sphere is made up of the surfaces of all the polygons, and that the surface of the sphere is measured by or 2 x 4 right angles, S+F-E is equal to 2. And from the equation S+F-E=2 it may be further shown that the sum of all the plane angles which contain the solid angles of any polyhedron is S-2 times 4 right angles. For, the plane angles of each face are together equal to twice as many right angles as the face has sides minus four right angles. Therefore, the plane angles of all the faces are together equal to twice as many right angles as all the faces taken together have sides, minus as many times four right angles as there are faces. But the number of sides of the faces is 2 E, for the reason before mentioned; and F is the number of faces. Therefore, the sum in question is equal to E-F times 4 right angles; that is, because S-2 = E-F, to S-2 times 4 right angles. Let A and B be any two points upon the surface of a sphere, and C any third point, which does not lie in the circumference of the great circle passing through A and B. Through the two points A, B, and the point C, there may be made to pass two equal and parallel small circles of the sphere. Let O be the centre of the sphere, and let c be the opposite extremity of the diameter CO c. Then it is evident (IV. 1.) that through the three points A, B, c there may be made to pass the small circle A B c, and through the point C the circle CDE parallel to A B c (IV. 11. Cor. 1.). Let P be the common pole of the two circles, and join P C, Pc. Then, because Cc is a diameter of the sphere, PC and Pc are arcs of the same great circle, and are together equal to a semicircumference. And, because the polar distances P C and Pc of the circles ABc and C D E are together equal to a semicircumference, the circles A B c and CD E, which have been made to pass through the two points A, B, and the point C, are equal to one another (3. Cor.). Next, let DB be any spherical arc which is terminated by these circles; and let it be cut by the great circle FGH, to which they are parallel, in the point G: DG shall be equal to GB. Join DP; let the great circle D G B be completed, and let its plane cut the plane of the great circle F G H in the diameter GOg, and the plane of the great circle DP in the diameter D O d. Then, because PD and Pd are together equal to a semicircumference, and that the point D is in the circle CD E, the point d is in the circle A B c, which is equal to CDE, and has the same pole with it (3. Cor.). Join OG, O B, Bd: and, because G O, Bd are sections of parallel planes F G H, A Bc by the plane DG Bd, GO is parallel to B d (IV. 12.); therefore the angle DOG is equal to the angle DdB, and the angle BOG to the angle OBd (I. 15.). But the angles OBd and OdB or DdB are equal to one another, because Od is equal to OB (I. 6.): therefore, also, the angle DOG is equal to the angle BOG (I. ax. 1.), and the arc D G to the arc G B (III. 12.). Therefore, &c. Cor. 1. It is shown in the latter part of the demonstration, that if there be two equal and parallel small circles, and if a great circle meets one of them in any point, it will meet the other in the opposite extremity of the diameter which passes through that point. Cor. 2. Hence if a great circle cuts one of two equal and parallel small circles, it will cut the other likewise; also if it touches one of them, it will touch gle ABC shall be equal to the triangle the other likewise. EBC. Let the circumference of the great PROP. 23. one Lunular portions of surface, which are contained by equal spherical arcs with the arcs of equal small circles of the same sphere, are equal to another; so also are the pyramidal solids, which have these portions for their bases, and their common vertex in the centre of the sphere. Let A B C, D E F be equal small circles, and let the spherical arcs AB, DE contain with the small arcs AC B, DFE the lunular surfaces ACB, DFE; then, if the arcs AB and DE are equal to one another, the surfaces ACB, DFE shall be likewise equal. Let P be the nearer pole of the circle ABC, and Q the nearer pole of the circle D E F, and join PA, PB, Q D, Q E. Then, because the circles are equal to one another, the polar distances PA, QD are likewise equal (3.): and, because the triangles PAB, QDE have the two sides PA, PB equal to the two sides QD, Q E, each to each, and the base AB equal to the base DE, the angle APB is equal to the angle DQE (15.). Therefore, if the pole P be applied to the pole Q, and the arc PA to the arc Q D, the arc PB will coincide with the arc QE, and the points A and B will coincide with the points D and E respectively, and the arc AB with the arc DE, and the small arc A C B with the small arc DFE; and therefore the lunular surface A C B coincides with the lunular surface D F E, and is equal to it. And because these surfaces may be made to coincide, it is evident that the pyramidal solids, which have these surfaces