/-•<": 7T. yy. cc- THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA GIFT OF Kenneth K. Noble HOLWES &00K CO, -^3 8, Main St. Los Angeles Erkata C'ORKIGE 11 1 with with- 13 6 that s tliat is 16 17-26 base etc. angle A by drawing the line AD. The triangle .456' will thus be di- vided into two triangles ABD and ACD, which will be equal (No. 11); and therefore the angles B and C opposite the common side J.i> will be equal. Q. E. D. Cor. The line that bisects the angle at the vertex of an isosceles trianirle, bisects the base of the tri- angle, and is perpendicular to it. 20 24 2(n— 2), or 2;i— 4. 2(r<— 2) right angles. 32 21 POHPOSITION PROPOSITION 33 25 DCB DOE 27 DCB DCE (54 14,-15,-16 and an equal included between equal angles, arc angle at ^, are equal (No.27,Cor'l; equal (No. 12); 20,-21 and an equal angle betv/een two equal angles, at^, 68 19 CD AD 87 15 EF EA 94 19 ao : AO Vo: VO .. 28 a^'.AO' W'.VG' 103 18 and 20 ? h 6 104 ^ 1 h 6 124 26 (No. 124, (No. 134, V Digitized by tine Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofgeometOObaymrich ELEMENTS GEOMETEY FOR THE USE OF BEGINNERS. BT JOSEPH BAYMA, S.J., Prof essor of Mathematics in Santa Clara CoUege, S. J., Santa Clara, California. SAN FBANCISCO : A. WALDTEUFEL, 737 MARKET STREET. 1885. Copyright, 1885, By a. WALDTEUFEL. GIFT CfM£3 CONTENTS. Preface, ...••••.••• 5 Preliminary notions, ..•••••• 7 BOOK I. Angles and triangles, ..•..•«• 9 BOOK II. Rectilinear areas, and relations of similar figures, ... 28 BOOK III. The circle and the regular polygons, . . . • » 43 BOOK IV. Geometric constructions, , , 69 BOOK V. Planes, and polyhedral angles, . • • • t • 71 8 709 4 CONTENTS. PAGE BOOK VI. Surfaces and volumes of polyhedrons, 86 BOOK VII. Surfaces and volumes of round bodies, 105 BOOK VIII. Spherical geometry, ......... 120 PKEFAOE. The present treatise, like the preceding one on Al- gebra, has been written for the instruction of the younger students of Santa Clara College. Among the many text-books of Geometry published in this coun- try which we have read or used in past years, some have been found too diffuse, others full of cumber- some matter, others again too comprehensive and too difficult for beginners. Comprehensive books are very useful in the hands of those whose minds are already formed ; but experience has taught us that a judicious parsimony proves more successful in encouraging the mental efforts of young beginners, amid the many dif- ficulties arising from the giddiness natural to their age, as well as from the number of their scholastic duties. It has been my purpose to offer to this class of students an elementary Geometry sufficiently devel- oped to embrace all the theorems that have a real importance, and yet sufficiently concise and plain to make it possible to learn it all within six or seven months in the first year of the mathematical curri- culum. The treatise, of course, contains nothing that 6 6 PREFAGJS. is really new ; it is not, however, a mere compilation. The order and method which I have followed, and the proofs which I have chosen, are such as will, in my opinion, materially lessen the labor, and at the same time foster the industry of the young, in the acquisi- tion of mathematical knowledge. Such, at least, is the goal I have striven to reach. J. B. ELEMENTS OF GEOMETRY. PEELIMINAKY NOTIONS. 1, Geometry is that branch of Mathematics which treats of lines, surfaces, solids, angles, and their manifold relations. A line is a length in space having no breadth and no thickness. A surface is an extension in length and breadth, but having no thickness. A surface when measured by a given unit is called an area. A solid is a magnitude having length, breadth, and thickness. Length, breadth, and thickness are called the three dimensions of geometric bodies. A solid when measured by a given unit is called a volume. An angle is the divergence, or difference of direction, of two lines, with reference to a common point of inter- section. 2, A line is nothing but the track of 2i. point moving in space. A point has no dimensions; but a moving point, by its movement, traces a line, whose length mea- sures the length of the movement. In fact, no line can be drawn but by moving from a point in space towards some other point in space in a continuous manner. Lines are either straight or curved. A straight lins is a line whose direction is one and the same throughout. 8 ELEMENTS OF A curved line, or a curve, is a line whose direction is continually changing by imperceptible degrees. Surfaces are either jplane or curved, A plane surface is one on which straight lines can be drawn in all direc- tions. A ciirved surface is one on which straight lines cannot thus be drawn. Angles are either right or oblique ; and oblique angles are either acute or obtuse. When one straight line meets another, it makes with it two angles, which are called adjacent angles. If these two angles are equal, they are called right angles; but if unequal, then the lesser is called an acute, and the greater an obtuse angle. . The lines which form the sides of an oblique angle are called oblique, whilst those which meet at right angles are said to hQ perjpendicular to each other. The point where two lines meet is called the vertex of the angle formed by them ; and the angle itself may be designated by a letter attached to the vertex. When, however, two or more angles are formed about the same point, then, to avoid confusion, every angle is designated by three letters, of which the middle one marks the ver- tex of the angle, and the other two the directions of the sides. GEOMETRY. BOOK I. ANGLES AND TRIANGLES. 3. Geometric quantities are quantities having dimen- sions in space. All sucli quantities are continuous, be- cause the lines which represent their dimensions {lengthy hreadth^ and thickness) are the traces of continuous move- ments. Lines may be compared with one another with regard both to their length, and to their relative position. Their position implies not only their distances, but also their mutual inclination, or the angle contained between them. It is by this simplest of all relations between straight lines that we must start our geometrical inves- tigations. PEOPOSITIOI^ I. 4:* If a straight line meets another^ the sum of the adjacent angles will he equal to two right angles. Proof. Let DC meet AB at C. Drawing CE perpen- dicular to AB, we have these two equations, DCA=ACE-\-ECD, BGD^ECB-ECD, which being added together give DCA-\-BCD = A CE-\- ECB. But the sum ACE ^ ECB is equal to two right angles. ^M- 10 ELEMENTS OF Therefore the sum DCA-\-BCD is equal to two right angles. Q. E. D. Cor. I. If one of the adjacent angles is right, the other also is right. Cor. II. The sum of all the angles formed about the same point on the same side of a line, is equal to two right angles. PEOPOSITIOIT 11. 5. If two straight lines intersect each other, the oppo- site, or vertical, angles will he equal. Proof. Let^^andi>^in- a, tersect at C. The angles A CD and ECB are called opposite or vertical y and so also are the angles A CE and DCB. ^w (Ko. 4), A CD + A CE = two right angles, ECB + ACE= two right angles ; Therefore ACD + A CE= ECB + A CE; and, bj re- duction, ACD = ECB. Q.E. D. It may be shown, in a similar manner, that A CE=DCB. Cor. The sum of all the angles that can be formed about a point is equal to four right angles. PKOPOSITIOIN^ III. ^. If a straight line intersects two pa/rallel lines, the sum of the two interior angles on the same side of the intersecting line is equal to two right angles. Proof. Two lines are said to be pa/rallel when they aEOMETRY. 11 lie in the same plane in a similar direction, that is, with ont any inclination towards each other. Hence parallel lines keep the same relative distance throughout their extent, and can never meet, how far soever produced. Let, then, AB and CD be parallel, and let FE in- tersect AB at G, and CD at H, Since GB and HD run in the same direction, they are equally inclined to the line FE. In other words, they make with it equal angles. Consequently BGF^DHG, or, adding the angle BGIl to both members of this equation, BGF^ BGH^ DHG -^BGR. But (:N'o. 4) the sum BGF+BGR is equal to two right angles. Therefore the sum 2>^6^-|"^^^^s ^^^^ equal to two right angles. Q. E. D. Cor. The alternate angles between parallel lines are equal. Alternate angles are those which lie at the oppo- site sides of the intersecting line, the one having for its sides the intersecting line and one of the two parallel lines, the other having for its sides the intersecting line and the other of the two parallel lines. Such are the an- gles A Gil and GIID, both interior, and such are also the angles FGB and CHE, both exterior. It is evident t\\2X AGII=GIID ; for both these angles are equal to the same angle FGB, 12 ELEMENTS OF PEOPOSITION ly. 7. Two lines will he pa/rallel if the sum of the two interior angles on the same side of an intersecting line , JiJ. he equal to two right angles, -"Jl ' ^ Proof Let the lines J.^ P / ^; and CD be intersected by ^ c, r :- a line FE so that the sum y^__ Qx/fl^ d of the angles BGII and GHD shall be equal to two q_ right angles. Then, since (Ko. 4) the sum of the ad- jacent angles BGH and BGF\^ also equal to two right angles, we shall have BGII\- GHD^zBGII+BGF, that is, GHD=zBGF, Now, this equality cannot subsist unless BG and HD run in the same direction, viz., unless they are parallel. Q. E. D. Cor. If the alternate angles between two lines are equal, the two lines are parallel. For, the alternate an- gles cannot be equal unless GHD = BGF. PKOPOSITIOIS' Y. 8. The sum of the three angles of any triangle is equal to two right angles. Proof Let ABC be a tri- angle. Produce AB to D, and from B draw BF paral- lel to AC. Then (ISTo. 6) the angles FBD and CAB will GEOMETRY. 13 be equal, as also QBE and AGB, which are alternate angles. Hence CAB-\-AGB^ CBA = EBD^ GBE^ CBA; and, as the second member of this equation is equal to two right angles (Ko. 4), it follows that the first mem- ber, too, that s, the sum of the angles of the triangle, is equal to two right angles. Q. E. D. Cot. 1. If any side of a triangle be produced, the ex- terior angle is equal to the sum of the two interior oppo- site. Thus CBD=CAB+ACB. Cor, II. If the sum of two angles in one triangle is equal to the sum of two angles in another triangle, the third angles will also be equal each to each. Cor. III. If one angle of a triangle be a right angle, the sum of the other two will also be equal to a right angle. Cor. lY. All the angles of a triangle may be acute ; but no triangle can have more than one right angle, or more than one obtuse angle. PEOPOSITIOiq" YI. 9, If two angles have their sides parallel, the two an- gles will he either equal or sujpjplementary . Proof. Let ^6^ be parallel to BD^ and AH parallel to BF. Then we have (]S"o. 6) HAC=FEC=FBD; that is, the angle FBD is equal to the angle HA C. Prolonging FB to G, the angle DBG will be supplementary to the angle DBF, or to its equal CAH; 14 ELEMENTS OF for, two angles are called supplementary when tlieir sum equals two right angles, as is the case^ith the angles DBE and DB G. Q. E. D. PEOPOSITION YII. 10, If two cmgles have their sides respectively jper- pendicular^ the two angles will l)e either equal or sup- plementary. Proof. Let the angles BAG and GDB have their sides respectively perpendicular. Producing GD till it meets the side AB in q jEJ we shall have two right-angled triangles, AGE and BDE, in the first of which the sum AEG-\- GAE is ^ B E equal to a right angle (No. 8, Cor. III.), and in the sec- ond the sum BED -f- BDE is also equal to a right an- gle ; and therefore AEG-^ GAE= BED + BDE. But AEG and BED are one and the same angle ; hence, by reduction, GAE = BDE. The angle BDG is sup- plementary to the angle BDE. Hence it is supplemen- tary also to the angle GAE. Q. E. D. PKOPOSITION YIII. • 11. Two triangles which have two sides, and the in- cluded angle of the one equal to two sides and the in- cluded angle of the other, each to each, are equal in all their parts. GEOMETRY, 15 Proof. In the triangles ABC and DEF, let AB = DE, AC=DF, and A=:D. Let the triangle DEF be placed upon ABC ^o that the side JDE shall coincide with its equal AB. Then, since I)=A^ the side i>i^will lie in the direction of A C ; and since DF—AC^ the point F will coincide with the point (7, and consequently the side EF will coincide with the side BC The triangles therefore coincide throughout : which means, in other words, that they are equal in all their parts. Q. E. D. PEOPOSITIOK IX. 12. Two triangles which have two angles^ and the in- cluded side of the one equal to two angles and the in- cluded side of the other, each to each, are equal in all their parts. Proof In the triangles ABC and DEF, let A=D, B = E, and AB — DE. Let the triangle DEF be brought upon J.^ (7 so that the side ^Z> shall coincide with its equal AB, Then, since Z>=:^, the side DF will lie in the direction of A C j and since E=: B, the side EF will lie in the direction of BC. The triangles therefore coincide throughout, and are equal in all their parts. Q. E. D. PEOPOSITIOK X. 13. In every triangle, one side is less than the sum of the other two sides. Proof Any side J. (7 of a triangle ABC exhibits the 16 ELEMENTS OF shortest patli from the point A to the point 6'/ for A C, being a straight line, leads di- rectly from A to C. It follows that the crooked path composed of the sides AB-^BC is longer than J. 6^. Q.E.D. A Cor. From the inequality AC of a given line BC. For this perpen- dicular must pass through the vertex A of an isosceles triangle whose base is bisected in D, and there can be but one direction from D to A. Cor. lY. Any point in a perpendicular drawn from the middle of a line is at equal distances from the two extremities of the line. For any such point can be the vertex of an isosceles triangle of which that line will be the base. PKOPOSITIOK XIL 16. In every triangle the greater angle is o^osite the greater side. Proof. In the triangle ABC, let AC>AB. On the greater side AC, take AD = AB, and draw BD, The triangle ABD will be isosceles, and consequently the angles ABD and b< ADB will be equal. But ADB = DCB + DBC (JSTo. 8, Cor. I.); there- fore also ABD = DCB + DBC; that is, ABD >DCB, and a fortiori ABC >DCB. Q.E.D. Cor, In every triangle the greater side is opposite to 18 ELEMENTS OF the greater angle. For in the triangle -^^(7, if ^ > (7, the side A C cannot be equal to the side AB, as the tri- angle is not isosceles ; and ^(7 cannot be less than AB, because then B should be less than (7, as we have joist shown. Therefore AC > AB. PKOPOSITIOlSr XIII. 17. Two triangles having their sides respectively equal, are equal in all their 'parts. Proof, Let J.^C'and J.(7i> be two triangles having the sides AB=zAI), BO=z DC, and AC=AC. Join them together by their longest sides A C, and draw BJ). The triangles ABB and BBC will both be isosceles. Hence the angles BBCmd BBC will he equal, as also the angles BCA and DC A, Hence also the angles J.^C'and ADC will be equal (ISTo. 8). And therefore the two triangles are equal in all their parts. Q. E. D. PKOPOSITIO:tT XIY. 18. If two triangles have two sides respectively equal, and the included angles unequal, the third sides will he unequal : viz., the greater side will he ojpposite the great- er a/ngle. Proof Let the two triangles ^^6^ and DBC\i2iYe two sides respectively equal, as AB=^BDy and BC = BC; GEOMETRY. 