*^M»M?fl;. '4'fe'i«"ii?* The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924001166341 ^'jments of geometry. 3 1924 001 166 341 ■ELEMEJSTTS OP aEOMETET. BY a^m:legendre. ADDITIONS AND MODIFICATIONS M. ArBLANCHET, "J ^LISVE OF THE POLTTECHNIO SCHOOL; DIBEOTOK OF STUDIES OF SAIKTE-BABBE, TRANSLATED FEOM THK ELEVENTH FRENCH EDITION, FRANCIS H.^MITH, A. M., BnPEBIirrENPENT AND PROFESSOR OF MATHEMATICS IW THE VIBOIHIA IIILITABT INSTITUTE, LEXINGTON, VIRGINIA. BALTIMORE: KELLY & PIET, PUBLISHEES, 174 Baltimokb Steeet. 1867. Entered accoiding lo Act of Congress, in the yenr 1867, By FRANCIS TI. SMITH, Id the Clerk's Office of the District Court of thQ United States for TirpinlL UiLT t rirr, fit n>i«| ^ALTinMV TO THE ALUMNI OF THE VIRGINIA MILITARY INSTITUTE, f lis max^ IS APPECTIONATBLT INSCRIBED THEIE FRIEND AND FORMER PRECEPTOR, THE TRANSLATOR. PREFACE. Pure Geometry may be considered to form the real foundation of mathematical instruction. Some acquaint- ance with Arithmetic and Algebra usually precedes its study ; but the time necessarily employed in gaining a practical facility in the application of the rules of these -branches, leaves little room for the cultivation of the reasoning faculties as they are exercised in Geometry. Geometry rests upon a few simple and self-evident truths ; and from these, by the rigid processes of deduc- tion, the student of mathematics is afforded a valuable mental discipline, which supplies an important corrective for some of the evils resulting from an exclusive devotion to analysis. Hence its importance as an essential branch of academic instruction. Geometry owes its origin, as a science, to the inventive genius of the Greeks. The demonstrations left by the Greek Geometers are models of accuracy, clearness, and elegance, and are admirably adapted to training the mind to habits of close reasoning and luminous arrangement. Soon after the revival of letters, the principal works of the Greek Geometers were translated in Italy by OommaTidine^ embracing fragments of ApoUonius' Conies and of the Collections of Pappus, Towards the close of the 16th century, Yieta restored the lost Tract on Tangencies, and 11 PREFACE. SiielUus reproduced the Plane Loci. Soon afterwards Viviani, the disciple of Galileo, supplied the 5th book of Apollonius. In pursuance of a plan to print at Oxford a complete series of the Greek Geometers, filling up the blanks from an inspection of the Arabic manuscripts, Dr. David Gregory edited the works of Euclid, and Dr. Halley, Apollonius; while Torelli's edition of Archimedes, pur- chased in Italy, was issued from the same press. Euclid is the founder of the ancient Greek Geometry. His great work, known as the Elements of Euclid, is still a text-book in most of the English schools. He appears to have lived in the time of the first Ptolemy, some 300 years before the Christian era, and is said to have been the founder of the Alexandrian mathematical school. His work on Geometry consists of thirteen books. Two others have sometimes been added, of which Hypsicles is sup- posed to have been the author. A complete edition of Euclid's Elements was published in Oxford, in 1703, in the original Greek, by Dr. Gregory. In 1824^1825, an edition of the first six books of Euclid was published in Berlin, in Greek,- with a Latin transla- tion, and copious notes forming a commentary on the text. This edition was edited by Camerer. In 1826-1829, another complete edition of the Elements was published in Berlin, in Greek, with a collection of various readings, by Ernest Frederick August. In 1781, .James Williamson published in Oxford the first of two quarto volumes of the Elements of Euclid, with dissertations, which was followed by. a second volume, printed in London, in 1788. This edition gives a close tri'uslation into English of the thirteen books of the original. Robert SimsOn, a distinguished Scotch mathematician, published a Latin edition of the Elements, in one quarto PREFACE. iil volume, in 1756, which he compressed, in 1760, into an English octavo. The English edition of Simson is a digest of the first six books, with the 11th and 12th of the origi- nal, materially modified to adapt them to the existing state of mathematical instruction. This edition has had a wide circulation, and constitutes the basis of the various editions of the Elements by Playfair^ Lardner, De Mor- gan^ Todhunter, and others, which have appeared, from time to time, to meet the demand for Euclid in the English schools. Many important additions were made to the digest of Euclid'^ Elements, by mathematicians of the 18th century, by which it has been corrected, simplified, and materially improved. The Elements of Geometry, by Thomas Simp- son, and similar works in the French language, by Clai- raut and Legendre. are the most deserving of notice. Simpson^s volume is brief but perspicuous. The Geometry of Clairaut is still shorter, and is designed to guide the student by deductive processes to the solution of math- ematical problems. But the Geometry of Legendre claims a higher position ; and is generally regarded as the best work on the elements of Geometry that has ever appeared. The work of Legendre has long been the standard text- book in the French schools. It was subsequently trans- lated for the English schools by Brewster ; and the trans- lation of Brewster, with modifications by Professor %Farrar, constitutes the basis of the American edition of Legendre, published by Professor Davies. The present translation is from the last French edition of Legendre, with ?idditions and modifications by M. A. Blanchet, an eleve of the Eoole PolyteoJinique, and di- rector of studies of Sainte-Barhe. M. Blanchet has materially improved the original text of Legendre, not only in the general arrangement of the work, but in the simple demonstrations he has given for IV PEEFAOE. the measures of the circle, cylinder, cone, and sphere. Valuable Appendices are also added, embracing The Theory of Transversals ; The Pole and Polar Line; Mdxhnuin Figures under a given perimeter ; together witli copious examples for the exercise of the student, in the demonsti-ation of Theorems and in the solution of Geometrical Problems. Til is translation of Blanchet's edition of Legendre's Geometry, constitutes one of the Mathematical Series of the Virginia Military Institute. The following works, embraced in this series, have already been published : SniitVs Elementary Arithmetic, for Beginners. " Smith's Arithmetic, for High Schools and Academies. Smiths Algebra. Smiths Bio€s Analytical Geometry. The Elements of Geometry will be immediately followed by a translation of Lefebure de Fourcy's Trigonometry, which will be published in the same volume with the Elements, as well as separately. VrRGIOTA MlLITAET INSTITUTE, Lexington, Va., March 1, 1867. ELEMENTS OF GEOMETRY. BOOK I. DEFiisrmoisrs. I. All bodies occupy in space a determinate position, which is called volume. II. The surf ace oi a body is the limit which separates it from the surrounding space. III. The locus, or place, in which the surfaces of two bodies meet, is called a line. I"V. A point is the locus or place in which two lines in- tersect each other. V. Volumes, surfaces, and lines, are considered inde- pendently of the bodies to which they belong. VI. The name of figures is given to volumes, surfaces, and lines. VII. Geometry has for its 6bj\ct the measurement of the extension of figures, and the study of their properties. VIII. A right line is an indetinite line which is the shortest distance between any two of its points. It is evident that we can only draw one right line from one point to another, and that, if two points of right lines coincide, the. lines themselves will coincide throughout their whole extent. 6 ELEMENTS OF GEOMETBT. IX. A Irokm line is a line composed of several right lines. X. Every line which is not a right line, nor a broken line, is a curve line. XI. A plane is a surface, in which, if any two points be taken at will, the right line joining them will lie wholly in the surface. XII. Surfaces, which are neither planes, nor composed of plane surfaces, are curved surfaces. XIII. The figure formed by two right lines, AB, AC, which cut each other in A, is called an angle. The point A is the vertex of the angle, and the lines AB and AC are its sides. The angle is sometimes designated by the letter A of the vertex of the angle; and sometimes by the three letters, BAC, or CAB, taking care to put the letter of the vertex in the middle. Two angles, A and a, are said to be equal when, if ap- A a plied one to tlie other, they may be made tocoincide. Thus, if we place the angle a on A, so that the side ab shall coincide with AB, if then the side ac takes the di- rection AC, the sides of the two angles li! ci s/ c\ will coincide, and the two angles will be equal to each other. An angle A is double, triple, &c., of the angle D, if it A D contains, between its sides, two, three, &c., angles equal to D. Angles may, therefore, be com- pared with each other as other magnitudes. XIV. Wlien a riglit line, AB, meets another right line, CP, so that the adjacent angles, SOOE f. t fiAC and BAD, are equal to each other, the line AB is . said to be perpendiculaf to the line CD, and the equal angles, BAG, BAD, are called right angles. It will be demonstrated that through any point A taken on the right line CD, we may always erect a perpendicular on this line, and that all right angles are equal to each other. Every angle greater than a right angle is called an ob- tuse angle, and every angle smaller than a right angle is called an acute angle. XY. Two lines are said to be parallel., when, being sit- uated on the same plane, they ^ will never meet, if produced in- definitely. The lines AB, CD, are parallel. , XVI. A ^?awej/?p'Mre is a plane terminated by lines. If the bounding lines are right lines, the space which they enclose is called a rectilinear figwre, or polygon, and the lines themselves taken together form the contov/r, or pe- rimeter. of the polygon. The lines AB, BC, CD, DE, and EA, taken together, form the perimeter of the polygon ABCDE. XVII. The simplest of all polygons is the polygon of three sides. It is called a triangle. A quadrilateral is a polygon of four sides ; a pentagon, one of five sides ; a hexagon, one of six sides, etc., etc. XVIII. When a triangle has all its sides equal, it is called an equilateral triangle / an isosceles triangle is a triangle which has two of its sides equal ; and a scalene triangle is a triangle which has all its three sides un- equal. The triangle ABC is equilateral / oho, having the sides ELEMEXTS OF GEOMETRy. db, ac equal, is isosceles, wliile the triangle def, all the sides of which are unequal, is a scalene triangle. XIX. The right angle triangle is a triangle which has one right angle. The side oppo- site the right angle is called the hypothenuse : thus, ABC is a right angle triangle, with a right angle at A ; and the side BC, opposite to the right angle, is the hypothenuse. XX. Among quadrilaterals we distinguish five— the square, the rectangle, the pwrallelogram,, the lozenge or rhombus, and the trapezoid. The square is a quadrilateral which has its sides equal, and its angles right angles: ABCD is a square. The rectangle is a quadrilateral, which has its angles right angles, without having its sides equal. ABCD is a rectangle. The parallelogram is a quadrilateral, which has its op- posite sides parallel : ahcd is a parallelogram. The lozenge or rhombus is a quadrilateral whose sides are equal, but its angles are not right angles. BOOK t. i) ABCD is a lozenge or rhombus. The trapezoid or trapezium is a quadrilateral, only two sides of which are paral- lel : abed is a trapezium. XXI. A diagonal is a right line which joins the vertices of two angles which are not adjacent to each other. XXII. The equilateral polygon is a polygon, the sides of which are equal ; and an equiangular polygon has all its angles equal. XXIII. Two polygons are equilateral when they have their sides equal,'each to each, and placed in the same order ; that is, following their perimeters in the same order, the first side of the one is equal to the first side of the other, the second of the one to the second of the other, and so on. The same definition applies to two equi- angular polygons. In both cases, the equal sides, or the equal angles, are called the homologous sides, or the ho- mologous angles. XXIV. A convex polygon is a polygon such, that if we produce any one of its sides, aU of the polygon will lie on the same side of tliis right line. The, polygon ABODE is a convex polygon. The perim- eter of a convex poly- B gon cannot be inter- sected by a right line in more than .two points; for, if a right line, IQ, cut the perimeter ABODE at the points M, ]Sr, P, Q, the side BO, which is cut by the right line in one of its intermediate points IS", would 1* lO ELEMENTS OF GEOMETRY. evidently have parts of the figure situated on both sides of its direction. XXY. Any two figures whatever, viz., volumes, sur- faces, or lines, are said to be equal, when they may be ap- plied, one to the other, and perfectly coincide. N. B. — In the four first books we shall consider plane fig- ures only, that is, figures drawn on a plane surface. Explanation of Terms and Signs. An Axiom is a self-evident proposition. A Theorem, is a truth which becomes evident by a pro- cess of reasoning, called a demdnstration. A Problem, is a question which is proposed, and which requires a solution. A Lemma is a subsidiary truth, which is employed in the demonstration of a theorem, or in the solution of a problem. The common name oi proposition is given, indifferently, to theorems, problems, and lemmas. A Corollary is a remark on one or more preceding prop- ositions, showing their connection, their utility, their re- striction, or their extension. An hypothesis is a supposition made, whether in the enunciation of a proposition, or in the course of the de- monstration. The sign = is the aign of equality: thus, the erpression, A = B, means that A is equal to B. To express that A is smaller than B, we write A < B. To denote that A is greater than B, we wri'.,e A > B. The sign -f , called plus, indicates addition. The sign — , called minus, indicates subtraction. Thus, A -i- B represents the sum of two quantities, A and B ; and A — B represents their difference, or what remains in taking B from A. / / BOOK I. 11 The sign x, indicates multiplication : thus, A x B rep- resents the product of A bj B. Instead of the sign x, we sometimes use a point ; thus, A . B is the same thing as A X B. The same product is also indicated without using any sign. Thus, AB ; but in using this last method, we must be careful to distinguish it from the line AB, which denotes the distance between two points, A and B. The expression A x (B + C — D), denotes the product of A by the quantity B + C — D. If we wish to mul- tiply A + B by A — B + C, we indicate the.produet thus, (A + B) X (A — B + C ) ; the quantities embraced within the parenthesis being considered as a single quantity. A number placed before a line or a quantity serves as a multiplier of this line or quantity : tlius, to express that the line AB is taken three times, we write 3 AB ; and to denote the half of the angle A, we write JA. The square of the line AB is written AB ; its cube by AB . The precise meaning of the square and cube of a line Will be explained in its proper place. The sign s/ indicates a. root to be extracted ; thus, •y/2 denotes the square root of 2 ; ^A x B represents the square root of the product A x B, or the mean propor- tional between A and B. Pboposition I. — Theoeem. Through a point taken, on a given, right line, we may erect one perpendicular on this line, and only one. In fact, let us suppose a right line AM at first coinciding with AC, to turn around the point A ; it will form two ad- jacent angles, MAC, MAB, of which one, MAC, from being very small, will go on increas- ing, while the other, MAB, at 12 ELEMENTS OF GEOMETRY. tirst greater than MAC, will constantly decrease until it becomes zero. The angle MAC, from being smaller than MAB, be- comes, as the line AM moves, greater than this angle ; consequently, there will be a position AM" for the moving line, where these two angles will be equal, and it is evi- dent there can be only one position where this equality can exist. Corollary. All Tight angles are equal. Let DC be perpendicular to AB, and HG perpendicular to EF; the angle DCB is equal to ^^ ° the angle HGF ; for, if we place the line EF B on AB, so that the point G fall on C, Gil will take the direction CD; otherwise two perpendiculars might be erected at the same point on a right line, which is impossible. Pkoposition II.— ^Theorem. Every right line CD, which meets another right line AB, makes with it two adjacent angles, ACD, BCD, the sum, of which is equaZ to two right angles. At the point C erect on AB the perpendicular CE. The angle ACD is the sum of the angles ACE, ECD ; also, ACD + BCD will be equal to the sum of the three angles, ACE, ECD, BCD. The first of these, ACE, is a right angle; the two others, ECD and BCD together, make a right angle. Hence, the sum of the two angles, ACD and BCD, which is equal to the sum of the three angles, ACE, ECD, BCD, must also be equal to two right angles. BOOK I. 13 CoroUary I. If one of the angles ACD, BCD be liglit, the other will be right also. Corollary II. If the line DE be per- pendicular to AB, reciprocally, AB will be perpendicular to DE. For, if DE be perpendicular to AB, it follows that the angle ACD is equal to its adjacent angle DCB, and that they are both right angles. But, if ACD be a right angle, its adjacent angle ACE is also a right angle. Hence, ACE = ACD, and AB is perpendicular to DE. Corollary III. The su m of the consecutive angles, BAG, CAD, DAE,EAF,formed on the same side of the right line BF, is equal to two right angles ; for their sum is equal to the two adjacent angles BAC, CAF. Pkoposii'ion III. — Theorem. If two adjacent angles, ACD, DCB, together equul two right angles, the two exterior sides, CA, CB, will form one and the same right line. For, if CB be not the prolongation of AC, let CE be the prolongation of AC ; then, the line ACE being a right line, the sum of the angles ACD and DCE will be equal to two right angles (Prop. II.) But, by hypothesis, the sum of the angles ACD and DCB, is also equal to two right angles ; then ACD + DCB = ACD + DCE. Taking from each of these equals, the angle ACD, there will remain the angle DCB, equal to the angle DCE, which is impossible. Hence, CB is the pro- longation of AC. u ELEMENTS OF GEOMETKT. Pkoposition TV. — Theorem. Jf two right lines, AB and DE, intersect each other, the opposite angles, formed at the vertex C, are equal. For, since the line DE is a right line, the sum of the angles ACD, ACE, is equal to two right angles : and since the line AB is right, the sum of the angles ACE, BCE, is also equal to two right angles : _3 hence ACD + ACE = ACE + c ^-^ " BCE. Taking from each the ~B same angle ACE, there will re- main the angle ACD equal to its opposite angle BCE. "We might demonstrate, in the same manner, that the angle ACE is equal to its opposite angle BCD. Scholium. Tiie lour angles formed around a point by two right lines wliich intersect each other, are together equal to four right angles ; for, the angles ACE, BCE, taken together, are equal to two right angles, and tlie two other angles, ACD, BCD, are also equal to two right angles. In general, if any number of right lines meet in a common point C, the sum of all the consecu- tive angles ACB, BCD, DCE, ECF, FCA, will be equal to four right angles ; for, if we form at the point C, by means of two perpendiculars, four right angles, their sum will evi- dently be equal to the sum of the suc- cessive angles ACB, BCD, &c. Pkoposition V. — Theokem. Two triangles are equal, when they have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each. Let the side AB be equal to tlio side DE,the side AC. BOOK 1. 15 to the side DF, and the angle A, inohided by the sides AB and AC, be equal to the angle D, included by the sides DE and DF ; then will the tri- angle ABC and DEF be equal. For, these triangles may be placed, one upon the other, so that they will exactly coincide. Thus, if we place the side DE on its equal AB, the point D will fall on A, and the point E on B ; but since the angle D is equal to the angle A, when the side DE is placed on AB, the side DF will take the direction AC. Besides, DF is equal to AC; hence the point F will fall on C, and the third side, EE, will coincide exactly with the third side, BC ; therefore, tlie triangle DEF is equal to the triangle ABC. Corollary. When three parts are equal in two triaii- gles, each to each, to wit, the angle A = D, the side AB = DE, and the side AC = DF, we may conclude that the three other parts are also equal; that is, the angle B = E, the angle C = F, and the side BC = EF. Peoposition YI. — Theoeem. Two triangles are equal, when they have two angles and the included side of the one, eq;ual to two angles and the included side of the other, each to each. Let the angle B be equal to the angle E, the angle C to the angle F, and the side BC, included by the D A angles B and C, equal to the side EF, in- cluded by the angles E and F ; tlien will the triangle DEF be equal to tlie trian-^le ABC, 16 ELEMENTS OF GEOMETBY. For, if we place the side EF on its equal BC, the point E will fall on B, and the point F on C. Since the angle E is equal to the angle B, the side ED will take the direc- tion BA, and the point D will be found somewhere on the line BA. Again, since the angle F is equal to the angle C, the line FD will take the direction CA, and the point D will be found somewhere on the side CA ; hence, the point D, which is at the same time on the lines BA and CA, will fall at their intersection A : the two triangles, therefore, coincide with each other, and are equal. Corollary. When two triangles have the three parts equal, viz., B = E, C = F, and BC=EF, we may con- clude that the three other parts are also equal; that is, AB=DE, AC = DF, and A = D. Proposition VII. — Theoeem. In every triangle, any one side is smaller than the sum of the other two. For, the right line BC, for example, is the shortest dis- tance from B to ; and hence BC is shorter than AB -|- AC. "We may remark also, that either side of a triangle is greater than the difference between the other two sides, for if we represent the three sides by a, b, and c, and suppose a the greatest side, we shall have a GFII ; then will F the side BC be greater than the side GH. Make the angle CAD = GFH, take AD = GF, and join C and D: _ the triangles ACD, GFH are equal, since they have two sides and the included angle of the one equal to two sides and the included angle of the other. (Prop. VI.) It suffices now to show that BC is greater than CD. Divide the angle BAD into two equal parts by the line' AE ; the right line will fall in the greater angle BAC; di-aw the line DE ; the two triangles, BAE, EAD, will he equal, having two sides and the included angle in each equal. Tlien BE = ED. But, in the triangle EDC, wo hnve CD < ED + EC. Keplacing ED by its equal BE, we have CD < BE 4- EC, or CD < BC. Reciprocally, if the sides AB, AC, of the triangle ABC, be equal to the two sides FG, FH, of the triangle FGH, and if, besides, the third side CB of the first triangle be greater than the third side GH of the second, the angle BAC will be greater than the angle GFH. For, if the angle BAC were smaller than the angle GFH, from what has been just demonstrated, CB would BOOK I. 19 be less than GH, which is contrary to the hypothesis ; and if the angle BAG be equal to the apgle GFH, we should have GB = GH (Prop. V.), which is also contrary to the hypothesis. ' Pkoi'Osition XI. — Theoeem. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles will he equal. Let the side AB = DE, AG = DF, BC = EF, then will the triangles ABC, DEF, be also equal, and we shall have the angle A = the angle » ^ D, B = E, and C = F. For, if the angle 'A were greater than the angle D, since the sides AB, AG are equal to T B the sides DE, DF, each to each, it follows, from the preceding theorem, that the side BG would be greater than EF; and if the angle A were smaller than tlie angle D, the side BC would be smaller than the side EF. But BC is equal to EF ; there- fore, the angle A cannot be greater nor less than the angle D. Hence, it niust be equal to it. We might pi'ove, in like manner, that the angle B = E, and the angle c =r. Scholiuih. We remark, that the equal angles lie oppo- site to the corresponding equal sides; thus, the equal angles, A and D, are opposite to the equal sides, BC, EF. Piti iposrrioN XII. — Tii k( iekm. In an isosceles triangle, the angles opposite the equal sides arc equal. 20 ELEMENTS OF GEOMETRY. Let tlie side AB = AC, then will the angle C = B. From the vertex A, draw AD to the mi'ddle point D of the base BC ; the two triangles ABD, ADC, will have the three sides equal, each to each : that is, AD common, AB = AC by hypothesis, and BD = DC by construction. Hence, by the preceding theorem, the angle B is equal to the angle C. Corollary. An equilateral triangle is, at the same time, equiangular. Scholium. The equality of the triangles ABD, ACD, proves at the same time that the angle BAD = DAC, and that the angle BDA = ADO. Hence, BDA and ADC are right angles. Therefore, the line drawn from the vertex of an isosceles triangle to the middle point of the base, is perpendicular to the base, and divides the angle at the vertex into tioo equal parts. In a triangle which is not isosceles, we take indifferently either side for the hose, and then its vertex will belong to the angle whicb is opposite to the base ; while in an isos- celes triangle, we take the side for the base which is not equal to either of the other sides. Proposition XHI. — Thkoeem. If two angles of a triangle be equal, tJie sides opposite to those angles will also be equal, and the triangle will be ^ isosceles. Let the angle ABC = ACB, then will the side AC be equal to the side AB. For, if AC be not equal to AB, let uS suppose AB to be greater' than AC. "c Take BD = AC and join DC. The angle DBC is, by hypothesis, equal to ACB; the two sides BOOK I. 21 DB, BC, are equal to AC and CB ; then, the triangle DBC would be equal to the triangle ACB. But the part can- not be equal to the whole ; and hence, AC and AB cannot be unequal, and the triangle ABC is therefore isosceles. Pboposition XIV. — ^Theoeem. Of two sides of a triangle, that side is greater which lies opposite to the greater angle ; and conversely, of two angles of a triangle, the greater angle is opposite to the greater 1°. Let the angle C > B, then will the side AB, opposite the angle C, be greater than the side AC, opposite the angle B. Make the angle BCD=B; then, in the triangle BDC, we shall have BD = DC. (Prop. XIII.) But the right line AC is shorter than AD + DC, and AD + DC = AD+DB=AB. Hence, AB is greater than AC. 2°. Let the side AB > AC ; then will the angle C, op- posite the side AB, be greater than the angle B, opposite the side AC. For, if C were less than B, it would follow, from what has just been demonstrated, that AB < AC, which is con- trary to the supposition. If C = B, we should have AB = AC (Prop. XIIL), which is also contrary to the hypothesis. Hence, as the angle C winnot be less than, nor equal to, the angle B, it must be greater. Pboposition XV. — TufeoKKM. From a given point without a right line, cnlij ktne per- pendicular can he let fall iipon that line. 22 ELEMENTS OF GEOMETET. Let lis suppose that, from the point A, we could let fall two perpendiculars on the line CD, viz., AE, AB ; pro- duce one of them AE to A', making EA'= AE, and join A'B. The tri- angle AEB is equal to the triangle A'EB, for the angles AEB, A'EB, are right angles ; the side AE = A'E, by construction ; and the side BE common. Hence, the angle ABE — EBA'; hut the angle ABE is a right angle, by supposition ; therefore, EBA' is a right angle also. But if the adjacent angles ABE -!- EBA'= two right angles, the line ABA' must be a right line, and we shall then have two right lines drawn between the points A and A',' which is impossible. Hence, only one perpendicular can be drawn from A to CD. Pboposition XVI. — Theorem. If from a point A without the right line DE, the per- pendicular AB he drawn to this line, and the several lines AE, AC, AD, (&o., to the different points of this line — 1°. The perpendicular will le shorter than any oblique line. 2°. Any two oblique lines AC, AE, drawn on different sides of the perpendicular, and at equal distances, BC, BE, will be equal. 3'^ Of two oblique lines, AC and AD, or AE and AD, drawn at will, that which is farther from the perpendicular will be the longer. Produce the perpendicular AB until BF = AB, 'and join EC, FD. 1°. Tlie triangle BCF is equal to the triangle BCA; for the right BOOK t. 23 angle CBF = CBA, the side BC is common, and the side BF — BA ; therefore, the third side CF is equal to the third side AC. (Prop. V.) But the right line ABF is shorter than the broken line ACF; and AB, the half of ABF, is shorter than AC, the half of ACF. Hence, 1°, thqj perpendicular AB is shorter than every oblique line. 2°. If we suppose BE = BC, the triangle ABE = ABC ; for, we have AB common, the right angle ABE — ABC, and BE = BC. Hence, the sides AE AC, are equal ; and 2°, two oblique lines equally distant from the perpendicular will be equal. 3"^. In- the triangle DFA, the sum of the lines AC, CF is less than the sum of the sides AD, DF (Prop. VIH.) ; and AC, the half of the line ACF, is less, than AD, the half of ADF ; consequently, 3°, the oblique line which is farthest from the perpendicular will be the longer. Corollary I. The perpendicular measures the true dis- tance of any point from a right line, since it is shorter than any oblique line. Corollary IT. From the same point we cannot draw three equal right lines to the same line ; for, if this were possible, we should have two equal oblique lines on the same side of the perpendicular, which is impossible. Pkoposition XYH. — Theorem. If, from the middle point C of the right line A£, we erect the perpendicular EF—1°, each point of the perpen- dicular will he equally distant from the two extremities of the line AB ; 2°, any point situated outside of the perpen- dicular will he unequally distant from the same extremities A and B. For, 1°, since AC =CB, the two oblique lines AD, DB are equally removed from the perpendicular ; and are 24 ELEMENTS OF GEOMETRY. therefore equal. The same is true of the two oblique lines AE, EB, and of the two AF, FB, &c. HencQ, 1°, every point of the perpendicular is equally distant from the extremities A and B. 2°. Take any point I outside of the perpendicular ; if we join lA, IB, one of the lines will cut the perpendicular at D, and drawing DB, we shall have DB = DA. But the right line IB is shorter than the broken line ID + DB, and ID + DB = ID + DA = lA ; therefore, IB < lA ; and hence, 2°, every point outside of the perpendicular is unequally distant from the extremities A and B. RemarTc. — A geometrical locus is a line, all the points of which possess a common property not possessed by all the other points of the plane. The line EF is the geometrical locus of the points equally distant from the points A and B. PEOPOSiTicftj XVIII. — Theorem. Two right-angled triangles are equal when they have the hypothenuse and a side of the one equal to the hypoth- cnuse and a side of the other, each to each. Let the hypothenuse AC = DF ; and the side AB = DE ; then will the right-angled triangle ABC be equal to the right-angled triangle DEF. The equality of these two triangles will evidently exist, if the third side BC is equal to the third side EF. Let us BOOK I. 25 suppose, if possible, that these, sides are not equal, and that BC is the greater. Take BG = EF, and join AG. The triangle ABG is equal to the triangle DEF ; for, the right angles B and E are equal, the side AB =:DE, and the side BG = EF. Hence, the two triangles are equal (Prop. V.), and we have AG = DF. But, by hypothesis, DF = AC. Hence, AG = AC. But the oblique line AC cannot be equal to AG (Prop. XVI.), since it is farther from the per- pendicular AB ; therefore, it is impossible that BC should be unequal to EF ; and hence, the triangle ABO is equal to the triangle DEF. ■'a' Peoposition XIX. — ^Thkoeem. Two rightrangled triangles are equal when they have the hypothenuse and an angle of the one equal to the hypoth- enuse and an angle of the other, each to each. Let AC = DF, and the angle A = D. Place the tri- angle DEF on the triangle ABC, so that DF shall coincide with its equal AC. Since the angle D = A, the line DE will take the direction AB, and at the same time FE will take the direction CB ; otherwise, we should have two perpendiculars to the line AB, from the point C. The point E will therefore fall on B, and the two triangles will coincide and be equal. Peoposition XX. — Theoekm. 1°. Every point M taken on the hisectrix* of an angle £AD, is equally distant from, the sides of this angle. * The Haectrix of au angle is tlie line which divides the angle into two equal parts. „ 26 ELEMENTS OF GEOMETE^'. 2°. Every point M taken in the angle BAD, and equally distant from the sides AB, AD, is a point of the hisec/rix of that angle. 1°. Draw, from the point M, the right lines MD and MC, respectively, perpend icnlar to AD and AB ; the right-angled triangles MAD, MAO are equal, for they have an hypothenuse MA common, and the angles MAD, MAC equal, by hypoth- esis. Hence, MD = MO. 2°. Reciprocally, if the perpendiculars MD, MO are equal, the right-angled triangles MAD, MAC will be still equal, having an hy- pothenuse MA common, and the sides MD, MO equal, by hypothesis. Hence, the angle MAD = MAC. There results from this, that any point taken in the angle BAD, outside of the bisectrix AM, will be unequally dis- tant from the two sides. Scholium. — The bisectrix of an angle is the geometrical Ipcns of the points situated in the interior of this angle, and equally distant from its two sides. V IB D THEOEY OF PAEALLELS. Proposition XXL — Theoeem. Two right lines, AC, BD, which are perpendicular to the same right line, are parallels. For, if they meet each other in M, for example, we should have two per- pendiculars drawn from the same point to the line CD, which is impossible. BOOK I. ■ 27. PROPOSITtON XXII. — ThKORKM. Through a given point we way always draw a parallel to a given right line. From the point A, draw AB perpendicular to BO, and draw AD perpendicular to AB ; the two perpendiculars, AD and BC, being both perpen- dicnlar to AB, will be parallels. We will admit, in the second place, as an evident proposition, that through a given point we can only draw one parallel to a given right line. Pkoposition XXIII. — Theoeem. If two right lines., CD, AB, are parallel, every right line, FH, perpendicular to one of them, AB, is perpen- dicular to the other, CD. It is at first evident that FH must meet CD, otherwise we could draw two parallels to CD through the point F. Finally, CD is perpendicu- lar to FH; for, if the line CD were oblique to FH, we might erect a perpendicular to FH, at the point H ; and we should thus have two right lines passing through the point H, parallel to AB, which is impossible. PKOPOsmoN XXIV. — Theorem. Two right lines AB, CD, parallel to a third line, EF, are parallel to each other. For, if the lines AB, CD met each other in a point M, we miglit draw two parHllcls ■^ to tlie right line EF, thmugh tlie point M, wliicli is iinpossible. >■ 28 ELEMENTS OF GEOMETRY. DEFINITIONS. When two right lines, AB, CD, are cut by a transversa], EF, there are eight angles formed at the points of intersec- tion G and H. The four angles (1), (4), (5), (S), Ijing between the right lines AB and CD, are called interior angles ; the four others are called exterior angles. The two angles (1) and (5), sit- I uated on different sides of the secant, interior and not adjacent, are called a^fernafe-iMferior angles. The two angles (8) and (2), situated on the same side of the secant, one interior, the other exterior and not adjacent, are called corresponding angles. Finally, the angles (2) and (6), situated on different sides of the secant, exterior and not adjacent, are called alternate- exterior angles. Peoposition XXY.— Theoeem. Two parallels form with a transversal — 1°. Equal alternate-interior angles. 2°. Equal alternate-exterior angles. 3°. Equal corresponding angles. i°. Interior angles on the same side of the secant, whose sum is equal to two right angles. 1°. Let the parallels AB, CD be cut by the transversal GH. From the point O, the middle of EF, draw OM perpendicu- lar to AB ; this line will be also perpendicular to CD. The right-angled triangles MOE, ONF are equal ; for. BOOK I. 29 the hypothenuse OE is equal to ihe hypothennse OF, by construction ; and the angles MOE, FO!N", being opposite vertical angles, are also equal. From the equality of these triangles, we conclude that the alternate-interior angles MEO, OFN, are equal. We see, also, that the angles BEF, EFC, are equal ; tor these angles are respectively the supplements* of the angles MEO, OFIST. 2°. The alternate-exterior angles GEB, CFH are equal ; for, they are" the opposite vertical angles to the equal angles MEO, OFJST. 3°. The corresponding angles GEB, EFD are equal ; for GEB is equal to AEF, and AEF = EFD. 4°. The sum of the angles BEF, EFD is equal to two right angles ; for, we have BEF 4- AEF = 2 right angles, and AEF = EFD. Peoposition XXVI. — Theoeem. Reciprocally^ if two right lines make with a transversal ; Equal alternate-interior angles; Equal alternate-exterior angles ; Equal corresponding angles ; Interior angles on ihe same side of the secamt, whose sum is equal to two right angles ; These right lines will he paral- lels. A -^^ B -^o^ -Lgi. ^jjg ^^^ j,jgj^j ^j^,gg ^^g^ CD be cut by the transversal GH ; ^ if the alternate-interior angles AEF, EFD are equal, AB will be parallel to CD. Otherwise, we might draw through the point E a line EI parallel to * The supplement of an angle is the difference between this angle and two right angles. 30 ELEMENTS OF GEOMETRY. CD ; but, in tliis case, the angles lEF, EFD, would be equal, as alternate-interior angles ; and as, bj' hypothesis, AEF =EFD, wo should have AEF = lEF, which is ab- surd. 2°. If the alternate-exterior angles GEB, CFH are equal, the opposite vertical angles AEF, EFD are also equal ; and, from what has just been demonstrated, AB will be parallel to CD. 3°. If the corresponding angles GEB, EFD are equal, as GEB is equal to AEF, we shall have AEF = EFD. Hence, AB is parallel to CD. 4°. If the sum of the angles BEF, EFD is equal to two right angles, as BEF -f AEF = 2 right angles, we conclude that AEF == EFD ; and hence, AB is parallel to CD. Proposition XXVII. — Theorem. Two angles which have their sides parallel are equal or supplementary. 1°. Let ABC, DEF be two angles whose sides are parallel and lying in the same direction. These angles ^A jj will be equal. In fact, the angles DLC, DEF are equal, as corresponding angles. But, for a like reason, DLC = ABC; and hence, ABC = DEF. 2°. Let the two angles ABC, MEN, have their sides parallel, but lying in a contrary direc- tion ; these angles will be equal ; tor MEN is equal to DEF, and DEF = ABC. 3°. Finally, if the two angles ABC, DEM have their sides parallel, BA and ED lying in the same direction, while the sides BC and EM lie in a contrary direction, feoot f. SJ then will the angles ABC and DEM be supplementary; for, DEM is the supplement of DEF, and DEF = ABC. Pkoposition XXYIII. — ^Theorem. If two angles have their sides perpendicular, each to each, these angles will he equal or supplementary. Let BAC, DEF be two angles, whose sides are respec- tively perpendicular to each other. Draw through the point A the line AI perpendicular to AB, and AH perpendicular to AC; the right lines AI, AH, will be respectively paral- lel to the right lines DE, EF, and lie in the same direction. Hence, the angle lAH is equal to DEF; but we have LAH + HAB = one right angle, and BAC + HAB also equal to one right angle; therefore, lAH = BAC. Scholium. — If we consider the angle formed by the right line EF, and the prolongation of DE, we shall see that the angle FEG is the supplement of the angle BAC. Pkoposition XXIX. — Theoehm. The sum of the three qngles of a triangle is equal to two right angles. Draw AE parallel to BC, and produce AC to D. The angles ACB, EAB are equal, being corresponding angles ^ formed by the parallels BC, ° AE, cut by the transversal AC. 32 ELEMENl'S OP GEOMETRf . The angles CBA, BAE are also equal, being alternate- interior angles of the parallels BC and AE, when cut by tlie secant AI^: hence, the sum of the three angles of the triangle is equal to the sum of the three angles CAB, BAE, EAD, formed around the point A, on the same side of the line AC. But tlie sum of these angles is equal to two right angles; therefore, the sum of the three angles of the triangle is equal to two right angles. Corollary I. In any triangle, there can only be one angle which is a right angle ; and, for a greater reason, there can only be one angle that is an obtuse angle. Corollary II. In every right-angled triangle, the sum of the two acute angles is equal to one right angle. Corollary III. When we know two angles of a trian- gle, or simply their sum, we may fiiid the third angle, by taking the sum from two right angles. Corollary I V. The exterior angle BAD, formed by the side BA and the prolongation of AC, is equal to the sum of the two interior angles CBA, BCA.. Pkoposition XXX. Tke sum of the interior angles of any convex polygon is equal to as many times two right angles as the polygon has sides, less two. Through one of the vertices A, draw the diagonals AC, AD, AE, AF. The polygon will thus be divided into as many triangles as there are sides in the polygon, less two ; for, these triangles may be considered as hav- ing for a common vertex the point A, and for bases the different sides of the polygon, — the two extreme triangles ABC, AGF, containing, each, two sides of the polygon. The sum of the angles of these trian- fiook i. 33 « gles IS evidently equal to the sum of the angles of the polygon ; and as the sum of the angles of the triangles is equal to as many times two right angles as there are trian- gles, the sum of the angles of the polygon will be equal to as many times two right angles as the polygon has sides, less two. If we represent by n the number of sides of the polygon, the sum of the angles will be 2 X (tj, - 2) or 2 71 - 4. Peoposition XXXI. — ^Theokem. The opposite sides and opposite angles of a parallelo- gram are .equal, respectively. Draw the diagonal BD ; the two triangles ADB, DBG have the common side BD ; besides, since AD, BC are par- allels, the angle ADB = DBG (Prop. XXVI.), and because AB, CD are parallels, the angle ABD = BDC ; hence the two triangles ADB, DBG are equal (Prop. VI.) ; and the side AB, opposite the angle ADB, is equal to the side DC, opposite the equal angle DBG ; and, in like manner, the third side AD is equal to the third side BG : therefore, the opposite sides of the parallelogram are equal. In the second place, from the equality of the same trian- gles, it follows that the angle A is equal to tlie angle C, and also tlie angle ADC, composed of the two angles ADB, BDC, is equal to the angle ABC, composed of the two equal angles DBG, ABD, hence the opposite angles of the parallelogram are equal. Corollary I. — Two parallels, AB, CD, comprehended between two other parallels, AD, BG, are equal. 2* 34 ELEMENT^ OF* GEOMETEY. Corollary II. — Two parallels are equidistant through- out their whole extent. For, CD and AB being par- allel, the perpendiculars IIF, and GE to AB, drawn through H and G, are parallel also, and will be equal, as lying be- tween two parallels. Peoposition XXXII. — Theorem. If, in any quadrilateral ABCD, the opposite sides are equal, so that we have AB = CD, and AD — BG, the equal sides will be parallel, and the figure will he a paral- lelogram. For, drawing the diagonal BD, the two triangles ABD, BDG will have the three sides equal, each to each ; they will therefore be equal ; and the angle ADB opposite the side AB, is equal to the angle DBG, opposite the equal side GD, hence (Prop. XXVI.) tho side AD is parallel to BG. Jn like manner, AB is parallel to CD, and the quadrilateral ABCD is a parallelogram. PEOPOsrrioN XXXIII. If two opposite sides, AB, CD, of a quadrilateral are equal and parallel, the two other sides will he equal and parallel also, and the figure ABCD will he a parallelo- gram. Draw the diagonal DB ; since AB is parallel to CD, the al- ternate-interior angles ABD, BOOK I. 35 EDO are eqnal (Prop. XXVt) ; besides, the side AB = DC, the side DB is common ; hence, the triangle ABD is equal to the triangle DBG (Prop. Y.) ; and we shall have the side AD = BO, the angle ADB = DBO, and conse- quently AD is parallel to BO, and the figure ABQD is a parallelogram. Peoposition XXXIV. — ^Theoeem. ' The two diagonals AG, DB of a parallelogram, divide each other into two equal parts : For, comparing the triangles ADO, OOB, we find the side AD = OB, the angle ADO ^ OBO (Prop. XXY.) ; and the angle DAO = OOB ; hence, these triangles are equal (Prop. VI.); and we have AO, the side opposite the angle ADO, equal to the side 00, oppo- site the equal angle OBO ; and we have also DO — OB. Scholium. In the case of the rhombus, the sides AB, BO being equal, the triangles AOB, OBO h3,ve the three sides equal, each to each, and are therefore equal ; hence, the angle AOB = BOO, and the diagonals will intersect each other at right angles. 36 ELEMENTS OF GEOMETRY. BOOK II. THE CIRCLE AND THE MEASURE OF ANGLES. DEFINITIONS. I. The circumference of a circle is a curve line, all the points of which are equally distant from a point within, called the centre. The circle is the portion of the plane bounded by the circumference. II. Every right line CA, CE, CD, &c., drawn from the centre to the circumference, is called a radius ; and every line, such as AB, which passes through the centre, and is ter- minated on both sides at the cir- cumference, is called a diameter. Hence, there may be an infinite number of radii in the same cir- cle, and they are all equal to each other ; and there may be also an infinite number of diameters in the same circle, and they will ako be equal, and each will be double of the radius. HL An arc is any portion of the circumference, sucli as FHG. The chord of an arc is the right line FG, which con- nects the two extremities of an arc, and subtends the arc. IV. The segment is the surface or portion of the circle embraced between the arc and the chord. N. B. The same chord FG will always correspond to two arcs, FHG, FEG, and also to two segments ; but ref- BOOK II. 37 erence is always made to the smaller, unless otherwise expressed. Y. A sector is the part of the circle comprised between an arc DE and the radii CD, CE, drawn to the extremities of this arc. VI. A right line is inscribed in a circle, when its ex- tremities are in the circumference, as AB. An angle is inscribed in a circle, when its vertex is in the circumference, and its sides are chords. An inscribed triangle, such as ABC, has the vertices of its three angles in the circumference, and its sides are chords. In general, an inscribed ^polygon is one, the vertices of whose angles are in the circumference, and M'hose sides are chords. Til. A secant is a line which cuts the circumference in two points : AB is a secant. VIII. A tangent is a line which has only one point com- mon with the circumference : CD is a tangent, and the point M is called the point of contact. IX. In like manner, two circumferences are tangent to each other, when they have only one point common. X. A polygon is circumscribed about a circle when its sides are tangent to the circumference. In the same case, we say also that the circle is inscribed in the polygon. 