triangle the sum of the segments of the base made by a perpendicular from the vertex, is to the sum of the other sides as the difference of those sides to the difference of the segments. Let ABC be a triangle, -and AD a perpendicular from the vertex to the base ; the sum of the segments BD, DC is to the sum of the sides AB, AC, as the difference of AB, AC to the difference of BD, DC. For (IL 5, cor. 4) the rectangle under the sum and difference of AB, AC is equivalent to the rectangle under the sum and difference of BD, DC; and therefore (V. 10, cor.) the sum of BD, DC is to the sum of AB, AC, as the difference of AB, AC to the difference of BD, DC. Scho. If the perpendicular fall within the triangle, the seg- 186 THE ELEMENTS OF ments make up the base, and their difference is less than the base; but if the perpendicular fall Avithout the triangle, as it does (second fig.) when one of the angles at the base is obtuse, the base is the difference of the segments, and their sum is greater than the base. Peop. VI. — Theor. — The rectangle under two sides of a tri- angle is to the rectangle under the excesses of half the perimeter above those sides, as the square of the radius to the square of the sine of half the contained angle. Let ABC be a triangle, and let s=i\{a-{-b+c) ; then be : (s — b) {s — c) : : R^ : sin'^A. Produce the less side AC through C, making AD equal to A AB ; join BD ; and draw AE, CF per- pendicular, and CG parallel to BD ; then (I. 24, cor. 2) AE bisects BD and the angle A. Now (II. 5, cor. 4) the rectangle under the sum and difference of BC, CD is equivalent to tlie rectangle under the sum and difference of BF, FD, that is, under BD and twice EF ; therefore the rectangle under half the sum and half the difference of BC, CD is equiva- lent to the rectangle BE.EF. But (Tkig. 1) AB : BE : : R : sin 1 A, and AC : CG or EF : : R : sin i A; whence (IV. 15) AC.AB : BE.EF, or ^ (BC + CD). i (BC-CD) : : R^ : sin^ A ; or, be : | (a+c — b). | (a+5— c) : : R^ : sin | A, because CD=c—b : and let 2s=a-\-b-\-c ; then, s — a=^ {b-\-c — a);s — b—i {a-\-c — b) ; and s — c—i {a+b — c) ; then we have be : (s—b) (s—c) : : R^ : suri A. Cor. Hence, taking R = l dividing the product of the means by the first extreme, and exti acting the square root, we find sin^A = Y- ^ 5 and it is plain that we should find in a similar manner, sin f o=\/ —- •, and sin 4- C = y ac V ab PLANE TRIGONOMETRY. 187 Prop. YII. — Theor. — The rectangle under two sides of a triangle is to the rectangle under half the perimeter and its ex- cess above the third side, as the square of the radius to the square of the cosine of half the angle contained by the two sides. Let ABC be a triangle, and let s=^ {a-^-b+c) ; then be : s {s—a) ::RM cos^A. Produce tlie less side CA, through A, making AD equal to AB ; join BD ; draw AE, CF perpen- d^ dicular, and AG parallel to BD. Then BD (I. 24, cor. 2) is bisected in E ; and the angle BAG being (I. 20) equal to the two equal angles D and ABD, each of them is equal to half the angle BAG, that is, half the angle A in the triangle ABC; and (I. 16) GAG is equal to D. Now, it would be shown, as in the pre- ceding proposition, that the rectangle under half the sum and half the diiference of DC, CB is equivalent to the rectangle BE.EF. But (Trig. 1) AB : BE : : R : cos ABE, or cos ^ A ; and AG : AG, or EF : : R : cos GAG, or cos ^ A ; whence (IV. 15), AG.AB : BE.EF, or i (DC+GB). i (DG-GB) : : R' : cos^A ; ovbc: \ {a+b-\-c). -J {b-\-c — a) : : R^ : cos^^ A, because DG=6+c. If 2s be the same as last proposition : be : 5 (5 — a) : : R*^ : cos'' ^ A. Cor. 1. Hence we find, as in the corollary to the preceding proposition, that cos-|A=^ — ^ -\ and it would be pi'oved in a similar man- ner, that costB = i/— ^ -i andcosfG=|/ — —~-. Cor. 2. From the sixth corollary to the definitions of trig- onometry, it is plain, when the radius is unity, if the sine of an angle be divided by its cosine, the quotient is its tangent. Hence, by dividing the expression for the sine of -j A in the corollary to the preceding proposition, by the value of its co- , , ,, , . , . /is — b) (s — e) Bine in the last corollary, we obtam tan ^ A=i/ j^ t — ; 188 THE ELEMENTS OF and we should obviously find in a similar manner, that tan|B: {s—a) {s—c) . As—a) (^ 'V s {s- ) and tan iC=/ (s-a) (s-b) -b) '"'-^""2 y s(s—c) ' Cor. 3. By dividing the values of tan ^^B, tan 4^C, in the pre- ceding corollary, each by that of tan |^A, we obtain tan iB 5 — a _ tan AC s — a TT= V and : ,— r= . tan +A s — o tan ^A 5 — c Prop. VIII. — Prob. — Given the radius of a circle^ and the cosine of an angle^ less than a right angle ; to compute the cosine of half the angle. Let CD, the cosine of ACB, less than a right angle, be given ; it is required to compute the cosine of its half Draw the chord AB, and perpendicular to it draw CFE ; draw also FG parallel to BB; then (III. 2, and Trig. defs. 5 and 7) AF or FB is the sine, and CF the cosine of ACE tlie half of ACB. Also (V. 2), DG is equal to GA, since BF is equal to FA, and DA=2DG; to each of these add 2CD; then CA+CDr3 2CG, and conse- quently CGi=i(CA + CD) ; or, if the radius be taken as unity, and the angle ACB be denoted by A, CG=r| (1+cos A). Again, in the similar triangles ACF, CFG, AC : CF : : CF : CG; whence (V. 10, cor. 2) CF-r=AC.CG; that is, cos 2 ^ A=| (1 +COS A). Hence, to compute cos | A, add 1 to cos A, take half the sum, and extract the square root. Scho. This proposition and the next afford means by which trigonometrical tables can be computed. Prop. IX.— Theor. — If A and B be any two angles, R : cosB:: sinA : i sin (A-B)+| sin (A-f B). Make AKC equal to A, and BKC, CKD each equal to B; draw BE, CF, DH, the sines of AKB, AKC, AKD ; join BD, and through the center K draw KNC ; then KN" is evidently the cosine of BKC. Draw also BML, NMG parallel to AK, DH. Now, in the similar triangles DLB, NMB, K n G F E A Bince DB is double of NB, DL (V. 3) is double of NM ; to PLANE TRIGONOMETRY. 189 DL add LII, BE, and to 2NM add what is equivalent, 2MG ; tlien DH+BE = 2NG ; wherefore NG is equal to half the sum of BE and DH, that is, to i sin (A— B) +^ sin (A-f B). Again, in the similar triangles CFK, NGK, we have (V. 3, and altern- ately) CK : NK : : CF : NG; that is, R : cos B : : sin A : isin (A— B)+isin(A + B). Cor. 1. Hence, if R = l, we have, by doubling the second and fourth terms, and by taking the products of the extremes and means, sin (A— B)4- sin(A + B)=;2 cos B sin A; whence, sin (A+B)=i2 cosB sin A— sin (A— B). Co7\ 2. If B=:A, the last expression becomes sirapl}"- sin 2 A = 2 sin A cos A. TRIGONOMETRICAL FORMULA. The lines hitherto considered may be computed for every conceivable angle, and they will each undergo a change of value when the angle passes through the gradations of magni- tude, hence they are the functions of the angle, a term imply- ing the connection between two varying quantities, that the value of the one changes with the value of the other, and they receive their values from the ratios or proportions arising from them and the angle. We have considered the numerical values only of these functions, and the angles from which they were deduced were all less than 180 degrees, which relate to plane angles and triangles. And we propose now to explain the processes for computing the unknown parts of rectilinear trian- gles, also the nature and properties of the angular functions, together with the methods of deducing all the formulae which express relations between them. When two diameters are drawn per- pendicularly to each other, they divide the circle into four equal parts called quadrants, which are first, second, third, and fourth quadrants, going from right to left, and the functions have certain algebraic values depending upon the particular quadrant in which the angle is. For instance, all the lines estimated from AC upward are positive, and from CA 190 THE ELEMENTS OF downward are negative ; from DB to the right are positive, and from DB to the left are negative. In the formulae the algebraic signs + and — are used, the former denoting posi- tive, and the latter negative. In the diagram it will be seen that the functions are all posi- tive in the first quadrant AEB ; that the sine, cosecant, and versed sine are positive, and the others negative in the second quadrant ; that the tangent, cotangent, and versed sine are positive, and the others negative in the third quadrant ; that the cosine, secant, and versed sine are positive, and the others negative in the fourth quadrant. Hence we can arrange them in the following table : Third Q. Fourth Q. — + FmsT Qttai). Second Q. Thir] Sine, + + Cosine, + Tangent, + — + Cotangent, + + Secant, + — — Cosecant, + + — Versed sine. + + + + + It is convenient to give different signs to the angles also. If we suppose the angles to be estimated from left to right, they are negative, and the sign of the angle will afiect the sign of its sine, but those of its cosine remain the same. From corollaries fourth, fifth, sixth, and seventh we can de- duce the following equations, when A denotes the angle and the radius is unity : Sin "A + cos 'A =1 (1) Sec=A=l + tan^A (2) Cosec^'A^l+cot'A (3) r_ . sin A TanA=: -r (4) cos A ^ ' ^ . cos A ig,\ QoiA=-r—. (5) sm A TanAxcotA = l (6) 1 cos A Sec A=-^ (7) PLANE TEIOONOMETRY. 191 Cosec A = -: T- (8) Sin A Ver, sin A=l — cos A (9) By the first proposition we have, in a right-angled triangle, radius : cos of either acute angle : : hyp. : side adjacent. Hence, in the following diagram : CB:0=CA cos C ; and DB:0-DA cosD, or CD=c>CA cos C— DA cos D. Dividing both members of the equation by CD, we have — 1=0=— =- cosC— ^=^=-cos D; hence (Trig, 2) OD kjiJ CA sin D - DA sin C , sin D ^ sin C pr^O- — T and 7=^^<2>=-^ — r; hence, lo-; — r-cosC — : — -r CD^^sin A CD sin A ' sin A sm A cos D, or sin Aosin D cos C— sin C cos D ; but the angle A is the difference of the angles ADB and ACB (I. 20) ; hence sin (D— C)=0=sin D cos C — sin C cos D. Thus, from the first and second propositions, by easy pro- cesses, we derive the formula for the sine of the difference of two angles, which is expressed in the following manner : The sine of the difference of any two angles is equivalent to the sine of the first into the cosine of the second, minus the cosine of the first into the sine of the second.* The formula for the sine of the sum of two angles can be derived from the preceding by substituting the negative for the positive value of the second angle, and bearing in mind that in estimating an angle from left to right, the algebraic sign of its sine is changed, and we get — The sine of the sum of any two angles is equivalent to the sine of the first into the cosine of the second, plus the cosine of the first into the sine of the second. The formula for the cosine of the sum of two angles can be derived from the preceding, by substituting the trigonometrical values of the functions when the sum of the angles is greater than a right angle, and remembering that the sine of an angle becomes the cosine of its complement, and we get — The cosine * The pupil would be much instructed by converting this and the fol- lowing expressions into their equivalent algebraic formula;. 192 THE ELEMENTS OF of the sum of two angles is equivalent to the cosine of the first into the cosine of the second, minus the sine of the first into the sifie of the second. By similar substitutions in the second formula, we get the formula for the cosine of the difference of two angles, expressed as follows: The cosine of the difference of two angles is equiva- le?it to the cosine of the first into the cosine of the second, ^>^ws the sine of the first into the sine of the second. The other corresponding formuliB are obtained by substitut- ing the respective equations for the trigonometrical fmictions derived from the fourth, fifth, sixth, and seventh corollaries for the values of the various functions ; for instance, to derive the formula for the tangent of the sum of two angles, we substi- tute, tanA = r in the second and third formulse, getting ' cos A ,, „, sin (A+B) sin A cos B 4- cos A sin B „ , ^^ t"" (^+^)=co4 At-B) = co.A cosB-sinAs hrB' ""^ '"^ ducing, we have — The tangent of the sum of two angles is equicalent to the tangent of the first, plus the tangent of the sec- ond, divided hy the square of the radius, minus the tangent of the first into the tangent of the second. And in a similar way we get the other formulae for the various functions. When the angle and radius are known, we can get the form- ula for the sine of double the angle by making the two angles equal in the second formula, and we have — The sine of twice an angle is equivalent to twice the sine of the angle into the cosine of the angle. In similar manner we derive the formula for the cosine of double the angle, and substituting the equation sin''A=l — cos' A in the third formula, we get — The cosine of twice an angle is equivalent to twice the square of the cosine of the an- gle, minus the square of the radius. Thus by substitution of the several equations in the respec- tive formulae, we can derive the other functions of double the ano-le when the ansfle and radius are known. From the formula for the cosine of double an angle, we can get by substitutions and reductions the formula for the sine of half an angle, and expressed as follows — The sine of half an angle is equivalent to the square root of half the difference of PLANE TRIGONOMETRY. 193 the radius and cosine of the angle. In a similar way we ob- tain — The cosine of half an angle is equivalent to the sqf/are root of half the sum, of the radius and cosine of the angle ^ and — The tangent of half an angle is equivalent to the sine of the angle divided hy the sum of the radius and cosine of the angle ; and so on for the other functions. By adding and subtracting the various formulae already men- tioned, we obtain a great number of consequences which are useful ; it will suffice to consider a few of them. From the first four we obtain — The sine of the sum of two angles added to the sine of the difference of the same angles is equivalent to twice the sine of the first into the cosine of the second. The siiie of the sum of two angles diminished by the sine of the difference of the same angles is equivalent to txcice the sine of the second into the cosine of the first. TJie cosine of the sum, of two angles increased by the cosine of the difference of the same angles is equivalent to twice the cos ne of the first into the cosine of the second. The cosine of the difference of two angles dimhiished by the cosine of the sum of the same angles is equioalent to twice the sine of the first into the sine of the second. These are very useful, because they change the pro- ducts of sines, cosines, and other functions from superficial into linear sines, cosines, etc. By the substitution of algebraic symbols into the preceding equations, we obtain certain analogies which give algebraic ex- pressions to so many theorems, as follows : Sum of sines : Dif. of sines : : tan of half Sum : tan of half Dif. Sum of sines : Sum of cos :: tan of half Sum : radius. Sum of sines : Dif of cos :: cot of half Dif : radius. Dif of sines : Sum of cos : : tan of half Dif : radius. Dif of sines : Dif of cos :: cot of half Sum : radius. Sum of cos : Dif of cos :: cot of half Sum : tan of half Dif. Sum of sines : sine of Sum : : cos of half Dif : cos of half Sum. Dif of sines : sine of Sum :: Sine of half Dif : sine of half Sum. INYESTIGATIONS OF THE METHODS OP COMPUTING TABLES OF SINES, TANGENTS, AND SECANTS. From what has been shown in relation to the previous form- 12 194r THE ELEMENTS OF nlae, it will be noticed that they all proceed from the first — and we derive the ^rst from the first and second propositions, namely, from the analogies that in a right-angled triangle : hypothennse : radius : : one of the legs : sine of opposite an- gle ; hence, when we know the length of the hypothenuse, leg, and radius, we can determine the sine of the angle opposite the leg. Then (Trig, def V, cor. 4) cos"A=l— sin'A ; that is, the square of the radius, minus the square of the sine of the angle, IS equivalent to the square of the cosine of the angle; then, square root of the difierence between the squares of the radius and sine is equivalent to the cosine of the angle; and the other functions are derived from the equations resulting from the fourth, fifth, sixth, and seventh corollaries of the definitions. Now, the chord of 60 degrees (III. 25, cor. 4) is equal to ra- dius, and when radius is unity, the cosine of 30° is 0.5 ; hence, sine of 30°= ^/(l — cos^30°)=: 4/.75, Bisecting this angle we get from the formula of cosine of half an angle, cos 15°= ■iVl+cos 30°, and seventeen such bisections give cos 1'^= .999999299 ; hence, sin V can be obtained. Some mathematicians divide 3.14159265358979, etc., into as many equal parts as there are seconds or minutes in 180°, and express the value of the sine of one second or minute by one of these equal parts, contending that the sine, chord, or arc of so small an angle differ very imperceptibly from each other; when this quantity is used, the cosine of one second becomes .999999957. It will thus be seen that the cosines obtained by these methods differ very little from each other, and for ordi- nary purposes either will give results sufiiciently accui'ate ; but when the greatest exactness is desired, the first method should be used, because the sine of an angle is a straight line and can never coincide with the arc which measures (I. def 19) the an- gle, however so small the angle be reduced, and (V. 25, scho.) 3.1415926, etc., is the approximate relation of the diameter and circumference. Hence, the first method is a pure deduction from geometrical and trigonometrical principles. When the angle is a right angle, the sine and cosine of the angle are equal, and the equations of the tangent and cotangent will give the values of those functions; and when the angle ex- ceed a right angle, the formula for the tangent of the sum of PLANE TRIGONOMETKY. 195 two angles can be used, making the right angle one of the angles and the excess the other angle. When the converging series are used, any angle less than 90'' is expressed by cc, and the formulae for the functions are: Sm a;=a; 1 -4- etc. 1.2.3^1.2.3.4.5 1.2.3.4.5.6.7^ x' X* a;* . ^ Cos x=:l 1 \- etc. 1.2^1.2.,3.4 1.2.3.4.5.6^ ^ , a' 2af 17a;' 62a;' Tan x=zx-\ 1 1 — - — -+-^ — --:+ etc. . 