for their bases, may likewise be made to coincide, and are equal to one another. Therefore, &c. PROP. 24. Spherical triangles, which stand upon the same base and between the same equal and parallel small circles, are equal to one another. Let the spherical triangles ABC, EBC stand upon the same base B C, and between the same equal and parallel small circles AEF, BCG: the trian circle to which both A EF and B CG are parallel cut the arc A C in H; join BH, and produce it to meet the circumference AEFin D (22. Cor. 2.), and join AD. Then, because every spherical arc which is terminated by the circles A EF, BC G is bisected by the circumference of the great circle to which they are parallel (22.), AH is equal to H C, and D H to H B; and because the triangles HAD, HCB have the two sides HA, HD equal to the two sides H C, HB, each to each, and the included angles equal to one another (8. Cor. 1.), the base AD is equal to the base C B (13.): and, for the like reason, AB is equal to CD: therefore, the triangles A B C, CDA have the three sides of the one equal to the three sides of the other, each to each, and are equal to one another (20.); so that the quadrilateral A B C D is double of the triangle AB C. In the same manner, the quadrilateral E B CF may be described, which is double of the triangle E B C, and has its sides EF, FC equal to the sides BC, EB, each to each. And, because the spherical arcs AD and EF are each of them equal to B C, they also, the chord AD is equal to the chord are equal to one another therefore, EF (III. 12. Cor. 1.) and the small arc AD to the small arc E F; and, if the small arc ED is added to, or taken from them, the small arc AE is equal to the small arc D F; and hence, again, because the chords of these equal small arcs are equal, the spherical arc AE is equal to the spherical arc D F (III. 12. Cor. 1.). And, because the triangles ABE, DCF have the three sides of the one equal to the three sides of the other, each to each, they are equal to one another (20.): but the lunular portions which are contained by the small arcs AE, DF with the equal spherical arcs AE, DF are likewise equal (23.): therefore, the remaining portions of the triangles A B E, D C F are equal to one another. Let each of these equals be taken from the whole surface included by the small arc A EF and the spherical arcs AB, BC, CF, and there remains the surface included by the small arc EF and the spherical arcs E B, BC, CF, equal to the surface included by the small are AD and the spherical arcs AB, BC, CD. To each of these equals add the equal lunular portions (23.) which are contained by the small arcs EF, A D, with the equal spherical arcs E F, AD; and the whole quadrilateral EBCF is equal to the whole quadrilateral ABCD. Therefore, because the halves of equals are equal, the triangle EBC is equal to the triangle A B C. Therefore, &c. Cor. Hence, if equal triangles A B C, EBC stand upon the same base B C, and the same side of it, the points A, E and B, C lie in the circumferences of two equal and parallel small circles. For, if the arc BE be produced to meet the circumference ADF in some point as E' (22. Cor. 2.), and if E' C be joined, the triangle E' B C will be equal to the triangle A B C, by the proposition, and therefore likewise equal to the triangle EB C, which is impossible, unless_the_point E' coincides with the point E. In fact, if the point E were supposed to fall between the two circles, it is evident that a triangle E' B C might be found having its vertex E' in the circumference A D F, and including the vertex E of the triangle E BC; or if E were supposed to fall within the circle AD F, a triangle E B C might be found having its vertex E' in the same circumference, and included within the triangle E BC; and in neither case could the triangle E' BC be equal to the triangle EBC, that is, to the triangle A B C, whereas, by the proposition, such triangle E' B C must be equal to the triangle A B C. Scholium. It may at first appear that the preceding proposition does not apply to all triangles which stand upon the base B C, and have their vertices in the circumference AEF; since the arc B A (for instance) may be produced to cut the circumference A E F a second time in a point A' which may be joined with the point C, and thus, apparently, a second triangle formed, which is greater than ABC: but, if the arc BA produced cuts the circumference A E F in a second point A', the arc A' A B will be a semicircumference (22. Cor. 1.), and therefore the triangle which is apparently contained by A' B, BC, and A/C, is not a triangle (def. 9.) but a lune. isosceles, having the side A B equal to the side AC: the triangle ABC shall have a less perimeter than the triangle DBC. Let P be the nearer pole of the circle ADE; and therefore, also, the more distant pole of the circle B C F. Join PA, and through the point A draw the spherical arc GH at right