38 ELEMENTS OF GEOMETRY. N. B. In general, we call a tangent to any curve the limit of the positions which a secant AB takes in turning around the point A, until the second point of intersection, B, B', &c., unites with the first A, If the curve is re-entrant, as the circle, and cannot be intersected by a right line in more than two points, it is evident that when the points of in- tersection are united in a single point, the line will only have one point in common with the curve, and we might therefore define a tangent to a curve as a right line which has only one point in common with the curve. But the first definition suits all curves; and even limiting it to the circle, has the advantage of pointing out remarkable analogies between many theorems. Peoposition I. — Theobem. Every diameter AB divides the circle and its circum- ference into two equal parts. For, if we apply the figure AEB to AFB, keeping the line AB com- mon, the curve line AEB must fall exactly on the curve line AFB, other- wise there would be points in the cir- cumference unequally distant from the centre of the circle, which is con- trary to the definition of the circle. BOOK II. 39 Proposition II. — Theokem. Every chord is smaller than the diameter. For, if at the extremities of tho chord AD, we draw the radii AC, CD, we shall have the right line AD < AC + CD, or AD < AB. Corollary. Hence, the diameter is the longest right line that can be in- scribed in a circle. Peoposition hi. — ^Thkoeem. A right line cannot meet a cirournference in more thO/n two points. For, if it met the circumference in three points, these points would be equally distant from the centre ; we should then have three equal right lines drawn from the same point to the same right line, which is impossible (Book I., Prop, xyi.) Proposition IV. — Theoeem. ' In the same circle or in equal circles, equal arcs are subtended hy equal chords / and, reciprocally, equal chords are subtended by equal a/rcs. The radius AC being equal to the radius EO, and the arc AMD to the arc ENG, the chord AD will be equal to the ^chord EG. Foi", the diameter AB being equal to the di- ameter EF, the semicircle AMDB may be applied to the semicircle ENGF, and the curve line AMDB will ■id ELEMENTS OF GEOMETRY. ex:ietly eciiucide with the curve line ENGF. But we have sii|)[)osed the portion AMD to be equal to the portion EXG ; then, the point D will fall on G; and the chord AD will be equal to the chord EG. Reciprocally, supposing AG = EG, if the chord AD = EG, we shall also have the arc AMD equal to the arc ENG. For, drawing the radii CD, OG, the two triangles ACD, EGG will have the three sides equal, each to each, to wit,' AC = EG, CD = GG, and AD = EG ; hence, these trian- gles are equal (Book I., Prop. XL) ; and the angle ACD = EGG. But in placing the semicircle ADB on its equal EGF, since the angle ACD = EGG, it is evident that -the radius CD will fall on the radius OG, and the point D will fall on G ; hence, the arc AMD is equal to the arc ENG. Proposition Y. — Thboeem. In the same circle or in equal circles, the greater chord is subtended hy the greater arc, and, reciprocally, the greater arc is subtended by the greater chord ; the arcs which are considered being always less than a sem,i-circufnference. For, suppose the arc AH > the arc ENG ; take the arc AMD = ENG, the chords AD, EG will be equal. Draw the radii DC, CH; the two sides AC, CH, of the tri- angle ACH, are equal to the two sides AC, CD of the triangle ACD; the angle ACH is greater than ACD ; hence, the third side AH will be greater than the third side AD ; and AH is greater than EG. Eeciprocally, if the chord AH > the chord EG, the arc • BOOK II. 41 AMI! will be greater than the arc ENG ; for, if AMH were equal to ENG, the chord AH would be equal to the chord EG, which is contrary to the hypothesis ; and if the arc AMH were smaller than EN6, the chord AH would be smaller than the chord EG, which is also contrary to the hypothesis. Soholium. We have supposed that the arcs were less than a semi-circumference. If they had been greater, the opposite property would have existed, that is, the greater arcs would have been subtended by less chords, and re- ciprocally. Pkoposition VI. — Theorem. The radius CG, perpendicular to a chord AB, divides the chord and its subtended arc AGB, each, into two equal parts. Draw the radii CA, CB ; these radii are, with respect to the perpendicular CD, two equal oblique lines ; hence they are equally distant from the per- pendicular (Book I., Prop. XVI.), and we have AD = DB. 2°. Since AD = DB, CG is- a per- pendicular at the middle point of AB ; hence, every point of this per- pendicular is equally distant froiii the extremities A and B (Bpok I., Prop. XVII.) The point G is one of these points, and hence AG = BG. But if the chord AG is equal to the chord GB, the arc AG will be equal to the arc GB (Prop. IV.) ; and therefore, the radius CG, perpendicular to the chord AB, divides the arc subtended by this chord into two equal parts at G. Scholium. The right line CG passes through the centre, through the middle of the chord, and through the middle 42 ELEMENTS OF GEOStETBY. of tlie arc. It is also perpendicular to the chord. But two of tliesc conditions are sufficient to determine the posi- tion of a right line ; and hence every right line which fultils two of tliese conditions will necessarily fulfil the other two. Thus, the perpendicular to the naiddle of the chord will pass through the centre of the circle, and through the middle of the arc ; and so on. Peoposition VII. — Theorem. Through three points A,B, C, not in the same right line, we may always pass a circumference, and only one. Join AB, EC, and through the middle points of these lines draw the perpendiculars DE, FG. These two per- pendiculars will meet; for, if they were parallel, the lines EA, BC, drawn through the point B perpendicularly to these parallels, would be in the pro- longation of each other, which is con- trary to the hypothesis. Now, the point of meeting, O, of the two lines DE, FG, being on the per- pendicular DE, is equally distant from the two points A and B ; and the same point 0, being on the perpendicn- ' lar FG, is equally distant from the two points B and C ; hence, tlie three distances OA, OB, 00, are equal, and the circumfei;ence described from as a centre, and with a radius OB, will pass through the three points A, B, 0. 2°. ISTo other circumference can pass through these three points, for, if this were possible, its centre would be found, at the same time, on the lines DE and FG, and these two right lines can only intersect each other in one point. Corollary I. The perpendicular erected at the middle point of AC, will pass through the point 0, since this point is equally distant from the points A and C ; hence, BOOK II. 43 the perpendiculars erected at the middle points of the sides of a triangle intersect in the same point. Corollary IT. Two circumferences cannot have more than two points common, without coinciding, one with the other. Peoposition VIII. — Theoeem. Two equal chords are equally distant from, the centre / and, of two unequal chords, the smaller is at a greater distance from, the centre. 1°. Let the chord AB = DE. Divide these chords into two equal parts by the perpendicu- lars CF, CG, and draw the radii CA, CD. The right-angled triangles OAF, DCG have the hypothenuse CA==CD. Besides, the side AF, the half of AB, is «qual to the side DG, the half of DE ; hence, these triangles will be equal (Bk. I., Prop. XVIII.), and the third side CF is equal to the third side CG ; therefore, 1°, the two equal chords AB, DE are equally distant from the centre. 2°. Let the chord be AH greater than DE, the arc AKH will be greater than the arc DME (Book II., Prop. V.) Take, on the arc AKH, the part ANB = DMB : draw the chord AB, and let fall on this chord the perpendicular CF, and also CI perpendicular to AH. It is evident that CF > CO, and CO > CI; hence, a fortiori, CF > CL But CF = CG, since the chords AB, DE are equal ; therefore, we have CG>CI; hence, of two unequal chords, the smaller is at a greater distance from the centre. u ELEMENTS OF GEOMETRY. Pkoposition IX.- — ^Theoeem. T/ie perpendicular BD, drawn at the extremity of the radius CA, is a tangent to the circumference. For, every oblique line CE is at a greater distance from the perpen- dicular CA ; hence, the point E is without the circle, and the line BD has no other point but A in common with the circumference. Hence, BD is a tangent. Reciprocally. The radius CA, drawQ at the point of contact of the tangent BD, is perpendicular to this tangent. For, all the points of this line, except the point A, being exterior to the circumferences, the radius CA will be the shortest line that can be drawn from the point C to the right line BD, and consequently will be perpendicular to this line. Corollary. Through a point A taken on a circumfer- ence, only oneAangent can be drawn. Pkoposition X. — Theorem. Two parallels, AB, DE, intercept equal arcs, MN, PQ, on the circumference. There may be three cases. 1°. If the two parallels are se- cants, draw the radius CH per- _B pendicular to the chord MP; it will, at the same time, be per- pendicular to its parallel NQ ; then, the point H will be, at the same time, the middle of the arc MHP, and of the arc NHQ. (Book II., Prop. VI.) And we shall have the arc MH = HP, and the arc NH = HQ. Hence, MH - NH = HP - HQ ; that is, MN = PQ. BOOK II. 45 2". If one of the parallels is a taugent, wliile the other is a secant, at the point of contact of the parallel tangent H draw the radius CH : this radius will be perpendicular to the tangent DE, by the preced- ing proposition, and also to its parallel MP. But, since CH is perpendicular to the chord MP, the point H is the middle of the Ilvc MHP. Hence, the arcs MH, HP, comprehended between the parallels AB, DE, are equal. 3°. Finally, if the two parallels DE, IL, are both tan- gents, draw the parallel secant AB ; we shall have, from -what has just been demonstrated, MH = HP, and MK = ■ KP ; and, therefore, the entire arc HMK — HPK; and we see, further, that each of these arcs is a semi-circumference. Pkoposition XI. — Theorem. If two circumferences have a common ^poird A outside of the line 00', which connects their centres, they have a sec- ond point A' situated on the perpendicular AB to 00', and at the same distance from the line as the point A. For A'B being equal to AB, the right lines CA, CA' ■will be equal, being oblique lines equally distant from the foot of the perpendicular CB to the right line AA.'. Hence, the circle described from the point C, as a centre, with CA as a ra- dius, will pass through the point A'. We also see, that the circle described from C as a centre, and CA as a ra- dius, must pass through the point A'. Oorollary 1. When two circumferences intersect, the line which connects their centres is perpendicular at the middle of the common chord. 0-^ ->C' 46 ELEMENTS OF GEOMETEY. Corollary II. If two circumferences are tangent, the point of contact is on the line of the centres; for, if not, the circumference would have a second point common, and tliey would, therefore, intersect. Two circumferences can only occupy, with respect to each other, five different positions. They may be exteiior or interior ; they may be tangent exteriorly or interiorly ; or, finally, they may intersect. Peoposition XII. — Theorem. If two circumferences are exterior .^ the distance lietween their centres is greater than the sum of the ra- dii. For, we have CC ^ CA + C'A' + AA'. Hence, CC > CA + CA'. Proposition XIII. — Theorem. If two circumferences are interior, the distance between their centres is less than the di'ffer- ence of their radii. A. For, we have CC'= CA - C'A' ^ AA'. Hence, CC < CA - CA'. Proposition XIV. — Theorem. If tv)o circumferences are tangent exteriorly, the distance lietween, thei/r centres is equal to the sum of the radii. fiooK ii. 47 For, the point of contact A, being on the line of the centres, we have, evidently, CC' = GA + AC'. Proposition X"V . — Thkoeem. If two ciroumferenoes are tangent interiorly, the distance ietween their cent/res is equal to the dif- ference of their radii. For, the point of contact A is on the line of the centres, and we have CC'= CA - C'A. Peoposition XVI. — Theorem. If two circumferences intersect, the distance ietween their centres will he, at the same time, less than the sum of the radii, and greater than their difference. For, joining the centres with one of the points of inter- section A, we will form a triangle, in which the line of the centres CC, and the radii CA, C'A will be the three sides. But we have seen that, in any triangle, one side is less than the sum of the other two, and greater than their difference. The reciprocals of the five preceding propositions are true, and may be demonstrated in the same manner. For example, if the distance between the centres is less 48 ELEMENTS OF GEOMETRY. than tlie sntn of the radii, and greater than tlieir differ- ence, the circumferences intersect each other; for, if they were exterior or interior, the distance between tlieir centres would be greater than the sum of their radii, or less than their difference ; and if they were tangent, the distance between the centres would be equal to the sum of the radii, or to their difference. Pkoposition XYII. — Theorem. In the same circle, or in equal circles, equal angles A CB, I)CE, whose vertices are at the centre, intercept equal arcs on the circumference. Reciprocally, if the arcs AB, DE are equal, the angles ACB, DCE will he also equal. For, 1°, if the angle ACB = DCE, these angles may be placed one on the other ; and as their sides are equal, it is evident that the point A will fall on D, and the point B on E. But then the arc AB must also fall on the ai-c DE ; for, -jT'^ ^ ^^ if these two arcs did not coincide, there would be, in one or the other of them, points unequally distant from the centre, which is impos- sible. Hence, the arc AB =DE; 2°. If AB =DE, the angle ACB will be equal to the angle DCE; for, if these angles are not equal, let ACB .be the greater, and let ACI = DCE ; we shall have, from what has just been demonstrated, AI = DE ; but, by hypothesis, the arc AB = DE. Hence, we shall have AI = AB, or a part equal to the whole, which is impos- sible. Therefore, the angle ACB = DCE. BOOK II. 49 Peoposition XYIII. — Theorem. In the same circle^ or in equal circles^ the angles at the centre are to each other as tlie arcs intercepted hetnoeen their sides. Let ACB, DCE be two angles at the centre of equal c „ circumferences. Suppose that the arcs AB and DE have a common measure, and that this measure is contained 7 times in AB, and 4 times in DE, the ratio of AB to DE will be \. If we join the several points of division of the two ares with the centres of the two circumferences, we see that the angle ACB will be divided into Y angles, which will be equal to each other, as lying upon equal arcs ; and the angle DCE will contain 4 of these equal angles. The ratio of the angles ACB and DCE will therefore be \ also. If the arcs AB and DE are incommensurable, divide the arc DE into 3 equal parts, for example ; and suppose the arc AB to contain 4 of these parts, with a remainder KB smaller than one of these parts. The ratio of AB to DE is greater than f and less than f. If, now, we join the centres C, F with the points of di- vision of the arcs, we see that the angle DFE is divided into three equal parts, and the angle ACB contains 4 of these parts, with a re- mainder KCB smaller than one of them ; the ratio of "K '^_^'^ the angles ACB, DFE is also, therefore, comprised between | and f. Hence, the ratios of ACB to DFE, and of AB to DE, are both com- prehended between | and |-. Now, dividing the arc into 3 60 ELEMENTS OF GEOMETRY. 10,100,1000... parts, we could prove, in like manner, that these ratios are embraced between two consecutive numbers of tenths, hundredths, &c.; therefore, these ratios are equal, for we have just proved that they are compre- hended between numbers whose difference may be made as small as we please. Measure of Angles. The measure of any magnitude is known, when we know the ratio of this magnitude to a unit of measwre of the same species. The right angle is used as the unit of measure for all angles, so that the measure of an angle becomes the ratio of this angle to one right angle. But the preceding tlieorem shows, that we may substi- tute for the ratio of two angles at the centre, the ratio of the arcs included between their sides. Thus, instead of comparing directly an angle with a right angle, we may compare the arc included between its sides with the fourth part of the circumference ; and it is in this sense that the angle at the centre has for its measure the arc included between its sides. To facilitate this comparison, we divide the circumfer- ence into 360 equal parts, called degrees, each degree into 60 minutes, and each minute into 60 seconds, &c. If the arc included between the sides of an angle at the centre contains 24 degrees, the measure of this angle will bp ^i or -*- Scholium. — Repeating, literally, the demonstration of the preceding theorem, we might prove that two sectors, taken in equal circles, are to each other as the arcs included by their sides. Pboposition XIX.' — Theorem. An inscribed angle BAD has for its measure the half of the arc BD included ly its sides. BOOK II. 51 Let us suppose the centre of the circle to be situated ■within the angle BAD ; draw the diameter AE, and the radii CB, CD. The angle BCE, ex- terior to the triangle ABC, is equal to the sum of the two interior angles, CAB, ABC. (Book I., Prop. XXIX.) But the triangle BAC being isosceles, the angle CAB = ABC. Hence, the angle BCE is double of BAC. The angle BCE, as an angle at the centre, is measured by the arc BE. Hence, the angle BAC will have for its measure the half of the arc BE. By like reasoning, the angle CAD will have for its measure the half of ED. Hence, BAC + CAD, or BAD, will have for a measure the half of BE + ED ; that is, the half of BD. Let us now suppose that the centre C is situated outside of the angle BAD ; then, drawing the diameter AE, the angle BAE will have for its measure the half of BE ; and the angle DAE, the half of DE. Hence, their difference, BAD, will have for its measure the half of BE, minus the half of ED or the half of BD. Hence, every inscribed angle is measured by half of the arc included by its sides. Corollary I. — All the angles, BAC.BDC, &c., inscribed in the same segment, are equal; for each is measured by the half of the same arc Corollary II. — Every angle in- scribed in a semi-circumference is a right angle ; for it has for its measure the half of the semi-circumference, or the fourth of the circumference. Corollary ///.—Every angle BAC inscribed in a seg- 52 ELEMENTS OF GEOMETRY. meiit greater than a semicircle, is an acute angle, for its measure is the half of the arc BOC, which is less than a seuii-circumference. Every angle inscribed in. a segment smaller than a semi- circle, is an obtuse angle ; for its measure is the half of the arc BAG, which is greater than a semi-circumference. Proposition XX. — Theorem. The angle BAG, formed ly a tangent and a chord, is measured hy half of the arc AMDC, included hy its sides. At the point of contact, draw the diameter AD ; the angle BAD is a right angle. (Bk. II., Prop. IX.) It is measured by half the semi-circumference AMD ; the angle DAG is measured by half the are DG ; then, BAD -f- DAG = BAG, will be measured by the half of AMD plus the half "of DG, or the half of the whole arc AMDG. It might also be shown that the angle GAE is measured by half of the arc AG. Peoposition XXI. — Theorem. The angle BA C, formed ly the two secants BE, A C, whose vertex is in the interior of the circumference, is measured hy the half of the sum of the two arcs included hy its sides and the prolongation of its sides. For, the angle BAG, exterior to the triangle AEG, is equal to the sum of the angles AEG, AGE, which have for b\ Yo their measures, respectively, the half of the arcs BC and DE. BOOK 11. 53 Peoposition XXII. — Theorem. The angle BAC, formed by the two secants AB, AC, whose vertex is outside of the circumference, is meas- ured hy half the difference of the arcs intercepted hy its sides. For, the angle A is equal to the difference of the angles BDC, ABD, which are measured, the first by the half of EC, and the sec- ond by the half of DE. The proposition is also true, when one of the sides of the angle, or when both, are tangents to the circumfer- ence ; and the same demonstration applies. Corollary I. The arc of the circle BA is the geometrical locus of the vertices of all angles equal to CDB, whose sides pass through the points C and B. For, 1°, every inscribed angle in the segment BAC is equal to CDB ; 2°, every angle CMB whose vertex is in the interior of the segment is greater than CDB ; for, if we produce CM to the circumference at N, and join ISTB, the angle CMB, exterior to the triangle' MNB, is greater than the interior angle MNB ; but the angle MNB = CDB. Hence, we have CMB > CDB. 3°. "We might, in like manner, show that every angle CGB, whose vertex is exterior to the segment, is smaller than CDB. Proposition XXIII. — Theokem. In every inscribed quadrilateral ABCD, the opposite angles are supplementary. 54 ELEMENTS OF GEOMETRY. a measure, A For, the opposite angles ADC, ABC, have together, for the half of the circumference ABCD. Eeciprocally, if, in any quadrilateral, two opposite angles are supplementary, this quadrilateral may be inscribed in a circumference. For, draw a circumference through the three points A, D, C ; the angle ADO is measured by half the arc AMC ; hence, the angle ABC, which is the supplement of the first, is measured by half the remaining arc ADC ; that is, it is equal to each of the angles inscribed in the segment AMC ; which could not be unless the point B was on the arc AMC. PEOBLEMS KELATING TO THE TWO FIEST BOOKS. JSfD PEOBLEM I. To divide a right line, AB, into two equal j/arts. From the points A and B, as centres, with a radius greater than the half of AB, describe two arcs, cutting each other in D; tiie point D will be equally distant from A and B. Lay off, also, above or below the line AB, a second point E, eqnally _R distant from A and B. Through the points D, E, draw the line DE. The line DE will divide the line AB into two equal jjarts at the point C. For, the two points D and E, being each equally distant from A and B, they must both lie on the perpendicular erected at the middle point of AB. But only one right Ve iOOK II. 55 line can be made to pass through two given points ; hence, the line DE will be the perpendicular which divides the line AB into two equal parts at the point C. D PROBLEM n. Through apomt A, on a given line SO, to erect a per- pendicular to this line. Take the points B and C, at equal distances from A ; from the points B and C as centres, and with a radius greater than BA, describe two arcs, cutting each other in D : draw AD, it will be the perpendicular required. For, the point D, being equally dis- — tant from B and C, belongs to the per- pendicular erected at the middle point of BC; hence, AD is that perpendicular. Scholium. — ^The same construction may be used to make a right angle, BAD, at any point A of a given right line BC. PROBLEM in. From a point A taken outside of a right line BD, to let fall a perpendicular on this line. From the point A, as a centre, and with a radius suffi- •ciently large, describe an arc which shall cut the line BD in the two points B and D : then, take a point E, equally distant from B and D, and draw AE. The right line AE will be the perpendicular required. For, the two points A and E are, each, equally distant from the points B and D ; and, therefore, the line AE is perpendicular to the line BD, at its middle point. ^ 56 ELEMENTS OF GEOMETRY. PEOBLEM IV. At the point A, on the line AB,to make an angle equal to a given angle K. From the vertex K, as a centre, and with any radius, describe the arc IL, terminated at the sides of the angle; from the point A, as a centre, and with a radius AB = KI, describe the indefinite arc BO ; at the point B, as a cen- o D tre, with a radius equal to the chord LI, describe an arc cutting the indefinite arc BO, at D; draw AD; then will the angle DAB be equal to the given angle K. For, the two arcs BD, LI have equal radii and equal chords ; they are, therefore, equal (Book II., Prop. IV.) ; hence, the angle BAD = IKL. PROBLEM V. To divide an angle or an arc into two equal parts. 1°. To divide the arc AB into two equal parts. At the points A and B, as centres, with the same radius describe two arcs, intersecting each other at D : through the point D and the centre C, draw CD. The right line CD will di- vide the arc AB into two equal parts at the point E. For, the two points C and D are, each, equally distant from the extremities A and B of the chord AB ; hence, the line CD is perpendicular at the middle of this chord ; and it divides the subtended arc AB into two equal parts at E. (Book IL, Prop. VI.) 2°. To divide the angle ACB into two equal parts. From the vertex C, as a centre, with any radius, describe the arc AB ; the line CD, already constructed, which divides the arc AB into two equal parts, will also divide the angle ACB into two equal parts. p E p V _.-•'' \ A &> BOOK It. 57 Schotium. — "We may, by the same construction, divide each of the halves AE, EB into two equal parts ; and thus, successively, divide a given angle or arc into four, eight, sixteen, &c., equal parts. PEOBLEM VI. Through a given point A, to draw a parallel to a given right line BO. From the point A, as a centre, and with a radius suflS- ciently large, describe the indefinite arc EO ; from the point E, as a centre, and with the same radius, describe the arc AF ; take ED = AF, and draw AD ; AD will be the parallel required. For, joining AE, the alternate-interior angles AEF, EAD are equal ; hence, the lines AD, EF are parallel. (Boot I., Prop. XXVI.) To construct this problem, practically, we more com- monly use an instrument called a square. The square is of the form of a right-angled triangle CDE ; we apply its hypothenuse on the line CD, to which we propose to draw a parallel through the point A, and we place the side CE against a fixed ruler CM : tlien, slipping the square along the ruler, un- til the hypothenuse reaches the point A, the right line AD' will be parallel to CD; for, the correspondent angles DCM, ACM are equal. PROBLEM VII. Two angles A and B of a triangle being given, to find the third angle. 3* 58 ELEMENTS OF GEOMETRY. Draw the indefinite line DEF ; at the point E, make the angle DEC = A, and the I, angle CEH = B ; the remain- ^^ -'""'^ ing angle HEF will be the re- F quired third angle ; for, these angles taken together make two right angles. PROBLEM vin. Ha/oing given two sides, B and G,of a triangle, and the angle A included hy these sides, to describe the tri- angle. Draw the indefinite right line DE ; at the point D make the angle EDF = A ; take DG = B, DH = C, and draw GH : DGH will be the triangle required. PEOBLEM IX. Having given one side and two angles of a triangle, to describe the triangle. .The two given angles will be either both adjacent to the given side, or one will be ad- jacent, and the other opposite. In the last case, find the third angle, by Prop. YII. ; we shall then have the two adjacent angles. Draw the riglit line DE equal to the given side. From the points D and E, draw the lines DF and EG, mak- ing with the line DE angles respectively equal to each of the adjacent angles : the two lines DF, EG, will inter- sect each other at H, and DEH will be the required triangle. PEOBLEM X. The three sides, A, B, C,of a triangle beiny given, to describe the triangle. BOOK II. 59 Draw DE equal to the side A. From the point E, as a centre, with a radius equal to the second side B, de- scribe an arc ; and from the point D, as a centre, with a radius equal to the third B side C, describe another arc, c cutting the first at F ; draw DF, EF : DEF will be the triangle required. In order that the problem may be possible, it is neces- sary that the circumferences described from the points D and E, as centres, shall intersect each other. This requires that the side DE shall be less than the sum of the two other sides, and greater than their difference. (Book II., Prop. XVI.) PKOBLEM XI. Having given two sides, A and B, of a triangle, and the angle G opposite one of the sides, B, to describe the triangle. There may be two cases ; 1°, if the angle C = a right angle, or an obtuse angle, make the angle EDF equal to. the angle ; take DE = A ; from the point E, as a centre, and with a radius equal to the given side B, describe an arc cutting the line DF at F; draw EF: DEF will be the triangle re- quired. It is necessary, in the first case, that the side B be greater than A ; for, the angle C, being a right angle, or an obtuse angle, is greater than either of the other angles of the triangle. Hence, the side opposite to it must also 60 ELEMENTS OF GEOMETRY. be the greatest side. 2°. If the angle C be acute, and B > A, tlie same construction will apply, and DEF is the triangle required. But if the angle C be acute, the side B < A, then the angle described from E, as a centre, with a radius EF = B, will cut the side DF in two points, F and G, situated on the same side of D : there will, therefore, be two triangles, DEF, DEG, which will satisfy the problem. Scholium. The problem will be impossible, in all cases, if the side B < than the perpendicular let fall from E on the line DF. PEOBLEM Xa. To find the centre of a circle, or of a gwen arc. Take any three points. A, B, C, in the circumference, nr in the given arc; join them by the right lines AB, BC ; divide these lines into two equal parts by the perpendicu- lars DE, FG ; the point O, in which these perpendiculars in- tersect, will be the centre re- quired. Scholium. — The same construc- tion applies, if it were required to pass a circumference through any three given points. A, B, C; and also to draw a circumference in which the sriven tri- angle ABC. shall be inscribed. If the three given points be in the same right line, the problem is impossible, since the perpendiculars to the lines AB, BO will be parallel. BOOK II. 61 PEOBLEM Xin. Through a given pmnt draw a tangent to a given circle. If the given point A be on the circumference, draw the radius CA; AD, D A / \ . perpendicular to CA, will be the tangent required. (Bk. II., Prop. IX.) If the point A be without the circle, join the point A and the centre by the right line CA ; divide the line CA equally, at ; from the point O, as a centre, with the radius OC, describe a cir- cumference intersecting the given circumference in B and D : AB and AD will each be the tangent required. For, the angle CBA, or CD A, being inscribed in the the semicircle, is a right angle. (Book II., Prop. XIX.) Hence, AB and AD are perpendicular to the radii CB and CD; and are, therefore, tangents. We see, that when the point is without the circle, two equal tangents may be drawn through the point A. They are equal, for the right-angled triangles CBA, CDA have the hypothenuse CA common, and the side CB = CD. Hence, they are equal (Book I., Prop. XYIII.) ; and we have AD = AB, and at the same time the angle CAD = CAB. PEOBLEM XIV. Inscribe a circle in a given triangle ABG. Draw the bisectrices AO, BO, of the angles A and B ; these right lines intersect each other at the point O, equally distant from the three sides AB, AC, BC. If, 62 ELEMENTS OF GEOMETRY. from the point 0, we let fall tlie perpendiculars OD, B OF, OE, on the sides of the triangle, these perpen- diculars will be equal, and the circumference de- scribed from the point O, A' — ^^p'^^ — ^^° as a centre, with OD as a radius, will be tangent to the three sides of the triangle. Remarli I. — The point 0, being equally distant from the sides BC, AC, belongs to the bisectrix of the angle C. Hence, the three hisectrices of the angles of a triangle meet in the same point. ItemarTt II. — If we draw the bisectrices of the two ex- terior angles MBG, BON, their point of meeting O' will be the centre of a circle tangent to the side BC, and to the other sides produced outwards. In like man- ner, we may find the centres 0", O'", of the two other tangent circumfor- ences to one of the sides of the triangle. and the prolongation of the other two sides. Four circumferences may, therefore, in drawn tangent to three given right lines. general, be PEOBLEM XV. On a right line AB., describe a segment capable of con- taining a given angle C; that is, a segment such, that all angles inscribed in it shall be equal to a given angle C. Produce AB towards D ; make, at the point B, the angle DBE = C; draw BO perpendicular to BE, and GO perpendicular at the middle point of AB ; from the point jlT of meeting 0, as a centre, with a ra- dius OB, describe a circle : AMB will be the segment required. K ' For, since BF is perpendicular at fiooK It. 63 the extremity of the radius OB, BF is a tangent, and the angle ABF is measnrei'd by half of the arc AKB (Book II., Prop. XX.) ; besides, tlie inscribed angle AMB is also measured by half of the arc. AKB. Hence, the angle AMB = ABF = EBD = C ; and, therefore, every angle inscribed in the segment AMB is equal to the angle C. Scholium. — If the given angle be a right angle, the re- quired segment would be the semicircle described on AB as a diameter. PROBLEM XTI. Draw a common tangent to two circumferences... 1°. Suppose the problem solved ; and let AA' be the common tangent exterior to the two circumferences. Draw the radii CA, C'A', to the points of contact, and the line C'JB parallel to AA'. The radii CA, C'A', being perpendicular to AA', will also be' perpendicular to the parallel C'B ; and hence, C'B will be tangent to a circumference described from the point C, as a centre, with a radius CB = CA — C'A'. We deduce, therefore, the following construction : de- scribe a circumference from the point C, as a centre, with a radius equal to CA — C'A', and draw through the point C' a tangent to this circumference. Knowing the point B, we draw the line CBA, and the line C'A', parallel to CA, and join AA' ; AA' will be the tangent required. The construction shows that there aj-e two solutions of the problem, since Ave may draw two tangents to the cir- cumference CB ; and the problem is only possible when we have CC > CA - C'A', or CC = CA - C'A', or, in 64 ELEMENTS OF GEOMETER. other words, when the circumferences are not interior to each other. 2°. Let it now be proposed to draw a common interior tangent to two circumferences, whose radii are CA, CM, and let AM' be the re- quired line ; draw the radii CA, CM', to the 1 points of contact, and the right line CB par- allel to AM'. The line AM' being perpendicu- lar to the radii CA, CM', CB will be perpendicular to these radii also ; it will therefore be tangent to a circumference described from the point C, as a centre, with a radius CB = CA + AB, or CA -f CM'. To construct the problem, we describe a circumference having C as its centre, and with a radius equal to the sum of the radii 6f the two given circumferences : we draw through the point C a tangent CB to this circumference. The remainder of the construction is the same as in the preceding case. This problem also has two solutions ; and it is only pos' sible when CC > CA + CM, or = CA + CM ; that is, when the circumferences are exterior, or tangent exte- riorly. PEOBLEM xvn. To find the greatest common measure to two given right lines AB, CD, and their numerical ratio. The greatest common measure to two lines cannot be greater than the less of the two lines CD ; but it would be A 1 E equal to CD, if this line was exactly contained in 01 JpHD the greater line AB. Lay feooK li. 65 o£P CD on AB, and suppose that we have AB = 20D + IB : the greatest common measure between AB and CD, is the same as that of the two lines CD and IB. Indeed, every common measure of AB and CD, dividing CD, will also divide AI ; and dividing AB, it will be ex- actly contained in the remainder IB, and will therefore be a common measure of CD and IB. Eeciprocally, every common measure of CD and IB will also divide AI and IB, and consequently AB ; and will therefore be a common measure of AB and CD. Thus, all common measures of AB and CD are the com- mon measures of CD and IB. Hence, the greatest com- mon measure will be the same for both. Lay off IB on CD, and suppose that we have CD = IB + KD ; we may prove, as before, that the greatest common measure of CD and IB is the same as that of IB and KD. Lay off KD on IB, and suppose that we have IB = 2KD ; KD will be the greatest common measure of the two lines AB and CD. We deduce from the preceding equalities : CD = 3KD, and AB = 8KD ; hence, the ratio of the two lines AB and CD is |. Remark. — "We have supposed'above, that in the series of operations we have a remainder equal to zero. We will prove that this is always so, if the two lines have a com- mon measure ; and that when they are incommensurable, we arrive at remainders smaller than any assignable quantity. For, let A, B be the two lines; /"i, r-,, /-j, r-^, &c., the suc- cessive remainders ; j', g",, q-i, q^, qi, &c., the several quo- tients. We shall have A = Bg-i -I- n B = r^q2 + Ti n = r^3 + n ^2 = nSi + ^i, &c., &c. 66 ELEMENTS OF GEOME^TE'^. But, r, is less than — ;for, ifB<^-j7-, a fortiori^ ry is less than — ; and if B^^^, then,?'i=A — B; hence,?*! ""CTo"* We shall have also, ^5 <;^ § ; tence, r^ <^ -. r, <^ -° ; hence, r-, <^ ^-g, &c., &c. We see that if the operation be indefinitely extended, we shall arrive at remainders as small as we please; and, con- sequently, if there is a common measure, we shall have a remainder equal to zero ; otherwise we should fall upon remainders smaller than the common measure, which is evidently absurd, from what has been said before. When the lines are incommensurable, we might, after a certain number of operations, neglect the last remainder ; the preceding remainder may then be taken as the greatest common measure, and will give an approximate value for the ratio. PKOBLEM XVIII. Two angles, A and JB, being given to find their common tneasure, if they have one, and their ratio in numbers. Describe, with equal radii, the arcs CD, EF, which serve to measure these angles. Proceed then to compare the A B arcs CD and EF, as in the /\ preceding problem ; for, an / \ arc may be laid off on an s / \ arc, as a right line is laid off obtain the common measure BOOK II. 67 of the arcs, if they have one, and their ratio in numbers. This ratio will be that of the given angles (Book II., Prop. XVIII,), and if DO be the common measure of the arcs, DAO will be that of the angles. If the two arcs are incommensurable, the angles will be incommensurable also, and we might obtain, as before, an approximate value for the ratio. 68 ELEMENTS OF GEOMETRY. BOOK III. OF THE MEASUEE AND SIMILAEITY OF POLYGONS. DEFINITIONS. I. The area of a figure is the ratio of its magnitude to that (if the unit of surface. ■ II. Two figures are said to be equivalent, when they have the same area. Two figures may be equivalent, although very dissimi- lar: thus, a circle may be equivalent to a square, or a triangle to a rectangle. Figures are said to be equal, when, being applied one to the other, they coincide in all their points: such are two circles with equal radii ; two triangles whose three sides are equal, each to each, &c. III. The altitude of a parallelogram is the perpendicu- j, (3 lar EF, which measures the distance between the two op- posite sides AB, CD, taken as iases. IV. The altitude of a triangle is the perpendicular AD drawn from the vertex of an angle A, on the opposite side EC, taken as the hase. V. The altitude of a trap- ezoid is the perpendicular EF, drawn between its two paral- lel sides AD, CD. BOOK III. 69 Proposition I. — Theorem. ' Parallelograms, whioh have eqxial bases and equal alti- tudes, are equivalent. Let AB be the common base of the parallelograms ABCD, ABEF; since they have the same altitude, tlie D ^l—O f upper bases DO, FE, will be situated on the same line parallel to AB. But we have, from the nature of a parallelo- gram, AD = BC, and AF = BE ; for a A B like reason, we have DC = AB, and, FE = AB ; then, DC == FE ; and subtracting DC and FE from the same line DE, the remainders CE and DF will be equal. It follows from this, that the triangles DAF, CBE, have their sides equal, respectively ; and they are consequently equal. But, if we take the triangle ADF from the quadrilateral ABED, there remains the parallelogram ABEF ; and if, from the same quadrilateral ABED, we take the equal triangle CBE, there remaiins the parallelogram ABCD ; hence, the parallelograms ABCD, ABEF, which have the same base and the same al- titude, are equivalent. Corollary. Every parallelo- gram ABCD is equivalent to the rectangle ABEF, which has the same base and the same altitude. Pkoposition II. — Theorem. Every triangle ABC is the half of the iMrallelogram ABCD, which has the same lase and the same altitude. For, the triangles ABC, ACD are equal. (Book I., Prop. XX XL) F D EC / / A B .0 ELEMENTS OF CliOMETuy, Corollary I. The triangle ABC is half of the rectangle BCEF, which has the same base BC, and the same alti- D tude AC ; for, the rectangle / BCEF is equivalent to the parallelogram ABCD. Corollary II. Triangles, which have equal bases and equal altitudes, are equivalent. Proposition III. — Theorem. Two rectangles of the same altitude are to each other as their hases. Let ABCD, AEFD, be two rectangles, which have AD for their common altitude : they will be to each other as their bases, AB, AE. Let us suppose, in the first place, that the bases AB, AE, are commensurable, and that they are to each other as the numbers 7 and 4. If we divide AB into 7 equal parts, AE will contain i of these parts. Erect at each point of division a perpendicu- lar to the base: we shall thus form 7 partial rectangles, which will be equal to each other, since they have equal bases and the same altitude. The rectangle ABCD will contain 7 of these partial rectangles, while the rectangle AEFD will contain 4. Hence the rectangle ABCD is to tlie rectangle AFED, as 7 is to 4, or as the base AB is to the base AE. If the bases AB, AE are incommensurable, we might prove, by the same reasoning applied (Book II., Prop. XVIIL), that the proposition is true. D P ! i i i A Z B Pkoposition IV. — Theorem. Any two rectangles are to each other as the product of their bases by their altitudes. BOOK III. 71 Let R, r, be the areas of two rectangles ; B, H,. the two dimensions of the iirst ; h, h, the two dimensions of the second ; assume a third rectangle R', which has the same base B as the first, and the same altitude h as the second : ' we shall have, by the preceding theorem, R H -R' B Multiplying these two proportions, member by member, and dividing the two terms of the first ratio by R', we have • Measure of a rectangle. — To measure a rectangle H, consists in finding its ratio to a certain rectangle r, which is taken as a unit of measure. The preceding theorem shows that we may obtain this ratio, by finding the number of times the lines B, H, h, h, contain the same unit, and then dividing the product B X H by the product J x A. Let B =^ 6 feet, H = 4 feet, J = 3 feet, A = 2 feet, we shall have — = - — ^ = 4. Thus, the rectangle R con- r 3 x2 ' ^ tains the rectangle r, taken as a unit, 4 times. Ordinarily, we take, as the unit of measure, the square whose side is equal to the ^Mlit of length. Reducing, then, the numbers h and A to unity, the proportion (1) becomes R_ B xH r ^ 1 ■ We see, therefore, that the ratio of the rectangle to the square, whose side is 1, is equal to the product of the num- bers which express how many times the base and altitude of the rectangle contain the linear unit. This is always 72 ELEMEl^TS OF GEOMETRY. expressed, wlieri we saj, that the measure of a rectangle is equal to the product of its base by its altitude. Pkoposition Y. — Theoeem. The area of any parallelogram is equal to the product of its hase hy its altitvde. For, tlie parallelogram ABCD is equivalent to the rect- angle ABEF, which has the same base AB, and the same J' n E c altitude BE. But the area of the rectangle has for its measure AB X BE ; hence the area of the ])Hrallelogram is equal to AB x BE, that is, to its base multi plied by its altitude. Corollary. — Parallelograms of the same base are to each other as their altitudes ; and parallelograms of the same altitude are to each other as their bases ; for A, B, C, be- ing any three dimensions whatever, we have, generally, A xC ^ A B X C ~ B" * Peoposition VI. — Theoeem. The area of a triangle is equal to the product of its hase hy the half of its altitvde. For, the triangle ABC is the half of the parallelogram ABCE, which has the same base BC, and the same alti- tude AD. But the area of the parallel- ogram is BC X AD, by the preceding theorem ; hence that of the triangle = I BC X AD, or BC x \ AD. Corollary. — Two triangles of the same altitude are to each other as their bases, and those of the same base are to each other as their altitudes. Book iti. 73 Proposition VII. — Theorem. The area of a trapezoid A BCD is equal to its altitude EF, rmdtiplied hy half the sum of its parallel lyases AB., CD. Through the middle point I, of the side CB, draw KL parallel to AD, and produce DC nntil it meets KL in K. D K c K In the triangles IBL, ICK, we have the side IB = IC; by construction, the angle LIB = CIK, and tlie angle IBL = ICK, since CK and BL are parallel (Book I., Prop. XXV.); E B therefore tlie triangles IBL and ICK are equal. (Book I., Prop. VI.) Hence the trapezoid ABCD is equivalent to the parallelogram ADKL, and has for its measure EF x AL. But we have AL = DK, and since the triangle IBL = KCI, the side BL = CK : then, AB' + CD = AL + DK = 2AL, and hence AL — half the sum of the two bases AB, CD. Therefore, the area of the trapezoid ABCD is equal to the altitude EF multiplied by half the sum of the bases AB, CD, which maybe expressed thus: area ABCD== Scholium. If through the point I, the middle of BC, we draw IH, parallel to the base AB, tlie point H will also be the middle of AD, for tlie figure AHIL is a paral- lelogram, as well as DHIK, since the opposite sides are parallel; we have then AH = IL, and Dil = IK. But IL = IK, since the triangle BIL = CIK ; therefore, AH = DH. , n- TTT AT AB+CD We may remark that tlie Ime HI = AL = ^ : therefore, the area of tlie trapezoid may also be expressed by EF X HI; that is, by the altitude of the trapezoid 4 G 74 ELEMENTS OF aEOMETR'S'. multiplied by the line joining the middle of the sides which are not parallel. Pkoposition YIII. — Tiieokem. If a line AC he divided into two parts, AB, BC, the square on the whole line AC will contain the square on one of the parts AB,plus the square on the other part BC, plus twice the rectangle of the two parts AB and BC ', that is, AC or (AB + BC)=' = Ab' + BC' + 2AB x BC. Construct the square ACDE ; take AF = AB, draw FG parallel to AC, and BH parallel to AE. The square 1- ji p ACDE is divided into four parts. The first, ABIF, is the square on AB, since AF = AB ; the second, IGDH, is the squar.e on BC ; for, since we have AC = AE, and AB = AF, the difference AC - "'^ AB is equal to the difference AE — AF, wliicli gives BC = ,EF. But on account of the parallels, IG = BC, and DQ^=f,%: hence, HIGD is equal to the sijUiire on BC. These two parts being taken from the wliole square, we have the two rectangles BCGI, EFIPI, each of which lias for its measure, AB x BC. Hence the square 0D, we have "bo + DO = 2 BG + 20G:. Hence, Ab'+ Bc'+ Ad'+ DC = 4BG + 40g'+ 4Ao'; 2 2 2 2 and since AC = 4A0, aftd BD = 4BG, we have, finally, A B + BC + AD + DC = 40G + BD + Ac! Corollary. If the quadrilateral be a parallelogram, the right line GO becomes zero; then, in every parallelogram., the. sum of the squares of its four sides is equal to the sum of the squares of its two diagonals. The reciprocal of this last theorem is also true. OF PROPORTIONAL LINES AND SIMILARITY OF FIGURES. Peoposition XVI. — ^Theorem. If a line be drawn jparallel to one of the sides of a tri- angle, it will divide the two other sides proportionally. BOOK III. 83 Join BE and DC; the^two triangles BDE, DEC have the same base DE; they have also the same altitude, since the vertices B and C are in a line par- allel to the base ; hence, these triangles are equivalent. The triangles ADE, BDE, with the '^ common vertex E, have the same alti- tude, and are to each other as their bases AD, DB ; and we have ADE _ AD BDE ~ DB" Th^ triangles ADE, DEC, with the common vertex D, have also the same altitude, and are to each other as their baSes AE, EC ; hence ADE _ AE DEC ~ EC But the triangle BDE = DEC ; hence, because of the common ratio in these two proportions, we conclude that AD DB AE EC" Corollary I. From this results, AD AE AD-t-DB AEH-EC- or and also. AD -H DB AE -f EC DB EC AB A hi AC AB "^'DB AC ~ EC" Corollary II. If two right lines AB, CD be intersected by any number of parallels, AC, EF, GH, BD, &c., the segments of the lines AB, CD will be proportional. 