of the one equal to one side of the other, and let the equal sides be made to coincide, so that the triangles A B C, EBC (in the figure of Prop. 24.) may represent the bases of the pyramids. Then the triangles ABC, E B C, being equal to one another, lie between the same equal and parallel small circles, and may be completed (as in Prop. 24.) into the quadrilaterals A B C D, E B C F. And it may be shown, as in the same proposition, that the triangles EA B, FDC have the three sides of the one equal to the three sides of the other, each to each, and therefore may be made to coincide (15.): wherefore, also, the pyramids, which have these triangles for their bases, may be made to coincide, and are equal to one another. And hence, as in Prop. 24., it was demonstrated, by the addition and subtraction of the lunular portions of surface, that the quadrilateral A B C D is equal to the quadrilateral EBCF, so, here, it may be demonstrated, by the addition and subtraction of the pyramidal solids (23.), which have these lunular portions for their bases, that the pyramidal solids, which have for their bases the quadrilaterals A B C D and E B C F, are equal to one another. But, because the triangles A B C, CDA have the three sides of the one equal to the three sides of the other, each to each, in the same order, they may be made to coincide (15.), and therefore the pyramids which have these triangles for their bases may be made to coincide, and are equal to one another; and each of them is the half of the pyramidal solid which has the quadrilateral A B C D for its base. And in the same manner it may be shown, that each of the pyramids on the triangular bases, EB C, CFE is the half of the pyramidal solid on the quadrilateral base EBC F. Therefore, because the halves of equals are equal, the pyramid upon the base ABC is equal to the pyramid upon the base EBC. Next, let the bases of the pyramids be any equal triangles A B C, KLM. be taken equal to the triangle A B C or KLM, and join K' M. Then, by the former case, the pyramids upon the bases A B C, K' L M' are equal to one another. And, because the triangle K'L M' is equal to the triangle K L M, the triangles K' M M and K K' M which have the common side K'M are likewise equal; therefore, by the first case, the pyramids, which have these triangles for their bases, are equal to one another; and, these being added to, or taken from, the pyramid upon the base KLM, the whole or remaining pyramid upon the base K' L M' is equal to the whole or remaining pyramid upon the base K L M, that is to the pyramid upon the base A B C. Lastly, let the bases of the pyramids P, P' be equal polygons. Let a triangle be found which is equal to one, and therefore also to the other of the polygons, and let Q be the pyramid which has this triangle for its base. Then, because this triangle is equal to the base of the pyramid P, it may be divided into triangles which are equal respectively to the triangles into which the base of P is divisible. And it has been already shown that pyramids which have equal triangles for their bases are equal to one another; and the sums of equals are equal; therefore, the pyramids P and Q are equal to one another. In the same manner it may be shown that the pyramids P' and Q are equal to one another. Therefore P is equal to P'. And it is evident that what has been shown with regard to spherical pyramids, being derived from coincidence, may be shown equally of their solid angles at the centre of the sphere. Therefore, &c. PROP. 28. Any two spherical pyramids are to one another as their bases; and the solid angles of the pyramids are to one another in the same ratio. Let P, P' be any two spherical pyramids, and let B, B' be their bases: the pyramid P shall be to the pyramid P as the base B to the base B'. For if the base B' be divided into any number of equal parts, then, because pyramids which stand upon equal bases are equal to one another, the pyramid P' will be divided into the same number of equal parts by planes passing through the arcs of division (27.); and if the base B contain exactly, or with a remainder, a certain number of parts equal to the former, the pyramid P will contain exactly, or with a remainder, the same number of parts equal to the latter (27.): therefore, P is to P' as B to B' (II. def. 7.) And the same may be said of the solid angles. Therefore, &c.7 Cor. Every spherical pyramid is equal to the third part of the product of its base and the radius of the sphere. For, if the whole sphere be divided into spherical pyramids, these pyramids will be to one another as their bases; and, therefore, any one of them is to the sum of all, as the base of that one to the sum of all the bases (II. 25. Cor. 3.); that is, any pyramid is to the whole sphere as its base to the surface of the sphere. Therefore, any pyramid is to the whole sphere as the product of its base and the radius of the sphere to the product of the whole surface and radius (IV. 26.); and because, in this proportion, the second term is equal to the third part of the fourth (V. 17.), the first term is likewise equal to the third part of the third (II. 19. and II. 13.). Scholium. Hence it appears that every solid angle is measured by the spherical surface which is described with a given radius about the angular point, and intercepted between its planes. For it is shown in the proposition that any two solid angles are to one another as these surfaces. This measure bears