Book VI. " parallelogram BE to the parallelogram EF;" nothing more fhould have been added but this, " and the rectilineal figure "ABC is equal to the parallelogram BE; therefore the recti"lineal KGH is equal to the parallelogram EF," viz. from prop. 14. book 5. But betwixt these two fentences he has inferted this; "wherefore, by permutation, as the rectilineal fi"gure 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 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 demonftrated in the elements as we now have them: But though this propofition, viz. the third of four proportionals is equal to the fourth, if the first be equal to the fecond, 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 said, the conclufion could have been immediately deduced without this fuperfluous ftep 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 fame blunder is found twice in the Greek text of prop. 23. book 11. and twice in prop. 2. b. 12. and in the 5. 11. 12. and 18th of that book; in which places of book 12. except the laft of them, it is rightly left out in the Oxford edition of Commandine's tranflation: And also that geometers may beware of making use of permutation in the like cafes; 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 feems, preoccupied many fo much, that they do not fufficiently understand the true nature of them. Befides, though the rectilineal figure ABC, to which another is to be made fimilar, may be of any kind whatever; yet in the demonstration the Greek text has triangle" instead of rectili "neal figure," which error is corrected in the above-named Oxford edition. PROP. XXVII. B. VI. The fecond cafe of this has 'as, otherwife, prefixed to it, as if it was a different demonftration, which probably has been done by fome unfkilful Librarian. Dr Gregory has right ly ly left it out: The fcheme of this fecond cafe ought to be Book VI. marked with the fame letters of the alphabet which are in the fcheme of the firft, as is now done. PROP. XXVIII. and XXIX. B. VI. These two problems, to the firft of which the 27th prop. is neceffary, are the moft general and useful 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 igno rantly left out by Tacquet and Dechales in their editions of the elements, who pretend that they are scarce of any ufe: The ca fes of these problems, wherein it is required to apply a rectangle which fhall be equal to a given fquare, to a given straight line, either deficient or exceeding by a fquare, as alfo to apply a rectangle which fhall be equal to another given, to a given ftraight 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 shall be equal to a given square, to a given ftraight line, deficient by a fquare: But the given fquare must not be greater than that upon the half of the given line. 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 must be equal, and this square, 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 muft be greater than C, acccording to the determination: Draw DE at right angles to AB, and make it equal to C; produce ED to F, fo that A EF be equal to AD or DB, H K F D G B and from the centre E, at C the diftance EF, describe a E circle meeting AB in G, and upon GB defcribe the fquare GBKH, and complete the rectangle AGHL; also join EG: And because AB is bifected in D, the rectangle AG, GB together with the square of DG is equal to (the fquare of DB, that is, of EF or EG, that is, a 5. 2. to) Book VI. to) the fquares of ED, DG: Take away the fquare of DG w from each of thefe equals; therefore the remaining rectangle a 6. 2. AG, GB is equal to the fquare 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 fquare upon the ftraight line C. Wherefore the rectangle AH, equal to the given fquare upon C, has been applied to the given ftraight line AB, deficient by the fquare GK. Which was to be done. 2. To apply a rectangle which fhall be equal to a given fquare, to a given straight 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 must 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 centre D at the distance DE defcribe a circle meeting AB produced in G; upon BG defcribe the fquare BGHK, and complete the rectangle AGHL. And because AB is bifected in D, and produced to G, the rectangle AG, GB together with the iquare of DB E KH D BG C is equal to (the fquare of DG, FA 3. To apply a rectangle to a given ftraight line which fhall 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 straight 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 square. Draw Draw AE, BF at right angles to AB, upon the fame fide of it, Book VI. and make AE equal to C, and BF to D: Join EF and bifect 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. 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 the rectangle C, D from the a 3. 3. 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 the b 6. 2. 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 EHF touches AB in L, and therefore the fquare of AL is equal to the rectangle EA, AH, that is, to the given rectangle C, D; and that which was required is done: But if EG, GL be unequal, EG must be the greater; and E H K C D G c 36. 3. A M L N B Q PO therefore the circle EHF cuts the ftraight 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 d LN, d 3. 3. and it has been proved that AL is 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, AH or the e Cor. 36. rectangle C, D: But the rectangle AN, NB is the rectangle 3. 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 ftraight line AB, deficient by the fquare BP. Which was to be done. Book VI. a 35. 3. 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 ftraight 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 fquare. E C D 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 centre G. at the diftance GE, defcribe 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 e. qual, and the rectangle EA, AH equal to the rectangle M G O P A L B N H F a 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 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 ftraight 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 first, as far as I know, who gave these constructions of the 3d and 4th problems in his Appollonius Batavus: And afterwards the learned Dr Halley gave them in the Scholium of the 18th prop. of the 8th book of Apollonius's conics reftored by him. The 3d problem is otherwife enunciated thus: To cut a given ftraight line AB in the point N, fo 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 4th problem is the fame with this, To find a point |