3 ^3.5^3=.5.7 3'.5.7.9^ Cotcc^---- — -^^- — -etc. , , a;' 5a;* 61a;* Sec a;=lH — --\ 1 \- etc. 1.2^1.2.3.4^1.2.3.4.5.6 1 X 1x^ S\x' Cosec x= 1 1 1 4- etc. a; ^1.2.3^3.4.5.6 3.4.5.6" Now, when the base of the Napierian logarithms is used, €= 2.7182818, and the following formulie will give the sine and co- sine of any angle x, from which the other functions can be obtained : Sm a;= -^ ; and cosa; = 2 ^^ 2 Having obtained the sine and cosine of any angle by either of the foregoing formulae, we can get the sine of twice the an- gle by the consequences fi-om adding and subtracting the first four formulae, page 193, from which we derive — 2 cos a; X sin x — sin = sin 2 ar, 2 cos X X sin 2 a; — sin a; = sin 3 x, 2 cos XX sin 3 a; — sin 2 a; = sin 4 a-, etc., etc., etc.,^ etc. Or by multiplying the first two formulae, page 191, and sub- stituting the value of the square of the cosine, we can deter- mine new formulae for further computation after having found the sines of x and 2 x. Sin (a;+2a;) sin (x— 2 x)=sin'a;— sin'2r ; hence, sin (a;+2a-) 196 THE ELKMENT8 OF sin (.c— 2 3-) = (sin x-\-s\n 2 x) (sin a:— sin 2 a;); or, sin (x — 2a'): sin X — sin 2x :: sin x+sin 2x : sin {x-{-2 x) ; applying lliia proportion, we have, Sin X : sin 2 a;— sin x : : sin 2 x + sin x : sin 3 x. Sin 2 X : sin 3 J— sin x : : sin 3a: + sin a: : sin 4 x. Sin 3 a; : sin 4 x — sin a; : : sin 4 .r+sin a; : sin 5 x. etc., etc., etc., etc. These last formulje will give the natural functions of the an- gles, but to avoid the operations of multiplication and divi- sion, and employ the simpler operations of addition and sub- traction, tables are constructed giving the logarithmic values of the several functions of the ansrles. As the sine and cosine of an anorle are the leers of a rieht- angled triangle, and the hypothenuse is the radius of the arc which measures the angle, for the convenience of logarithms the hypothenuse or radius is considered as 10,000,000,000, and its logarithm is 10. The sines, cosines, tangents, and cotangents are the only- functions put in the tables, as the other functions are easily found from them. TRIGONOMETRICAL PROBLEMS. The principles which have been thus established, enable us to solve all the elementary cases of plane trigonometry. Now, of the three sides and three angles of a triangle, some three, and those not the three angles, must be given to determine the tri- angle (I, 14, scho.), and the resolution of plane triangles may therefore be reduced to the three followinsr cases : I. When a side and the opposite angle, and either another side or another angle are given ; II. When two sides and the contained anijle are aiven ; III. When the three sides are c^iven o' The FIRST CASE is solved on the principle (Trig. 2) that the sides are proportional to the sines of the opposite angles. Thus, if A, B, a be given, add A and B together, and take the sum from 180°; the remainder (I. 20) is C. Then b and c will be PLANE TRIGONOMETRY. 197 found by the following analogies : sin A : sinB : : « : Jy and sin A : sin G : : a : c. If, again, a, b, A be given, we compute B "by the analogy, a : 6 : : sin A : sin B. Then, C is found by subtracting the sura of A and B from 180°, and c by the analogy, sin A : sin C When in this case two unequal sides, and the angle opposite to the less, are given, the angle opposite to the greater (Tkig. defs, cor. 8) may be either that which is found in the table of Bines or its supplement ; and thus the problem admits of two solutions (I. 3, case 4). If in this case one of the anHes be a riuht anole, the solution is rather easier; as, by the second corollary to the definitions, the sine of that angle is equal to the radius. The same con- clusion may also be obtained by means of the first propo- sition. To exemplify the solution of this case,* let az=l3 yards, 5= 15 yards, and Ar=53° 8', — to resolve the triangle; and the operation by means of logarithms will be as follows: As a 13 1.113943 : b 15 1.176091 : : sin A 67° 8^ 23^ 9.903108 : sinB 9.965256 or 112'=' 37' As sin A 9.903108 : sin C 59» 29' 9.935246 : : a 14 1.113943 : c 1.146081 As sin A 9.903108 ; sin C 14" 15' 9.391206 :: a 4 1.113943 : c 0.602041 * For logarithmic computations the pupil is referred to the Tables now in preparation by Prof. Docharty, of the College of the City of New York, or the Tables computed by Prof Davies, of tlie United States 198 THE ELEMENTS OF In these operations, to find the fourth term, the second and third terms are added together, and the first is taken from the sum. This may be done very easily, in a single operation, by addins: the fisrures of the second and third terms successively to what remains after taking the right-hand figure of the first term from 10, and each of the rest from 9, and rejecting 10 from the final result. Thus, in the first operation, we have 8 and 1 are 9, and 7 are 16; then 1 and 9 are 10, and 5 are 15, etc. It is still easier, however, when the quantity to be sub- tracted is a sine, to use the cosecant, and when it is a cosine, to use the secant, each diminished by 10, and then to add all the terms together. The reason of this is evident from the na- ture of logarithms, and from the fifth corollary to the defini- tions of Trigonometry. In like manner, when the number to be subtracted is a tangent, or cotangent, we may use in the former case a cotangent — in the latter, a tangent, subtracting in each case 10, either at first or afterward. This example evidently belongs to the doubtful case; and hence we have two values for each of the quantities B, C, and c ; and therefore two analogies are requisite for finding the values of c. The SECOJTD CASE is solved by means of the third and fourth propositions. Thus, if a, h, C be taken, take C from 180°, and (I. 20) the remainder is A 4-B. Take the half of this, and then, by the third proposition, as a-\-h : a—h : : tan| (A + B) : tan \ (A— B). This analogy gives half the difference of A and B ; and (II. 12, scho.) by adding this and i (A + B) together, A, the greater angle, is obtained, while B is found by taking \ (A — B) from-|(A+B). The remaining side will be calculated (Tkig. 4) by means of either of the following analogies, and "by employing both, an easy verification of the process is ob- tained ; as cos i (A— B) : cos ^ (A + B) ::a\h:c; and sin \ (A— B) : sin | (A + B) wa-h-.c. When the given angle C is a right angle, the solution is most easily efiected by means of the first proposition of Trig- Military Academy, or those of Frof. Loomis, of Yale College, New llaven, Connecticut. PLANE TRIGONOMETRY. 199 onometry; the oblique angles being obtained by the analogy, a : b :: U : tan B, or cot A ; and the hypotheinise either by c. the analogy, R : sec B : : a : c, or sin A : R : : a : As an example, As a-\-b 99.98 1.999913 : a — b 14.V8 1.169G74 : : tan ^ (A+B) 61° 37'i 10.267498 :tan^(A— B) 15° 18'i 9.437259 Hence A = 76° 56', and Bi=46o 19' As cos i (A— B) 15° 18'i 9.984311 : cos i (A+B) 9.676913 : : a+b 1.999913 C 49.20 1.692515 As sin I (A— B) 9.421626 : sin i (A + B) 9.944411 :: a-b 1.169674 : c 49.26 1.692459 Half the difference of A and B is here taken as 15° 18'^. When determined accurately, however, it is found to be 15° 18' 23", Hence the cause of the slight difference in the loga- rithm of c, as obtained by the two different analogies. It is plain that after A and B are computed, c might also be found by means of the first case, by either of the analogies; sin A : Bin C : : a : c, and sin B : sin C : : b : c. The foregoing method, however, is much preferable. The THIRD CASE may be solved by means of the fifth, sixth, or seventh proposition. Thus, assuming a (see the figures for the fifth proposition) as base, we have a to b-\-c as b—c,OT- c—b to a fourth proportional. If this be less than BC, it is the difference of the segments BD, DC, in the first figure ; and if half of it and half of the base be added together, the sum will be the greater segment, while the less will be found by taking half that proportional from half the base. If the fourth pro- 200 THE ELEMENTS OF portional be greater than the base, it is the sura of the seg- ments in the second figure, and, as before, the segments are the sum and difference of half the proportional and half the base. Then, by resolving, by the first case, the two riglit-angled tri- angles ADB, ADC, in which there are given the hypothenuses AB, AC, and the legs BD, CD, the angles B and ACD will be obtained, which, in the first figure, are two of the required an- gles ; while in the second, the angle C is the supplement of ACD. Again : by adding the three sides together, and talcing half the sum, the value of 5 is obtained; and, in applying the sixth proposition, the sides containing the required angle are to be separately taken from sy but, in applying the seventh, only the side opposite to the required angle is to be subtracted; while if all the three sides be subtracted successively, another mode of solution is furnished by the second and third corolla- ries to the seventh proposition. This last method is preferable to any of the others, when it is necessary to determine all the angles ; and if they be all computed by means of it, the cor- rectness of the operation is ascertained by trying whether their sum is 180°. To exemplify the last of these methods, let a=Qld, b—53ly and c=429 ; to compute the angles. Here, by adding the three sides together, we obtain 1645, the half of which, 822.5, is s. Then, by taking from the three sides successively, we find s — a= 143.5, s — &= 285.5, and s — c=393.5. The rest of the operation, the subtraction in the first part of which may be performed in the manner pointed out in the example for the first case, is as follows : 8 8 — a 822.5 143.5 2.156852 ) s h 285.5 2.455606 8 C 393.5 2.594945 2) 19.978563 n ^A 44° 17'i 9.989281 A=88° 35' PLANE TRIGONOMETRY. tan i A 9.989281 ) ^^^ log{s — a) 2.156852) 12.146133 ) , . c subt. log{s — b) 2.455606) 201 tan IB 26° V'i 9.690527 B = 520 15' 12.146133)^^^^ log {s — c) 2.594945 ) tan i C 19° 35' 9.551188 C = 39° 10' In the first part of this operation, the halving of the loga- rithm serves for the extraction of the square root. The remain- der of the work consists in adding together tan ^ A and log {s — a), and subtractingAog {s — b) and log (s — c) successively from the sum. This method of solution is remarkably easy, requiring for the entire operation only four logarithms to be taken from the tables ; and affording at the same time a most satisfactory verification by the addition of the three angles, when found. The preparatory part of the process also admits of an easy verification, as the sum of the three remainders s — «, s — b, s — c is equal to the half sum. Prob. I. — Let it be required to find the height of an accessi- ble object AB, standing on a horizontal plane. On the horizontal plane take a station C, and measure with a line, a chain, or any such instrument, the dis- tance CB to the base of the object ; and with a quadrant, a theodolite, or other angular in tru- ment, measure the angle BCA, called the angle of elevation. Then, since B is a right angle, the height AB will be found (Trig. 1) by the analo- gy,^R : tan C : : CB : BA. This gives the height of A above CB, the horizontal line passing through the eye of the observer ; and tlierefore to find the entire height, AB must be increased by the height of hia 202 THE ELEMENTS OF eye above the base or the object. The like addition must be made in every problem of this kind, when the angle of eleva- tion above the horizontal line is given. pROB. n. — To find the height of an object AB, stanaing on a horizontal plane, but inaccessible ow account of the uneven- ness of the ground near its base, or the intervefition of obsta- cles. In a straight line passing through the base of the object take two stations C, D ; and measure CD, and the two angles of elevation BCA, BDA. Then (I. 20) CAD is the difference of ACB, ADB ; and (TFao. 2) sin CAD : sin D : : DC : CA. Again (Trig. 1), R : sin ACB : : AC : AB ; whence AB will be found. The computation will be rendered rather more easy by mul- tiplying together (IV. 15) the terms of the two analogies, and dividing the third and fourth terms li^'the result by AC ; as by this means we get the analogy li X sin CAD : sin D x sin BCA : : DC : AB. Hence, to find the logarithm of AB, to the loga- rithm of DC add the logarithmic sines of D and BCA, and from the sum take the sine of CAD and the radius. Prob. Ill, — To find the distance oftxno objects A a7idB on a horizontal plane. This may be effected in different ways according to circum- stances. 1. A base AC may be taken, terminated at one of the ob- jects. The angles A and C, and the side AC are then measured ; and the required distance AB is found by the analogy, sinB : sin C : : AC : AB. 2. This method fails if the objects A and B be not visible from one another, as then the angle A can not be measured. In this case, a station C may be taken as before, from which both A and B are visible. Then, having measured the angle C, and the sides AC, BC, we compute the distance AB by means of the second case of trigonometry. PLANE TKIGONOMETKY. 203 3. When from inequalities in the ground, or other causes, the preceding methods are inapplicable, the solution may be effect- ed in the following manner: Measure a base CD, such that A and B are both visible from each of its extremities ; measure, also, the two angles at C, and the two at D. Then, by the first case in trigonometry, -we compute AC in the triangle ACD, and BC in the triangle BCD ; from which, and from the contained angle ACB, AB is computed by means of the second case. The operation may be verified by computing AD, BD, by the first case, and thence AB by the second case. Prob. IV. — Let AFB be a great circle of the earth, supposed to be a sphere; E a point in the diameter B A produced, EF a straight line touching the circle in F, and ED a straight line in its plane, per2Jendicular to AB; it is required to compute the angle DEF, and the straight line EF. Draw the radius CF. Then, since (III. 12) CFE is a right angle, we have (hyp. and I. 20, cor. 3) DEC=CEF+ECF." Take away CEF, and there remains DEF=:ECF. Now (III. 21) EF- = BE.EA. Hence EF will be found by adding AE to AB, multiply- ing the sum by EA, and extracting the square root. To find CE, add AE to the radius AC. Then (Trig. 1) CE : EF : : R : sin ECF, or sin DEF ; or CE : CF : : ~ Pv : cos ECF, or cos DEF. 8cho. These examples have been selected from the " Ele- ments of Plane Trigonometry," by Prof Thomson, of the Uni- versity of Glasgow, since they exhibit in the simplest manner the elementary principles of trigonometrical computation. 4* THE ELEMENTS OF SPHERICAL TRIGONOMETRY. DEFINITIONS. 1. Spherical Trigonometry explains the processes of cal- culating the unknown parts of a spherical triangle when any three parts are given; and certain formulae derived from Plane Trigonometry are employed to express the relations between the six parts of a spherical triangle. 2. A spherical triangle is that portion of the surface of a sphere bounded or contained by the arcs of three great circles intersecting each other ; the spherical triangle being formed by three planes passing through the sphere, and intersecting each other, each angle of the triangle (VI. 24) is contained by the tangents of the sides at their point of intersection, and is measured by the arcs of great circles described from the ver- tices as poles, and limited by the sides of the triangle produced if necessary. Also, the angles of a spherical triangle vary (VL 24, cor. 2) between two and six right angles. 3. The spherical angle being contained by the tangents of the sides at their point of intersection, the properties of the spherical triangle are explained by means of Plane Trigonome- try, and its analogies are applied to imaginary rectilineal tri- angles, the sides of the spherical triangle being regarded functions of rectilineal angles having the sides of the spherical triangle as arcs measuring (I. def 19) them. Spherical Trig- onometry treats of the angles at the apex of a triangular pyra- mid; but Plane Trigonometry treats oi plane angles ^ therefore Spherical Trigonometry treats of solid angles. 4. Let ABC be a spherical triangle, and H the center of the sphere ; the angles of the tnangle are equal to the angles in- cluded by the planes IIAB, HAC, and IIBC (VI 24),"which are the angles formed by the planes at the apex of a triangular pyramid, and the arcs AB, BC, and CA measure the angles on the planes at the apex of the pyramid AHB, BHC, AHC re- SPHERICAL TRIGONOMETRY. 205 spcctively. And we can represent the side opposite the angle A by a, the side opposite the angle B by b, and the side op- po>ite the angle C by c. On the line HA take any point as L, and draw perpendiculars as FL, LG to HA. Then, GLF will be equal to the angle A, LEG equal to B, and FGL equal to C ; and the sides CB, BA, and AC of the spherical triangle ABC will measure the angles CHB, AHB, and AHC respectively ; hence these an- gles are denoted by a, c, b. 5. If FG be joined in the triangles FHG and FLG, we will have (Plane Trig., 6 cor.), HF'+HG'-FG* cos BHC=cos a= cos FLG = cos A= 2HFxHG LF=+LG=-FG* 2LGxLF • Reducing and subti-acting second from the first, we will have, 2 [cos a (HFxHG)— cosA (LGxLF)]=.2HLl Dividing both members by 2 (HFxHG), we get. cos a— cos A LGxLF HL HL HFxHG" HF^HG" Since regular and similar polygons have their perimeters proportionate to their apothems, and circles have their circum- ferences proportionate to their diameters (V. 14, cor. 3), the sine of an angle is the ratio of the radius, or the hypothenuse of a liglit-angled triangle to the perpendicular from the vertex of the right angle to the hypothenuse. „ LG , , LF . HL HL Hence we get ^- = sin 5, g^ = sm e, g-, = cos c, jj^ = cos b. Substituting and ti-ansposing, we derive the formula, cos a=cos b cos c+sin b sin c cos A, ) cos b = cos. a cos c+sin a sin c cos B, /• (1) cos c=cos a cos 6+ sin a sin b cos C. ) 206 THE ELEMENTS OF Or, The cosine of either side of a spherical triangle is equal to the product of the cosiyies of the other two sides increased by the product of their sines into the cosine of the an^le included hy them. The three formulae show the relations between the six parts of a spherical triangle such, that if any three of them be given, the other three can be determined. 