which is contained by six triangles, one of them by three Book XI. triangles the common vertex of which is the point G, and their bases the straight lines AB, BC, CA, and by three other triangles, the common vertex of which is the point E, and their bases the same lines AB, BC, CA: The other solid is contained by the same three triangles the common vertex of which is G, and their bases AB, BC, CA: and by three other triangles of which the common vertex is the point F, and their bases the same straight lines AB, BC, CA: Now the three triangles GAB, GBC, GCA, are common to both solids, and the three others EAB, EBC, ECA, of the first solid, have been shown equal and similar to the three others, FAB, FBC, FCA, of the other solid, each to each: therefore these two solids are contained by the same number of equal and similar planes: But that they are not equal is manifest, because the first of them is contained in the other: Therefore it is not universally true that solids are equal which are contained by the same number of equal and similar planes. 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, 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, ༣ BOOK XI. 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 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. 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 B 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 ex- Book XI. cess 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. THERE may be innumerable solid angles, all unequal to one 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 1, and its corollary, find a 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 two of the angles Z. BOOK XI. 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 right angles: But A and B together are greater than the angle E; wherefore the doubleof A, 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 right angles, 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, 23. 11. D, E, each to eacha: And, by prop. 26, 11, at the point F, in the straight 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, be greater than the third, and also every two of the three C, D, M, be greater than the third: And, as in the preceding part, it may be demonstrated, that N 2 23.11. the three A, B, M, are less than four right angles, as also Book XI. that the three C, D, M, are less than four right angles. Make therefore a solid angle at N contained by the three plane angles ONP, PNQ, ONQ, which are equal to A, B, M, each to each: And by prop. 26, 11, make at the point N, in the straight line ON, a solid angle contained O P by three plane angles, of which one is the angle ONQ equal to M, and the other two are the angles QNR, ONR, which are equal to the angles C, D, each to each. Thus, at the point N, there is a solid angle contained by the four plane angles ONP, PNQ, QNR, ONR, which are equal to the angles A, B, C, D, each to each. And that the two solid angles at the points P, N, each of which is contained by the above-named four plane angles, are not equal to one another, or that they cannot coincide, will be plain by considering that the angles GFK, ONQ, that is, the angles E, M, are unequal by the construction; and therefore the straight lines GF, FK, cannot coincide with ON, NQ, nor consequently can the solid angles, which therefore are unequal. 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.D. 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 equality of two solid angles, whereof each is contained by three plane angles which are equal to one another, each to each, is only assumed, and not demonstrated. |