84 ELEMENTS OF GEOMETRY. For, let be the point in which the lines AB, CD meet : in the tri- angle OEF, the line AC being par- allel to the base EF, we shall have 0E_ AE OF ~ CF' In the triangle OGH, we shall D T, • 1 M, OE GE " have, in like manner, -=-= = ===. Hence, from the common ratio, we have AE_GE CF ~ FH" f We might prove in the same way, that ~r^ = g^, &c. Peoposition XVII. — Theoeem. Reciprocally^ if the sides AB, AC of a triangle ABC are cut proportionally iy the line DE, so that we have AD ' AE jy^ = -^yy, the line DE will he parallel to the hase BC Fur, if DE is not parallel to the base BC, suppose DO A to be parallel ; then, by the preceding theo- AD AO rem, we shall have ED OG o But, by Iiypothesis, gp = g^; hence, we ^should have -^^ = ^^, which is impossible, Miice, on the one side, the antecedent AE is greater th^n AO, and, on the other, the consequent EC is less than OC ; therefore, the parallel to BO, drawn through the point D, cannot differ from DE ; hence DE is parallel to BC. AE 1 AE > BD AD~ CE AE' or AD BD~ AE CE' BOOK III. 85 Scholium. The same conclusion would be reached, if , , , AB AC ^ , . we had supposed -JY) = "xy ; lor, this proportion would . ■ AB - AD AC - AE give or, •Peoposition XVni. — Theorem. 1°. The bisectrix AB of the angle A of the triangle ABO dvuides the base BO into two segments, BD, DC, proportional to the two sides AB and AC. 2°. The bisectrix AF of the exterior angle CAE cuts the base produced at a point F, so that the two segments BF, OF are also proportional to the same two sides of the triangle. 1°. Through the point C draw CE parallel to AD, and produce it until it meets BA produced in E. In the triangle BCE, F G D ^ the line AD being paral- ^ lei to the base, we have the proportion (Book II., Prop. XYI.), BD_ AB DC ~AE" But, the triangle ACE is isosceles ; for, since AD and CE are parallel, the angle ACE = DAC, and the angle AEC = BAD ; but, b}^ hypothesis, DAC = BAD ; hence, the angle ACE = AEC, and consequently, AE = AC. Putting AC in the place of AE, in the preceding propor- tion, we shall have BD^AB DC AC" 86 ELEMENTS OF GEOMETRY. 2°. Draw CG parallel to AF ; in the triangle BAF, we have BF_AB FC ~ AG* The triangle AGO is isosceles, as may be proved by the preceding reasoning ; hence AG = AC, and we have BF_ AB FC ~ AC Corollary. If the point A moves in the plane in .such a manner that the ratio of AB to AC is constantly equal to — , the bisectrices of the angles BAG, EAC will always T?T) BF pass through the points D and F; since the ratios pjp)-7rp must always be equal to — Besides, the right lines AD, AF, bisectrices of the two adjacent angles, are perpendicular to each other. Hence, the point A, in all its positions, will be on >the circumfer- ence described on FD as a diameter. From which it follows : That the geometrical loans of jpoints whose dis- tances from two given points B and G are in a given ratio, is a circumference of a circle. DEFINITION. Similar triangles are those which have their angles equal, and the homologous sides proportional. Those sides are called homologous which lie opposite to equal angles. In general, similar polygons are those which have their angles equal, each to each, and the homologous sides pro- portional. I In polygons, those sides are called homologous -which are adjacent to equal angles. BOOS: III; 87 Pboposition XIX. — Theobem. Two equiangularsbriaingles have their homologous sides jproportional. Let ABO, DEF be two triangles which have their angles equal, each to each, viz., A = D, B = E, C = F ; then will the homologous sides be pro- portional, and we shall have AB_AC_BC '^ DE DF~EF' Take AG = DE, AH = DF, and join GH; the trl^ angles AGH, DEF, will be equal, having two sides and an included angle, in each, equal. Hence, the angle AGH will be equal to the angle E. But we have E = B ; therefore, the angle B = AGH, and GH is parallel to BC ; and we have the proportion -j-^ — xtt" Draw HL parallel to AB ; we shall have the proportion AC_BC AC _ BC AH BL'°''aH~GH' for the right lines BL, GH are equal as parallels compre- hended between parallels. Comparing the last proportion with the preceding, we deduce from it, on account of the common ratio, AB^AC^BC AB^AC_BC AG AH GH' °^ DE " DF ~ EF' Corollary. If two triangles have two angles equal, each to each, the triangles will be similar, for the third angle in each will be equal, and the triangles will be equi- angular. Proposition XX. — Theorem. I'wo triangles which have their sides proportional are equiangular. 88 ELEMENTS OF GEOMETRY. Let|^ = A? =^, then will the triangles ABC, DEF be equiangular, and having A = D, B = E, C = F. Take AG = DE, AH = DF, \. and join GH ; we have, by- hypothesis, the proportion, AB AC w .1 =r^ = =rrr; ; or, what is the same DE DF ' ' AB A O thing, -To~ira' Sence, it follows that GH is parallel to BC, and consequently the angle AGH = ABC. The triangles ABC, AGH, being equiangular, we have, by the preceding theorem, AB__ AC_ BC AG ~ AH ~ GH' But, by hypothesis, we have AB DE AC ^DF BC ^EF" And since AG = DE, and AH = DF, we have GH = EF ; the triangles AGH, DEF, having their three sides equal, each to each, are therefore equal ; and hence the angle DEF = AGH = ABC, the angle DFE = AHG = ACB, and the angle D = A. ^ Scholiurn I. We remark that the eqvial angles of two triangles lie opposite to the proportional sides. Scholium II. We see, by the last two propositions, that, in triangles, the equality of the angles is a conse- quence from the proportionality of the sides, and recip- rocally ; so that, if either of these conditions is fulfilled, the two triangles will be similar. This is not the case in figures of more than three sidts ; for, in the case of quad- rilaterals, we may, without cb fging the angles, vary the proportion between the sid - or, without varying the BOOK III. 89 sides, cliange the angles. Hence, the proportionality of the sides does not necessarily follow from the equality of the angles, and vice versa. We see, for example, that if we draw the line EF paral- lel to EC, the angles of the quadrilateral AEFD are equal to those of the quadri- lateral ABCD ; but the proportion of the sides is different : so, witliout varying the four sides AB, 'EC, CD, AD, we may cause the point B to approach to or recede from the point D, which would necessarily alter the angles. Scholium 111. The two preceding propositions, which really constitute but one, joined to that of the, square on the hj'pothenuse, are the most important propositions in geometry. Taken alone, they are nearly sufficient for all the applications of geometry, and for the solution of all problems. The reason is, that all figures may be divided into triangles, and any oblique triangle may be divided into two right-angled triangles. So that the general properties of triangles contain, by implication, those of all other figures. Peoposition XXI. — Theokem. Two triangles^ which have an angle in each equal, and the equal angle included l)y projportional sides, are similar. Let the angle A = D, and suppose we have ^tt? = T=rr^ > DJii Dr then will the triangle ABC be similar to the triangle DEF. Take AG = DE, and draw GH parallel to EC : the angle AGH will be equal to the angle ABC (Book I., Prop. XXV.) ; and the ij E F .triangle AGH will be equiangu- 90 ELEMENTS OF GEOMETRY. AT) A p lar with tlie triangle ABC; and we shall have ttt = rTT" But, by hypothesis, =7^^ = ^y^ ; and, by construction, AG = DE : hence, AH = DF. The two triangles AGH, DEF, having then two sides and the included angle in each equal, are therefore equal. But the triangle AGH is sim- ilar to ABC: hence, its equal DEF is also similar to ABC. Proposition XXH. — ^Theoeem. Two tnangles which have their sides parallel or perpen- dicular to each other, each to each, are similar. Let A, B, C, be the angles of one of the triangles, and A', B', C, the angles of the other triangle. We know that two angles which have their sides paral- lel or perpendicular to each other, are equal or supple- mentary (Book I., Prop. XXVII., XXVIII.) We can only then make one of the three following hy- potheses : 1°- A 4- A' = 2 right angles, B + B' = 2 right angles, C -I- C = 2 right angles. 2°. A + A' = 2 right angles, B + B' = 2 right angles, C = C'. 3° A = A', B = B', and consequently, C = C. But, in the first hypothesis, the sum of the angles of the two triangles would be equal to six right angles. In the second hypothesis, the sum of the angles of the two triangles would be greater than four right angles. But as both of these results are impossible, the third hypothesis is the only one that is admissible; hence, the two triangles are equiangular and similar. Hemark. The homologous sides of the two triangles are the sides which are parallel or perpendicular to each other. BOOK III. 91 Proposition XXIII. — Thkoekm. A polygon being given, we can always construct in it a second polygon — stick, that the two polygons shall ie com- posed of the same number of similar triangles, and these triangles similarly situated. For, let ABODE be the given polygon ; draw from the vertex A the diagonals AC, AD; then, taking arbitrarily, on the side AB, the point b, draw be parallel to BC, and cd parallel to CD, \q and finally, de parallel to DE ; the trian- gles Kbc, Acd, &c., will be respectively similar to the triangles ABC, ACD, E^' &c., &c. ; and the polygons ABCDE, Abode, which might be placed in any manner whatever with respect to each other, will be composed of the same number of triangles, similar, each to each, and similarly situated. Peoposition XXIV. — Theorem. The polygons ABODE, abcde, composed of the same number of triangles, similar, each to each, and similarly situated (as in the last theorem), have theangles equal, each to each, and the homologous sides proportional, and are tlierffore similar. For, the similarity of the tri- angles ABO, abc gives the angle BOA = bca ; the triangles ACD, acd, being similar, give the an- gle "ACD = acd: hence, we con- clude that BCD = bed, and so on. Besides, we shall have, on ac- count of the similarity of the same triangles, the follov/ing equal ratios : 92 ELEMENTS OF GEOMETRY. AB _ EC _ AC _ CD _ AD _ DE ^ AE_ ah ~ be ~ ao cd ad de, ae ' therefore, the two polygons have their homologous sides proportional, and are similar. Peoposition XXV. — Theorem. Reciprocally, two similar polygons may he divided into the same number of triangles, similar, each to each, and similarly situated. In the polygon ABCDE, draw, from the vertex of the same angle A, the diagonals AC, AD to the other angles. In the other polygon, FGHIK, draw, in like manner, from the angle F, homologous to A, the ^i diagonals FH, FI, to the other angles. Since the polygons are similar, the angle ABC is equal to its homologous angle FGH, and the sides AB, BC, are also proportional to the AB BC sides FG and GH ; so 'that we have -=^^ = y^^-j. It follows from this, that the triangles ABC, FGH, have an angle in each equal, and these angles included by two proportional sides ; they are, therefore, similar (Book II., Prop. XXI.) ; hence, the angle BCA is equal to GHF. Taking the equal angles BCA, GHF from the equal angles BCD, GHI, the remaining parts, ACD, FHI, will be equal. But since the triangles ABC, FGH, are simi- , , AC BC , ^ lar, we have ^pTrr = prf ; and because the two polygons . ., , BC CD , AC CD are similar, we have Qg = -^T ! ^'®°'=^' pJJ == hT" BOOK III. 93 But we have already seen that the angle AOD = Fill ; hence, the triangles ACD, FHI, have an angle in each equal, and these angles included by proportional sides ; they are, therefore, similar. We might, in the same way, show the similarity of all the ^other triangles, whatever the number of the sides of the polygon ; hence, two simi- lar polygons are composed of the same number of similar triangles, which are similarly situated. Remwrk. The decomposition may be made in many ways, by drawing the diagonals through any two homol- ogous vertices. There results, therefore, that in two simi- lar polygons, two homologous diagonals CE, HK, are to each other, as two homologous sides ; for, they are the homologous sides of similar triangles ODE, HIK, which form parts of the two polygons. Proposition XXYI. — Theorem. The lines AF, AO, dic.^ drawn from the vertex of a triangle to any points of the iase, divide the iase -BC and any line DF drawn parallel to the hase into pr&portional parts, and we shall have DI_J[K_KL BF ~ EG ~ GH' , • For, since DI is parallel to BF, the triangles ADI, ABF, are equiangular, and we have the proportion -jj= = -j-=j. Likewise, IK being parallel to FG, we have — ^ = -rrpr ; and, from the AF EG' ' ' common ratio, we get DI _ IK BE ~ EG" 94 ELEMENTS OF GEOMETRY. ^ IK KL We might show, in the same manner, that =p^ — grj > &c. Therefore, the line DE is divided at the points I, K, L into parts proportional to the base BC, at the points F, G, H. Corollary. If BC be divided into equal parts, at the points F, G, H, &c., the parallel DE will likewise be di- vided into equal parts at I, K, L. Pboposition XXVII. — Theorem. If., from the right angle A of a right-am,gled triangle we draw AD perpendicular to the hypothenv^e — 1°. The two partial triangles ABD, ADC will he simi- lar to each other, and similar to the whole triangle ABC ; 2°. Each side, AB, AC, will he a mean proportional hctween the hypothenuse BC and the adjacent segment of the hypotJienuse, BD or DC ; 3°. The perpendicular A D will he a mean proportional hetween the two segments, BD DC ; 1°. The triangles BAD, BAG have the angle B common, and the angle BDA = BAG, because they are right angles. Hence, the third angle BAD is equal t^ the third angle C. The two triangles are, therefore, equiangular and similar. We could prove, in like manner, that the triangle DAG is similar to the tri- angle BAG. Hence, the three triangles are equiangular and similar. 2°. Since the triangles BAD and BAG are similar, their homologous sides are proportional. The side BD in the small triangle is hotnologous to BA in the large, because they are opposite to tlie equal angles BAD, BGA ; the hypothenuse BA of tlie small triangle is homologous to Book in. 96 the hypothenuse EG of the large. We may, therefore, torm the proportion p-^ = -^^■ in like manner, we have -r-r^ = -^fr^. ' AG BC Therefore, 2°, each of the sides AB, AC is a mean pro- portional between the hypothenuse and the segment adja- cent to that side. 3°. Finally, the similar triangles ABD, ADC give, for their homologous sides, the proportion -^jr = -=rp. Hence, 3°, the perpendicular AD is a mean proportional between the segments BD, DC of the hypothenuse. Scholium. The proportion -po^-op gives, by placing the product of the extremes equal to that of the means, AB=BDxBC. We have, likewise, AC = DCxBC. Hence, AB + AC = BD x BC + DO x BC. The second member is the same thing as (BD + DC) x BC = BC x BC = BC . Hence, AB + AC = BC , which shows that the square on the hypothenuse BC is equal to the sum of the squares on the two sides AB, AC. We have thus a second demonstration of this important theorem, by a process quite different from that pursued (Prop. XI.) ; and we see that, strictly speaking, the proposition of the square on the hypothenuse is a consequence from the proportionality of the sides in the equiangular triangles. Corollary. If, from any point A of the semi-circum- ference, we draw the two chords AB, AC to the extrem- ities of the diameter BC, the triangle BAC will be right-angled at A. (Book II., Prop. XIX.) Hence, 1°, the per- pendicular AD is a mean propor- ' tional ietween the two segments of the 06 ELEMENTS OF GEOMETRY. diameter BD, DO ; or, iu other words, AD is equal to the rectangle BD x DC. 2°. The chord AB is a mean proportional between tJie diameter BO and the segment BD ; or, AB = BD x BC. We have also, in like manner, AC = CD x BC. Hence, Ab" BD = ; and if we compare AB with BC , we shall AC DC have a = , and also, , — ; ratios which have BC BO BC BC already been determined in Cor. III. and IV., Prop. XL Pkoposition XXYIII. — Theoeem. Two triangles which have an angle in each equal, are to each other as the rectangles of the sides containing the equal angle; that is, the triangle ABO is to the triangle ADE, as the rectangle AB x AO is to the rectangle AD X AE. Draw BE ; the two triangles ABE, ADE having the common vertex E, have the same alti- tude, and are to each other as their bases AB, AD (Cor., Book II., Prop. „,. „ ABE AB. , VL) Hence, -^^^= -^ ; we have, > . ,., ABC AC ^E mhke manner, -j^ = -^ Multiplying these two proportions together, we have, omitting the com- ^^^ ABC AB X AC monterm ABE, ^pg^ ^P^^^ . BOOK III, 97 Pboposition XXIX. — Theorem. Two similar triangles are to each other as the squares of their homologous sides. Let the angle A = D, and the angle B = E. Since the angles A and D are equal, we have, by the preced- ing theorem, ABC _ AB X AC DEF ~ DE X DF' which may be written, AC DF' But the two triangles being similar, we have the pro- ,. AC AB ^ portion ^='jw- Hence, ABC DEF^ AB AB 'de^de AB DE Pboposition XXX. — Theorem. The contours or perimeters of similar polygons are to each other as the homologous sides ; and thevr areas' as the squares of these sides. For, 1°, since, from the nature of similar figures, we AB BP CD have =^ = j^ttt = tjT ' &c., we deduce, by the composition 98 'ELEMENTS OF GEOMETRY. ,,, , ,. AB + BC+CD+&C. AB ,., of these equal ratios, j^q^.qh +HI +& ^ ^ FG' "'' demonstrates the first part of the theorem. 2°. Since the triangles ABC, FGH are similar, we liave 2 ABP AC by the precedina; theorem =— =v = 2 ; and, from the •^ ^ ^ FGH FH similar triangles ACD, FHI, we have sllso, AGD_ AC FHI fg' Comparing these two proportions, and omitting the common ratio, we have ABC _ ACD FGH ~ FHI * In like manner, we could show that ACD _ APE . FHI ~ FIK ' and so on, for a greater number of triangles. From this series of equal ratios, we have, by composition, ABC + ACD + ADE ABC ^ Ac" _ Ab' =2- FGH + FHI +FIK FFt yR FG Hence, the areas of two similar polygons are to each other as the squares of the homologous sides. Peoposition XXXI.— Theorem. The segments of two chords, AB, CD, which intersect within a circle, are reciprocally pi'oportional ; that is, AO_CO DO ~ OB' BOOK III. 99 Join AC and BD. The triangles AGO, BOD, have the opposite vertical angles at O equal; also, the angle A is equal to the angle D, being inscribed in the same seg- ment (Book II., Prop. XIX.): for a like reason, the angle C = B ; and the two triangles being equiangular, are therefore similar, and we have the pro- portion AOCO DO~OB' • Corollary. From this proportion, we deduce AO x OB = DO X CO, Hence, the rectangle of the injao seg- ments of one of the chords is equal to the rectangle of the two segments of the other. Pkoposition XXXII. — Theorem. Jff', from a point 0, taken without a circle, we draw the secants OJS, OC, terminated in the concave arc £C, the secants will he reciprocally proportional to their exterior segments ; that is, we shall have -p^ = ^=r^. For, joining AC, BD, the triangles OAC, OBD, having the angle O common, and the angle B = C (Book II., Prop. XIX.), are equiangular, and therefore similar; and their homologous sides give the proportion OB_OD 00 OA Corollary. From this proportion, we deduce OA x OB == 00 x OD. Hence, the rectangle of one secant and its exterior segment, is equal to the rectangle of the other secant and its exterior segment. 100 • ELEMENTS OF GEOMETRY. SeJiolium. We may remark that this proposition has a close analogy to the preceding ; and it only differs in this, that the two chords AB, CD, instead of intersecting within the circle, intersect without. Pkoposption XXXIII. — Theobem. If, from a point 0, taken without a circle, we draw a tangent OA, and a secant OC, the tangent will he a mean proportional between the secant and its exterior segment, and we shall have 7=-^ = 7=-=;. OA OD For, joining AD and AC, the triangles OAD, OAC have the angle O common ; the angle OAD, formed by the tangent \v and chord (Book II., Prop. XX.), is yy^^\^"'-\ measured by the half of the arc / N. \ AD, and the angle C has the same '^ \\ measure. Hence, OAD = C ; and — — -^c the two triangles being equiangular \ / will be similar, and give the pro- V y r OC OA ^ ^—2 \ ^ ^ ^ portion ^y^ — ^=r=r. HencG, OA = OC X OD. Scholium. This proposition may be deduced from the preceding, by considering the tangent OA as the limit of the positions which a secant takes in turning around the point 0. FEOPosmoN XXXIV. — Theorem. In every triangle ABC, the rectangle of the two sides A B, AC is equal to the rectangle of the diameter CE of the circumscribed circle, and the perpendicular AD let fall on the third side : so that we shall home AB x AC = CE X AD. ■ BOOK III. 101 For, joining AE, the triangles ABD, AEG are right- angled, one at D, the other at A. Besides, the angle B = E : hence, these triangles are similar, and thev ^, .. AB AD give the proportion -^ = -^. Hence, AB X AC = CE X AD. Corollary. If we multiply these equal quantities bj' the same quantity BC, we shall have AB X AC X BC = CE X AD X BO. But AD x BC is double of the area of the triangle ABC (Book II., Prop. VI.) : hence, the product of the three sides of a triangle is equal to the area of the triaiigle multiplied hy twice the diameter of the circumscribed circle. The product of three lines is sometimes called a solid^ for a reason which will be shown hereafter. Its value is easily conceived, by supposing that the lines are reduced to numbers, and multiplying the numbei-s together as is required. Scholium. We may show also, that the area of a tri- angle is equal to its perimeter multiplied hy half of the. radius of tlie inscj^hed circle. For, the triangles AOB, BOC, AOC, which have their common vertex at 0, have, for tlie common altitude, the radius of the inscribed circle. Hence, the area of the three triangles will be equal to the sun\ of the bases AB, BC, AC, multiplied by half the radius OD. There- fore, the area of the triangle ABC, which is composed of these small triangles, is equal to its perimeter multiplied by half the radius of the inscribed circle. 102 ELEMENTS OF GEOMETRY. Peofosition XXXV. — Theorem. If^ in a triangle ABC, we divide the angle A into two equal parts, ly the line AD, the rectangle of the sides AB, AC will "be equal to the rectangle of the segments BD, DO of the third side, jjlus the square of the secant AD ; and we shall have BA x AC = Ad' + BD x DC. Pass a circumference through the three points A, B, C ; produce AD to the circumference, and join CE. The triangles BAD, EAO, are sim- ilar; for, by hypothesis, the angle f BAD = EAC. Besides, the angle B = E, since they are both measured by half of the arc AC : these tri- angles are therefore similar, and the homologous sides give the proportion -^ — -^-~ ; and wo have BA X AC =^ AE X AD. But AE = AD -f- DE, and multiplying both members by AD, we have AE x AD = AD -I- AD X DE ; besides, AD x DE = BD x DC. Hence, finally, BA x AC = Ad" + BD x DC. Proposition XXXVI. — Theorem. fn every inscribed quadrilateral A BCD, the rectangle of tlie diagonals AC, BD is equal to the sum of the rect- angles of the opposite sides ; so that we have AC X BD == AB X CD -F AD X BC. Draw the right line Bl, making the ylo angle CBI equal to the angle ABD, and produce it until it meets AC at I. The angle ADB = BCI, because they are inscribed in the same segment BOOK Hi. 103 AOB; besides, the angle ABD = OBI, by construction;, hence, the triangles ABD, IBC are similar, and give the proportion -^=j5^; and AD x BO = CI x BD. (1) The triangle ABI is also similar to the triangle BDC ; for, the angle ABD being equal to CBI, if we add DBI to both, we shall have ABI = DBO. Besides, the angle BAI = BDC, being inscribed in the same segment; hence, the triangles ABI, DBC are similar, and the homologous AB AI sides give the proportion ^^^^ = -^^ 5 and AB x CD = AI x BD. (2) Adding the equalities (1) and (2) together, and observ- ing that AI X BD + CI x BD = (AI + CI) x BD = AC x BD, we shall have AD x BC + AB x CD = AC x BD. PEOPosmoN XXXVII. — ^Thkoeem. Iti every quadrilateral which cannot ie inscribed in a circle, the rectangle of the diagonals is less than the sum of the rectangles of the opposite sides. For, through the three points A, B, C, pass a circumfer- ence which will not contain the fourth vertex D : make the angle ABI = DBC, and the an- gle BAI = BDC. The right line AI cannot coincide with AC, since the point D not being on the circumfer- ence, the angle BDC is different from BAC; finally, join I and C. The triangles ABI, BDC, equian- gular by construction, give the pro- portion -jy = =^ ; hence, AI x BD = AB x DC. (1) The triangles ABD, IBC are also similar ; for, if from the equal angles ABI, DBC, we take the common angle 104 ELEMENTS OF GEOMETRY. DBI, there will remain the angle ABD — IBC ; again, because the triangles ABI, DBG are similarj we have the AB RT proportion =j5-p- = ^,7=; : the triangles ABD, IBC have, then, an angle in each equal, and the equal angles included by proportional sides ; they are, therefore, similar, and we have 42 = IS ; lience, IC x BD = AD x BO. (2) Adding together (1) and (2), we have BD X (AI + IC) = AB X DC + AD X BC. And since AI + IC > AC, we have BD X AC < AB X DC + AD X BC. Scholium. We conclude that if, in a quadrilateral, the rectangle of the diagonals is equal to the sum of the rect- angles of the opposite sides, this quadrilateral may be inscribed in a circle. Pboposition XXXVIII. — Theoeem. The diagonals of an inscribed quadrilateral are to each other as the sums of the rectangles of the sides which vieet in their extremities. The quadrilateral ABCD may be divided into the two triangles ABC, ADC, by the diagonal AC; but, desig- nating by E the radius of the cir- cumscribed circle, we have (Book III., Prop. XXXIV.), AB X BC X AC = 4K X ABC, c and AD X DC X AC = 4E X ADC. Adding these equalities, we have AC X (AB X BC + AD X DC) = 4R x ABCD. BOOE m. 105 But, if we divide the quadrilateral into triangles, by the diagonal BD, we should also have BD X (AB X AD + BC x DC) = 4R x ABGD; hence, AC X (AB X BC + AD X DC) = BD (AB x AD+BC x DC), which gives the proportion AC _ AB x AD + BC X DC BD~ABxBC + ADxDC" PEOBLEMS RELATING TO BOOK IH. --K PEOBLEM I. Divide a given right line into parts proportional to given lines. 1°. Let it be proposed to divide the line AB into five equal parts. Through the extremity A, draw the indefi- A nite right line AG, and assuming any ar- bitrary length AC, lay off AC five times on AG. Join the last point G with the extremity B, of the given line AB, by the line GB, then draw CI parallel to GB ; AI will be the fifth part of the line -■M AB ; so that laying oft' AI five times on AB, the line AB will be divided into five equal parts. For, since CI is pai-allel to GB, the sides AG, AB are cnt proportionally at C and I. (Book II., Prop. XVI.) But AC is the fifth part of AG ; hence, AI is the fifth part of AB. 2°. Let it be i-equired to divide the line AB into parts proportional to the given lines P, Q, E. Through the - L m ELEMENTS OF G^EOMETRi^ extremity A, diiuv the indefinite line AG, take AC = P, K B CD=:Q, DE=R,join tlie extremities E and B, and through the points C, D draw the parallels CI, DK, to E^*5 EB; the intercepted parts AI, IK, KB, of the line AB, will be proportional to the given lines P, Q, R. For, since CI, DK, EB are parallel, the parts AC, CD, DE are proportional to AI, IK, KB (Book 11., Prop. XVI.) ; and, by construction, AC, CD, DE are equal to the given lines P, Q, R. PROBLEM n. To find a fourth proportional to three given lines A, £, C. Draw the two indefinite ' lines DE, DF, making any angle witli each other. On DE take DA = A, and DB = B, on DF take DC = C ; join AC, and through ilie "g point B, draw MX pai- B/ \x 1 !c allel to AC ; DX will be the fourth proportional required : for since BX ^F is parallel to AC, we have ,, ,. DA DC the proportion ^ = j^, but the three terms of this proportion, DA, DB, DC, are equal to the three lines A, B, C; hence DX is the foui-!h proportional to A, B, C. Corollary. "We could find also a third proportion ,-.1 in the two lines A, B; for it will be the same as the f<.nr;h proportional to the three lines A, B, C. BOOK III. 107 PEOBLEM in. To find a mean proportional ietween two given Unes A,B. On the indefinite line DF, take DE =: A, and EF = B ; on the whole line DF, as a diameter, describe the semi- circumference DGF ; at the point E, erect the perpendicular EG on the diameter, meeting the circum- ference in G : EG will be the mean proportional sought. For, the per- pendicular GE, drawn from a point of the circumference to the diameter, is a mean propor- tional between the segments of the diameter DE, EF (Book n., Prop. XXVn., Cor.); but these segments are equal to the given lines A and B. Second Construction. Take DF = A, DE — B, and describe a circum- ^' ' ference on DF as a diameter ; draw EG perpendicular to DF, and join D and G : GD will be a mean proportional between A and B. Third Co^istruction. Take 00 = A, OD = B ; through the points D and C, pass any circumference whatever, and through the point O, draw a tangent, OA, to this circumference : the line OA will ^' ' be a mean proportional between A and B. PEOBLKM IV. Thrmigh a given point ^, in the given angle BGD^ to draw the line BD, so that the parts AB, AD, compre- 108 ELEMENTS OF GEOMETEY. liended between the point A and the two sides of the angle, may he equal. Tlirougli the point A, draw AE parallel to CD ; take Q BE = CE, and through the points B and A draw BAD : BAD will be the line sought. For, AE being parallel to CD, we ^ have 1^ = ?^- But BE - EC, and EC AD therefore BA = AD. PROBLEM V. To make a square that shall he equivalent to a given parallelogram or a given triangle. 1°- Let AB be the base of the given parallelogram, DE Ti its altitude, and X the side of the required square: we must have X^ = AB X DE ; AB_ X ^'■' X "DEi Hence, the side X is a mean proportional between AB and DE. 2°. We see, in like manner, that the side of the square which is equivalent to a given triangle, is a mean proportional between the base of the triangle and half of its altitude. PEOBLEM TI. To make on the line AD a rectangle ADEX, which shall he equivalent to a given rectangle ABFC. BOOK III. 109 Let AX be the unknown altitude of the rectangle ADEX: since the two rectangles are assumed to be equivalent, we have AD X AX = AB X AC ; which gives ,, .• AD AC the proportion -^ = -^ The required line AX, is therefore a fourth proportional to the three lines AD, AB, AC. PROBLEM Vn. To find two right lines which shall ie in the same ratio as the areas of two rectangles. Let A, B be the altitude and base of the first rectangle ; C, D, those of the second. One of the required lines may be chosen arbitrarily ; let it be equal to A, and let X be the second line. From the conditions of the problem, we must have A xB _A X' CxD Hence, X = C xD C xD X A A xB B Therefore, the required line X is a fourth proportional to the three lines B, C, D. PROBLEM VIII. To C07i8truct a triangle equivalent to a given polygon. Lot ABCDE be the given polygon. Draw the diagonal CE, cutting ofi'from the given poly- gon the triangle CDE : through the point D, draw DF parallel to CE, and produce it until it meets the prolongation AE in F. Join CF; and the polygon ABCDE 110 ELEMENTS OF GEOMETRY. will be equivalent to the polygon ABCF. which has one side less than the given polygon. For, the triangles CDE, CFE, have the common base CE ; they have also the same altitude, since their vertices, D, F, are situated on the line DF, parallel to the base. Hence, these triangles are equivalent. Adding to each the figure ABCE, we shall have the polygon ABODE equivalent to the polygon ABCF. We may, in like manner, cut off the angle B, by sub- stituting for the triangle ABC the equivalent triangle AGC, and thus the pentagon ABDE will be changed into an equivalent triangle GCF. The same process will apply to every other polygon ; for, by successively diminishing the number of its sides, we will fall ultimately upon an equivalent triangle. Scholium. We have seen that every triangle may bo converted into an equivalent square (Prob. v.) ; so that we may always find a square that shall be equivalent to a given rectilinear figure. This is what is called squaring the recti- linear figure ; or, in other words, finding its quadrature. The problem of the quadrature of the circle consists in finding a square equivalent to a circle whose diameter is known. PROBLEM IX. To find a square which shall he equivalent to the sum or difference of two given squares. Let A and B be the sides of the given squares. 1°. Let it be required to find a square equal to the sum of these squares ; draw the two indefinite lines ED, EF, at right angles to each other ; take ED=A, andEG = B ; join DC : DG will be the side of the required square fiOOS III. Ill For, the triangle DEG being rectangular, the square on DG is equal to the sum of tlie squares on ED and EG. 2°. If it be required to find a square equal to the dif- ference of the given squares, construct, as before, the right angle FEH ; take GE equal to the smaller of the sides A and B ; from the point G, as a centre, and with a radius GH equal to the other side, describe an arc cutting EH in H : the square on EH will be equal to the difference of the squares on the lines A and B. For, the triangle GEH being rectangular, the hypothe* nuse GH = A, and the side GE = B. Hence, the square on EH is equal, &c. Sclwlium. We may also find a square equal to the sum of any number of squares; for the construction which reduces two to one, may be employed to reduce three to two, and these to one, and the same for othera. The same principle would apply if it were required to take some of the squares from the sum of the others. FKOBLEM X. To construct a square which shall he to a given square ABGD^ as the line M is to the line N. On the indefinite line EG, take EF = M, and FG = N. On EG as a diameter, describe a semi-circumference, and at the point F, erect a perpendicular FH to the diameter. ' From the point H draw the chords HG, HE ; on the first, take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG : HI will „ be the side of the required square. Ml 1 For, since the lines KI, GE are p.r.„el,wehave 5L=^; 112 from which we get ELEMENTS OF GEOMETET. Hf he' HK HG But, in the right-angled triangle EHG, we have also (Book III, Theorem XI.), he' _ EF _ M Sg fg n" Hence, HI M HK N But HK = AB ; hence, the square on HI is to the square on AB as M is to N. PROBLEM XI. On the side FO, homologmis to AB, to construct a poly- gon similar to the given polygon ABCDJE. In the given polygon draw the diagonals AC, AD ; at the point F, " make the angle '^-—""''^^ GFH = BAG, and at the point G, the angle FGH = ABC : the lines FH, GH, will intersect each other in H ; and FGH will be a triangle similar to ABC. Likewise, on FH, liomologons to AC, construct the triangle FIH, similar to ADC, and on FI, homologous to AD, construct the triangle FIK, similar to ADE. The polygon FGHIK will be the required poly- gon, which will be similar to ABCDE. For, the two polygons are composed of the same num- ber of triangles similar to each other, and similarly sit- uated. BOOK III. 113 PEOBLEM XII. Two similar jpolygons being given, to construct a similar polygon which shall he equal to their sum or their differ- ence. Let P and Q be the areas of the given polygons ; A and B two homologous sides of these polygons : let X be the area of the required polj'gon, x the side of this polygon liomologous to A and B. Similar polygons being to each other as the squares of the homologous sides, we shall have ^-%. Hence, ^ - ^' Q B'' ' P + Q A'' + B^ P A^ We have also ^ — ~zr'' ^^^ sibee X = P + Q, the two last proportions have the three first terms common. Hence, Thus, the side x is the hypothenuse of a right-angled triangle, the sides including the right angle being A and B. Knowing the side x, the problem is reduced to the preceding. If the polygon X = P — Q, we should still have the pro- portions P A^ „ P A^ ^,. Hence, Q B^" —'"-^' p _ Q A^ - B^" P A^ We have, besides, = =: — ; from which we conclude that a' = A^ - Bl PEOBLEM Xni. To construct a polygon similar to a given polygon, and which shall he to this polygon in the ratio of m to n. 114 ELEMENTS OF GEOMETRY. I Let P be tlie area of the given polygon, A one of its sides ; let X be the area of the required polygon, and x the side homologous to A: we shall have, by the conditions of the problem, p = — ; and, since the polygons are similar, X a^ we shall have t> = -to- P A^ Hence, -ri = — i We shall then determine the side x by Problem x. PEOBLKM XIT. To construct a polygon similar to P and equivalent to Q. Let A be a side of the polygon P, and x the homologous side of the required polygon X. P A^ Because the polygons are similar, we shall have ^ = ,2 ' X X- P A^ and since X must be equivalent to ^, pr = —^. If we find two squares M'^ and N^ equivalent to P and M^ A^ MA Q, we shall have =rFr, = -r; hence, :ir= = — . W X^^ ' N 83 The line x will then be a fourth proportional to the three lines M, N, A. PROBLEM XV. To construct a rectatxgle equivalent to a given square C, and the sum of wliose adjacent sides shall he equal to the given line AB. On AB, as a diameter, describe a semi-circumference; draw the line ED parallel to the diameter AB, and at a distance from it equal to the side of the given square C. From the point E, in which tlie parallel meets the circum- BOOK III. 115 ference, let fall the perpendicular EF : AF and FB will be the sides of the required rectangle. For, their sum is equal to AB ; and their rectangle AF x FB, is equal to the square of EF, or to the square of AD ; hence, this rect- angle is equivalent to the given square C. Scholium. It is necessary, in *■ B order that the problem be possi- ble, that the distance AD shall not exceed the radius; that is, that the side of the square C shall not exceed the half of the line AB. PEOBLEM XVI. To construct a rectangle which shall he equivalent to a square C, and the difference of whose adjacent sides shall he equal to the given line AB. On the given line AB, as a diameter, describe a circum- ference ; at the extremity of the diameter, draw the tan- gent AD equal to the side of the square C ; through the point D and the centre O draw the secant DE : DE and DF will be the adjacent sides pf the required rectangle. For, 1°, the difference of their sides is equal to the diameter EF, or AB. 2°. The rectangle ' DE x DF is 2 equal to AD ; hence, this rectangle will be equivalent to the given square C. PROBLEM XVII. To divide a line AB into mean and c.rtr:me ratio , 116 ELEMENTS OF GEOMETEY. tltai is, into two jparts such, that the greater shall he a mean proportional ietween the whole line and the other part. At tlie extremity B of the line AB, erect the perpendic- ular BC, equal to the half of AB ; from the point C, as a centre, and with a radius CB, describe a circum- ference ; draw AC cut- ting the circumference in D, and take AF = AD : the line AB will be divided at the point F, in the manner required. For, produce AC until it meets the circumference in the second point E ; the line AB being tangent, we have the proportion AE AB AB AD' or, AE - AB AB - AD AB AD ■ But AB = DE ; we have then, AE - AB = AD = AF ; we have, also, AB - AD = AB — AF = FB. And, finally, we have the proportion AF_ FB AB ~~ AF' Scholium. Let AB = a ; we have AF = AD = AC— CD. But, AC = / Ab'+ BC = Vd' "^4 ~ a — CD is equal to 5- ; therefore, AF = Jv5--^ = ^(^^-l). BOOZ TV. 117 BOOK lY. EEGTJLAR POLYGONS, AND THE MEASURE OF THE OIEOLE. DEFi^riTioisrs. I. A regular jocHygon is one which is at the same time equilateral and equiangular. There may be regular polygons of any number of sides. For, if we conceive a circumfer- ence divided into m equal parts, and that the consecutive points of division. A, B, C, &c., are j oined by right lines, a polygon of m sides will be formed. The sides of this polygon will be equal, as chords of equalarcs ; and the angles A, B, C will be equal inscribed 'angles, which subtend equal portions of the circumference. The equilateral triangle is the regular polygon of three sides, the square that of four sides, and so on. n. Isojperimetrical polygons are polygons which have equal perimeters. Peoposction I. — Theokem. Two regular polygons of the same number of sides are similar figures. Take, for example, the two regular hexagons, ABCDEF, 118 ELEMENTS 07 GEOMETRY. dbcdef; the sum of the angles is the same in both, and is equal to 8 right angles. (Book I., Prop. XXX.) The angle A is the one- sixth part of 8 right angles, and so is the angle a. Hence, the angles A and a are equal, and the same is true of the angles B, h, and C, c, &c. Besides, since, from the nature of these polygons, the sides AB, BG, CD, &c., are equal, as well as ah, bo, cd, &c., we shall have the propor- tions AB_BG_CD. ab be cd ' Hence, the two polygons have their angles equal, and their homologous sides proportional ; they are therefore similar. Corollary. The perimeter of two regular polygons of the same number of sides are to each other as their homol- ogous sides ; and their areas are to each other as the squares of their homologous sides. Peoposition II. — Theorem. Every regular polygon may be inscribed in a circle, or ■may be circumscribed about it. Let ABODE be any regular polygon ; pass a circum- ference through the three points A, B, C ; let O be its centre, and OP the perpendicular let fall on the middle of the side BO; join AG and OD. The quadrilateral OPOD, and ^r> the quadrilateral OPBA, may be placed one upon the other ; for, the side OP is common, the angle 'E OPO = OPB, since they are right angles; then the side PC BOOK IV. 119 will coincide -with its equal PB, and the point C will fall on B. Besides, from the nature of the polygon, the angle PCD = PBA. Hence, CD will take the direction BA ; and since CD = BA, the point D will fall on A, and the two quadrilaterals will entirely coincide. The distance OD is therefore equal to AO, and consequently, the cir- cumference which passes through the three points A, B, C, will also pass through the point D ; but by like reasoning, we might show that the circumference which passes through B, C, D, passes also through the following vertex E, and so on. Hence, the same circumference which passes through the points A, B, C, passes through all the vertices of the angles of the polygon, and the polygon is inscribedin the circumference. In the second place, all the chords AB, BO, CD, &c., are equal chords of the same circumference ; they are, therefore, equally distant from the centre. (Book H., Prop. VIH.) Hence, if from the point O, as a centre, and with a radius OP, we describe a cir- cumference, this circumference will be tangent to the side BO, and to all the other sides of the polygon, at the middle point of each; and the polygon will be circumscribed about the circle, or the circle inscribed in the polygon. Scholium I. The point 0, the common centre of the in- scribed and circumscribed circles, may be regarded as the centre of the polygon ; and for this reason, we call the angle at the centre, the angle B formed by the two radii drawn to the extremities of the same side AB. Since all the chords, AB, BC, &c., are equal, it is evi- dent that the angles at the centre are equal, and that thus the value of each angle is found by dividing four right angles by the number of sides of the polygon. Scholium II. To inscribe a regular polygon of a given number of sides in a given circumference, it is only neces- sary to divide the circumference into as many equal parts as the polygon has sides. 