L shall be equal to a given rectangle, exceeding by a square. Let AB be the given straight line, and the rectangle C, D, the given rectangle; it is required to apply a rectangle to AB equal to C, D, exceeding by a square. Draw AE, BF, at right angles to AB, on the contrary sides of it, and make AE equal to C, and BF equal to D: join EF and bisect it in G; and from the centre G, at the distance GE, describe a circle meeting AE again in H: join HF, and draw GL parallel to AE; let the circle meet AB produced in M, N, and upon BN de E scribe the square NBOP, C and complete the rectangle D- ор EAB, AB, and HF are A B parallels and thereforeAH M N and BF are equal, and the rectangle EA, AH, equal H to the rectangle EA, BF, that is, to the rectangle C, D: and because ML is equal to LN, and AL to LB, therefore MA is equal to BN, and the rectangle AN, NB to MA, AN, that is * to . 35. 3. the rectangle EA, AH, or the rectangle C, D: therefore the rectangle AN, NB, that is, AP, is equal to the rectangle C, D; and to the given straight line AB, the rectangle AP has been applied equal to the given rectangle C, D, exceeding by the square BP. Which was to be done. Willebrordus Snellius was the first, as far as I know, who gave these constructions of the 3d and 4th Problems, in his Apollonius Batavus: and afterwards the learned Dr. Halley gave them in the Scholium of the 18th Prop. of the 8th Book of Apollonius's Conics restored by him. The 3d Problem is otherwise enunciated thus : to cut a given straight line AB in the point N, so as to make the rectangle AN, NB, equal to a given space: or, which is the same thing, having given AB the sum of the sides of a rectangle, and the magnitude of it being likewise given, to find its sides. And the 4th Problem is the same with this : to find a point N in the given straight line AB produced, so as to make the rectangle AN, NB, equal to a given space: or, which is the same thing, having given AB the difference of the sides of a rectangle, and the magnitude of it, to find the sides. 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 Book 5. as Clavius 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 there 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. XXXII. B. VI. 31. 1. If two triangles which have two sides of the one, &c. Let GAF, HFC, be two triangles which have two G E! H and let it meet GF produced in K: because AG, KC, are each of them parallel to FH, they are parallel * to one an B K other, and therefore the alternate angles AGF, FKC, are equal: and AG is to GF, as (FH to HC, that is *) CK to KF; 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 22d, as follows: * 30. 1. • 34. 1. * 6.6. * 14. 1. *Cor.19.5. * 32. 6. 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 parallelograms ABCD, AEFG, have the angle BAD common to both, and be similar, and similarly placed : ABCD, AEFG are about the same diameter. Produce EF, GF, to H, K, and join FA, FC; then because the parallelograms ABCD, A EFG are similar, DA is to AB, as GA to AE: wherefore the remainder DG A G D to the remainder EB, as GA to AE: but DG is equal E F H to FH, EB to HC, and AE to GF: therefore as FH to HC, so is AG to GF; and FH, HC, are parallel to AG, B C Next, let the parallelograms KFHC, GFEA, which 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, à štuxe, “any whatever”, are left out in the demonstrations 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 äga, 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 Μεγάλη Σύνταξις, p. 50. PROP. B. C. D. B. VI. These three propositions are added, because they are frequently made use of by geometers. * 32. 6. 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 where 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 number 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 nor 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 to 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 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 he called the first Def. of the third Book, is really a theorem in which those 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 bas 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 shewn by many examples; the following is sufficient : ab Let there be any plane rectilineal figure, as the triangle ABC, and from a point D within it draw* the * 12. 11. 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 contain right angles; therefore the base EB is equal * to the base FB; * 4. 1. in the same manner bila ad op EA is equal to FA, Curie Garland herine and EC to FC: Baieme Rislu Tol and in the triangles sta/Arisinopiurett EBA, FBA, EB, be on on ou BA, are equal tod / E road et FB, BA, and the bytku ale base EA is equal to ask A the base FA; where lo fore the angle EBA abil // D is equal to the bildes* 8.1. angle FBA, and the B triangleEBA equal* Vida OU to the triangle FBA, To benies and the other angles jednoj How tsbro equal to the other F Go egoiste angles; therefore • 4. 1. |