6. Then (VI. 24, cor. 1), if we denote the angles and sides of the spherical triangle polar to ABC, respectively, by A', B'', C^, a\ h\ c', we will have, «'=180°— A, d'=180°— B, c'=180O— C, A'==180°— a, B'r=l80°— 6, C"=180O— c. Since any of the formulae (l) is applicable to polar spherical triangles, we have, after substituting the respective values and changing the signs of the terms, other formulae : cos A=sin B sin C cos a — cos B cos C, \ cos B=sin A sin C cos h — cos A cos C, \ (2) cos C=sin A pin B cos c — cos A cosB. ; Or, The cosine of either angle of a spherical triangle is equal to the product of the sines of the two other angles into the co- sine of their included side, diminished by the product of the cosines of those angles. V. Transposing the first and second formulae (l) we get, cos a — cos h cos c=sin h sin c cos A,^ cos b — cos a cos c=:sin a sin c cos B ; respectively adding and subtracting, we get, cos a+ cos 6— cos c (cos a+ cos 5)=sin c (sin 5 cos A+ sin a cos B), cos a — cos &+COS c (cos a— cos b) =m\ c (sin b cos A— sin a cos B) ; which can be put in the forms (1 — cose) (cos a -|- cos J) = sin c (sin b cos A -|- sin a cos B), (1 +COS c) (cos a— cos J) = sin c (sin b cos A — sin a cos B) ; multiplying these equations together, substituting for (1 — cos') its value (sin'), and for (cos') its value (1 — sin'), and dividing by Bill* c, we have, SPHERICAL TRIGONOMETRY. 207 COS* a— cos' 5=sin'' 5— sin' h sin" A— sin' a+sin* a sin' B; then, since (1— sin'a)— (1— sin= h) =sin' h — sin' «, we have, cos' a — cos* 6=sin' h — sin' a, and we get, sin" h sin' Am sin' a sin' B ; extracting tlie sqiiare root, sin 5 sin A = sin a sin B, or ' sin A sin a sin B sin b' sin A derive And from first and third formulae (l) we From the second and third formu- sin C sin c sin C sin c \i?) lae we derive —. — t^_ . ,. sin ±> sin Or, In every spherical triangle^ the sines of the angles are to each other as the sines of their opposite sides. 8. Taking the third formula (l), and substituting for (cos h) its value as expressed in the second, and for (cos' a) its value (1— sin' a), and dividing by sin a, we will have, cos c sin a=sin c cos a cos B + sin b cos C. But (Spher. Trig. 7) we have sin b~ ._ ^, — ; substituting for sin C cose sin b its value, and dividing by sin c, we get, -. sin a=C08 a cosB4 sin B cos C sin C _, cos But -r- =COt, sin Hence we can derive, by similar processes, cot a sin b=cot A sin C + cos b cos C, cot a sin c=cot A sin B+cos c cosB, cot b sin a=cot B sin C + cos a cosC, cot b sin c=cot B sin A+cos c cos A, cot c sin a=:cot C sin B+cos a cos B, cot c sin b=cot C sin A+cos b cos A. } (4) The formulae (1) are the fundamental analogies of Spherical Trigonometry, from which all the others are derived, which others are more adapted for logarithmic computations. 208 THE ELEMENTS OF THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES BY LOGARITHMS, The following equations give the unknown parts of a right- angled spherical triangle when C is the right angle and any two other parts are known. There are six cases. Let C be the right angle, and c be the hypothenuse. Case 1. Given a and b to find c, A, and B ; , . tan a _, tnn b cos e=cos a cos o; tan A= r; tan l3= . ■^ tan o tan a Case 2. Given c and a side b to find a, A, and B ; cos c . tan b . „ sin & cos a= r; cosA=: — — : sin 13=^ . cos b tan c sin c Case 3. Given aside a and opposite angle A to find others; . , tan a . „ sin a . _ cos A sm 0= — r ; sin C=-. — -; sin B= -. tan A sm A cos a Both acute or both not acute. Case 4. Given a side a and adjacent angle B to find others; tan 6=sin a tan B ; cot c=:cot a cos B ; cos A = cos a sin B. Case 5. Given the hypothenuse c and an angle A to find others ; sin « = sine sin A ; tan 5i=tan c cos A; cot B = cose tan A. Case 6. Given the oblique angles A and B to find others ; cos A - cos B . _ cos a=—. — ^rr; cos o=— — - ; cos c = cot A cot B. sin B sin A Napier's circtdar parts are much the simplest method of re- solving right-angled spherical triangles; they are the two sides about the right angle, the complements of the oblique angles, and the complement of the hyjiothenuse. Hence there are five circular parts; the right angle not being a circular part, is supposed not to separate the two sides adjacent to the right angle ; therefore these sides are regarded adjacent to each other, so that when any two parts are given, tlieir corre- sponding circular ])aits are also known, and these with the re- quired part constitute the three parts under consideration ; therefore these three paias will lie together, or one of tliem SPHERICAL TRIGONOMETRY. 209 ■will be separated from both tlie others. Hence one part is known as the middle part; and when three parts are undei- con- sideration, the parts separated by the middle part are caUcd t\\G adjacent parts ; and the parts separated liuni tlie middle parts are called the opposite parts. IS'ow, assume any part in B the diagram for the middle part, and using the formulae (1) •when the other parts are the ojiposite parts, and we get, 27ie sine of the middle part is equal to the product of the cosines of the opposite parts. Then, assume again any part for the mid- dle, and use the formulae (2) when the other parts are the ad- jacent parts, and we get, The sine of the middle pa. t is equal to the pyroduct of the tangents of the adjacent parts. Hence "We derive the five following equations : sin a=tan h tan (90"— B)=cos (90° — A) cos (90° — c;, sin 6=tan a tan (90°— A)=:cos (90°— B) cos (90°— c), sin (90°— A)=tan b tan (90°— c)=cos (90°— c) cos c/, sin (90° — c)=tan (90°— A) tan (90°— B)==cos a cos 6,. sin (90°— B)— -tana tan (90°— c)=cos 6 cos (90° — A). The a7za7o^/es of Xapier are derived from the formulce (l) by eliminating the cosines of any of the sides, reducing and chang- ing to linear sines and cosines (Plane Trig. p. 193), and we bave, cos I {a-\-h) : cos | (a — b) : : cot | C : tan | (A + B), sin \ (a+b) : sin -^ (a—b) : : cot -^ C : tan ^ (A— B), cos -^ (a-f c) : cos | {a—c) : : cot ^ B : tan ^ (A+C), sin ^ {a+c) : sin -^ (a—c) : : cot | B : tan ^ (A — C), cos ^ {b-hc) : cos -^ (b—c) : : cot ^ A : tan ^ (B + C), sin ^ {b+c) : sin -^ {b—c) : : cot | A : tan | (B — C), 14 210 THI. ELEMENTS OF The same proportions applied to tlic triangle polar to ABC, with accents omitted, we have, cos I (A + B) : cos i (A— B) : : tan | c : tan i {a+b), sin i (A+B) : sin i (A— B) : : tan -J c : tan i (a-b), cos i (A+C) : cos i (A— C) : : cot ^ & : tan ^ (a+c), sin i (A+C) : sin | (A— C) : : cot i 6 : tan ^ (a— c), cos i (B + C) : cos ^ (B— C) : : cot i a : tan i (6+c), sin i (B + C) : sin i (B— C) : : cot i a : tan | (*— c). The same ambiguity that there is between plane triangles (I 3, Fourth Case) exists also between spherical triangles, which may be avoided by remembering that every angle and gide of a spherical triangle are each less than two right angles, and that the greater angle is opposite to the greater side, and the least angle is opposite to the least side; and conversely. Quadrantal spherical triangles are such which have one side equal to ninety degrees ; hence they can very easily be solved by formulae for right-angled spherical triangles. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES BT LOGARITHMS. Case 1. Given the three sides to find the angles. Ain {s—fi) sin (5— 5) sin (»— c) Find s=i (a+6+c) ;let M=|/ ■ ^-'^ ^ 5 then, tan iA = ^-^^^^3:^; tan i B=^^^^^-; tan i C= ]\I ein (.« — c)' Case 2. Given two sides a and b and the included angle C, to find others. Tan^(A+B)=^^^^i^4NotiC; tan^A-B^'-i^^-^ ■^^"^^^^^^ cosi(a+6) ' ' Biui(u+6) cot \ C. But A=i (A+B)+i (A-B) ; B=| (A+B)-i (A-B) ; sin c=sin a^!^=sin b ^^, or find an angle cot 9=tan o sin A sin 15 cos a sin (J+(j)) cos C ; cos 0= -. • ' sin (J) BPOEKICAL TKIGONOMETKY. 211 Case 3. Given the sides a and b and an angle opposite to one of them, to find others. Find cot (p=tan b cos A, and tanj^^cos b tan A; then, sin , , . cos a sin (p . ^ . . sin b . , , . . {c+(p)=i — ; sm JL> = sm A -; — : sin (c +>-)=: cot a tan ^ ' cos 6 ' sin a ^ ^' fiinx- Case 4. Given the angles A and B and the inchiaed side, to find others. Find tan (p=cos c tan A, and tan p^=cos c tan B ; , tan c sin (p , tan c sin y then, tan a— . .,, , — -; tan 6 = - . , . , t; p_cos A cos (B + (p) _ cosB cos (A+y) cos 9 cos -/^ Case 5. Given A and B and a side opposite one of them, to find others. Find tan (p=:tan a cos B ; cot ;^=:C08 a tan B; ein irzsin a sin B sin A ■ sin (<3— (p)=cot A tan B sin 9, • //-I \ cos A sin Y sm (C — y) = — ^. ^ ^' cosB Case 6. Given the three angles A, B, C, to find the sides. Take • S=|(A+B + C);andN=,4/ — - ""'*''' ^ ' cos 10 — J cos (S— A) cos (S— B) cos (tS— C) ; then, tan \ a—1^ cos (S — A) ; tan| &=N' cos (S — B); tan \ c=N cos (S — C). * THE SURFACE OP A SPHERICAL TRIANGLE. Let ABC be a spherical triangle, AC=: DF, BC=FE, ABC=DEF. S=surface of ACB, 5=surface of hemis- phere— BHDC—AGEC—DCE=0 R'— (lune AHD — S) — (lune BGE— S)— (lune CDFE-S) =eR'(,-j 80 180""" W + ^^- 212 BPHKRICAL TRIGONOMETKT. 180*^, or equivalent to the sum of the three angles above 180°; hence, its spherical excess is sometimes taken as the measure of the ti'iangle. Or in terms of its sides, formula given by L'Huiller, tan ^ E= V [tan ^ s tan ^ {s — a) tan ^ (s — b) tan ^ {s — c)]. EXERCISES IN ELEMENTARY GEOMETRY, AND m PLANE AND SPIIEKICAL TKIGONOMETRY.* DEFI^nXIOXS. 1. Lines^ avgUs^ and spaces are said to be given in magni- tude, when they are either exhibited, or when the method of finding them is known. 2. Points, lines, and spaces are said to be given in position, ■which have always the same situation, and which are actually exhibited or can be found. 3. A circle is said to be given in magnitude when its radius is oiven ; and \w position, when its center is given. Mao-nitudes, instead of being said to be given in magnitudey or given in position, are often said simply to be given, when no ambiguity arises from the omission. * For these Exercises I am indebted to Thomson's Euclid (Belfast), they being judiciously selected by that eminent writer, and their presentation here is a valuable acquisition to an American school text-book. I would gladly acknowledge my obligations for many propositions in this volume, but they being culled for more than two thousand years from the best writers on Geometry, and being so much modified by each succeeding age, that it is impossible at this day to attribute tliem to their rightful authors, and many of them being more or less contained in every work on the subject, they have become public property. What I have introduced myself will be well recognized by every student of Geometry, and my only apolugy is, the desire to advance the cause of Truth. 214 EXEBCISK8. 4. A ratio is said to be giveti when it is the same as that of two cciven maGrnitudes. 5. A rectilineal figure is said to be given in species, when its several anarles and the ratios of its sides are iriven. 6. When a series of unequal magnitudes, unlimited in num- ber, agree in certain relations, the greatest of them is called a ma/x'mum ^ the least, a minimum. Thus, of chords in a given circle, the diameter is the maxi- mum ; and of straiiiht lines drawn to a given straight line, from a given point without it, the minimum line is the perpen- dicular. 7. A line which is such that any point whatever in it fulfills certain conditions, is called the locus of that point. 8cho. 1. Several instances of loci have already occurred in the preceding books. 1. Tlius, it was stated in the fifth corollary to the fifteenth proposition of the first book, that all triangles on tlie same base, aiid between the same pai'allels, are equivalent in area; and hence, if only the base and area of a triangle be given, its vertex may be at any point in a straight line parallel to the base, and at a distance from it which may be determined by ap])lying (11. 5, scho.) to half the base a parallelogram equiva- lent to the given area ; and therefore the parallel is the locus of the vertex. Here the conditions fulfilled are, that straight lines drawn from any point in the parallel to the extremities of the given line, form with it a triangle having a given area. 2. It was stated in the first corollary to the eighteenth proposi- tion of the third book, that all anHes in the same segment of a cinOe are equal ; and hence, if only the base and vertical angle of a triangle be given, the vertex may be at any point of the arc of a segment described on the base, in the manner pointed out in the nineteenth pi'oposition of the third book; that arc, therefore, is the locus of the vei-tex. 3. It will be seen in the sixth proposition of these Exercises that straight lines drawn from any point whatever in the circumfer- ence of the circle ABGC to the points E, F, have the same ratio — that of EA to AF. Hence, therefore, if the base of a triangle, and the ratio of the sides be given, the locus of the ■vertex is the circumference of the circle described in the man- EXEUCISES. 215 ner pointed out in the corollary to this proposition ; unless the ratio be that of equality, in which case the locus is evi- dently a perpendicular bisecting the straight line joining the points. 4, It follows likewise, from the fifth corollary to the twcnty- fomth proposition of the first book, that when the base of a tri- angle and the difference of the squares of the sides are given, if the point D be found (II. 12, scho.) in the base BC, or its continuation, such that the difference of the squares of BD, CD is equivalent to the difference of the squares of the sides; and if through D a perpendicular be drawn to ]3C, straight lines drawn from any point of the perpendicular to B, C will have the difference of their squares equivalent to the given difference; and hence the perpendicular is the locus of the vertex, when the base and the difference of the squares of the sides are given. 5. It will appear in a similar manner from the corollary to the twelfth proposition of the second book, that if BC the base of a triangle, and the sum of the squai'es of the other sides AB, AG be given, the locus of the vertex is the circumference of a circle described from D, the middle point of the base as center, and with the radius T>A. To find DA, take the diagonal of the square of BD as one leg of a right-angled triangle, and for its bypothenuse take the side of a square equivalent to the given sum of the squares of AB, AC ; then the diagonal of the square described on half the remaininoj le<x of that rigcht-aniiled trian- gle will be the radius of DA. The proof of this is easy, de- pending on the third and fourth coi'ollaries to the twenty-fourth jDroposition of the first book, and on the corollai-y to the twelfth proposition of the second book. Scho. 2. In discovering loci, as well as in other investigations in geometry, the stuflent is assisted by what is termed geomet- rical analysis ; of the nature of which it may be proper here to give some explanation. Take this proposition : If a chord of a given circle have one extremity given in positior), and if a segment terminated at that extremity he taken on the ch rd, produced if necessary^ such that the rectangle under the segment and chord may be equivalent to a given space ; the locus of the point of section is a straight I i?ie given inp)osition. 216 EXERCISES. Let AB be the diameter of the circle and AC a chord of the circle. If, in the proposition, instead of being informed that the locns is a strai!j;ht line, we were required to find what tlie locus is, we might proceed in the following manner: Lft D be any point in the required line, so that the rectangle ACAD is equivalent to the given space ; and having drawn the diameter AB, find E, so that the rectangle AB.AE may be equal to ACAD, and therefore E a point in the required line ; and join DE, BC Then (V. 10, cor.) AB : AC : : AD : AE. Hence (V. 6) tho triangles DAE, BAC, having the angle A common, are equi- angular; and thereiore AED is equal to ACB, which is a right angle. The point D is therefore in a perpendicular passing through E; and in the same manner it would be t^hown, that any other point in the required line is in the perpendicular ; that is, the perpendicular is its locus. The investigation just given is called the analysis of the projjosition, while the solutions hitherto given are called the synthesis or compositio7i. In analysis we commence by sup- posing that to be effected which is to be done, or that to be true which is to be proved ; and, by a regular succession of con- sequences founded on that supposition, and on one another, we arrive at something which is known to be true, or which we know the means of effecting. Thus, in the second corollary to the seventeenth proposition of the sixth book, the conclusion obtained for the area of the circle is shown by the third corol- lary to be consistent with the proportion established by Ar- chimedes between the cone, sphere, and cylinder, and also consistent with the geometrical truth in regard to the sur- faces of the sphere and cylinder. Plence, analysis takes into consideration this consistence, and confirms the second corol- lary from its agreement with established truths of geometr3^ Again: in the corollary to the twenty-fourth proposition of the fifth book, since circles are in propoition to the squares of tlieir radii, the quadrant ACB is equivalent to the semicircle ADC, we have (I. ax. 3) the triangle ABC equivalent to th^ crescent ADC. Now, when we api)ly the conclusion derived by the second corollary to the seventeenth proposition of the sixth book to the above, we find a perfect agreement ; taking EXERCISES. ^17 the circle as three times square of radius, we have quadrant ACB equivalent to J AB'; hence (I. ax. 1) the semiciicle ADO is equivalent to | AW. But (VI. 17, cor, 2) the segment AC of the quadrant ACB is equivalent toi AB"; therefore (I. ax. 3) we have the triangle ABC and the crescent ADC each equiva- lent to ^ AB', thus showing the agreement between the second corollary to the seventeenth proposition of the sixth book, and the established truth relating to the crescent or lune. Also, we have (VI. 17, cor. 2) the hemisphere generated by the quadrant BNP equivalent to the solid generated by the trapezium BSNP, and we have the solid generated by the figure BTNP common ; thei-efore (I. ax. 3) the solid generated by the segment BT is equivalent to the solid generated by the figure TSN. Now, the solid generated by the segment BT is a part of the hemisphere; hence its contents are computed by the same radius as the hemisphere ; the solid generated by the figure TSN is a part of the solid generated by the trapezium BSNP ; hence its contents are also computed by the same radius as the hemisphere. Therefore we obtain by analysis the ti-uth, that