PRO P. LXXIX. LXXX. LXXXI. PR () P. LXXXII. PROP. LXXXVIII. XC. j first have no distinct meaning, and the third which he fays is the beft, tho' it contains a true Propofition which is the go. in this Edition, has no connexion in the least with the Greek text. and it is strange that Dr. Gregory did not obferve, that if Prop. 86. was changed into this, the Demonstration of the 86. must be cancelied, and another put in its place. but, the truth is, both the Enuntiation and the Demonftration of Prop. 86. are quite entire and right, only Prop. 87. which is more simple, ought to have been placed before it; and the deficiency which the Doctor juftly oblerves to be in this part of Euclid': Data, and which no doubt is owing to the carelellness and ignorance of the Greek Editors should have been supplied, not by changing Prop. 86. which is both entire and necessary, but by adding the two Propofitions which are the 88. and go. in this Edition. PROP. XCVUI. C. These were commuvicated to me by two excellent Geometers, the firft of them by the Right Honourable the Earl Stanhope, and the o'her by Dr. Matthew Stewart; to which I have added the Demonstrations. Tho' the order of the Propositions has been in many places changed from that in former Editions, yet this will be of little difadvantage, as the antient Geometers never cite the Data, and the Noderns very rarely. As S that part of the Composition of a Problem which is its Construction may not be fo readily deduced from the Analysis by beginners; for their fake the following Example is given in which the derivation of the several parts of the Construction from the Analysis is particularly shewn, that they may be afifted to do the like in other Problems. PROBLEM. Having given the magnitude of a parallelogram, the angle of which ABC is given, and also the excess of the square of its fide BC above the square of the side AB; To find its fides and describe it. The Analysis of this is the same with the Demonstration of the 87. Prop. of the Data. and the Construction that is given of the Problem at the end of that Proposition, is thus derived from the Analysis. Let EFG be equal to the given angle ABC, and because in the Analysis it is said that the ratio of the rectangle AB, BC to the parallelogram AC is given by the 62. Prop. Dat. therefore from a point in FE, the perpendicular EG is drawn to FG, as the ratio of FE to EG is the ratio of the rectangle AB, BC to the parallelogram M А. 3 BPD C F G L O 0 HN AC by what is shewn at the end of Prop. 62. Next the magnitude of AC is exhibited by making the rectangle EG, GH equal to it, and the given excefs of the square of BC above the square of BA, to which excess the rectangle CB, BD is equal, is exhibited by the rectangle HG, GL. then in the Analysis the rectangle AB, BC is faid to be given, and this is equal to the rectangle FE, GH, because the rectangle AB, BC is to the parallelogram AC, as (FE 10 EG, that is as the rectangle) FE, GH to EG, GH; and the parallelogram AC is equal to the rectangle EG, GH, therefore the rectangle AB, BC is equal to FE, GH. and consequently the ratio of the rectangle CB, BD, that is of the rectangle HG, GL, to AB, BC, thas is of the straight line DB to BA, is the same with the ratio (of the rectangle GL, GH to FE, GII, that is) of the straight line GL to FE, which ratio of DB to B A is the next thing said to be given in the Analysis. from this it is plain that the square of FE is to the square of GL, as the square of BA which is equal to the rectangle BC, CD is to the square of BD, the ratio of which spaces is the next thing faid to be given. and from this it follows that four times the square of FE is to the square of GL, as four times the rectangle BC, CD is to the square of BD; and, by Composition, four times the square of FE together with the fousre of GL is to the square of GL, as four times the rectangle BC, CD together with the square of BD, is to the square of BD, that is [8. 6.] as the square of the straight lines BC, CD taken together is to the square of BD, which ratio is the next thing said to be given in the Analysis. and because four times the square of FE and the square of GL are to be added together, therefore in the perpendicular EG there is taken KG equal to FE, and MG equal to the double of it, because thereby the squares of MG, GL, that is, joining ML, the square of ML is equal to four times the square of FE and to the square of GL. and because the square of ML is to the square of Gl., as the square of the straight line made up of BC and CD is to the square of BD, therefore [22. 6.] ML is to LG, as BC together with CD is to BD, and, by Co.npofition, ML and LG together, that is, producing GL to N, so that ML be equal to LN, the straight line NG is to GL, as twice BC is to BD; and by taking GO equal to the half of NG, GO is to GL, as BC to BD the ratio of which is said to be given in the Analyíls. and from this it follows, that the rectargle HG, GO is to HG, GL, as the square of BC is to the rectangle CB, BD which is equal to the rectangle HG, GL, and therefore the square of BC is equal to the rectangle HG, GO, and BC is contequently found by taking a mean proportional betwixt HG and CO, as is fald in the Construction, and because it was fhewn that GO is to GI, as BC to BD, and that now the three first are found, the fourth BD is found by 12. 