120 ELEMENTS OF GEOMETRY. Scholium III. If in an arc we inscribe a series of equal chords, the figure thus formed is called' a portion of a regular polygon. This portion has the principal properties of regular polygons ; it has equal angles, it may be in- scribed within or circumscribed about a circle ; but it only forms a part of a regular polygon, strictly speaking, when the arc subtended by one of its sides is an aliquot part of the circumference. Peoposition hi. — Pkoblem. To inscribe a square in a given circumference. Draw two diameters, AC, ^D, cutting eacli other at right angles; join the extremities A, B, C, D, and the figure ABCD will be the inscribed square ; for, the angles AOB, BOG, &c., being equal, the chords AB, BC, &c., are equal. Scholium. The triangle BOG being right-angled and isosceles, we have (Book II., Prop. III.), BC_2^. BO l' hence, the side of the inscribed square is to the radius as the square root of 2 is to unity. Peoposition IV. — Peoblem. To inscribe a regular hexagon, and also an equilateral triangle, in a given circumference. BOOE IV. 121 of the inscribed hexagon to I Let us suppose the problem solved, and let AB be a side if we draw the radii AO, OB, the triangle AOB will be equi- lateral. For, the angle AOB is the sixth part of four right angles ; and in taking the right angle for ■unity, we shall have AOB=|^=:f . The two angles ABO, BAO, of the same triangle, are together equal to 2 — f = 4 ; and as these angles are equal, each is equal Hence, the triangle ABO is equiangular, and there- fore equilateral ; and, therefore, the side of the inscribed hexagon will be equal to the radius of the circle. It follows from this, that to inscribe a regular hexagon in a given circumference, we have only to lay off the radius six times on the circumference, and join by right lines the points of division. The regular hexagon ABCDEF, being inscribed, if we join the vertices of the alternate angles we shall inscribe an equilateral triangle. Scholium. The figure ABCO is a parallelogram, and also a rhombus, since AB = BC = CO =^ AO : therefore (Book III., Prop. XY.), the sum of the squares of the diagonals AC and BO, is equal to the sum of the squares of the sides. That is, AC -t-BO =AB +BC +C0 -+-A0 =4AB =4:B0. Taking BO from each member, there remains AO = 3B0. Hence, AC BO^- 3 AC v/3 ^1' °^'bo" 1 Therefore, the side of the iyiscribed equilateral triangle is to the radius as the square root of 3 is to unity, 6 122 ELEMENTS OF QEOMETEf . Peoposition V. — Problem. To inscribe a regular decagon in a circle. Suppose the problem solved, and let AB be a side of the inscribed decagon. The angle at the centre, AOB, is equal to -^ or to f ; the sum of the two angles OBA, OAB, is therefore equal to 2 right angles — f , or equal to |. Hence, each of the equal angles OBA, OAB, is equal to f. Draw the bisectrix BM, of the angle OBA; the triangle MOB is isosceles, since the angles MOB, OBM are each equal to J ; and we have, therefore, OM = MB. The triangle BAM is also isosceles ; for the angle MBA = f , and the BAM to | : the third angle, AMB, must necessarily be equal to |. Hence, AB = BM = MO, and we have (Book IH., Prop. xym.), BO _ MO AO _ OM BA~AM' ""^ '■^^^^^' OM ~ AM- We see, then, that the radius AO is divided at the point M into mean and extreme ratio, and that the greater seg- ment is equal to the side of the inscribed decagon. Remarli. The side of the inscribed decagon, in a circle whose radius is R, is equal to — ^^ — (Book HI., Prob. XVn., Scho.) Corollary I. If we join the vertices of the alternate angles of the regular decagon, we form the regular pen- tagon. Corollary II. AB being the side of the decagon, let AL be the side of the hexagon : then the arc BL will be, BOOK IV. 123 with respect to the circumference, ^ — j^ = rs- Hence, the chord BL will be the side of the regular polygon of 15 sides. We see, at the same, time, that the arc CL is the thii-d of CB. Scholium. A regular polygon being inscribed, if we bisect equally the arcs subtended by its sides, and draw tlie chords of the half arcs, we shall form a new regular polygon of double the number of sides. Thus, if the given inscribed regular polygon be a square, the continued bisection of the arcs subtending the sides pf the inscribed polygons will enable us succes- sively to inscribe the regular polygons of 8, 16, 32, &c., sides. If we make use of the regular inscribed hexagon, we may from it, in the same manner, form regular poly- gons of 12, 24, 48, &c., sides ; and from the decagon, de- duce the regular polygons of 20, 40, 80, &c., sides ; and from the pentedecagon, polygons of 30, 60, 120, &c., sides. Peoposition VI. — Peoblem. Having given the regular inscribed polygon AJ3CD, <&c., to circumscribe a similar polygon about the same cir- cumference. At the point T, the middle point of the arc AB, draw the tangent GH : it will be parallel to AB. (Book XL, Prop. X.) Draw tangents at the middle of the other arcs BO, CD, &c. : these tangents will form the regular circumscribed )m polygon GrHIK, &c., which will be similar to the inscribed poly- gon. It is easy to see, in the first place, that the three points O, B, H are in a right line ; for 124 ELEMENTS OF GEOMETRY. the triangles OTH, OHN have the common hypothenuse OH, and the side OT = ON : they are therefore equal. (Book I., Prop. XIX.) Hence, the angle TOH = HON; and the line OH passes through B, the middle of the arc TN : for a like reason, the point I is on the prolongation of OC, &c. But since GH is parallel to AB, and HI to BC, the angle GHI = ABC ; likewise, HIK = BCD, &c. Hence, the angles of the circumscribed polygon are equal to those of the inscribed polygon. Again, by reason of the same parallels, we have GH_OH HI_OH AB~OB' B0~ OB' GH HI Hence, AB = BC" But AB = BC : hence, GH = HI. For the same reason, HI = IK, &c. Hence, the sides of the circumscribed polygon are equal to each other ; and therefore the poly- gon is regular, and similar to the inscribed polygon. Corollary 1. Eeciprocally, if we have the circum- scribed polygon GHIK, and we wish to inscribe a similar regular polygon ABC, &c., it is only necessary to draw to the vertices G, H, I, &e., of the given polygon, the lines OG, OH, &c., meeting the circumference in the points A, B, C, &c. Join afterwards these points by the chords AB, BC, &c. : we shall form the required inscribed poly- gon. We might also, in the same case, join the points of contact T, N", P, &c., by the chords TN", NP, &c. ; and we should equally form an inscribed polygon which would be similar to the circumscribed polygon. Corollary II. We may always circumscribe about a given circle all the regular polygons which may be in- scribed in this circle, and reciprocally. BOOK IV. 125 the triangle Oil [ iON; Peoposition VII. — Theorem. The area of a regular polygon is equal to its perimeter multiplied hy half the radius of the inscribed circle. Let GHIK, &c., be the regular polygon. The triauglc GOH has for its measure GH x iOT; has for its measure HI x but ON = OT: hence, the triangles GOH + OHT, have for their meas- ure (GH + HI) X |0T. Proceed- ing in the same manner with the other triangles, we see that the sum of all the triangles, or the whole polygon, has for its measure the sum of the bases GH, HI, IK, &c. ; or, the perimeter of the polygon multiplied by 5OT, half the radius of the in- scribed circle. Scholium. The radius of the inscribed circle OT, is the perpendicular let fall from the centre of the circle on one of the sides of the polygon, and is called also the apothegm of the polygon. Pkoposftion Vin. — ^Theorem. The perimeters of two regular polygons of the same num- ber of sides are to each other as the radii of the inscribed or circumscribed circles; and their areas, as the squares of these radii. Let AB be a side of one of the given polygons, its centre ; OA will be the radius of the circum- scribed circle, and OD, perpendi- cular to AB, will be the radius of the inscribed circle : again, if ab be the side of a similar regular polygon, its centre; oa and od 126 ELEMENTS OF GEOMETRY. will be the radii of the circuiiisci-ibed and inscribed circles. The perimeters of the two polygons are to each other as the sides AB and ah\ but the angles A and a are equal, each being half of the angle of the polygon ; the angles B and h are also equal ; hence, the triangles ABO, abo, are simi- lar, as well as the right-angled triangles ADO, ado ; therefore, AB _ AO ^ DO ab ao do ' Hence, the perimeters of the two polygons are to each other as the radii AO, ao, of the circumscribed circles, and also, as the radii DO, do, of the inscribed circles. The areas of tliese polygons are to each other as the squares of the homologous sides AB, ah ; they are conse- quently to each other, also, as the squares of the radii AO, ao, of the circu inscribed circles, or as the squares of the radii OD, od, of tiif in.scribed circles. DEFINrnONS. I. A variable quantity is a quantity which takes succes- sively different states of magnitude. II. A limit is a fixed magnitude towards which a varia- ble quantity may approach indefinitely near, without reaching it. III. Arithmetic and Geometry present numerous ex- amples oi variable quantities and of limits. Thus, the angle of a regular polygon of m sides has for its value 2m - 4 4 — 2 right angles . m ° ° m But, if we suppose the number of sides of the polygon to be indefinitely increased to infinity, the value of the angle will continually increase ; and as ?«, may become so large that the fraction ^j maybe smaller 4.han any assignable quan- SOOK IV. 12^ tity, we conclude that the successive values of the variulU angle of the polj-gon will have for a Ivnit 2 right angles. Again, if we take c, the middle point of the right line ^ c 0" ^^' '"'^®° ^'■> *^^® niiddle of cB, 1 1 i — f — |B and so on, the lines Ac, Ac', Ac", &e., &c., will have for a limit AB. And we miglit indefinitely multiply such illus- trations. , lY. It is evident, that if the factors a, h, c, of a pro- duct have for li7nits A, B, C, the product a xl x g will have for a limit A x B x C. p V. Let ABCD be a polygon in- scribed in a circumference : the perimeter of this polygon is less '{g than the length of the circumfer- ence ; for each side is less than the corresponding arc. Let us take on the arcs AB, BO, CD, &c., points of division F, G, H, E, &c., and draw the chords AF, FB, BG, &c. ; we shall have inscribed a second polygon, with a perimeter greater than the pe- rimeter of the first polygon. If now, we take new inter- mediate points of division, we shall have a third polygon, whose perimeter will still be greater than the preceding ; and so on, indefinitely. The perimeter of these successive polygons will contin- ually approach the length of the circumference, and we may admit as a self-evident proposition, that if the num- ber of sides of the polygon be suiBciently great, the difl:er- ence between the length of the circumference and the perimeter of the polygon will be less than any assignable quantity ; or, in any other words. The length of the cir- cumference will he the limit toioards icliich the jieriineter of an inscribed polygon tends, as the number of its sides is indefinitely increased. 128 ELEMENTS OF GEOMETET. We may remark, also, that the areas of the successive polygons, which are all less than the area of the circle, differ from it less and less ; and, if we admit that this dif- ference may become less than any assignable quantity, we may conclude that the area of the circle is the limit of the area of an inscribed polygon, the number of whose sides increases indefinitely. YI. It result^ evidently from what has just been said, that every property which belongs to the perimeter or area of an inscribed polygon, what- ever be the number of its sides, is equally applicable to the length and the area of the circle. Thus, for example, the perimeter of every inscribed polygon being less than the perimeter of a polygon en- veloping the circumference, we may conclude that the length of the circumference is itself less than the perimeter of every circumscribing polygon. When we inscribe in a circle regular polygons, the num- ber of sides of which continually increase, the apothegms increase, since the sides of the poly- gons become smaller, and are, there- fore, more distant from the centre. These apothegms have for a limit the radius of the circle. For, let AB be the side of a regular inscribed poly- gon, OC its apothegm, and OB the radius, we have, in the triangle BCO, OB - OC < CB. But CB, the half of AB, may become as small as we please ; hence, a fortiori, OB — OC may become less than any assignable quantity. BOOK IV. 129 Peoposition IX. — Theoeem. 1°. The Gvrcwmferences of circles are to each other as their radii. 2°. The areas of circles are to each other as the squares of their radii. 1°. Inscribe two regular similar poly- gons in the circumferences whose radii are OB and CA. Let P, jp be the perimeters of these polygons ; and de- signate by R and r the radii OB and CA, and by and c, their circumfer- ences; we shall have (Book IV., Prop. VIII.), P ^ E p r' But, this proportion existing, whatever be the number of sides of the polygon, will be equally applicable to limiting circumferences, and we shall have 2 = 5. (1) r ' 2°. Let C, c' be the areas of the same circles ; S, s, the areas of two similar inscribed regular polygons ; we shall have s ')^' And as this proportion is true, whatever be the number of sides of the polygon, we shall have for the limiting circles C^_E^ c' ~ r"' Scholium. From the equality (1) we deduce 2R 2r Hence, the ratio of a circumference to its diameter is the 130 ELEMENTS OF GEOMETRY. saiiK! for all circiimt'erenees. This ratio, which is ordina- rily designated by tt, is incoinmensurable, and can only Le determined approximately.' Its value in decimals is TT = 3.1415926535897932, &c. "We shall soon give an elementary method for calculating, approximately, the value of rr. Knowing the value of tt, we are enabled to calculate the length of a circumference whose radius is known ; for, from the equality Q -— = 7r, we deduce C — 27rE. Exam.ple. E = 18.35 ; taking for -n, the approximate value 3.14, we have C = 2 X 3,14 X 18.35 = 115.2380 feet. DEFINITIONS. Similar arcs, sectors, and segments, are those which he- long to equal angles at the centre. Pboposition X. — Theoeem. 1°. Sim,ilar arcs are to each other as their radii. 2°. Similar sectors are to each other as the squares of tJieir radii. Let AB, DE, be similar arcs ; AC, OD, the radii. 1°. We have (Book II., Prop. XVII 1.), arc BA _ angle C circ. AC 4 right angles" Likewise, < \ arc DE angle O ' circ. OB A right angles' But, since angle C = angle O, we have arc BA _ circ. AC _ AC • arc DE ~ circ. OD ~ OD" BOOE IV. 131 2°. We have, in like manner, for the similar sectors ACB, DOE, sect. AGE ^ angle C sect. DOE _ angle circ. AC 4 right angles' circ. DO ~ 4 right angles" Hence, 2 sect. ACB _ circ. AC _ CA ; sect. DOE circ. DO OD* Peoposition XI. — ^Theoeem. The area of a circle is equal to its cirmimference multi- plied iy half its radius. In the circle whose radius is OA, inscribe a regular polygon. Let P be the perimeter of this polygon, S its area ; we have S = P X ^OC. But the area of a circle is the limit of the areas of the inscribed regular poly- gons, the number of whose sides in- creases indefinitely ; we shall have then the measure of the circle, in seek- ing the limit of the product P X ^OC. Now, P has for its limit the circumference OA, and OC has for a limit OA ; therefore, Area circle OA = circumference OA x ^OA. MemarTc. Eepresent by K the radius of the circle ; we have Area of circle = SttE x -^ = tR. Application. Let E = 3 feet, and take tt = 3.1415 ; we have Area circle =28.2735 square feet. Corollary. The area of a sector is equal to the arc of this sector multiplied by the half of the radius. 132 ELEMENTS OF GEOMETRY. For, the sector ACB is to the entire circle as the arc AMB is to the entire circumference -^ ABD (Book IL, Prop. XVIIL), or, as Af! AMB X ^AC is to ABD X ^. But, the entire circle = ABD x 2 AO ; hence, the sector ACB has for its measure AMB X ^AO. ^ Aj)plication. Let AC — 12 feet, and suppose the arc AMB to contain 60°. To find the length of this arc, we have the proportion arc AMB _ 60 2nR " 360' Hence, 27rE X 60 7r X R tt X 12 arc AMB :47r. 360 3 3 We have then, the sector ACB =4Tr x6 = 24:n = 75.3960 square feet. Proposition XII. — Pkoblem. Knowing the side AB of a regular inscribed polygon, and the radius OC of the circle, to calculoM the side AG of the regular inscribed polygon having double the number of sides. Let AB = a, 00 = E, and AC = c ; draw AD and AO : we have, in the right-angled triangle CAD, o AO^ = CD X 01, or c'' = 2E x 01. But, 01 =: CO - 01 = E - OL Be- sides, in the right-angled triangle AOI, we have 01 = ^^^ - Al'= \/e^ - Hence, 01 = E - sjs? - ^. BOOK IV. 133 And consequently, c^ = 2K x [R — y'B?— — ). (1) Eeciprocally, we may calculate a, when c is known ; and it is only necessary to resolve the equation (1) with respect to a; ; we obtain for the value a in terms of c, ; « g2 • (^) To make application of the formula (1), let us suppose a to be the side of thp hexagon ; a will then be equal to E, and we shall have for the side of the regular inscribed dodecagon, To apply formula (2), take o equal to the side of the decagon, and let us determine the side of the regular pentagon. We have (Book lY., Prop. V.), c = v"^^"^) ; Hence, we conclude, 5l(1:i1:^]4E^_]^::l^:^1 ^, a^= ^, ^=^(10-2 A). Hence, we have for the side of the regular pentagon, E Remark. By adding the square of the radius to the square of the side of the decagon, we find that the sum E='(6-2 v^5) ^ E^a0j-^2v^. "'"4 4 ' that is, is equal to the square of the side of the regular pentagon. Therefore, the side of the regular inscribed 134 ELEMENTS OF GEOMETEY. pentagon will he the hypothenuse of a nght-angled triangle which has, for the sides containing the right angle, the radius and the side of the decagon, Peoposition XIII. — Peoblem. Knowing the side of a regular polygon and the radius of the circumscribing circle, to find the side of a similar circumscribing polygon. Let AB = a,GA = R, and EF = x. The similarity of the triangles EOF, E M r ACB gives the proportion EF_CE AB~OA" ,, , , , CE CM But we have also -^^rr- = tvft- CA CD Hence, on account of the common ratio, we have _EF^CM »__R_ AB CD ' '"'' a ~ CD" ^^ In the right-angled triangle ACD, we have also, CD =V'CA -AD ='^W --; 4 X E Hence, "^ |/r2_^; 4 2aE and, consequently, x = ^ ^ v^4R^ - a' Peoposition XIV. — Peoblem. Knowing the side AB of a regular polygon of m sides, and the radius CA of the circumscribing circle, to find the area of the polygon. SOOK IV. 135 Let AB = a, CA =E, S = area of the polygon. "We have (Book lY., Prop. VII.) S = maX^- But, CD = Vb,'-^ = ^V4R^ - a'. Hence S _ maV 4:Bi' — a' Application. Let us find the area of the regular hex- agon. We have a = E, m = 6 ; therefore, S- BE/iE^ - E^ SE'vT 4 2 Remark. "We might also, by means of the same given M quantities, calculate the area of the inscribed regular polygon of Id \^ 2m sides. For, let M be the middle of the arc AB, and draw AM ; the area of the required polygon S', is composed of 2m triangles equal to ACM. AD Exa But, Hence, ACM = CM X S' = 2ot X 2 4 ■ E X a wEflS Let us find, as an application, the area of the regular inscribed dodecagon. a = E, 7?i = 6 ; hence, S' ,= —^ ■ :3E^ Proposition XV. — Problem. Saving given the radius CD = E* and the apothegm, CA = r^of a regular polygon, to calculate the radiits R' * The radius of a polygon ia only an abridged method of expressing the radius of the circumscribing circle. 136 ELEMENTS OF GEOMETEt" . and the apothegm r' of a regular isoperimetrical polygon, having double the number of sides. Let BD be the side of the given polygon ; C the centre of this polygon. Produce the apothegm CA until it meets the circumscribing circumference at I, and draw BI, ID ; BID will be the angle at the centre of the required polygon ; for, it is the half of BCD. If now we let fall the perpendic- ular CK on BI, and draw KE parallel to BD, KE will be the half of BD, and will represent the side of the new polygon : IK will be the radius, and HI the apothegm. U + r ^ ^ , ^^ lA CI + CA But we have IH = — -- = -, ; or, r = • Finally, in the right-angled triangle CKI, we have (1) -v^irx7=\/Ex^- (2) IK = IC X IH, E Scholium. It is easy to see, by the figure, as well as from the formulae, that r' is greater than r, and that, on the contrary, E' is less than E, so that the difference between the radius and the apothegm in the new polygon is less than in the given polygon. If we transform, in like manner, the second polygon into a third, then the third into a fourth, and so on, we shall ultimately arrive at a polygon, in which the difference between the radius and the apothegm will be less than any given quantity. Indeed, in the triangle BCA, we have , BC - CA < BA, or E— r < BA. But BA is the half of the side of the polygon, and this side may be made smaller than any given quantity, by doubling indefinitely the number of its sides. Hence E — r may be made less than any assignable quantity. BOOK IV. 137 Peoposition XYII. — Problem. To find the approximate value of the ratio of the eir- cii^mf&rence to the diameter. We have, from the definition, tt = „' — . (1) And it has been shown that V = ■ — ^^p — • (2) From these formulae result four methods of determining the value of tt. 1°. Formula (1) enables us to calculate the radius when we know the length of the circumference. 2^. Formula (1) also gives the value of the circumference, when the radius is known. 3°. Formula (2) makes known the area of the circle, when we know the radius ; and 4°. Formula (2) determines the radius, when we know the area of the circle. We will explain the two first methods ; and we propose, in the first place, to calculate the radius of a circumference the length of which is 4. Let us construct a square, and taking the ^side of this square for unity, its perimeter will be 4. Let R bo the radius of the circumscribing circle, and r the apothegm of the square ; we have Now we may transform this square into a regular octa- gon, having the same perimeter ; and applying formulae (1) and (2) of the preceding problem, we shall find for the values of the radius and apothegm of the octagon, xA + vT , , 1+V2" j>, ^V 1-11^, and r' = —^^ We might, in like manner, calculate the values of Rj, »'2, of the regulai- isoperimetrical polygon of 16 sides; and 138 ELEMENTS OF GEOMETRY. continuing this operation indefinitely, we should arrive at a polygon, the perimeter of which would still be equal to 4, and for which E„, r„ would differ from each other by a quantity less than any given quantity. But the circumferences described with E„ and r„ as radii, are one greater, and the other less than 4. The radius of the circumference equal to 4, is therefore com- prehended between K„ and /•„ ; and may be calculated to any degree of approximation desired. If the values of E„ and ?•„ are estimated in decimals, it is evident the common decimals will belong to the required radius. If we limit the calculation to a polygon of 8192 sides, the following table presents the successive values of the radius and of the apothegm, for polygons of 4, 8, 16, .... 8192 sides. No. OF SlDBS. Apothegms. Eadh. 4 r, =0.5000000 El =0.7071068 8 7-2 =0.6035534 Ea =0.6532815 16 7-3 =0.6284174 E3 =0.6407289 32 r, =0.6345731 E, =0.6376435 64 n =0.6361083 E5 =0.6368754 128 r, =0.6364919 Eg =0.6366836 256 Tj =0.6365878 E, =0.6366357 512 7-8 =0.6366117 Eg =0.6366237 1024 r, = 0.6366177 Eg =0.6366207 2048 Tio = 0.6366192 Eio = 0.6366199 4096 ni= 0.6366196 En = 0.6366197 8192 7-12 = 0.6366196 Ei2 = 0.6366196 Limiting the calculation with the polygon of 8192 sides, for which the values of E12 r,2 coincide for seven places of decim-als, we have 0.6366196 for the approximate value of BOOK IV. 139 the radius of the circle, the circumference of which is equal to 4, the perimeter of the given square. And since ■"■ = ^p ' ; we have, by eubstituting these values for cir- cumference and radius, '-'''''' 3.1415926.,. i 1.2732392 for the approximate value of the ratio of the circumference to the diameter. Archimedes found ^■^- as an approximate value of tt, and this value is often used in practice. Metius found a nearer approximation in the fraction f-^-f . Peoposition XVII. — Peoblem. Having given the perimeters p, P, of two regular simir lar inscyrihed and circumscribed polygon's, to calculate the perimet&rs p', P', of two regular inscribed and circum- scribed polygons having double the number of sides. Let AB, EF, be the sides of the polygons, the perim- eters of which are respectively p, P; and let m be the number of sides of each polygon. Draw the chord AM, and at the points A and B draw the tangents AP, BQ ; finally, draw the right line PC. AM and PQ will be the sides of the inscribed and circumscribed polygons of 2m sides, the perimeters of which are p' and P'. From Prop. VIII., we have P _ CE _ CE . p " OA ~ CM ' 140 ELEMENTS OF GEOMETBT. and since CP is the bisectrix of the angle EOM, we have also FE_ CE PM ~ CM' Hence, from the common ratio, we have ,T_ PE ^~PM' From which, by composition, we get P+j9 _ PE + PM ^ EM I 2^ ~ 2PM ~ PQ* But the lines EM, PQ, are contained in the perimeters P, P', 2m times. Hence, 2p P'* And we have for the perimeter of the circumscribing poly- gon with 2m sides, 2Px^ F+p ^ > To determine p', we remark, that the two triangles PMN, MAjD, are equiangular and similar, and give the proportion AM_ PM AD MJSr" But the lines AM, AD, are contained 2ot times in the pe- rimeters p' and p ; and the lines PM, MN, are contained 4m times in P' and^'. Hence, we have P' P' — = — , ; and therefore^' — vP' xp. (2) Corollary. Formulae (1) and (2) give the second method of calculating the value of tt. Let us explain this method : Assume the radius of a circle equal to the linear unit 1, Inscribe and circumscribe a square to this circle. The BOOK IV. 141 side of the inscribed square will be ^2, and that of the circumscribed square will be 2 ; and their perimeters will be respectivelyj> = -i V2, and P — 8. Substituting these values in (1) and (2), we deduce the values of the perim- eters of the inscribed and circumscribed octagons ; and by the aid of the octagons, we may, in like manner, calculate the perimeters of the regular inscribed and circumscribed polygons of 16, 32, 64, &c., sides. But we know that in this series of operations, the perimeters of the polygons will approach nearer and nearer the circumference of the circle, and by continuing the calculation to any degree of approximation desired, we shall obtain the length of the circumference. Dividing the length by 2, we shall have the value of the number *, corresponding with that before found. By similar means we might explain the two other methods of calculating the approximate value of tt. 143 ELEMENTS OF GEOMETRY. EOOK V. '"-■- : - OF THE PLANE AXD EIGHT LINE, CONSIDERED IN SPACE. ' .■- DEFINITIONS. / I. A right line h- j)e?'pendicula?' to a plane, m- lien, it is pei'pendiculai- to eyery, right line drawn through ii^ foot ^ in the plane. Keciprocally, the plane is pcrjpendlcular to. , the Uuti. . ■ The foot of tlie perpendicular is the poipt in which the perpendicular meets the plane. , ,, . II. Aright line, \& : par allelioM /plane, y;]iQn it cannot ■ meet the plane, although both are produced indefinitely. Reciprocally, the plane is said to \>q parallel toihe line. III. Two planes ai'e parallel to each other, when, if both be produced indefinite!}^, they cannot meet. IV. In the reiiresentation of planes in figures, we are compelled to give limits to them ; but it must alwa)',s be remembered, that, unless otherwise stated, they must be conceived as indefinite in extent. Pkoposition I. — Theoeem. IFe meiy always pass a plana throng Ji ivio right lines which intersect each other ; and toe can pass lut one. Let AB, AC be two light lines inter- secting each other in A. ^Ve nmy con- ceive a plane which shall contain the line AB ; if, now, we move this plane around AB, until it pass through the point C, the line AC, whicli has two of its points in this plane, will be entirely in this plane, and the position of the plane will Ik; dettn-mined in space. BDOE V. lis Corollary L A ti-i angle ABC, or three points A, B, 0, not in 'tliu same right line, will de- termine the position of a plane. Cor'oUahj II. Two parallels, AB, CD, al'S6 determine the posi- " tib'n of a '[ilane; for ^ve kndw that A.^ .B two parallels are ill'the'fekilie pkn'e, "^ ' ' ' ' ' " ' p' and tWo diffei'eilt' planes catiflot ' ' cohtain these lines, SlVlC'e' each must contiiln- two points 'df AB, 'and' one point of CD'; thalt is'to say,'tiiree'J)oi'n'ts not ill' thfesame"i'ight' line. ' PkOPOSITJON II.— TflEOKJEM- . The •inter sect ion of two planes is a right -line. -.. For, if, among the points common to the two. planes,, we fiad. three . not. in this, right, line, the two planes, passing each through' these points, form one and the/same .pl^ne, which is contrary, to. the supposition. ' Pkoposition in. — Theoeem. ...... ..... ^ If a rigTitline AP he ferjjendicular to two lines PS^ PU, at iJieir point of intersection in tlie plane MN, ii'will he perpendicular to any line PQ, drawn through its foot in the same plane, and will also he pe)pe7idicidar to the plane. Draw, in the plane MlST, a right line Be, cuttin* the three lines PB, PQ,PC : produce AP until PA' = AP, and join the points A and A' with the points B, Q, C. The line PC being perpendicular to AA', at its middle point, the oblique lines CA, CA' are equal; 144 ELEMENTS OF GEOMETRY. for a like reason, BA = BA'. Hence, the triangles BOA, BOA' are equal, as they have a common side BO, and the other sides are equal, each to each. If now, the triangle BCA' be revolved around BC, as an axis, until it coincides with its equal BCA, the point A' will fall on A ; the point Q being in the axis, will remain fixed, and the line QA' will coincide with QA. Hence, since PQ has two of its points, P and Q, equally distant from the extremities of the line AA', it is perpen- dicular to AA'. As the same is true for every other right line drawn in the plane MN, through the point P, the line AA' is also perpendicular to the plane MN. Pkoposition IY. — ^Thkoeem. Throitgh a given point in a plane we may draw a per- pendicular to this plane, and we can draw hut one. 1°. Let us suppose the given point O to be situated in the plane MN. Take any line whatever, AB ; draw two planes through ^ this line, and in these planes draw the perpen- diculars BC, BD, to the line AB, at B. By the .c preceding theorem, the ""^■--^j, line AB will be perpen- dicular to the planes of the two lines BC, BD. Move this figure in such a man- ner, that the plane CBD shall coincide with the plane MIf, the point B falling upon O. The lines BC, BD, will take the position OC, OD, and the line BA will take the posi- ^ tion OF, evidently perpendicular to the plane MN. In the second place, let us suppose that the given point, F, is situated outside of the plane MN. Having constructed the figure ABCD, as before, move BOOK V. 145 / -C N it until the plane BCD coincides with the plane MIT ; so that the line BA shall at the same time pass tlirough the point F. The line BA will take the position FO, and will be perpendicular to the plane MIST. A B 2°. Through a given point 0, on the plane MN, only one perpendicu- lar can be drawn to this plane. For, if we could have two, such as OA, OB, draw through these lines a plane, and let 00 be the intersec- tion of this plane with the plane Then, the two lines OA, OB would be perpendic- ular to 00, at the same point and in the same plane, which is impossible. In like manner, it is impossible to let fall two perpendiculars upon a plane, from a point without; for, if FA, FB were these perpendiculars, the triangle FAB would have two of its angles right angles, which is impossible. ME". Pkoposition v. — Theorem. Through a given point we may draw a plane perpen- dicular to a given right line / and we can draw hut one. ^ 1°. Let us suppose that the given point is situated on the right line AB. Conceive any two planes to be drawn through AB, making any angle with each other ; and in these planes, draw CD and CE perpen- dicular to tlie line AB. The plane MD, which contains tlie lines CD and CE, will evidently be perpendicular to AB. 7 14(J ELE^fENTS QF GEOMETRY. Xmv, any otliei' ylaua illl, jjassing, through the ])a)-t (\ caiiDot be perpendicular to AB ; for, it' it were, any plane whatever drawn through the line AB, would intersect. the jilanes j\ID and MH, in two right lines. CX, CG, which would both be perpendicular to._AB, at the same ])oint and in the same plane, which is impossible. 2''. Let us now suppose the point C to be situated out- side of the line AB. Draw CD perpendicular to AB, and in a plane passing through AB, draw DE perpendicular to AB. The plane MN", of the right lines 01), DE, will be perpendicular to AB. FiiialLv, any other plane MP, passing tln-diigh the point C, cannnt be perpen- dicular to AB ; for if this wero possible, tlie plane. ABC would intersect the plane MP in a right line CG diffei'eiit from CD; we should then liBve two perpen- diculars drawn from C to AB, which is impossible. Corollary. All the right lines BG,BD, BE^ drawn- through any point B of the lino AB, perpendicular to AB, arc in the same plane, which is also perpendicular to AB. For, the plane MN, of tlie lines BC, BD, is perpendicular to AB. It is sufficient then to show that this plane contains all other perpendiculars. But if BE were not in the plane ilX, the plane drawn through the two right Htics AB, BE, would intersect the plane MJST in the riglit lin(> BII ; and we should have two perpendiculars, BE, BII. drawn from the same point, and in the same plane, to the line AB, whiqh is ijnpossible. BOOE V. 147 Proposition VI. — Theorem. , If from (t point A, outside of tJie plane MN, we draw to this plane thepcYpcndicidarAP, and the oblique lines AD,AC,AE— 1°. T})S peipendicular is shorter than any oblique line y ., 3"". .The oblique lines equally distant from the perpen- dicular, are equ.al; 3*^, Of two- oblique lines unequally distant from the foot of the perpendicular, that which is farther from the per- pC'iul 'ocular is the greater. 1°. The ti'iangle' APC is right-angled at P; lieuce, the obIi<^ue line AC, oppo- site the right angle, is greater than the perpendicular AP. 2°. The angles APC, APD being right angles, if we suppose PC — PD, the tri- angles APC, APD will be equal, since thej' have two sides and the included angle in each equal;' hence, AC = AD. ,, 3^. If the distance PE is gi'eater than PD, take PB = PD, and joiu.AB; we shall have AB = AD. But AB < AE ; hence, AD < AE. Corollary. All the fequal oblique ' lines ABj AC, AD, &c., meet the plan^MN, in the circumference described from: the' foot of the pei-pendicular P as a centre. There- fore, having given any point A outside of a plane, to find the point in which the perpendicular from A will meet this plane, it is only necessary to take any three points, B, C, D, on this plane, equally distant from A, and find the; c^entre of the circumference passing through. these three points ; this centre will be the required point P. .,._._ J 14S ELEMENTS OF GEOMETRy. Proposition VII. — Thkokem. - Let AP he any perpendicular to the plane MN^ and BC any right line situated in this plane ; if from the foot of the perpendicular P, we draw PD perpendicular to BC, the line AD vnll he perpendicular to BC. Take DB = DC, and join PB, PC, AB, AC; since A DB = DC, the oblique line PB = PC ; and with respect to the perpendicular AP, since PB = PC, the oblique line AB = AC ; hence, the line AD has two of its points A Jn and D equally distant from the extremities B and C; therefore AD is perpendicular at the middle point of BC. Corollary. We see, also, that BC is perpendicular to the plane APD, since BC is perpendicular, at the same time, to the two right lines AD, PD. Pkoposition VIIL — Theoeem. If the line AP He perpendicular to the plane MN, every line DE parallel to AP will he perpendicular to the same plane. Pass a plane through the parallels AP, DE, its intersec- tion with the plane MN will be PD ; in the plane MN draw BC perpendicular to PD, and join AD. By the corollary of the pre- ceding theorem, BC is perpen- dicular to the plane APDE ; hence, the angle BDE is a BOOK V. 149 right angle ; but the angle EPD is a right angle also, since AP is perpendicular to PB, and DE is parallel to AP ; therefore, the line DE is perpendicular to the two lines DP, DB, and is perpendicular to^lheir plane. Corollary I. Reciprocally, if the right lines AP, DE are perpendicular to the plane MN, they are parallel ; for, if they are not, draw through the point D a parallel to AP ; this parallel will be perpendicular to the plane MN ; then through the same point D two perpendiculars would be erected to the same plane, which is impossible. Corollary II. Two lines A and B, parallel to a third line C, are parallel to each other; for conceive a plane perpendicular to the line C ; the lines A and B being par- allel to C, will be perpendicular to the same plane ; hence, by the preceding corollary, they will be parallel to each other. It is understood that the three lines are not in the same plane, as this condition has already been examined. Pkoposition IX. — Theorem. Through any point A in space, only one parallel can he drawn to the right line CD. For, a parallel to CD, drawn through the point A, is situated in the plane passing through A this point and the right line CD ; and we know that in a plane only one par- c D allel can be drawn to a given line through a given point. PjiOPOsmoN X. — ^Theorem. If the line AB is parallel to the line CD, drawn in the plane MN', it will he parallel to this plane. For, if the line AB, which is in the plane ABCD, met the 150 ELEMENTS OF GEOMETEY. plane MJf, it could only meet it in some point of the line CD, the common intersection of the two planes ; but, AB cannot meet CD, since it is parallel to it. Therefore, it cannot meet the plane MIST, and must be parallel to it. - Corollary I. If a right line AE is parallel to a plane MN, every plane ABCD drawn through AB, will inter- sect MN" in the line CD parallel to AB. For, the lines AB, CD, being in the same plane ABCD, if the line AB met the line CD, it would meet the plane MN, which is contrary to the hypothesis. Corollary II. If, through a point C on a plane MN, parallel to the right line AB, we draw CD parallel to this line, this parallel will be in the plane MN. For, otherwise, the plane drawn through the line AB and the point C would intersect MN in a line CE parallel to AB, and we should thus have two parallels drawn through the same pouit to a given right line, which is impossible. PKOPOsrrioN XL — Thkoeem. Two planes, MN, PQ, perpendioular to the same right line AB, are parallel to each other. For, if they intersect, let be one of the points of their ^ common intersection, and join -;—^ n OA, OB; the line AB, per- """" \ pendicular to the plane MN, is perpendicular to the line OA, drawn through its foot in this plane. For the same ■^'^ reason, AB is perpendicular A B m\ C _^^ V. \n \ i'r'""=c fioo£ V. ISI to Bo ; therefore, OA and OB would be two perpendic- ulars drawn from the same point O to the same right line, which is impossible. Hence, the two planes MN , PQ cannot meet, and are, therefore, parallel. Proposition XII. — Thboeem. The intersections EF, GH^ of the parallel planes Mlf, PQ,iy a third plane EG, are parallel. For, if the lines EF, GH, situated in the same plane, are not parallel, they will meet, if produced ; then, the planes MN", PQ, in which they are situated, will also meet, and could not, therefore, be parallel, which is contrary to the hypothesis. Peofosition XIII. — Theorem. Through a point A we may draw a plane pwrallel to the plane MN j and we can draw hut one. 1°. From the point A, draw AB perpendicular to MN ; \F through the point A draw the plane PQ perpendicular to AB ; the plane PQ will U be parallel to MN. 2^. If there could be a second plane, such as AF passing through A, and par- \jj allel to MN, any plane whatever drawn -^ through AB would intersect the planes MIST, PQ, AF, in the right lines BD, AG, AE. Bat the lines AG, BD are parallel as the intersections of two parallel planes by a third plane; for a like reason AE would be parallel to BD, which is absurd. 152 ELEMENTS OP GEOMETRY. Proposition XIV. — Theorem. The line AB, perpendicular to the plane MN, is perpen- dicular to the plane PQ, parallel to MN. Draw any line AD in the plane MIST, and through AB and AD pass a plane BAD ; this plane will intersect the plane PG : otherwise, through A, two parallel planes to PQ would be drawn. The' intersection BC of the planes BAD and PQ, will be parallel to AD. (Prop. XII.) But the line AB, perpendicular to the plane MN, is perpendicular to the right line AD ; it is also perpendicular to its par- allel BC. We also see that AB will be perpendicular to any right line passing through its foot in the plane PQ ; and ia, therefore, perpendicular to this plane. !^N ^ Proposition XV. — Theorem. Two planes. A, B, parallel to a third plane C, are parallel to each other. Draw DF perpendicular to the plane C ; this line is perpendicular to the planes A and B, by the preceding theo- rem ; hence, these planes are parallel, since they are perpendicular to the same right line. i Proposition XVI. — Theorem. The parallels EG, FH, comprehended between two paral- lel planes MN', PQ, are equal. BOOK V. 153 Throngli the parallels EG, FH, pass the plane EGHF, intersecting the parallel planes in EF and GH. The lines EF and GH are parallel to each other (Prop. XII.), as well as the lines EG, FH ; hence, the figure EGHF is H a parallelogram, and EG = FH. Corollary. It follows from this, that two parallel planes are evei^ywhere equidistant ; for, if EG and FH are per- pendicular to the two planes MN, PQ, they will be paral- lel to each other, and are, therefore, equal. PsoposriioN XVII. — Theorem. If two angles CAE, DBF, not in the same plane, have their sides parallel and lying in the same direction, they wiU he equal, and their planes will he parallel. Take AG = BD, AE = BF, and join CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABCD is a parallelogram ; hence, CD is eqnal and parallel to AB. For a like reason, EF is equal and parallel to AB ; and we have also CD equal and parallel to EF. The figure CEFD is, there- fore, a parallelogram, and the side CE is equal and paral- lel to DF ; hence, the triangles CAE, BDF have their sides equal, each to each ; therefore, the angle CAE = DBF. In the second place, the plane ACE is parallel to the plane BDF ; for, suppose that the plane parallel to BDF, through the point A, met the lines CD, EF, at points G and H, diflerent from the points C and E, then (Prop. XVI.) the three lines AB, GD, FH would be eqnal. But the three lines AB, CD, EF are equal ; hence, CD = GD, 7» 154 ELEMENTS OP GEOMETRY. and FH — EF, which is absurd ; therefore, the plane ACE is parallel to BDF. Corollary. If two parallel planes MN, PQ are met by two other planes. CABD, EABF, the angles CAE, BDF, formed by the intersections of the parallel planes, will be equal ; for, the intersection AC is parallel to BD, and AE to BF ; hence, the angle CAE == DBF. Pkoposition XYIII. — Thkokem. Mr lO 7^ 2 Two right lines, included ietween three parallel planes, are cut proportionally hy these planes. Suppose tliat the line AB meets the parallel planes MN, PQ KS, at A, E, B, and that the line CD meets the same planes in C, F, D ; then we shall have AE_ BE CF ~ FD' Draw AD, meeting the plane PQ, in G, and join AC, EG, GF, BD ; the intersections EG, BD, of the parallel planes PQ, KS, by the plane ABD, are parallel (Prop. XII.) ; hence, AE_AG EB ~ GD ' ■ \ Again, since the intersections AC, GF, are parallel, we have AG_ CF GD FD ■' Hence, from the common ratio, we have AE_ CF EB ~ FD" BOOK V. 156 DEFINITIONS. The prelection of a poini on a plane, is the foot of the perpendicular let fall from this point en the plane, a, tn, h are the projections, respectively, of the points A, M, B, on the plane MN. The projection of a line AMB on a plane, is the line amb, formed by joining the projections of all the points of the line AMB. Peoposition XIX. — Theokem. The projection of a right line on a plane is a right line. From any point A of the line AB, let fall the perpen- dicular Aa on the plane ES, and through the lines AB, Aa, draw the plane BAa, cut- ting the plane ES in ah. K, through the points M, N, ifec, of the line AB, perpen- diculars be let fall on the plane RS, they will all be parallel to Aa, and will be situated in the plane BA CBD GFH 3 4^ hence, CABD _ CBD GEFH GFH' If the two diedral angles had no common measure, we should see, by reasoning already presented, that the pro- portion would still subsist. Scholium. From this theorem, it follows, that if we wish to measure a diedral angle D, that is, to find the ratio of this angle to a diedral angle taken for unity (the right diedral angle, for example), it would be sufficient to find the ratio of the plane angle of D to a right angle. Proposition XXVI. — Theorem. If the right line AP he perpendicular to the plane MN, every plane APB drawn through it, will also lie perpendicular to the plane MN. Let BC be the intersection of the pliines AB, MN : if in the plane MN we draw PD perpendicular to BP, the line AP, being perpendicular to the [\ A \ • c 1 ^ \^ \ BOOK V. 161 plane MN, will be perpendicular to each of the two right lines EC, PD ; but the angle APD, formed by the two perpendiculars PA, PD, at the common intersection BP, measures the angle of the two planes AB, MN. Hence, since this angle is a right angle, the two planes are per- pendicular to each other. Scholium. When three right lines AP, BP, DP are perpendicular to each other, each of these right lines is perpendicular to the plane of the two others, and the three planes are perpendicular to each other. Pboposition XXYII. — Theokem. If the plane AB is perpendicular to the plane MW, and we draw, in the plane AB, the line PA perpendicular to the common intersection PB, PA will he perpendicular to the plane MN. For, if in the plane MN we draw PD perpendicular to PB, the angle APD will be a right angle, since the planes are perpendicu- lar to each other. Hence, the line AP is perpendicular to the two lines PB, U PD, and is therefore perpendicular to their plane Ml!^. Corollary. If the plane AB is per- pendicular to the plane MN, and through the point A of the plane AB, we let fall a perpendicular on the plane ME", this pei-pendicular will be iu the plane AB ; for, if not, we might draw in the plane AB a perpendicular AP to the common intersection BP, which would, at the same titne, be perpendicular to the plane ME ; and we should then have two perpendiculars drawn from the same point A to the plane MN, which is impossible. 