Book VI PROP. XXXI. B. VI. In the demonstration of this, the inversion of proportionals is twice neglected, and is now added, that the conclusion may be legitimately made by help of the 24th prop. of B. 5, as Clavins had done. PROP. XXXII. B. VI. The enunciation of the preceding 26th prop. is not general enough; because not only two similar parallelograms that have an angle common to both, are about the same diameter? but · likewise two similar parallelograms that have vertically opposite angles, have their diameters in the same straight line: but their seems to have been another, and that a direct demonstration of these cases, to which this 32d proposition was needful: and the 32d may be otherwise and something more briefly demonstrated as follows: PROP. XXXIÍ. B. VI. If two triangles which have two sides of the one, &c. E A B G K D a 31 1. H b 30. 1. Draw CK parallel to FH, and let it meet GF produced in K-: because AG, KC are each of them parallel to FH, they are parallel to one another, and therefore the alternate angles AGF, FKC are, equal and AG is to GF, as (FH to HC, that is) CK to KF ; c 34, 1. wherefore the triangles AGF CKF are equiangular, and the angle AFG equal to the angle CFK: but GFK is a straight line, therefore AF and FC are in a straight line. The 26th prop. is demonstrated from the 32d, as follows: If two similar and similarly placed parallelograms have an angle common to both, or vertically opposite angles ; their diameters are in the same straight line. First, Let the parallelogram ABCD, AEFG have the angle BAD common to both, and be similar and similarly placed; ABCD, AEFG are about the same diameter. d 6. 6.* e 14. 1. Book VI Produce EF, GF, to H, K, and join FA, FC: then because the parallelograms ABCD, AEFG are similar, DA is to AB, as GA to AE: wherefore the re a Cor. 19. mainder DG is to the remainder 5. E EB, as GA to AE: but DG is equal b 32. 6. straight line". A G K Ꭰ . H Next, Let the parallelograms KFHC, GFEA, which are similar and similarly placed, have their angles KFH, GFE vertically opposite; their diameters AF, FC are in the same straight line. Because AG, GF are parallel to FH, HC; and that AG is to GF, as FH to HC; therefore AF, FC are in the same straight line." PROP. XXXIII. B. VI The words "because they are at the centre," are left out, as the addition of some unskilful hand. In the Greek, as also in the Latin translation, the words a Tuxe any whatever," are left out in the demonstration of both parts of the proposition, and are now added as quite necessary; and in the demonstration of the second part, where the triangle BGC is proved to be equal to CGK, the illative particle aga in the Greek text ought to be omitted. The second part of the proposition is an addition of Theon's, as he tells us in his commentary on Ptolemy's Mean Zuvraćes, p. 50. PROP. B. C. D. B. VI. These three propositions are added, because they are fre quently made use of by geometers. DEF. IX. and XI. B. XI. The similitude of plane figures is defined from the equality 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 coincide, 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 Book XI. Book XI. called the 1st def. of the 3d book is really a theorem in which these circles are said to be equal, that have the straight lines from their centres to the circumferences equal, which is plain, from the definition of a circle; and therefore has by some editor been improperly placed among the definitions. The equality 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 e 12. 11, ABC, and from a point D within it draw 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, FB, 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 contain right angles; therefore b 4. 1. the base B is equal to the base FB; in the same manner EA is equal to FA, and EC to FC: and in the triangles, EBA, FBA, EB, BA are equal to FB, BA, and the base EA is equal to the base FA, wherefore the angle c 8. 1: EBA is equal to the angle FBA, and the triangle EBA equalb to the triangle FBA, and the other angles equal to the other angles; therefore these triangles are" B E G F 46. similar"; in the same manner the triangle EBC is similar to d1. def. 6. the triangle FBC, and the triangle EAC to FAC; therefore Book XI. there are two solid figures, each of which is contained by six triangles, one of them by three 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 manifést, 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, B, C |