PROP. XXIV. B. VI. It seems that some unskilful editor has made up this demonstration, as we now have it, out of two others; one of which may be made from the 2d Prop. and the other from the 4th of this Book. For after he has, from the 2d of this Book, and composition and permutation, demonstrated, that the sides about the angle common to the two parallelograms, are proportionals, he might have immediately concluded, that the sides about the other equal angles were proportionals, viz. from Prop. 34, B. 1, and Prop. 7, B. 5. This he does not, but proceeds to show, that the triangles and parallelograms are equiangular : and in a tedious way, by help of Prop. 4, of this Book, and the 22d of Book 5, deduces the same conclusion: From which, it is plain, that this ill-composed demonstration is not Euclid's: these superfluous things are now left out, and a more simple demonstration is given from the 4th Prop. of this Book, the same which is in the translation from the Arabic, by help of the 2d Prop. and composition; but in this the author neglects permutation, and does not shew the parallelograms to be equiangular, as it is proper to do for the sake of beginners. PROP. XXV. B. VI. It is very evident, that the demonstration which Euclid had given of this proposition has been vitiated by some unskilful hand: for, after this editor had demonstrated, that “ as the rectilineal figure ABC is to " the rectilineal figure KGH, so is the parallelogram “ BE to the parallelogram EF”; nothing more should have been added but this, “and the rectilineal figure “ ABC is equal to the parallelogram BE: therefore “the rectilineal KGH is equal to the parallelogram EF,” viz. from Prop. 14, Book 5. But betwixt these two sentences he has inserted this, “wherefore, by permu“ tation, as the rectilineal figure ABC to the parallelo gram BE, so is the rectilineal KGH to the pa“ rallelogram EF”; by which it is plain, he thought it was not so evident to conclude, that the second of four proportions is equal to the fourth from the equality of the first and third, which is a thing demonstrated in the 14th Prop. of B. 5, as to conclude that the third is equal to the fourth, from the equality of the first and second, which is no where demonstrated in the Elements as we now have them : but though this proposition, viz. the third of four proportionals, is equal to the fourth, if the first be equal to the second, had been given in the Elements by Euclid, as very probably it was, yet he would not have made use of it in this place, because, as was said, the conclusion would have been immediately deduced without this superfluous step by permutation: this we have shewn at the greater length, both because it affords a certain proof of the vitiation of the text of Euclid; for the very same blunder is found twice in the Greek text of Prop. 23, Book 11, and twice in Prop. 2, Book 12, and in the 5th, 11th, 12th, and 18th of that Book; in which places of Book 12, except the last of them, it is rightly left out in the Oxford edition of Commandine's translation: and also that geometers may beware of making use of permutation in the like cases; for the moderns not unfrequently commit this mistake, and among others Commandine himself in his commentary on Prop. 5, Book 3, p. 6. b. of Pappus Alexandrinus, and in other places: the vulgar potion of proportionals has, it seems, pre-occupied many so much, that they do not sufficiently understand the true nature of them. Besides, though the rectilineal figure ABC, to which another is to be made similar, may be of any kind whatever: yet, in the demonstration, the Greek text has “ triangle" instead of “rectilineal figure”, which error is corrected in the above-named Oxford edition. PROP. XXVII. B. VI. The second case of this has äraws, otherwise", prefixed to it, as if it were a different demonstration, which probably has been done by some unskilful librarian. Dr. Gregory has rightly left it out: the scheme of this second case ought to be marked with the same letters of the alphabet which are in the scheme of the first, as is now done. PROP. XXVIII. and XXIX. B. VI. These two problems, to the first of which the 27th Prop. is necessary, are the most general and useful of all in the Elements, and are most frequently made use of by the ancient geometers in the solution of other problems; and therefore are very ignorantly left out by Tacquet and Dechales in their editions of the Elements, who pretend that they are scarce of any use: the cases of these problems, wherein it is required to apply a rectangle which shall be equal to a given square, to a given straight line, either deficient or exceeding by a square; as also to apply a rectangle which shall be equal to another given, to a given