162 ELEMENTS OF GEOMETRY. Proposition XXVIII. — Theorem. If two planes AB, AD are perpendicular to a third plane MN, their common in- tersection AP will he perpen- dicular to this third plane. For, if from a point A, taken on the intersection, we let fall a perpendicular on MN, this perpendicular must be found at the same time in '■'^ the plane AB, and also in the plane AD. Hence, it is the common intersection of these two planes. DEFINITIONS. I. A solid angle, or a polyedral angle, is the figure formed by several planes intersecting each other in a com- mon point. II. The intersections of the planes are called the edges of the solid angle ; and the point in which the edges tneet is the verttx of the solid angle. The angles formed by tlie edges are called the faces or plane angles of the solid angle. III. A triedral angle is a solid angle formed by three plane angles. IV. Tliose solid angles only are considered, which are situated on the same side of their faces, when produced. Such angles are called convex solid angles. V. A solid angle SABCD being given, if we produce the edges SA, SB . . . beyond the vertex A, we form a new solid angle SA'B'C'D', which is said to be symmetrical with the first. It is evident that this new solid angle has the same piano angles and the same diedral angles as the first. Neverthe- BOOK V. 163 less, these two solid angles cannot, in general, be placed one upon the other, so that they may coincide ; for, if we make the face D'SA' coincide with its equal ASD, so that the edges of the two solid angles may be situated on the same side of the common face, we see that setting out from the edge SD, and making the revolution of the two solid angles, the plane angles ^ and the diedral angles will be found in an inverse order. pEOPosrrioN XXIX. — Thkoebm. Two triedral angles are eg[ual in all their parts when they have a diedral angle in each equal, and the two faces including the diedral angle equal, each to each. Let ASB = DTE, BSC = ETF, and the diedral angle SB equal to the di- edral angle TE. Place the angle ASB on its equal DTE ; since the di- edral angles SB and TE are equal, the plane ETF may be applied to the plane BSC, and as the angles ETF and BSC are equal, the edge TF will take the direction SC ; the two triedral angles will therefore coincide, and be equal in all their parts. If the equal faces of the two triedral angles were in- versely situated with respect to the equal diedral angles, we might apply the triedral angle T on the symmetrical anffle SABC, and we should be led to the same conclusion. 164 ELEMENTS OF GEOMTCTRY. Peoposition XXX. — ^Theorem. Two triedral angles are equal in all their parts when they have two diedral angles and the included face of the one equal to two diedral angles and the included face of the other, each to each. Let ASC = DTF, the diedral angle SA = TD, and the B' diedral angle SC = TF. Place thefaeeDTF on its equal ASC; since the diedral an- gles SA and TD are eqnal, as well as the diedral angles SO and TF, the, planes DTE, FTE may be respectively applied to the faces ASB, CSB ; hence, the edgeTE will coincide with SB, and the two triedral angles will coincide and be equal in all their parts. If the equal diedral angles were inversely disposed, with respect to equal faces, we might apply the triedral angle T to the symmetrical of the triedral angle S, and the same result would be reached. pBOPOsniON XXXI. — ^Theorem. If two triedral angles have their faces equal, each to each, the diedral an- gles opposite to the equal faces will he equal, also. Let ASB = DTE, ASC = DTF, BSC = fETF. Take the six equal lengths SA, SB, SC, TD, TE, TF, and BOOK V. 165 draw the lines AB, AC, EC, DE, DF, EF. The isosceles triangles SAB, DTE are equal, since they have an angle in each included by equal sides, equal ; the same equality also exists between the triangles SBC, TEF, and the tri- angles SAC, TDF. Finally, from the equality of these triangles, the triangles ABC, DEF having their sides equal, are also equal. Now, through any point M on the edge SA, draw, in the faces SAB, SAC, the perpendiculars MN, MP to SA ; these perpendiculars will meet the sides AB and AC, since the triangles SAB, SAC, being isosceles, the angles at the bases, SAB, SAC, are acute ; finally, draw NP. Let us now take DG = AM, and repeat, in the second triedral angle, the preceding construction. The right-angled triangles AMN, DGK are equal, since they have tlie side AM = DG, and the acute angle MAN" = GDK ; hence, AN = DK, and MN = GK. "We see, also, that MP= GH, and AP = DH. We see, also, that the triangles PAN, HDK, having an angle included between equal sides equal, are equal ; hence, NP = KH. Therefore, the triangles NMP, K'GH, have the three sides equal, each to each ; hence, the angle NMP, wliich measures the diedral angle SA, is equal to the angle KGH, which measures the diedral angle TD. Scholium. If the two solid angles have, besides, their faces similarly disposed, they will be equal by superpo- sition ,' if the faces are not similarly situated, the solid angles will be symmetrical. PKOFOsrrioN XXXII. — Theokem. If, through any point F, taken on the edge of a diedral angle AB, we erect a perpendicular, FG, to the face AC, on the same side of the plane ABC as the face ABD ; and to the face ABD a perpendicular FE, on the same side of 166 ELEMENTS OF GEOMETRY. the plane ABD as the face ABC, the angle EFG is the supplement of the plane angle which measures the diedral angle. For, the perpendiculars EF, FG, to AB, determine a plane perpendicular to AB, "which will intersect the two faces of the di- edral angle in the right lines FI, FH, also perpendicular to AB ; and the angle IFH measures the diedral angle. But FG being perpendicular to the plane ABC, we have GFI equal to one right angle. We have, for a like reason, EFH = one right angle ; hence, we have GFI + EFH = two right angles, or EFG + IFH = two right angles. Proposition XXXIII. — Thkoeem. If, at the vertex of a triedral angle SABC, we erect a perpendicular to each face, on the same side of the plane of this face as the third edge, the triedral angle thus formed will he supplementary to the given triedral angle. Note. — Two triedral angles are supplementary, when the plane angles of the one are supplements of the angles which measure the diedral angles ''I 1^ of the other. Let SC be perpendicular to the face ASB, on the same side of the plane ASB as the edge "1 SC ; also, SB perpendicular to the face ASC, on the same side of the face ASC as the edge SB; and iinally, let SA' be perpendicular to the plane CSB on the same side of this plane as the edge SA. 1°- The line SC is perpendicular to the plane ASB, and is on the same side of this plane as the face CSB ; the line BOOE V. 167 SA' is perpendicular to tlie plane CSB, and is situated on the same side of this plane as the face ASB ; hence, the angle C'SA' is the supplement of the angle corresponding to the diedral angle SB. We see, also, that the angle CSB' is the supplement of the angle which corresponds to the diedral angle SA, and that the angle A'SB' is the supplement of the angle which measures the diedral angle SC. Hence, 1°, the plane angles of SA'B'C are the sup- plements of the diedral angles of SABC. 2°. The line SA', being perpendicular to the plane CSB, is perpendicular to SC ; the line SB', perpendicular to the plane CSA, is perpendicular to SC : hence, the line SC is perpendicular to the plane A'SB'. Besides, SC being perpendicular to the plane ASB, and being on the same side of this plane as SC, the angle CSC is acute ; and since SC is perpendicular to A'SB', and makes an acute angle with SC, we conclude that SC is on the same side of the plane A'SB' as SC. In like manner, we see that SB is perpendicular to the plane A'SC, and is on the same side of this plane as SB', and that SA is perpendicular to the plane CSB', and is on the same side of this plane as SA'; in other word^i, the triedral angle SABC has been constructed by means of SA'B'C, just as the triedral angle SA'B'C has been con- structed by means of SABC. Hence the plane angles of SABC are supplements of the diedral angles of SA'B'C. Proposition XXXIV. — ^Thkoeem. If two triedral angles have their diedral angles ngual, each to ea^h, they have also equal faces. Let S and S' be the two given triedral angles ; T and T' their supplementary triedral angles. Since S and S' have their diedral angles equal, the triedral angles T, T' have their faces equal, each to each, 168 ELEMENTS OF GEOMETRY. and consequent!)', their diedral angles are equal. Finally, tlie triedral angles T, T' having their diedral angles equal, the triedral angles S, S' will have their faces equal. Scholium. If the equal faces of the two triedral angles are similarly disposed, the triedral angles will be eqnal, by sujaerposition ; otherwise, they will be symmetrical. fEOPOsiTioN XXXV. — Theorem. If a solid angle is formed iy three plane angles, the sum of any two of these plane angles will he greater than the third. Let the solid angle S be formed by the three plane angles ASB, ASC, BSC, and let us suppose that the angle ASB is the greatest angle of the three ; then we shall have ASB < ASC + CSB. In the plane ASB, make the angle BSD = BSC. Draw the right line ADB, and taking SC = "c" SD, join AC, BC. The two sides BS, SD, are equal to the two sides BS, SC ; the angle BSD = BSC ; hence, the two triangles BSD, BSC are eqnal, and BD = BC. But, we have AB < AC + BC ; taking BD from one side, and its equal BC from the other, there will remain AD < AC. The two s-ides AS, SD are equal to the two sides AS, SC ; the third side AD is less than the third side AC ; hence, the angle ASD < ASC. Adding BSD = BSC, to each side, we shall have ASD + BSD, or ASB < ASC + BSC. I Proposition XXXYT. — Theoeem. The sum of the plane angles which form, a convex solid angle is always less than four right angles. BOOK V. 101 Intei-sect the solid angle S by a plane ABODE, so as to meet all the edges ; from a point O, taken in this plane, draw to the vertices A, B, C, D, E, the lines OA, OB, 00, OD, OK. The sum of the angles of the triangles ASB, BSC, &e., foi-ined around the vertex S, is equal to the sum of the angles of an equal Dumber of triangles AOB, BOG, &c., formed around the vertex 0. But at the point B, the sum of the angles ABO, OBC, which, taken together, make the angle ABC, is less than the sum of the angles ABS, SBO (Prop. XX XV.) ; also, at the point C, we have BOO + OCD < BOS + SOD ; and the same for all the angles of the polygon ABODE. Hence, the sum of the angles at the bases of the triangles which have for a common vertex O, is less than the sum of the angles at the bases of the triangles which have their common vertex at S ; hence, by compensation, the sum of the angles formed around the point O is greater than the sum of the angles formed around the point S. But tlie sum of the angles around O is equal to four right angles ; hence, the sum of the angles which form the solid angle S is less than four right angles. Pkopositiou XXXYII. — Theorem. 1°. In every triedral angle, the sum of the three diedral angles is less than six and greater than two ri^ht angles. 2°- The smallest diedral angle, augmented hy two right angles, is greater than the sum of the other two. 1°. Let a, b, c be the three diedral angles of the given triedral angle ; and let A, B, C be the faces of the sup- 8 170 ELEMENTS OF GEOMETRY. ]ilt'iiii;ntary triedml angle, we have a = 2 right angles — A, h = 2 right angles — B, c = 2 right angles — C ; hence, by addition, a + h-\-c = Q right angles — (A + B + C). Farther, the sum A + B + C is greater than o, and less than 4 right angles ; hence, the sum (a + & + c) is less than 6 right angles and greater than two. 2°. a, i, c, being the diedral angles of the given triedral angle, and a the smallest, the faces of the supplementary triedral angle will be respectively 2 right angles — a, 2 right angles — 5, 2 right angles — o, of which 2 right an- gles—a, will be tlie greatest; hence, we shall have, by Theorem XXXV., 2 light anj^les— « < '2 right angles — 5 + 2 right angles— c; adding h +a + a to both sides, and subtracting 2 right angles, we have & + c < 2 right angles + a. Proposition XXXVIII. — Theoebm. In order that we may form a solid triedral a7)gle with three yiven plane faces, it is necessary and it is siijficicnt that the sum of the three faces shall he less than 4 riyJU angles, and that the greatest face shall he less than the sum. of the other two. We have already seen that these conditions are neces- sary ; we have now to prove that they are sufficient Let BSC, ASB, DSC be the three given faces, which are supposed to be placed on the same plane ; and let BSC be the greatest. From the point S, as a centre, with any radius SA, describe a cii- cumference, and draw the perpen- diculars Aa, 'Dd, from the points A and D to the lines SB and SC. The angle BSC beiu'j- the greatest of the three, BC will be greater than either of the HOOK V. 171 arcs BA, CD ; and since Ba = BA, we see that the point a is between B and C, and the same with d. Besides, we have, by hypothesis, BSC < ASB + CSD ; and, consequently, BC < AB + CD. Hence, since Ba = BA, and Qd = CD, the point a is on the arc BC, on the right of d. Finally, the sum of the three given faces being less than four right angles, the point D will be situated beyond the point C, on the circumference passed over from the point A in the direction ABC. The point d will then be situated between A and a, and the point a between d and D; hence, the chords Aa, Dris7ns which have equal bases and equal altitudes are equal. For, having the side AB = ab, and the altitude BG = JA, the rectangle ABGF will be equal to the rectangle abgf\ the same is true of the rectangles BGHC, bghc : thus, the three planes which form the solid angle B are equal to the three planes which form the solid angle b ; hence, the two prisms are equal. Peoposition II. — Theorem. In, every parallelopipedon the apposite faces are equal and parallel. From the definition of this solid, the bases ABCD, EFGH are equal parallelograms, and their sides are paral- lel ; it remains now to prove that the same is true for two opposite lateral faces, such as AEHD, 4/ i L I BFGG. Now AD is equal and parallel to BG, since the figure ABCD is a parallelogram ; for a like reason, AE is equal and parallel to BF. Hence, the angle DAE is equal to the angle CBF (Book Y., XVII.), and the plane DAE is parallel to CBF ; hence, also, the parallelogram DAEH is equal to the parallelogram CBFG. In like manner, we could prove that the opposite parallel- ograms ABFE, DGGH are equal and parallel. BOOK VI. 177 Corollai'y. Since the parallelopipedoa is a solid com- posed of six planes, whose opposite faces are equal and parallel, it follows that any face and its opposite ma}- be taken for the bases of the parallelopipedon. Scholium. Having given three right lines, AT3, AE, AD, passing through the same point A, and making with each other any given angles, we may construct on iliom a parallelopipedon. It is only necessarj- to draw through the extremity of each line, a plane parallel to the ]ilane of the other two lines : that is, through the point B, a plane parallel to DAE ; and through the point D, a plane paral- lel to BAE ; and through the point E, a plane parallel to BAD. The mutual intersection of these planes will form the required parallelopipedon. Peoposition III. — Theorem. The diagonals of a parallelopipedon mutually divide each other into two equal parts. Fur, let the diagonals EC, AG be drawn : since AE is equal and parallel to CG, the figure AEGC is a parallel- ogram. Hence, the diagonals EC and AG of this parallelogram divide each other into two equal parts. We might prove, in like manner, that the diagonals EC and DF di- Tide each other equally. Hence, the four diagonals divide each other mutually into two equal parts, at the same point, which may be considered as the centre of the parallelopipedon. Peoposition IV. — Theorem. In eve^'y prism, ABCI, the sections NOPQR, STVXY, dec, made ly parallel planes, are equal polygons. 8* 178 ELEMENTS OF GEOMETEY. Fur, the sides NO, ST are included between the paral- lels NS, OT, which are edges of the prism : hence, NO = ST. For a like reason, the sides OP, PQ, QE, &c., of the section NOPQR, are equal, respectively, to the sides TV, YX, XY, &c., of the section STVXY. Besides, the equal sides being at the same time parallel, it fol- lows that the angles NOP, OPQ, &c., of the first section, are equal, respect- ively, to the angles STY, T7X, &c., of the second. Hence, the two sec- tions NOPQR, STVXY are equal polygons. Corollary. Every section made in a prism by a plane parallel to the base of the prism, is equal to the base. Pkoposition v. — Theoeem. The plane which jpasses through the opposite eiUji s J^£, Z>ir, of the parallelopipedon AG, divides the parallelopip- I don into equivalent triangular prisms. Through the vertices B and F, draw the planes Bar/f, Yigh perpendicular to the edge BF, meetin[» the three other edges AE, DH, CG, ;f the t^uine parallelopipedon, in the points a, ^/, c, for one plane, and e, A, g, for the othi'i'. The sections Bac?c, Yehg will be equal parallelograms. They are equal, because they are made by planes perpendicular to the same right line, and consequently parallel : they are parallelograms, because two op- posite sides, aB, dc, of the same sec- tion, are the intersections of two par- allel planes ABFE, DCGH with the same plane. BOOK VI. 179 For a like reason, the figure BaeF is a parallelogram, as well as the latei-al faces BFgrc, cdhg, adhe of the solid BadcFehg. Hence, this solid is a prism ; and this prism is a right prism, since the edge BF is perpendicular to the plane of the base. If, now, we divide the right prism BA into two right triangular prisms aBdeFh, BdcFAg, by the plane BFHD, the oblique triangular prism ABDEFH will be equivalent to the right triangular prism oBdeFh. These two prisms having a common part, ABDAeF, it will be sufficient to prove that the remaining parts, viz., the solids BaADd, FeEHA, are equivalent. Since ABFE, aBYe are parallelograms, the sides AE, ae, which are both equal to their parallel BF, are equal to each other ; and taking away the common part Ae from each, there remains Aa — Ee. We might prove also that DJ = HA. ITow, place the two solids BaKDd^ FeEHA, one on the other, 60 that the base FeA shall fall on its equal Bad: then, the point e falling on a, and the point A on c?, the edges eE, AH will fall on their equals aA, ^ZD, since they are perpendicular to the same plane Ba^. Hence, the two solids will coincide in all their parts,"and the oblique prism BADFEH is equivalent to the right prism BadYeh We could prove, in a like manner, that the oblique prism BDCFHG is 'equivalent to the right prism Bi^cFA^. But the two right prisms Ba^^FcA, Bdc?hg are equal to each other, since they have the same altitude BF, and their bases Bad^ Bdc are halves of the same parallelogram. Hence, the two triangular prisms BADFEH, BDCFHG being equivalent to equal prisms, are equivalent to each other. Corollary. Every triangular prism ABDHEF is the half of the parallelopipedon AG, constructed on the same solid angle A, with the same edges AB, AD, AE. 180 ELEMENTS OF GEOMETET. Peoposiiion VI. — Theorem. If two parallelopipedons AG, AL have a common iase ABCD, and their upper bases in the same plane and between the same parallels, they will he equivalent. llie triangular prism AEIDHM is equal to the triangu- lar prism BFKCGL. For, AE = BF, being opposite sides n G M L of a parallelogram : for the same reason AI = BK: finally, the angles EAI, FBK, having their sides parallel, are equal. Hence, the triangles EAI, FBK are equal. The parallelograms AH, BG are equal, being opposite faces of a parallelopipedon : the parallelograms IH, GK are equal, for EH = FG, EI = FK, and the angle HEI = GFK. The three faces which form the solid angle at the point E, are therefore equal to the three faces which form the solid angle at F: besides, the solid angles E, F, having their plane angles equal, each to each, and similarly situated, are equal. Hence, the prisms AEIDHM, BFKCGL are equal. But, if the prism AEM be taken from tlie solid AL, there will remain the parallelopipedon AIL; and if from the same solid we take the prism BFL, there will remain the parallelopipedon AEG. Hence, the two parallelopip- edons AIL, AEG are equivalent to each otiier. Peoposition YII. — Theorem. Two parallelopipedons having the same base and the same altitude are equivalent. BOOK VI. 181 Let ABCD be the common base of the two parallelopip- edons AG, AL ; since they have the same altitude, their upper bases EFGH, IKLM, will be in the same plane. Further, the sides EF and AB are eqnal and paral- lel, as well as the sides IK and AB ; hence, EF is equal and parallel to IK; for a like reason, GF is equal and parallel to LK. Let the sides EF, HG, and the sides LK, IM, be produced, until they form, by their intersections, the parallelogram NOPQ; it is evi- dent that tliis parallelogram will be equal to each of the bases EFGH, IKLM. If, now, we. conceive a third paral- lelopipedon, having the same lower base ABCD, with NOPQ as its npper base, this third paralleh)pipedon will be equivalent to the parallelopipedon AG (Book YL, Prop. YL), since, having the same lower base, the upper bases are in the same plane, and between the same parallels CQ, FN. For a like reason, this third parallelopipedon would be equivalent to the parallelopipedon AL ; hence, the two parallelopipedons AG, AL, which have the same base and the same altitude, are equivalent to each other. PEOPosmoN VIII. — Thkoeem. Mieiy parallelopipedon may be changed into an equiva- lent rectangular parallelopipedon, which shall have the same altitude and an equivalent hose. Let AG be the given parallelopipedon ; from the points A, B, C, D, drawAL BK, CL, DM, perpendicular to the 182 ELEMENTS OF GEOMETRY. plane of the base : we shall thus form the parallelopipedpn AL equivalent to the parallelopipedon AG, and the lateral faces of which will be rectangles. If, now, the base ABCD is a rectangle, AL will bo the rectangular parallelopipedon which will be equivalent to the given parallelopipedon AG. But if ABCD is not a rectangle, draw AO and BN perpendicular to CD, and then OQ and NP perpendicular to the base : a solid ABNOIKPQ will be formed, which will be a rectangular parallelopipedon ; for, by construc- tion, the base ABNO, and the opposite face, IKPQ, are rectangles; the lateral faces are also rectangles, since the edges AI, OQ, &c., are perpendicular to the plane of the base ; hence, the solid AP is a rectangular parallelopipe- don. But the two parallelopipedons AP, AL, may be considered as having the same base ABKI, and the same altitude AO ; they are, therefore, equivalent: hence, the parallelopipedon AG, which had first been changed into an equivalent parallelopipedon AL, is now converted into an equivalent rectangular parallelopipedon AP, which has the same altitude AI, and the bases ABNO, ABCD being also equivalent. Pkoposition IX. — Theoeem, I'wo rectangular parallelopipedons AO, AL, having the same iase ABCD, are to eaoh other as their altitudes AE, AL SOOE VI. l83 E H \ ^ \ G ] M \ 2 \ L y K D \ \ i 3 C In the 'first place, suppose that the altitudes AE, AI are commensurable, and that they are to each other as the numbers 15 and 8. Divide AE into 15 equal parts, AI will contain 8 of them ; through the points of division X, y, s, &c., draw planes paralliel to the base. These planes will divide the solid AG into 15 partial parallelopip- edons, which will be equal to each other, since they have equal bases and equal altitudes : the bases are equal, because they are sections made by planes parallel to the base (Book YI., Prop. Y.) ; the alti- tudes are equal," because they are the equal divisiohs Aa;, ccy, ys, &c., of the line AE. But of these 15 equal par- allelopipedons, 8 are contained in AL; hence, the solid AG is to the solid AL as 15 is to 8, or, in general, as the altitude AE is to the altitude AI. If the altitudes AE and AI were incommensurable, we might prove, by the method explained in Book 11., Prop. XYIII., that their ratio would always be the same as that of the parallelopipedons. Remark. In a rectangular parallelopipedon, if a, J, c are three contiguous edges, and one of them be taken for the altitude of the parallelopipedon, the two others will form the two dimensions of the base. Peoposition X. — ^Theokem. Two rectangular parallelopipedoTis having the same alii- tude are to each other as their bases. Let P, P' be the two rectangular parallelopipedons ; let a, h, c be the thi-ee dimensions of the first, a, b', c', those of the second. Construct a third rectangular parallelopipedon P", the dimensions of which are a, J, o'. 18i ELEMENTS OF GEOMETRY. The parallelopipedous P and P", having two ■common dimensions a and J, are to each other as their altitudes c and c', and we have J__c_ P" ~ c' For a like reason, we have P^_^ P' V Multiplying these proportions together, member by mem- ber, and dividing the two terms of the first ratio by P", we have P' 5' X c^ ^^ We- know, also, that the bases B, B', of the two parallel- opipedous, are to each other as the products 5 x c, and V X c' ; hence, PBOPosrrioif XI. — Theorem. Any two rectangula/r parallelopipedons are to each other as the product of their, bases iy their altitudes, or, as the product of their three dimensions. Let PI be the altitude of the parallelopipedpn P, a and i the two dimensions of its base B. Let H' be the altitude of the parallelopipedon P', and a' and b' the two dimensions of its base B'. Let P" be a third parallelopipedon, having H for its altitude and B' for its base. , The parallelopipedous P, P", having the same altitude, are to each other as their bases, by the preceding theorem, and we have P" B'* SooE VI. 185 The pai-allelopipedons P" and P', having the same base, are to each other as their altitudes (Book VI., Prop. IX.), P" H and we have ^, = — . Multiplying these proportions together, and dividing the two terms of the first ratio by P", we have, ?-=?-^ (1) P' B'xH'" ^' We know, also, that the bases B, B' are to each other as the products a xi; hence, BxH _ axbx H B'xB.'~ a' xb' X B'' and we conclude that P axh X H P' a' xh' X H' (2) MEASURE OF THE EECTANGULAE PABALLELOPIPKDON. To measure a rectangular parallelopipedon P, is to find its ratio to a certain rectangular parallelopipedon P' taken for unity. But, the proportion (2) shows, that to obtain this ratio, it is necessary to determine the value of a, i, iZj a', h', S', as compared with the same linear unit, and to divide the product of the three first numbers by the product of the three others. The calculation is much simplified if we take for the unit of volume P', the cube whose edge is the linear unit ; for, in this case, the numbers whicli represent a', h', H' are reduced to unity ^ and the proportion (2) becomes P _ axlxR P'~ 1 * ' from which we see, that the measure of a rectangular jtaral- lelqpipedon is equal to the jproduct of its three dimensions. l86 ELEMENTS OF GEOMETEY. "VVe ina}' remark that the product a xl shows how many times the base B of the parallelopipedon P contains the square constructed on the linear unit. The measure of a rectangular parallelopipedon is, there- fore, also equal to the product of its base by its altitude; taking for the unit of area the square constructed as the linear unit, and for the unit of volume the cube constructed on the same unit. Applications. 1°. Let a = 2.51 feet, 5 = 3.25 feet, H = 2.45 feet, the measure of the rectangular parallelopipedon will be P = 2.51 X 3.25 X 2.45 = 19.985875 solid feet. 2°. Let B = 25.51 square feet, and H =12.5 linear feet, the rpeasure of the rectangular parallelopipedon will be P = 25.51 X 12, 5 = 318.875 solid feet. Peoposition XIL — Theorem. The measure of any parallelopipedon, and, in general, the measure of any prism, is equal to the product of its hase hy its altitude. For, 1°, any parallelopipedon is equivalent to a rec- tangular parallelopipedon which has the same altitude and ati equivalent base. (Book VL, Prop. VIIL) But the measure of the rectangular parallelopipedon is equal to its base multiplied by its altitude ; hence, this measure also applies to any parallelopipedon. 2°. Every triangular prism is the half of the parallelo- pipedon which has the same altitude, and a base equiv- alent to double the base of the triangular prism. (Book VL, Prop. V.) But the measure of the parallelopipedon is equal to the product of its base by its altitude ; hence, the measure of the triangular prism is equal to the product of its base (which is half of that of the parallelopipedon) by its altitude. BOOK VI. 187 3°. Any prism whatever may be decomposed into as many triangular prisms of the same altitude as we may form triangles from the polygon which is its base. Eut the measure of each of the triangular prisms is equal to the product of its base by its altitude ; and, since the altitude is the same for all, it follows, that the sum of all the par- tial prisms will be equal to the sum pf all the triangles which form tlieir bases, multiplied by the common altitude. Hence, the measure of any prism is equal to the product of its base by its altitude. Corollary. If, we compare two prisms which have the same altitude, the product of the bases by the altitudes will be as their bases ; hence, two prisms of the same alti- tude are to each other as their bases y and, for a like reason, two prisms of the same base are to each other as their dli{- tudes. PBOPOsmosr XIII. — Theoeem. If avyram.id SABCDE be cut by a plane parallel to its base — 10. The edges SA, SB, SC, a7id the altitude SO, will be divided proportionately at a,b, c . . . . and o ; 2°. The section abode will be a polygon similar to the base ABODE. For, 1°, the planes ABC, dbo, being parallel, their inter- sections AB, db, by a third plane SAB, will be parallel (Book V., Prop. XII.) ; hence, the triangles SAB, sah, are similar, and we have the proportion SA SB sa sb We shall also have SB SC sb ~ SG 188 ELEMENTS OF GEOMETRY. And SO on for all the triangular faces of tlie pyramid. Hence, all the edges, SA, SB, SO, &c., are cut proportion- all}' at a, b, c, &c. The altitude SO is also divided in the same proportion at o ; for, BO and bo are parallel, and we S0_ SB Ho " ab' have SB BC Sb be' AB BC ab ~ bo' BC CD be "■" cd 2°. Since ab is parallel to AB, be to BC, cd to CD, &c., the angle aba = ABC, the angle bed = BCD, &c. Besides, since the triangles SAB, Sa& are similar, we have the pro- portion — J- = -—r- ; and from the similar triangles SBC, ab DO SJc, we have Hence, We shall also have -7— = — t> &c. Hence, the poly- ■gons ABODE, abede have their angles equal, each to each, and their homologous sides proportional ; they are, there- fore, similar. Corollary. Let SABCDE,TXYZ be two pyramids which have the same al- titude, and vifhose bases are situa- ted in the same plane. If these pyramids be in- '^ tersected by a plane drawn par- allel to the planes of the bases, the sections abode, xyz will be formed, and these sections will be to each other as the bases ABODE, XYZ. BOOE VI. 189 For, the polygons ABODE, ahcde, being similar, wo have ABODE AB But, Hen ce, abode r ab AB SB SO ah "" SS ~ So' ABODE 2 SO abode So hall liave XYZ SO 2 ^J/^ So' > ABODE XYZ Hence, , ^ _ abode xyz If the bases are equivalent, the sections made at the same distance from the bases will be equivalent also. Peoposition XIV. — Theorem. Two triangular pyramids which have equivalent bases and equal altitudes, are equivalent. Let SABO, sabc be two pyramids, having their bases ABC, ahc equivalent, and situated in the same plane. 190 ELEMENTS OF GEOMETRY. Let TA be the common altitude of the pyramids. If these pyramids are not equivalent, let sabc be the smaller, and let Ax be the altitude of a prism, which, constructed on the base ABC, will be equal to their difference. Divide the common altitude AT into equal parts, each of which is smaller than Aa;, and let h be one of these parts ; through the points of division of the altitude draw planes parallel to the plane of the bases: the sections made in the two pyramids by these planes will be equiv- alent (Book yi., Prop. XIII.), viz., DEF equivalent to def, GHI to ghi, &c. On the triangles ABC, DEF, GHI, &c., as bases, construct the exterior prisms which have for their edges the parts AD,'DG, GK, &c., of the edge SA; and, likewise, on the triangles def, ghi, Mm, &c., taken as bases, construct, in the second pyramids, the interior prisms, having for their edges the corresponding parts of the edge sa : all of these partial prisms will have h for their common altitude. The sum of the exterior prisms of the pyramid SABC is greater than this pyramid; the sum of the interior prisms of the pyramid sabo is smaller than this pyramid ; hence, the diiference between the sum of the exterior and the sum of the interior prisms must be greater than the difference between the two pyramids. Now, commencing with the bases ABC, abc, the second ex- terior prism DEFG is equivalent to the first interior prism ■ def a, since their bases are equivalent ; and they have the same altitude Ic. For a like reason, the third exterior prism GIIIK is equivalent to the second interior prism ghid, the fourth exterior to the third interior, and so on, to the last in both. Hence, each of the exterior prisms of the pyramid SABC, with the exception of the first ABCD, has its cor- I'csponding interior prism in the pyramid sohc. Therefore," the prism ABCD is the ditferciice between the sum of the exterior prisms of the pyramid SABC, and the sum of the BOOK VI. 191 interior prisms of the pyramid sdbc; but the difference between these two sums is greater than the difference between the two pyramids ; hence, the prism ABCD taust be greater than the prism ABCX : but this is impossible, since the prism ABCX, having the same base ABC, and an altitude Ao; greater than k, must be greater than the prism ABCD. Hence, the supposed inequality of the two pyramids cannot exist ; and therefore, two pyramids with equivalent bases, and equal altitudes, are equivalent. Pkoposition XV. — Theoekm. Every triangular pyrarmd is the third part of tlie tri- angular pnsm which has the same hose and the same altitude. Let SABC be a triangular pyramid, and ABODES a triangular prism, having the same base and the same altitude. If the pyramid SABC be taken from the given prism, there will remain the solid SACDE, which may be consid- ered as a quadrangular pyramid with its vertex at S, and which has for its base the parallelogram ACDE; draw the diagonal OE ; the plane SCE, drawn through CE, will divide the quadrangu- lar pyramid into two triangular pyramids SACE, SDCE. These two pyramids have for a common altitude the per- pendicular let fall from the vertex S on the plane ACDE ; they have equal bases, since the triangles ACE, DCE are the halves of the same parallelogram ; tlierefore, the two pyramids SACE, SDCE are equivalent ; but the pyramids SDCE, SABC have equal bases, ABC, DES ; they have also the same altitude, for this altitude is the distance be- tween the two parallel planes ABC, DES ; hence, the 192 ELEMENTS OF GEOMETRY. two pyramids SABC, SDCE are equivalent ; but we have shown that the pyramid SDCE is equivalent to the pyra- mi(i SAGE; hence, the three pyramids SABC, SDCE, SACE, of which the prism ABD is composed, are equiv- alent to each other. Therefore, the pyramid SABC is the third part of the prism ABD, which has the same base and the same altitude. Corollary. The measure of a triangular pyramid is equal to the third part of the product of its hasehy its alti- tude, or to the product of its hose hy one third of its alti- tude. Peoposition XVI. — Theoeem. Every pyramid S ABODE has for its rneasure the third part of the product of its base hy its altitude ; or, the pro- duct of its iase hy one third of its altitude. For, drawing the planes SEB, SEC through the diago- nals EB, EC, we will divide the polygonal pyramid SABCDE into triangular pyramids, which will all have the common al- titude SO. But, by the preceding theorem, the measure of each of these pyramids is obtained by multiplying the bases ABE, BCE, CDE by the third of the altitude SO ; hence, the sum of the triangular pyramids, or the polygonal pyramid SABCDE, ^° will have for a measure the sum of the triangles ABE, BCE, CDE, or, the polygon ABODE, multiplied by ^SO ; therefore, every pyramid has for its measui'e the third part of the product of its base by its altitude. Corollary I. Every pyramid is the third part of the prism having the same base and the same altitude. BOOK VI. 193 Cm'ollary II. Two pyramids of the same altitude are to each other as their bases; and two pyramids having the same base are to each other as their altitudes. Scholium. We may calculate the solid content of any polyedral body by decomposing it into pyramids ; and this decomposition may be made in several ways. One of the sirnplest methods is to pass planes through the vertex of the same solid angle : we shall then have as many pyra- mids as there are faces in the polyedron, except those which form the solid angle through which the planes are drawn. These pyramids might themselves be decomposed into triangular pyramids, by dividing their bases into triangles. DEFINITION. I. A truncated pyramid \s. that part of a pyramid in- cluded between its base and any section made by a plane cutting the pyramid. When the cutting plane is parallel to the base of the pyramid, the truncated pyramid is called a frustum of a pyramid. n. The altitude of the frustum of a pyramid is the perpendicular distance between the planes of its parallel bases. Peoposition XVII. — Theorem. If a pyramid be cut ly a plane parallel to its base, the measure of the frustum of the pyramid is equal to the sum of three pyramids which have for their common oLti- fude the altitude of tlie frustum, and for their bases, the upper hase of 'the frustum, the lower base of the frustum, and a mean proportional between the two bases. Let SABCDE be a pyramid cut by the plane abd, drawn parallel to its base ; let TFGH be a triangular pyramid 9 194 ELEMENTS OF (JEOMETKY. witli an equal altitude and equivalent base to those of the pyramid SABCDE. "We may suppose the two bases sit- uated in the same plane, and then the plane ahd will, if produced, deter- mine in the trian- gular pyramid a section fgli at the 'F same distance from the common plane of the bases; there- fore, the section fgh is to the section ahcdo as the base FGH is to the base ABD (Bk, VI., Prop. XIII.) ; and since the bases are equivalent, the sections will be also. The pyramids Sabcde, Tfgh, are therefore equivalent, since they have the same altitude and equivalent bases. Tlie whole pyramids SABCDE, TFGH are equivalent, for tin; same reason ; hence, the frustums ASDdab, and YGYLhfy are equivalent, and, therefore, it is only necessar/ to provii the proposition in the case of the frustum of a triangiibir pyramid. Let FGHA/"^ be the frustum of a triangular pyra- mid ; through the points F, g, H, draw the plane FiyH, cutting off from the frnstuin the triangular pyramid ^FGII. This pyramid has for its base the lower base of the frustum ; and for its altitude the altitudi; ^)h of the frustum, since the vertex is in the plane of the upper base fgh. Taking away this pyramid, there will remain the quadrau- BOOK VI. 195 galar pyramid ^AHF, the vertex of which is g. and yAHF its base. Through the points /", ^, H, draw the plane /g'H, dividing the quadrangular pyramid into two trian- gular pyramids J'F/'H, gfKB.. The pyramid g/"/iH has for its base the upper base of the frustum ; and for an alti- tude the altitude of the frustum, since its vertex H is in the lower base. Thus, we have two of the three pyramids of which the frustum is composed. "We have now to con- sider the third, g'¥fK. If we draw ^K parallel to fF, and conceive a new pyr- amid to be formed with K as its vertex, and F/"!! as its base, the pryamids/THK, g^f^ will have the same base F/H; they will also have the same altitude, since the ver- tices g and K are situated on the line ^K parallel to F/", and consequently parallel to the plane of the base : hence, these pyramids are equivalent. But the pyramid ./FKH may be considered as having its vertex at f, and it will then have the same altitude as the frustum. Its base FKH is a mean proportional between the bases FGrB[,/^A; for, tlie triangles FHK, fgh have an angle equal, F —f, and a side equal, FK =fg- We have, therefore, FHK FH We have also fgh fh- FHG FG FG FHK FK fg ' ! But the similar tiiangles FGH, fgh give the proportion, j FG^FH fg . fh' FGH _ FHK Hence, j^gg ~ 'fyT' Therefore, the base FHK is a mean proportional between the two bases FGH,/^/i, and hence, the measure of a frus- tum of a triangular pyramid is equivalent to three pyra- 196 ELEMENTS OF GEOMETRY. mids which have for their common altitude the altitude of the frustum, and for their bases, the upper base of the frustum, the lower base of the frustum, and a mean pro- portional between the two bases. Peoposition XVIII. — Theorem. If a triangular jprism whose iase is ABC, be cut ly a plane DJES, not parallel to its base, the solid ABGDES, which results, will be equivalent to three pyramids whose vertices are D, E, S, and whose common base is ABO. Through the three points S, A, C, pass the plane SAC, cutting off from the truncated prism ABODES the triangular pyramid SABC ; this pyramid has for its vertex the point S, and ABC for its base. Eemoving the pyramid SABC, there will remain the quadrangu- lar pyramid SACDE, which has S for its vertex and ACDE for its base. Through the three points S, E, C, draw the plane SEC, dividing the quadrangular pyramid into .two triangular pyramids SAGE, SCDE. The pyramid SAEC, which has the triangle AEC for its base and the point S for its vertex, is equivalent to a pyr- amid EABC which has AEC for its base and the point B for its vertex. For, these two pyramids have the same base; they have also the same altitude, since the line BS, being parallel to each of the lines AE, CD, is parallel to their plane ACE. Hence, the pyramid SAEC is equiv- alent to the pyramid EABC, which may be considered as having ABC for its base and the point E for its vertex. The third pyramid SCDE may be changed into A8CD ; for, these two pyramids have the same base SCD ; they BOOK VI. 197 Lave also the same altitude, since AE is parallel to the plane SCD. Hence, the pyramid SCDE is equivalent to ASCD. "We may now change the pyramid ASCD into ABCD, for these pyramids have the common base ADD ; they have also the same altitude, since their vertices S and B are situated in a plane parallel to the plane of the base. Hence, the pyramid SCDE, equivalent to ASCD, is also equivalent to ABCD ; and the last may be considered as having ABC for its base, and the point D for its vertex. Therefore, the truncated prism ABCDES is equivalent to the sum of three pyramids which have for a common base ABC, and for their vertices, respectively, the points ■D, E, S. Corollary. If the edges AE, BS, CD are perpendicular to the plane of the base, they will be at the same time the altitudes of the three pyramids which compose the trun- cated prism : so that the measure of the truncated prism may be expressed by ^ABC x AE + ^ABC x BS + -J-ABC X CD; or which may be reduced to JABC X (AE + BS + CD). SYMMETRY OF FIGUEES. Two points are symmetncal with respect to a plane, when this plane is perpendicular at the middle point of the right line which joins these two points. This plane is called the plane of symmetry. Two figures are symmetrical with respect to a plane, when every point of one of the figures has its symmetrical point on the other figure. Pkoposition XIX. — ^Theorem. A right line AB has for a symmetrical line anotlier right line. 198 ELEMENTS OF GEOMETKY. Take on the given right line two points A and B, and determine tlieir symmetrical points A', B', by drawing from the points A and B per- pendiculars on MN, and pro- ducing these perpendiculars, distances equal to AC, BD, respectively. Draw A'B' and CD. To show that every point of the line AB has its symmet- rical point on A'B', draw 01 perpendicular to MN, and produce it until it meets A'B'. If we revolve the quadrilateral ACBD around CD as an axis, until it coincides with CA'B'D, the angles ACD, A'CD being right angles, CA will take the direction CA' ; and since CA = CA', the point A will fall on A'. For a like reason BD will coincide with DB' ; and AB will then coincide with A'B'. Further, since OIC, O'lC are right angles, 01 will take the direction 10' ; and the point 0, as it must be found at the same lime on A'B' and on 10', will fall at 0'. We have then 01 = 10' ; and hence, O' is symmetrical with 0. Corqllwry. The same demonstration proves that the line AB, which connects two points A and B, is equal to the right line A'B', which connects their symmetrical points. The PEOPOsrriON XX. — Theokem. angle of two right lines AB, A G is equal to the angle formed hy their sym- metrical lines A'B', A'O'. We remark, in the first place, that the point of meet- ing A, of the two lines AB, AC, lias for its symmetrical point, the point A', since the symmetrical of A must BOOK vr. 1§9 be found at the same time on A'B' and on A'C. Take, on AB and AC, two points B and C, and let B' and C be their symmetrical points : draw BC, B'C. The triangles ABC, A'B'C have their sides equal, each to each ; therefore, the angle BAG = B'A'C I PROPOSITION XXI. — Theorem. A plane has for its symmetncal surf ace another jplane^ and these two planes form equal ajigles with the plane of symmetry. Let AB be the intersection of the plane MAB with the plane of symmetry ABC, and draw through AB a plane ABM', which shall form with the plane of sym- metry the same angle as the plane MAB. It is required to prove that every point P, of the plane ABM, has its sym- metrical on ABM'. Draw Yp perpendicu- lar to ABC, and produce it until it meets the plane ABM' in P' : then draw^I perpendicular to AB, and join PI, P'l. The two right lines PI, P'l are perpendicular to AB ; and the angles Pip, P'lp are equal, as measuring the equal diedral angles MABC, M'ABC. The right-angled tri- angles PI^, P'lp, are therefore equal, since they have the side \p common, and an acute angle equal : hence, P/? = P'j9. Therefore P' is symmetrical with P. RemarTc. If the given plane is parallel to the plane of symmetry, its symmetrical plane will be also parallel to ABC, and at the same distance from it. 200 ELEMENTS OF GEOMETRY. Proposition XXII. — Theoeem. The dledral angle formed hy two planes ABC, ABD, is equal to the angle formed ly their symmetricals A'B'C\ A'B'D'. We remark, in the first place, that the common inter- section AB of the two planes ABC, ABD, has for its sym- metrical the common inter- section A'B' of the two planes A'B'C, A'B'D'. At the point B, form the plane angle CBD, which measures the diedral angle AB. Form also at the point B', the symmetrical of B, tlie angle C'B'D', measuring tlio diedral angle A'B'. The line BD, in the plane ABD, will have for its symmetrical a right line passing through the point B' and situated in the plane A'B'D'. Besides, since BD is perpendicular to AB, the right line symmetrical with BD will be perpendicular to A'B'. (Book VI., Prop. XX.) Hence, B'D' will be this symmetrical. We will see, also, that B'C is symmetrical with BC ; and therefore the angle CBD = C'B'D. (Book VI., Prop. XX.) Proposition XXIII. — Theorem. Two polyedrons symmetrical with respect to a plane, luive, 1°, their faces equal, each to each I 2°, their homolo- gous solid angles symmetrical. 