19 and let the angle ABC be greater than the angle DBG. The two triangles being placed so that their shortest sides BG shall coincide, join AD ; and from B draw BE per- pendicular to AD^ and finally from the point F^ where BE intersects AG^ draw FD. The triangles ABD and AFD will be isosceles, because ABz=.BD and AF=FD. ISTow, A G= AF+ FG= FD + EG. But FD +FG > GD (No. 13). Therefore AG>GD. Q. E. D. PKOPOsiTioj^ xy. 19. TTie shortest line from ajpoint to a straight line is the ^perpendicular from thatjpoint to the line. Proof Let (7 be a point, and AB a straight line. Draw GD perpendicular, and GE oblique, to AB. In the resulting triangle EGD, GE is greater than GD., because it is opposite a greater angle (No. 16); and this will be true also of any other oblique line, as GE, GG, etc. Hence the perpendicular GD is the shortest line that can be drawn from G to the straight line AB. Q. E. D. Gor. I. The perpendicular is the true distance from a point to a straight line. 20 ELEMENTS OF Cor, II. The oblique lines are greater or smaller ac- cording to their greater or lesser obliquity. Thus CG > CF^ as the angle CFG opposite to CG is greater than the angle CGF opposite to CF. Cor. Ill Lines of equal obliquity have equal length, and are sides of isosceles triangles. They meet the line to which they are drawn, at equal distances from the foot of the perpendicular. Cor. I V. From a point to a line, and also from a line to a point, only one perpendicular can be drawn. For there is only one shortest line. PKOPOSITIOK XYI. 20, In any jpolygon, the sum of the interior angles is equal to twice as many right angles, less four, as the fig- ure has sides. Proof Let ABCDE be any polygon. From any vertex A draw diagonals, as AC, AD ; the polygon will thus be divided into triangles, every one of which will contain two right angles. If the polygon has n sides, the tri- angles will be 71 — 2 ; hence the sum of all the angles will be 2 (ti — 2), or 2/i — 4. Another proof. From a point within the polygon, draw OA, OB, DC, etc. The polygon will be divided into as many trian- gles as it has sides. The sum of all the angles will therefore be equal to twice as many right an- gles as there are sides. But the CfEOMETRY, 51 angles about 0, whose sum is equal to four riglit angles, do not belong to the polygon. Hence, taking them away, the angles of the polygon will amount to twice as many right angles, less four, as the figure has sides. Q. E. D. PROPOSITIOlSr XYII. 21.7/^ the opposite sides of a quadrilateral are equal, they are parallel, and the figure is a parallelogram. Proof. Let AB CD he a quadrilateral, in which AB=z CD, and AD = BO. Draw the diagonal AC; the tri- angles A CD and A CD, as having equal sides, will be equal in all their parts (No. 17). Hence the angles DAC and ^6^J., opposite to the equal sides DC and AD, will be equal, and therefore the sides AD and DC will be parallel (No. 7, Cor.) Simi- larly, the angles A CD and DA C, opposite to the equal sides AD and DC, will be equal, and therefore the sides AD and DC will be parallel. Q. E. D. Cor. I. In all parallelograms, the opposite sides, as also the opposite angles, are equal. For, the diagonal will make equal alternate angles between the parallel sides ; and consequently will divide the figure into two equal triangles, in which the equal sides will be parallel and opposite. Cor. II. If two parallel lines intercept two other par- allel lines, they form a parallelogram. Cor. III. The diagonals of a parallelogram bisect each other. For, if AD CD is a parallelogram, the 22 ELEMENTS OF diagonals make the alternate angles ABD DC A— CAB. The trian- gles AOB and DOC have therefore two equal angles / about the equal sides AB / .^.-•••- 0\ and DC^ and consequently /--•••-" X^ they are equal in all their ^ parts. Hence AO—OC 2.-^^ BO— OD. Cor. IV. The diagonals of a rhombus bisect each other at right angles. For the rhombus is a parallelogram whose sides are all equal ; whence it follows that the two diagonals are the bases of isosceles triangles. Hence A C, which bi- sects BD^ is perpendicular to BD (No. 14, Cor.) BDC, and B GEOMETRY, 23 BOOK II RECTILINEAR AREAS, AND RELATIONS OF SIMILAR FIGURES. 23. In measuring surfaces, the unit of measure is a square having for its side the unit of length. This unit is arbitrary. Let the square a^ be our unit of surface. Then any rectangular surface ABGD will be measured by ascertaining how many times the surface a? is contained in it. If, for instance, the base AB is equal io five units of length, and the altitude AD is equal to six of the same units, then the sur- . . face ABGD will contain six I I I I I I I **' times a row of fiA)e square units, ' that is, 30 units of surface, which are obtained by mak- ing the product of the base AB into the altitude BC ; and then we say that the area of ABGD, as measured by such a unit, is = 30. We have supposed AB and BG to contain the unit of length a an entire number of times. Let now m and n be any two numbers, and let us assume AB=zma, BG:=zna, Then ABxBG=mna\ and AB BG m X n ITow, it is plain that these equations must be true, whether m and n are entire or fractional, and whether they are commensurable or incommensurable with a. For we have in all cases 24 ELEMENTS OF AB m ma TYh BC n na n Hence the area of a rectangle is always expressed by the product of its base into its altitude, whatever may be their relation to the unit of measure. Surfaces that are not rectangular can also be measured by any square unit, as we are going to show in this sec- ond book. PEOPOSITIOJST I. 23. Parallelograms having equal bases between the same parallels have equal areas. Proof. Let ABCD be a parallelogram. Prolong CD towards E^ and construct on AB another parallelogram, ABFE, by drawing any two parallel lines AE and BF. The two parallelograms will have equal areas. For the area of ABCD is equal to the area of the whole figure ABDE minus that of the triangle A CE\ and the area of ABFE is equal to the area of the whole figure ABDE minus that of the triangle BED. J^ow, the triangles ACE and BED are equal (No. 11). Therefore the expression of the area of the two parallelograms is exactly the same ; and consequently those two areas are equal. Q. E. D. Cor. 1. The area of a paral- lelogram is equal to the base multiplied by the perpendicular height. For construct upon the same base and between the same parallels a rectangular parallelo- GEOMETRY, 25 gram ABFE ; the areas ABCD and ABEF will be equal. But ABFE= AB X BF (No. 22) ; therefore ABDG^ABy^BFz^ABy^DH. Got. II. As the diagonal divides the parallelogram into two equal triangles (Ko. 21, Cor. I.), the area of a triangle is ^ equal to one-half the area of the y^ \ parallelogram, having the saine x \ hase and the same height. Thus ^ ^ area ABG^ \ABx GD. Gor. 111. Triangles having equal bases and equal heights have also equal areas ^ for the expression of both areas is the same. Gor. 1 V. The area of a trapezoid (a quadrilateral, of which two sides are parallel) equals one-half the sum of its P ^5 parallel sides multijplied hy their / ^^ i\ distance. For if ABGD is a / ^■■■■^ \ \ trapezoid, it may be divided by a ^ H — b diagonal A G into two triangles, J.^(7and ADG, having parallel bases AB and i>(7and a common altitude GH. Hence the area of the trape- zoid will be expressed by \ABy. GH-^ \ GDx GH, or GH (^^+^"^ )» PEOPOSITIOIsr II. 34. The square described on the sum of two lines is equal to the sum of the squares m^ade on each line, jplus twice the rectangle contained by the two lines. Proof Let AB and ^6^ be the two lines, and A GDF 26 ELEMENTS OF the square described on their sum A C. Draw BE par- allel to CD, and complete the square ABHG by drawing GI parallel to ^ ED A G. The area A CDF will thus be the sum of four areas, viz. : ABGII G = AB\ HEFG^GHy^EH^AB xBC,EIBC^HBxBCz=^ABx BG, and EIDE=:zBG\ Conse- quently the square of the sum AB + BGis> AB''-\-'iABxBG\BC\ Q.E.D. H peopositio:n hi. 25. The square described on the difference of two lines is equal to the sum of the squares made on each line, minus twice the rectangle contained hy the two lines. Proof Let AB and BGh^ the two lines, and A CDE the square described on their difference AG. Construct on AB the square ABFG, and on BG the square GBIK, and pro- duce ED to H. The area of the square A GDE will be equal to the area of the whole figure, minus the two areas GEJIFaind DHIK. JS'ow, the area of the whole figure is ABFG-\- GBIK, or AR^BG\ and the two areas GEHF and DHIK are P A C' GEOMETRY. 2l both equal to ABxBG. Therefore the square of the difference AB — BG is AB'-2ABxB0+Ba\ Q.E.D. PKOPOSITIO]^ lY. 36. The difference of the squares described on any two lines is equal to the rectangle of their sum hy their difference. Proof Let AB and- AG h^ the two lines, and ABBE, A GHF, their squares. The difference between these two squares is the sum of the two rectangles FGDE and GBGH. 1^0 w, FGDE=^FG x FE=ABx BG, and GBGH^GHxBG^AG XBG. Hence their sum is ABxBG+AGxBG=::{AB-\-AG)BG, which, because BG=zAB—AG, is the same as {AB -\- A G) {AB -AG)\ and therefore AB'- AG'^ {AB + AG) {AB - A G). Q. E. D. PKOPOSITIOE" Y. 2*7. The squa/re made on the hypotenuse of a right- angled triangle equals the sum of the squares made on the other two sides. Proof. By the term hypotenuse is meant the side op- posite the right angle. Let, then, AB be the hjpote- 2S ELEMENTS OF nuse of the triangle ABC. Make the squares ABHK, CBDE.ACFG, and draw CL perpendicular to HK, Joining GB and CH^ the triangles E GAB and CAH will be equal, as thej have two equal sides en- closing an equal angle ; for the angles GAB and CAH are both composed of a right angle and of a common an- gle CAB. But the triangle GAB has the same base and the same altitude as the square A CFG ; hence its area is equal to one-half the area of the same square. On the other hand, the triangle 6ZA^has the same base and the same altitude as the rectangle AILH ; and thus its area is equal to one-half the area of the rectangle. Therefore twice the triangle GAB^ that is, the square A CFG, equals twice the triangle ACH, that is, the rectangle AILR, In the same manner, joining AD and CK, it will be proved that the triangles ABD and CBK are equal ; also that ABD is equal to one-half the square CBDE, and that CBK is equal to one-half of the rectangle BKLIy and consequently that the square CBDE and the rectangle BKLI are equal. But the two rectangles AILH and BKLI, taken to- GEOMETRY. . 29 gether, make the square ABKH, Therefore the squares AGFG and GBDE^ taken together, equal the square ABKH, Accordingly, AB'z=.AG'^BG\ Q. E.D. Cor. Two right-angled triangles, in which the hyjpote- nuse and one side are respectively equal, are equal in all their parts. For let the two triangles be ABC and A'B'C\ AB and A'B' being the hypotenuses. Then we shall have in the one A:B' = BC'+ AC' ; whence AG'^ AB'- BG\ and in the other A'B''^ B'C'-^A'G"; whence A' C''=A'B''-B'C'\ If, then, the hypotenuses AB and A^B^ are equal, and the sides BC and B^G^ are also equal, the third sides A^G^ and AC must also be equal; which shows that the two triangles are equal (No. lY). PROPOSITIOI^ YI. 2S, In any triangle, the square of a side opposite an acute angle equals the sum of the squares of the other two sides, minus twice the ^product of one of these sides into the projection of tJie other upon its direction. Proof. Let ABC be a triangle, ^ in which AC i^ opposite the acute y\ angle B. Draw CD perpendicu- y^ lar to AB. In the right-angled y^ triangle A CD, we have (Prop. Y.) / AC'zz^ AB'+ Cl>\ A 30 ELEMENTS OF But (Eo. 25) Al>'z=^ {AB-DBf^ AB'- 2AB x DB-\-DB\ and (Ko. 27, Cor.) CD''^ CB'- I)B\ Hence, substituting and reducing, AG^=AB''-\- CB'- "iABxBB; DB being the so-called jprojection oi BG upon AB, Q. E. D. PEOPOsiTioN yn. ^9. In any obtuse-angled triangle^ the square of the side opposite the obtuse angle equals the sum of the squares of the other two sides ^ plus twice the prodxtct of one of these sides into the projection of the other upon its direction. Proof Let ^^6^ be a C triangle, in which AG i^ ' ^ opposite the obtuse angle ^^ / B. Draw GD perpendicu- ^^ / lar to AB produced. In ^^ / the right-angled triangle '^ 1. AGD^ we have AG" z::^ AD'' -\- GD\ But AD'' =3 {AB-\-BDY = AB'-\- 2AB x BD+BD% and GD'=GB'-BD%' hence, substituting and reducing, AG'z=:AB'+ GB'+2ABxBD; BD being the projection of the side BG on the direc- tion of the other side AB. Q. E. D. GEOMETRY. 31 PEOPOSITIOlNr YIII. 30. In any triangle^ the sum of the squares described on two sides is equal to twice the square of one-half the third side^ plus twice the square of the line drawn to the middle jpoint of that side from the vertex of the ojpjposite angle. Proof. In the triangle ABC bisect AB at E and join CE. Draw also GH perpendicular to AB. Then in the triangle ACE we shall have (JSTo. 29) AG'^AE'^CE' -\-2AExEH, and in the triangle BCE we shall have (E"o. 28) BG'=: BE'+ CE'- 2BEx EH. Adding together these two equations, and observing that AE and BE are equal, we shall have AC^ + BG'' = 2AE' + 2(7X^ Q. E. D. Cor. The sum of the squares of the four sides of a jparallelogram is equal to the sum of the squares of its diagonals. For drawing AD and BD parallel to BG and A G respectively, and prolonging CE to the point Z>, the figure will be a parallelogram, of which AB and CD are the diagonals. Now, the equation just found, multiplied by 2, gives 32 ELEMENTS OF wliicli may be written thus, AO'+£D'+BC'+A^'= AB'+ CD\ PEOPOSITIOI^ IX. 31, Parallelograms Jiamng equal altitudes are to one another as their hases / and, if they have equal bases, are to one another as their altitudes. Proof. Let P designate the area of a parallelogram having a base h and an altitude h. Let also P' designate the area of another parallelogram having a base V and an altitude A'. We shall have (:N'o. 23, Cor. I.) P:=hXh, P'^l'y.h'; hence P\ P::hxh\ h'xh'. If A = A^then P'.P'wl'.y, and if l-=^V, then P \P'\\h\h'. Q. E. D. Cor. Triangles having equal altitudes are to each other as their hases / and triangles homing equal hases are to each other as their altitudes. For triangles are semi-parallelograms, and \P:^P:'.P'.P'. POEPOSITION X. 32. In any triangle, a line parallel to one of the sides divides the other sides into ^proportional jparts. Proof. In the triangle ABC, draw PE parallel to GEOMETRY. 33 BC. Draw BE, The triangles BDE and DAE, with a common vertex E, will have the same alti- tude; hence they will be propor- tional to their bases (No. 31, Cor.) Therefore BDE: DAE:: BD : AD. ]N"ow draw DC. The triangles ODE and DAE, with a common ^ vertex Z>, will have the same alti- tude; hence they, too, will be proportional to their bases. Therefore DOE: DAE:: GE\EA, But, as DE and BG are parallel, the triangles BDE and DOE, which have a common base DE, have also the same altitude, viz., they are equal. Hence, in the two proportions just found, the first ratios are equal. Consequently th e last ratios also are equal, viz. : BD : AD :: CE : EA. Q.E.D. 33. Conversely, If a line divides two sides of a triangle into proportional parts, that line is paral- lel to the third side of the triangle. For the two ratios BD : AD and CE : EA cannot be equal, nor can they form a proportion, unless the ratios BDE : DAE and DC^: DAE, with which they are strictly connected, be also equal, that is, unless the triangles BDE and DCB be equal. Now, these triangles have a common base DE I they must therefore have an equal altitude ; and consequently their vertices B and C must be on a line parallel to their base DE, 34 ELEMENTS OF \ PEOPOSITIOlSr XL /^' 34. Similar (that is, equiangular) triangles have their corresjponding sides proportional. Proof. In tlie triangle ABC, draw DE parallel to AB. Then the triangle CDE will be equiangular with ABC, and we shall have AD \ DC w EB \ EC, whence, bj composition, AD-]-DC\DC:\EB+EC\ EC, that is, AC'.DCwBC'.EC. (1) If we now draw EF parallel to A C, we shall have also BF :FA::BE :EC, whence, by composition, BF+FA :FA:: BE+EC : EC, that is, BAiFA :: BC : EC, or, because AF= DE, BA:DE::BC :EC (2) Hence, comparing the two proportions (1) and (2), AC:DC::BC :EC::BA : DE. Therefore the corresponding sides of any two equiangu- lar triangles ABC and DEC are proportional. Q. E. D. J^ote. The corresponding sides are those which are op- posite to equal angles, and are called homologous, be- cause they form equal ratios. GEOMETRY. 35 35. Conversely, Triangles whose sides are jproj^or- tional are equiangular and therefore similar. For let there be between the sides of two triangles ABC and abo the proportion AB:ab::AG'.aG::BC\hG. Taking on ^^ a length AD =:ab, and on AG ?i length AE— ac, and drawing DE, we shall have AB : AD :: AC : AE. Now, the existence of this proportion requires that DE be parallel to BG (No. 33). Therefore DE and BG are parallel, and the triangles ABG and ADE are equian- gular. Hence we have also AG'.AEwBG'.DE, that is, AG\ ao :: BG : DE. But we have already AG : ac :: BG : ho ; therefore DE= ho. Thus the three sides of ADE are respectively equal to the three sides of aho; and therefore these two trian- gles are equal in all their parts. But ADE and ABG are equiangular. Therefore abo and ABG, too, are equiangular. Q. E. D. Gor. I. Triangles whose sides are respectively paral- lel are equiangular and similar. Gor. II. Triangles having an equal angle between two proportional sides are equiangular and similar. 36 ELEMENTS OF PEOPOSITION^ XII. 36. Triangles whose sides are respectively perpendi- cular a/re equiangular and similar. Proof. Let ABC and DEF have tlieir sides mutual- ly perpendicular, that is, DE perpendicular to AB^ EF to BC,2indiJDF to AC. Pro- long DE, EF, ED till they meet AB, BC, and AC re- spectively. Then we shall have three quadrilaterals AHDG, HEIB, FICG, in each of which two angles will be right and two other angles will be supplementary ; for the sum of the four an- gles must be equal to four right angles (No. 20). Thus we have GAH -\- GDH— two right angles; but we have also (Ko. 4) EDF-\- GDH = two right angles. Hence GAH= EDF. And, in the same manner, it will be found that HBI is equal to DEF, and ICG equal to DEE. Q. E. D. PEOPOSITIOlSr XIIL 37. In any triangle, the line that hi sects one angle divides the opposite side into parts proportional to the adjacent sides. Proof. Let ABC be any triangle. Draw CD bisect- ing the angle (7, and produce J. 67 by a length CE= CB, GEOMETRY. 