c 8. 1. Book XI. EA is equal to the base FA; wherefore the angle EBA is G b 4. 1. EBA equal to the triangle FBA, and the other angles equal to the other angles; there E fore these triangles are 4. 6. similar d : In the same A D B F Cor. From this it appears, that two unequal solid angles may be contained by the same number of equal plane angles. For the solid angle at B, which is contained by the four plane angles EBA, EBC, GBA, GBC is not equal to the solid angle at the same point B which is contained by the four plane angles FBA, FBC, GBA, GBC; for this last contains the other; and each of them is contained by four plane angles which are equal to one another, each to each, or are the self same, as has been proved: And indeed there may be innumerable solid angles all unequal to one another, which are each of them contained by plane angles that are equal to one another, Book XI. each to each: It is likewise manifest, that the before-mentioned solids are not similar, since their solid angles are not all equal. And that there may be innumerable solid angles all unequal to one another, which are each of them contained by the same plane angles disposed in the same order, will be plain from the three following propositions. PROP. I. PROBLEM. THREE magnitudes A, B, C being given, to find a fourth such, that every three shall be greater than the remaining one. Let D be the fourth: therefore D must be less than A, B, C together; of the three A, B, C, let A be that which is not less than either of the two B and C: And, first, let B and C together be not less than A; therefore B, C, D together are greater than A; and because A is not less than B; A, C, D together are greater than B: In the like manner, A, B, D together are greater than C: Wherefore in the case in which B and C together are not less than A, any magnitude D which is less than A, B, C together, will answer the problem. But if B and C together be less than A; then, because it is required that B, C, D together may be greater than A; from each of these taking away B, C, the remaining one D must be greater than the excess of A above B and C: Take therefore any magnitude D, which is less than A, B, C together, but greater than the excess of A above B and C: Then B, C, D together are greater than A; and because A is greater than either B or C, much more will A and D, together with either of the two B, C be greater than the other; and, by the construction, A, B, C are together greater than D. Cor. If besides it be required, that A and B together shall not be less than C and D together, the excess of A and B together above C must not be less than D, that is, D must not be greater than that excess. PROP. II. PROBLEM. FOUR magnitudes A, B, C, D being given, of which A and B together are not less than C and D together, and such that any three of them whatever are greater than the fourth ; it is required to find a fifth magnitude E such, that any two of the three A, B, E shall be greater than the third, and also that any two of the three C, D, E shall be greater than the third. Let A be not less than B, and C not less than D. Book XI. First, Let the excess of C above D be not less than the excess of A above B: It is plain, that a magnitude E can be taken which is less than the sum of C and D, but greater than the excess of C above D; let it be taken; then E is greater likewise than the excess of A above B; wherefore E and B together are greater than A; and A is not less than B; therefore A and E together are greater than B: And by the hypothesis, A and B together are not less than C and D together, and C and D together are greater than E; therefore likewise A and B are greater than E. But let the excess of A above B be greater than the excess of C above D: And because, by the hypothesis, the three B, C, D are together greater than the fourth A; C and D together are greater than the excess of A above B : Therefore a magnitude may be taken which is less than C and D together, but greater than the excess of A above B. Let this magnitude be E; and because E is greater than the excess of A above B, B together with E is greater than A : And as in the preceding case, it may be shown that A together with E is greater than B, and that A together with B is greater than E; therefore, in each of the cases, it has been shown, that any two of the three A, B, E are greater than the third. And because in each of the cases E is greater than the excess of C above D, E together with D is greater than C; and, by the hypothesis, C is not less than D: therefore E together with C is greater than D; and, by the construction, C and D together are greater than E: Therefore any two of the three C, D, E are greater than the third. PROP. III. THEOREM. 1 THERE may be innumerable solid angles all unequal to on another, each of which is contained by the same four plane angles, placed in the same order. Take three plane angles, A, B, C, of which A is not less than either of the other two, and such, that A and B together are less than two right angles: and by Problem I, and its corollary, find the fourth angle D such, that any three whatever of the angles A, B, C, D be greater than the remaining angle, and such, that A and B together be not less than C and D together: And, by Problem 2, find a fifth angle E such, that any two of the angles A, B, E be greater than the third, and also that any of the two angles C, D, E be greater than the third : And because A and B together are less than two right angles, the double of A and B together is less than four Book XI. right angles : But A and B together are greater than the angle E ; wherefore the double of A and B together is greater than the three angles A, B, E together, which three are consequently less than four right angles; and every two of the same angles, A, B, E, are greater than the third ; therefore, by Prop. 23, 11, a solid angle may be made contained by three plane angles, equal to the angles A, B, E, each to each. Let this be the angle F, contained by the three plane angles GFH, HFK, GFK, which are equal to the angles A, B, E, each to each : And because the angles C, D together are not greater than the angles A, B together, therefore the angles C, D, E are not greater than the angles A, B, E: But these last three are less than four rightangles, as has been demonstrated; wherefore also the angles, C, D, E are together less than four right angles, and every two of them are greater than the third ; therefore a solid angle may be made, which shall be contained by three plane angles equal to the angles C, D, E, each to each a : And by Prop. 26, 11, at the point F, in the straight a 23. 11. line FG, a solid angle may be made equal to that which is contained by the three plane angles that are equal to the angles C, D, E: Let this be made, and let the angle GFK, which is equal to E, be one of the three ; and let KFL, GFL be the other two which are equal to the angles C, D, each to each. Thus there is a solid angle constituted at the point F, contained by the four plane angles GFH, HFK, KFL, GFL, which are equal to the angles A, B, C, D, each to each. Again, Find another angle M such, that every two of the three angles A, B, M may be greater than the third, and also every two of the three C, D, M greater than the third : And, as in the preceding part, it may be demonstrated, that the three A, B, M are less than four right angles, as also that the three C, D, M, are less than four right angles. Make Book XI. therefore a a solid angle at N, contained by the three plane w angles ONP, PNQ, ONQ, a 23. 11. which are equal to A, B, M, each to each: And by Prop. N R P And, because from the four plane angles A, B, C, D, there can be found innumerable other angles that will serve the same purpose with the angles E and M; it is plain that innumerable other solid angles may be constituted which are each contained by the same four plane angles, and all of them unequal to one another. Q. E. Ď. And from this it appears, that Clavius and other authors are mistaken, who assert that those solid angles are equal which are contained by the same number of plane angles that are equal to one another, each to each. Also it is plain, that the 26th Prop. of Book 11. is by no means sufficiently demonstrated, because the quality of two solid angles, whereof each is contained by three plane angles which are equal to one an. other, each to each, is only assumed and not demonstrated. PROP. I. B. XI. The words at the end of this, “ for a straight line cannot 5 meet a straight line in more than one point,” are left out, as an addition by some unskilful hand, for this is to be demonstrated, not assumed. Mr Thomas Simpson, in his notes at the end of the second edition of his Elements of Geometry, p. 262, after repeating |