an obvious analogy to the measure of a plane angle, as stated in the Scholium at Book III. Prop. 13. The angular unit of the latter measurement is the right angle; into four of which the whole angular space about any point in a plane is divided by two straight lines drawn at right angles to one another; and each right angle is measured by a quadrant, or fourth part of the circumference of a circle described about that point with a given radius. In like manner, if, through a point in space, three planes be made to pass at right angles to one another, they will divide the whole angular space about that point into eight solid right angles, each of which (a solid angular unit) is measured by an octant, or eighth part of the surface of a sphere described about that point with a given radius. Thus, then, the plane angles, dihedral angles, and magnitude of a solid angle, which has three or a greater number of faces, are represented, upon the surface a sphere described about the angular point, by the sides, angles, and surface of a spherical triangle or polygon: and whatever has been stated with regard to the latter may be understood likewise of the former. It is shown (for instance) in Prop. 13. that if two solid angles, each of which is contained by three plane angles, have two plane angles of the one equal to two plane angles of the other, each to each, and likewise the dihedral angles contained by them equal to one another, the remaining plane and dihedral angles of the one shall be equal to the remaining plane and dihedral angles of the other, each to each. And, generally, all questions which relate to solid angles will be placed in the clearest view before us, when we contemplate only their representation and that of their parts, upon the surface of the sphere. SECTION 4.-Problems. In the solution of the following problems, it is assumed, 1. That of any two points, which are given upon the surface of a sphere, the direct distance may be obtained; and, 2. That from any given point as a pole, with any given distance less than a semicircumference, a circle may be described upon the surface of a sphere. We may observe, however, that the first assumption is only used in Prob. 1 to obtain the diameter of the sphere; and, therefore, if the diameter is supposed to be given, it may be dismissed as unnecessary. Constructions made within the sphere are excluded. PROP. 29. Prob. 1. To find the diameter of a given sphere ABCD. From any point A, as a pole, with any distance AB, describe the circle B CD; a d B A in the circumference BCD take any three points B, C, D; describe the plane triangle bed with its sides bc, cd, bd equal to the direct distances BC, CD, B'D respectively; find e the centre of the circumscribing circle (III. 59.), and join e b; from the point e draw ea perpendicular to eb, and from the centre b, with a radius equal to the direct distance BA describe a circle cutting e a in a; produce a e to f, and from b draw bf perpendicular to ba (I. 44.) to meet af in f: af shall be equal to the diameter of the sphere. For, if O be the centre of the sphere, and the diameter A O F be drawn cutting the plane B C D in E, and EB be joined, A F will be the axis (def. 3.), and therefore (1.) the point E the centre, of the circle B C D, and the angle A E B will be a right angle. Also, EB is equal to e b, because the triangles B CD, bcd may be made to coincide, and, then, the points E, e being, each of them, the centre of the same circumscribing circle, will likewise coincide. Therefore, because in the right-angled triangles AEB, ae b, the hypotenuse A B is equal to the hypotenuse ab, and the side EB to the side e b, the angle B A E is equal to the angle bae (I. 13.). Join AB, BF, OB. Then, because O B is equal to the half of A F, the angle ABF is a right angle (I. 19. Cor. 4.); and, because, in the right-angled triangles ABF, abf, the side AB and angle BAF are equal respectively to the side ab and angle baf, the hypotenuses A F and af are equal to one another (I. 5.), and af is equal to the diameter of the sphere. Therefore, &c. Cor. Hence, a sphere being given, the quadrant of a great circle may be found; for it is equal to the quadrant of the circle which is described upon the diameter af. PROP. 30. Prob. 2. Any point A being given upon the surface of a sphere, to find the opposite extremity of the diameter which passes through that point. From the pole A, with the distance of a quadrant, (29. Cor.) describe a spherical arc PQ (5.); and, from any two points P, Q of this arc, as poles, with the same distance, describe two A great circles passing through A, and cutting one another again in the point a. Then, because the planes of any two great circles cut one another in a diameter of the sphere, the points A, a are opposite extremities of a diameter. Therefore, &c. PROP. 31. Prob. 3. To join two given points A, B upon the surface of a sphere. From the pole A, with the distance of a quadrant (29. Cor.), describe a great circle, and from the pole B, with the same distance, describe a great circle cuting the former in the point P. From P as a pole, with the same distance, describe the arc A B. Then, because AB is the arc of a great circle (5.) described between the points A, B, it joins those two points on the surface of the sphere (def. 5.). Therefore, &c. Cor. In the same manner, any spherical arc being given, the great circle may be completed of which it is a part. PROP. 32. Prob. 4. To