37 Draw EB ; the triangle EGB will be isosceles, viz., the angles QBE and CEB will be equal. Now, the exte- rior angle ACB is equal to the sum of the two angles CBE and CEB Q^o, 8, Cor. I.); therefore ACB, or 2DGB, is equal to 2 CBE, and DCB=z CBE. But these last two angles are al- ternate angles between CD and BE. Hence CD and BE are parallel (ISTo. 7, Cor.), and the triangles ACD and AEB are similar; and therefore ^i> : DB :: AC : CE ; or, since CE and (7^ are equal by construction, AD'.DBwAC \ CB. Q.E.D. PKOPOSITION XIY. 38. Triangles which have one angle equal are to each other as the rectangles of the including sides. Proof. Let the triangles ABC and CDE have one equal angle (7, and conceive them to be so placed that the equal angles ACB and DCE shall be vertically op- posite. Then join AE. .The triangles A CB and A CE will have a common vertex A, and consequently (ISTo. 31, Cor.) they will be to each other as their bases. Thus ACB: ACE:: BC : CE. The triangles ACE and ECD will also have a common 38 ELEMENTS OF vertex JE, and consequently tliej, too, will be to eacli other as their bases. Hence ACE \DCE\\AG : CD. Multiply these two proportions, term by term, and can- cel the common factor A CE ; the result will be ABC: CDE::ACxBC: CDxCE, Q.E.D. PEOPOSITIOISr XY. 39. In a right-angled triangle^ the perpendicular drawn from the right angle to the hyjpotenuse^ divides the triangle into two triangles similar to the given tri- am^gle and to each other. Proof. Let ^^6^ be a triangle right-angled at (7, and let CD be the perpendicular on the hypotenuse AB, Designate the angle DCB by a, and the angle DCA by^. In the triangle J. j& (7 the sum A-\-B of the oblique angles is equal to a right angle (No. 8, Cor. III.); also in the triangle DCB the sum ^ + a of the oblique angles is equal to a right angle ; and similarly in the triangle DCA the sum ^ _|_ ^ of the oblique angles is equal to a right angle. We have therefore -4-f^ = ^-f « = ^ + /5; whence J. = a, Bz=z^. And thus it appears that the triangles ABC, ADC, BDC are mutually equiangular. They are therefore similar to one another. Q. E. D. GEOMETRY. 39 Cor. I. Each side ahout the right angle is a meanjpro- portional hetween the hypotenuse and the adjacent seg- ment. For in the similar triangles ABC and ACD we have AB\AC\\AC\AD, and in the similar triangles ABC and BDC, we have AB'.BGwBC'.BD. Cor. II. The perpendicular CD is a mean propor- tional hetween the segments of the hypotenuse. For in the similar triangles ADC and BDC we have AD: CD:: CD : DB. Cor. III. From Corollary I. we have AU'^ABxAD, and BC'^ABxBD. By adding these two equations, we obtain AC' + BC'= AB {AD +DB) = AB% a result identical with that obtained above (^o. 27). PEOPOSITION XYI. 40. Similar triangles a/re to one another as the squares of their homologous sides. Proof. Let ABC and ^ abc be any two similar /fC triangles. Draw CD and / ! \ X cd perpendicular to the / 1 \ /\ homologous sides AB / | \ / \\ and ah respectively. The ^^ g ^g J- ^ — \ area of ABC will be ■J-J.J&X CD^ and the area of ahc will be ^ahxcd (No. 31, Cor.) Hence 40 ELEMENTS OF ABC : abo :: ABxCD : ahxcd. But since the triangles A CD and acd are also similar (for A = a, and ABO = ado, and consequently ACB= acd ), we have also OB: cd ::A0 : ac :: AB : ab ; hence, substituting AB : ah in the place of OB : cd in the above proportion, we have ABO lahc:: AB' : d>\ Q. E. D. PKOPOSITION XYIL 41, The perimeters of similar polygons are to each other as any two homologous sides / and their areas are to each other as the squares of any two homologous sides. Proof Similar poly- gons are polygons which can he divided into the same numher of simi- lar tria/ngles similarly placed. Hence all the sides and diagonals of the one are proportional to the corresponding sides and diagonals of the other. ABOBE and ahcde be two similar polygons, shall have Let We AB ah BO he OB BE AE cd de ae whence, by an algebraic rule regarding equal ratios, we obtain GEOMETRY. 41 AB+BC^CI)+DE+EA ^AB _BG _ ah -\- 1)0 -\- cd -{- de -\- ea ah ho But the first member of this equation is evidently the ratio of the two perimeters. Hence Perim. ABODE: Perim. abode:: AB : ab::BC\ ho::... "We have, similarly, for the areas of the triangles of each polygon, ABC _ AGP _ APE _ AB^ _ BO^ _ ahc aed ade ab^ Sc"^ Whence, by the algebraic rule just mentioned, we have ABC -^r AGP -\- APE _ AB^ _ BG^ _ abo -\- acd + ade ~ ^^ bd" ~ ' ' ' But the first member of this equation is evidently the ratio of the two polygonal areas. Hence AresiABCPE: 2ire^abcde::AB' :ab'::BG' : bo' And these results are equally true, whatever the number of sides allotted to the similar polygons. 42 ELEMENTS OF BOOK III THE CIRCLE AND THE REGULAR POLYGONS. 43, The circle is a plane figure, bounded by a curved line, every point of which is equally distant from a point within, called the centre. The bounding line is called the circumference J and any straight line drawn from the centre to a point of the circumference is called a ra- dius. A line drawn through the centre and terminating at both ends in the circumference is called a diameter. An a/rc is any part of the circumference ; a chord is a straight line joining the extremities of an arc. The part of the circle included between an arc and its chord is called a segment of the circle ; whereas the part of the circle included between two radii is called a circular sec- tor. Thus, in the annexed figure, is the centre of the circle, DM a radius^ BD a diameter^ CN a chords the portion CHN of the circle a segment^ the portion DMA a sector. 43. The circumference of a circle is usually divided into 360 equal parts, called degrees f each such degree is again divided into 60 parts called minutes ; and each minute subdivided into 60 parts called seconds. It fol- GEOMETRY. 43 lows that, if we draw two diameters A C and BD per- pendicular to each other, the arcs AB, BC^ CD^ DA (which are evidently equal) will each be equal to 90 de- grees; and consequently the arc of 90 degrees corre- sponds to, and is taken as the measure of, a right angle at the centre of the circle. So also arcs greater or less than 90 degrees correspond to angles greater or less than the right angle, and are taken as their measure. Thus, if the arc AM is equal to 30 degrees, the angle MO A, to which it corresponds, is an angle of 30 degrees. Degrees, minutes, and seconds are distinguished by special marks : tlius, to designate 30 degrees, 4 minutes, and 40 seconds, we write 30° 04' 40'^ ; and to designate 25 minutes and 50 seconds, we write 0° 25' 50''. PEOPOSITIOK I. 44. A radius perpendicular to a chord, hisects the chord, and also the arc subtended by it. Proof. Let AB be a chord, and 01) a radius perpen- dicular to it. Draw the radii OA and OB. Then the right- angled triangles AOC and BOG will have the sides AO and BO equal, and CO com- mon. They are therefore equal in all their parts (^o. 27, Cor.) Hence AG~BC; that is, the chord is bisected. Q. E. D. Again, the angles AOD and BOD are equal, and the sides AO, BO, DO are equal ; therefore the chords AD and BD are equal, as also are 44 ELEMENTS OF the arcs AD and BD, wliicli correspond to the equal angles. Q. E. D. Cor. I. In the same circle, equal arcs are subtended hy equal chords. For if the arcs AD and DB are equal, the angles AOD and BOD are also equal (No. 43), and the triangles ADO and BDO will be equal (No. 11). Hence the chords AD and BD will be equal. Cor. II. In the same circle, equal chords suhtend equal arcs, and greater chords subtend greater arcs. For if the chords AD and DB are equal, the angles AOD and BOD are equal, as also the corresponding arcs ; and if the chord AD were greater than the chord BD, the angle A OD would be greater than the angle BOD (No. 18) ; and consequently the arc AD would be greater than the arc BD, PEOPOSITION II. 45. In the same circle, equal chords are equally dis- tant from the centre, and of unequal chords the greater is nearer the centre. Proof Let AB and ^4^ be two chords. From the centre draw the radius OA, as also the lines OP, OQ respectively perpendicular to AB and A C. Then we shall have OP = (92' - AP", and Oq = OA'- AQ\ Now, if the chords AB and AC are equal, then a^OMETRT. 45 AF'=:AQ' Q^o.U); and therefore OF' = 0Q% smd OF = OQ. If the chords are unequal, suppose AO^AB. Then AQ'> AF" ; and therefore OQ the measure of the an- gle BOG, one-half of the same arc BG\^ the measure of the angle BA G. Jn the second case, as in the angle BAD, drawing OD and OB, we have one-half of BG for the measure of the angle BA G, and one-half of GD for the measure of the angle GAD. And since BAD is the sum of the two angles, its measure will be the sum of their mea- sures, that is, half the arc BD. In the third case, as in the angle DAE, since DAJS=: E AG— DAG, the measure of DAE will be the diffe- rence of the measures of EA G and DA G, that is, half the arc DE. Q. E. D. Gov. I. All the angles in- scribed in the same segment of a circle are equal: as BAF, BGF, BDF, BEE. For they are all measured by the same measure, viz., by one- half of the arc BE. Gor. II. All the angles in- scribed in a semicircle are right angles. For they are all measured by half the semi-circumference, that is, by 90°, which is the measure of a right angle (No. 43). 48 ELEMENTS OF PKOPOSITION YI. 49. Arcs intercepted hy jparallel chords, or ly a tan- gent and ajparallel chord, are equal. Proof. Let AB and CD be two parallel chords. Join AD. The alternate angles BAD and CDA will be equal ; hence their measures will be equal. J^ow, BAD is measured by half the arc BD, and CDA by half the arc A C. Therefore BD and A C are equal. ISTow, let ET be a tangent parallel to the chord CD. Since the radius OE is per- pendicular to the tangent and to the chord CD, the arc CED will be bisected in E i^o. 44). And therefore the arcs intercepted bj a tangent and a parallel chord are equal. Q. E. D. r.X xn V PEOPOSITION YIL 50. If two chords intersect within a circle, the angle thus formed is measured hy half the sum of the includ- ed arcs. Proof. Let AB and CD be two chords intersecting at E. Draw^i^'parallel to 6^i>. Then the angles FAB and DEB will be equal (No. 6, Cor.) But the measure of FAB is one-half of the arc FB, that is, one-half of FD-^DB, or one-half of AC+ DB {Eo. 49), which is the sum of the included arcs. Q. E. D. SfEOMETRY, 49 PKOPOSITION yiii. 51. The angle formed hy two secants meeting without the circle^ is measured hy half the difference of the included arcs. Proof Let AB and CB be two secants meeting at B. Draw Di^ parallel to BC. The angles ABF and ABC will be equal (]S"o. 6) ; hence the measure of ADF, that is, one-half of AF, will also be the measure of ABC. ^ui AF=AC-FG =^(7— Z>^, which is the difference of the arcs inter- cepted. Hence, etc. Q. E. D. PEOPOSITION IX. 52. The angle formed hy a tangent and a chord is measured hy one-half of the intercejpted arc. Proof Let J.r be a tangent, and AB a chord. Draw the radius OA^ and the line OG perpendicu- lar to the chord. Then the angles A 6/and TAB will be equal (No. 39). But the angle AOG is measured by its corre- sponding arc, that is, by one-half of AB. Hence the angle TAB^ too, is mea- sured by one-half of AB. Q. E. D. 50 ELEMENTS OF PKOPOSITIOlSr X. 53. The angle formed by a secant and a tangent is measured hy half the difference of the arcs intercejpted. Proof. Let the secant BT meet the tangent AT 2ii the point T. From (7, where the secant cuts the circle, draw CD parallel to AT, Then the an- gles BTA and BCD will be equal (No. 6). Now, the angle BCD is measured by one-half of BD, that is, by one-half of BA —AD, or by one-half of BA~-AC; for AC=:AD (No. 49). Therefore the angle BTA also is measured by BA ^AC. Q.E.D. PKOPOSITION XI. 54. When two chords intersect, the ^product of ths seg- ments of the one equals the product of the segments of the other. Proof Let AB and CD be two chords intersecting at K Join AC ^ndBD. The triangles ACi: and BDi: will be similar ; for the an- gles A and D have the same measure, as also the angles C and B. Hence AE'.EDwCE'.EB; therefore AExEB= CExED. Q. E. D. GEOMETRY. 51 PKOPOSITIOJSr XII 55. Two secants meeting at a jpoint without a circle are to each other inversely as their ex- terior segments. Proof. Let AG ?c^d. BC he two secants meeting at C, and DC, EG their exterior seg- ments. Join AE and BD. The triangles AGE and BGJ) \ will be similar; for their an- gles have respectively the same measures. Hence the propor- tion AG \BG\\GE : GD Q. E. D. PKOPOSITION XIII. 56. 7/^ a secant and a tangent meet at a point, the tangent is a mean proportional hetween the secant and its exterior segment. Proof. Let the tangent AT and the secant BT meet at T, and let GT be the exterior seg- ment of the secant. Join AB and A G. The triangles ABT and A GT will be similar ; for, besides a common angle T, they have GBAz=:z GAT (No. 52); and consequently the third an- gles BAT and A GT y^iW also be equal. Hence BT\AT\\AT\ GT. Q.E.D. 52 ELEMENTS OF PKOPOSITIOJSr XIY. 57. The side of a regular hexagon inscribed in a cir- die is equal to the radius of the circle. Proof Let ABCDEFhQ a regular inscribed hexagon. Draw the radii OA, OB, etc. Since each side subtends the sixth part of the circumfer- ence, each angle AOB^BOC, etc., will be =60°. On the other hand (]^o. 20), the an- gles at the vertices A, B, O, etc., are each =120° ; hence 0^^=60°, OBA=:60°, etc., and all the triangles are equi- angular and equilateral ; accordingly AB =iAO, etc. Q. E. D. PEOPOSITIOK XY. 5S, The area of a regular jpolygon is equal to half the jproduct of its perimeter hy the aj^othem. Proof. Let ABCDEF be a regular polygon, its centre, OH its apothem, or the radius of the inscribed circle. Drsiw A, OB, 00, etc. The polygon will be thus divided into triangles, whose bases will be the sides of the polygon, and whose altitude will be equal to the apothem. But the area of any triangle is equal to half the product of its base by its altitude (No. 23). Therefore the area GEOMETRY. 53 of the polygon, which is the sum of the areas of such triangles, is equal to half the product of the perimeter into the apothem. Q. E. D. r^^^PKOPOsiTioisr xyi. 59. The areas of an inscribed and a circumscrihed regular polygon of ^n sides can he expressed in terms of the areas of the inscribed and circumscribed regular polygons of n sides. Proof. Let p be the area of the inscribed, and P that of the circumscribed, polygon of n sides ; B E A D c and let p' be the area of the inscribed, and P' that of the circum- scribed, polygon of 2?i sides. Let OIIKhQ one of the n triangles which compose the area^/ if we represent the area of OHKhj a, we shall have na=:p. Let OBG be one of the n triangles which compose the area P ; if we represent the area of OBC by J, we shall have nb = P. Let GAIT he one of the 2n triangles that compose the area p\ Kepresenting the area GAB by x, we shall have 2nx :=p\ Let GDE be one of the 27* triangles which compose 54 ELEMENTS OF the area P' ; if we represent the area ODE by y, we shall have 2ny—F', JSTow, wehave OMxMH=a^ OAxAB=l; hence OM.ABxOA, MH=. ah. But the similar triangles OMII and OAB give OM'.MHwOA \AB; hence OM .AB^OA, MH, Therefore {OA . MH)'^ ah. But OA . MII=.^x ; hence ^x^—ab^ and consequently 4^V= ^« . nJ =J^Py and 2?ia? = y jpP ; and, since 2/ia? = ^', hence Such is the expression of the area of the inscribed poly- gon of 2?2/ sides in terms of the areas of the polygons of n sides. And now, since OE bisects the angle A OB^ we have (No. 37) AE:EB::OA: OB, But OA: OB:: OM: OH::OM: OA; and therefore AE:EB::OM: OA, hence AE : AE-\-EB : : OM : 0M-\- OA, , . ^E _ OM that IS, AB- OM-^OA: Multiply and divide the first member of this equation by OA, and the second member by MH, We shall then have GEOMETRY. 55 AExOA OMxMH ABxOA ~ OMxMH-^OAxMR But AE.OA = y, AB . 0A= h, OM. MH= a, OA . MH^^x; -, i> y a therefore ^ = — r'lr ' . 2ny na hence pr-z = r~^ — 2no na -f- 2nx But 2ny = P\ 2nb == 2P, na =p, 2nx = p\ Therefore Such is the expression of the area of the circumscribed polygon of 2?i sides in terms of the areas of the poly- gons of n sides. Q. E. D. PEOPOSITIOlSr XYIL 60, The approximate value of the area of a circle^ whose radius =1, is 3.14159265. Proof. The circle may be considered as a regular poly- gon having an innumerable multitude of sides of an ex- tremely small dimension.