vchen equivalent solids are generated by equiva- lent surfaces, the generating surfaces are up n equal radii, a truth corresponding to the truth established by the second coroUaiy to the seventeenth proposition of the sixth book, tJiat equivalent surfaces upon the same radius loill generate equiva- lent solids. The synthesis then commences with the conclusion of the analysis, and retraces its sevei-al steps, making that pre- cede which before followed, till we arrive at the required con- clusion. Therefore the demonstrations given in the second corollary to the seventeenth proposition, book sixth, obtain the conclusion from which the analyses precede. From this it ap- pears that analysis is the instrument of investigation ; while svnthesis affords the means of communicatino- what is already known ; and hence, in the Elements of Euclid, the synthetic method is followed throughout. What is now said will receive further illustration from the solution of the following easy problem. Given the perimeter and angles of a triangle, to construct it. Analysis. — Suppose ABC to be the required triangle, and produce BC both ways, making BD equal to BA, and CE to 218 EXERCISES. CA ; then DE is given, for it is equal to the sum of the throo sides AB, BC, CA ; that is, it is equal to the given perimeter. j^ Join AD, AE. Then (I. 1, cor. 1) the angles D and DAB are equal, and therefore each of them is half of ABC, because (T. 20) ABC is equal to both. The angle D therefore is given ; and in the same manner it may be shown that E is given, being half of ACB. Hence the triangle ADE is given, because the base DE, and the angles D, E are given ; and ADE being given, ABC is also given, the angle DAB being equal to D, and EAC equal to E. Composition. — Make DE equal to the given perimeter, the angle D equal to the half of one of the given angles, and E equal to the half of another; draw AB, AC, maldng the angle DAB equal to D, and EAC to E; ABC is the triangle re- quired. For (T. 1, cor. 2) AB is equal to BD, and AC to CE. To these add BC, and the three, AB, BC, CA, are equal to DE, that is, to the given perimeter. Also (I. 20) the angle ABC is equal to D and DAB, and is therefore double of D, since D and DAB are equal. . But D is equal to the half of one of the given angles; therefore ABC is equal to that angle; and, in the same manner, ACB may be proved to be equal to another of the given angles. ABC therefore is the triangle required, since it has its perimeter equal to the given perimeter, and its angles equal to the given angles. It is impossible to give rules for effecting analyses that will answer in all cases. It may be stated, however, in a general way, that when sums or differences are concerned, the corre- sponding sums or differences should be exhibited in the as- sumed figure ; that in many cases remarkable points should be joined ; or that tlirough them lines may be drawn perpendicu- lar or parallel to remarkable lines, or making given angles with them ; and that circles may be described with certain radii, and from certain points as centers; or touching certain lines, or passing tlirough certain points. Some instances of analysis will be given in subsequent propositions ; and the student will EXERCISES. 219 find it useful to make analyses of many other propositions, such as several in the Exercises. 8. A jyorism is a proposition affirming the possibility of find- ing such conditions as will render a certain problem indeter- minate, or capable of innumerable solutions. Scho. 3. Porisms may be regarded as liaving their origin in the solution of j^roblenis, which, in particular cases, on account of pccidiar relations in their data, admit of innumerable solu- tions ; and the proposition announcing the property or relation which renders the problem indeterminate, is called a porism. This will be illustrated by the solution of the following easy problem. Through a given point A, let it be required to draw a straight line bisecting a given parallelogram BCDE. Suppose AFG to be the required line, and let it cut the sides BE, CD in F, G, and the diagonal CE in H. Then from tho equivalent figures EBC, FBCG take FBCII, and the remaining triangles EHF, ClIG are equal. Now, since (I. 16 and 11) these triangles are equiangular, it is evident that they can be equal in area only when their sides are equal ; wherefore II is the middle point of the diagonal. The construction, therefore, is effected by bisecting the diagonal EC in H, and drawiu"' AFHG. For the triangles CHG, EIIF are equiangular, and since CH, HE are equal, the triangles are equal. To each of them add the figure FBCH ; then the figure FBCG is equiv- alent to the triangle EBC, that is, to half the parallelogram BD. Now, since the diagonal CE is given in magnitude and position, its middle point II is given in position, and therefore H is always a point in the required line, wherever A is taken. Hence, so long as A and II are different points, the straight line AHG (I. post 1) is de- termined. This, however, is no longer so, if the given point A be the intersection of the diagonals, that is, the point H, as in that case only one point of the required line is known, and the problem becomes indetermhiate, any straight line whatever, through H, equally answering the conditions of the problem; and we are thus led by the solution of the problem to the con- 220 EXERCISES. elusion, that in a parallelogram a point may he found, such that a)iy straight line wliatever drawn through it, bisects the parallelogra7n / and this is a porism. The seventy-sixtii pi'oposition of the Exercises, when con- sidered in a particular manner, affords another instance of a porism ; as it appears that if a circle and a point D or E be given, another point E or D may be found, such that any circle whatever, desciibed through D and E, will bisect the circum- ference of the given circle; and this may be regarded as the indeterminate case of the problem, in which it is required, through two given points, to describe a circle bisecting tlie cir- cumference of another given circle, — a problem which is always determinate, except when the points are situated in the man- ner supposed in the proposition. 9. Isopjeritnetrical figures are such as have their perimeters, or bounding lines, equal. 10, The general problem of the tangencies, as understood by the ancients, is as follows: Of three points, three straight lines, and three circles of given radii, any three being given in posi- tion ; it is required to describe a circle passing through the points, and touching the straight lines and circles. This gen- eral problem comprehends ten subordinate ones, the data of which are as follows: (l.) three points; (2.) two points and a straight line ; (3.) two points and a circle ; (4.) a point and two straight lines ; (5.) a point, a straight line, and a circle ; (G.) a point and two circles; (7.) three straight lines; (8.) two straight lines and a circle ; (9.) a straight line and two circles; and (10.) three circles. The first and seventh of these are the second and fifth corollaries of the twenty-fifth proposition of the third book. If a circle be continually diminished, it may be regarded as becoming ultimately a point. By being continually enlarged, on the contrary, it may have its curvature so much diminished that any portion of its circumference may be made to differ in as small a degree as we please from a straight line. Viewing the subject in this light, we may regard the first nine of the problems now mentioned, as comprehended in the tenth. Thus, we shall have the first, by supposing the circles to become in- finitely small ; the seventh, by supposing them infinitely great; EXEHCISES. 221 the fifth, by taking one of them infinitely small, one infinitely great, and one as a circle of finite magnitude; and so on with regard to the othei'S. These views of tlie subject tend to illus- trate it ; but they do not assist in the solution of the problems. Hcho. 4. In tiie fifth problem, the straight line may fall with- out the circle, may cut it, or may touch it ; the point may be without the circle, withfn it, or in its circumference ; or it may be in the given straight line, or on either side of it ; and it will be an interesting exercise for the student, in this and many other problems, to consider the variations arising in the solu- tion from such changes in the relations of the data, and to de- termine what relations make the solution possible, and what render it impossible. It may also be remarked, that in many problems there will be slight variations in the proofs of differ- ent solutions of the same problem, even when there is no chi"ige in the method of solution; such as in the present in- stance, when the required circle is touched externally, and Avhen internally. Thus, while in one case angles may coincide, in another the corresponding ones may be vertically ojiposite; and the reference luay sometimes be to the coJiverse of the first corollary and sometimes to the converse of the second corollary of the eighteenth proposition, book third. It is, in general, un- necessary to point out these variations, as, though they merit the attention of the student, they occasion no difficulty. PROPOSITIONS. Prop. I. — Tiieor. — If an angle of a triangle, he bisected by a straight line^ which likewise cuts the base,, the rectangle con- tained by the sides of the triangle is equivalent to the rectangle contained by the segments of the base,, together with the square of the straight line bisecting the angle. Let ABC be a triangle, and let the angle BAC be bisected by AD; the rectangle BA.AC is equal to the rectangle BD.DC, together with the square of AD. Describe the circle (III. 25, cor.) ACB about the triangle ; produce AD to meet the circumference in E, and join EC. Then (hyp.) the angles BAD, CAE are equal ; as are also (IIL 18, cor. 1) the angle B and E, for they are in the same segment; 222 EXERCISES. thej-efore (V. 3, cor.) in the ti-iangles ABD, AEC, as BA : AD :: EA : AC; and consequently (V. 10, cor.) the rectangle BA. AC is equivalent to EA.AD, that is (II. 3), ED.DA, together with the square of AD. But (III. 20) the rectangle ED.DA ia equivalent to the rectangle BD.DC ; therefore the rectangle BA.AC is equivalent to BD.DC, together with the square of AD ; wherefore, if an angle, etc. Scho. From this proposition, in connection -with the fourth proposition of fifth book, we have the means of computing AD, when the sides are given in numbers. For, by the fourth, BA: AC :: BD : DC; and, by composition, BA+AC : AC : : BC : DC. This analogy gives DC, and BD is then found by taking DC from BC. But by this proposition BA. AC = BD.DC + AD-; therefore from BA.AC take BD.DC, and the square root of the remainder will be AD, In a similar manner, from the fourth proposition of the fifth book, and the second proposition of these Exercises, the line bisecting the exterior vertical angle may be computed; and from the third proposition of these Exercises, in connection with the eleventh or twelfth of the second book, the diameter of the circumscribed circle may be computed, when the sides of the triangle are given in numbers. Prop. II. — Theor. — If an exterior angle of a triangle be bisected by a straight line, which cuts tlie base produced ; the rectangle contained by the sides of the triangle, and the square of the bisecting line are together equivalent to the rect- angle contained by the segments of the base intercepted between its extremities and the bisecting line. Let, in the foregoing diagram, the exterior angle ACF of the triangle BAC be bisected by HC ; the rectangle BC. AC and EXERCISES. 223 the square of HC are together equivalent to the rectangle BH.AII. Describe the circle (III. 25, cor.) ABEC about the triangle BAG; produce HC (I. post. 2) to E, and join EA. Then, since (l>yp.) the angles FCII and HCA are equal, their supple- ments FCE and ACE (T. def. 20 and ax. 3) are also equal ; and (III. 18, cor. 1) B and E are equal. Therefore (V. 3, cor.) in the triangles BCH and EAC, BC : CH : : EC : AC, and con- sequently (V. 10, cor.) the rectangle BC. AC is equivalent to EC.CII. To each add square of CII; therefore BC.AC+CIP are equivalent to EC.CH+Cff; or (II. 3) BC.AC + CIP are equivalent to EH.CH ; or (III. 21, cor.) BC.AC + CIP are equivalent to BH. AH. Therefore, if an exterior angle, etc. Prop. III. — Theor. — If from an angle of a triangle a per- pendicular be drawn to the basej the rectangle contained by the sides of the triangle is equivalejit to the rectangle contained by the perpendicular and iAe diameter of the circle described about the triangle. Also, in the foregoing diagram, let ABC be a triangle, AL the perpendicular from the angle A to BC ; and AE a diameter of the circumscribed circle ABEC; the rectangle BA.AC is equivalent to the rectangle AL.AE. Join EC. Then the right angle BLA is equal (III. 11) to the angle EC A in a semicircle, and (HI. 18, cor. 1) the angle B to the angle E in the same segment ; therefore (V. 3, cor.) as BA : AL : : EA : AC; and consequently (V. 10, cor.) the rectangle BA.AC is equivalent to the rectangle EA.AL. If, therefore, from an angle of a triangle, etc. Prop. IV. — Theor. — The rectam^le contained by the diago- nals of a quadrilateral inscribed in a circle, is equivalent to both the rectangles contained by its opposite sides. Let ABCD be a quadrilateral inscribed in a circle, and join AC, BD ; the rectangle AC.BD is equivalent to the two rect- angles AB.CD and AD.BC. Make the angle ABE equal to DBC, and take each of them from the whole angle ABC ; then the remaining angles CBE, ABD are equal; and (HI. 18, cor. l) the angles ECB, ADB are equal. Therefore (V. 3, cor.) in the triangles ABD, EBC, 224 EXERCISES. as BC : CE BD.CE Again BD : DA; wlience (V. 10, cor.) BC.DA= in the triangles BAE, BDC, because (const.) the angles ABE, DBC are equal, as also (III. 18, cor. 1) BAE, BDC; therefore (V. 3, cor.) as BA : AE : : BD : DC; ^\ hence (V. 10, cor.) BA.DC==BD.AE. ' Add these equivalent rectangles to the equivalents BC.DA and BD.CE; then BA.DC + BC.DA = BD.CE + BD.AE, or (II. 1) BA.DC + BC.DA=BD.AC. There- fore, the rectangle, etc. Cor. 1. If the sides AD, DC, and consequentlj^ (TIT. 16, cor. 1) the arcs AD, DC, and the angles ABD, CBD be equal, the rectangle BD.AC is equivalent to AB.AD together with BC. AD, or (II. 1) to the rectangle under AD, and the sum of ABand BC. Hence (V. 10, cor.) AB-f-BC : BD : : AC : AD or DC. Cor. 2. If AC, AD, CD be all equal, the last analogy be- comes AB+BC : BD : : AD : AD ; whence AB + BC=BD. Hence in an equilateral triangle inscribed in a circle, a straight line drawn from the vertex to a point in the arc cut oiF by the base is equal to the sum of the chords drawn from that point to the extremities of the base. Prop. V. — Theor. — The diagonals of a qtiadrilateral in- scribed in a circle, are proportional to the sums of the rectan- gles contained by the sides meet'nr/ at their extremities. Let ABCD be a quadi-ilateral inscribed in a circle, and AC, BD its diagonals; AC : BD : : BA.AD-h BC.CD : AB.BC+AD.DC. If AC, BD cut one another perpendicu- larly in L, then (Ex. 2) AK being the diame- ter of the circle, AL. AK = BA.AD, and CL. AK=:BC.CD; whence, by addition (I. ax. 2), AL.AK + CL.AK, or (II. 1) AC.AKz= B.\.AD+BC.CD. In a similar manner, it would be proved that BD.AK = AB.BC Hence ACIAK : BD.AK, or (V. 1) AC : BD : • BA.AD + BC.CD : AB.BC + AD.DC. But if AC be not perpendicular to BD, draw AEF perpen- + AD.DC. EXERCISES. 225 dicnlar and CF parallel to BD, and DGH perpendicular and BH parallel to AC. Then, because EF is equal to the perpen- dicular drawn from C to BI), and GH equal to the one drawn from B to AC ; it would be prov- ed as before, that AP\AIv = BA. AD + BC.CD, and I)n.AK = AB.BC + AD.DC. Hence, AF. AK : DH.AK, or (V. l) AF : DII :: BA.AD-^BC.CD : AB. BC + AD.DC. But the triani^les AFC, DIIB are equiangular, liav- in<x the rio;ht anojes Fand H, and the angles ACF, DBH, each equal (I. 16) to ALD; therefore (V. 3) AF : AC : : DH : DB, and alternately AF : T3IT : : AC : DB. Hence (IV. 7) the foregoing analogy becomes AC : BD : : BA.AD + BC.CD : AB.BC-hAD.DC. Wherefore, the diagonals, etc. Scho. From this proposition and the last, when the sides of a quadrilateral inscribed in a circle are given, we can find the ratio of the diagonals and their rectangle, and thence (V, 15) the diagonals themselves. Also, if the sides be given in num- bers, we can compute the diagonals. Thus, let the sides taken in succession round the figure be 50, 78, 104, and 120. Then, the ratio of the diagonals will be that of 50 x 784-104 x 120 to 60x120-1-78x104; that is, 16880 to 14112, or 65 to 56, by- dividing by 252. Again, the rectangle of the diagonals is 50 x 104-1-78x120, or 14560. But similar rectilineal figures are as the squares of the corresponding sides, and consequently the sides are as the square roots of the areas ; therefore, taking 65 and 56 as the sides of a rectangle, we have its area equal to 3640; and ^ 3640 is to 4/14560, or -^Z 3640 is to |/ (4x3640), that is, 1 : 2 : : 65 : 130 : : 56 112 ; so that 130 and 112 are the diagonals. Prop. VI. — Theor. — If in a straight line drawn through the center of a circle^ and on the same side of (he center, (wo povits be taken so that the radius is a mean proportional her- tween their distances from, the cei^ter / two straight lines drawn 15 226 EXEECISES. from, those points to any point whatever in the circumference^ are proportional to the segments into which the circumference divides the straight line intercepted leticeen the same points. Let ABC l)e a circle, and CAE a straight line drawn throngh its center D ; if ED : DA : : DA : DF, and if BE, BF be drawn from any point B of the circumference ; EB : BF : : EA : AF. Join AB, BD, Then, since DB is equal to DA, we have (hyp.) ED : DB :: DB : DF. The two triangles EDB, BDF, therefore, have their sides about the common angle D proportion- al ; wherefore (V. 6) the angle E is equal to FBD. Now the angle liAD is equal (I. 20) to the two angles E and EBA, and also (I. 1, cor.) to ABD ; wherefore E and EBA are (I. ax. 1) equal to ABD. From these take the equal angles E and FBD, and (I. ax. 3) the remaining angles EBA, ABF are equal ; and therefore (V. 4) in the triangle EBF, EB : BF : : EA : AF. If, therefore, in a straight line, etc. Cor. Join BC, and produce EB to G. Then, since ABC is (III. 11) a right angle, it is equal to the two EBA, CBG. From these equals take the equal angles ABF, EBA, and the remainders FBC, CBG are equal ; and therefore (V. 4, 2d case) EB : BF : : EC : CF. But it has been proved that EB : BF : : EA : AF ; therefore (IV. 1) EA : AF : : EC : CF. Hence, if the segments EA, AF be given, the point C may be determined by the method shown in the third corollary to the sixteenth proposition of the first book; and the circle ABC may then be described, its diameter AC being determined. Scho. The circle may also be determined in the following manner: Since (hyp.) ED : DA : : DA : DF, by division, EA : DA : : AF : DF ; whence, alternately and by division, EA— AF : AF : : AF : DF. Hence DF is a third proper- tional to the difference of E A, AF, and to AF, the less ; and thus the center D is determined. From the last analogy also we obtained (IV. 11) EA— AF : EA : : AF : AD; an anal- ogy which serves the same purpose, since it shows that the ra- EXERCISES. 