1°. Let A, B, C, D be the vertices of a face of one of the polyedrons : we know that their symmetricals A', B', C, D' are in the same plane. (Book VI., Prop. XXI.) BOOK Vl. 201 \ ^ Further, the polygons ABCD, A'B'C'D' are equal ; for they have their angles and sides equal, each to each. (Book VI., Pr9p. XX. and XIX.) 2°. Two homologous solid angles B and B' have their , faces equal. (Prop. XX.) Their diedral angles are also equal, each to each. (Prop. XXII.) If, now, we make the face A'B'E' coincide with its equal ABE, so that the other edges of the two solid angles fall on the same side of the common face, we observe that the other plane angles of the two solid angles are disposed in an inverse order. Hence, the solid angle B' is symmetrical with B. Corollary 1. "We conclude, that a polyedron P has only a single symmetrical. For, let P' and P" be two sym- metrical polyedrons of P, constructed with respect to two different planes of symmetry; the faces of these poly- edrons are equal to each other, as being respectively equal to the faces of the polyedron P. Besides, their solid angles being symmetrical with the solid angles of P, will be equal to each other. Hence, the polyedrons P' and P" may be placed one upon the other, so as to coincide throughout : they are therefore one and the same poly- edron. Corollary II. If we decompose a polj'edron P into tri- angular pyramids, which have all for a common vertex one of the vertices of the polyedron, a symmetrical pyra- mid may be formed, in the symmetrical polyedron P', which will correspond to each of the pyramids in the given polyedron. 9* 202 ELEMENTS OF GEOMETRY. We see, thus, that two symmetrical polyedrous may be divided into the same number of tetraedrous which will be symmetrical, each to each. Scholium. Two polyedrons wliich have their faces equal, each to each, and their solid angles symmetrical, are always called symmetrical, whatever be the position they have with respect to each other: but it is proper to remark, that the symmetry exists no further than in the form of the solids. Note. Homologous solid angles are those whose ver- tices are symmetrical. Peoposition XXIV. — Theorem. Two symmetrical polyedrons are equivalent. For, two symmetrical polyedrons may be decomposed into the same number of sym- metrical tetraedrons ; and it only remains to prove that two sym- metrical tetraedrons are eqniv- }^ alent. Let SABC be a tetraedron ; construct its symmetrical, tak- ing as the plane of symmetry one of the faces ABC (Prop. XXIIL, Cor. L); the two te- traedrons SABC, S'ABC are equivalents, for they have the same base ABC, and their altitudes SO, S'O are equal. Two points A and A' are symmetrical with respect to a , third point O, called the centre A O A i ' ' — ^ ' " — ^ of symmetry, when the right line which joins these points is divided into two equal parts at the point O. BOOK VI. 203 Two figures are symmetrical with respect to a point 0, when every point of the one has its symmetrical in the othePi We may establish for symmetry with respect to a point, similar theorems to those which have just been explained. These are left as an exercise for the pupil. 0¥ THE SIMILAErrr OF FIGTJEES. Similar polyedroiis are those which are bounded by the same number of faces, similar each to each, and whose homologous solid angles are equal. Homologous solid angles are those which are formed by similar faces. The homologous lines of two similar polyedrons are those which connect homologous vertices. Pkoposition XXV. — Theoeem. If the edges TF, TG, TH, of the ietraedron TFOE, le divided in the same ratio at the^pointsf g, h, and the lines fff} fK ff^j ^^ joined, the ietraedron Tfgh thus formed is similar to the first. For, the triangles Tfg, TFG are similar, having an angle in each equal, and the sides includ- ing the equal angle proportional ; for a like reason, T^A is similar to TGH, and T/A to TFH. Besides, the lines fg, gh, being parallel to FG and GH, respectively, the plane fgh is parallel to the plane FGH, and the triangle y^A is similar to ^° the triangle FGH. (Book VI., Prop, xm.) Finally, the two homologous solid angles G, g, are equal ; for, since theii- faces are similar, their plane angles are equal. 20i ELEMENTS OF GEOMETRY. eacli to each, and these plane angles are similarly situated ; hence, the tetraedrons have their faces similar, and their liomologons solid angles equal, and they are, therefore, similar. Scholium. Wq may remark that two similar tetrae- drons have their homologous edges proportional. Eeciprocally, when two tetraedrons have their edges proportional, and similarly disposed, they are similar ; for, from the proportionality of the edges, we deduce at once the similarity of the faces ; and the faces being similar and similarly situated, the homologous solid angles are equal, since they have their plane angles equal, each to each, and similarly situated. Peoposition XXVl. — Theoebm. .Two tetraedrons SABC, TDEF are similar^ when they have a diedral angle in each equal, and their equal angle included lyy similar faces, similarly situated. Suppose the diedral angle SB = TE ; the triangle SAB similar to TDE, and SBC similar to TEF. The solid angles S and T are equal, since they have an equal diedral angle in each, included by two equal faces, and similarly situated. Hence, the angle ASC-DTF. Further, since the triangles ASB. and DTE are similar, and also SBC and TEF, we have the proportions SB_ AS J SB_SC. TE ~ DT' TE ~ TF ' BOOK VI. 205 Hence, AS^SC DT ~ TF" The triangles ASO, DTF are therefore similar, having an angle in each equal, included by proportional sides. We might see, in like manner, that the solid angles B and E are equal, and that ABC is similar to DEF. Fi- nally, the solid angles A and C are respectively equal to the solid angles D and F, having the plane angles equal, each to each, and similarly situated ; hence, the tetraedrons are similar. Peoposition XXVII. — Theorem. Two nmilar polyedrons may ie decomposed into the same numier of similar tetraedrons, similarly situated. Decompose into triangles the faces of the polyedron SEFGDABO, not adjacent to the vertex S; these trian- gles will be the bases of the tetraedrons, which will have S for their common vertex, and will together form the given polyedron. Decompose, in like manner, the faces of the polyedron sefgddbc, not adjacent to s, the homologous vertex to S, and join the point s with the vertices of these triangles; this second polyedron will be decomposed into tetraedrons : it is required to show that these tetraedrons are respect- ively similar to those which form the iirst polyedron. 206 ELEMENTS OF GEOMETET. If we compare the tetraedrons SDCA, sdoa, we see that the triangles SDA, CDA are respectively similar to the triangles sda, cda, since the faces EDAS, edas are similar, and also the faces CDAB, cdab ; besides, the diedral angle DA is equal to the diedral angle da, since the faces of the two polyedrons are equally inclined ; hence, the two te- traedrons SDCA, sdoa are similar, having a diedral angle in each equal, and the equal angles included by similar faces, similarly situated. If we pass to the tetraedrons SDCF, sdcf, we see that the triangles SDC, sdc are similar, as homologous faces of similar tetraedrons : FDC is also similar to fdc, since the polygons FEDC, fedc are similar. Further, the diedral angles FDCA, fdca, are equal, by hypothesis ; and the diedral angles SDCA, sdoa are equal, on account of the similarity of the tetraedrons SDCA, sdca : hence, the diedral angles FDCS, fdcs are equal, being the differences between equal diedral angles. Hence, finally (Book VI., Prop. XXVI. ), the tetraedrons SDCF, sdof a,ve similar; and so for all the others. Remarh I. "We remark that the preceding decomposi- tion may be effected by beginning with any two homolo- gous vertices. Remark II. "We deduce from the theorem just demon- strated, that in two similar polyedrons, two right lines A, a, which join homologous vertices, are proportional to two homologous edges B, i of the two polyedrons. We shall have, then, — = — a c Besides, in the similar polyedrons, the homologous edges are proportional, on account of the similarity of the faces ; we have also, — = -^ ; CO and finally, ~ ~ T' BOOK vi. m Proposition XXVIII. — Theokkm. Two polyecLrons^ composed of the same number of similar tetraedrons, similarly situated, have their faces similar, each to each, and their homologous solid angles equal, and are consequently similar. Let SABC, SADC, SDCF ... be the pyramids of which the first polyedron is composed : sahc, sadc, sdcf, the pyra- mids -which form the second. 1°. The triangles DCA, CAB, which form a face of the firet polyedron, are respectively similar to the triangles doa, cab in the second, on account of the similarity of the teti-aedrons. Besides, the triangles DCA, CAB being in the same plane, the triangles dca and cab are also in the same plane. For, since the tetraedrons SCAD and scad are similar, as well as SABC and sdbc, the diedral angles SCAD, SCAB are respectively equal to the diedral angles scad, scab ; but the sum of the first two is equal to two right angles : hence, the sum of the last two is also equal to two right angles. Therefore, the polygons DCBA, dcba are similar, as being composed of the same number of similar triangles, similarly situated; and the same is true with the other faces, taken two and two. 2°. We see, in the second place, that the diedral angle BA, which is the sum of tlie diedral angles CSAD, CSAB, 208 ELEMENTS OF GEOMETRT. is equal to the diedral angle «a, the sum of the diedral angles csad, csdb, respectively equal to the first; and that, in general, two homologous diedral angles of the two poly- edrons are equal, as being the sums of homologous diedral angles of similar tetraedrons. Hence, the two homologous solid angles A and a are equal, for j;hey have their faces equal, each to each, simi- larly situated, and equally inclined. Scholium. The demonstration which has just been made, justifies the 'definition which has been given of similar polyedrons ; for we may always form polyedrons which shall be composed of tjie same number of similar tetraedrons, similarly situated. For, decompose the polyedron SABDEFG into triangu- lar pyramids, having S for their common vertex ; and let SBDC, SADB, SDAE ... be the tetraedrons of which the polyedron is composed. If we divide the edges, be- ginning at S, proportionally, at the points a, 5, c, d, &c., the tetraedrons shdo, sadh . . . will be respectively similar (Book YI., Prop. XXV.) to the tetraedrons SBDC, SADB . . . and will be similarly situated : their sum will, therefore, compose the second polyedron, which will be similar to the first, by the pre- ceding theorem. This second . polyedron might then be placed in any position whatever with respect to the first. I Peoposition XXIX. — ^Theoeem. Two similar tetraedrons are to each other as the cubes of their homologous edges. B00£ VI. 209 Since the tetraedrons are similar, we may place the smaller on the greater, in such a manner that they may have the common solid angle S ; and tlien the bases ale, ABC will be paral- lel, since the edges SA, SB, SC are divided proportionally at the points «, 5, c. Draw SO perpendicular to the plane of the base ABC. The tri- angles ABC, dibc are similar, and give the proportion ABC^AB^ (1) ahc oh We have also, AB_SA ^^^' SO __ SA ah sa^ so sa Hence, on account of the common ratio, we have SO so Multiplying the proportions (1) and (2), and dividing the terms of the first ratio by 3, we have 4?-(') ABC X ISO ahc X ASo AB ah But ABC X ^SO is the measure of the tetraedron SABC ; and ahc x ^So is the measure of the tetraedron S«5c. Hence, two similar tetraedrons are to each other as the cubes of their honjologous edges. Peoposition XXX. — Theorem. Two similar polyedrons are to each other as the cubes of their homologous edges. 210 ELEMENTS OP GEOMETRY. For, the two similar polyedrons may be decomposed into the same number of similar tetraedrons. Let T, T', T", &c., be the tetraedrons which form the polyedron P; t, t', t" . . . the tetraedrons of ^: let A, A', A" ... be the edges of the tetraedrons T, T', T", &c. ; and a, a', a", &c., the homologous edges in the tetraedrons t, if, t", &c. We shall have T A' T' A" T"' A"^ V il ~ „/3' i'lZ ~ ^"3' *°- And since the homologous edges of similar polyedrons are proportional, we shall have T _ T' _ T" t~ t'~ t'" T + r + T", &c. _ P _ A' BOOK VII. 211 BOOK YII. I THE SPHEEE. I DEFINITIONS. L The sphere is a solid terminated by a curved surface, all the points of which are equally dis- tant from a point within called the cent/re. We may consider the sphere to be generated by the revolution of the semi- circle DAE about its diameter DE ; for the surface described by the curve DAE will have all of its points equally dis- tant from the centre C. n. The radius of the sphere is a right line drawn from the centre to a point of the surface : th^ diameter or axis is a right line passing through the centre, and terminated on both sides at the surface. AH radii of the same sphere, or of equal spheres, are equal. In like manner, all diameters of the same sphere, or of equal spheres, are equal. The diameter is double of the radius. m. A plane is tangent to a sphere, when it has only one point in common with the surface. IV. Two spheres are tangent, when their surfaces have only one point common. Peoposition I. — ^Theoeem. Every section of the sphere made hy a plane is a circle. Let AMB be the section made by a plane in the sphere 212 ELEMENTS OF GEOMETEY. whose centre is C. From the point C draw CO perpen- dicular to the plane AMB, and the lines CM, CE, CB to. the several points of the curve AMB which ter- minates the section. The oblique lines CM, CN, CB, are equal, since they are radii of the sphere; they are, therefore, equally distant from the perpendicular CO. Hence, all the lines OM, ON, OB are equal ; and, there- fore, the section AMB is a circle, having O as its centre. Corollary I. If the section pass through the centre of the sphere, its radius will be equal to the radius of the sphere. Circles whose planes pass through the centre of the sphere are called great circles. All other circles of the sphere are called small circles. All great circles of the same sphere, or of equal spheres, are therefore equal. Corollary II. Two great circles divide each other into two equal parts ; for, their common intersection, passing through the centre, is a diameter. Corollary III. Every great circle divides the sphere and its surface into two equal parts ; for, if after separat- ing the two segments, we apply them on the common base, the two surfaces would coincide in all their points, else there would be points on the two surfaces unequally dis- tant from the centre, which is impossible. Corollary I Y. The centre of a small circle and that of the sphere are on .the same right line perpendicular to the plane of the small circle. Corollary Y. Small circles equally distant from the centre of the sphere are equal. Of small circles unequally distant from the centre of the sphere, the greater is at a shorter distance from the centre. Corollary Yl. Through two given points on 'the sur- face of a sphere, we may pass an arc of a great circle ; for BOOK VII. 213 the two given points and the centre of the sphere give three points which fix the position of a plane. If the two given points were on the extremities of a diameter, these two points and the centre would be on a right line, and an infinite number of great circles might be drawn through them. Corollary Yll. The position of a small circle on the surface of a sphere would be determined by three points of its circumference. Peoposction II. — Theoeem. Eveiy jplane perpendiaular to a radius of a sphere at its extremity is tangent to the sphere. Let FAG be a plane perpendicular to the radius OA, at .jG its extremity. If we take any point on this plane, such as M, and join OM and AM, the angle 0AM will be a right angle ; and hence, the distance OM will be greater than OA. Tlie point M is, therefore, outside of the sphere, and as this is the case with every other point of the plane FAG, it follows, that the plane has only one point, A, common with the surface of the sphere, and is, therefore, tangent to it. (Def. 3.) Reciprocally, every tangent plane FAG is perpendicu- lar to the radius OA at the piiint of contact. For, if we join the centre with any point M of this. plane, OM will be greater than the radius OA, since M is exterior to the sphere. OA is, therefore, the shortest line that can be drawn from the centre to the plane FAG, and is perpendicular to this plane. Corollary. Through a given point on the surface of a sphere only one tangent plane can be drawn. 214 ELEMEKTS OF GEOMETRY. Proposition III. — Theorem. The intersection of two spheres is a circle^ the plane'of which is 'perpendicular to the line connecting their centres, and the centre of the circle is on this line. Through the line OC, which joins the centres of the two spheres, draw any plane whatever. This plane intersects the two spheres in great circles 'which intersect each other at the points A and A', symmetrical with respect to the line OC. If, now, the semi- circles DAE, GAH, revolve around OC as an axis, these two semicircles will generate the surfaces of the two spheres, and the point A will describe their line of intersection. Further, in this revolution, the right line AI does not change its magni- tude, and is all the time perpendicular to OC ; hence, the intersection of the two surfaces is a circumference, the centre of which is I, and the radius AI, and the plane of which is perpendicular to OC. Remark. The two spheres will be exterior or interior, they will be tangent exteriorly or interiorly, or finally secants, according as the two circles DA A', GAA' are exterior or interior, tangent exteriorly or interiorly, or secants. There will then exist for each of these positions of the two spheres, the same relations between the distances of the centres, and the radii of the spheres, as between the corresponding positions of two circles. DEFINITIONS. I. The angle 'between two arcs of great circles is the die- BOOK vir. 2l5 dral angle formed by the planes of these circles. The arcs of great circles are the sides of the angle, and the inter- section of these arcs is the vertex of the angle. II. A spherical triangle is a portion of the surface of the sphere bounded by the arcs of three great circles. These arcs form the sides of the spherical triangle, each of which is always supposed to be less than a semi-circum- ference. The angles formed by these arcs are the angles of the triangle. III. A spherical triangle is rectangular, isosceles, equi- lateral, &c., in the same cases as a rectilinear triangle. lY. A spKencdL polygon is a portion of the surface of the sphere bounded, by the arcs of great circles. The arcs of the great circles form the sides of the spherical polygon. A spherical triangle is a spherical polygon of three sides. Convex spherical polygons only are considered ; these are such, that the plane of each side leaves the polygon on the same side of its direction. PEOPOsmoN IV. — Theorem. In every spherical triangle ABC, either side is less than the sum of the other two. Let O be the centre of the sphere, draw the radii OA, OB, OC. If the planes AOB, AOC, COB be supposed to be drawn through 7^ these radii, they will form at a solid angle, and the angles AOB, AOC, COB will have for their measures the sides AB, AC, BC of the spherical triangle ABC. But each of these plane angles which form the solid angle is less than the sum of the other two ; hence either o side of the triangle ABC is less than the sum of the other two. 216 ELEMENTS OF GEOMETET. PEOP<'>srrioN V. — Theoeem. The sum of the three sides of a spherical t/riangle is less than the circumference of a great circle. Let ABC be any spherical triangle ; produce the sides AB, AC, until they meet again in D. The arcs ABD, ACD will be semi-circumfer- ^ ences, since two great circles always divide each other equal- ly. (Book YII., Prop. I.) ^But in the triangle BCD, we have ^ ^ the side BC < BD + CD (Book VII., Prop. IV.) ; adding AB + AC to both members, we have AB + AC + BC < ABD + ACD, which is less than a circumference. Hernarh. In order to construct a spherical triangle with three given sides, it is necessary and it is sufficient that the sum of the three sides be less than a circumference, and that the greatest side be less than the sum of the other two ; for, these are the conditions necessary to construct a solid angle, the three faces of which have for their meas- ures the three given sides. If we place the vertex of this solid angle at the centre of the sphere, the faces of the solid angle will form on the surface of the sphere the required spherical triangle. Proposition VI. — Theorem. The suin of the sides of a spherical polygon is less than a circumference of a great circle. Let ABCDE be a spherical poly- gon. From the centre of the sphere 0, draw the radii OA, OB, 00, OD OE ; we shall form at O a convex solid BOOK 217 atigle, the plane angles of wliich, AOB, AOC, &c., have for their measures the arcs AB, BC, CD, &c. But, the sum of the plane angles which form the solid angle is less than i right angles ; hence, the sum of the arcs AB, BC, &c., is less than a circumference. DEFINITIONS. I. The pole of a circle of the sphere is the extremity of the diameter perpendicular to the plane of the circle. II. Every circle of the sphere has two poles. m. All circles of the sphere, the planes of which are parallel, have the same poles. Peoposition VII. — Theoeem. Every point of the circumference FWO of a circle of the sphere is equally distant from the pole of this circle. F*']-, if we draw the radii OF, ON, OG, in the circum- ference FNG, and also the right lines DF, DIST, DG, the right-angled triangles DOF, DOl^, DOG, &c., will be equal ; for they have the side DO common, and the lines OF, ON, OG are equal, as radii of tlie same circle ; hence, DF = DN=DG. We see, also, that the arcs of the great circles FD, DIST, DG are equal, because they are subtended by equal chords, and the planes of these great circles are perpendicular to the plane of the circle FJSTG, for they all pass through the line DO, whicii is perpendicular to the plane of this circle. The reasoning given for the circle FNG is evidently ap- plicable to the pole of the great circle AMB ; but, in this case, the right angles DCA, DCM, DCB, being at the 10 21S ELEMENTS OF GEOME'TET. centre of the great circles DAE, DME, &c., the arcs DA, DM, DB are quadrants. Scholium. The properties of the poles enable us to trace on the surface of the sphere arcs of a circle with the same facility as on a plane surface. For this purpose we use a pair of dividers, called spjier- ical compasses, and such a disposition is given to the two legs as permits them to be inclined under any angle. It is evident that if one of the points of the compasses be placed at D, and the other at F, while the latter moves around D, the extremity F will describe the circumference FNG. If we wished to describe a great circle from the point D, as a pole, it would be necessary to make the distance between the two points of the compasses equal to the chord of a quadrant ; and to have this chord we must know the radius of the sphere. PfiOPOsiTioN VIII. — Peoblem. Having given a sphere, to find its radius. With any arbitrary opening of the compasses AC de- scribe on the surface of the sphere a circumfer- ence ODE ; mark the points C, D, E on tJiis circumference, and mea- sure with the compasses the rectilinear distances CD, DE, CE; finally, construct, on a plane, a triangle with CD, DE, CE, as its sides; the radius of the circle circumscribing this triangle will be the radius of the circle CDE. Now, conceive a great circle ACBE to be passed through BOOK VII. 219 the diameter AB, and that the right lines CA, CB, and CO are drawn. In the right-angled triangle CAO, we know the hypothenuse AC and the side CO ; we might then construct on a plane a ti'iangle C'A'O' equal to CAO ; besides, the line CB being perpendicular to CA, if we draw CB' perpendicular to A'C, the line A'B' = AB will be the diameter of the sphere. Pkoposition IX. — Problem. To trace on the surface of a sphere the circumference of a great circle that shall pass through two given points A and B. From the points A and B, as poles, with an opening in the com- passes equal to the chord of a quad- rant, describe two great circles, in- tersecting each other in P; the point P will be the pole of the arc of the great circle AB, and will serve to describe this arc. Peoposition X. — Problem. From a point A on the surface of a sphere, to draw a great circle perpendicular to a given great circle. From the point A as a pole, with an opening in the compasses equal to the chord of a quadrant, describe a great circle cutting the circle CMD in S. Then, with the point S as a pole, with an opening SA, describe the great circle AM ; it will be perpendicular to CMD. (Eook YIL, Prop. YII.) 220 ELEMENTS OF GEOMETRY. Proposition XI. — Theoeem. The shortest path from the point A to the point B on the surface of a sphere is the arc of a great circle less than a semi-circumference joining these points. The demonstration of this theorem is founded npon the two following lemmas, which we will first explain. Lemma I. The shortest path from the pole P of the cir- cumference ABD to any point of this circumference, is also the shortest path from the point P to any other point of the same circumference. For, draw through the pole P the arcs of great circleef PA, PB, to the points A and B of the circumference ABD; and let PGB kD be the shortest path on the surface of the sphere from the point P to the point B. If the hemisphere which is on the right of the great circle PBE be re- volved about PE until the great circle PBE coincides with the great circle PAE, the arc PB will coincide with its equal AP, and the hemisphere PBEHD will coincide with the hem- isphere PBEHA ; but in this position, the line PGB will still represent the shortest path from P to B, and it will, therefore, equally be the shortest path from P to A. Lemma II. Let AB, AO he two arcs of great circles., each less than a semi-drcumference; and let AC < AB, ^ the shortest path from A to C will he less than that from A to B. For, describe a circumference CM from A as a pole, cutting the arc AB at a point I between A and B ; and let AMB be the shortest line between A and B. This line will meet the circle CI in a point M, and the line AM will BOOK VII. 221 be the shortest path from A to M ; for, if there were a shorter path between these points, AMB would not be the shortest path from A to B, which is contrary to the hy- pothesis. Besides, from the preceding lemma, the shortest path from A to M is also the shortest path from A to C. Hence, the shortest path from A to C is less than that from A to B, since AC = AM, and AM < AB. Now, to apply these lemmas to the enunciated theorem, let AB be the arc of a great circle, less than a semi- circumference, joining the points A and B; and suppose that there exists, outside of this arc, a point C on the shortest line between A and B. Draw the arcs of great circles AC, BC, and take AD = AC. We shall have (Book VII., Prop. IV.) AB < AC-f CB. Taking AD = AC from both, there remains DB < BC. But, by Lemma I., the shortest path from A to D is the same as that from A to C. Hence, since the point C belongs to the shortest line from A to B, the distance from C to B must be shorter than that from D to B, which is absurd, from Lemma II., since BC > BD. Hence, no point of the shortest distance between A and B can be outside of the arc AB ; hence the arc AB is the shortest path. Proposition XII. — ^Theorem. The angle BAC, formed hy the arcs of two great circles, has for its measure the arc of the great circle B C, described from A as a pole, and limited hy the sides of the angle. For, draw the radii OB, 00 ; the arcs AC, AB, being quad- rants, the angles AOB, AOC, will be right angles ; and BOC will be the plane angle corresponding to 222 ELEMENTS OF GEOMETBT. the diedral angle DBAC ; but the angle at the centre BOO ■ is measured by the arc BO. Hence, the arc BO is also the measare of the diedral angle DBAC, which is the angle of the two arcs of great circles. Corollary. The angles of spherical triangles may be compared with each other, by comparing the arcs of great circles described from their vertices as poles, and limited by the sides of the spherical triangles produced. Thus, it is easy to make an angle equal to a given angle. Scholium. The opposite vertical angles AOO and BOJST, are equal ; for, each is the angle formed by the two )|b planes ABC, OOK We see, also, that the sum of the two adjacent angles AGO, GOB, formed by the intersection of two arcs, is equal to two right angles. DEriNITIONS. K from the vertices A, B, 0, of a spherical triangle ABO, as poles, the arcs of three great circles EF, FD, DE be described, these arcs will form a second triangle DEE, which is called the polar triangle of the triangle ABO. The vertex homblogous to A, is determined by the intersection of the arcs described from B and 0, ■ as poles. These arcs, it is true, in- tersect each other in two points, but it is only necessary to take that which is on the same side of the side BO that A is, and the same with the other vertices, SOOE tti. ^%% Pkoposition XIII. — Thkoeem. If the spherical triangle ABC has DEF for its polar triangle, reciprocally, ABO will he the polar triangle of DEF. For, the point A being the pole of the arc EF, the dis- tance AE is a quadrant ; the point C heing the pole of the arc DE, the distance CE is 'also a quadrant ; hence, the point E is at a quadrant's distance from each of the points A and C ; it is, therefore, the pole of the arc AC. We might show, in like manner,, that D is the pole of the arc EC, and F the pole of the arc AB ; hence, ABC is the polar triangle of DEF. Peoposition XIV. — Theorem. Having given two polar triangles ABC, DEF, the meas- ure of an angle in either of these triangles will he a semi- circumference, minus the side opposite to it in the other triangle. Produce, if necessary, the sides AB, AC, until they meet EF in G and H. Since the point A is the pole of the arc GH, the angle A will have for its measure the arc GH. But the arc EH is a quadrant as well as GF, since E is the pole of AH, and F is the pole of AG ; hence, EH + GF =^ semi-circum- ference ; but EH -f GF is.the same thing as EF -f GH ; hence, the arc GH which measures the angle A is equal to a semi-circumference minus the side EF. In like manner, 224 ELEMENTS OF GEOMETRY. the angle B will have for its measure ^ circf. — DF, aud the angle C, I circf. — DE. This property is reciprocal in both triangles, since each is described in the same manner, one by the other. Thus, the angle D has for its measure the arc MI ; but MI + EC = MC + BI = ^ circf. ; hence, the arc MI, which meas- ures the angle D — ^ circf. — BC. In like manner, the measure of E will be ^ circf. — AC, and that of F will be ^ circf. - AB. Scholium. If from the centre of the sphere radii be drawn to the vertices of the triangles ABC, DEF, we shall form two triedral angles, the plane angles of which have for measures the sides of the splierical triangles, and the diedral angles of which correspond with the angles of the given triangles. -But, it results from the theorem which has just beeu demonstrated, that in these two triedral angles, the diedral angles of one are supplements of the plane angles of the other, and reciprocally; therefore, these triedral angles are supplementary. DEFINITIONS. Let ABCD be any spherical polygon ; draw from the centre of the sphere radii to the vertices of this polygon, and produce them until they meet the sphere on the opposite side at A', B', C, D', and describe the arcs A'B', B'C, A'D', CD'. The solid angles formed at are symmetrical ; consequently, their plane angles and their diedral an- gles are respectively equal. Hence, the spherical polygons ABCD, A'B'C'D' have their parts also equal. Still, these polygons cannot be placed one upon the other ; for, if we place the side CD' on its equal CD, 60 that the other sides of the two polygons shall fall on the same side of CD, the point D' will fall at C, and in following the polygons in the same direction, commencing with C, the sides and angles will be presented in an in- verse order. These spherical polygons are called symmetrical, what- ever positions be given to them on the surface of the sphere. Peoposition XY.^Theoeem. Two spherical triangles, on the same sphere, or on equal spheres, are equal in all their parts, when they have two sides and the included angle of the one, equal to two sides and the included angle of the other, each to each. Let the side AB = EF, the side AC = EG, and the an- gle BAG = FEG. The triangle EFG may be applied to the triangle ABC in the same manner that plane triangles have been ap- plied which had two sides and the included angle in each equal. Then all the parts of the triangle EFG will be equal to those of the triangle ABC ; that is, besides the parts supposed to be equal, we shall have BC = FG, the angle ABC = EFG, and the angle ACB = EGF. If the equal sides of the two triangles were inversely disposed with respect to the two equal angles, we might apply EFG to the symmetrical of ABC, and we should have the same result. Proposition XVI. — ^1'heoeem. Two triangles, on the same sphere, or on equal spheres, are equal in all their parts, wh^n they have two angles and 10* 226 ELEMENTS OF GEOMETRY. the included side of the one, equal to two angles and tlie included side of the other, each to each. For, one of these triangles may be applied to the other, or to its symmetrical, as plane triangles were applied. (Book I., Prop. VI.) Peoposition XVII. — Theoeem. If two triangles, on the same sphere, or on equal spJieres, have three sides of the one equal to three sides of the other, mch to each, the angles opposite equal sides will also he equal. Join the centres 0, O' of the spheres with the vertices of the two triangles : we shall thus form two triedral angles, the plane an- gles of which being measured by the sides of the spherical tri- angles, are respect- ively equal. But we have shown that in this case the diedral angles opposite to equal faces are equal. Hence, in the two spherical triangles, the angles opposite to the equal sides are equal. Peoposition XVIII. — Theoeem. Jn every isosceles . spherical triangle the angles opposite to the equal sides are equal. Let *the side AB — AC ; we shall have the angle C = B. For, if from the vertex A we draw the arc AD to the middle point D of the base, the two triangles ABD, ADC will have the three sides equal, each to each ; viz., AD common ; BD = DC, BOOK VII. 227 and AB = AC. Hence, by the preceding theorem, these triangles will have equal angles, and we shall have B — G, The same demonstration proves that the angle BAD = DAG, and that the angle BDA = ADC. Hence, BDA and ADG are right angles : therefore, if from the vertex of an isosceles spherical triangle an arc he drawn to tht middle point of the lase^ it will ie perpendicular to the base, and will divide the vertical angle into two equal parts. Scholium. It follows, from this theorem, that the sym- metrical of an isosceles spherical triangle is €qual to it by superposition. Pkoposition XIX. — Theokem. Jf two angles of a spherical triangle ie equal, the sides opposite the equal angles are also equal. Let the angle B = G ; then will AG = AB ; for, if the side AB is not equal to AG, let AB be the greater : take BO = AC, and join OC. The two sides BO, BC are equal to the two sides AC, BC. The angle OBC is equal to ACB. Hence, the triangles BOG, ACB have their other parts equal (Book VII., Prop. XV.) ; and we have the angle 0GB = ABC. But the ailgle ABC = ACB, by hypothe- sis : hence, we shall have OOB = ACB, which is impos- sible. Therefore, the side AB cannot be different from AC. Hence, AB, AC, are equal. Peoposition XX. — ^Teceoeem. In every spherical triangle, the greater side lies opposite to the greater angle i and reciprocally, the greater angle lies opposite to the greater side. 228 ELEMENTS OF GEOMETRY. 1°. Let angle A > B ; make the angle BAD = B : we shall have AD = DB, by the last theorem. But AD + A DC is greater than AC : substituting DB for its equal AD, we shall have DB + DC > AC ; or BC > AC. 2°. If we suppose BC > AC, the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC = AC ; and if we had BAC < ABC, we should have BC < AC ; both of which results are contrary to the supposition. Hence, the angle BAC is greater than ABO. Peoposition XXI. — Theokem. If two triangles on the same sphere, or on equal spheres, have the three angles of the one equal to the three angles of the other, each to each, the sides which lie opposite to the equal angles will also he equal. Let A and B be the two given triangles, P and Q their polar triangles. Since the angles in the triangle A are respectively equal to those of the triangle B, the sides in the polar triangle P will be respectively equal to the sides of the polar triangle Q. (Book VII., Prop. XIY.) But if the polar triangles P and Q have their sides equal, each to each, their angles will also be equal (Book VII., Prop. XVII.) ; and finally, if the polar triangles P and Q have their angles equal, each to each, the sides of their polar triangles A and B will also be equal, each to each (Book Vn., Prop. XIV.), which was to be proved. Scholium. This proposition does not hold good in plane triangles ; for, in plane triangles, we can only conclude that the sides are proportional, when the angles are equal, each BOOK VII. 229 to each. The reason for this difference between rectilinear triangles and spherical triangles may be readily accounted for. In this proposition, as well as in Prop. XV., XVI., XVII., and XX., it is expresslj'^ stated that the given tri- angles are traced on the same sphere, or on equal spherei^ But similar arcs are proportional to their radii. Hence, on two equal spheres, two triangles cannot be similar without being, at the same time, equal. It is not, therefore, sur- prising tliat the equality between the sides should result from the equality between the angles, in two spherical triangles. It would be otherwise, if the triangles were traced on unequal spheres ; for then, the angles being equal, the tri- angles would be similar ; and the homologous sides would be to each other as the radii of the spheres. Proposition XXII. — Theoeem. 1°. The sum of the angles of a spherical triangle is greater than tv:o and less than six right angles. 2°. The smallest angle augmented hy two right angles is greater than the sum of the other two. 1°. For, the measure of an angle of a spherical triangle is a semi-circumference minus the opposite side of the polar triangle. (Book VII., Prop. XIV.) Hence, the sum of the three angles has for a measure three semi-circumferences minus tlie sum of the three sides of the polar triangle. But this last sum is greater than zero and less than a cir- cumference; taking it, then, from three semi-circumfer- ences, the remainder will be less than three semi-circtim- ferences and greater than one semi-circumference. Hence, the sum of the angles of a spherical triangle is less than six and greater than two right angles. Corollary. A spherical triangle may have two or three 230 ELEMEKTS OF GEOMETBY. of its angles right angles, or "two or three of its angles obtuse angles. If the triangle ABC have two of its angles right angles, viz., B and C, the vertex A will be the pole of the base BC ; and the sides AB and AC will be quadrants. Such a triangle is called a U-reotangular triangle. If the triangle has all three of its angles right angles, it is called a tri-rectangula/r tri- angle. Each of the sides of this triangle is a quadrant. When MN is a quadrant, AMN is a trl-rectangnlar triangle. The tri-rectangular triangle is contained eight times in the sur- face of the sphere. 2°; Let A, B, C be the angles of the triangle, and A < B and < C. The sides of the polar triangle will be respectively 180° — A, 180° — B, 180° - C (Book VII., Prop. IV.) ; but we have 180° - A < 180° - B + 180° - C. Adding A + B + C to both members, we have B + C<180° + A. "With three angles A, B, C, which fulfil the conditions enunciated in this theorem, we may form a spherical tri- angle ; for, these conditions are necessary and sufiScient to construct a triedral angle with A, B, C as the three diedral angles. Scholium. "We have supposed that the spherical tri- angles have always their sides less than a semi-circumfer- ence. We may remark, however, that there are spherical triangles some of the sides of which are greater than a semi-circumference, and certain angles greater than two right angles. BOOK VIl. 231 For, if we produce the side AC an entire circumference ACE, that portion of the hemisphere which remains, after ^ taking away the triangle ABC, is yf" ^\ a new triangle which may also be / \ designated by ABC, and of which AL__Jg \ the sides are AB, BC, and AEC. "We see that the side AEC is greater than a semi-circumference ; and the opposite angle B >• two right angles. We exclude such cases as this from the definition of spherical triangles, because the resolution of such triangles, and the determination of these greater sides and angles, may always be reduced to the resolution of triangles, which are embraced in the definition. "We see, also, that if we know the angles and sides of the tri- angle ABC, we shall know the angles and sides of the triangle which remains when ABC is taken from the hemisphere. DEFmmONS. 1. A lune is that portion of the surface of the sphere bounded by the semi-circumferences of two great circles. n. A spherical ungula or wedge, is that portion of the volume of the sphere bounded by two semi great circles, and having the lune formed by the same circles for its in. A spherical jpyramiid is that portion of the volume of the sphere bounded by the planes of a solid angle, having its vertex at the ceiitre, and the base of which is the spherical polygon formed by the same planes. "When two spherical polygons coincide, the pyramids to which they belong will also coincide. IV. Symmetrical spherical pyramids are those which have symmetrical spherical polygons for their bases. 