* For both the inscribed and ♦ It has often been objected that a curve cannot be made up of straight lines. But this stale objection is easily disposed of. To say that a curve consists of straight lines oi. finite length is certainly absurd ; but to say that a curve consists of straight infinitesimal elements is to state a clear and necessary truth, about which mathe maticians could never have hesitated, had they kept in view the fact, so clearly pointed out by Sir Isaac Newton, that all lines are traced by continuous movement. An infinitesimal line is traced by an infinitesimal movement, that is, by the linking of two consecutive points. Now, between two consecutive points there is no room for a third point, without which no curvature can be conceived. Hence no infinitesi- mal line can be a curve. And therefore all the infinitesimal elements of curves are essentially rectilinear. It is therefore perfectly rational to consider the circle as a regular polygon having an infinite multitude of InflniteBimal sides. 56 ELEMENTS OF the circumscribed polygons, by doubling again and again the number of their sides, tend, the one by increasing, the other by diminishing, to assume equal dimensions, so that the difference of their areas can be reduced to an exceedingly small quantity, and the area of the one can be considered practically equal to the area of the other. A fortiori either of them can be taken as the area of the circle which is intermediate between them. Is'ow, starting from two squares, the one inscribed within, and the other circumscribed about, a circle whose radius is unity, the formulas above found (]S'o. 59) give us the following areas : Number of Sides. Inscribed Polygons. Circumscribed Polygons. 4 2.0000000 4.0000000 8 2.8284271 3.3137085 16 3.0614674 3.1825979 32 3.1214451 3.1517249 64 3.1365485 3.1441184 128 3.1403311 3.1422236 256 3.1412772 3.1417504 512 3.1415138 3.1416321 1024 3.1415729 3.1416025 2048 3.1415877 3.1415951 4096 3.1415914 3.1415933 8192 3.1415923 3.1415928 16384 3.1415925 3.1415927 From this table, which might be further extended, it is obvious that the area of the circle whose radius = 1, will be very approximately expressed by the number 3.14159265. The exact area cannot be expressed in finite terms, as its calculation involves a process in in- /J' GEOMETRY. 57 fiiiitum ; but it can be approached nearer and nearer indefinitely. It is usually designated by the Greek let- ter n^ whose value has been found by the infinitesimal calculus to be Tz = 3.1415,926535897932384626433832795 . . . Oor. The area of a circle whose radius is r will be 7rr\ For, as all circles are similar, their areas are to one another as the squares of the radii, which are homolo- gous lines. PEOPOSITIOISr XYIII. 61. The circumference of a circle, whose radius =z 1, is 27r. Proof Let dbcdmn be a circle. Draw the radii Oa, Oh, Oc, Od, . . . , dividing the circle into a great num- ber of very small equal sectors. The sum of all these sectors will be the area tc of the circle. Now, if the arcs ab, he, cd, . . . are exceed- ingly small as compared with the radius, the area of each ' sector will be expressed, like the area of a triangle, by half the product of its base into its altitude, the base being the arc, and the altitude the radius. Hence the area of the circle whose radius is =1 will be the sum of the areas ^ah, ihc, \cd, ... or \{ah-\-hG-\-cd-{- . . . ). But the sum of all the small arcs is equal to the whole cir- cumference. Therefore, designating the circumference by (7, we have ;r=|(7/ and consequently (7=2;r. Q.E.D. 58 ELEMENTS OF Cor. 1. The circumference of a circle whose radius is r will be 'ircr. For, as all circles are similar, their linear dimensions are to one another as the radii, which are ho- mologous lines. Cor. II. The number n expresses the length of a semi- circumference having the radius =1. Hence -^ ex- presses the length of the arc of one degree in that cir- cumference. Thus, when /" = 1, we have 1°= 0.017453292, also l'r= 0.000290888, and r=: 0.00000484:8. GEOMETRY. 59 BOOK lY GEOMETRIC CONSTRUCTIONS. 62. As Geometry cannot be understood without dia- grams, every student of Geometry should know how to construct all the needed figures, and how to ascertain that they are accurately constructed. To this end we give here the solution of a number of problems on geo- metric constructions, as a practical application of the theorems above established. PEOBLEM I. 63. To Insect a given straight line. Solution. — Let AB be the given line. From A and B, as centres, with a radius greater A.^ than one-half of AB describe arcs intersecting at E and F, and draw EF. This line will bisect AB at the point C (]S"o. 15, Cor. I.) % PEOBLEM II. 64. To erect a perpendicular to a line at a given, point. Solution. — Let AB be the line, and C the given point. 60 ELEMENTS OF ^ Lay off the equal dis- tances CD and CE^ and from D and £J, as cen- tres, with a radius great- er than one-half of DE ^ g- describe arcs intersect- ing at E, and draw EC. This line will be perpendicular to AB at the point Q^o. 14, Cor.) PKOBLEM III. 65. To erect a perpendicular at the end of a given line. Solution. — Let AB be the given line. Take any point without the line, and from 0, as centre, with a radius OB describe a semi- circle, and from the point C where the line AB is intersected draw the diameter CD. Then draw DB. perpendicular to ^^ at its extremity B (]S: o. 48, Cor. IL) Or else, measure three lengths from B to any point (7 on the given line, then from (7, as centre, with a radius of ^yq lengths, and from B^ as centre, with a radius of four lengths, describe arcs This line will be GEOMETRY. 61 intersecting at D. Draw DB. This line will be per- pendicular to tlie line AB at its extremity B. For since CD'— 25, GB'— 9, and BD'z^ 16, and 25 = 16+9, (7Z> will be tlie hypotenuse of a right-angled triangle (No. 27). PKOELEM lY. 66. i^^m ^ point without a line to draw a perpen- dicular to the line. S'olution. — Let BD be the line, and A the given point. From ^, as a cen- tre, describe an arc cutting BD in two points ^ and F. Bisect the chord EF at (7, and draw A C, which will be the perpendicular required (No. 14, Cor.) \E F/ PROBLEM Y. 67. To construct on a line an angle equal to a given angle. Solution. — Let CAB be the given angle, arid MN 2, given line. From A^ as a centre, with any radius AB describe the arc BC. Then, with the same radius, f.rom Jf as a centre describe an arc PO. Finally from P, as a centre, with a radius equal to the 62 ELEMENTS OF chord JBG, draw an arc cutting PO in Q, and draw MQ. The angle FMQ will be the angle required (]^o. 17). PKOBLEM YI. 68. To bisect a given arc^ or a given angle. Solution. — Let ACB be the given arc, and its centre. Draw the chord AB^ and bisect it by drawing OG, The arc AB will also be bisected at C (N^o. 44). To bisect a given angle A OB^ from 0^ as a centre, describe an arc AB^ and bisect it. The angle, too, will be bisected. PEOBLEM YII. 69. Irom a given point to draw a line parallel to a given line. Solution. — Let C be the given point. From any point A of the given line AB, with J. (7 as radius, draw pf ^ an arc CB. Then from C, as a centre, with the same radius draw an arc AD, andmake^Z>=:^(7. Draw CD. This will be the required Kne (Xo. 6, Cor.) PKOBLEM Yin. •70. To find the centre of a given a/rc of circle. GEOMETRY. 63 Solution. — Let A CB be the given arc. Draw two chords, AB and AC^ and erect perpendiculars at their middle points D and E. The point 0^ where the perpendiculars intersect, will be the centre of the arc^6^^(Ko. 44). PEOBLEM IX. 11. From a given jpoint to draw a tcmgent to a given circle. Solution. — Let A be the given point, and the cen- tre of the given circle. Join A 0, and bisect it '"" in C. From (7, as a centre, with the radius CO^ describe a circle OB AD. This circle will intersect the given circle at the two points B and D. Draw AB and AD. Both these lines will be tangents to the given circle. For, joining DO and BO, the angles ADO and ABO will be right, as being both inscribed in a semi-circumference (ISTo. 48, Cor. IL) Hence AB and AD will be perpendicular to the radii OB and OD respectively ; and therefore they are tangents to the circle (^N^o. 46). 64 ELEMENTS OF PKOBLEM X. 73. To inscribe a circle in a given triangle. Solution. — Let ABC be the given triangle. Bisect the angles A and B by the lines AO and BO meeting at the point y and from let fall the per- pendiculars OD, OE^ PF on the three sides of the triangle. These perpendiculars will all be equal. For the triangles A OE and A OF, having a common hypotenuse A and an equal angle at A, are equal (No. 27, Cor.) ; hence 0E= OF; and similarly the trian- gles BOF and DOB, having a common hypo- tenuse OB and an equal angle at B, are equal ; whence OF^=^OD. Con- sequently the point is equally distant from the three sides of the triangle ; and if, from as centre, with the radius OD we describe a circle, this circle will be the inscribed circle required. PKOBLEM XL 73. To divide a line into jparts jprojportional to other given lines. GEOMETRY, 65 Solution. — Let A, B^ and C be three given lines, and MN the line to be di- vided. From the ex- tremity M draw a line MS making an acute angle with MN^ and take upon it MLz=zA, Then jom IN. draw HQ , and parallel to IN. and LP The parts J/P, PQ^ QN oi the given line JfiT will be pro- portional to the given lines A, B, C (N'o. 32). PKOBLEM XII. 74. To construct a fourth projportional to three given lines. Solution. — Let A, B, and C be the given lines. Draw two lines making an angle at J/, take MZ=zA, ZK^Con one line, and MNz=zB on the other line. Join IN, and from ^draw ICO parallel to IN; NO will be the line required (No. 32). PROBLEM XIII. 75. To construct a 7nean proportional hetween two given lines. Solution. — Let A and B be the given lines. Draw 66 ELEMENTS OF ML =^A-\-B, and take LN^A,MNz=.B. On ML as a diameter de- scribe a semicircle MOL. From the point iV^ erect NO perpendicular to ML. This perpendicu- lar will be the mean pro- A portional required. For, g the angle MOL being a right angle (No. 48, Cor. II.), ONi^ a mean proportional between the segments of the hypotenuse ML (No. 39, Cor. II.) PKOBLEM XIY. 76, Through a point within an angle^ to draw a line intersecting the sides of the angle at equal distances from the given point. Solution. — Let be the given point, and BAG the given angle. Draw OD parallel to the side AC oi the angle, and on the other side take DE=i AD. From ^draw a line EF passing through the point 0. The points E and F will be equally distant from the given point ; for OF and OF are to each other in the same ratio as ED and DA., which are equal by construction (No. 32). GEOMETRY. 67 PEOBLEM XY. 77. To divide a line in extreme and mean ratio. Solution. — Let AB be the given line. To divide it in extreme and mean ratio is to divide it into two parts, 01 which the greater shall be a mean proportional be- tween the whole line and the remaining smaller part. At the extremity B of the given line AB draw BO per- pendicular to AB^ and make it equal to one-half of AB. From (? as a centre, and with OB as radius, describe a circumference BBC, and draw a secant AG passing through the centre 0. From ^, as a centre, with a ra- dius equal to the exterior part [ AD of the secant, describe an ' arc BE intersecting the given line at E. Then AE and EB will be the parts required. For (Ko. 56) we shall have CA :AB::AB: whence CA -AB : ABwAB- or, because ABz=z CD by construction, CA-CD:AB::AB-AD that is, AE'.ABwEB : AE. AD, -AD AD, AD, PEOBLEM XYI. 78. To inscribe an equilateral triangle in a given circle. Solution. — Let ABCDEF be the given circle, and 68 ELEMENTS OF its centre. Draw the chords A£, BG, CD, DE, EF, FA, each equal to the radius OA. The polygon thus formed will be a regular hexagon (No. 57). ^oin BD, BE, 'i^ will be the tri- angle sought. The side BE is perpendicu- lar to the radius OA, by which it is bisected at / (No. 21, Cor. lY.) Sr" OA=r, we have BE^^r"— ir'z^-j-; hence Now, making BI^ 2 V^ ; therefore BE= rVS. PKOBLEM XYII. *79. To inscribe a regular octagon in a given circle. Solution. — Let AB aiid CD be two diameters of the given circle, perpen- dicular to each other. The chords ^(7, CD, DB, and BC will form a square. Bisect these chords, and their arcs, by drawing OF, OF, OG, and OH, and draw the chords CE, FA, AF, . , . These chords will be the sides of the regu- lar octagon inscribed in the circle. GEOMETRY, 69 The side AC oi the square ADJBOis equal to r V~2. As to the side of the octagon, we have j:e'z= ap-\-ep^ aI'+ {oe- oiy ; and because u4./=:J-46^ — ^ fore , and OI=^AI, there- AE hence .=- + (,_^^)-=.(|+>_v-,+|). AE'= r' (2 - \^\ and AE= r |/2- V2. PEOBLEM XYIII. 80. To inscribe a regular decagon in a given circle, /Solution. — Let OA be the radius of the given circle. Divide it in extreme and mean ratio (No. 77), and let 00 he the part which is a mean proportional between OA and A 0. Draw a chord AB ^=^00 ; such a chord will be the side of the in- scribed regular decagon. For, since AB ^:^ 00, the proportion OA : 00 ::0C : ^^ becomes OA :AB::AB:A0, which shows that the triangles OAB and BA are simi- lar; and therefore the latter, as well as the former, is isosceles. Hence AB^BG^ and consequently also 70 ELEMENTS OF BG^:^OC^ and the triangle OCB is also isosceles. We have, then, on the one hand, the angle COB equal to the angle CBO, and, on the other, the angle COB equal to the angle CBA j hence the angle OB A is equal to the double of the angle A OB. The sum of the angles of the triangle A OB is therefore A0B-\-^A0B-^2A0Bz=im''\ hence ^ (9^ === :^ = 36° ; and as 36° is the tenth part of the circumference, the chord AB will be the side of a regular inscribed poly- gon of ten sides. The side AB is determined bj the proportion r:ABr:AB:r- AB, from which we obtain AB''=^ r"— r . AB ; hence V6-r AB GEOMETRY. 71 BOOK Y PLANES, AND POLYHEDRAL ANGLES, 81. A plane is a surface on which straight lines can be drawn in all directions. It is evident that a straight line will lie wholly in a plane, whenever two of its points are in the plane. When two planes are parallel, they can never meet ; but when they are inclined towards each other, they meet in their common intersection, and form a dihedral angle (that is, an angle made by two faces), which is measured by the plane angle formed by two lines drawn in the two planes perpendicular to their intersection. The intersection is called the edge of the angle. If three or more planes intersect one another, they will form a trihedral or Oi polyhedral angle, whose faces will be the portions of the planes lying between the edges. The point where the edges meet is called the vertex of the angle. PEOPOSITION I. 82. Three points, not in a straight line, determine the position of a plane in space. Proof. Let J., ^, and C" be three given points. Join AB, and con- ceive a plane passing through, and revolving around, AB. When the plane reaches the point C, it has a 72 ELEMENTS OF definite jDosition ; and tlnis the three points determine the position of the plane. Q. E. D. Cor. I. Two straight lines^ which intersect, determine the jposition of a plane. Cor. 11. Two parallel lines determine the position of a plane. PEOPOSITIO]^ 11. 83. The intersection of two planes is a straight line. Proof Let AB and CD be two planes, and E and F two points common to both planes. Draw the straight line EF. Because this line has two points in the plane AB, the whole line will be in that plane ; and because it has^two points in the plane CD, it will lie wholly in this plane. ]N^ow, the two planes have no common line but their intersection. Hence the straight line EF, which is in both planes, is the intersection of the two planes Q. E. D. PEOPOSITIOlSr III. 84. J. straight line perpendicular to two straight lines at their point of intersection, is perpendicular to the plane of those lines. Proof Let BC and DE be two lines intersecting at P, and let AP be perpendicular to them. Through P draw any line P^ in the plane of PE and PC, and through the point Q draw a line EC so that QC and QE shall be ecjual (Ko. 76). And now draw GEOMETRY. 73 CA, EA, and QA. The base EC of the triangle AEG being bisected by A Q, we shall have Q^o. 30) AC'+ AE'^ 2J^^+ ^CQ\ In like manner, the base EG of the triangle PEG being bisected by PQ-, we shall have also PG'-\-PE'= 2P^^+ 267^'. Subtracting this from the preceding equation, we have J^«_ FC'+ AE'- PE'= 2AQ'- 2PQ\ But AG'- PG'=z AP% and AE'- PE'= AP' ; hence 2AP':= 2AQ'- 2PQ' ; whence AQ'= AP'+ PQ' ; which shows (No. 2Y) that the triangle APQ is right- angled at P, and therefore AP is perpendicular to ^^/ and since i^^ is any line drawn through P in the plane of J5(7and DE, it follows that AP, if perpendicular to these two lines, is perpendicular to every other line drawn through P in their plane, viz., AP is perpen- dicular to the plane PEG. Q. E. D. PEOPOSITIOlSr lY. S^, The perpendicula/r is the shortest distance from a jpoint to a plane. Proof. From the point A let fall AP perpendicu- lar to the plane JO^. Then all the oblique lines, as A G^ AD, AE, drawn from A to the same plane will be great- er than AP ; for they will 74 ELEMENTS OF be hypotenuses of right-angled triangles, of which AP is a side. Hence AP is the shortest line from A to the plane. Q. E. D. Cor. I. Oblique lines equally inclined, or equally di- verging from the perpendicular, are equal. For they are the hypotenuses of equal triangles. Cor. II. The oblique lines AD and A C, meeting the plane at equal distances from the foot of the perpendi- cular, are equal. Cor. III. From a given point to a given plane only one perpendicular cam, be drawn. PEOPOSITIOK Y. 86. If from the foot of a perpendicular to a plame a line be drawn meeting at right angles a straight line of that plane, a line joining the point of their in- tersection with any foint of the perpendicular will be perpendicular to the line of the plane. Proof Let AP be perpendicular to the plane JfiT, and BC the given line at right angles with PQ. Talie QC=QB, and draw BP, CP, BA, CA, and AQ. The triangles PBQ and PCQ will be equal, for they have equal sides about a right angle ; hence PC^PB. The triangles APB and APC will also be equal, for they, too, have equal sides about a right GEOMETRY. 