Book VI. "figure ABC to the parallelogram BE, fo is the rectilineal KGH " to the parallelogram EF;" by which, it is plain, he thought it was not fo evident to conclude that the fecond of four proportionals is equal to the fourth from the equality of the first and third, which is a thing demonftrated in the 14. Prop. of B. 5. as to conclude that the third is equal to the fourth, from the equality of the first and fecond, which is no where demonftrated in the Elements as we now have them. but tho' this Propofition, 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 ufe of it in this place, because, as was faid, the conclufion could have been immediately deduced without this fuperfluous ftep by Permutation. this we have fhewn at the greater length, both because it affords a certain proof of the vitiation of the Text of Euclid, for the very fame blunder is found twice in the Greek Text of Prop. 2 3. B. 1 1. and twice in Prop. 2. B. 12. and in the 5. 11. 12. and 18. of that Book; in which places of B. 12. except the laft of them, it is rightly left out in the Oxford Edition of Commandine's Tranflation. and alfo that Geometers may beware of making ufe of Permutation in the like cafes, for the Moderns not unfrequently commit this miftake, and among others Commandine himfelf in his Commentary on Prop. 5. B. 3. p. 6. b. of Pappus Alexandrinus, and in other places. the vulgar notion of proportionals, has, it feems, pre-occupied many fo much, that they do not fufficiently understand the true nature of them. Befides, tho' the rectilineal figure ABC, to which another is to be made fimilar, may be of any kind whatever, yet in the De monstration the Greek Text has " triangle” instead of " rectilineal "figure," which error is corrected in the above-named Oxford Edition. PROP. XXVII. B. VI. The fecond Cafe of this has dxxc, otherwife, prefixed to it, as if it was a different Demonstration, which probably has been done by fome unfkilful Librarian. Dr. Gregory has rightly left it out. the fcheme of this fecond Cafe ought to be marked with the fame 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 27. Prop. is neceffary, are the most general and ufeful of all in the Elements, and are most frequently made ufe of by the antient Geometers in the folution 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 ufe. the Cafes of thefe Problems, wherein it is required to apply a rectangle which shall be equal to a given fquare, to a given ftraight line, either deficient or exceeding by a fquare; as alfo to apply a rectangle which fhall be equal to another given, to a given straight line, deficient or exceeding by a fquare, are very often made ufe of by Geometers. and on this account, it is thought proper, for the fake of beginners, to give their conftructions, as follows. 1. To apply a rectangle which fhall be equal to a given fquare, to a given straight line, deficient by a fquare. but the given fquare muft not be greater than that upon the half of the given line. Let AB be the given straight line, and let the fquare upon the given straight line C be that to which the rectangle to be applied must be equal, and this fquare by the determination, is not greater than that upon half of the ftraight line AB. L Bifect AB in D, and if the fquare upon AD be equal to the fquare 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, fo that EF be equal to AD or DB, and from the center E, at the distance EF defcribe a circle meeting a A H K F D G B C E Book VI. AB in G, and upon GB defcribe the fquare GBKH, and com plete the rectangle AGHL; alfo join EG. and becaufe AB is bifected in D, the rectangle AG, GB together with the fquare of DG is equal to (the fquare of DB, that is of EF or EG, that is a. 5. zi to) the fquares of ED, DG. take away the fquare of DG from each of thefe equals, therefore the remaining rectangle AG, GB is equal to the fquare of ED, that is, of C. but the rectangle AG, GB is the rectangle AH, becaufe GH is equal to GB. therefore the rec Book VI. tangle AH is equal to the given fquare upon the ftraight line C. wherefore the rectangle AH equal to the given fquare 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 fquare, to a given ftraight line, exceeding by a fquare. Let AB be the given ftraight line, and let the fquare upon the given ftraight line C be that to which the rectangle to be applied muft be equal. Bifect AB in D, and draw BE at right angles to it, fo that BE be equal to C, and, having joined DE, from the center D at the distance DE defcribe a circle meeting AB produced in G; upon BG defcribe the fquare BGHK, and complete the rectangle AGHL. .6.2. KH A D D B G C and because AB is bifected in D, 3: To apply a rectangle to a given ftraight line which shall be equal to a given rectangle, and be deficient by a fquare. but the given rectangle muft not be greater than the fquare upon the half of the given ftraight line. Let AB be the given ftraight line, and let the given rectangle be that which is contained by the ftraight lines C, D, which is not greater than the fquare upon the half of AB. it is required to apply to AB a rectangle equal to the rectangle C, D, deficient by a fquare. Draw AE, BF at right angles to AB, upon the fame fide of it, and make AE equal to C, and BF tó D. join EF and bisect it in G, and from the center G, at the distance GE describe a circle meeting AE again in H; join HF and draw GK parallel to it, and GL Book VI. parallel to AE meeting AB in L. Because the angle EHF in a femicircle 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; alfo EK is equal to KH, and a. 3. 3. the rectangle C, D, from the determination, is not greater than the fquare of AL the half of AB, wherefore the rectangle EA, AH is not greater than the fquare of AL, that is of KG. add to each the fquare of KE, therefore the fquare of AK is not greater than b. 6. 2. the fquares of EK, KG, that is than the fquare of EG; and confequently the straight line AK or GL is not greater than GE. Now, if GE be equal to GL, the circle EHFtouches AB in L, and therefore the fquare of AL is equal to the H rectangle EA, AH, that is to the given rectangle C, D; and that which was required is done. but if EG, GL be un c E IK Q D G M LN F c. 36. 3. PO e equal, 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 defcribe the fquare NBOP, and complete the rectangle ANPQ. because ML is equal to LN, and it has been proved that AL is d. 3. 3. equal to LB, therefore AM is equal to NB, and the rectangle AN, NB equal to the rectangle NA, AM, that is to the rectangle EA, e.Cor. 36. 3, 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. A 4. To apply a rectangle to a given ftraight line that shall be equal to a given rectangle, exceeding by a fquare. Let AB be the given straight line, and the rectangle C, D the Book VI. 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 fides of it, and make AE equal to C, and BF equal to D. join EF and bifect it in G, and from the center 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 defcribe the fquare NBOP, and complete the rectangle ANPQ, because the angle EHF in a femicircle is equal to the right angle EAB, AB and HF are parallels, and therefore AH and BF are equal, and the rectan E C D. G ОР L B N gle EA, BF, that is to the rectan H F gle C, D. and becaufe ML is equal to LN, and AL to LB, therefore gle EA, AH equal to the rectan-. MA is equal to BN, and the rectangle AN, NB to MA, AN, a. 35. 3. that is to the rectangle EA, AH or the rectangle C, D. there fore the rectangle AN, NE, 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 fquare BP. Which was to be done. Willebrordus Snellius was the firft, as far as I know, who gave thefe conftructions of the 3. and 4. Problems in his Apollonius Batavus. and afterwards the learned Dr. Halley gave them in the Scholium of the 18. Prop. of the 8. B. of Apollonius's Conics reftored by him. The 3. Problem is otherwife enuntiated 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 fame thing, Having given AB the fum of the fides of a rectangle, and the magnitude of it being likewife given, to find its fides. And the 4. Problem is the fame with this, To find a point N in the given straight line AB produced, fo as to make the rectangle AN, NB equal to a given space. or, which is the fame thing, Having given AB the difference of the fides of a rectangle, and the magnitude of it, to find the fides. |