Book XI. DEF. IX. and XI. B. XI. The similitude of plane figures is defined from the equality of their angles, and the proportionality of the sides about the equal angles; for from the proportionality of the sides only, or only from the equality of the angles, the similitude of the figures does not follow, except in the case when the figures are triangles: the similar position of the sides which contain the figures, to one another, depending partly upon each of these: And for the same reason, those are similar solid figures which have all their solid angles equal, each to each, and are contained by the same nuinber of similar plane figures : For there are some solid figures contained by similar plane figures, of the same number, and even of the same magnitude, that are neither similar por equal, as shall be demonstrated after the notes on the 10th definition : Upon this account it was necessary to amend the definition of similar solid figures, and so place the definition of a solid angle before it: And from this and the 10th definition, it is sufficiently plain how much the Elements have been spoiled by unskilful editors. DEF. X. B. XI. SINCE the meaning of the word “ equal” is known and established before it comes to be used in this definition ; therefore the proposition which is the 10th definition of this book, is a theorem, the truth or falsehood of which ought to be demonstrated, not assumed; so that Theon, or some other editor, has ignorantly turned a theorem, which ought to be demonstrated, into this 10th definition. That figures are similar, ought to be proved from the definition of similar figures; that they are equal, ought to be demonstrated from the axiom, “Magnitudes that wholly coin" cide are equal to one another;" or from Prop. A. of Book 5, or the 9th Prop, or the 14th of the same Book, from one of which the equality of all kinds of figures must ultimately be deduced. In the preceding books, Euclid has given no definition of equal figures, and it is certain he did not give this: For what is called the first def. of the third book, is really a theorem'in which thosc circles are said to be equal, that have the straight lines from the centres to the circumferences equal, which is plain from the definition of a circle; and therefore has by some editor Boox XI. been improperly placed among the definitions. The equa lity of figures ought not to be defined, but demonstrated : Therefore, though it were true, that solid figures contained by the same number of similar and equal plane figures are equal to one another, yet he would justly deserve to be blamed who would make a definition of this proposition, which ought to be demonstrated. But if this proposition be not true, must it not be confessed, that geometers have, for these thirteen hundred years, been mistaken in this elementary matter? And this should teach us modesty, and to acknowledge how little, through the weakness of our minds, we are able to prevent mistakes, even in the principles of sciences which are justly reckoned amongst the most certain; for that the proposition is not universally true, can be shown by many examples: The following is sufficient: Let there be any plane rectilineal figure, as the triangle • 12. 11. ABC, and from a point D within it draw a the straight line DE at right angles to the plane ABC; in DE take DE, DF, equal to one another, upon the opposite sides of the plane, and let G be any point in EF; join DA, DB, DC; EA, EB, EC; FA, FB, FC; GA, GB, GC: because the straight line EDF is at right angles to the plane ABC, it makes right angles with DA, DB, DC, which it meets in that plane; and in the triangles EDB, FDB, ED and DB are equal to FD and DB, each to each, and they 4. l. contain right angles; therefore the base EB is equal to the base FB; in the to contents equal to FA, and EC 53,15 01043 bu FA; wherefore the an• 8. 1. gle EBA is equal e to D € * দ গ WHT 54. 6. angles; therefore these 1. Def triangles are similard: In the same manner the triangle EBC is similar to the triangle FBC, and the triangle EAE to FAC; therefore there are two solid figures, each of which is contained by six triangles, one of them by three Boox 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 seif-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, 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 Č, 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 Cand 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. 1 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 B K M H Η greater than the third, and also that any two of the angles Z. |