232 ELEMENTS OF GEOMETBY. Proposition XXIII. — Theokem. The surface of a lune is to the surface of a sphere as the angle of the lune is to four right angles ; or as the arc which measures this angle is to the circumference. Let us suppose, in the first place, that the arc MN is com- mensurable with MNPQ, and that in dividing the cir- cumference into 48 equal parts, the arc MN contains 5 of these parts. The ratio of the arc MN to the circumference is -^. Pass through the diameter AB and the points of division a series of planes; we shall thus form on the surface of the sphere 48 lunes, all equal to each other, as having the same angle ; and 5 of these luneS; will be contained in AMBN. The ratio of the lune AMBN" to the surface of the sphere will therefore be -f-^, and will be the same as that of the arc MN to the circumference MNPQ. If the arc MN be incommensurable with the circumfer- ence, we might prove, bj known course of reasoning, that the surface of the Inne will always be to the surface of the sphere, as the arc MN is to the circumference. Measure of the Lune. Let L, L' represent the surfacea of two lunes, having A and A' as their respective angles, and S be the surface of the sphere to which they belong. We have, by this theorem, L A , L' A' -=_, and g^ = -. Hence, j-, = j,- (1) If, therefore, we wish to estimate the value of the sur- BOOK vii. 233 face of any lune by comparing it with that of an assumed lune taken as the unit of measure, we see, by the aboTe proportion, that it is only necessary to find the ratio between the angles of these lunes. If we take the lune whose 8(ngte is a right angle for the anit of measure, proportion (1) becomes E' = r (^) Thus, the ratio of any lune to the right lu^e, is equal to the ratio of its angle to a right angle : helice, when L' =1, the measure of any lune is equal to its angle. If we take for the unit of measure of any portion of the spherical surface the tri-rectangular triangle T, which is half of the right lune, we shall have, by substituting 2T for L', in (2), ^ = ^. Hence, multiplying each member by 2, we conclude that L_2A T~ 1 ■ "Which shows that the ratio of the surface of a lune to that of the tri-rectangular triangle, is equal to the ratio of double the angle of the lune to one right angle : or, ia other words, the measure of the surface of the lune is equal to double its angle multiplied hy the tri-rectangular tri- angle / or simply double the angle, when the tri-rectangular triangle is counted as unity. Scholium. We see, also, that the volume of a spherical ungula is to the volume of the sphere as its angle is to four right angles, and that the ungula has for its measure thje angle of the ungula, in taking for the unit of volume the right ungula, and for the unit of angle the right angle ; or, in other words, double of its angle, in taking for the unit of volume the tri-rectangular pyramid which is half the right ungula, and for the unit of angle the right ^ngl«j. S84 ELEMENTS OF GllOMETR'?. Pkoposition XXIV. — Thkokem. Two symmetrical spherical triangles a/re equal in surface. Let ABC be a spherical triangle, and A'B'C its sym- metrical. Take the pole P of the small circle passing through the points A, B)C. The cir- cumscribing circle to the triangle ABC must be a small circle of the sphere, for if it were a great circle, the three sides AB, BC, AC, would be in the same plane, and the triangle be reduced to one of its sides. The arcs of the great circles PA, PB, PC, are equal. Draw the diameter POP', and the arcs of the great circles P'A', P'B', P'C. The angles POA, P'OA' are equal, because they are opposite vertical angles ; hence, the arc PA = P'A'. For a like reason, PB = P'B', and PC = P'C; and since PA = PB = PC, we have P'A' = P'B' = P'C. The triangles APC, A'P'C have the three sides of the one equal to the three sides of the other, each to each ; they are also isosceles. Hence, they may be placed one on the other, and are equal. We see, in like manner, that the triangle CPB is equal to the triangle CP'B', and tliat the triangle APB is equal to A'P'B'. Hence, the triangle ABC, which is the sum of the triangles APC, CPB, APB, is equal, in surface, to the triangle A'B'C, which is the sum of the triangles A'P'C, CP'B', A'P'B'. Remark. "We see, in like manner, that two symmetrical spherical pyramids, which have ABC, A'B'C for their bases, are equivalent. . Pkoposition XXV. — Theorem. If two great circles A OB, COD intersect each other on the hemisphere AOCBD, the sum of the apposite tria/ngles BOOK VII. 235 AOC, BOD loill be equivalent to the lime which has BOD for its angle. For, in producing the arcs OB, OD until they meet at ]Sr, in the other hemisphere, OBN will be a semi-circum- ference, as well as AOB : tak- ing OB from each, we have BN = AO. For a like rea- son, we have DN = CO, and BD = AC. Hence, the two triangles AOC, BDIST have their sides equal, each to each ; besides, they are sym- metrical. Hence, they are equal in surface (Book VII., Prop. XXI-Y.), and the sum of the triangles AOC, BOD is equivalent to the lune OBISTDO, whose angle is BOD. Scholium. It is also evident, that the two spherical pyramids whicb have for their bases the triangles AOC, BOD will be together equivalent to the spherical ungula which has BOD for its angle. Pboposition XXVI. — ^Theorem. The surface of a spherical triangle has for its measure the excess of the sum of its three angles above two right angles multiplied hy the tri-rectangular triangle. Let ABC be the given triangle. Complete the great circle AB, and produce the arcs ACj BC until they meet this great ^^ — — ' \ circle. We have, evidently, ABC + BCE = Lune A, ABC-f ACD=LuneB; and by the preceding theorem, we have ABC -h DOE ^ Lune C. 286 ELEMENTS OF GEOMETRY. Adding these equations, and observing that the sum of these six triangles exceeds tlie hemisphere by twice the triangle ABC, we shall have 2ABC + ^ sphere = Lune A + Lune B + Lune C. Hence, taking J sphere from each member and dividing by 2-, we have . -p^ Lune A + Lune B + Lune G — h sphere ABO- 2 Dividing each member by T, the surface of the tri- rectangular triangle, we have ABC _ Lune A + Lune B + Lune C — ^ sphere ~T 2T ■ -p Lune A _ A Lune B B Lune C C _ , ' ^T — ~ V ~2f T' ~W~ ^' T ^^^^^ VIL, Prop. XXIIL) ; and i-^— = 2. Hence, t^ = ^i+l+C_:i_?andABC = (A + B + C-2)xT. Hence, a spherical triangle has for the measure of its sur- face the excess of the sum of its angles above two right angles multiplied by the tri-rectangular triangle; or when T = 1, is equal to the excess of the sum of its angles above two right angles. Scholium I. If the angles of the triangle are given in degrees of the arcs which measure them, we take 180*^ from their sum, and the ratio of -this difference to 90° will show what paii of the tri-rectangular triangle, taken as the unit of measure, the given triangle is. Application. Let A = 70° 10', B = 60° 20', and C = 79° 30'. Taking 180° from the sum of these arcs, we find the difference to be 30°, and to obtain the ratio of 30° to 90°, we reduce the degrees to minutes in both cases, and BOOK VII. 237 ■we have, by dividing the first number by the second, jfffff = g-, as the value of the area of the spherical triangle in terms of the tri-rectangular triangle. It is, therefore, 5- of the ti'i-rectangular triangle. Scholium II. We might show, in the same manner, that a spherical triangular pyramid has for its measure the excess of the sum of the angles of its base above two right angles multiplied by the tri-rectangular pyramid ; or, tak- ing for the unit of volume the tri-rectangular pyramid, and for the unit of angle the right angle, it is equal to the excess of the sum of the angles of its base above two right angles. Peoposition XXYn. — Theorem. The surface of a spherical polygon has for its meas- ure the excess of the sum of its angles above two right angles taken as many times as the polygon has sides less two, multiplied iy the tri-rectangular trianyl •. From the vertex A draw the diagonals AC, AD ; the polygon ABODE will be divided into as many triangles as the polygon has sides less two. But the measure of each triangle is the excess of the sum of its angles above two right angles multiplied by the tri-rectangnlar triangle ; and it is clear that the sum of all the angles of the tri- angles is equal to the sum of the angles of the polygon. Hence, the area of the spherical polygon is equal to the excess of the sum of its angles above two right angles, taken as many times as the poly- gon has sides less two, multiplied by the tri-rectangular triangle. 23S ELEMENTS OF GEOHETBY. Calling P the area of the polygon, S the snm of its angles, and n the number of its sides, we have for the measure, p = |S - 2 (n - 2) } X T = (S - 2n + 4) X T. If T be taken as unity, the value of P becomes P = (S _ 2>t + 4). :606k; viii. 239 BOOK VIII. THE CYLESTDEE— THE CONE— THE MEASURE OF THE SPHERE. DEFraiTIONS. I. A cytmder is the solid generated by tke revQlutjon of a rectangle ABCD around one of its sides AB as an axis. In this revolution, the sides AD, EC being constantly perpendicular to AB, describe the equal circles BPH, CGQ, called the hoses of the cylinder, and the side CD generates the laieraZ surface: CD is called the generatrix, and the fixed line AB is called tke (^^ of the cylinder. Every section KLM, made in the cylin- der by a plane perpendicular to itg axis, is a circle equal to each of the b^ses: for, while the rectangle ABCD revolves around AB, the perpendicular IK de- scribes a circle equal to the base; and this circle is the section made by a plane perpendicular to the axis, through the point I. Every section PQGH made by a plane drawn through the axis, is a rectangle which is double the generating rectangle. II. A cone is the solid generated by the revolution of the light-angled triangle SAB around one of its sides SA as an axis. 240 ELEMENTS OP GEOMETRY. In this revolution, the side AB describes the circle BDCE, which is the base of the cone, and the hypothenuee SB generates the lateral surface of the cone. The point S is called the vertex of the cone, SA the axis or alti- titde, and SB the slant height or apothegm. Every section HKFI made by a plane perpendicular to the axis, is a circle; and every section SDE made by a plane passed through the axis is an isos- celes triangle, which is double of the generating triangle. III. If from the cone SGDB we take the cone SFKH, formed by a plane parallel to the base, the solid CBHF is called a frustum of the cone. This frustum of the cone may be conceived to be gen- erated by the revolution of the trapezoid ABHG, in which the angles A and G are right angles, around the axis AG. The fixed line AG is called the axis or altitude of t\i& frus- tum; the circles BDC, HFK are the bases; and BH is its slant height or side. IV. Two cylinders or two- cones are similar, when their axes are to each other as the diameters of the bases. Y. If, in the circle A CD, taken as the base of a cylinder, we inscribe a polygon ABODE, and on this polygon, as a base, construct a right prism having the same altitude as the cylinder, the prism is said to be inscribed in the cylinder, or the cylinder circumscribed about the prism. It is evident that the edges AF, BG, CH, &c., of the prism, being perpendicu- BOOK vra. 241 lar to the plane of the base, are included in the lateral surface of the cylinder. Hence, the prism and cylinder touch each other along these edges. We may admit, as a self-evident proposition, that if the number of sides of the inscribed polygon be indefinitely increased, the difference between the lateral surface of the cylinder and the lateral surface of the prism will become less than any assignable quantity; or, in other words, the lateral surface of a cylinder is the limit of the lateral surface of an inscribed prism, the number of whose faces inG7'eases indefinitely. We may also admit, that the volume of a cylinder is the limit of the volume of an insc7'iied prism, the nutnber of whose faces increases indefinitely. Let ABC be a circle forming the base of a cone, and ABCD an inscribed polygon. Construct on the polygon ABCD as a base, a pyramid, having S for its vertex : the edges SB, SC, SD of this pyramid, are in the lat- eral surface of the cone ; and this pyramid is said to be inscribed in the cone, or the cone circumscribed about the pyramid. We may admit, as a self-evident proposition, as before, that the laf' eral surface of the cone is the limit of the lateral surface of an in^ scribed pyramid, the number of whose faces is indefinitely increased; and also, that the volume of the cone is the limit of the volume of the same pyramids. PeopositIon I. — Theorem. The volume qf a cylinder has for its measure the pi'O' duct of its hose by its altitude. 242 ELEMENTS OF GEOMETRY. Inscribe a regular polygon in the base of the cylinder, and construct on this polygon, as a base, the right prism, having an altitude equal to that of the cylinder. Let B = area of the base of the prism, V its volume, and H = altitude of the cylinder or prism. We have V = B X H. Bat, the volume of the cylinder is the limit of the volume of an inscribed prism, the number of whose sides is indefinitely increased. We shall, therefore, have the volume of this cylinder by taking the limit of the product B X H. But the area B of the inscribed polygon has for a limit the area of the circle which forms the base of the cylinder, and the factor H is constant. Hence, Vol. cylinder = area circle x H. Corollary I. Cylinders of the same altitude are to each other as their bases ; and cylinders of the same base are to each other as their altitudes. Corollary II. The volumes of similar cylinders are to each other as the cubes of their altitudes, or as the cubes of the diameters of their bases. For, the bases are to each other as the squares of their diameters ; and since the cylinders are similar, the diam- eters of the bases are as the altitudes : hence, the bases are as the squares of the altitudes ; therefore, the bases multiplied by the altitudes, or the volumes of the cylin- ders, are as the cubes of the altitudes. Scholium. Let R = radius of the base of the cylinder, H = its altitude, the area of the base of the cylinder will be ttR^, and the volume of the cylinder will be ttE^ x H == ttE^H. BOOK VIII. 2i3 Pboposition II. — Theorem. The lateral surface of a right prism has for its measure the perimeter of its iase multiplied by its altitude. For, this surface is equal to the sum of the rectangles AFGB, BGHC, CHID, &c., of which it is composed ; but the altitudes of these rectangles AF, EG, CH, &c., are equal to the altitude of the prism ; their bases, taken together, make the perimeter of the base of the prism. Hence, the sum of these rectangles, or the measure of the lateral surface of the prism, is equal to the perimeter of its base mul- tiplied by its altitude. Corollary. If two right prisms have the same altitudes, the areas of their lateral surfaces will be to each other as the perimeters of their bases. PEOPosrrioN ILL — ^Theoeem. 2^ lateral surface of a cylinder has for its measure the circumference of its base multiplied by its altitude. Inscribe in the base of the cylinder a regular polygon, and construct on this polygon, as a base, a right prism, having the same altitude as the cylinder. Let P = the perimeter, S = measure of the lateral surface of the prism, and H the common altitude of the cylinder and prism. We have S = P X H. But the lateral surface of the cylinder is the limit of the lateral surface of an inscribed prism the 2i4 ELEMENTS OF GEOMETEY. number of whose faces increases indefinitely ; hence, we shall have the measure of the surface of the cylinder by taking the limit of P x H. But the limit of the perim- eter P is the circumference of the base of the cylinder, and the factor H is constant ; therefore, S = circumference of the base x H. Scholium. Representing by R the radius of the base, the measure may be expressed S = 2:rR x H. I Peoposition IV. — Thkokem. The volume of a cone has for its measure the product of its hose hy one-third of its altitude. Inscribe in the base of the cone a regular polygon, and construct a pyr- amid having the polygon for its base, and the vertex of the cone for its ver- tex. Let B — area of the inscribed polygon, V = volume of the pyramid, and H = the altitude of the cone : TT we have V = B x — . o But the volume of the cone is the limit of the volume of the inscribed pyramid, the number of whose faces is indefinitely increased. The area B of the base of the pyramid has for a limit the area of the circle TT which forms the base of the cone, and the factor — is con- o Btant. Hence, TT V = Vol. of cone = area of base x — . o Corollary. A cone is one-third of the cylinder, having the same base and the same altitude. Hence, it follows: BOOK VIII. 245 1°- That the volumes of cones of equal altitudes are to each other as their bases. 2°. That the volumes of cones of equal bases are to each other as their altitudes. 3°. That the volumes of similar cones are to each other as the cubes of the diameters of their bases ; or as the cubes of their altitudes. Scholium. Let II = radius of the base of the cone, H = its altitude, the volume of the cone = ttR^ x ^H, or Peoposition Y. — Theokem. The volume of the frustum of a cone, having AO, DP, as the radii of its iases, and PO its altitude, has for its measure \-nOV (AO + DP + AO x DP). Let TFGH be a triangular pyramid having the same altitude as the cone SAB, and whose base FGH is equivalent to the base of the cone. Let the bases be sup- posed to be placed on the same plane ; then the vertices S and T will be at equal dis- tances from the plane of the bases, and the plane EPD produced will make the section IKL in the pyramid. Now, the section IKL is equivalent to the upper base DE ; for, the bases AB, DE are to each other as the squares of the radii AO, DP, or as the squares of the altitudes SO, SP. The triangles FGH, IKL are to each other as the squares of the same altitudes. Hence, the circles AB, DE are to each other as the triangles FGH, IKL. But, by hypothesis, the 246 ELEMENTS OF OEOMETEY, triangle FGH is equivalent to the circle AB ; therefore, the triangle IKL is equivalent to the circle DE. Now, the base AB x 3SO is the volume of the cone SAB, and the base FGH x |S0 is the volume of the pyr- amid TFGH ; hence, since the bases are equivalent, the volume of the pyramid is equivalent to that of the cone. For a like reason, the pyramid TIKL is equivalent to the cone SDE. Hence, the frustum of the cone ADEB is equivalent to the frustum of the pyramid FGHIKL. But the base FGH, equivalent to the circle whose radius is 2 AO, has for its measure tt x AO ; and we have also for 2 the measure of the base IKL, tt x DP ; and the mean pro- portional between rr x AO and tt x DP is tt x AO x D. Hence, the volume of the frustum of the pyramid, or that of the frustum of the cone, has for its measure ^OP X (ttAo'+ttDP + 77 x AO X DP), or, what is the same thing, ^tt x OP x (Ao' +DP' + AO x DP). Peoposition VI. — Theorem. The lateral surface of a regular pyramid SAB CD E has for its measure the perimeter of its hase multiplied by half the apothegm* SI. For, the lateral surface of the regular pyramid is composed of the equal isosceles triangles SOD, SBC, SAB, &c. But we have SI SOD = CD X SBC = BC X SAB = AB X 2 SI SI , &c. * The apothegm is the perpsflcJiciJ^r drawn fj-oift the vertex S to one of the Bides of tjie base, BOOK VIII. 247 Hence, we have, by addition, for the measure of the lateral surface of the pyramid (CD + BC + AB + AE + ED) x Peoposition Vn. — ^Theoeem. The lateral surface of a cone has for its measure the circumference of its base multiplied by half its slant height. Inscribe in the base of the cone a regular polygon ABCD, and construct on it, as a base, the regular pyramid having the point S for its vertex. Let P be the perimeter of the polygon, S the lateral surface of the pyramid, and SI the slant height or apothegm; we shall have S^ 2 " S = P X But S has for its limit the lateral surface of the cone ; P has for its limit the circumference OE ; and it is easy to see that SI has for its limit SE, for SE - SI < EI, and we know that EI may become less than any assignable quantity, if the number of sides of the polygon be suffi- ciently great. Hence, Lateral surf, cone = circf. OE x SE Scholium. Let L be the slant height of the cone, R the radius of its base ; we shall have for the measure of tlie lateral surface of the cone, in terms of w, and the radius of the base, 2ffR x ^ L, or nE x L. 248 ELEMENTS OF GEOMETRT. PROPOsrnoN VIII. — Theorem. The lateral surface of the frustum of a cone ADEB has for its measure its slant height AD multiplied hy the half sum of the circumferences of its two bases AB, DE. In the plane SAB which passes through the axis 80, draw AF perpendicular to SA, and make it equal to the circumference which has AO for its radius ; join SF, and draw DH parallel to AF. Since the triangles SAO, SDC are similar, we shall AO SA have D(5 = "SD' and because the triangles SAF, SDH are similar, we shall AF _SA DH SD" AF _ AO circf. AO DH~ have Hence, DO circf. DC But, by construction, AF = circf. AO ; hence, DH = circf. DC. Now, the triangle SAF, which has for its measure AF x ^SA, is equivalent to the lateral surface of the cone SAB, which has for its measure circf. AO x|SA. For a like reason, the triangle SDH is equivalent to the lateral sur- face of the cone SDE. Therefore, the measure of the fioofi viii. 249 lateral surface of the frustum ADEB is equivalent to the measure of the trapezoid ADHF. But the measure of the (ATi' -+• T)TT\ ) ; hence, the lateral sur- face of the frustum of the cone ADEB has for its measure AD multiplied by half the sum of the circumferences of its two bases. Corollat'y. Through the point I, the middle of AD, draw IKL, parallel to AB, and IM parallel to AE. We might prove, as above, that IM = circf. IK. But the trapezoid ADHF =: AD x IM ^ AD x circf. IK. Hence, ■we have also for the measure of the lateral surface of the frustum of a cone, its slant height multiplied hy the cir- cumference of a section made at equal distance from the two bases. Scholium. If a line AD situated on one side of 00, and in the same plane, revolve around 00 as an axis, the sur- face described by AD will have for its measure . „ /circf. AO + circf. DC \ AD X (- ), or AD X circf. IK, the lines AO, DO, IK being perpen- diculars let fall from the extremities, and from the middle of the line AD, on the axis 00. For, if we produce AD and 00 until they meet in S, it is evident that the surface generated by AD is that of the frustum of a cone of which OA and DO are the radii of the bases, S being the vertex of the cone. Hence, the surface generated by AD will have the same measure. This measure would still hold good when D coincided with S, which would change the frustum into the whole cone ; and also, when AD is parallel to the axis, which would give a cylinder. In the first, DO would be zero ; and in the second, DO = AO = IK. 11* 250 Elements oi* Geometry. Peoposition IX. — ^Theorem. The surface generated hy a portion, of a regula/r polygon ABCD revolving about a diameter FG, has for its meas- ure the circumference of the inscribed circle multiplied by th£ projection MQ of the perimeter of the polygon on the diameter FG. The point I being the middle of AB, and IK being a perpendicular to the axis FG from the point I, we have for the measure of the surface generated by AB, Surface AB = AB x circf. IK. Draw AX parallel to the axis : the triangles ABX, OIK will have their sides perpendicular, each to each. Hence they are similar, and give the proportion AB _ AB _ 01 _ circf. 01 AX MN IK " circf. IK' Hence, AB X circf. IK = MN X circf. 01. And therefore. Surface AB = MN x circf. 01. In like manner, we shall have, Surface BC ^NP X circf. 01; and, Surface CD = PQ x circf. 01. Hence, by addition, Surface ABCD = (MN + KP + PQ) X circf^ 01 =: MQ X circf. 01. { Remarh "We are to understand by surface AB, surface AC, &c., the surfaces generated by the revolution of AB, BC, &c., around a given axis. Corollary. If we consider a regular polygon of an even number of sides, and that the axis FG passes through two * BOOK VIII. 251 opposite vertices, the entire surface generated by the revolution of the semi-polygon FACG, will be equal to its axis multiplied by the circumference of the inscribed circle. This axis FG is, at the same time, the axis of the circumscribing circle. DfiFTNITIONS. I. ji spherical sone is a portion of the surface of the sphere included between two parallel planes. The sec- tions made by these planes on the surface of the sphere form the bases of the zone. One of the planes may be tangent to the sphere ; and in this case the zone has but one base. II. The altitude of the zone is the distance between the planes of its two bases. III. While the semi-circum- ference DAE, revolving around the diameter DE, gene- rates the surface of the sphere, the arc EH, revolving around the same axis, generates the surface of the zone. PBOPosmON X. — ^Theoeem. The area of a spherical sone is equal to its altitude mul- tiplied hy the oiroumference of a great circle. Let the zone be generated by the revolution of the arc AB about the diameter MN. Inscribe in»the arc AB a portion of a regular polygon 252 ELEMENTS OF GEOMETRY. ADCB, and designate by S the surface generated by ADCB revolving around MN; we shall have, S = circf. 01 X EF. But we may admit, as a self-evir dent proposition, that the surface gen- erated by the arc AB is the limit of the surface generated by a portion of the regular inscribed polygon, the number of whose sides is increased indefinitely. Hence, we shall have the measure of the surface of the zone by taking the limit of circf. 01 X EF. But circf. 01 has for its limit circf. OM, and the factor EF is constant. Hence, Surface zone AB = circf. OM X EF. Cmollary I. In equal spheres, two zones are to each other as their altitudes. Cmollary II. The surface of the sphere may be re- garded as a zone the altitude of which is equal to the diameter ; and hence, the measure of the surface of the sjphere is equal to the circumference of a great circle mul- tiplied hy its diameter. Let R = radius of the sphere ; we shall have Surface sphere = 2TrE X 2R = 4TrRl Hence, the surface of the sphere is equal to four great circles. Corollary III. The surface of the spnere being thus ' estimated in terms of the square constructed on the linear unit, we may readily determine the measures of lunes, spherical triangles, and spherical polygons, in terms of the same unit ; since we have already found, in Book VH., their values in terms of the tri-rectangular triangle, which is the eighth part of the surface of the sphere; BOOK VIII. 253 Pkoposition XL — Thkohkm. The volume of the solid generated ly a triangle revolving around an axis in the same plane and passing through one of the vertices of the triangle, has for its Tneasure the sur- 'face described hy the side opposite to this vertex multiplied by the third of the altitude corresponding to this side. 1°- Let us suppose, in the first place, that the triangle CAB revolves about one of its sides CB. The volume generated by CAB is equiva- lent to the sum of the cones generated by the right-angled triangles CAD, ADB : we have, therefore. Vol. CAB = ittAD X CD + iff X ad x DB = ^ttAD X CB. (1) But if we draw CG perpendicular to AB, we have CG X AB = CB X AD ; for these two products measure the double area of the triangle CAB : substituting in equation (1) CG X AB for its equal AD X CB, we have Yol. CAB =l7rAD X CG X AB -= ^CG x ttAD x AB. But ttAD X AB is the measure of the lateral surface of the cone generated by AB. Hence, Yol. CAB — surface AB X iCG. 2°. Let us now suppose that the triangle CAB re- volves about a CD passing right line through one of its vertices : produce AB until it meets the axis at D ; and let fall the perpendicular CE on AB. 254 ELEMENTS OF GEOMETRY. We shall have vol. CAB = vol. CAD - vol. CBD. But vol. CAD = surf. AD x ^CE ; aDd vol. CBD = surf. BD x ^CE. Hence, vol. CAB = (surf. AD - surf. BD) x ^CE ; and finally, vol. CAB = surf. AB x ^CE. 3°- The preceding demonstration supposes that the side AB meets the axis CD ; let us now examine the case in which these lines are parallel. Draw AE and BD perpendicular to the axis CD, and CF perpendic- ular to AB ; we shall have, vol. CAB = vol. CAE + vol. ABDE - vol. CBD. But vol. CAE ' vol. ABDE Itt X AE X CE. : TT X AE X ED. vol. CBD = in X Ae' X CD. Adding the two first equations and subtracting the third, we have vol. CAB = 77AE '(J-CE + ED - J-CD). And since CD = CE + ED, we have, vol. CAB = n . TEx fED = ^AE x 27r x AE x ED. vol. CAB = iCF X surf. AB. Bemark. "We indicate bj vol. CAB, vol. CAD, the volumes generated by the revolution of the triangles CAB, CAD about an axis. Proposition XII. — Theorem. If a polygonal sector A ODA on the same aide of the diameter FG, revolve about this diameter as an axis, the solid generated will have for its measure the surface gen- erated iy the jperimeter A BCD multijolied ly one-third of the apotJiegm 01. BOOK VIlI. 255 For, the volume generated by the sector AOD is equiva- lent to the sum of the vol- umes generated by the equal isosceles triangles AOB, BOC, COD. But the vol- ume AOB = surface AB x iOl (Book YIII., Theorem XL); and vol. BOC = surf. BC x i^OI. vol. COD = surf. CD x ^01. Hence, by adding, we have vol. AOD = J 01 (surf. AB + BC -f CD) : surf. ABCD. ^01 X DEFINITIONS. I. A spherical sector is the solid generated by the revo- lution of a circular sector FCH around the diameter DE, while the semicircle DHE generates the sphere. The spherical sector is bounded by the zone generated by the arc FH. II. A segment of a sphere is a part of the volume of the sphere included between the sections made by two parallel planes. The sections form the iases of the segment ; and the altitude of the segment is the perpen- dicular distance between the planes of its bases. Peoposition Xni. — ^Theobem. A spherical sector has for its measure the zone which forms its base, rriultiplied by the third of the radius of the sphere. 256 ELEMENTS OF GEOMETRY. Let AOB be the circular sector which generates the spherical sector in its revolution around MN. Inscribe in the arc AB a portion of a regular polygon ADCB ; and call V the volume generated by the poly- gonal sector ADCBO ; we shall have V = surf. ADGB x^OI. But we may admit, as a self-evident proposition, that the volume of the spherical sector is the limit of the volume generated by the polygonal sector, when the number of sides of the portion of the polygon inscribed in the arc AB is increased indefinitely. We shall find, therefore, the measure of the spherical sec- tor, by taking the limit of the expression, surface ADCB x ^01 : but surface ADCB has for its limit the zone AB, and OM is the limit of 01. Hence, spherical sector : : zone AB x ^OM. Scholium I. If the circular sector which generates the spherical sector become equal to a semicircle, the volume generated will be that of the sphere : but in this case the zone which forms the base of the spherical sector becomes the surface of the sphere. Hence, the measure of the volume of a sphere is equal to its surface multiplied hy one-third of its radius. Scholium II. Let R be the radius of tlie sphere, H the altitude of the zone which forms the base of the spherical sector : the zone has for its measure 27rR x H. Hence, the spherical sector has for its measure 2TrE x H x ^K = §7rE^ X H. In the case in which the spherical sector becomes the volume of the^ sphere, we have H=2R. Hence, the fiOOK VIII. 257 measure of the volume of the sphere becomes ItR^ x 2E = AttEI If D represent the diameter of the sphere, we have D D^ R =-^. Hence, E' = — ; and the vplume of the sphere becomes I^tt— = ^ttD'. o Pkoposition XIV. — ^Thkoekm. Tlie surface of the sphere is to the total surface of the circumscriiing cylinder, including its two hoses, as 2 is to 3. The volumes of these two bodies are to each other in the same ratio. Let MPNQ be the great circle of the sphere ; ABCD tlie circumscribing square. If the semicircle PMQ, and the rectangle PADQ, revolve around the diameter PQ, the semi- circle will generate the sphere ; and the rectangle the cylinder circum- scribing the sphere. The altitude AD of this cylinder is equal to the diameter PQ : the base of the cylinder is equal to the great circle, since the diameters AB and MN are equal. Hence, the lateral surface of the cylinder has for its measure the circumference of the great circle multiplied by its diam- eter. But this measure is the same as that of the surface of the sphere (Book YIII., Prop. X.) ; therefore, the sur- face of the sphere is equivalent to the lateral surface of th^ circumscribing cylinder. But the surface of the sphere is equivalent to four great circles. Hence, the measure of the lateral surface of the circumsci-ibing cylinder is also four great circles. If we 17 &58 ELEMENTS OP GEOMETRi?. add to this measure the two bases of the cylinder, which are equal to two great circles, the total surface of the cir- cumscribing cylinder will he equivalent to six great circles. Hence, the surface of the sphere is to the surface of the circumscribing cylinder as 4 is to 6, or as 2 is to 3. In the second place, since the base of the circumscrib- ing cylinder is equal to a great circle, and its altitude to the diameter, the volume of the cylinder will be equal to the great circle multiplied by the diameter. But the volume of the sphere is equal to four great cir- cles multiplied by the third of the radius (Book VIIL, Prop. XIII.) ; which reduces to one great circle multiplied by -3- of the radius, or | of the diameter. Hence, the vol- ume of the sphere is to that of the circumscribing cylinder as 2 is to 3 ; and, therefore, the volumes of these bodies are to each other as their surfaces. Soholium. If we suppose a polyedron having all of its faces tangent to the sphere, this polyedron might be con- sidered as composed of pyramids having all their vertices at the centre of the sphere, and for their bases the different faces of the polyedron. It is evident that all of these pyr- amids will have for their common altitude the radius of the sphere ; so that each pyramid will be equal to the face of the polyedron wliich forms its base multiplied by one- third of the radius. Hence, the volume of the polyedron will be equal to its surface multiplied by one-third of tlie radius of the inscribed sphere. We see by this, that the volumes of polyedrons circum- scribing a sphere are to eacli other as the surfaces of those polyedrons. Thus the property just demonstrated for the circum- scribing cylinder is common to an infinite number of other bodies. We might also remark that the areas of polygons cir- cumscribing the circle are to each other as their perimeters. BOOK VIII. 259 Proposition XV. — ^Thkobem. The solid generated by the circular segment BM.D re- volving around a diameter A CG exterior to this segment, has for its ineasure the sixth part of the circle which has for its radius the chord JBD of the segment, multiplied by ihe projection EF of the chord BD on the axis AC. For, we have vol. BMD = vol. CDMB - vol. CDB. But vol. CDMB ^ttxGB xEF(1) vol. CDB = iCI X surf. DB. And since surf. DB = 27r x CI x EF, we have vol. CDB = |7r x~Cl' x EF. (2) Taking equation (2) from (1), we have, vol. BMD = |7T X EF (CB - CI ). Besides, in the triangle CBI, we have, BD CB - CI =BI = -p ; 4 BD 2 hence, vol. BMD = frr x EF x — = Itt x BD x EF. Scholium. The solid genei-ated by the segment BMD is to the sphere which has BD for its diameter, as ^Tt x BD X EF is to iTT X BD ; or, as EF is to BD. 260 ELEMENTS OF GEOMETRY. Peoposition XVI. — ^Theorem. Emry segment of a sphere included hetweeu two paral- lel planes has for a measure the half sum of its bases mul- ilplied hy its altitude, plus the solidity of the sphere whioh has this altitude for a diameter. Let BE, DF be the radii of the bases of the segment, EF its altitude ; we have, from the preced- ing theorem, vol. BMD = Itt X BD' X EF. f We have also (Book YIIL, Prop. V.), vol. BDFE = It: X EF X (Be' + 151^ + BE X DF). -Q Hence, the segment of the sphere -which is the sum of these two volumes has for its measure ^TT X EF X (2BE + 2DF + 2BE x DF + BD ). But, drawing BO parallel to EF, we shall have DO =DF - BE =D0 = DF - 2DF x BE + BE ; and also, BD = BO +D0 = EF + DF - 2DF x BE + BE . Substituting this value of BD in the expression of the segment, we shall have, after reduction, for the solidity of the segment, i77 X EF X (3Be'+ 301^+ Ef'), an expression which may be decomposed into two parts ; one. ^77 X EF X (3Be'+ SDf") ; or, EF (— BE + TT X DF 2 ) BOOK VIII. 261 is the half sum of the bases multiplied by the altitude of the segment ; the other, ^n EF , represents the solidity of the sphere which has EF for its . diameter. Hence, the theorem is proved. Corollary. If one of the bases reduces to zero, the above measure reduces to EF x — (- IttEF , which shows, that the solidity of a spherical segment with one hose is equivalent to half the cylinder with the same hose and the same altitude, ^lus the sphere which has this altitude for its diameter. 262 ELEMENTS OF GEOMETBT. PRACTICAL EXERCISES FOR THE STUDENT IN PLANE GEOMETRY. THEOREMS, 1. The figure formed by joining the middle points of the sides of a quadrilateral by right lines, is a parallelogram. 2. If from a point, taken within an equilateral triangle, perpendiculars be drawn to the three sides, the sum of these perpendiculars is constant. Examine the theorem, when the given point is without the triangle. 3. If through the point of con- tact A of two tangent circles we draw any two secants BE', CO', prove that the lines BC, B'C are parallel. 4. In every quadrilateral cir- cumscribing a circle, the sum of two opposite sides is equal to the sum of the two other sides. The reciprocal is also true. 5. If a circle be tangent to the two sides of a given angle A, and the tangent BEG be drawn, termin- ated in the two sides of the angle ; prove, 1°, that the perimeter of the" triangle AEG is constant, at what- ever point of the arc the tangent MEN be drawn ; 2°, that the angle BOG is also constant. 6. If the feot of the three alti- tudes of a triangle be joined, two and two, a new triangle will be formed, in which the bi- sectrices of the angles are the altitudes of the first triangle. PeactiCal exercises. 263 ?. The feet of the altitudes of a triangle and the middle points of its three sides are on the same circumference. 8. Having given a quadrilateral, if tangent circles be drawn to three Consecutive sides, the centres of the four circles thus drawn will form a quadrilateral which may be inscribed in a circumference. 9. The bisectrices of the angles formed by the opposite sides of a quadrilateral which may be inscribed in a cir- cumference, intersect each other at right angles. 10. If from any point of a circle circumscribing a given triangle perpendiculars be drawn to the three sides of the triangle, the feet of the perpendiculars will be in the same right line. 11. If we construct on the two sides AB, BC, of a tri- angle ABC, any parallelograms ,,?-... J. ABFE, BCDL, and produce *V I /'^^^ EF and LD, until they meet in ' /§C ^^^\^ O, and join OB, and thus cou- e/ / ^"^v.^^^ ^^ struct a parallelogram on AC, Ar~~ ^ -^ with the adjacent side AG / / equal and parallel to OB, the G' 'h parallelogram AGHC will be equivalent to the sum of the two others. Deduce from this theorem that of the square upon the hypothenuse. 12. The three altitudes of a triangle intersect each other in a common point. 13. The lines which connect the vertices of a triangle with the middle points of the opposite sides intersect each other in a common point. 14. The common point of intersection of the altitudes of a trianffle, and of the lines drawn from the vertices to the middle of the opposite sides, and the centre of the circum- scribing circle, are in the same right line ; and the distance between the first two points is double of that between the two last. 264 ELEltElfTS OF GEOltii'fftr. 15. If from a given point we draw to a circle two se- cants which are perpendicular to each other, the sum of the squares of the chords will be constant. 16. When three circles intersect each other, two and two, the three chords of intersection intersect each other in the same point. 17. K from the point A, the middle of the arc BO, we draw the two secants AFD, AGE, the four points DFGE are on the same p / / I \ n circumference. 18. When three circles are tangent, two and two, the tangent drawn at the points of contact intersect in a common point. 19. The sum of the squares of the diagonals of a quadrilateral is double the sum of the squares of the lines which join the middle of the opposite sides. 20. If in a triangle we draw a series of parallels to the base, and also draw diagonals to each of the • trapezoids thus formed, the points of intersection of these diagonals will be on the right line which connects the vertex of the triangle with the middle point of the base. 21. Prove that in an inscribed quadrilateral the product of the perpendiculars let fall from a point of the circum- ference upon two opposite sides is equal to the prorduct of the perpendiculars let fall from the same point to the other sides. 22. If from a point, taken within a regular polygon of m sides, we draw perpendiculars to all the sides, the sum of -these perpendiculars is equal to m times the radius of the inscribed circle. 23. K from all the vertices of a regular polygon we draw perpendiculars on any right line passing through the centre of the polygon, the sum of the perpendiculars PRACTICAL JEXMCISES. Pj]5 which fall on one side of this line is equal to the sum of the perpendiculars which are on the other side. 24. Prove that if a circle be revolved in another circle which is fixed in position, and which has a double radius, the circles being constantly tangent to each other, any point of the first circle will describe in this revolution a right line. 25. Prove that the three right lines which connect the vertices of a triangle with the opposite vertices of the equi- lateral triangles constructed on the three sides, intersect each other at the same point. 26. Prove that the sum of the perpendiculars let fall on the sides of a triangle from the centre of the circumscrib- ing circle is equal to the radius of this circle, plus the radius of the inscribed circle. 27. Prove that, in a trapezoid, the sum of the squares of the diagonals is equal to the sum of the squares of the opposite sides not parallel, plus twice the rectangle of the parallel bases. 28. Prove that if, in a triangle ABC, three right lines drawn from the vertices Inter- ' sect each other in the same point O, we have ') OD OE OF _ ad''"be ■''bf ~ ' 29. Two concentric circumferences being given, prove that the sum of the squares of the distances of any point whatever of one of the circumferences to the extremities of the diameter of the other is constant. 80. Two quadrilaterals are equivalent when their diagonals are equal, and make equal angles with each other. 12 S()6 ELEMESftS OS GiJOtfETEf . GEOMETRICAL LOOT. 1. Find the locus of points such that the sum of th^ distances of each of them to two given right lines shall bef ■equal to a given line. 2. Find the locus of points such that the difference of the distances of each of them from two right lines shall be icgual to a given line. ■3. Find the geometrical locus of the centres of circles ■paesimg through two given points. 4. iBiiivd the geometrical locus of the centres of circles having, a tjofwen radius and tangent to a given circle. .S. JUraw ^.arallel lines through all the points of a given '■«ii?flQmferenee, and take on each parallel a given length-: tfind the locus of the extremities of these lines. 6. Find the locus of the middle points of the chords of », circle passing through a given point. 7. Find the locus of points such that the feet of perpen- diculars let fall from each upon the three sides of a tri- angle shall be on the same right line. 8. Find the locus of points from which the tangents drawn to a circle intersect each other under a given angle. 9. Find the locus of points the distances of which to two right lines shall be in a given ratio. 10. Find the locus of points 'such that the sum or the difference of the squares of their distances from two given points shall be equal to a given square. 11. Having given two circles, iind the locus of points such that the tangents drawn from these points to the two circles shall be equal. 12. Draw through the point A a right line AM, terminating in the circumference ; and divide this line at the point N, in such PRACTICAL EXERCISES. a manner that we have AM A^ m = — : find the locus of the , , AB, m that we have -r-p^ = — ; A(J n points N". 13. Having drawn the line AM through the given point A, and terminating in the circumference 0, take on this line a point N , such that AM x AN = K^ : find the locus of the points N. Solve the two preceding problems, substituting for the cii'cumference a right line. 14. Through a point A, draw the line AB terminating in the given line XT ; draw AO such that the angle BAC shall be equal to a given angle, and or AB X AC = K^ : find the locx^s of the ^ ^ ^ points C. Solve the problems when XY is replaced by a circum- ference. 15. Find the locus of the points from which two given circles are seen under a given angle. 16. Let a right line of a given length slide on two right lines taken at right angles to. each other : find the locus of the middle points of the hypothenuses of the triangles thus formed. 17. Having given an equilateral triangle, find the locus of points such that the distance of one of them from one of the vertices of the equilateral triangle shall be equal to the sum of the distances from the same point to the two other vertices. 18. Through a point A, taken in the plane of a circle O, draw a secant AC; and at B and C draw the tangents BD, BC; find the 2fiS ELEMENTS OF GEOMETRY. locus of the points D. — The locus is a right line DE per- pendicular to the diameter passing through A. ] 9. Find the locus of points such that the sum of the squares of their distances from the vertices of an equilat- eral triangle shall be equal to a given square. 20. Solve the last problem, replacing the equilateral triangle by any regular polygon. 21. Find the locus of points such that the sum of the squares of their distances from the sides of a regular poly- gon shall be equal to a given square. 22. Let AB be a diame- ter^of the circle BO : draw the secant BCD, and take CD = BC. Join the point D -with the centre of the circle, and the point C with the point A : find the locus of the point M of intersec- tion of the lines AC, OD. 23. From any point A on the diameter BD pro- duced, draw the tangent AC, and the bisectrix of the angle CAO, and let fall the perpendicular OM on AN : find the locus of the point M. 24. Through a point of the hypothenuse BC of the right-angled triangle ABC, draw any secant DOE; draw the circles OBE, OCD : find the locus of the point M in which these two circles meet. PRACTICAL EXEECTSES. 26a / 9 ^ B / / ] D 25. Having given a circle BO aud a diameter AB, draw any radius 00, and CD perpendicular to AB, and take OM = CD : find the locus of the point M. 26. In a quadrilateral ABCD, we have given AB, BC, AC, and CD. Find, 1°, the geometrical locus of the middle point of the diagonal BD ; 2°, the geo- metrical locus of the middle of the i-ight line EF, which joins the middle of the two diagonals. 27. Find the locus of all the points so situated that the sum of the distances of one of these from three given right lines shall be equal to a given right line. PKOBLEMS. 1. Through a given point draw a right line equally dis- tant from two given points. 2. Having given two points A and B, find on the line A j5 ST a point P, such that the angles APS, BPT shall be equal. ^3. Through a given point draw a right line cutting two parallel lines in such manner that the part of this line comprised between these two parallels shall be equal to a given line. i. Construct a square, knowing the diflFerence between the diagonal and the side of the square. 5. Construct a triangle, knowing the base, the angle opposite, and the sum or the difference of the other two sides. 6. Describe a circle of a given radius : — 270 ELEMENTS OF aKOMETRT. 1°. Passing through two given points. 2°. Passing through one point and tangent to a right line. 3°. Tangent to two right lines. 4°- Tangent to a right line and oirclo. 5°- Passing through a point and tangent to a circle. 6°. Tangent to two circles. 7. Draw in a circle a right line through a given point, and such that the intercepted chord shall be equal to a given line. 8. Describe a eii'cle tangent to a circle and a right line at a given point. 9. Construct a circle tangent to a circle at a given point, and passing through a given point. 10. Construct a triangle equal to a given triangle, the sides of whicli shall pass through three given points. 11. Having given two circumferences which intersect each other, draw, through one of their points of intersection, a right line MIST, such that 'the distances MN, comprised between the two points of intersection of this right line with the two circumferences, shall be equal to a given right line. 12. Draw through the uoint A the right line MAN, so that AM = AN. 13. Di'aw through the point A the right line MAN, so MA m that -.-Ti. = — AiST n ■ * 14. Having given two circumferences, draw a secant parallel to a given line, so that the sum of the chords shall be equal to a given line. 15. Construct a quadrilateral, knowing two opposite angles, the diagonals and the anglobutween the diagonals. 16. Having given two circles, find a point such that the PRACTICAL EXERCISES. 2Yl tangents drawn to this circle shall be equal and make a given angle. 17. Having given the arc CD and the diameter AB, find on the circumference a point P, such that drawing the lines PD, PC, we shall have OM = ON. 18. Construct a triangle, knowing the three lines drawn from the ver- tices of the angles to the middle ~p points of the opposite sides. « 19. Construct a triangle, knowing the three altitudes. 20. Construct a triangle, knowing the angles and the perimeter. 21. Construct a triangle, knowing the base, the altitude, and the ratio between the two other sides. 22. Construct a triangle, knowing the base, the altitude, and the rectangle between the two other sides. 23. Two right lines being given, produced until they meet, draw through a given point a right line to the com- mon point of intersection. 24. Find in a triangle a point such that joining it with the three vertices, the triangle will be divided into three equivalent triangles. 25. Describe a circle passing through a point, and tan- gent to two given circles. " 26. Describe a circle which shall be tangent to three given circles. 27. Construct a trapezoid, knowing the angles and the diagonals. 28. Having given three concentric circumferences, con- struct a triangle similar to a given triangle, whose three vertices shall lie on these circumferences. 