15 angle ; hence A£ = AC. Consequently the triangle ABC is isosceles, and the line AQ which bisects its base BC is perpendicular to BO (ISTo. 14, Cor.) Q. E. D. Cor. The line BCis perpendicular to the plane of the triangle APQ ; for it is perpendicular to the two lines P^ and ^(2 (No. 84). PKOPOSITIOK YL 87. If one of two jparallels is perpendicular to a plane, the other also is perpendicular to that plane. Proof. Let the line DQhQ parallel to the perpendicu- lar AP. If we pass a plane through these Hnes, its in- tersection with the plane J/TT will be PQ. Through the point Q draw in the plane MN iho, line BC perpendicu- lar to P Q, and join A Q. Then BQ^ being perpendicular to PQ and to AQ (No. 86), will be perpendicular to the plane APQD (No. 84); and there- fore BQD is a right angle. Now, DQP is also a right angle ; for AP and DQ are parallel. Therefore DQ \s> perpendicular to the two lines PQ and QB lying in the plane MN ; and conse- quently (No. 84) it is perpendicular to the same plane. Q. E. D. Cor. I. If two lines are perpendicular to the same plane., they are parallel. For, \i DQ were not parallel to AP, we could draw through Q another line parallel to AP, which would be perpendicular to the plane MN^ ^6 ELEMENTS OF and then there would be two perpendiculars at the same point, which is impossible. Cor. II. If two straight lines are parallel to a third^ they will he parallel to each other. For a plane perpen- dicular to one of them will be perpendicular to the other two. PEOPOSITIOJSr YII. 88. 7f t^ straight line is parallel to any line of a pla/ne, it is parallel to the plane. Proof. Let AB be parallel to the line CD lying in the plane MN. The lines AB and CD, being paral- A b lei, lie on a plane ABCD, and therefore AB cannot meet J/IZV except in some point of the line CD. But this is impossible, because AB and CD are parallel. Hence AB can nowhere meet the plane MN ; and con- sequently this line is parallel to the plane. Q. E. D. PROPOSITION YIII. 89. If two planes are perpendicular to the same straight line, they are parallel to each other. Proof, Two planes are parallel which can never meet. But two planes per- pendicular to the same line AB can never meet ; for, if they met at any point 0, there would be a triangle GEOMETRY. 77 ABO with two right angles GAB and DBA, which is impossible. Q. E. D. PEOPOSITIOJS' IX. 90. If ajplane intersects two j^arallel jplcmes^ the lines of intersection will he parallel. Proof Let JO^and P^ be two parallel planes intersected by a plane ABDC. If the intersections AB and CD are not parallel, they will meet somewhere ; for they are in the same plane. But if these lines meet, then the parallel planes MN and PQ will meet, which is impossible. Hence the lines AB and CD are parallel. Q. E. D. PEOPOSITIOlSr X. 91. If two lines are jpar allele they are equally in- clined to any jplane passing through them. Proof. Let AB and CD be two parallel lines, and MN" a plane passing through them at A and C. Take AB= CD, and draw AC and BD. The quadrilateral ABDC V7\\\ be a parallelo- gram (:N'o. 21) ; hence BD= AC. Let fall ^^ and PP perpendicular to the plane, and draw JEF. The quadri- lateral BDEF will be a par- allelogram; hence BE=DF, and EF=BD = AC. Draw now AF and CF. The quadrilateral AEFC will 78 ELEMENTS OF also be a parallelogram ; and therefore AE= CF. Hence the two triangles ABE and CDF have equal sides, and the angles BAE and DCF^ opposite to the equal sides BE and DF, are equal. Q. E. D. * pkopositio:n^ xi. 93. Parallel lines hetween pa/rallel jplanes are equal. Proof, Let AB and CD be two parallel lines included between the parallel planes MN and PQ. The plane ABDC oi the two lines will intersect the two planes in the lines A C and BD^ which will be parallel (]Sro. 90). Hence ABDC will be a parallelogram ; and consequently AB=CD. Q.E.D. PEOPOSITION XII. 93. Two angles^ situated in different jplanes^ whose sides are jparallel and similarly directed^ are equal j and their jplanes are ^parallel. Proof. Let BAC and EDF \iQ two angles, whose sides are parallel and similarly directed. Take AB = DE and AC=DF, and draw ^(7 and EF. Since AB and DE are equal and parallel, we may, by joining AD and BE, form a parallelogram ABED i also, since A C and DF are equal and parallel, A CFD will be a paral- u D / /'\vl' / . / \ \ 1_ / /°" -^ y GEOMETRY, 79 lelogram; hence BE=AD^CF. But if BE=CF, BEFG will also be a parallelogram ; and therefore BG = EF. Thus the triangles ABC and DEF will be equal ; and accordingly the angles BA C and EDF will be equal. Q. E. D. Again, since AB and DE are parallel, they cannot meet ; also A C and J)F, being parallel, cannot meet ; hence the plane determined by ^^ and AG cannot meet the plane determined by BE and BE; that is, the planes of the angles BA G and EDF are parallel. Q.E.D. Gov. Two planes, as AGED and BGFE, intersecting two parallel planes, form equal angles, as A GB^ BEE, by their intersections with them. PROPOSITIOIS' XIII. 94, Straight lines^ if cut by parallel j>lanes, will he divided into jprcyportional jparts. Proof, Let any two straight lines AB and GB be cut by the parallel planes MN^ PQ, JST. Draw AB. Since PQ and /ST are parallel, the lines EF and BB, in which they are met by the plane ABB, are parallel (No. 90). Hence the triangles ABB and AEF are similar, and we have AE:BE::AF:BF And since JfiV and PQ are parallel, the lines A G and EG, in which they are met by the M /• n7 P_ \ 1 / = ^r1° / S \ 1 ^ /' ^'/ 80 ELEMENTS OF plane ADC, are parallel, and the similar triangles A CD and FGD will give AF \DF\\ CG\DG, Combining these two proportions, we obtain AE\BE\\Ca\DG. Q.E.D. PEOPOSITIOIT XIY. 95. ^ a straight line is jperjpendiculaT to a plane^ all planes passing through that line will he jperjpendi- cular to that 'plane. Proof, Let AP be perpendicular to the plane MN, and let a plane BE pass through AP. In the plane MN draw the line PD perpendi- cular to the intersection EC oi the two planes ; then, because ^P is perpendicular to the plane MN, the angle APD will be a right angle ; and this angle, because it is formed by two perpendiculars to the inter- section BC oi the two planes, is the measure of the divergence of the two planes. Hence the two planes are at right angles ; and thus every plane passing through a line perpendicular to a plane, is per- pendicular to that plane. Q. E. D. PKOPOSITION XY. 96. If two planes are perpendicular to each other. GEOMETRY. 81 any line drawn in one of them jperpendicular to their intersection^ will he jperj^endicular to the other. Proof. Let the planes MN and BE be perpendicular to each other, and let the line AP^ drawn in the plane BE^ be perpendicular to the inter- section BC. Then, since the two planes are perpendicular, APD will be a right angle; and therefore AP is perpendi- cular to the two lines BC and PD at their intersection. It is therefore perpendicular to their plane MN, Q. E. D. M i\^.^ E /. ^^ 1/ PKOPOSITION XYI. 97. The intersection of tioo planes perpendicular to a third plane is perpendicular to that plane. Proof Let the planes BE and BE be perpendicular to the plane MN, and let J.P be their intersection. If from P we erect a perpendicular to the plane MN, such a perpen- dicular must be in the plane BE, and also in the plane JDE (IS'o. 96), and is, therefore, their common intersection AP. Q. E. D. Cor. A plane perpendicular to two other planes is per- pendicular to their intersection. 82 ELEMENTS OF ' PEOPOSITIO]^ XYII. 98. The sum of any two jplane angles formed hy the edges of a trihedral angle is greater than the third angle. . Proof Let AYB, BVC, and OVA be the three plane angles, and let A VjB be the greatest. In the plane A VB make an angle BYD equal to BVC, and draw AB at pleasure. Take YC= YD, and draw A C and CB. The triangles CYB and B YD will be equal (N^o. 11) ; and therefore BD =BG. But (:^ro. 13) AC-\-BC>AD-^DB; hence, cancelling the equal parts BC and DB, we have AC> AD; and consequently (ISTo. 18) AYC>A YD; and, adding the equal angles B YC and B YD, AYO+BYOAYD + BYD, that is, AYC+BYOA YB. Q. E. D. PEOPOSITIOIS' XYIII. 99. The sum of the plane angles around a polyhedral angle is always less than four right angles. Proof. Let T^be the vertex of any polyhedral angle. By passing a plane through its edges YA, YB, YC, . . . GEOMETRY. • 83 we shall form a set of triangles AYB, BYC, , . , \ and, if from any point in the plane ABC ... we draw the straight lines OA, OB, 00, . . . we shall form another set of trian- gles ^6>^, ^0(7, . . . Let n be the number of the triangles of each set. The snm of the angles of all the triangles of the first set will be 180° X w, and will consist of two parts, viz., of the angles about F", the sum of which we may designate by (F), and of the angles YBA, YBO, YOB, . . . around the base. We may then write (Y) + {YBA + YBO+YOB+ . . . ) = 180°x^. The sum of the angles of all the triangles of the second set is also 180° X n, and consists of two parts, viz., of the angles about 0, the sum of which we may designate by {0), and of the angles OBA, OBO, OOB, . . . "We may therefore write {0) + {OBA+OBO+OOB+ . . . ) = 180°x^. Hence (Y) + {YBA + YBO+YOB+ . . . ) = {0) + {OBA + OBO+OOB+ . . . ). ^Now, if the sum YBA-{-YBO+ ... is greater than the sum OBA-\-OBO-{- . . . , it is obvious, from this equation, that ( Y) must be less than (0). But we have proved (No. 98) that YBA + YBO is greater than ABO, that is, YBA + YBO> OBA + OBO. And, since the same is true also of the other angles around the 84 ELEMENTS OF polygon ABC . . . , we conclude that ( V) is less than (6^). But (6^) represents four right angles. Hence ( V) is less than four right angles. Q. E. D. PEOPOSITIOK XIX. 100. If the jplcme angles at the vertex of a trihedral angle are equal, each to each, to the plane angles at the vertex of another trihedral angle, the planes of the equal angles are equally inclined to each other. Proof Let J[f and iTbe the vertices of the trihedral angles, and let AMC = DNF, AMB=DNE, BMC= ENF. Draw AB and AC, both perpendicular to AM. Then will the angle BA C measure the inclina- tion of the two planes MAB2.v.^MAC Layoff ND equal to MA, and draw DE and DF both perpendicular to ND. Then will the angle EDF measure the inclination of the two planes NDE and NDF. Now, the right-angled triangles MAB and NDE have the sides ^JtTand DN equal by construction, and the angles AMB and DNE equal by supposition. Hence AB = DE and MB = NE. Similarly, the right-angled triangles MA C and NDF have the sides AM and DN equal, and the angles AMC and DNE equal. Therefore AC—DF and MC=zNF. GEOMETRY. 85 Finally, the triangles CMB and FMJE have the equal angles CMB and FNE comprised between equal sides. Hence these triangles are equal, and therefore CB—FE. The consequence is, that the triangles ABC and DEF are equal, and that the angles CAB and FDE, which measure the inclination of the respective planes, are also equal. And the same conclusion holds with regard to the inclination of the other planes; that is, the planes of equal angles are equally inclined. Q. E. D. 86 ELEMENTS OF BOOK YL SUBFACES AND VOLUMES OF POLY- HEDBONS, 101. Polyhedrons are geometric solids bounded by polygons. These polygons are cdll^di faces of the poly- hedron ; the lines in which they meet are called edges; and the points in which the edges meet, vertices of the polyhedron. Begular polyhedrons are those whose faces are equal and equally inclined, as the regular tetrahedrons, hexa- hedrons, octahedrons, dodecahedrons, and icosahedrons. All the other polyhedrons are irregular, because they have not equal faces equally inclined ; to this class be- long the irregular jpmms smd pyramids, A prism is a polyhedron, of which two faces are equal and parallel polygons, called bases, the other faces being parallelograms. A prism whose bases are parallelograms, is called a parallelepipedon.^ A pyramid is a polyhedron bounded by a polygon called the base and by triangles meeting at a common point, called the vertex of the pyramid. All the other irregular polyhedrons may be considered as made up of prisms and pyramids arranged in a cer- tain order. * Our countrymen write parallelojnpedon ; but the Greek word is TrapoAAijA- firineSov, and I see no reason why it should be corrupted. GEOMETRY. 87 PEOPOSITION I. 102. The convex surface of a right prism is equal to the perimeter of the 'base multiplied hy its alti- tude. Proof Let ABODE — I hQ a right prism, viz., a prism of wliich the lateral edges are perpendicular to the basis. Its convex surface is the sum of all the lateral surfaces, that is, of the rectangles AG.BH, CI, DK.^u^ EF. ]^ow, these rectangles have the same altitude, viz., the altitude of the prism. Hence the sum of their surfaces is {AB^BC^ CD+DE+EF) AF, B which is the perimeter of the base multiplied by the al- titude. Q. E. D. Cor. If two right prisms have the same altitude, their convex surfaces are to each other as the perimeters of their bases ; and if they have the same base, their con^ vex surfaces are to each other as their altitudes. PEOPOSITIOK II. 103. The opposite faces of any parallelepipedon are equal, and their planes are parallel. Proof. Let ABCP—G be a parallelepipedon. Its 88 ELEMENTS OF bases A BCD and EFGH are equal and parallel (]N"o. 101). Hence EF^AB =D G ^HG, and HE^ABz^^CB^ GF. Consequently (JSTo. 21, Cor. II.) EA = FB^ CG^DH; hence DAEH^GBFG, and ABFE^DCGIL Q. E. D. Since ^D and 6^^, as also EA and ^i^^, are parallel, hence (Ko. 93) the planes AEHD and BFGG are par- allel. In like manner, it may be shown that the planes ABFE and DCGH are parallel. Q. E. D. Cor. In a rectangular parallel- epipedon^ the square of a diagonal equals the sum of the squares of the three edges which meet at the sa/me vertex. For, let ^^ be a diagonal of the parallelepiped on, and AC 2^ diagonal of the base. Then AD''= nC' + AG'' ; and because AC'— AB'-\- BG\ hence c AB'^ BC'Ar ^^'+ ^^'= i>(7'+ EC'-\- BC\ PEOPOSITIOlSr III. 104. If a plans he passed diagonally through a parallelepipedon^ it will divide it into two equal tri- angular prisms. GEOMETRY. 89 Proof. Let AG \)q 2. parallelepipedon, and BDHF the diagonal plane which divides it into two prisms. The tri- hedral angles of the two prisms, as formed by equal plane angles, will be re- spectively equal (No. 100), the triangular bases will be equal (No. 21), and the edges also equal. Hence the two prisms are equal in all their parts. Q. E. D. PEOPOSITIOlSr lY. 105. Two parallelepipedons having a common lower hase., and their upper bases hetween the same parallels, are equal in volume. ^ Proof. Let the parallelepipedons J-<9 and J.JIf have the common lower base ABCD, and their upper bases EFGH and KLMN be- tween the same parallels FK^id^GN. The volume of ^ 6r is equal to the whole volume of the figure, minus the prism AEK—N; and the volume of ^J^is equal to the wliole volume of the figure, minus the prism BLF— G. But these two prisms are equal ; for their bases AEK and BLF have AK^=. LB. AE— BF^ and the angle ^J.^ equal to the angle LBF^ and therefore (No. 11) they are equal; on the 90 ELEMENTS OF Other hand, the faces ^i^iiTTand BCML, also AD HE and BCGF, and lastly KEHIST 2.ndi LFGM are respec- tively equal. Hence the two prisms are equal in all their parts. Hence whether from the volume of the whole figure we subtract the one of the prisms or the other, the remainder will be the same. But by subtract- ing the prism AEK—N^ the remainder is the parallel- epipedon AG ; and by subtracting the prism BLF—G, the remainder is the parallelepipedon AM. Therefore the two parallelepipedons are equal in volume. Q. E. D. PEOPOSITIOJS" Y. 106. Two parallelepvpedons having a common base cmd the same altitude, are equal in volume. Proof. Let the parallelepipedons AR and A G have the common bfese A BCD and the same altitude. Be- cause they have the same altitude, their upper bases will lie in the same plane. Prolong EF and GH, as also PS and QE. These prolongations will form a parallelogram KLMN equal to ABCD and to EFGH. Now, if a third parallelepipedon be con- structed on ABCD with the upper base KLMN, its volume will be equal to that of A G, because the lines KF and NG are parallel, and will also be equal to that of AR^ because the lines PN GEOMETRY. 