66 Boox VI. " 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 parallelogram "BE, so is the rectilineal KGH to the parallelogram 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 shown 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 5, 11, 12, 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 notion of proportionals has, it seems, preoccupied 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 dλaws, otherwise, prefixed to it, as if it was 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 Book VI. 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; 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 follows: 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, according to the determination : Draw DE at right angles to AB, and make 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, ID GB C and upon GB describe the square GBKH, and complete a Book VI. 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 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 rectangle AG, GB, together with the square of 1 6. 2. DB, is equal to (the square of DG, or DE, that is to) the squares of EB, BD. From each of these equals take the square of DB; therefore the remaining rectangle AG, GB, is equal to the FA KH BG C 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 Book VI. of it, and make AE equal to C, and BF 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 GK parallel to it, and GL parallel to AE, meeting AB in L. C D a a Because the angle EHF in a semicircle is equal to the right angle EAB, AB and HF are parallels, and AH and BF are parallels; wherefore AH is equal to BF, and the rectangle EA, AH, equal to the rectangle EA, BF, that is, to the rectangle C, D: And because EG, GF are equal to one another, and AE, LG, BF parallels; therefore AL and LB are equal, also EK is equal to KHa and the rectangle * 3. 3. C, D, from the determination, 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 each the square of KE; therefore the square of 6. 2. AK is not greater than the squares of EK, KG, that is, than the square of EG, and consequently the straight line AK or GL is not greater F than GE. Now, if GE be equal to GL, the circle EHF touches AB in L, and therefore the square of AL isc equal to the rectangle EA, AH, that is, to the given rect- H angle C, D, and that which was required is done: But if EG, 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 LM is equal to LN, and it has been proved that AL is equal to 3. 3. LB; therefore AM is equal to NB, and the rectangle AN, NB, equal to the rectangle NA, AM, that is, to the rectanglee EA, AH, or the rectangle C, D: But the rectangle⚫ Cor. 36. 3 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. A K G € 36. 3. F M L B Q PO 4. To apply a rectangle to a given straight line that shall be equal to a given rectangle, exceeding by a square. BOOK VI. 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. F 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 describe the square BNOP, and complete the rectangle ANPQ; because the angle EHF in a semicircle is equal to the right angle EAB, AB and C D— G O P HF are parallels, and there A L B fore AH and BF are equal, M N and the rectangle EA, AH, equal to the rectangle EA, H BF, that is, to the rectangle C, D: And because ML is equal to L N, and AL to LB, therefore MA is equal to BN, 35. 3. and the rectangle AN, NB, to MA, AN, that is a to 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. |