29. Through a given -point on an angle, draw a right Ime so tliat the product of the segments comprised between the point' and each of the sides of the angle shall be equal to a given square. 272 ELEMENTS OF GEOMETEY. 30. Through a given point, in the plane of a circle, draw a right line, so that the distances from this point to the points of intersection of the right line and circle shall be to each other in the ratio of m to n. 31. Through a given point and the centre of a circle, pass a circumference so that the common chord shall be equal to a given right line. > PEACTICAL EXERCISES IN GEOMETEY OF THEEE DIMENSIONS. THEOREMS. 1. If a right line be perpendicular to a plane, every plane parallel to this line will be perpendicular to this plane. 2. Two parallel right lines make equal angles with the same plane. 3. If, in a triedral angle, two of the plane angles are equal, the opposite diedral angles will be equal, and re- ciprocally. 4. In a triedral angle, the greater face lies opposite the greater diedral angle, and reciprocally. 5. The three planes perpendicular to the faces of a tri- edral angle, drawn through the three bisectrices of the plane angles of this triedral angle, intersect in the same right line. • 6. The three planes drawn through the edges of a tri- edral angle, perpendicularly to the opposite faces, intersect each other in the same right line. 7. The three planes through the edges of a triedral angle and the bisectrices of the opposite faces, intersect each other in the same right line. 8. If through the vertex of a triedral angle a perpen- Practical exercises. 273 dicnlar be drawn in each lace to the opposite edge, these three perpendiculars wiLl he in the same plane. 9. In every tetraedron, the lines -which join the middle points of the opposite edges mutually divide each other into two equal parts. 10. Two tetraedrons which have a solid angle equal, are to each other as the product of the edges which comprise the equal solid angle. 11 . The plane which bisects a diedral angle of a trian- gular pyramid, divides the opposite edge into two seg- ments proportional to the adjacent faces. 12. If a tetraedron contains a tri- rectangular solid angle, the square of the opposite face will be equal to the sum of the squares of the three others. 13. The volume of a truncated parallelopipedon has for its measure the product of its base by the perpendicular let fall from the centre of the upper base upon the lower. 14.. In a tetraedron, the lines which join each vertex with the point of intersection of the lines drawn from the vertices of the angles of the opposite faces to the sides opposite will intersect in the same point. 15. The perpendicular let fall from the common inter- section is the last theorem on a plane, is the mean propor- tional between the perpendiculars let fall from the four vertices of the tetraedron on the same plane. 16. Every plane passing through the middle of tlie two opposite edges of a tetraedron divides this tetraedron into two equivalent parts. 17. Through four points not in the same plane we may draw the surface of a sphere, and we can only draw one. 18. "We may inscribe a sphere in every tetraedron. 19. When three spheres intersect each other, two and two, the planes of the three circles of intersection inter- sect each other in a right line perpendicular to the plane of the three centres. 274 elemejSTts of oeometeY. 20. When three lines, at right angles to each other, in- tersect a sphere, and pass through the same point, the sum of the squares of the chords formed is constant. GEOMETEIOAL LOCI AND PKOBLKMS. In plane geometry, we have defined a geometrical locus to be a line containing all the points of the plane possess- ing a comriion property. So likewise in geometry of three dimensions, we define a geometrical locus to be the series of points in space which satisfies one or two given condi- tions. This geometrical locus may be a line on a surface. Thus, the sphere is the geometrical locus of points sit- uated at the same distance from a given point ; and the perpendicular drawn to the plane of a circle at its centre is the geometrical locus of points equally distant from all the points of the circumference. 1. Find the locus of points equally distant from two given points. 2. Find the locus of points equally distant froin three given points. 3. Find the locus of points equally distant from two given planes. 4. Find the locus of points equally distant from three given planes. 5. Find the locus of points in space equally distant from two right lines situated on the same plane. 6. Find the locus of points in space equally distant fi-om three lines in the same plane. 7. Find the locus of points such that the sum of the dis- tances of each from two given planes shall be equal to a given line. 8. Find the locus of points such that the sum of the squares of the distances of each from two given points shall be equal to a given square. 9. Find the locus of points such that the diflFerence of PRACTICAL EXEEaSES. 275 the squares of the distances of each from two given points shall be equal to a given square. 10. If two right lines, not in the same plane, be inter- sected by a series of parallel planes, and lines be drawn joining the points of intersection of the planes with the given lines, find the geometrical locus of the points, which divides all the lines which connect the points of intersec- tion, in the ratio of m to n. 11. Having given two right lines at right angles to each other, and not in the same plane, let lines of a given length be drawn between the two given lines ; find the geomet- rical locus of the middle of these lines. 12. Calculate the volume generated by the revolution of a regular hexagon around one of its sides. 13. Find the volume generated by the revolution of a regular semi-decagon whose side is a around the diameter of a circumscribing circle. 14. Find the surface of the earth in square feet. 15. Find the measure of a pyramid, when we take for the unit of volume the sphere which has for its radius the linear unit, and for the unit of surface the circle which has for its radius the linear unit. 16. The angles of a spherical triaugle are respectively, 58° 18', 64° 8', 82° 4', the radius of the sphere is 6 yards ; determine the surface of the spherical triangle in square yards. 17. Find the volume of a spherical segment, with one base, having an altitude of four feet, and situated on a sphere whose radius is nine feet. 18. The radii of the bases of a frustum of a cone are 3 yards and 5 yards, and its slant height 7 yards ; find the convex surface of the frustum, and the volume of tlie frus- tum. 276 ELEMENTS OF fi^OMETRT. ArPENDIX TO BOOK III. THEOEY OF TRANSVERSALS. A Transversal is a right line which intersects a system of many lines. THEOREM I. If the three sides of a triangle, produced if necessary, arc cut hy a transversal, each side will lie divided into two segments, and the six segments of the three sides will le so divided, that the product of the three segments which have no com,mon extremity, will he equal to the product of the three other segments. Let ABC be the given triaLgle, and DEF the transversal. Throuffli ^^ the point C draw CG parallel to ^ the side AB, and jiroduce it until it meets the transversal. The simi- lar ti-iangles ADE, EGO, give the proportion AE AD , . , . -Qg = -QQ ; which gives AE X GG = GE X AD. (1) The similar triangles GGF, FBD, give the proportion GF QO BF ^ BD ' ^®"'^^' CF X BD = CG X BF. (2) Multiplying the equations (1) and (2), and dividing out the common factor CG, we have AE X CF X BD = AD X BF X CE. APPENDIX TO BOOK III. 277 THEOREM n. Reciprocally, if three, points taken, in even number, on the sides of a triangle, and, in uneven number, on the pro- longations of the sides, determine six segments, such that the product of three which are not consecutive is equal to the product of the other three, the three points will he on the same right line. Let ABC be the given triangle, and D, E, F the three points, so taken, that we have AD x BF x CE = BD X CF X AE. (1) Then will the three points D, E, F be on the same* right line ; for, if the right line DE meet the line BC in a point F', different from F, we should have, by the last theorem, AD X BF' X CE = BD X CF' x AE. (2) Dividing equation (1) by (2), and suppressing the com- mon factors, we have Hence, a proposition which is evidently absurd, unless the point F' coincides with the point F. Hemarl:. This theorem enables us to recognize, in a very simple manner, when three points are on the same right line, as we shall illustrate by an example. LEMMA I. Two circumferences A and B being given, if two paral- lel radii A2f, -SiV he drawn in the same direction, the right line iry, which joins their extremities, will meet the line of their centres at a point C, which will be the same what- ever he the direction of the parallel radii AM, JSJV. BF CF BF' ~ CF" BF BF' - CF BF BG -CF'~BF"°' BC" BF . "BF" 278 ELEMENTS OF GEOMETKr. For, the similar triangles AMC, BNC give the propor- tion AC_AM BC~ BN' Hence, and AC - BC _ AM -EN BO BN ' AB AM - BN BO BN ■ But, AB, AM — BN, and BN, are constant quantities ; hence BO will also have a constant value. The point C is called the centre of direct similitude of the two circumferences, and its position on the line of the' AC AM centres is determined by the proportion -^^ = "rn^' ' ^® see, also, that this point coincides with the point of inter- section of the common exterior tangents. LEMMA II. Having given two circumferences A and £, if we draw two parallel radii AM, BN, in opposite directions, the right line MN will meet the line of tlie centres at a point C, which will he the same, whatever le the direction of the radii AM, BN. For, the similar triangles AMC, BNC give the propor- tion AC_AM CB ~ BN' APPENDIX TO BOOK ill. m The point C divides the line AB in the ratio of the two radii ; and hence this point is constant. The point C is called the centre of inverse similitude of the two circumferences. "We see, also, that it is the point of intersection of the common interior tangents. THEOEEM m. The centres of direct similitude of three circles, taken two and two, are on the same right line. Let A,' B, C he the centres of the three circles ; R, R', R", their radii ; and let M, N, P be the centres of direct similitude of these E' circles, taken two and two. "We have AM_E BM ~ R" BP_R' CP R'" CN_'Rr Aisr R ■ Multiplying these proportions, member by member, we have AM X BP X CN R X R' X R" Hence, BM X CP X AN R' X R" X R AM X BP X ON .= BM x CP x AN. 280 ELEMENTS OP GEOMETRY. But the lines AM, BP, CIST are, with respect lo the tri- angle ABC, three segments of its sides, having a common extremity ; and BM, CP, AN form three segments of the same sides which are not consecutive ; and since the pro- duct of the three first segments is equal to the product of the three last, we conclude that the points M, N, and P are on the same right line. I THEOEEM rV. If the sides of a plane polygon, produced if necessary, he intersected hy a transversal, there will he formed on each side two segments such that the product of tlie segments which' have not common extremities will he equal to the product of all the others. A Let there be, for example, the penta- gon ABODE; and let it be intersected by the transversal mnpqr : then we m n p q r s 't shall have An X Br X Cm xDq xEt = Bn x Cr x Dm x Eq x At. For, draw from one of the vertices A of the polygon diagonals to all the other vertices, and produce them until they meet the transversal. The triangles ABC, ACD, AED considered separately, as cut by the transversal, give An X Br X Gp = Bn X Cr x Ap. , Ap X Cm xDs — Qp x Dm x As. ; ' A« X D^ X Ei! = Ds X E^' X A;;. Multiplying these equations, and suppressing the com- mon factors, we have An X Br X Cm x Dq x Et — Bn X Cr x Dm x Eq y At. APPENDIX TO BOOK III. 281 THEOREM V. If the different portions of a right line he projected on a plane, each projection loiil be equal to the corresponding portion of the line multiplied h/ a constant nurnber. Project on the plane MIST, the portions AB, CD of tlie right line AE, and let ah, cd be these projections : draw DF and BH parallel to Ew. The similar triangles CDF, ABEL give the pro- portion DF_BH DC ~ AB' And since dc = DF, and la = BH, dc _ ah DC~AB* dc Tlien, if =^ = m, which gives dc = DC x m, we shall have. -^ = m, and ah = AB x m. THEOEEM VI. If the sides of a warped polygon, produced if necessary, he cut hy a transversal pla/ne, there will he on each side two segments, such that the product of all those segments which have nx>t a common extremity will he equal to the product of all tJie others. {A warped polygon is a polygon whose sides are not in the saine plane.) Let abcde be the given polygon, and let m, n, o, p be 282 ELEMENTS OF GEOMETRY. the points in which the sides db, io, cd, de, and ea meet the transversal plane. Project this polygon on a plane per- pendicular to the transversal plane. Let ABCD be this projection, and let SN be the inter- section of the plane of projection with the transTersal plane. If we produce the sides of the polygon ABCD until they meet SN", we shall have, by Theorem IV., AExBLxCMxDE=:BExCLxDMxAK. (1) But AE and BE are the projections, on the same plane, of the segments am, Im of the given polygon : we have, therefore (Theorem Y.), AE = am X C ; BE = Jwi x C. We have, in like manner, BL = 5n X C ; GL = cnx C. CM= CO X C"; DM= (?o X C". DE = , P B Q aM x^5 X SN" = MN x apx^h; hence, aM. „ pb S5 The same triangle a^h, cut by the transversal ^q, gives a'RxhqX^'E = NE x aqx^J,-^ hence, :jnpY7 X — x — — = 1. (2) ' NR aq Sb ^ ^ 286 ELEMENTS OF GEOMETRY. x^lxSN hence, Comparing (1) and (2), we have oR ''l^R"aq" Sb' But, the line ab being divided harmonically, at the points J? and q, we have J)b _bq _ op aq ' oM _ oR DEFINITION. If we produce the opposite sides of a quadrilateral ABCD until they meet in M and N, the right line MN is called the third diagonal of the quadrilateral, and the simple quadrilateral ABOD when united to the quad- rilateral DMBIS^, which is not convex, forms what is called a complete quadrilateral. THEOREM I. Saah diagonal of a complete quadrilateral is divided harmonically by the other two. Let ABCDMN beacompleteqnad- rilateral ; let AC, BD, MN, be its diagonals, then will the diagonal MN, for example, be cut harmonically by the other two at the points P and APPENDIX TO BOOK 111. 28T Q. For, the triangle DMN, cut by the transversal ACQ, gives QM X NO X DA = QN X CD X AM; , QM^NC^DA , ... ^'''''^ QN^CD^AM = 1- (') In the same triangle, the lines PD, CM, NA, drawn from the vertices, and intersecting each' other in B, give PM X NC X DA = PlSr X CD X AM ; PM NC DA , ,„, hence, pj^^cD^AM^l- ^'^ Comparing (1) and (2) we deduce, PM_QM PN " QN" We might prove, in like manner, that each of the diag- onals AC, BD is divided harmonically by the other two. 1>fi'^ ELEMElSfTS OF GEOMETRY. OF THE POLE AND OF THE POLAR LINE, WITH RESPECT TO A SYSTEM OF TWO RIGHT LINES. THEOEEM XI. Two right lines Oy, Ox leing given, and also a point P, if we draw, through this point, a secant line PBA, and determine the conjugate harmonic Q of the point P, with respect to the line AB, the geometrical locus of the point Q, in making the secant PBA vary, %aill ie the fourth iranch of the harinonic pencil, of which OP, OA, OB are the three first. For, if we draw the lines OP, OQ, the figure OPBQA forms an harmonic pen- cil. Every secant Yhqa will therefore be divided harmon- ically by the pencil ; hence, every point of the locus is found on the right line OQ. The right line OQ is called the polar line of the point P, with respect to the two right lines Oy, Ox, and is the locus of the conjugate harmonics of the point P, with re- spect to these two lines. THEOEEM XII. Two right lines Oy, Ox, leing given, and a point P, if we draw, through this point, two secants PBA, PB'A', and the lines BA', B' A, the locus of the point of intersec- tion Q of these two lines, when we cause the secants to vary, will he the right line OQ, which is the polar line of the point P, with respect to the lines Oy, Ox. AfrPENBiX TO fiOOE III. For, let us consider the com- plete quadrilateral OBQA, B'A', the diagonal B'A' will be cut harmonically by the two others OQ, AB ; hence, the four lines OP, Oy, OC, Ox, form an harmonic pencil, since the line PB'C'A' is divided harmonically; and the line OC is the polar line of the point P, with respect to the two lines Oy, Ox. But, every point Q, of the geometrical locus, is situated on OC ; therefore, the geometrical locus sought is the polar line of the point P, with respect to the two right lines Oy, Ox. Scholium. We deduce from this theorem the means of resolving, with the ruler only, the following problem : Having given a right line AB, and a point P, on this line, to find the conjugate harmonic of P with respect to AB. For this purpose, join any point whatever, such as S, with the three given points ; c/ \ \ \ draw through the point P any line PDC; draw AD and BC intersecting each ^ B " other in 0, and join SO : the line SO will intersect AB in the required point Q. This results evidently from the theorem which has just been demonstrated. 13 290 ELEMENTS OF GEOMETRT". OF THE POLE AND THE POLAR LINE WLTH, EESPECT TO THE CIRCLE. THEOEEM XIII. If^ through a point P, tahen in the plane of a circle, any secant PAB he drawn, and the conjugate harmonic Q, of the point P, he determined _ with respect to AB, the gemnetrical locus of the point Q, when the secant PAB is varied, is a right line. (The right line which forms the geometrical locus of the point P, is called the polar line of the point P with respect to the circle, and the point P is called the pole.) Draw the diameter Vh, and construct the conjugate har- monic q of the point P, with respect to the diameter ah : join Qq, and draw OM per- pendicular to AB. The four harmonic points P, A, Q, B, give the proportion PB^QB PA QA' Hence, PE -f PA _ QB + QA PA ~ QA • And, since QB -f QA = AB = PB - AP, we have PBjI-PA ^ PBj-PA PA ~ QA ■ And by composition, we have PB-fPA ^ PA _ PA 2PB PA-fQA PQ" ^^ APPENDIX TO BOOK III. 291 But, the point M being the middle point of AB, we have PA + PB = 2PM. And the proportion (1) becomes 2PM PA PM PA or. 2PB PQ' ' PB ~ PQ- Hence, PQ X PM = PB X PA. (2) We might prove, in like manner, that P^ X PO = Pa X P5. But PA X PB = Pa X PJ. Hence, PQ x PM = P^ x PO. Or, PQ_ PO P^ ~ PM" The two triangles PQg', PMO, having an equal angle P included between proportional sides, will be similar, and since the angle PMO is a right angle, the angle Pg'Q is also a right angle ; hence, the geometrical locus described by the point Q is a right line perpendicular to Pa5 at the point q, the conjugate harmonic of the point P with respect to ab. Corollary. The position of the point q is determined by the equation P^ X PO = Pa X P5. (2) But, Pa X PJ = (PO + Oa) (PO - Oa) - Po'-"0^ Besides, P^ = PO - O^. Therefore, the relation (2) becomes (PO - 0^) PO = Po'_ W; Or, PO X 0^ 3= Oa. This relation shows that if PO become equal to Oj-. 292 ELEMENTS OF GEOMETRY 0(2 will be equal to PO ; therefore, the polar line of the two points P or q passes through the other point. It follows also from this relation, that if the point P approach the point a, the point q approaches it also ; and if PO becomes equal to Oa, Oq will be equal to 0« also ; hence, the polar line of a point on the circumference is the tangent at this point.' If PO becomes smaller than Oa, Oq becomes greater than the radius, and the polar line is exterior to the circle ; finally, if PO = O, Oq becomes infinitely great. I THEOEEM XrV. When a point P moves on a right line ST, the polar line of this point passes, always through the pole of the line ST. Draw the line OP, and deter- mine the point Q, so that OQ X OP = OA . The line QN, perpendicular to OP, will be the polar line of the point P, and it will also contain the pole 'of the line ST. Draw OM perpendicular to ST, cutting NQ in N ; the quadrilateral MNQP may be inscribed in a circumference, since two opposite angles are right angles. We shall have then OM x ON = OP X OQ. And since OP x OQ = OA^, we shall have OM x ON" = OA.'; Hence, the point N is the pole of ST. Corollary I. "We see, by the above demonstration, that if the line ST revolve around the p)oint P, the pole of this line will always le found on the polar line of the point P. Corollary II. The right line which connects the poles of two right lines is the polar line of their point of inter- APPENDIX TO BOOK III. ]93 section. For, from Corollary I., the pole of this right line must be found at the same time on the two given lines. Corollary III. If two polygons ABCD, abed, having the same number of sides, drawn in the plane of a circum- ference, are such that the vertices A, B, C, D are the poles of the sides of the others ; reciprocally, the vertices a, i, c, d of the second will be the poles of the sides of the first. By virtue of this property, the two polygons are called reciprocal polar lines to each other. r — THEOREM XV. If through a point P, taken in the 'plane of a circle, we draio a secant PAB, and the tangents AM, BM at the points A and B, the geometrical locus described hy the point 21, when the secant varies, will he the polar line of the point P For, the tangent AM is the polar liue of the point A ; the tangent BM is also the polar line of the point B. It follows that the point of intersection M of the two tangents is the pole of the line AB. But the point P is on the line AB ; hence, the polarline of this point passes through the point M. Corollary I. The polar line of a point M, taken without the circle, is the chord of contact AB of the angle which has this point for its vertex. Corollary II. If a polygon ABCD circumscribe the cir- cle, the polygon MXPG, form- ed by joining the consecutive points of contact, will be the reciprocal polar line of the poly- gon ABCD. 294 ELEMENTS OF GEOMETHY. THEOREM XVI. If, through a point P, taken in the plane of a circle, the secants PAB, PCD le drawn, and we draw the lines AC, BD, intersecting each other in If, and the lines BG, AD, intersecting each other in N, tlie geometrical locus of the points MandN, when iJie secants PAB, PCD are varied, is a right line, which is the polar line of the point P. Draw the line MN, cutting AB and CD in a and c. By reason of the theorem of the complete quadrilateral MANB, CD, the secants AB, CD will be cut harmonically, the first /'^^^'"cf^l ^ — ~~~'" — ^ t) at P, a, and the second at P, c ; the line ac is, therefore, the polar line of the point P. But the points M and N are on this line; hence, the geometrical locus of the points M and N is the polar line of tlie point P. Remark I. If the given point IST be exterior to the cir- cumference, in drawing any two secants CB, AD, then joining AC, BD on one side, and AB, CD on the other, it would be required to find the geometrical locus of the point M, or of the point P ; but this locus would always be the polar line of the point N. For, we have just seen that tlie point P is the pole of the line MN ; for a like i-eason, the point M is the pole of the line PN. Hence, the point of intersection of the lines MN, PN, is the pole of the line PM. Remark II. The preceding theorem furnishes the method of constructing the polar line of a point by means of a ruler only. If the point P is exterior to the circumference, the polar line MN is the chord of contact of the circumscribed APPENDIX TO BOOK III. 295 angle whose vertex is at P. The intersection of this line ■with the circle will determine the points. of contact of the tangents which we may draw to the circumference through the point P. THEOEBM XVII. In every inscribed hexagon., the points of intersection of the ojyposite sides are in the same right line. Produce the three sides BC, DF, AE, which are not con- secutive; these sides determine, by their intersections, the tri- angle QKP, which is cut by the transversals EFM, ABL, DCN ; we have, therefore, by the fundamental theorem of transversals, the equalities QE X EM X PF =EE x PM x QF. QA X KB X PL = RA X PB x QL. QNxEC xPD=ENxPC x QD. Multiplying these equalities together, and noting that QE X QA = QF X QD, PF x PD = EB X BC = EE X EA, PB X PC, wehave EM xQN xPL = EN xPM x QL. Hence, the three points N, L, M are in the same right line, which is the transversal of the triangle QEP. Remark I. The preceding theorem would be equally true, if the inscribed hexagon were not convex. 296 ELEMENTS OE GEOMETRY. RemavTc II. The theorem, holding good wliatever be the magnitudes ot" the sides, is equal- ly applicable when any of the sides of the hexagon re- duce to zero. But, in this case, they must be replaced by tangents to the circumference. From this result many other theo- rems, which the pupil may readily enunciate and prove. We may limit ourselves to the following : If a triangle he inscribed in a circle, the points of inter- section of the tangents drawn at each vertex with the op- posite sides are in the same right line. THEOEEM XVIII. In every circumscribed hexagon, the diagonals wJdch connect the opposite vertices intersect each other at the same point. Let GHLMNO be the circumscribed hexagon ; join the consecutive points of con- tact : we shall form an in- scribed hexagon ABCDEF, which is the reciprocal polar line of GHLMNO. The point of intersection of the sides BO, EF, which we will designate by P, will be the pole of the diagonal HIST; likewise, the point of intersection of the sides AB, ED, which we will designate BOOK vin. 29ir by R, will be the pole of the line GM. Finally, the point of intersection of the lines AF, CD, which we will desig- nate by S, wil' be the pole of the line GL. But the three points P, 11, 'o are in the same right line ; hence, their polar lines HN, GM, OL intersect each other in the same point. 298 ELEMENTS OF GEOMETEY. APPENDIX TO BOOK IV. DEFINITIONS. I. A maximum quantity h the greatest among all those of the same species : a minimum, is the smallest. Thns, the diameter of the circle is the maximum among all lines which connect two points of the circumference ; and the perpendicular is a minimum among all right lines drawn from a given point to a given line. II. Isoper'imetrlcal figures are those which are equal in perimeter. Pkoposition I. — Theoeem. Of all triangles formed with two given sides malcing with each other an arhitrary angle, the maximum is that in which the two given sides maJce a right angle. Let the two triangles BAC, BAD have the side AB common, and the side AC = AD : if the angle BAC is a right angle, the triangle BAC will be greater than the triangle BAD, in which the angle A is acute or obtuse. For, the base AB being the same, the two triangles BAC, BAD are fo each other as the altitudes AC, DE : but the perpendicular DE is shorter than the oblique line AD, or its equal AC. Hence, the triangle BAD is smaller than BAC. 1- I i Pkoposition IL— Theoeem. The circle is greater than any plane figure having an equal perimeter. APPENDIX TO BOOK IV. 299 1°. It is, at first, evident that there may be an infinite number of figures having a given perimeter, which have different forms and diflferent areas : but we see, also, that these areas cannot be indefinitely increased. There are, therefore, among figures of a given perim- eter, one or more which may be maximum. 2^. Every figure which encloses a maximum area in a given perimeter is convex. For, let AMBlSr be an enclosed line not convex : if we cause the re-entrant part AMB to re- volve around the points A and B so that it occupies the position AM'B, the figure AM'BlSr will have the same pe- rimeter as the first, and will enclose a greater area. 3°- Let AMBN be a maximum figure under a given perimeter: every right. line AB which divides the perim- eter into two equal parts, will also divide the area into two equivalent parts ; for, if the por- tion ANB were greater than AMB, when ANB revolves about AB, so that it tabes the position AN'B, the figure AifEN would have the same perimeter as AMBIT, and would have a greater area: AMBN would not, then, be a maximum. It follows, also, from what has been said, that if AMB!N" be a maximum figure, AN'BN is a maximum also ; and we see that in this last figure, every perpendicular ITN' to AB is divided by this line into two equal parts ; so that the triangles ANB, AN'B are equal. This being established, if the angles ANB, AIST'B are not I'ight angles, we might enlarge simultaneously the area of the triangles ANB, AJSF'B without changing the 300 ELEMENTS OF GEOMETRY. magnitude of the sides AN, NB, AN', N'B, or the mag- nitude of the segments APN, NQB, AP'N"', N'Q'B; the common base AB alone would change. But, by this variation, the area of the figure would be augmented without any change being made in the perimeter; which is contrary to the hypothesis. Hence, the angles N and N' are right angles ; and furthermore, the point N is on the curve ANB. Hence, this curve is a semi-circumfer- ence. Thus we see that if a line divides a maximum figure into two equal parts, each of these parts will be a semi- circle. Hence, the whole figure is a circle. Pkoposition hi. — Theorem. Among aU plane figures which have ihe same area, the circle has the smallest perimeter. For, if any figure whatever, whose area is A, had a smaller perimeter than the circle with the same area, we might, by the preceding theorem, transform it into a circle with the same perimeter, and with an area B > A. This second circle would have a greater area than the first, and a less perimeter, which is absurd. Peoposition IV. — Theorem. Every polygon of M sides which contains a maximum area in a given perimeter, is convex. For, let AEDCFB be a polygon of m sides, which con- tains a re-entrant angle AED. If we make the re-entrant part AED revolve around the line AD, so that it takes the position AE'D, the polygon AE'DCFB will have the same perime- ter as the first, and a greater area : the polygon AEDCFB cannot, therefore, APPENDIX TO BOOK IV. 301 be maximum among all those of the same perimeter and the same number of sides. Peoposition V. — Lemma. Every polygon ABODE which contains a re-entrant angle, may he transformed into a polygon with a greater area, and having the same perimeter, and one side less. For, if we produce AB, and join all the points of this B J line produced with the point D, the 7 sum BI + ID will increase, continu-' ouslj, to infinity. There is, therefore, a point I, where we have BI + ID = BC + CD. We thus obtain a polygon ABIDE evidently greater than ABODE, which has the same perimeter,_ and one side less. Peojposition VI. — Theorem. Of all isoperimetrical polygons of the same number of sides, the regular polygon is the greatest. We will prove successively that, if a polygon has not its sides equal and its angles equal, it cannot be a max- imum among the isoperimetrical figures of the same num- ber of sides. 1°. Suppose that the polygon ABODE has m sides; and let AB < BO. Take on BO the point M sufficiently near to C that we have always AB < BM. Make afterwards AMB' = BAM: ,take MB' = AB, and join AB'. The triangle ABM is equal to the triangle AB'M. We conclude from this, that we might substitute for the polygon 302 ELEMENTS 0.? GEOMETEY. ABODE the polygon AB'MCDE, without changing the perimeter or the area : only this new polygon would have in -\- 1 sides, and besides, a re-entrant angle; for, AB being less than BM, we have the angle BMA < BAM or B'MA. But this polygon may be transformed into another with m sides, having the same perimeter and a greater area. Hence, ABODE could not be the maximum polygon among isoperimetrical figures of the same numhei- of sides. 2°- Let us sujipose that in the polygon AI^ODII of ?n sides, -we have the angle A > B. Take a point M near B enough to B that the angle MAH be greater than AMC. Make, also, the angle MAB' = AMB : )c take AB' = MB, and join MB'. The tri- angle MAB' is equal to ABM, and the polygon AB'MCDH has the same ai-ea and the same perimeter as ABODII: but it has m + l sides and a re-entrant angle; for, the sum AMC + AMB being equal to two right angles, we have MAH -f MAB' > two right angles. Hence, the polygon might be transformed into another of m sides, and of the same perimeter and a greater area; and therefore ABODH would not be a maximum. Peoposition VII. — Theorem. Of all the polygons having equal areas and the same nuviher of sides, the regular polygon has the least perim- eter. For, if an irregular polygon of m sides, whose area is A, had a less perimeter than the regular polygon of the same area and of the same number of sides, we might transform it into a regular isoperimetrical polygon of to sides, having an area B > A ; this second regular polygon would then APPENDIS TO BOOS: if. 303 iiave the same number of sides as the first, witli a less perimeter and a greater area, which is absurd. ! Pjroposition YIII. — Thkoeem. Of two regular polygons of equal perimeter, that which has the greatest number of sides is the greatest. For, let ABCDEF be a regular polygon of 6 sides. If we take a point I on one of the sides, we may consider this polygon as an irregular polygon of 7 sides, in which the sides IC, ID would make an angle equal to two right angles : but this polygon is less than the F E regular polygon of 7 sides, and of the same perimeter. Hence, &c., &c. 304 ELEMENTS OF GEOMETRY. APPEiNDIX TO BOOKS VI. AND VII. COJSrSTKUCTION OF REGULAE POLYEDRONS. Peoposition I. — Peoblem. Having given one of the faces of a regular polyedron, or only its edge, to construct the jpolyedron. This problem presents five cases, which may be solved successively. 1°- Construction of the tetraedron. Let ABC be the equilateral triangle forming one of the faces of the tetra- edron. At the point 0, the centre of this triangle, erect the perpendicular OS to the plane ABC: let this per- pendicular terminate at the point S, so that AS = AB : join SB, SC, and the pyramid SABC will be the re- quired tetraedron. For, since OA, OB, OC are equal, the oblique lines SA, SB, SC are equally removed from the perpendicular SO, and are therefore equal. One of them SA = AB : hence, the four faces of the pyramid SABC are triangles, which are equal to the given triangle ABC. Besides, the solid angles of this pyramid are equal to each other, since they are formed each with three equal plane angles ; and therc- jj p fore the pyramid is a regular tetraedron. 2°. Construction of the hexaedron. 'Let ABCD be a given square. On the base ABCD . construct a right prism having an altitude AE equal to AB. It is evident that the faces of this prism are equal squares, and tliat its solid angles are equal to each other, each \ E \ n \ ? \ \ \ Appendix to books vi. and vii. 30S being formed with three right angles. Hence, this prism is a regular hexaedi-on, or cube. S*^- Construction of the octoedron. Let AMB be a given equilateral triaugle. On AB describe the square ABCD : at the point O, the centre of this square, erect tlie per- pendicular TS to its plane, terminated above and below this plane at T and S, so that OT = OS = AO : join, SA, SB, TA, etc. The solid SABCDT will be formed, composed of two quadrangu- lar pyramids SABCD, TABCD, having ABCD for a common base : this solid will be the regular octoedron re- quired. For, the triangle AOS is right-angled at O, as well as the triangle AOD : the edges AO, OS, OD are equal: therefore, these triangles are equal. Hence, AS = AD. In like manner we might prove that the other right- angled triangles AOT, BOS, COT, &c., are equal to the triangle AOD : therefore, the edges AB, AS, AT, &c., are equal to each other; and consequently the solid SABCDT is comprised under eight equal triangles, each equal to the given equilateral triangle ABM. Besides, the solid angles of the polyedron are equal to each other : for example, the angle S is equal to the angle B. For, it is clear that the triangle SAC is equal to the tri- angle DAC, and also the angle ASC is a right angle ; therefore, the figure SATC is a square equal to the square ABCD. But if we compare the pyramid BASCT with tlie pyramid SABCD, the base ASCT of the first may be placed on the base ABCD of the second ; then, the point O beino- a common centre, the altitude OB of the first wi'l coincide with the altitude OS. of the second, and the 306 ELEMENTS OF GEOMETRY. two p3'ramids will be identical ; therefore, the solid angle 8 is equal to the solid angle B ; and hence, the solid SABCDT is a regular octoedron. Scholium. If the three equal lines AC, BD, TS are perpendicular to each other, and intersect each other at their middle points, the extremities of these lines will be the vertices of a regular octoedron. 4°. Construction of the dodecaedron. Let ABODE be a given regular pentagon ; let ABP, CBP be two plane angles, equal to the angle ABO : with these plane angles form the solid angle B, and let K be the mutual inclination of two of these planes. Form, in like manner, at the points 0, D, E, A, solid angles, equal to the solid angle B, and situated in like manner. The plane OBP will be the same as the plane BOG, since they are both inclined the same quantity K to the plane ABOD. We may, then, in the plane PBOG, describe the pentagon BCGFP equal to the pen- tagon ABODE. If we do the same in each of the othei- planes GDI, DEL, &c., we shall have a convex surf'iieo PFGII, &c., composed of six regular equal pentagons, each inclined to its adjacent pentagon the same quantity K. Let pfgh be a second surface equal to PFGH, &c. ; these two surfaces may be joined in such a manner as to form a single continuous convex surface. For, the angle opf, for example, may be joined with the two angles OPB, BPF, to make the solid angle P equal to the angle B, and in doing this there will be no change in the inclination of the planes BPF, BPO, since this inclination is such as is required to form the solid angle. But at the same time that the solid angle P is formed, the side pf will be ap- APPENDIX TO BOOKS VI. AND VII. 307 plied to its equal PF, and the point F will be found united with the three plane angles PFG, pfe, efg^ which form a solid angle equal to each of the angles already formed. This will be done without any change taking place in the n o state of the angle P, or of that of the surface efgh, &c. ; for, the planes PFG, efp^ already united at P, make with each other the given angle K, as well as the planes efg, efp. Continuing thus, we see that the two sur- faces will adjust themselves mutually to each other, form- ing a sini^le .continuous surface, re-entrant on itself; this surface will be that of a regular dodecaedron, since it is composed of twelve equal regular pentagons, and all its solid angles are equal to each other. 6°. Construction of the icosaedron. Let ABC be one of its faces ; it is necessarj', in the first place, to form a solid angle with five planes equal to the plane ABC, and equally inclined, each to its adjacent plane. For this pur- pose, on B'C, equal to BC, construct the regular pentagon B'C'H'I'D'; at the centre of this pen- tagon erect a perpendicular to its plane, terminating in A', so that B'A' = B'C'; join A'C, A'H', AT, A'D', and the solid angle A', formed by the five planes B'A'C, C'A'H', &c., will be the required solid angle. For, the oblique lines A'B', A'C, &c., are equal; one of them, A'B', is equal to the edge B'C; hence all the triangles B'A'C, C'A'H', &c., are equal to each other and to the given triangle ABC. Furthermore, it is evident that the planes B'A' ', 308 ELEMENTS OF GEOMETEY. C'A'H', &c., are equally inclined, each to its adjacent; for, the solid angles B',C', &c., are equal to each other, since each is formed by two angles of equilateral triangles and one angle of a regular pentagon. Let us designate by K the inclination of the two planes in which the equal angles exist ; the angle K will be, at the same time, the inclination of the planes which compose the solid angle A', each to its adjacent plane. Now, if we make, at the points A, B, C, solid angles equal each to the angle A', we shall have a convex sur- face DEFG, &c., composed of ten equilateral triangles, each inclined to its adjacent, the angle K; and the angles D, E, F, &c., of its contour will unite alternately three and two angles of the equilateral triangles. Let us now suppose a second surface equal to the surface DEFG, &c. : these two surfaces might be mutually adapted by joining each triple angle of the one to the double angle of the othor ; and since the planes of these angles make with each otliei- the inclination K necessary to form a quintuple solid angle equal to the angle A, there will be no change effected by this junction in the form or relative position of either surface ; and the two thus united will form a single con- tinuous surface, composed of twenty equilateral tiiangles. This surface will be that of the regular icosaedron, since, besides, it has all its solid angles equal to each other. Proposition II. — Peoblem. To find the inclination of two adjacent faces of a regular polyedron. This inclination is deduced directlj'' from the construe- APPENDIX TO BOOKS VI. AND VII. 309 tion which has just been given of the five regular poly- edrons ; to -which it is necessary to add this problem of descriptive geometry, by which, having given the three plane angles which form a solid angle, we determine the angle which two of these planes make with each other. Jn the tetraedron. Each solid angle is formed by three angles of equilateral triangles ; it is therefore necessary to find, by the problem referred to, the angle which two of these planes make with each other : this angle will be the inclination of two adjacent faces of the tetraedron. In the hexaedron. The angle of two adjacent faces is a right angle. In the ootoedron. Form a solid angle with two angles of equilateral triangles and a right angle ; the inclination of the two planes in which the angles of the triangles are, will be that of two adjacent faces of the octoedron; In the dodecaedron. Each solid angle is formed with three angles of regular pentagons ; thus, the inclination of the planes of two of these angles will bd that of two ad- jacent faces of the dodecaedron. In the icosaedron. Form a solid angle with two angles of equilateral triangles and an angle of a regular pentagon ; the inclination of the two planes in which the angles of the triangles are, will be that of two adjacent faces of the icosaedron. Peoposition m. — Peoblem. Having given the edge of a regular polyedron, to find the radius of the inscribed sphere and that of the circum- scribed sphere to the jpolyedron. It is necessary, in the first place, to prove that every regular polyedron may be inscribed in a sphere, and may also be circumscribed about a sphere. 310 ELEMENTS OF GEOMETRY. cedilla: problem. Let AB be the common edge of two ad- jacent faces; let and E be the centres of these two faces, and CD, ED the per- pendiculars let fall from their centres on the common edge AB, which will meet the edge AB at the middle point D. The two perpendiculars CD, DE make with each other a known angle, which is equal to the inclination of the two adjacent faces, determined by the pre- But, if in the plane CDE, perpendicular to AB, we draw the indefinite perpendiculars CO and EO to CD and ED, intersecting in 0, the point will be the centre .'if the inscribed sphere and that of the circnm- Bcrihed sphere, the radius of the first being OC, and that of the second OA. For, since the apothegms CD, DE are equal, and the hypothenuse DO common, the right-angled triangle CDO is equal to the right-angled triangle ODE, and the perpen- dicular OC is equal to the perpendicular OE. But, AB being perpendicular to the plane CDE, the plane ABC is perpendicular to CDE, or CDE to ABC; besides, CO, in the plane CDE, is perpendicular to CD, the common in- tersection of the planes CDE, ABC ; hence, CO is per- pendicular to the plane ABC. For a like reason, EO is perpendicular to the plane ABE ; therefore, the two per- pendiculars CO, EO, drawn to the planes of the two ad- jacent faces through the centres of these faces, intersect in a common point, and are equal. Let us now suppose that ABC and ABE represent any two other adjacent faces: the apothegm CD will remain always of the same magni- tude, as well as the angle CDO, the half of CDE ; hence, the right-angled triangle CDO and its edge CO will be equal for all the faces of the polyedron ; therefore, if from the point 0, as a centre, and with OC as a radius, we de- APPENDIX TO BOOKS VI. AND VII. 311 scribe a sphere, this sphere will be tangent to all the faces of the polyedrons at their centres, for the planes ABC, ABE will be perpendiculars at the extremity of a radius, and the sphere will be inscribed in the polyedron, or the polyedron circumscribed about the sphere. Join OA, OB ; since CA = CB, the two oblique lines OA, OB, being equally removed from the perpendicular, are equal ; this will be the case with any two lines drawn from the centre to the extremities of the same edge ; hence, all these lines are equal. Therefore, if from the point 0, as a centre, and with a radius OA, we describe a spherical surface, this surface will pass through the vertices of all the solid angles of the polyedron, and the sphere will circumscribe the polyedron, or the polyedron inscribed in the sphere. This being assumed, the solution of the given problem may be effected without any diffi- culty, as follows: Having given the edge of a face of the polyedron, construct this face, and let CD be its apothegm. Find, by the pre- cedina; problem, the inclination of two adjacent faces of the poly- edron, and make the angle CDE equal to this inclination. Take DE equal to CD, draw CO and EO perpendicular to CD and ED ; these two perpendiculars intersect in a point 0, and CO will be the radius of the sphere inscribed in the polyedron. On the prolongation of DC, take CA equal to the ra- dius of the circle circumscribing a face of the polyedron, and OA will be the radius of the sphere circumscribing this polyedron. Scholium. .We may draw from the preceding proposi- tions miny consequences. 313 ELEMENTS OF GEOMETRY. 1°. Every regular polyedron may be divided into as many regular pyramids as the polyedron has faces ; the common vertex of these pyramids will be the centre of the polyedron, which is at the same time the centre of the in- scribed and circumscribed spheres. 2°- The solidity of a regular polyedron is equal to its surface multiplied by one-third of the radius of the in- scribed sphere. 3°. Two regular polyedrons of the same name are sim- ilar solids, and their homologous dimensions are propor- tional ; hence, the radii of the inscribed and circumscribed spheres are to each other as the edges of these poly- edrons. 4°. If we inscribe a regular polyedron in a sphere, the planes drawn through the centre and the different edges of the polyedron will divide the surface of the sphere into as many equal and similar spherical polygons as the poly- edron has faces. THE END.