227 dius of the circle is a fourth proportional to the difference of the segments EA, AF, and to those segments themselves. Prob. VII, — TnEOR. — The perpendiculars drawn from the three angles of any triangle to the ojjposite sides, intersect one another in the same point. If the triangle be right-angled, it is plain that all the perpen- diculars pass through the right angle. But if it be not right- angled, let ABC be the triangle, and about it describe a circhi; then, B and C being acute angles, draw ADE perpendicular to BC, cutting BC in D, and the circumference in E; and make DF equal to DE ; join BF and produce it, if necessary, to cut AC, or AC produced in G ; BG is perpeixlicular to AC. Join BE ; and because FD is equal to DE, the angles at D right angles, and DB common to the two triangles FDB, EDB, the angle FBD is equal (I. 3) to EBD ; but (III. 18, cor. 1) CAD, EBD are also equal, because they are in the same segment ; therefore CAD is equal to FBD or GBC, But the angle ACB is common to the tv o triangles ACD, BCG; and therefore (I, 20, cor. 5) the remaining angles ADC, BGC are equal ; but (const.) ADC is a right angle; therefore also BGC is a right angle, and BG is perpendicular to AC, In the same manner it would be shown that a straight line CII, drawn through C and F, is perpendicu- lar to AB. The three perpendiculars therefore all pass through F ; wherefore, the perpendiculars, etc, Scho. This limitation prevents the necessity of a different ease, which would arise if the perpendicular AD fell without the triangle. If the angle A be obtuse, the point F lies with- out the circle, and BF, not produced, cutsAC produced. The proof, however, is the same, and it is very easy and obvious. Another easy and elegant proof, of which the following is an outline, is given in Garnier's "Reciproques," etc., Theor. III., page 78 : Draw BG and CH perpendicular to AC and AB ; join GH, and about the quadrilaterals AHFG and BHGC de- 8cribe circles, which can be done, as is easily shown ; draw also AFD. Then the angles BAD, BCH are equal, each of 223 EXERCISKS. them being equal (TIT, 18, cor. 1) to HGF ; and the angle ABO "being coniinoii, ADB is equal (I. 20) to BHC, and is therefore a right angle. Prop. VTTT. — Theor. — From AB, the greater side of the tri- angle ABC, cut off AD equal to AC, and join DC; draw AE bisecting the vertical angle BAC, andjoiji DE ; d aw also AF perpendicular to BC, and DG parallel to AE. Then (1.) the angle DEB is eqtdvalent to the difference of the angles at the base, ACB, ABC ; or of BAF, CAF, or of AEB, AEC ; and DE is equal to EC ; (2.) the angles BCD, EAF are each equiv- alent to half the same difference ; (3.) ADC or ACD is equiva- lent to half the sum of the angles at the base, or to the comple- ment of half the vertical angle ; (4.) BG is equivalent to the difference of the segments BE, EC, made by the line bisecting the vertical angle. ^ 1. In the triangles AED, AEC, AD, AC are equal, AE com- mon, and the contained angles equal ; therefore (I. 3) DE is equal to EC, the angle ADE to ACE, and AED to AEC. But (I. 20) because BD is produced, the angle ADE is equivalent to B and BED; therefore BED is the differ- ence of B and ADE, or of B and ACB. Also BED is the difference of AEB, AED, or of AEB, AEC. Again : ABF, BAF are equivalent to ACF, CAF, each pair being (L 20, cor. 3) equivalent to a right angle. Take away ABF; then, because the difference of ACF, ABF is BED, there re- mains BAF equivalent to BED, CAF ; that is, BED is the difference of BAI\ CAF. 2. The difference BED is equivalent (T. 20) to the two angles ECD, EDC, which (I. 1, cor.) are equal ; therefore ECD is half of BED. Again : in the triangles AlID, AIIC, because AD, AC are equal, AIT common, and the contained angles equal, DII is equal (I. 3) to IIC, and the angles at H are equal, and are therefore right angles. Then, in the triangles AEF, CEII, the angles AFE, CUE are equal, and AEC common ; therefore EXERCISES. 229 (T. 20, cor. 5) the article EAF is equal to ECU, which has been proved to be equivalent to the half of BED, 3, Since the anole BAG is common to the triangjles ABC, ADC, the angles ADC, ACD are (I. 20) equal to ABC, ACB ; and therefore, since ADC, ACD are equal, each of them is equivalent to half the sum of ABC, ACB ; also either of them, ADC, is the complement ofDAIi, half the vertical angle, since AHD is a right angle. 4. Because DH, HC are equal, and HE, DG parallel, GE is equal (V. 2) to EC; and therefore BG, the difference of BE, GE, is also the difference of BE, EC. Sclw. 1. It is easy to prove without proportion, that if AB (in the figure for the above proposition) be bisected in D, the straight line DE parallel to BC bisects AC, and that the trian- gle ADE is a fourth of ABC. For (I. 15, cor. 5) the triangles BDC, BEC are equal. But (I. 15, cor. 5) BDC is half of ABC ; and therefore BEC is half of ABC, and is equal to BEA. Hence the bases AE, EC are equal ; for if they were not, the triangles ABE, CBE (I. 15, cor. 6) would be unequal. Again : because AD, DB are equal, the triangle ADE is (I. 15, cor. 5) half of ABE, and therefore a fourth of ABC. Conversely, if DE bisect AB, AC, it is parallel to BC. For (T. 15, cor.) the triangles BDC, BEC are each half of ABC; and these being therefore equal, DE is parallel (I. 15, cor. 5) to BC. Hence, it is plain (I. 14) that the straight lines joining DE, DF, EF divide the triangle ABC into four equal triangles, similar to the whole and to one another; and that each of these lines is equal to half the side to which it is parallel. Scho. 2. Instead of cutting off AD equal to AC, AC may be produced through C, and by cutting off, on AC thus produced, a part terminated at A, and equal to AB, and by making a con- struction similar to that of the foregoing proposition, it will be easy to establish the same properties as those above demon- strated, or ones exactly analogous. Prop. IX. — Prob. — Giveii the base of a triangle, the differ- ence of the S'des, and the difference of the angles at the base; to construct it. 230 EXERCISES. Make BC equal to the given base, and CBD equal to half the difference of the angles at the base ; from C as center, at a distance equal to the difference of the sides, describe an arc cutting BD in D ; join CD and produce it ; make the angle DBA equal to BDA ; ABC is the required triangle. For (I. 1, cor.) AD is equal to AB, and the difference of AC, AD, or of AC, AB, is CD ; and (Ex. 8) since AD is equal to AB, CBD is equal to half the difference of the angles at the base. The triangle ABC, therefore, is the one required, as it has its base equal to the given base, the difference of its sides equal to the given difference, and the difference of the angles at the base equal to the given dif- ference of those angles. Method of C\miputation. In the triangle BCD there are given BC, CD, and the angle CBD ; whence the angle C can be computed ; and the sum of this and twice CBD is the angle ABC. Then, in the whole triangle ABC, the angles and the Bide BC are giveti ; whence the other sides may be computed ; or, one of them being computed, the other will be found by means of the given difference CD. Prop. X.— Prob. — Given the segments into which the base of a triangle is divided by the line bisecting the vertiecd anglQ and the difference of the sides ; to construct the triangle. Construct the triangle CED, having the sides CE, ED equal to the given segments, and CD equal to the given dillerence of the sides; produce CE, and make EB equal to ED; bisect the angle BED by EA, meeting CD produced in A, and join AB ; ABC is the required, triangle. A For, in the triangles AEB, AED, BE is equal to ED, EA common, and the angle BEA equal to DEA ; therefore (I, 3) BA is equal to DA, and the angle EAB to EAD. Hence, ABC is the re- quired trianale; for CD, the difference of its sides, is equal to the given differ- ence, and BE, EC, the segments into which the base is FXERCISE8. 231 divided by the line bisecting the vertical angle, are equal to the given segments. Meth d of Computation. The sides of the triangle CDE are given, and therefore its angles may be computed; one of which and the sni)plement of the other are tlie angles C and B. Then, in the triangle ABC, the angles and BC are given, to compute the remaining sides. OTHERWISE, Since (V. 4) CE : EB : : CA : AB, we have, by division, CE— EB : EB : : CA— AIJ : AB ; which, therefore, becomea known, since the first three terms of the analogy are given; and thence AC will be found by adding to AB the given dit- ference of the sides. Prop. XT. — I V.ob. — Given the base of a triangle^ the vertical angle, and the difference of the sides ; to construct the tri- angle. Let MNO be the given vertical angle; produce ON" to P, and bisect the angle MNP by NQ. Then, make BD equal to the difference of the sides, and the angle ADC equal to QNP; from B as center, with the given base as ra- dius, describe an arc cutting DC in C ; and make the angle DCA equal to ADC ; ABC is the required tri- angle. For (const.) the angles ACD, ADC are equal to MNP, and there- fore (I. 20 and 9) the angle A is equal to MN). But (I. 1, cor.) AD is equal to AC, and therefore BD is the difference of the sides AB, AC ; and the base BC is equal to the given base ; wherefore ABC is the triangle required. Method of Computation. In the triangle CBD, the sides BC, BD, and the angle BDC, the supplement of ADC or MNQ are given ; whence the other angles can be computed. The rest of the operation will proceed as in the ninth proposition of these Exercises. Q N 232 EXERCISES. Prop. XII. — Prob. — Given one of the angles at the base of a triangle, the difference of the sides, and the difference of the segments into which the base is divided by the line bisecting the vertical angle; to construct the triangle. Construct the triangle DBG, liaving the angle B equal to the given angle, BD equal to the difference of the sides, and BG equal to the difference of the segments ; draw DC perpendicular to DG, and meet- ^'^ '^ ing BG produced in C; produce BD, and make the angle DCA equal to CDA ; B G E c ABC is the triangle required. For it has B equal to the given angle, and the differ- ence of its sides BD equal to the given difference; and if AHE be drawn bisecting the angle BAC, it bisects (I. 3) CD, and is perpendicular to it ; it is therefore parallel to DG, one side of the triangle CDG; and, bisecting CD in II, it also (V, 2) bisects GC in E. Hence BG, the difference of BE, GE is also the difference of BE, EC, the segments into which the base is divided by the line bisecting the vertical angle. Method of Computation. In the triangle DBG, BD, BG, and the angle B are given ; whence (Trig. 3) we find half the difference of the angles BGD, BDG, which is equal to half the angle C. Then (by the same proposition) we have in the tri- angle ABC, tan I (C-B) : tan ^ (C+B) : : c-b or BD : c+ b ; whence the sides c and b become known, and thence BC by the first case. For (I. 20) BEA=r| A+C, and consequently BEA— | A = C. But (I. 16) BGD = BEA, and BDG^BAE^^A; and there- fore BGD— BDG =C. Prop. XIII. — Prob. — Given the base of a triangle, the verti- cal angle, and the sutn of the sides ; to construct it. Make BD equal to the sum of the sides, and the angle D equal to half the vertical angle ; from B as center, with the base as radius, describe an arc meeting DC in C ; and make the angle DCA equal to D ; ABC is the required triangle. For (I. 1, cor.) AD is equal to AC, and therefore BA, AC are equal to BD, the given sum. Also (I. 20) the exterior angle BAC is equal to the two D and ACD, or to the double of D, EXERCISES. 233 because D and ACD are equal ; therefore, since D is half the given vertical angle, BAG is equal to that ansjle. The triancjle ABC, therefore, has its base equal to the given base, its vertical angle equal to the given one, and the sum of its sides equal to the given sum ; it is therefore the triangle required. Scho. Should the circle neither cut nor touch DC, the prob- lem would be impossible with the proposed data. If the circle meet DC in two points, there will be two triangles, each of which will answer the conditions of the problem. These tii- angles, however, will differ only in position, as they will be on equal bases, and will have their remaining sides equal, each to each. This problem might also be solved by describing (III. 19) on the given base 15C a segment of a circle containing an angle equal to half the vertical angle ; by inscribing a chord BD equal to the sura of the sides; by joining DC; and then proceeding as before. The construction given above is prefer- able. / Method of Computation. In the triangle BDC, the angle D, and the sides BC, BD are given ; whence the remaining angles can be compute 1 ; and then, in the triangle ABC, the angles and the side BC are given, to compute the other sides. Prop. XIV. — Prob. — Given the vertical angle of a triangle^ and the segments into which the line iisectiiig it divides the base ; to construct it. In the straight line BC, take BH and CII equal to the seg- ments of the base ; on BC describe (III. 19) the segment BAC containing an angle equal to the vertical angle, and complete the circle ; bisect the arc BEC in E ; draw EHA, and join BA, CA ; ABC is the re- quired triangle. For (III. 16, cor. 1) the angles BAII, CAH are equal, because the arcs BE, EC are equal ; and therefore the triangle ABC manifestly answers the conditions of the ques- tion. 234 EXEECISF.8. Scho. The construction might also be effected by describing on BH and CH segments each containing an angle equal to half the vertical angle, and joining their point of intersection A ■with B and C. Another solution may be obtained by the principle (V. 4) that BA • AC : : BII : HC. For if a trian- gle be constructed having its vertical angle equal to the given one, and the sides containing it equal to the given segments, or having the same ratio, that triangle will be similar to the required one ; and therefore on CB construct a triangle equi- angular to the one so obtained. Method of Computation. Join BE, and draw ED perpen- dicular to BC. Then BD or DC is half the sum of the sear- nients BII, HC, and DH half their difference ; and BD is to DM, or twice BD to twice DH, as tan DEB to tan DEH. Now it is easy to show that BED is half the sum of the angles ABC, ACB, and HED half their difference ; and therefore these angles become known; and BC being given, the triangle ABC is then resolved by the method for the Hrst case. Cor. Hence, we have the method of solving the problem in wliich the base, the vertical angle, and the ratio of the sides of a triangle are given^ to construct it. For (V. 4) the sides be- ing proportional to the segments BH, HC, it is only necessary to divide the given base into segments proportional to the sides and then to proceed as above. * Prop. XV. — Pbob. — Given the base^ the perpendicular^ and the vertical angle of a triangle / to construct it. Make BC equal to the given base, and (HI. 19) on it de- scribe a segment capable of containing an angle equal to the vertical angle ; draw AK parallel to BC, at a distance from it equal to the given perpendicular and meeting the arc in A ; join AB, AC; A13C is evidently the tiiangle re- quired. Method of Compiitat'on. Draw the perpendicular AD, and parallel to it draw LGH, through the center G; join BG, AG, AH. Now, iiuce AH evidently bisects the anglu BAC, the angle HAD or EXERCISES. 235 n is (Ex. 8) equal to half the difference of the ansrles ABC, ACB, and therefore (III. 1 0) AGK is the wliole difference of those angles. Then, in the riglit-angled triangle BP'G, the angles and BF being known, FG can be computed ; from which and from AD or KF, KG becomes known. Now, to the radius BG or AG, FG is the cosine of BGF, or BAG, and KG the cosine of AKG; and therefore FG : KG : : cos liAC : cos AGK. Hence the angles ABC, ACB become known, and thence the remaining sides. iSc/io. Should the parallel AK not meet the circle, the solu- tion would be impossible, as no triangle could be constructed having its base, perpendicular, and vevlic:il angle of the given magnitudes. If the parallel cut the cinrle, there will be two triangles, either of which will answer the conditi(m of the ques- tion. They will differ, however, only in position, as their sides will be equal, each to each. If the parallel touch the circle, there will be only one triangle ; and it will be isosceles. Pkop. XVI. — Prob. — Given the base of a triangle, the ver- tical angle, and the radius of the inscribed circle ; to construct the triangle. Let GHK be the given angle ; produce GH to L, and bisect LHK by HM; on the given base BC desciibe (III. 19) the seg- ment BDC, containing an angle equal to GHM ; draw a straight line parallel to BC, at a distance equal to the given ra<lius, and meeting the arc of the segment in D; join I)B, DC; and make the angles DBA, DCA equal to DBC, DCB, each to each ; ABC is the required triangle. L ii g Produce BD to E, and drawDF perpen- dicular to BC. Then, since (const.) the angle BDC is equal to GHM, the two jj DBC, DCB are equal (I. 20 and 9) to LHM, and therefore (const.) ABD, ACD are equal to KHM. But (1. 