angles to PA: then, because the arc PA, being less than a quadrant, is less than any other arc which can be drawn from P to G H (18. Cor. 1.) and that PD is equal to PA, the arc G H will cut the arc BD in some point H which is between D and B. From B draw the arc B G perpendicular to GH, and produce it to meet CA produced in the point K; and join HC, H K, PB, PC. Then, because the spherical triangles PA B, PAC have the three sides of the one equal to the three sides of the other, each to each, the angle PA B is equal to the angle PAC (15); but the angle PAG is equal to the angle PAH, because they are right angles; therefore, the remaining angle BAG is equal to the remaining angle CAH or (8.Cor. 1.) KAG. And, because the triangles BGA, KGA have two angles of the one equal to two angles of the other, each to each, and the interjacent side GA common to both, G K is equal to G B, and AK to AB (14.). Therefore, because the triangles H G B, HGK have two sides of the one equal to two sides of the other, each to each, and the included angles HGB, HGK equal to one another, H K is equal to HB (13.). But in the triangle HKC, the side KC is less than H K and HC together (9.) therefore, A B and A C are together less than H B and H C. And HB, HC together are less than DB, DC together, because HC is less (9.)than DC and D H together. Much more, therefore, are AB and AC together less than DB and DC together. Therefore, if BC is added to each, the perimeter of the triangle A B C is less than the perimeter of the triangle D B C. Therefore, &c. Cor. Hence, of all triangles which are upon the same base, and have equal perimeters, the isosceles has the greatest area. For triangles, which have the same or a greater area than the isosceles, have greater perimeters. PROP. 26. If two spherical triangles have two sides of the one equal to two sides of the other, each to euch, and the angle which is contained by the two sides of the first equal to the sum of the other two angles of that triangle, but the angle which is contained by the two sides of the other not so; the first triangle shall be greater than the other. Let the spherical triangles A B C, DEF have the two sides AB, BC equal to the two sides DE, E F, each to each, and let the angle ABC of the first tri angle be equal to the sum of the other two angles B C A, B A C, but the angle DEF of the other not equal to the sum of the other two angles E FD, EDF: the triangle ABC shall be greater than the triangle DE F. Draw Because the angle ABC is equal to the sum of the two angles B C A, BAC, let it be divided into the parts GBC, GBA, equal to them respectively, by the arc B G which meets AC in G. the arc G H, bisecting the angle A G B, and let it meet AB in H. Then, because the angles GB C, GBA are equal to the angles G C B, G A B respectively, GC and GA are equal, each of them, to GB (11.), and therefore, also, equal to one another; and, because in the triangles GHA, GHB the two sides GA, GH are equal to the two sides GB, GH, each to each, and the included angles equal to one another, AH is equal to HB, and the angle A H G to the angle BHG (13.). Now, through the two points B, C and the third point A let there be made to pass two equal and parallel small circles BCK and ALM (22.). And, because in arcs A B and A C, terminated by these circles, the points of bisection H and G lie in the circumference of the great circle to which they are parallel (22.), the arc G H is an arc of that great circle. Make the angle CB L equal to the angle FE D, and let the arc BL cut the arc GH in N, and the circumference ALM in L (22. Cor. 2.); and join L C, NA. Then, because the triangles N HA, NHB have two sides of the one equal to two sides of the other, each to each, and the included angles NHA, NH B equal to one another, the arc N A is equal to NB (13.); but NB is equal to NL, because the arc GH produced bisects every arc A L which is terminated by the circles ALM, BCK; therefore N A is equal to N L: but N A and N B are together greater than AB (9.); therefore, LB is likewise greater than A B. Therefore, if BQ be cut off from the arc BL produced, equal to A B or DE, the point Q will be between L and B. Join QC: and, because the triangles Q B C, DEF have the two sides QB, B C equal to the two sides DE, E F, each to each, and the included angles equal to one another, the third sides QC, DF are equal to one another (13.); and, consequently (20.), the triangle QBC is equal to the triangle DEF. But LB C is greater than QBC; and A B C is equal to LB C, because they are upon the same base BC, and between the same equal and parallel small circles (24.). Therefore, the triangle ABC is greater than the triangle DEF. Therefore, &c. Cor. Two given finite spherical arcs AB, BC, together with a third indefinite, inclose the greatest surface possible, when placed so that the included angle ABC may be equal to the sum of the other two angles