91 and QM are parallel (No. 105). Hence the volumes of the parallelepipedons AB and A G are equal. Q. E. D. Cor. Any ohlique jparallelejpipedon can he trans- formed into a right jparallelejpijpedon homing the same base and the same altitude^ with no change of volume. PEOPOSITION YI. 10*7. The volume of a rectangular jparallelepi^edon is measured hy the product of its three dimensions. Proof Let AD be a rectangular parallelepipedon, and a^ the unit of measure adopted for volumes. This unit is the cube of a unit of length. To express the volume of the parallelepi- pedon it suffices to state how many times it contains the unit of measure. [NTow, divide the three dimensions of the parallelepipedon bj the unit of length, and as- sume AB^^pa, BG = qa, CD = ra. Then the unit of measure a"" will be contained in the parallelepipedon pqr times ; for a row of p units laid down from A to B, and repeated q times from B to G, will form a first Isijerpqa' on the base of the parallelepipedon ; and this layer must be taken r times in ascending from G to D Q. E. D. From what has been said above (IS'o. 22) it can be shown that this proposition is true whether p, q, and r / / / D / f C / / ^ / ^ / / / / / / / / / / / 92 ELEMENTS OF be entire or fractional, and commensurable or incommen- surable with the unit of measure. PKOPOsiTioisr yii. 108. A right parallelepipedon can he i/i^ansformed into a rectangular parallelepipedon of the same alti- tude^ without change of volume. Proof Let ABCD—G be a right parallelepipedon having for its base the parallelo- gram ABCD. Through AE and BF pass the planes A Q and BP perpendicular to the plane AF. Their intersections with the other planes will determine the rectan- gular parallelepipedon ABMN —P, equal in volume to the given parallelepipedon, because it has an equal altitude, and a base of equal area (Ko. 106). Q. E. D. Cor. I. Since any oblique parallelepipedon can be transformed into a right parallelepipedon having the same base, the same altitude, and the same volume, and since any right parallelepipedon can be transformed into a rectangular parallelepipedon having an equal base, an equal altitude, and an equal volume, it follows that the vohime of a/ny jparallelejpijpedon can he expressed hy the jproduct of the three dimensions of a rectangidar paral- lelepipedon having the same altitude and an equal hase. Cor. II. Parallelepipedons having equal hases are to each other as their altitudes; and parallelepipedons having equal altitudes are to each other as their hases. c \ H 3 \^ > F M y N / y D • y GEOMETRY, ds For, let P and P' be two parallelepipedons, having the bases B and B'^ and the altitudes A and A' respectively. Then P= A . B, P'^ A', B' ; whence P\P'\\ A.B:A\B\ If B=:B\ this proportion becomes P : P' :: A : A\ If, on the contrary, A=A', the proportion becomes P.P'v, B\B\ Cor. III. Triangular prisms are semi-parallelepipedons (No. 104), and therefore their volume is measured hy the product of their triangular hase (which is one-half of the base of the parallelepipedon) into their altitude. If they have equal bases, they are to each other as their altitudes / and if they have equal altitudes, they are to each other as their bases. pkopositio:n" yiii. 109. The volume of any prism is equal to the pro- duct of its base by its altitude. Proof. Let ABCDEF — L be any prism a lateral edge BH draw the planes BN, BM, BL dividing the prism into triangular prisms. These prisms will have the same altitude ; hence the sum of their volumes, that is, the volume of the given prism, will be equal to the sum of their bases multiplied by the common altitude (InTo. 108, Cor. Ill); that is, the whole volume equals the whole base multiplied by the altitude. Q. E. D. Through N ; 1 / \ / ■^ / \ ■-•■/ \ / .. •• / ^ E k F - /■'. / \ i ..^ Y '-•■ .••■■' / ^^ 3 c 94 ELEMENTS OF PKOPOSITIOJST IX. 110. Jf a jpyramid he cut hy a jdane parallel to its hase, the edges and the altitude will he divided propor- tionally^ and the section will he a polygon similar to the Proof. Let the pyramid Y—ABC^ wliose altitude is O V, be cut by a plane ahc par- allel to the base ABC. Then ah and AB, he and BO, ac and A Oy as also ao and A 0, will be parallel (IS'o. 90). Hence the angles ahc and ABOy hca and BOA, cah and CAB are respec- tively equal (No. 93), and the two polygons are similar. The triangles AOV and ao V are also similar, as also are the tri- angles ah V and AB V. Hence ah: AB::aV: AV::ao : AO. Q. E. D. Cor. I. The area of any section parallel to the hase of the pyramid is proportional to the square of its distance from the vertex. For area ahc : area ABC \\ aV '. AB"": ao AO' Cor. II. If tvw pyramids stand on the same plane, and have equal altitudes, the sections made through them hy a 'plane parallel to that of their hases will he propor- tional to the hases. For if B and B' are the bases of the two pyramids, and H their altitude, and if h and h' are the two sections, and h their distance from the ver- tex, then GEOMETRY. 95 B:h::ir : h\ and B' :h'::H' : h'; and consequently B : h :: B' : h' . Cor. III. If two jpyramids have the same altitude and equivalent bases, any sections made though them at equal distances from the hases, or from the vertices, will have an equal area. PEOPOSITION X. 111. The convex, or lateral, surface of a right jpyra- mid is equal to half the product of the slant height into the perimeter of the base. Proof. In a right pyramid the base is a regular poly- gon, and the lateral faces are equal isosceles triangles. The surface of one of these triangles is equal to half the product of its base into its altitude, which altitude is the slant height of the pyramid. Hence the lateral surface of the pyramid, which is equal to the sum of the sur- faces of these triangles, is expressed by half the product of the perimeter of the base into the slant height of the pyramid. Q. E. J). Cor. If a right pyramid be truncated by a plane par- allel to its base, the convex surface of the remaining frustum will be equal to the sum of the surfaces of all the lateral faces. Such faces are trapezoids, and each of them is equal to half the product of its altitude into the sum of the parallel sides. Their sum, or the surface of the frustum, is therefore half the product of tlie slant height into the sum of the perimeters of the parallel 96 ELEMENTS OF PEOPOSITIOI^ XL 112. Two triangular pyramids ha/ving equal alti- tudes and equivalent bases, are equal hi volume. Proof. Let Y-ABG and Y'-A'B'C be two pyra- mids having equal altitudes and equivalent bases. Di- vide their altitude into equal parts, and through the points of division pass planes parallel to the plane of the bases. The sections made by each of these planes will be respectively equal (No. 110, Cor. III.) On ABC, DEF, GHl, KLM as lower bases con- struct as many exterior prisms between the parallel planes. Their altitudes w411 be equal. On D'E'F'^ G'H'I', K'L'M' as upper bases construct as many in- terior prisms. They will have equal altitudes, and their bases will be respectively equal to the bases DEF^ GUI, KLM of the prisms in the other pyramid. They will therefore be equal each to each. The difference between the sum of the exterior prisms made in the pyramid Y^ABC and the sum of the interior prisms made in the pyramid F'— ^'^'6^' is therefore the prism whose base is ABC. This difference will become less and less GEOMETRY. 97 in proportion as our prisms have a less and less altitude ; and therefore if the altitude of the pyramids be divided into an indefinitely increasing number of parts, and if an infinitely increasing number of exterior and inte- rior prisms be constructed, having indefinitely decreas- ing altitudes, the difference between the sum of the exterior and interior prisms will be a prism of an inde- finitely small altitude constructed on ABC as a base ; that is, a volume less than any assignable volume. But the two pyramids evidently differ less from each other than the two sums of said exterior and interior prisms. Therefore the two pyramids have volumes, between which no difference is assignable ; which is to say, that the two pyramids are equal in volume. Q. E. D. PEOPOSITIOJS- XII. 113. Any triangular jprism can he divided into three equal triangvla/r pyramids. Proof. Let ABC—F be a triangular prism. Draw the planes A CE and DCE. The prism will be divided into three pyramids. The pyramids J. ^(7 —E and BEE— C are evident ly equal, as they have equal bases and altitudes. The pyra- mids ACB-E and BCE -E are also equal, as their bases are halves of the same parallelo- gram, and their vertex is at the same point E. Therefore the three pyramids are equal in volume. Q. E. D. Cor. I. The volume of a triangular jpyramiid is equal to one-third of the jproduct of its hase into its altitude. 98 ELEMENTS OF Cor. IT. The volume of any pyramid is equal to one- third of the product of its hase into its altitude. For any pyramid can be divided into a number of triangular pyramids, whose total volume will be one-third of the product of the sum* of their bases into their common al- titude ; and the sum of their bases is the base of the pyramid which they compose. PKOPOSITIOIsr XIII. 114. The volume of the frustum of a pyramid is equal to the sum of the volumes of three pyramids whose common altitude is the altitude of the frustum^ and whose hases are the lower hase of the frustum^ the up- per hase of the frustum, and a mean proportional he- tween them. Proof 'LQiABCDE-a be the frustum of a pyramid whose vertex is Y. The vol- ume of this frustum is equal to the volume of the pyramid ABODE— V minus the vol- ume of the pyramid ahcde — Y. Make ABCDE=.P and ahcde =^, also YO—H, and Yo—h, and call F the volume of the frustum. Then (ISTo. 113, Cor. 11.) F=\{Py.E:-py.h). But (No. 110, Cor. I.) !P P'.:h': H'; hence j?=P Z- H' GEOMETRY. 99 and therefore But H'-h'^ {S-h) (jy'+ ^A + h?) ; and consequently ■ H-h (P.H'+ PHh + PA'\ or F= 7a 7 I^ow, P-^^=:^, and P-^ is a mean proportional be- tween P and P-^a ? that is, between P and ^. There- fore and since H—h is the altitude of the frustum, it follows that the volume of the frustum is equal to the sum of the volumes of three pyramids of the same height, whose bases are the lower and the upper bases of the frustum, and a mean proportional between them. Q. E. D. PKOPOSITIO]^ XIY. 115. Similar triangular prisms are to one a/nother as the cubes of their homologous edges or linear dimen- sions. Proof. The prisms ABC—Fdm^ ahc—f if similar, 100 ELEMENTS OF have similar faces between homologous edges, equally inclined to their Hence, letting fall DO and do perpendicular to ^^6^ and dbc, and drawing AG and ao, the triangles A OD and aod will be similar; for, besides the right an- gles at and J.6^ and dao equal, Consequently DO\do\\AD\ad. On the other hand, from the similarity of the figures we have ABC\ ahcwAR : al^wAD'; '^ ; hence, multiplying these proportions term by term, we have ABCxJyO : abcxdo :: AD' : S'/ but ABCxDO and aboXdo are the volumes of the two prisms; therefore Prism ABC-F : Prism abc -/ : : AD' : ad". Q. E. D. Cor, I. Any two similar jprisms are to each other as the cubes of their homologous dimensions. For their bases are similar polygons containing the same number of similar triangles, similarly placed ; and consequently these prisms consist of similar triangular prisms, which are to each other as the cubes of their homologous di- mensions ; and as, for equal ratios, the sum of the ante- cedents is to the sum of the consequents as each antece- dent is to its consequent, the whole volume of one prism GEOMETRY. 101 is to the whole volume of the other as each triangular prism is to its similar one. Cor. II. Similar jpyramids are to each other as the cubes of their homologous dimensions. For they are similar parts of similar prisms. PROPOSlTIOISr XY. 116, The volume of a wedge is equal to the volume of a pyramid having for its hase a right section of the wedge, perpendicular to its hach, and an altitude equal to the sum of its three parallel dimensions. Proof. The wedge is a volume bounded by a rectangle ABCD called the hacJc, two E F equal trape- zoids ABEF, DGEF called faces, and two equal triangles AED, BEG called ends. The line EF, in which the faces meet, is called the edge. The volume is composed of three parts, viz., of a rectangular prism comprised be- tween the two sections EGH, FKL, perpendicular to the back of the wedge, and of the two equal pyramids AHGD-E and LBGK-F. Let AB—l, EE^l\ AD=h, and the altitude EI of the triangle EGH (which is also the altitude of the wedge) = h. The expression of the volume Y of the wedge will be 102 ELEMENTS OF V= prism GIIE-F-\- two pyramids AHGB -E. Now, the volume of the prism is \bhy,HL — \hlil' ; and the volume of the two pyramids is \hy^AD {AII-\-LB) = \hh {l-V), Therefore F= ^l)hV-{- \bh {l-V) ; which may be put under the form 2 V 3 7 ' and this is the volume of the pyramid above mentioned. Q. E. D. When the length I of the back is less than the length V of the edge, the volume will be equal to that of the right prism EGII-Fmhms that of the two equal pyramids JIABG-Esind BLKC —F. And therefore which again may be reduced to the form And the same result applies also to a wedge in which I = V . For then we have y-'i^^)-^^^^ which is the expression of the volume of the rectangular prism whose base is — , and whose altitude is I. GEOMETRY, 103 IK PEOPOSITIOlSr XYI. 117. The volume of a prismoid is equal to the vol- ume of a pyramid of the sam^e altitude, having for hase half the sum of its parallel faces, plus twice the middle section hetween the parallel faces. Proof A prismoid is a frustum of a wedge. Its faces ABCB and EFGH 2.yq paral- lel rectangles ; and its altitude is KI perpendicular to the parallel faces. Drawing a dia- gonal plane BGFE, the vol- ume of the prismoid will be divided into two wedges ; and therefore, putting AB = I, EF=.l\ DA = h, EII=l', and Kl = A, we shall have (ISTo. 116) ^ hh/2i+r\ . h'h/2i'+i\ or Y=^ (2U + 2bT+ U+ 1% which may be written thus : F= ^ {U -f l'l'+ M-{- VT) + ^ (J^ + hT). Draw the middle section LMON, and make LM = m^ LN = n. As this section is midway between the bases, we have- 2m = Z-f Z', 271 = l + h\ and ^mn = {l^l') {h-\-l)% or - 4m7i = Z5 + VV+ W+ Vh ; 104 ELEMENTS OF Substituting this in the expression of the volume, we have or which is the volume of the pyramid above mentioned. Q. E. D. GEOMETRY, 105 BOOK YII 8UBFACE8 AND VOLUMES OF ROUND BODIES. 118. Round bodies are those geometric bodies wbicb can be described by a plane figure revolving about a fixed line called the axis of revolution. Thus a right- angled triangle revolving about one of its sides describes a cone / a rectangle revolving about one of its sides de- scribes a cylinder ; a semicircle revolving about its dia- meter describes a sphere, and so forth. As every point of the revolving figure describes the circumference of a circle, the determination of the surfaces and volumes of round bodies must be based on the relations existing be- tween straight lines and circular curves. The unit of measure for all surfaces is, of course, a square construct- ed on a unit of length, and the unit of measure for all volumes is a cube constructed on the unit of length ; this unit of length being always a straight line. Now, round surfaces and volumes cannot be measured by any straight unit, unless they be considered as made up of a multi- tude of straight consecutive elements, a little inclined to one another. Hence, in treating of round bodies, we must keep in mind that it is only by regarding the circle as a regular polygon having an exceedingly great multi- tude of sides exceedingly small, that its area and its cir- cumference become expressible, in terms of rectilinear measures, to any desired degree of approximation, as we have explained above (IsTos. 60, 61). 106 ELEMENTS OF PEOPOSITION I. 119, The convex surface of a right cylinder is equal to the product of the circumference of its hase into its altitude. Proof. A right cylinder is what a right prism be- comes when its base is a regular poly- gon having an innumerable multitude of sides. Its convex surface is there- fore expressed, like that of every right prism (No. 102), by the product of its altitude into the perimeter of its base, the perimeter being, in this case, the cir- cumference of a circle. Q. E. D. Let AB=^h be the altitude of the cylinder, and CA = r the radius of the base. Then the convex surface willbe/(S=2;r7'X>^. PKOPOSITIO]?