20) BDC is equal to BEC, ECD, or to BAC, ABD, ACD, because (I. 20) BEC is equal to BAC, ABD. Therefore BAC, ABD, ACD are equal to GHM; from the former take ABD, ACD, and from the latter KHM, which is equal to 236 KXERCISE8. them, and the remainders BAG, GHK are equal. It is plain, also (I. 14), tliat perpendiculars drawn IVoni D to AB and AC would be each equal to DF ; and therefore a circle described from D as center, with DF as radius, would be inscribed in the trian<ijle ABC ; and BC being the given base, and A being equal to the given vertical angle, ABC is the required tri- angle. The method of computation is easily derived from that of the preceding proposition. Scho. Should the parallel to BC not meet the arc of the seg- ment, the solution would be impossible, as there would be no triangle which could have its base, its vertical angle, and its inscribed circle of the given magnitudes. If the parallel be a tangent to the arc, the radius of the inscribed circle would be a maximum. Hence, to solve the problem in which the base and the vertical angle are given, to construct the triangle, so that the inscribed circle may be a maximum, describe the seg- ment as belbre, and to find D bisect the arc of the segment. The rest of the construction is the same as before ; and the tri- angle will evidently be isosceles. Prop. XVII. — Prob. — Given the three lines drawn from the vertex of a triangle, one of them 'perpend cnlorto the base, one bisecting the base, and one bisecting the vertical angle y to con- struct the triitngle. Take any straight line BC and draw DA perpendicular to it, and equal to the given perpendicular; fi-om A as center, with radii equal to the lines bisecting the vertical angle and the base, describe arcs cutting BC in E and F, and draw AEH and AF ; through F draw GFII perpendicular to BC, and draw AG making the angle IIAG equal to II, and cutting HG in G; from G as center, with GA as radius, describe a circle cutting BC in B and C; join AB, AC ; ABC is the triangle required. For (HI. 2) since GFII is perpendicu- lar to BC, BC is bisected in F ; and (III. 17) the arcs BII, HO are equal. Therefore (III. 16, cor. 1) the angles BAII, CAH EXERCISES. 237 are equal. Hence, in the triangle ABC, the perpendicular AD, the line AE bisecting the veitical angle, and the line AF bisecting the base, are equal to the given lines. Therefore ABC is the triangle required. Scho. It" the three given lines be equal, the problem is inde- terminate ; as any isosceles triangle, having its altitude equal to one of the given lines, will answer the conditions. 3fethod of Computation. Through A draw a parallel to BC, meeting HG ))rodnced in K, Then, in the right-angled triangle ADE, AE, AD being given, DAE, or H, may be computed ; the double of which is AGK ; and AK or FD being given, AG, GK can be found, and thence GF. Hence, if GB were drawn, it and GF being known, BF, and the angle BGF, or BAC, can be computed. The rest is easy; DAE, half the difference of B and C, being known. Analysis. Let ABC be the required triangle, AD the per- pendicular, and AE, AF the lines bisecting the vertical angle and the base. About ABC describe (HI, 25, cor. 2) a circle, and join its center G, with A and F, and produce GF to meet the circumference in H. Then (HI. 2) GFH is perpendicular to BC, and (IH. \1) the arcs BH, HC are equal. But (HI. 16) the equal angles BAE, CAE at the circumference stand on equal arcs; and therefore AE being produced, will also pass through H; and the point H, and the angle GHA and its equal HAG are given. Hence also the center G and the circle are given, and the method of solution is plain. Prop. XVIH. — Prob. — Given the base of a triangle^ the ver- tical angle, and the straight line bisecting that angle y to con- struct the triangle. On the given base BC describe (HI. 19) the segment BAC capable of containing an angle equal to the given vertical an- gle, and complete the circle ; bisect the arc BEC in E, and join EC; perpendicular to this, draw CF equal to half the line bisecting the vertical angle, and from F as center, with FC as radius, describe the circle CGII, cutting the straight line pass- in through E and F in G and H ; make ED equal to EG, and draw EDA; lastly, join AB, AC, and ABC is the required triangle. 238 EXEKCI8E8. For the triangles CEA, CED are equiangular, the angle CEA being tomnion, and BCE, CAE being each equal to BAE. Therefuie AE : EC : : EC : ED, and (V. 9, cor.) AE.ED= EC=. But (ITI. 16, cor. 3, and 21) HE.EG or HE.ED=EC and therefore AE.ED^HE.ED; whence AE=:HE, and (I. ax. 3) AD=GH = '2CF. AD is therefore equal to the given bisect- ing line, and it bisects the angle BAC. Hence ABC is the re- quired triangle. Method of Computation. Draw EL perpendicular to BC, and join CH. Then BCE is equal to BAE, half the vertical angle A ; and therefore, to the radius EC ; CL is the cosine of ■§ A, and CF is the tangent of CEF to the same radius ; where- fore, to any radius, CL : CF, or BC : AD : : cos ^ A : tan CEF or cot EFC ; and hence the angle H, being half of EEC, is known. Also ECU is the complement of II, because ECF is a right angle, and FCH equal to H. But (Tgig. 2) EC : EH or EA : : sin II : sin ECU, or cosH ; or (Trig. defs. cor. 6) EC : EA : : It : cot H. Also in the triangle ACE, EC : EA : : Bin -^ A : sin ACE ; whence (IV. 7) R : cot H : : sin ^ A : sin ACE ; whence ACE may be found ; and if from it, and from ABE, its supplement (BE being supposed to be joined) BCE be taken, the remainders are the angles at the base. Analysis. Let ABC be the required triangle, and let AD, the line bisecting the vertical angle, be produced to meet the circumference of the circumscribed circle in E; join also EC. Then (III. 19) the circumscribed circle is given, since the base and vertical angle are given ; and the arc BEC is given, as are also its half EC, and the chord EC. Now the triangles AEO, EXERCISES. 239 DEC are equiangular ; for the angle CEA is common, and (IH. 18, cor. 1) BCE is equal to BAE or EAC. Hence AE : EC : : EC : ED, and theiefore AE.ED=ECl Hence (IH. 21) it is evident that if EC be made a tangent to a circle, and if through the extremity of the tangent a line be drawn cutting the circle, so that the part within the circle may be equal to AD, DE will be equal to the external part ; whence the con- struction is manifest. Prop. XIX. — Proh. — Given the straight liius drawn from the three angles of a trian;^ le to the points of bisection of the opposite sides / to construct the triangle. Trisect the three given lines, and describe the triangle ABC having its three sides respectively equal to two thirds of the three given lines ; complete the parallelogram ABEC, and draw the diagonal AE ; produce also CB, making BE equal to BC ; and join FA, FE; AFE is the required triangle. Produce AB, EB to G, H. Then (H. 14, cor.) AE, BC bisect each other in D, and therefore FD is equal to one of the given lines, for BD is one third of it, and FB two thirds. Again : because FB, BC are equal, and HB pai-allel to AC, FA is bisected in H, and HB is half of AC or BE. Hence, HE is equal to another of the given lines, and it bisects FA. In the same manner it would be proved, that AG is equal to the remaining line, and that it bisects FE. Hence FAE is the triangle required. Method of Computation. BD, which is a third of one of the given lines, bitects AE, a side of the triangle ABE, in •which the sides AB, BE are respectively two thirds of the two remaining lines. Then (I. 24, cor. 6, and II. 12, cor.) 2AD^= AB'+BE*— 2BD'; whence AD, and consequently AE may be found ; and in the same manner the other sides may be com- puted. Prop. XX.— Prob, — Given the three perpendiculars of a tri- angle y to construct it. 240 EXERCI8E8. A- B- C- D- E Let A, B, C be three given straight lines; it is required to describe a triangle having its three perpendiculars respectively eqnJil to A, B, C. Take any straight line D, and describe a triangle EFG. hav- ing the sides FG, P'E, EG third i)ro- portionals to A and 1), B and D, C and D; and draw the perpendiculars EH, GL, FK. Then the rectangles rG.A,EF.Bare equal, each being equal to the square ofD; and therefore EF : FG : : A : B. But in the similar triangles EHF, GLF, EF : FG : : EH : GL; where- fore EH : GL : : A : B ; and in the same manner it would be proved, that EH : FK : : A : C. Hence (IV. 7) if EH be equal to A, GL is equal to B, and FK to C ; and EFG is the triangle requii-ed. But if EH be not equal to A, make EM equal to it, and draw NMO parallel to FG, and meeting EF, EG, produced, if neces- sary, in N and O ; ENO is the required triangle. Draw OP perpendicular to EN. Then EM : EH : : EO : EG, and OP GL : : EO : EG ; whence (IV. 7 and alternately) EM : OP : EH : GL ; or, by the foregoing part, EM : OP : : A : B wherefore (IV. 7) since EM is equal to A, OP is equal to B and it would be proved in a similar manner, that the perpen- dicular from N to EO is equal to C. Method of Computation. By dividing any assumed num- ber successively by A, B, C, we find the sides of the triangle EFG, and thence its angles, or those of ENO; whence, since the perpendicular EM is given, the sides are easily found. Or, when the sides of EFG are found, its perpendicular EH may be comj)Uted in the manner pointed out in the note to tlie twelfth proposition of the second book. Then EH : A : : FG: NO : : EF : EN : : EG : EO. Prop. XXI. — Prob, — Given the sum of the legs of a right- angled triangle, and the sum of the hypothenuse, and the per- pendicular to it from the right angle; to construct the triangle. EXERCISES. 241 Let tho sum of the legs of a right-angled triangle be equal to the straight line A, and the sum of the hypothenuse and perpendicular equal to BC ; it is required to construct the tri- angle. Find (I. 24, cor. 4) a straight line the square of which is equal to the excess of the square of BC above that of A, and cut off BD ~ equal to that line ; on DC as diam- eter describe a semicircle, and draw EF parallel to DC ate a distance equal to BD ; join either point of intersection, E, with D and C ; DEC is the required triangle. Draw the perpendicular EG, which (const.) is equal to BD. Then (II. 4) BC^ or AHEG^ = (DC-fEG)^=DC'^ + EG=+ 2DC.EG; whence A^=DC^ + 2DC.EG. Also (DE + EC)^= DE^ + EC^+2DE.EC=DC^-|-2DC.EG, because (III. 11, and I. 24, cor. 1) DC'=DE^+EC^ and DC.GE^DE.EC, each being equal to twice the area of the triangle DEC. Hence (DE4- EC)- = A-; wherefore (I. 23, cor. 3)bE + EC=A; and DEC is the triangle required. Method of Computatiofi. From the construction, we have BD or EG= V(BC'—A'). Then DC = BC-BD; by halving which we get the radius of the semicircle ; and if from the square of the radius drawn from E, the square of EG be taken, and if the square root of the remainder be successively taken from the radius, and added to it, the results will be the seg- ments DG, GC ; from which, and from EG, the sides (I. 24, cor. 1) are readily computed. Prop. XXII. — Prob. — Given the base of a triangle, the per- pendicular, and the difference of the sides ; to construct it. Make AB equal to the given base, and parallel to it draw CD, at a distance equal to the given perpendicular; dravvBDP perpendicular to CD, and make DF equal to DB ; from A as center, with a radius AE equal to the given difference, describe the circle ELN ; through B, F describe any circle cutting ELN" in L, N, and let G be the point in which a straight line drawn through L, N cuts FB produced ; draw the tangent GK, and 16 242 EXEECISES. draw AKM cutting CD in M; join BM; and it k ^ident from the second corollary to tbe ninth proposition of the third book, that AMB is the required triangle* Method of Computation. From M as center, with MK as radius, describe a circle, and by the corol- lary referred to, it will pass through B and F, Join AG and produce BA to H. Then the rectangle FG.GB=AG^-AK^ each being equal to the square of GK ; that is, FB.BG+BG==AB^-fBG^— AK^ Take awayBG^; the:i FB.BG=AB^-AK==(AB+AK) (AB — AK)=HB.EB. Hence BG becomes known. Then, in the two right-angled triangles GBA, GKA, the angles at A can be computed, and their difference is the angle MAB in the re- quired triangle. The rest is easy, if the perpendicular from M to AB be drawn. Prop. XXIII. — Prob. — Given the base, the area, and the •ratio of the sides of a triangle; to construct it. Let AB be the given base, and (V. 3, and scho.) find the points C, D, so that AC, CB, and AD, DB are in the ratio of the sides ; on CD as diame- ter describe the semicircle CED, and (II. 5, scho.) to AB apply a parallelogram BF double of the given area ; let FG, the side of this oppo- site to AB, produced if neces- sary, cut the semicJircle in E, and join EA, EB j EAB is the re- quired triangle. For (Ex. 6, cor.) AE is to EB as AC to CB ; that is, in the given ratio. Also (1. 15, cor. 4) AEB is half of the parallelo- gram BF, which is double of the given area. Therefore AEB is on the given base, is of the given area, and has its sides in the given ratio. Scho. If FG, or FG produced, do not meet the semicircle, EXERCISE8. 243 the problem is impossible ; if it cut it in E and E^, there will be two triangles essentially different, each of which will answer the conditions of the problem ; if it touch the semicircle, there will be only one triangle, and it will be the greatest possible with the base of the given magnitude, and the sides in the given ratio ; and hence we have the means of solving the problem in which it is required to construct a triangle on a given base, having its sides in a given ratio, and its area a maximu7n. Method of Computation. Join the center H with E, and draw the perpendicular EK. Then, let m '. n : '. AC : CB, and consequently m, : n : : AD : DB ; then, from these, by composition and division, we get m+n : 7i : : AB : CB, and m—n '. n : : AB : DB; whence DC and its half, the radius of the circle, become known. EK also is found by dividing double of the area by AB. Then, in the triangle EKH, KH can be found, and thence AK and KB ; and, by means of them and EK, the sides EA, EB may be computed. If E'' be taken as the vertex, the method of computation is almost the same, and is equally easy. Prob. XXIV. — Prob. — Given the base of a triangle, the ver- tical angle, and the rectangle of the sides ; to construct it. On the given base describe a segment containing an angle €qual to the given angle; to the diameter of the circle of which this segment is a part, and to the lines containing the given rectangle, find a fourth proportional; this proportional (Ex. 3) is the perpendicular of the triangle ; and the rest of the solu- tion is effected by means of the fifteenth proposition of these Exercises. Prop. XXV. — Prob, — To divide a given triangle into two parts in a given ratio, by a straight line parallel to one of the sides. Let ABC be a given triangle ; it is required to divide it into two parts in the ratio of the two straight lines, m, n, by a straight line parallel to the side BC. Divide (V. 3, scho.) AB in G, so that BG : GA : : m : w, and between AB, AG find (V. 11) the mean proportional AH' 244 EXERCISES. draw HK parallel to BC ; ABC is divided by HK in the man- ner required. For (V. 14, scho.) since the three straight lines AB, All, AG are propor- tionals, AB is to AG as the triangle ABC to AHK ; whence, by division, BG is to AG, or (const.) m is to w, as the quadrilat- eral BCKH to the triangle AHK. In practice, the construction is easily and elegantly effected, when the triangle is to be divided either into two or more "^ parts proportional to given lines, by divid- n ing AB into parts proportional to those lines, and through the points of section drawing perpendiculars to AB, cutting the arc of a semicircle described on AB as diameter; then by taking lines on AB, terminated at A, and severally equal to the chords drawn from A to the points of section of the arc, the points on AB wall be obtained through which the parallels to BC are to be drawn. The reason is evident from the foregoing proof in connection with the corollary to the eighth proposition of the fifth book. Cor. Hence a given triangle can be divided into two parts in a given ratio, by a straight line parallel to a given straight line. Also, a triangle can be divided into two parts in a given ra- tio, by a straight line drawn through a given point in one of the sides; and a given quadrilateral can be divided into tw^o parts in a given ratio, by a straight line jsarallel to one of its sides. Prop. XXVI. — Peob. — From a given point in one of the sides of a given triangle, to draw two straight lines trisecting the triangle. . Let ABC be the given triangle, and D the given point ; then, if BD be not less than one third of BC, to BD, to a third part of BC, and to the perpendic- ular from A to BC, find a fourth pro- portional, and at a distance equal to it EXERCISES. 245 draw a parallel to BC, cutting BA in E, and join ED ; the tri- angle BED is evidently (V. 10, cor., and I. 15, cor. 4) a third part of ABC ; and CDF will be constructed in a similar man- ner, if CD be not less than a third of BC. If DC, one of the segments, be less than a third of BC, the triangle BDE is constructed as before, but the rest of the pre- ceding solution fails, as the second parallel would fall above the triangle. In this case, cut off BA between E and A, a part equal to BE, and call it EG; then, if DG be joined, the trian- gle ABC is trisected by DE, DG. Scho, It is easy to see that this method may be readily ex- tended to the division of a triangle into more equal parts than three, or into parts proportional to given magnitudes, by straight lines drawn from a given point in one of the sides. Prop. XXVII. — Theor. — If the sides of a right-angled tri- angle be continual proportionals, the hypothenure is divided in extreme and mean ratio by the perpendicular to it from the right angle / and the greater segment is equal to the less or re- mote side of the triangle. Let ABC be a triangle right-angled at A, and let AD be per- pendicular to BC ; then if CB : BA : : BA : AC, BC is divided in ex- ■*■ treme and mean ratio in D, and BD is equal to AC. For (hyp.) CB : BA : : BA : AC, and (V. 8, cor.) CB : BA : : BA : BD ; therefore BA : AC : : BA : BD, and AC is equal to BD. Again (V. 8, cor.), BC : CA : : CA : CD, or BC : BD : : BD : DC, and therefore (V. def 3) BC is divided in extreme and mean ratio in D. Scho. Conversely, if BC : BD : : BD : CD, and if BAC be a right angle, and DA perpendicular to BC; CB : BA : : BA : AC, and BD is equal to AC. For (hyp.) CB : BD : : BD : CD, and (V. 8, cor.) CB : CA : : CA : CD ; wherefore BD^ is equal to CA', each (V. 9, cor.) being equal to the rectangle BC.CD, and therefore BD is equal to CA. Again (V. 8), CB : BA : : BA : BD or AC. 246 EXERCISES. Prop. XXVllL — Prob. — Given the angles and diagonals of a parallelogram ; to construct it. On one of the diagonals describe a segment of a circle con- taining an angle equal to the given angle at either extremity of the other ; from the middle point of this diagonal as center, with half the other diagonal as radius, describe an arc cutting the arc of the segment ; through the extremities of the first diagonal draw four straight lines, two to the intersection of the arcs, and two parallel to these ; the parallelogram thus formed is easily proved to be the one required. Prop. XXIX. — Prob. — Given the vertical angle of a trian- gle^ and the radii of the circles inscribed in the parts into which the triangle is divided by the perpendicular y to construct the triangle. Take any straight line ABC, and through any point B draw the perpendicular BD ; make BA, BC equal to the given radii, and let E, F be the angular points, remote from B, of squares described on AB, BC ; join EF, and on it describe the segment EDF, containing an angle equal to half the given vertical angle ; let the perpen- dicular cut the arc EDF in D, and join DE, DF ; draw DG, DH making the angles EDG, FDH respectively equal to EDB, FDB; DGH is the required triangle. For (I. 14) perpendiculars drawn from E to DB, DG are equal, and each of them is equal (const.) to the perpendicular from E to GB. Each of them therefore is equal to the given radius AB and a circle described from E at the distance of one of these is inscribed in the triangle DGB. In the same manner it would be shown, that a circle described from F as center, with the other given radius, would be inscribed in DBII. Hence, since the angle GDH is double of EDF, GDII is equal to the given vertical angle, and the triangle GDH answers the conditions of the question. Scho. The preceding solution is strictly in accordance with EXEECI8E8. 