bisect a given spherical arc AB. the distance A B describe From the pole A, with a circle, and from the pole B with the same distance cle cutting the former cir- in E. See Book I. Prop. 43. Therefore, &c. A C B It must here be observed that although the points C, D are determined by the intersection of small circles, the the sides of the spherical triangles in the arcs CA, CB, CD, DA, DB, which are demonstration, are portions of great circles; and the same remark applies to some subsequent problems. Cor. In the same manner, a spherical arc CD may be drawn which shall bisect any spherical arc A B at right angles. PROP. 33. Prob. 5. To draw an arc, which shall be perpendicular to a given spherical arc AB, from a given point C in the same. From CB, or CB produced, cut off CP equal to a quadrant, and from the pole P, D A. B P B A In A B take any point B; make A C equal to A B, and join BC; from the poles B, C, with the common distance B C, describe two spherical arcs cutting one another in D, and join AD (31.). AD is the bisecting arc required. See Book I. Prop. 46. Therefore, &c. PROP. 36. Prob. 8. At a given point A, in a given arc AB, to make a spherical angle equal to a given spherical angle C. From the pole C, with the distance of a quadrant, describe a circle cutting the sides of the given angle in the points G\ F E PROP. 38. Prob. 10. Through two given points A, B, and a third point C on the surface of a sphere, to describe two equal and parallel small circles; the points A, B, C not lying in the circumference of the same great circle. Join AB, and draw D E bisecting it at right angles in the point D (32. Cor.); find A' the opposite extremity of the diameter which passes through A (30.); join A' C (31.), and bisect it at right angles with the arc FP which meets DE in the points P, P (32. Cor.); join PA, PC, and from the pole P with K the triangles PDA, PDB have two sides of the one equal to two sides of the other, each to each, and the included angles PDA, PDB equal to one another (13.). And, for the same reason, if PA and PC are joined, PA will be equal to PC. But PÅ and PA are together equal to a semicircumference, because A A' is a diameter of the sphere. Therefore, PA and PC are together equal to a semicircumference; and, consequently (3. Cor.), the parallel circles AGH, CK L are equal to one another. Also, because PA is equal to PB, the circle A G H passes through the point B. Therefore, &c. PROP. 39. Prob. 11. To describe a triangle which shall be equal to a given spherical polygon, and shall have a side and adjacent angle the same with a given side AB and adjacent angle B of the polygon. First, let the given polygon be a quadrilateral A B C D. Through the two points A, C, and the B E third point D, describe two equal and parallel small circles (38.), and let the arc B C, which cuts one of them in C, be produced to cut the other in E (22. Cor. 2.), and join AE, AC (31.). Then, because the triangles A CD, ACE lie between the same equal and parallel small circles, they are equal to one another; and, therefore, the triangle ABC being added to each, the triangle ABE is equal to the quadrilateral A B C D. F Next, let ABCDEF be the given polygon. Join AC, AD. Make, as in the former case, the triangle ADG equal to the quadrilateral ADEF, the triangle ACH equal to the quadrilateral A A C-D G, and the triangle ABK equal to the quadrilateral ABCH. It is evident that the triangle A B K is equal to the polygon ABCDEF. B Ꮐ E H D K And, in each case, the triangle described has the same side AB and angle B with the given polygon. Therefore, &c. PROP. 40. Prob. 12. Given two spherical arcs AB and Q, which are together less than a semicircumference; to place them so, that, with a third not given, they may contain the greatest surface possible. Bisect AB in H (32.), and produce HA to P, so that HP may be equal to a quadrant; from the pole P, with the distance PB, describe the small circle BCK, and from the pole B, with the distance Q, describe a circle cutting the circle BCK in C. Join AC, BC. The triangle A B C shall be the triangle required. Б For, if, with the distances PA, PH, there are described from the pole P the small circle ALM and the great circle G, the circles A L M and B C K will be HGN, the latter cutting the arc AC in equal to one another, because the distances PA, PB are together equal to twice the quadrant PH (I. ax. 9.), that is, to a semicircumference (3. Cor.); and they are parallel because they have the same poles; and HGN is the great circle to which they are parallel; therefore, AG is equal to GC (22.). But, because PH is at right angles to H G (5.); and, P is the pole of the great circle HGN, because the triangles GHA, GHB have two sides of the one equal to two sides of the other, each to each, and the included angles G HA, G H B equal to (13.). Therefore, because in the isosceles one another, G B is equal to GA or G C triangles GA B, GBC, the angles GBA, GBC are equal to the angles GAB, GCB respectively (11.), the whole angle the sum of the two angles CAB, ACB ; A B C of the triangle A B C is equal to wherefore the triangle ABC is the greatest that can be formed with the two sides A B, B C, or the greatest that can be formed with the given sides A B and Q (26. Cor.). Therefore, &c. PROP. 41. Prob. 13. Through a given point A to describe a great circle, which shall touch two given equal and parallel small circles BCD, EFG. Find the point P which is the com P H K F |