straight line, deficient or exceeding by a square, are very often made use of by geometers: and, on this account, it is thought proper, for the sake of beginners, to give their constructions as follow : 1. To apply a rectangle, which shall be equal to a given square, to a given straight line, deficient by a square: but the given square must not be greater than that upon the half of the given line. Let AB be the given straight line, and let the square upon the given straight line C, be that to which the rectangle to be applied must be equal, and this square by the determination is not greater than that upon half of the straight line AB. Bisect AB in D, and if the square upon AD be equal to the square upon C, the thing required is done: but if it be not equal to it, AD must be greater than C, I K according to the determination : draw DE at right angles to AB, and make D GB it equal to C; produce ED to F, so that EF be equal to AD or DB, and from the centre E, at the distance EF, describe a circle meeting AB in G, and upon GB describe the square GBKH, and complete the rectangle AGHL; also join EG: and because AB is bisected in D, the rectangle AG, GB, together with the square of DG, is equal to (the square of DB, that is, of EF or EG, that is, to) the squares of ED, DG: take away the square of DG from each of these equals; therefore the remaining rectangle AG, GB, is equal to the square of ED, that is, of C: but the rectangle AG, GB, is the rectangle AH, because GH is equal to GB; therefore the rectangle AH is equal to the given square upon the * 5. 2. 7 • 6.2. straight line C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given straight line AB, deficient by the square GK. Which was to be done. 2. To apply a rectangle which shall be equal to a given square, to a given straight line, exceeding by a square. Let AB be the given straight line, and let the square upon the given straight line C, be that to which the rectangle to be applied must be equal. Bisect AB in D, and draw BE at right angles to it, so that BE be equal to C; and having joined DE, from the centre D at the distance DE, describe a circle meeting AB produced in G; upon BG describe the square BGHK, and complete the rectangle AGHL. And because AB is bisected in D, and produced to G, the F. rectangle AG, GB, together with the square of DB, is equal * to the square of DG, or DE, that is, to) the squares of T A B G EB, BD. From each of these equals take the square of DB; therefore the remaining rectangle AG, GB, is equal to the square of BE, that is, to the square upon C. But the rectangle AG, GB, is the rectangle AH, because GH is equal to GB. Therefore the rectangle AH is equal to the square upon C. Wherefore the rectangle AH, equal to the given square upon C, has been applied to the given straight line AB, exceeding by the square GK. Which was to be done. 3. To apply a rectangle to a given straight line which shall be equal to a given rectangle, and be deficient by a square. But the given rectangle must not be greater than the square upon the half of the given straight line. Let AB be the given straight line, and let the given rectangle be that which is contained by the straight lines C, D, which is not greater than the square upon the half of AB; it is required to apply to AB a rectangle equal to the rectangle C, D, deficient by a square. Draw AE, BF, at right angles to AB, upon the same side of it, and make AE equal to C, and BF to D; join EF, and bisect it in G; and from the centre Because the angle EHF in a semicircle is equal to parallels; therefore AL and LB are equal, also EK is * 3. 3. equal to KH* and the rectangle C, D, from the de termination, is not greater than the square of AL, the half of AB; wherefore the rectangle EA, AH, is not greater than the square of AL, that is, of KG: add to * 6.2. each the square of KE; therefore the square * of AK C K G M L N B D, and that which was required is done: but if EG, Q Ρ Ο GL, be unequal, EG must be the greater: and therefore the circle EHF cuts the straight line AB: let it cut it in the points M, N, and upon NB describe the square NBOP, and complete the rectangle ANPQ: because ML is equal to * LN, and it has been proved that AL is equal to LB ; there fore AM is equal to NB, and the rectangle AN, NB, *Cor. 36.3. equal to the rectangle NA, AM, that is, to the rectangle* EA, AH, or the rectangle C, D: but the rectangle AN NB, is the rectangle AP, because PN is equal to NB: therefore the rectangle AP is equal to the rectangle C, D; and the rectangle AP, equal to the given rectangle C, D, has been applied to the given straight line AB, deficient by the square BP. Which was to be done. 4. To apply a rectangle to a given straight line, that * 36. 3. * 3. 3. |