^ II. 120. The volume of a right cylinder is equal to the jproduct of the area of the hase into the altitude. Proof The volume of any right prism is equal to the product of its altitude into the area of its base (No. 109). But a right cylinder is nothing but a right prism whose regular base has an innumerable multitude of sides. Therefore the volume of a right cylinder is equal to the product of its altitude into the area of its base. Q. E. D. Let h be the altitude, and r the radius of the base ; then the volume of the cylinder is Y— itr' X A. GEOMETRY. 107 PKOPOSITION III. 121. The convex surface of a right cone is equal to half the product of the circumference of its hase into its slant height. Proof The convex surface of any right pyramid is equal to half the product of the peri- meter of its base into its slant height (N"o. 111). But the right cone is no- thing but a regular right pyramid whose base has a circumference for its perimeter. Therefore the convex sur- face of a cone is equal to half the pro- duct of the circumference of its base into its slant height. Q. E. D. If AB = 7' is the radius of the base, and B V= s the slant height, the convex surface of the cone will be S=2Kr X ^=^7:rXs. PKOPOSITIOX lY. 122. The volume of a cone is equal to the product of the area of its hase into one-third of its altitude. Proof The volume of a right pyramid is equal to the product of the area of its base into one-third of its alti- tude (No. 113) ; and the cone is a right pyramid whose base is a regular polygon of an innumerable multitude of sides. Hence the conclusion. Q. E. D. Let h be the altitude of the cone, and r the radius of its base ; then its volume will be F = ^nr"^ X h. 108 ELEMENTS OF PEOPOSITIO]Sr Y. 123. The convex surface of the frustum of a cone is equal to half the product of its slant height into the sum of the circumferences of the parallel hases. Proof The convex surface of the frustum of a pyramid is e^jual to half the product of its slant height into the sum of the peri- meters of the parallel bases (No. Ill, Cor.) When the perime- ters become circumferences of circles, the frustum of the pyra- mid becomes the frustum of a cone. Hence, etc. Q. E. D. Let r and r' be the radii of the bases, and s the slant height ; then the convex surface of the frustum will be S^r.(T-\-T')s. PEOPOSITIOlSr YI. 124. The volume of the frustum of a cone is equal to the sum of the volumes of three cones whose common altitude is the altitude of the frustum, and whose hases are the lower hose of the frustum, the upper hase of the frustum, and a mean proportional between them. Proof Tliis proposition has been proved (No. 114) for any frustum of a pyramid. And therefore the same is true for the frustum of a cone, which is nothing but the frustum of a right pyramid whose base is a regular polygon of an infinite multitude of sides. Q. E. D. GEOMETRY. 109 Let T and r' be the radii of the two bases, and h the altitude of the frustum. Its volume will be PEOPOSITIOJST YII. 125. If a regular semi-polygon he revolved about its axis, the surface generated hy the semi-^jperimeter will he equal to thejproduct of the axis into the circumference of the inscribed circle. Proof. Let ABGDEF be a regular semi-polygon, AF\i'& axis, and OM its apothem, or the radius of the inscribed circle. From the extremi- ties of any side, as BC, draw BG and CH perpendicular to AF, and BN perpendicular to CH. The sur- face generated by ^6^ is the convex surface of the frustum of a cone, whose bases are 27rX^-^and 27rX BG, and whose slant height is BG ; hence such a surface (Ko. 123) is equalto^(7X7r((7^+^6^). From the middle point Jf of BG draw Ml perpendicular to AF ; then GH^BG — '^MI,^^^ the expres- sion of the surface becomes BGy. 2nMI. Now, from the similar triangles GBJSF and MGl we have BG : BJST: : GM : MI: : 2;r6)Jf : 2;rJ[//; 110 ELEMENTS OF hence BCy.'I'kMI^BNy^'ItiOM ; - but BNz=. GH ; therefore the generated surface will be GEy.'lnOM; that is, the surface generated by any side is equal to the product of the projection of that side on the axis into the circumference of the inscribed circle. Hence the surface generated by the entire semi-perimeter is equal to the product of the sum AF of the projections of its sides into the circumference of the inscribed circle. Q. E. D. Cor. The surface generated hy any portion of the senni-jperimeter is equal to the jproduct of its projection into the drcumference of the inscribed circle. PEOPOSITION YIIL 126. The surface of a sphere is equal to the product of its diameter into the circumference of a great circle. Proof. The surface of the sphere is generated by the revolution of a semi- circumference around its diameter. Now, a semi-circumference is nothing but a regular semi- perimeter of a numberless multitude of sides ; its projec- tion is its diameter; and the radius of the inscribed circle is nothing else than its own radius. Therefore (No. 125) the surface of the sphere is equal to the pro- duct of its diameter into the circumference of a circle of the same diameter. Q. E. D. Cor. 1. The surface of the sphere whose radius is r is Sz=^^Tzry,%r-=.^:Tzr^. It is therefore equal to four times the area of a great circle. GEOMETRY. Ill Cor. II. The surface gene- rated by any arc of a semi- circle, is equal to the product of its projection on the diam- eter into the circumference of a circle of the same diameter (No. 125, Cor.) Such a sur- face is called a zone. Thus, if the arc AG revolves around the diameter MN^ it generates the zone A BCD, whose alti- tude is EF, and whose area is ^izry^EF. Cor. III. Two zones on the same sphere are to each other as their altitudes. PEOPOSITION IX. 1!37. The volume generated by a triangle revolving around its hase is equal to the jproduct of one-third of its base into the area of the circle described by the vertex opposite the base. Proof. Let the triangle ABC vqyoIyq around its base ^^ as an axis. Draw from C the line CD perpendicular to the axis. The portion ADC of the triangle will generate a cone whose volume (No. 122) will be \n . CD'XAD, and the portion BCD will generate a cone whose vol- ume will be \n.CD'xDB. 112 ELEMENTS OF Hence the total volume generated will be \7i, CD' {AD-\-DB\ or \7iCD'X AB, If the perpendicular CD falls on the prolongation of the base AB, the result is still the same. For in this case the volume generated bj the triangle ABC will be the difference between the volumes of the cones gene- rated bj A CD and BCD ; that is, \7: . CD' {AD-DB\ or \'kCD' X AB, Q. E. D. PKOPOSITIOlSr X. 128. If an isosceles triangle revolves about a line passing through its vertex, the volume generated will he equal to the jproduct of one-third of the altitude of the triangle into the surface generated by its base. Proof. Let^^ be the base, (7 the vertex, and 6Tthe altitude of an isosceles triangle, and suppose, ' first, the base AB to be parallel to the axis of revolution- CD. Draw AP and BQ perpendi- cular to the axis. The volume generated by the triangle will be equal to the volume of the cylinder generated by the rectangle APQB, minus the volumes of the two cones generated by the two equal triangles APC and BQC. The vol- ume of the cylinder is tiCPx AB, and that of the two cones is ^nAl^'X PC+irrBQ'X CQ, or IttCI'X AB. Hence the volume generated by the triangle ABC is GEOMETRY. 113 F=|;r(7Px AB=\CIxAB X ^nCI, where AB X 2yr(7/ expresses the surface generated by the base AB, Suppose, in the second place, the base AB to be inclined to the axis CD. Produce AB till it meets the axis at D. The vol- ume generated by the triangle ABC will be equal to the volume q generated by the trian- gle A CD minus the volume generated by the triangle BCD. We have therefore (Ko. 127), AE and BF being the altitudes of these two triangles, Y=^7:AE' XCD- \'kBF' X CD, that is, Y=.\tz {AE'- BE') X CD. Draw IG perpendicular and BH parallel to CD. Since AE'- BF'^ {AE^ BF) {AE- BF) = 2IG x AR, the expression of the volume becomes V=i7:.JG.AR.CD. But the right-angled triangles ABII and CID, which are similar, give AH : AB :: CI : CD • hence AH.CD = AB .CI; and consequently Y=\Cly.'^7zlGxAB, where ^nlGxAB is the surface generated by AB, 114 ELEMENTS OF Suppose, tliirdlj, tlie base f^ AB to be so inclined that /\ the point B falls on the / \\ axis CD. Draw AE and /^--•'^T iX IG both perpendicular to ^^^ ^ ^ |^ g- CD. The volume generat- ed by the revolution will be (No. 12Y) Y=:^jcAE'X CB. But, since the triangles AEB and CIB are similar, we have AE \AB\\C1\ CB; hence AE . CB ^ AB . CI. Therefore F= \7zAE . AB . CI. But AE= 21 G ; and consequently * V=\Clx27:IGxAB, where ^nlG X AB is the surface generated by AB, Hence the theorem enunciated is true in all cases. Q.E.D. PEOPOSITIOISr XL 129. The volimie generated hy a regular semi-poly- gon revolving about its axis is equal to the product of one-third of its apothem into the surface generated hy the semi-perimeter. Proof Let ABCDEEhe a regular semi-polygon, and 01 its apothem. If we draw BO, CO, ... so as to di- vide it into isosceles triangles, each triangle will, by its GEOMETRY. 115 revolution, generate a volume equal to the product of one-third of its al- titude into the surface generated by its base (No. 128). And, since all these triangles have for their altitude the apothera of the semi-polygon, the total volume generated will be ex- pressed by the product of one-third of the apothem into the whole sur- face generated by the semi-perimeter. Q. E. D. Cor. The volume generated by any portion OB CD of the semi-polygon is equal to the product of one-third of the apothem into the surface generated by the corre- sponding portion of the semi-perimeter. PEOPOSITION XII. ,130. TJie volume of a sphere is equal to the j^roduct of its surface into one-third of its radius. Proof. The volume of the sphere is generated by the revolution of a semicircle about its diameter. Kow, a semicircle is nothing but a regular semi-polygon of a numberless multitude of sides, whose apothem is nothing else than its own radius. Therefore (No. 129) the vol- ume of the sphere is equal to the product of one-third of its radius into the whole surface of the sphere. Q. E. D. Cor. 1. The surface of the sphere being ^nr^ (No. 126, Cor. I.), its volume will be F: ~3~' 116 ELEMENTS OF Cor. 11. Any circular sector, that is, any portion of a semicircle bounded by two radii, will generate a volume equal to the product of one-third of the radius into the sur- face, or zone, generated by the arc of the sector. Such a voliJme is a spherical sector. Thus ABODG is a spherical sector generated by the revolution of the circular sector GAB about the diameter EF. The por- tion AOG oi the sphere forms no part of the volume of this spherical sector. Cor. III. The volumes of different spheres are to one another as the cubes of their radii, or of their diameters. PEOPOSITIOE" XIII. 131. The volume generated hy a circular segment revolving about a diameter exterior to it, is equal to a cone having for its hase tlie circle constructed on the chord of the segment as a diameter, and for its altitude tivice the j^ojection of that chord on the axis of revo- lution. Proof. Let A CB be the segment of a circle. Bisect the chord AB in D ; draw the radii AG and BG, and let fall the perpendiculars AH, BG, and DI on the axis GEOMETRY. iir EF ; draw also ^iV^ perpendicular to BG, and OD per- pendicular to the chord AB. The volume generated by the segment A CB will evidently be equal to the volume generated by the sector ACBO, minus the volume generat- ed by the isosceles triangle ABO. Hence (ISTos. 130 and 128) Y=27:AOxGHx^AO -^TiDIxABx^BO. But from the similar triangles OBI and BAN we have Dl\ OD::AN:AB, whence JDIx AB=ODx A]Sr= OB x GH ; hence, by substitution and reduction, F= |;r . GR {AG'- BO') = \7z.GH, AB% ^ 2GH or X 7rAB\ which is the volume of a cone whose altitude is ^GH and whose base has the radius AB. Q. E. D. PKOPOSITION^ XIY. 132. The volume of a sjpherical segment is equal to the volume of a sphere whose diameter is the altitude of the segment^ plus the volumes of two cylinders whose bases are the bases of the segment, and whose altitude is half that of the segment. Proof. Let A CB be the arc of a circle. Bisect the 118 ELEMENTS OF chord AB in D. Draw AH, BG, and 7>/ perpendicular to the diame- ter SF. If the area ACBGH re- volves about JEF. it will generate a spherical segment, whose volume will consist of the volume generated by the circular segment ACB, and of the conical frustum generated by the trapezoid ABGH. Hence (Kos. 131 and 124) 2GH V: TT .AD' GH + ^Tzi^AW^BG'-^^AIIy^BG), !N'ow, since AB^'^AB, we have AB'=: iAB'= i {AW'+ BIT'), and because Jjr'= GH' and BW'= BG'+AH'- 2BG X AH, hence AB'=i {GH'+AH'+BG'- ^AHxBG). Substituting this value of AB' in the expression of V, we find ^ GH /GH'+AH'+BG' ^AHxBG+AH'-l-BG'-AHxBG^ ; which, by reduction, becomes -^ GH /SA7r+SB~G'+GIB\ GEOMETRY. 119 whence, putting 26^^/ instead of GH, we shall finally obtain AttGI* y^ —3— + TzAH'x GI+7cBG'x GI, which is the sum of the volumes designated in our pro- position. Q. E. D. Cor. If the spherical segment has but one base, then AH becomes zero, and Gil becomes GE. Hence the expression of the volume will become that is, a spherical segment having hut one hase is equal to a sphere whose diameter is the altitude of the segment, plus a cylinder having for its hase the hase of the seg- ment, and for its altitude half the altitude of the seg- m£nt. 120 ELEMENTS OF BOOK YIII 8PHEBICAL GEOMETRY, 133. In spherical geometry we have to deal with an- gles, triangles, and polygons the sides of which are arcs of circles whose planes pass through the centre of the sphere, and which are styled great circles. The angle made by two arcs of great circles at their intersection is a dihedral angle ; for it measures the inclination of the planes of the two circles. A spherical polygon is a por- tion of the surface of the sphere bounded by arcs of a great circle. If bounded by three arcs only, it will be a spherical triangle ; if bounded by two arcs only, it will be a lune^ and the two arcs will then be two semi-circum- ferences. The solid bounded by a lune and two semi- circles meeting in a common diameter is called a spheri- cal wedge. A spherical pyramid is a solid bounded by a spherical polygon and sectors of great circles, the centre of the sphere being the 'vertex, and the polygon the hase of the pyramid. Such are the main subjects to be inves- tigated in the present book. PEOPOSITION I. 134. The angle formed hy two arcs of great circles is equal to that formed hy their tangents at their jpoint of intersection. Proof. Let the angle MAN be formed by the two GEOMETRY. 121 arcs AM and AI^. The tangent AP drawn in the plane A OJV, and the tangent A Q drawn in the plane A OM^ are both perpendicular to the radius OA which lies in the intersection of the two planes. Hence the angle of the two tangents is the measure (No. 81) of the inclination of the two planes. It is therefore the measure of the angle MAN deter- mined by the arcs lying in those planes. Q. E. D. Cor. I. The arc MW of a great circle perpendicular to ^6^ is also the measure of the angle MAN. For, draw MO and WO ; these lines will be parallel to AQ and AP respectively. Hence the angle MON is equal to the angle QAP ; and therefore the arc MN is their common measure. Cor. II. The vertical angles formed at the intersection of two arcs are equal, for they are measured by the ver- tical angles of their tangents, which are equal. PEOPOSITIOK II. 135. Any section of a sphere made hy a plane is a circle. Proof. Let EFD be a plane section of the sphere. Draw the radius OP perpendicular to the plane of the section, and let C be the point where it pierces the 122 ELEMENTS OF plane. From the centre of the sphere draw radii to any points of the section, as D and F. The tri- angles 6>(7i> and 6> 67^ will have CO common, OB and Oi^ equal, and the angle in C right. Hence (No. 27, Cor.) the third side CD of the one will be equal to the third side CF of the other. The point C is therefore equally distant from any points taken in the section ; and consequently the section is a circle. Q. E. D. Cor. I. The section is greater or smaller according as it is nearer to, or farther from, the centre of the sphere. The greatest section passes through the centre of the sphere and is called a great circle; all the other sec- tions are called sinall circles. Cor. II. Every great circle divides the sphere and its surface into two equal parts. Cor. 111. The centre of the sphere and the centre of any small circle are in a straight line perpendicular to the plane of the circle. Thus the diameter PQ which passes through the centre of any circle EFD is perpen- dicular to its plane, and the points/^ and Q are ih.^ j^oles of the circle ; for a pole of a circle is a point on tlie sur- face of the sphere equally distant from all the points of the circumference of the circle. Now, this is the case with the points P and Q; for, if we conceive chords to be drawn from P to i>, to E^ to F^ etc., all these chords GEOMETRY. 