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by beginners to be less useful than the rest, because thy Cantot fo readily five how they are to be made use of in the fol:tion of Problems; on this account the two following Problems are added, to thew that they are equally useful with the other Prcs pofitions, and from wliich it may eafi.y be judged that many order Problems depend upon thcie Propofitions.
O find three firaight lines such, that the ratio ci
the first to the second is given ; and if a given straight line be taken from the second, the ratio of the remainder to the third is given ; also the rectangle contained by the first and third is given.
Let AB be the first straight line, CD the second, and EF the third, and because the ratio of AB to CD is given, and that if a
given straight line be taken from CD, the ratio of the remainder to 4. 15. Dai. EF is given ; therefore a the excess of the first AB above a given
straight line has a given ratio to the third EF. Let BH be that gie ven fraight line, therefore AH the excess of
A H B В AB above it has a given ratio to EF; and conf quently 6 the rectangle BA, AH has a C G D given ratio to the rectangle AB, EF, which
last reangle is given by the Hypothefis; E F c. s. Dat. therefore © the rectangle BA, AH is given,
and Bil the excess of its fides is given ; K NML O 4. 85. Dat. wherefore the sides AB, AH are given 4. and
because the ratios of AB to CD, and of AH to LF are given; CD and EF are given.
The Compofition. Let the given ratio of KL to KM be that which AB is required to have to CD; and let IG be the given straight line which is to be taken from CD, and let the given ratio of KM to KN be that which the remainder must have to E!'; also, let the given rectangle NK, KO be that to which the rectangle AB, EF is required to be equal. find the given straight line BH which is to be taken from AB, which is done, as plainly appears from P:op. 2.4. Dat
by making as KM to KL, so GD to HB. to the given straight line BH apply a rectangle equal to LK, KO exceeding by a square, c. 29. 6. and let BA, AH be its sides. then is AB the first of the straight lines required to be found. and by making a: LK to KM, fo AB to DC, DC will be the second. and lastly, make as KM to KN, fo CG to EF, and EF is the third.
For as AB to CD, so is HB to GD, each of these ratios being the same with the ratio of LK to KM; therefore? Al is to CG, f. 19.5. as (AB to CD, that is, as) LK to KM; and as CG to EF, fo is KM to KN; wherefore, ex aequali, as All to CF, fo is LK to KN. and as the rectangle BA, AH to the rectangle DA, EF, fo is the 5. 1.6. rectangle LK, KO to the rectangle KN, KO. and, by the ConIruction, the rectangle BA, AH is equal to LK, KO, therefore h h. 14.50 the rectangle AB, EF is equal to the given rectangle NK, KO. and AB has to CD the giren ratio of KL to KM; and from CD the given straight line GD being taken, the remainder CG has to EF the given ratio of KM to KN. Q. E. D.
PRO B. II.
the first to the second is given ; and if a given straight line be taken from the second, the ratio of the remainder to the third is given ; also the sum of the squares of the first and third is given.
Let AB be the first straight line, BC the second, and I'D'the third. and because the ratio of AB to BC is given, and that if a given straight line be taken from BC, the ratio of the remairder to BD is given; therefore the excess of the first AB above a given a. 24.1 straight line has a given ratio to the third BD. let AE be that given straight line, therefore the remainder EB has a given ratio to BD. let BD be placed at right angles to EB, and join DE, then the triangle EBD is o given in species; wherefore the angle BED b. 44. Dat. is given. let AE which is given in magnitude be given also in position, and the straight line ED will be given in position. join AD, c. 22. Vat. and because the sum of the squares of AB, BD, that is d, the square d. 17. 1. of AD is given, therefore the straight line AD is given in magnitude ; and it is also given e in position, because from the given e. 24. l'at. point A it is drawn to the straight line ED given in position. therefore the point D in which the two straight lines AD, ED given in
f. 28 Dat. position cut one another is given f. and the straight line DB which 8. 33. Dat. is at right angles to AB is given 6 in position, and AB is given in
position, therefore f the point B is given. and the points A, D are h. 29. Dat. given, wherefore h the straight lines AB, BD are given. and the i. 2. Dat. ratio of AB to BC is given, and therefore i BC is given.
The Compofition. Let the given ratio of FG to GH be that which AB is required to have to RC, and let i IK be the given straight line which is to be taken frou BC, and let the ratio which the remainder is required to have to BD be the given ratio of HG to GL, and picce GL at right angles to FX, and join LT, LH. Next, as HG is to
H K GF, fo make HK to AE ; produce AE to N so that AN be the Straight line to the square of which the sum of the squares of AB, 1.D is required to be cqual; and make the angle NED equal to the angle GFL. froin the center A at the distance AN ducribe a circle, and let its circumference meet ED in D, and draw DB perpendicular to AN, and DM making the angle BDM cqual to the angle GLII. lastly, produce BM to C so that MC be equial: Blk. then is AB the first, BC the fecond and BD the third of the Iti uch: lines that were to be found.
Por t'i trianglez EB!), FGL, as alío DBM, LGH being equie angular, as ID to 1), 10 is TG to G!; and as DD O DM,ois IG OCH; theriore, ex quali, as LB to B'i, tois, Gis Gil, and to i:) AL to II Kur MC; whicefore k AB is to BC, 25 Al to lik, that is, as FG tu GH, that is, in the given ratin. ? from the trisht line BC tubing NiC which is equal to the gia fraicht line HX, the remainder DM has to BD the ghen tatud KG to GL. and the fun of the squares of AB, BD is equal to the square of AD or AN which is the giren force. Q. L. D.
I beliere it would be in vain to try to deduce the precedias Ceasudica from an ulcebraical Solution of the Problem.
k. 12. 5.