6. it was likewise Mewa that LG is to FE, or GK, as DB to BA, and the three first are now found, and there. by the fourth PA. make the angle ABC equal to EFG, and corplete the parallelogram of which the fides are AB, BC, and the construction is finished ; the rest of the Compofiioa contains the Demonstratiop. AS S that part of the Composition of a Problem which is its Conftruction may not be fo readily deduced from the Analysis by beginners; for their fake the following Example is given in which the derivation of the several parts of the Construction from the Analysis is particularly shewn, that they may be alifted to do the like in other Problems. PROBLEM. Having given the magnitude of a parallelogram, the angle of which ABC is given, and also the excess of the square of its fide BC above the square of the side AB; To find its fides and describe it. The Analysis of this is the same with the Demonstration of the 87. Prop. of the Data. and the Construction that is given of the Problem at the end of that Proposition, is thus derived from the Analysis. Let EFG be equal to the given angle ABC, and because in the Analysis it is faid that the ratio of the rectangle AB, BC to the parallelogram AC is given by the 62. Prop. Dat, therefore from a point in FE, the perpendicular EG is drawn to FG, as the ratio of FE to EG is the ratio of the rectangle AB, BC to the parallelogram А. BPD C F G L O HN AC by what is shewn at the end of Prop. 62. Next the magnitude of AC is exhibited by making the rectangle EG, GH equal to it, and the given excess of the square of BC above the square of BA, to which excess the rectangle CB, BD is equal, is exhibited by the rectangle HG, GL. then in the Analysis the rectangle AB, BC is said to be given, and this is equal to the rectangle FE, GH, because the rectangle AB, BC is to the parallelogram AC, as (FE to EG, that is as the rectangle) FE, GH to EG, GH; and the parallelogram AC is equal to the rectangle EG, GH, therefore the rectangle AB, BC is equal to FE, GH. and consequently the ratio of the rectangle CB, BD, that is of the rectangle HG, GL, to AB, BC, that is of the straight line DB to BA, is the same with the ratio (of the rectangle GL, GH to FE, GII, that is) of the straight line GL to FE, which ratio of DB to B A is the next thing izid to be given in the Analysis. from this it is plain that the squire of FE is to the square of GL, as the square of BA which is equal to the rectangle BC, CD is to the square of BD, the ratio of which spaces is the next thing said to be given. and from this it follows that four times the square of FE is to the square of GL, as four times the rectangle BC, CD) is to the square of BD; and, by Composition, four times the square of FE together with the square of GL is to the square of GL, as four times the rectangle BC, CD together with the square of BD, is to the square of BD, that is [ 8. 6.] as the square of the straight lines BC, CD taken together is to the square of BD, which ratio is the next thing faid to be given in the Aralysis. and because four times the square of FE and the square of GL are to be added together, therefore in the perpendicular EG there is taken KG equal to FE, and MG equal to the double of it, because thereby the squares of MG, GL, that is, joining ML, the square of ML is equal to four times the square of FE and to the fquare of GL. and because the square of ML is to the square of Gl., as the square of the straight line made up of BC and CD is to the square of BD, therefore [ 22. 6.] ML is to LG, as BC together with CD is to BD, add, by Co.npofition, ML and LG together, that is, producing GL to N, so that ML be equal to LN, the straight line NG is to GL, as twice BC is to BD; and by taking GO equal to the half of NG, GO is to GL, as BC to BD the ratio of which is said to be given in the Analysis. and from this it follows, that the rectarg'e HG, GO is to HG, GL, as the square of BC is to the rectangle CB, BD which is equal to the rectangle HIG, GL, and therefore the square of BC is equal to the rectangle HG, GO, and BC is consequently found by taking a mean proportional betwixt HG and CO, as is faid in the Construction. and because it was fhewa that GO is to CI, as BC to BD, and that now the three first are found, the fourth BD is found by 12. 6. it was likewise mewn that LG is to FE, or GK, as DB to BA, and the three fiiit are now found, and thereby the fourth PA. make the angle ABC equal to EFG, and complete the parallelogram of which the fides are AB, BC, and the construction is finished; the rest of the Composition contains the Demonstratiop. |