of the triangle ABC. of III. 39. Cor. with regard to the This conclusion is analogous to that greatest possible area which can be enclosed by two given finite straight lines with a third indefinite; for, although the angle ABC included by the spherical the sum of the other two angles of the arcs is not a right angle, it is equal to triangle A B C, which is the case also with the right angle of a right-angled plane triangle (I. 19. Cor. 3.). Scholium. In a triangle ABC which has one of its angles ABC equal to the sum of the other two, the containing sides AB, BC are together less than a semicircumference. For G, the middle point of AC, is the pole of a small circle passing through A, B, and C, because G A, G B and GC are equal to one another: and, because a spherical arc (9. Schol.) is the shortest distance between two points along the surface of the sphere, the great arcs AB and BC are together less than the small arcs A B and B C, that is, than the semi-circumference of a small circle: much more, therefore, are AB and BC together less than the semicircumference of a great circle. In fact, when the given sides A B and B C are together equal to a semicircumference, BC coincides with BR, the polar diameter of the circle BCK, and because AC is always bisected by the circumference of the great circle HGN, AG and GC are in this case quadrants, and consequently (19. Cor. 1.) A GC is at right angles to A B; so that ABC is a lune, not a triangle, and equal to a fourth part of the surface of the sphere. And when the given sides A B and B C are together greater than a semicircumference, it is evident, without reference to the figure, that a triangle may be found which shall have A B, BC for two of its sides, and differ by as little as we please from half the surface of the sphere. For, in this case, if A B, BC are placed in the circumference of the same great circle, that is, at an angle equal to two right angles, the arc AC which completes the circle will be less than a semicircumference; and if A B, BC are placed at any angle less than two right angles, the arc AC will be still less (18.); therefore, if A B, BC are placed at any angle less than two right angles, A C being joined will complete a spherical triangle (def. 9.), and this triangle will differ less and less from the hemispherical surface, as the angle A B C approaches to two right angles. In a triangle ABC of this kind, the angle A B C, which is equal to the sum of the other two, is always greater than a right angle, because the three together are greater than two right angles (10). It is worthy of notice, also, that the triangle B'A C, which is the difference between the lune BAB' C, and the triangle ABC has its surface always equal to a fourth part of the surface of the sphere; for it is measured by the difference of 2 B and 2 B-2 R, that is, by 2 R, and the sphere's surface by 8 R (21. Cor. 1.). It may be observed that a similar course of reasoning to that which is exhibited in Book III. Prop. 40, 41, 42, and 43, founded upon this proposition, will lead to the same conclusions with regard to the surfaces included by spherical polygons and circles of the sphere, which are there stated with regard to the areas of rectilineal figures, and circles on a plane surface: viz. 1. Of all spherical polygons, contained by the same given sides, that one contains the greatest portion of spherical surface, which has all its angles in the circumference of a circle. 2. A circle includes a greater portion of the spherical surface than any spherical polygon of the same perimeter. 3. The lunular surface, which is included by a spherical arc and a small arc, is greater than any other surface which is included by the same perimeter, of which the same spherical arc is a part. A We may also infer from Prop. 25., that, of all spherical polygons having the same number of sides and the same perimeter, the greatest is that which has all its sides equal and all its angles equal. For if a spherical polygon ABCDE have not all its sides equal, and A B, A E be two adjoining sides which are unequal, a greater A B C D E may be found with the same perimeter by describing upon the base B E the isosceles triangle A'BE, which has the same perimeter with AB E. And, as above, B if a spherical polygon have not all its angles lying in the circumference of a circle, a greater may be found with the same sides. Therefore, none is greatest, but that which has all its sides equal, and all its angles lying in the circumference of a circle, that is, which has all its sides equal and all its angles equal; and, since there is evidently some greatest, the greatest is that which has all its sides equal and all its angles equal. PROP. 27. C D E Spherical pyramids, which stand upon equal bases, are equal to one another; so, likewise, are their solid angles. First, let the bases of the pyramids be equal triangles, which have one side |