24:T the enunciation, taken in its limited sense. There will be in- teresting variations, however, if we regard the given circles, not merely as inscribed., but as those which touch all the sides of each of the right-angled triangles, either internally or exter- nally. These variations will be obtained by giving the squares on the radii every possible variety of position in the four right angles formed by the intersection of AC, DB ; and the solution will obtain complete generality by taking into consideration both the points in which BD cuts the circle of which EF is a chord. Prop. XXX. — Theo^, — The area of a triangle ABC ts equal to half the continued product of two of its sides, AB, BC, a?id the sine of their contained angle B, to the radius I. Draw the perpendicular AD. Then (Trig. 1, cor.) AD = AB X sin B. Multiply by BC, and take Liiif the product ; and (I. 23, cor. 6) we have the area equal to ^ AB x BC X sin B. Cor. Hence (I. 15, cor. 1) the area of a parallelogram is equal to the continual product of two contiguous sides, and the sine of the contained angle. jScho. From this proposition, and from the third and sixth corollaries to the twenty-fifth proposition of the third book, we can derive neat algebraic expressions for the radii of the four circles, each touching the three sides of a triangle. Thus, by dividing the expression for the area by s, we find, according to the third corollary, that the radius of the inscribed circle is /(s — a) (s—b) (s—c) T ... , T 'T equal to ,{/ —^ —-^ ~ . In like manner, by dividing the expression for the area successively by s — a, s—b, s — c, we find, according to the sixth corollary, that the radii of the cir- cles touching a, b, c, externally, are respectively, /s{s-b){s-c)/s{s-a){s_-c)^ and ^ ^ (s-a) (s-b) ^ r s—a y s—b r s — o By taking the continual product of these four expressions, and contracting the result, we get s {s — a) (s — b) (s — c), which. is equal to the square of the area ; and hence, by expressing 248 EXERCISES. this inwards, we have the following remarkable theorem : The continual product of the radii of the four circles^ each of which touches the three sides of a triangle, or their prolongations, is equal to the second power of the area, which is proven by the next proposition. Prop. XXXI. — Theor. — Let a, b, c he the sides of a trian- gle, and s half their sum ; the area is equal to the square root of the continual product ofs, s— a, s — b, and s — c. It was proved in the second corollary to the ninth proposi- tion, Plane Trigonometry, that the sine of twice any angle is twice the product of the sine and cosine of the angle. Hence, by multiplying together the values of sin-^A and cos^A, given in the corollaries to the sixth and seventh propositions, Plane Trigonometry, and doubling the result, we get sin A= 2 \/\s (s-a) (s-b) (s-c)] ^^ , , -j^ -. Now, by the precedmg proposi- tion, the area of a triangle is found by multiplying the sine of one of its angles by the sides containing it, and taking half of the product; multiplying, therefore, the value now found for sin A, by be, and taking half the product, we find the area to be V[s {s — a) {s — b) (s — c)]. This proposition is much used in surveying coasts and harbors. Prop. XXXTI. — Prob, — A semicircle ACB being given, and other semicircles being described as in the diagram, / it is re- quired to find the sum of the areas of all those inscribed semi- circles. Cii'cles (V. 14), and consequently semicircles, are as the squares of their diameters or of their radii. Now the square of GD is half the square of DF or CF, and therefore the semicircle DFE is half of ACB. For the same reason HGK is half of DFE; and universally, each semicircle is half of the one in which it is in- scribed. Hence the entire amount will be the sum of the infinite series iACB-f-J ACB+i ACB+^ ACB+etc. ; and EXERCISES. 249 therefore (IV. 19) i ACB-i ACB : ^ACB :: iACB : ACB, llie required sum ; and it thus appears that the sum of all the inscribed semicircles is equivalent to the given semicircle. Pkop. XXXIII— Theor.— J;z any triangle, the center of the circumscribed circle, the point in which the three perpendiculars intersect one another, and the point of intersection of the straight lines drawn from the angles to bisect the opposite sides, lie all in the same straight line. Let ABC be a triangle, and let the two perpendiculars AD, CE intersect in F ; bisect AB, BC in H, G, and draw AG, CH intersecting in K ; draw also GI, HI perpendicular to BC, BA, and intersecting in I. Then (Ex. 7) F is the intersection of the three perpendiculars, K (III. 1, cor. 2) the intersection of the three lines drawn from the angles to bisect the opposite sides, and (III. 25, cor. 2) I is the center of the cir- cumscribed circle. Join FK, KI ; FKI is a straight line. Join HG; it is (V. 2 and 3) parallel to AC, and is half of it. Also the triangles ACF, GHI are (I. 16, cor. 3) equiangular, and therefore GI is half of AF. So likewise (III. 1, scho.) is KG of KA. Hence the two triangles AKF, GKI have the alternate angles KAF, KGI equal, and the sides about them proportional ; therefoi-e (V. 6) the angles AKF, GKI are equal, and (1. 10, cor.) since AKG is a straight line, FKI is also a straight line. Scho. It is plain (V. 3) that FK is double of KI. We have also seen that AF is double of GI. Hence it appears, that the distance between any of the angles and the point of intersec- tion of the three perpendiculars is double of the perpendicular drawn from the center of the circumscribed circle to the side opposite to that angle. Prop. XXXIV. — Theor. — Straight lines drawn from the angles of a triangle to the points in which the opposite sides touch the inscribed circle, all pass through the same point. 250 EXERCISES. Let ABC be a triangle, and D, E the points in which the sides AB, AC touch the inscribed circle ; draw BFE, CFD ; draw also AFG cutting BC in G ; G is the point in which BC touches the inscribed circle. If possible, let another point K be the point of contact, and draw DH,DI parallel to BC, CA. Then in the similar tri- angles FDI, FCE, FD : DI : : FC : CE, or CK ; and in the similar triangles FDH, FGC, DH : DF : : GC : FC ; from which and from the preced- ing analogy we get, ex cequo^ DH : DI : : CG : CK. Again, BD : DI : : BA : AE or AD : : BG : DH. Hence, altern- ately, and by inversion, BG : BD : : DH : DI : whence (IV. 7) BG : BD or BK : : CG : CK, or alternately, BG : CG : : BK : CK; and by composition, BC : CG :: BC : CK ; and there- fore CG, CK are equal ; that is, G and K coincide, and AFG passes through the point in which BC touches the circle. Prop. XXXV. — Theor. — 171 a triangle, the sum of the per- pendiculars draion from the center of the circumscribed circle to the three sides is equal to the sum of the radii of the in- scribed and circumscribed circles. Let ABC be a triangle, having its sides bisected in D, E, F, by perpendiculars meeting in G, the center of the circum- scribed circle ; the sura of GD, GE, GF is equal to the sum of the radii of the inscribed and circumscribed circles. Join GA, GB, GC, and DE, EF, FD. Then, putting a, b, c to denote the sides opposite to the angles A, B, C, we have (V. 2 and 3) FE^^a, FD=i6, and DE=ic; and (HI. 11) Bince AEG, AFG are right angles, a circle may be described EXERCISER. 251 about the quadrilateral AEGF. For a like reason circles may- be described about BDGF and CDGE. Hence (Ex. 4) FE. AG=AF.GE + AE.FG; or, by doubling, a.AG=c.GEi-b.FG. In the same manner, it would be shown, since AG, BG, CG are equal, that ^>.AG=c.GD + a.FG, and c.AG = a.GE + ^..GD. Hence, by addition, (a-|-5+c)AG = (a+c)GE+(ff + 5)GF + (54-c)GD, Now 5.GE is evidently equal to twice the triangle AGO, c.GF equal to twice AGE, and a.GD equal to twice BGC ; also, denoting the radius of the inscribed circle by r, we have (IH. 25, cor. 3), r{a+b-\-c) equal to twice the area of the triangle ABC, and consequently r{a-\-b-i-c)=b.GE + c.GF + a.GD. Hence, by addition, {a+b-\-c)AG-\-r{aA-b-t-c) = (a+5+c)GE+(a+5+c)GF+(a+6+c)GD; and consequently AG4-r=GE+GF+GD. Cor. Since, by the scholium to proposition thirty-third of these Exercises, the parts of the three perpendiculars of the trianffle, between their common intersection and the three an- gles, are respectively double of GD, GE, GF, the sum of those parts of the perpendiculars is equal to the sum of the diameters of the inscribed and circumscribed circles. Prop. XXXYI. — Theor, — If on the three sides of any tri- angle equilateral triangles be described^ either all externally^ or all internally, straight lines joining the centers of the circles inscribed in those three triangles form an equilateral triangle. On the three sides of any triangle ABC, let'^the equilateral triangles ABD, BCF, CAE be de- scribed externally, and find G, H, K, the centers of the circles de- scribed in those triangles ; draw GH, HK, KG ; GHK is an equilat- eral triangle. Join GA, GB, HB, HC, HF, KC, AF. Then the angle FBC is two thirds of a right angle, and the an- gles GAB, GBA, FBH, BFH each one third. The triangles FBH, ABG are therefore similar, and (V. 3) FB : BH : : BA : BG ; whence, alternately, FB : BA : : 252 EXERCISES. HB : BG; that is, in the triangles FBA, HBG the sides about the angles FBA, HBG are proportional; and these angles aj-e equal, each of thorn being equal to the sum of the angle ABC and two thirds of a right angle. Hence (V. 6) these triangles are equiangular ; and therefore (V. 3) FB or BC : FA : : BH : HG; and it would be shown in the same manner, by means of the triangles ACF, KCH, that FC or BC : FA : : CH : HK ; therefore (IV. 1) BH : HG : : CH : HK. But BH is equal to CH, and therefore (IV. 6) HG is equal to HK ; and it would be demonstrated in a similar manner, that HG, HK are each equal to GK. If the equilateral triangles were described on the other sides of the lines AB, BC, CA, the angles ABF, GBH would be the difference between ABC and two thirds of a right angle ; but the rest of the proof is the same. Sc/io. If ABC exceed aright angle and a third, the sum of it and two thirds of a right angle is greater than two right angles. In that case, the angles ABF, GBH, understood in the ordinary- sense, are each the difference between that sum and four right angles. If ABC be a right angle and a third, the sum becomes two right angles, and FB, BA are in the same straight line, as are also HB, BG. In this case it is proved as before, that FB : BA : : HB : BG ; and then (IV. 11) FB or BC : FA :: HB : HG. The rest of the proof would proceed as above. It may be remarked that if an equilateral triangle be de- Bcribed on a stj'aight line, and if on the two parts into which it is divided at any point, other equilateral triangles be described, lying in the opposite direction, the lines joining the centers of the three equilateral triangles will also form an equilateral tri- angle. The connection of this and the proposition will be per- ceived by supposing two angles of the triangle continually to diminish, till they vanish, as the triangle may thus be conceived to become a straight line. TO BE PROVEN. 1. The least straight line that can be drawn to another etraight line from a point without it, is the perpendicular to it ; of others, that which is nearer to the perpendicular is less than EXERCISES. 253 one more remote ; and only two equal straight hnes can be drawn, one on each side of the perpendicular. 2. Of the triangles formed by drawing straight Hnes from a point within a parallelogram to the several angles, each pair that have opposite sides of the parallelogram as bases, are half of it. 3. If, in proceeding round an equilateral triangle, a square, or any regular polygon, in the same direction, points be taken on the sides, or the sides produced, at equal distances from the several angles, a similar rectilineal figure will be formed by joining each point of section with those on each side of it. 4. If the three sides of one triangle be perpendicular to the three sides of another, each to each, the triangles are equi- angular. 5. A trapezoid, that is, a trapezium having two of its sides parallel, is equivalent to a triangle which has its base equal to the sum of the parallel sides, and its altitude equal to their per- pendicular distance. 6. Given the segments into which the line bisecting the ver- tical angle divides the base, and the difference of the angles at the base ; to construct the triangle, and compute the sides. 7. Within or without a triangle, to draw a straight line par- allel to the base, such that it may be equivalent to the parts of the other sides, or of their continuations, between it and the base. 8. Given the perpendicular of a triangle, the diffei*ence of the segments into which it divides the base,and the difference of the angles at the base ; to construct the triangle. 9. In the figure for the first corollary to the twenty-fourth proposition of the first book, prove that CD is perpendicular to AH, or a line from C to F is perpendicular to CD. 10. The angle made by two chords of a circle, or by their continuations, is equal to an angle at the circumference stand- ing on an arc equivalent to the sum of the arcs intercepted be- tween the chords, if the point of intersection be within the circle, or to their difference, if it be without ; (2) the angle made by a tangent and a line cutting the circle, is equal to aa angle at the circumference on an arc equivalent to the differ- ence of the arcs intercepted between the point of contact and 254 EXEKCIPHa, the other line; and (3) the angle made by two tangents is equal to an angle at the circumference standing on an arc equivalent to the difference of those into which the circumfer- ence is divided at the points of contact. Cor. If a tangent be parallel to a chord, the arcs between the point of contact and the extremities of the chord are equal. 11. Given the sum of the perimeter and diagonal of a square ; to construct it. 12. On a given straight line describe a square, and on the side opposite to the given line describe equilateral triangles lying in opposite directions ; circles described throiigh the ex- tremities of the given line, and through the vertices of these triangles, are equal. 13. To inscribe an equilateral ti'iangle in a given square. 14. Given the angles and the two opposite sides of a trape- zium ; to construct it. 15. In a given circle to place two chords of given lengths, and inclined at a given angle. 16. In the figure for the twenty-fifth proposition of the third book, if AM, BM be joined, the angle AMB is half of ACB. 17. If, in proceeding in the same direction round any trian- gle, as in 3, points be taken at distances from the several angles, each equal to a third of the side, the triangle formed by joining the points of section is one third of the entire triangle. 18. To describe a square having two of its angular points on the circumference of a given circle, and the other two on two given straight lines drawn through the center. Show that there may be eight such squares. 19. Given the vertical angle of a triangle, and the segments into which the base is divided at the point of contact of the in- scribed circle ; to describe the triangle, and compute the sides. 20. If any three angles of an equilateral pentagon be equal, all its angles are equal. 21. Given two sides of a triangle, and the difference of the sefrments into which the third side is divided by the perpen- dicular from the opposite angle ; to construct the triangle. 22. Given the vertical angle of a triangle, the line bisecting it, and the perpendicular; to construct the triangle. 23. From a given center to describe a circle, from which a EXERCISES. 265 straight line, given in position, will cut off a segment contain- ing an angle equal to a given angle. 24. In a given triangle to inscribe a semicircle having its center in one of the sides. 25. Through three given points to draw three parallels, two of which may be equally distant from the one between them. 26. Given an angle of a triangle, and the radii of the circles touching the sides of the triangles into which the straight line bisecting the given angle divides the triangle ; to construct it. 27. In a rhombus to inscribe a square. 28. Given the lengths of the two parallel sides of a trapezoid, and the lengths of the other sides ; to construct it. 29. Given one of the angles at the base, and the segments into which the base is divided at the point of contact of the in- scribed circle ; to describe the triangle. 30. To draw a tangent to a given circle, such that the part of it intercepted between the continuations of two given diam- eters may be equal to a given straight line. 31. To draw a tangent to a given circle, such that the part of it between two given tangents to the cii'cle may be equal to a given straight line. 32. Given the vertical angle of a triangle, the difference of the sides, and the difference of the segments into which the line bisecting the vertical angle divides the base j to construct the triangle. 