123 will be equal, as being hypotenuses of equal triangles, and consequently all the arcs PB, PE^ PF^ . . . will also be equal. And in a similar manner the arcs QP, QE^ QF^ . . . will be equal. Cor. 1 V. The poles of a great circle lie at 90 degrees from every point of the great circle; and any arc PGQ passing through the poles of a great circle AGB is per- pendicular to the great circle (E^o. 95); and similarly any arc perpendicular to a great circle passes through its poles. PKOPOSITIOI^ III. 136. Any side of a spherical triangle is less than the sum of the other two. Proof Let ^^6^ be a spherical triangle situated on a sphere whose centre is 0. Draw OA, OB, 00. The trihedral angle will be composed of three plane angles, of which any one will be less than the sum of the other two (^o. 98). And since these angles are measured by the arcs AB, AC, BC, any of these arcs is less than the sum of the other two. Q. E. D. PROPOSITION lY. 13 7* If from the vertices of a spherical triangle, as poles, arcs of great circles he descrihed forming a spheri- cal triangle, the vertices of this second triangle will he poles of the sides of the first triangle. ■ Proof. Let ^^6^ be the given triangle. From^,^, 124 ELEMENTS OF C as poles describe the arcs EF, FD^ DE. Since A is the pole of EF^ the dis- tance AE is a quadrant; and since C is the pole of DE^ the distance CE is a quadrant also ; hence the point E is at the distance of a quadrant from the arc A C, and therefore it is the pole of this arc (No. 135, Cor. ly.) It may be proved in the same manner that F is the pole of AB, and D the pole of BG. Q. E. D PEOPOSITIOJSr Y. 138. Any angle of a spherieal triangle is measured hy the sujpjplement of the side ojyjposite to it in its jpolar triangle. Proof. Let ABCh% the given triangle, and DEF its polar triangle, that is, a tri- angle whose vertices are the poles of the sides of ABC. Prolong the sides AB^ A C, and BC till they meet the sides of DEF. The angle A will be measured by the arc GH (No. 124, Cor. I.) Now, since EJI=:90° and GF=:GJI+RF= 90°, we have EF=ER+IIF=90°+{90°-GR\or GR=1S0°-EF, and therefore ^ = 180° - EF GEOMETRY, 125 And in a similar manner it may be shown that ^ — 180°- DF, and C^ 180°- DK Q. E. D. Cor. From the definition of polar triangles it is ob- vious that if a triangle A'B'C is polar to the tri- angle ABG, then ABC is also polar to A' B' C . Hence calling «, J, G and a' ^ h\ c' the sides opposite to the an- gles ^, ^, G2.xi^A',B\C', we shall liave not only JL=zl80°-^', ^ = 180°- but also A'=i 180°- a, B'= 180°- 1, (7'=:180°-C. PKOPOSITIOlSr YI. 139. The sum of the sides of a spherical triangle is always less than the circumference of a great circle. Proof Let ABC be a spherical triangle. The sides AB and AC produced will meet at a point D diametri- cally opposite to A; hence the two arcs ABD and A CD together are the circumference of a great circle. But {^o. 136) BC is less than BD-\- CD ; therefore the sum ^J?-j-^^+^^ is less than the circumference of a great circle. Q. E. D. 126 ELEMENTS OF PEOPOSITION yii. 140. Symmetrical triangles are equal in all their jparts. Proof. Triangles are called symmetrical when they have sides respectively equal, but disposed in opposite ways, so that one triangle cannot be made to coincide with the other by superposition. Let ABC and BCD be two sym- metrical triangles having AB^=^ BD, AC— CD, and the third side BC oi the one placed on the equal side BC oi the other. Draw the radii AG, BO, CO, DO. The ra- dii AG, BO, CO will be the edges of a trihedral angle having its ver- tex at 0, whilst the radii BO, DO, CO will be the edges of a second trihedral angle having also its ver- tex at 0. As the plane angles formed by these edges are respectively measured by equal arcs, they are respec- tively equal ; hence (No. 100) the planes of the equal angles are equally inclined to each other. But (No. 134) tlie angles made by these planes are equal to the corre- sponding angles of the spherical triangles. We have, therefore, ^^(7= ^i> 6/, ABC=DBC, and ACB — DCB ; and consequently the two triangles are equal in all their parts. Q. E. D. Cor. If the two triangles have angles respectively equal, their sides will also be respectively equal. For if the angles are respectively equal, the planes which form them are equally inclined to each other, and their edges form in equal trihedral angles, having equal faces, that is, composed of plane angles respectively GEOMETRY, 127 equal. JS'ow, these plane angles are measured by the opposite arcs, that is, by the sides of the two triangles. And therefore the sides opposite to equal angles will be equal, and the two triangles will be equal in all their parts. PKOPOsiTiojsr yiii. 141. Two spherical triangles having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, are equal in all their parts. Proof. Let AB G and DEF have the sides AB -DE, and AC^=.DE, and the included angles A and D equal. Since A — D, if the side DE of the second triangle be superposed on the side AG oi the first, the side DE will necessarily coin- cide with the side AB, and the' point E will fall on B ; and consequently the side EE will lie on the side BG. The two triangles are therefore equal in all their parts. If the triangles can- not be superposed owing to their symmetrical position, as is the case with the triangles ^^6^ and DEG, then by constructing the triangle DEE symmetrical to DEG, and by superposing it on ABG, we first show the equali- ty of i>^i^ with ABG, then, because DEE is equal to its symmetrical DEG, we conclude that DFG also is equal to ^^(7. Q. E. D. 128 ELEMENTS OF PEOPOSITIOlSi IX. 142. Two spherical triangles having two angles and the included side of the one equal to two angles and the included side of the other ^ each to each, are equal in all their jparts. Proof Proceeding as in the last proposition, place the side DF Vi^oi\ its equal AC. If the angles A and D are equal the side DE will lie upon AB, and if the angles F and (7' are equal the side FE will fall upon BC, and the intersection of the arcs DF and FF will coin- cide with the intersection of AB and BC, and therefore the triangles will entirely coincide. If the triangles can- not be superposed, as is the case with ABC and DFG, make the triangle DFF symmetrical to DFG ; super- pose DFF on ABC, as above, and from its equality with ABC, it will be proved thati>i^6^ also must be equal to ABC Q. E.D. PKOPOSITIOI^ X. 143. Two spherical triangles^ lohose sides are all re- spectively equal, are equal in all their jparts. Proof. If the two triangles are symmetrical, it has al- ready been shown (No. 140) that equal angles will be opposite the equal sides. If the two triangles are not symmetrical, they may be superposed so as to coincide throughout ; for, as their sides are equal to those of the symmetrical triangles, so also their angles will be equal to those of the symmetrical triangles (No. 140); and therefore they will be equal in all their parts. Q. E. D. GEOMETRY. 129 PEOPOSITION XI. 144. In any isosceles sjpherical triangle the angles opposite to the equal sides are equal i and conversely. Proof. Let ABC be an isosceles triangle liaving AB::^AC. Draw an arc of a great circle from the ver- tex A to the middle point D of the base BC. The triangles ABD and ACD will have their sides respec- tively equal ; hence the angles B and C opposite to the same side AD will be equal (No. 143). Q. E. D. Conversely, if the angles B and C are equal, the opposite sides ^6^ and AB must be equal. For, if AB^ for instance, were greater than A (7, it might be shortened by a portion^^', so as to leave BE^^AG ; but in such a case, drawing the arc EG^ we would have in the triangles ABG and EBG the sides AG and BE equal by construction, the side ^6^ common, and the in- cluded angle ^6/5 equal to the included angle ^^5 (7. Hence (Xo. 141) the triangles J.^(7and EBG would be equal in all their parts, and therefore the angle A GB would be equal to the angle EGB, that is, the whole would be equal to one of its parts. This being impos- sible, it is impossible to admit that one of the > two sides is greater than the other. Q. E. D. PKOPOSITIO:^ XII. 145. In any spherical triangle the greater side is opposite to the greater angle ; and conversely. Proof. Let ^^{7 be a spherical triangle in which the 130 ELEMENTS OF angle A is greater than the angle B. Draw the arc AD so as to make the angle DAB equal to the angle B. Then ABD will be isosceles, and AD = BD. N^ow, the sum AD + CD is greater than A C (No. 136). Therefore the sum BD -f- CD^ or the side BG, is greater than AG. Q. E. D. Conversely, if j56^> J.(7, the angle A is greater than the angle B. For, if ^ were not greater than B, it would be either less or equal. Now, it cannot be equal, so long as the opposite sides are unequal (No. IM) ; and it cannot be less, for then BG should be less than AG, as we have just proved. PEOPOSITION XIII. 146. Two spherical triangles having all their angles respectively equal, are equal in all their parts. Proof. If the two triangles are symmetrical, it has al- ready been shown (No. 140, Cor.) that equal sides are opposite their equal angles. If the two triangles are not symmetrical, they may be superposed so as to coincide throughout; for, as their angles are equal to those of the symmetrical triangles, so also their sides will be equal to those of the symmetrical triangles (No. 140, Cor.) ; and therefore they will be equal in all their parts. Q. E. D. Gor. Two spherical triangles cannot he similar with- out being equal. GEOMETRY. 131 PKOPOSITIO]^ XIY. 147. The sum of the angles of a sjpherical triangle is greater than two right angles, and less than six right angles. Proof. JuQtABC2.ndLDEF be two polar triangles. We liave seen that the angle A is measured by 180°— EF, the angle B by 180°- DF, the an- gle C by 1S0°-I)F. Hence each angle is less than 180°; and therefore their sum is less than 3x180°, or six right an- gles. On the other hand, since A + B+C=SxlSO''-{FF+J)F+I)F), and the sum EF-\- DF-{- DE is necessarily less than 360° (No. 139), it follows that A-^B-^G is greater than 3x180°— 360°, that is, is greater than 180°, or two right angles. Q. E. D. Cor. I. A spherical triangle may have two, and even three, right angles; also two, and even three, obtuse angles. Cor. II. If a spherical triangle is tri-rectangular, its sides will all be quadrants of a great circle. For if in a triangle AB C we have A = 90°, B = 90°, C=z 90°, then in its polar triangle we must have (No. 138, Cor.) «'=180°-^=:90°, 5'= 180°- ^=90°, c'— 180°- ^=:90°, and thus, the sides a\ V, c' being quadrants, the angles 132 ELEMENTS OF A', B\ C of the triangle A'B'C will be tlie poles of this triangle as well as of the triangle ABC. Ilence the two polar triangles are, in this case, one and the same triangle, and a^a\ h = h\ c = c^ / that is, all the sides are quadrants. Conversely, if all sides be quadrants, the triangle will be tri-rectangular. Cor. III. Four tri-rectangular triangles make up the surface of a hemisphere, and eight the entire surface of the sphere. Hence the area of a tri-rectangular triangle is expressed (No. 126, Cor. I.) by T^ ^Tzr". Cor. lY. If a spherical triangle is bi-rectangular, the sides opposite to the right angles will be quadrants of a circle ; for, since they are perpendicular to the third side, they must pass through the pole of that side (No. 135, Cor. lY.), and their intersection will take place at the pole it- self. Thus if ABC is bi-rectan- gular, the sides AB and A (7, op- posite to the right angles, are quadrants, and A is the pole of the side BC. Conversely, if the sides AB and AC are quadrants, the angles C and B will be right angles; for the planes AOB and AOG will be perpendicular to the plane BOC PEOPOSITION XY. 148. The surface of a lune is to the surface of the sphere as its angle is to four right angles. Proof, Conceive the surface of the sphere as made up GEOMETRY. 133 of 360 lunes having each an angle of 1°. Then 360 such lunes will be equal to the surface of the sphere. Hence designating one of these lunes by Z, and the surface of the sphere by S^ we shall have or Z:a^:: 1:360. Let us have now any lune ABECA^ and suppose it to contain m times the lune I; m being any entire or any fractional number whatever. Its angle will also be m times the angle 1°, that is, m degrees. The preceding proportion will then be- come that is, the surface of the lune is to that of the sphere as the angle of the lune is to four right angles. Q. E. D. Cor, If we denote the area of a tri-rectangular triangle by T, the area of the lune by Z, and the angle of the lune by J., the right angle being denoted by 1, we shall have Z:8r::J.:4: hence L=:2AxT; that is, the surface of a lune is equal to a tri-rectangular triangle multiplied by twice the angle of the lune. As 90° is here made =1, the value of the angle A must be the expression of how many times A degrees con- tain 90°. 134 ELEMENTS OF PKOPOSITIOJSr XYI. 14:9, The area of a spherical triangle is equal to its spherical excess multiplied hy a tri-rectangular triangle. Proof. By spherical excess we mean the excess of the sum of the angles of a spherical triangle over two right angles. This excess is calculated by taking the right angle as the unit of measure of spherical angles ; and consequent- ly the number of degrees contained in the spherical ex- cess must be divided by 90° ; and the quotient, which is the true value of the spherical excess, is to be considered an abstract number or coefficient. Let, then, ABC be any spherical triangle drawn on the hemisphere DEHFGK. Prolong the arcs AB^ A (7, BG till they reach the section of the hemi- sphere. Then the ver- tical triangles ABE and AGF taken to- gether will be equal to a lune of the angle A^ the triangles BFII and BBK taken together will be equal to a lune of the angle B^ and the triangles GGK and GEH taken together will be equal to a lune of the angle G. The sum of the (Ko. 148, Cor.) three lunes will be 2^xr+2^X^+2(7xr=2r(^ + ^-f ^). Now, this sum exceeds the surface 4T of the hemisphere GEOMETRY. 135 by twice the triangle ABC ; for this triangle has been taken three times instead of one time. We have there- fore 2ABC=2T{A + B-\-C)-4.T. Dividing the equation by 2, and reducing, we have ABC=:{A-{-B+C-2)T, A -\- B -\- C — '^ being the spherical excess. Q. E. D. Cor. Since the area of the triangle has been deduced from the area of the lunes, in whose expression an angle of 90° was taken as a unit, the number 2 in the expres- sion of the spherical excess represents 180 degrees. Thus, if in a given triangle we have A — 86° 10^, B — 69° 20', ^=100° 15', the value of the spherical excess will be expressed by 86° 10'+ 69° 20'+ 100° 15'- 180° 90° ' the quotient 0.841666 . . . being an abstract number. PROPOSITIOlSr XYII. 150, The area of a spherical polygon is equal to its spherical excess multiplied hy a tri-rectangular triangle. Proof. A spherical polygon ABCDE can always be divided into triangles by drawing diago- nal arcs. If the polygon has n sides, it will be divided into n— 2 triangles, every one of which will be equal to its spherical excess multiplied by a tri-rectangular 136 ELEMENTS OF triangle (IS'o. 149). Hence tlieir sum, or the area of the polygon, will be equal to S being the sum of all the angles. Q. E. D. PROPOSITION XYIII. 151. The volume of a sj)herieal wedge is equal to the product of its base [the Iwie) into one-third of the radius. Proof. The volume of the sphere being conceived as made up of spherical wedges, just as its surface is conceived as made up of as many lunes, we easily perceive that the wedge is to the volume of the sphere as its base, or lune, is to the surface of tlie sphere. Hence, W being the wedge, AiKr^ W \ \L\ ^7:r'; which is readily reduced to Wz Z . ^. Q. E. D. o PROPOSITION XIX. 152. The volume of a spherical pyramid is equal to the ^product of its hase {a spherical polygon] into one- third of the radius. Proof Let ABCD — be a spherical pyramid. Di- GEOMETRY. 137 viding the area of the base ABCD into a very great number of exceedingly small areas, the spherical pyra- mid may be conceived as a sum of pyramids of an alti- tude equal to the radius of the sphere, and having for their bases the small areas of which the polygon ABCD is composed. These areas, when exceedingly small, do not differ from plane surfaces ; and therefore (No. 113, Cor. II.) the volume of each small pyramid will be equal to the product of its small base into one-third of the radius. Consequently the spherical pyramid, which is the sum of all the small pyramids, will be equal to the sum of all the small bases multiplied by one-third of the radius. But the sum of all the small bases equals the base ABCD. There- fore the volume of the spherical pyramid is equal to the product of its base into one-third of the radius of the sphere. Q. E. D. 14 JJAY Ubh RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 12 loy'54/0 REC'D LD OCT3 0'64-4P|V r.!i^^iSlH%V Univ^^g&n,.