33. Given any three of the circles mentioned in the fifth ■corollary to the twenty-fifth proposition of the third book ; to describe the triangle. 34. A straight line and a point being given in position, it is required to draw through the point two straight lines inclined at a given angle, and inclosing with the given line a space of given magnitude. 35. From two given straight lines to cut off equal parts, each of which will be a mean proportional between the remainders. 36. A square is to a regular octagon described on one of its sides, as 1 to 2 (1-f V2). 37. In a given triangle to inscribe a parallelogram of a given area. 256 EXEECI8E8. 38. In a given circle to inscribe a parallelogram of a given area. 39. Through a given point between the lines forming a given angle, to draw a straight line cutting off the least possi- ble triangle. 40. To divide a given straight line into two parts, such that the square of one of them may be double of the square of the other, or may be in any given ratio to it. 41. To produce a given straight line, so that the square of the whole line thus produced may be double of the square of the part added, or in any given ratio to it. 42. Given the area of a right-angled triangle, and the sura of the legs ; to construct it. 43. Given the area and the difference of the legs of a right- angled triangle ; to construct it. 44. Given one leg of a right-angled triangle, and the remote segment of the hypothenuse, made by a perpendicular from the right angle ; to construct the triangle. 45. Given the base of a triangle, the vertical angle, and the side of the inscribed square standing on the base ; to describe the triangle. 46. Given the base of a triangle, and the radii of the inscribed and circumscribed circles ; to construct the tri- angle. 47. To draw a chord in a circle which will be equal to one of the segments of the diameter that bisects it. 48. In a given circle to draw a chord which will be equal to the difference of the parts into which it divides the diameter that bisects it. 49. If two sides of a regular octagon, between which two others lie, be produced to meet, each of the produced parts is equivalent to a side of the octagon, together with the diagonal of a pquare, described on the side. 50. The perimeter of a triangle is to the base as the perpen- dicular to the radius of the inscribed circle. 51. From a given point without a given circle to draw a straight line cutting the circle, so that the external and inter- nal parts may be in a given ratio. 52. Each of the complements of the parallelograms, about EXERCISES. 257 the diagonal of a parallelogram, is a mean proportional between those parallelograms. 53. Given the ratio of two straight lines, and the difference of their squares ; to find them. 54. The square of the perimeter of a right-angled triangle is equivalent to twice the rectangle under the sura of the hy- pothenuse and one leg, and the sum of the hypothenuse and the other. 55. The quadrilateral formed by straight lines bisecting each pair of adjacent sides of a quadrilateral is a parallelogram, which is half of the quadrilateral ; and straight lines, joining the points in which the sides of that parallelogram are cut by the diagonals of the primitive figure, form a quadrilateral simi- lar to that figure and equivalent to a fourth of it. 66. From three given points as centers, and not in the same straight line, to describe three circles each touching the other two. Show that this admits of four solutions. 57. To add a parallelogram to a rhombus, such that the M'hole figure may be a parallelogram similar to the one added. 58. A straight line being given in position, and a circle in magnitude and position, it is required to describe two equal circles touching one another, and each touching the straight line and the circle. 59. The squares of the diagonals of a quadrilateral are to- gether double of the squares of the straight lines joining the points of bisection of the opposite sides. 60. In a given rhombus to inscribe a rectangle having its sides in a given ratio. 61. If, throu2rh the vertex and the extremities of the base of a triangle, two circles be described intersecting one another in the base or its continuation, their diameters are proportional to the sides of the triangle. 62. To draw a straight line cutting two given concentric circles, so that the parts of it within them may be in a given ratio. 63. From a given point, within a given circle, or without it, to draw two straight lines to the circumference, perpendicular to one another, and in a given ratio. "When will this be im- possible ? It 258 EXERCISES. 64. If a straight line be divided in extreme and mean ratio, the squares of the whole and the less part are together equiva- lent to thi-ee times the square of the greater, 65. If a straight line be cut in extreme and mean ratio, and be also bisected, the square of the intermediate part, and three times the square of half the line are equivalent to twice the square of the greater part. 66. If the hypothenuse of a right-angled triangle be given, the side of the greatest inscribed square, standing on the hy- pothenuse, is one third of the hypothenuse. 67. To divide a given semicircle into two parts by a perpen- dicular to the diameter, so that the radii of the circles inscribed in them may be in a given ratio. 68. To draw a straight line parallel to the base of a triangle, making a segment of one side equal to the remote segment of the other. 69. In the figure for the tenth proposition of the second book, the square of the diameter of the circle, passing through the points F, H, D, is six times the square of the straight line join- ing P^D. VO. Given the base, the area, and the sum of the squares of the sides of a triangle ; to construct it. 71. If, from the extremities of the hypothenuse of a right- angled triangle as centers, arcs be described passing through the right angle, the hypothenuse is divided into three segments, such that the square of the middle one is equivalent to twice the rectangle of the others. 72. On a given hypothenuse to describe a right-angled tri- angle, such that the difference between one leg and the adja- cent segment of the hypothenuse made by a perpendicular from the right angle, may be a maximum, 73. On a given hypothenuse to construct a right-angled tri- angle, such that one segment of the hypothenuse made by the perpendicular from the right angle, may be equivalent to the Bum of the perpendicular and the other segment. 74. The square of DH (see figure for III. 24) is equivalent to the rectangle CF.BG ; and if the circles touch one another ex- ternally, DH is a mean proportional between their diameters. Also the square of DH is equivalent to the rectangle CG.BF. EXERCISES. 259 75. On a given straight line to describe an isosceles triangle, having the vertical angle treble of each of the angles at the base. V6. If in the diameter of a circle and its continuation two points be taken on the opposite sides of the center, such that the rectangle under their distances from the center may be equivalent to the square of the radius, any circle whatever de- scribed through these points bisects the circumference of the otlier circle. '77. To find a point from which, if straight lines be drawn to three given points, they will be proportional to three given straight lines. 78. Given the segments into which the base of a tnangle is divided by two straight lines trisecting the vertical angle ; to construct it. 79. If one diagonal of a quadrilateral inscribed in a circle be bisected by the other, the square of the latter is equivalent to half the sum of the squares of the four sides. 80. To divide a straight line into- two parts, such that the squares of the whole and one of the parts may be double of the square of the other part. 81. Given the segments into which the base of a triangle is divided by the straight line bisecting the vertical angle, to construct the triangle so that its angle adjacent to the greater segment may be either of a given magnitude, or a maximum, and in each case to compute the remaining sides and angles. 82. To draw a straight line bisecting a given parallelogram, so that if it be produced to meet the sides produced, the ex- ternal triangles will have a given ratio to the parallelogram. 83. Through a given point to draw a straight line, which, if continued, would pass through the point of intersection of two given inclined sti'aight lines, without producing those lines to meet. 84. If, from any point in the circumference of the circle de- scribed about an equilateral triangle, chords be drawn to its three angles, the sum of their squares is equivalent to six times the square of the radius of the same circle, or to twice the square of a side of the triangle. 85. Given the difference of the angles at the base of a trian- 260 EXERCISES. gle, the difference of the segments into which the hase is divided by the perpendicular, and the ratio of the sides ; to construct the triangle. 86. Given the base and vertical angle of a triangle, to con- struct it so that the line bisecting the vertical angle may be a mean proportional between the segments into which it divides the base. 87. Given two sides of a triangle, and the ratio of the base and the line bisecting tlie vertical angle j to construct the tri- angle. 88. If the vertical angle of a triangle be double of one of the angles at the base, the rectangle under the sides is equivalent to the rectangle under the base, and the line bisecting the ver- tical angle. 89. If a straight line be divided into parts, which, taken in succession, are continual proportionals, and if circles be de- scribed on the several parts as diameters, a straight line which touches two of the circles on the same side of the straight line joining their centers, will touch all the others. 90. Given the segments into which the base is divided by the straight line bisecting the vertical angle, and the angles which that straight line makes with the base ; to construct the triangle. 91. To divide a given circle into two segments, such that the squares inscribed in them may be in a given ratio. 92. Through a given point in the base of a given isosceles triangle, or its continuation, to draw a straight line such that the lines intercepted on the equal sides, or their continuations between that line and the extremities of the base, may have one of the equal sides as a mean proportional between them. 93. Through a point in tlie circumference of a given circle, to draw two chords, such that their rectangle may be equivalent to a given space, and the chord joining their other extremities equal to a given straight line. 94. In any triangle the radius of the circvmiscribed circle is to the radius of the circle which is the locus of the vertex, when the base and the ratio of the sides are given, as the dif- ference of the squares of those sides is to four times the area. 95. The difference of the sides of a triangle is a mean pro- KXERCI8K8. 261 portional between the clifFerence of the segments into which the base is divided by the perpendicular, and the difference of those into which it is divided by the line bisecting the vertical angle. 96. Let the angles of a pai*allelogram which has unequal sides be bisected by straight lines cutting the diagonals, and let the points of intersection be joined ; the figure thus formed is a parallelogram, which has to the proposed parallelogram the duplicate ratio of that which the difference of the unequal sides of the latter has to their sum. 97. A circle and a point being given, it is required to de- scribe a triangle similar to a given one, having its vertex at the given point, and its base a chord of the given circle. 98. Given the three points in which the sides of a triangle are cut by the perpendiculars from the opposite angles ; to con- struct the triangle. 99. Given the ansjles of a triansrle, and the leno;ths of three straight lines drawn from the angular points to meet in an- other point ; to construct the triangle. 100. Given the base of a triangle, and the ratio of its sides; to construct it, so that the distance of its vertex from a given point may be a maximum or minimum. 101. To divide {^circle into two segments, such that the sum of the squares inscribed in them may be equivalent to a given space. 102. Through a given point, with a given radius, to describe a circle bisecting the circumference of a given circle. 103. With a given radius to describe a cn-cle bisecting the circumferences of two given circles. 104. In a right-angled triangle, the rectangle under the radius of the inscribed circle, and the radius of the circle touching the hypothenuse and the legs produced, is equivalent to the area. So, likewise, is the rectangle under the circles touching the legs externally, and the continuations of the other sides. 105. If three straight lines be continual proportionals, the 6um of the extremes, their difference, and double the mean will be the hypothenuse and legs of a right-angled triangle. 106. From two given centers, to describe circles having their 263 EXERCISES. radii in a given ratio, and the part of their common tangent, between the points of contact, equal to a given straiglit line. 107. A straiglit line and two points equally distant from it, on the same side, being given in position, it is required to draw through the points two straight lines forming with the given line the least isosceles triangle possible, on the side on which the points ai"e. 108. To describe a circle touching a diameter of a given cir- cle in a given point, and having its circumference bisected by that of the given one. 109. If an angle of a triangle be 60^, the square of the oppo- site side is less than the squares of the other two by their rect- angle ; but if an angle be 1 20°, the square of the opposite side is greater than the squares of the others by their rectangle. 110. In the figure on page 251, prove that the three straight lines joining AF, BE, CD are all equal. 111. The chord of 120° is equal to the tangent of 60°. 112. The sines of the parts into which the vertical angle of a triangle is divided by the straight line bisecting the base, are reciprocally proportional to the adjacent sides. Show from this how a given angle may be divided iuto two parts, having their sines in a given ratio. 113. The diameter of the circle described about any triangle is equivalent to the product of any side and the cosecant of the oj^posite angle. 114. In any triangle ABC, the radius of the inscribed circle . , sin A B sin A C . sin i A sin ^^ C IS equivalent to a. t—: > to o. r-f, — > or to ^ cos Y A cos ^- 1> sin 4- A sin 4 B £ n . -i i * /• c. r-^ — ; or, finally, to the cube root oi cos -J O abc sin ^ A sin ^ B sin -J C tan ^ A tan | B tan ^ C. 115. Given the sum of the tangents, and the ratio of the se- cants, of two angles to a given radius ; to determine the angles geometrically and by computation. 116. Find an angle, such that its tangent is to the tangent of its double, in a given ratio ; suppose that of 2 to 5. 117. If a spherical triangle be right-angled at C, and any two of the other parts be made s -.ccessively 108° 42' and 87° 33' 19", what will be the values o the remaining parts? ADDENDUM. 263 118. Let any three parts of an oblique-angled spherical trian- gle be successively 87° 45^ 96° 57' 48'^, and 106° 53' 13''; what will be the values of the other parts? 119. When the three sides of a spherical triangle be respect- ively, 34° 39' 44", 78° 27' 49", and 134° 15' 23", what will be the surface of the triangle when the radius of the sphere is 16 ? "What will be the base of the triangular pyramid subtended by those sides, and what will be the surface and. solidity of the spherical pyramid with its apex at the center of the sphere ? 120. If the foregoing values be the magnitudes of the angles of a spherical triangle when the radius of the sphere is 16, what will be the base of the triangular pyramid subtended by the sides of the spherical triangle ? What will be the surface of the spherical triangle, and the surface and solidity of the spherical j^yramid with its apex at the center of the sphere ? 121. If the earth be regarded a sphere with 7973.8798-f miles for its diameter, and a spherical pentagon be measured on its surface having 341.78 miles, 309.25 miles, 278.64 miles, 173.97 miles, and 97 miles for its sides ; and the angles contained by those sides be respectively 74° 34' 19", 107° 09' 51", 41° 0' n'", 85° 17' 09", and 76° 41' 35", what will be the surface of the spherical pentagon ? and what will be the solidity of the pyramid having the spherical pentagon for its base and its apex at the center of the earth ? ADDENDUM, Illustrating the Third Proof for the Second Corollary TO THE Seventeenth Proposition of the tixTii Book, Elements of Euclid and Legendre. Circles are to one another as the squares described upoft their diameters (V. 14), consequently squares are to one an- other as the circles described (V. 14, cor. 2) upon their sides; that is, there is a ratio of equality between the circle and squares which have the same straight line for their respective diameter, side, and diagonal • therefore the circle has the same 264 ADDENDUM. arithmetical proportion to the inscribed square having the diameter for its diagonal, as the circumscribed square having the same diameter for its side has to the circle. If 10 be the diameter of the circle, and x ^6 the area of the circle, 100 will be the area of the circumscribed square, and 50 will be the area of the inscribed square — and we have the arithmetical propor- tion, 100, X, 50; and the geometrical proportion, 100 : Y : : 50 : -5-. From the first proportion, we derive 100— x=x— 50. From the second proportion, we have 100-x = 2 (50- I). .-. x-50 = 2 (50-|x) = 100-x; or, 2p(;=:150, hence p^ = 75. The arithmetical proportion gives, 2x=150 or pf=75 Substituting this value for -/^ in the above proportions, \,lpl get 100 : 75 : : 50 : 371; and 100—75 = 75—50; and 100— 75 = 2(50— 371); and 100, 75, 50; /. s ^ 150 ^ or, 2 (75) = 150, or 75= =75. Li Hence, the circle is the arithmetical mean between the squares, circumscribed and inscribed about it ; or three fourths of the circumscribed square; or three times square of the ra- dius. (See Exercises, def 7, scho. 2). Thus we have the area of the circle expressed by a finite quantity instead of the irrational quantity (V. 25, scho. 1), giving the approximate area only of the circle. THE- END. 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