PROB, II. To find three straight lines such, that the ratio of 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 BD 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 24. dat, the remainder to BD is given; therefore the excess of the first AB above a given 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 44. dat, is given in species; wherefore the angle BED is given; Let AE which is given in magnitude, be given also in posi tion, as also the point E, and the straight line ED will be 32. dat. given in position: Join AD, and because the sum of the 47. 1. squares of AB, BD, that is, the square of AD is given, therefore the straight line AD is given in magnitude; and it 34. dat. is also given in position, because from the given point A it" is drawn to the straight line ED given in position: There fore the point D, in which the two straight lines AD, BD, 28. dat. given in position, cut one another, is given: And the straight 33. dat. line DB, which is at right angles to AB, is given in position, and AB is given in position, therefore the point B is given: 29. dat. And the points A, D are given, wherefore the straight lines ig. dat. AB, BD are given: And the ratio of AB to BC is given, and therefore BC is given. The Composition. LET the given ratio of FG to GH be that which AB is required to have to BC, and let HK be the given straight line which is to be taken from BC, and let the ratio which the E BNM remainder is required to have to BD be the given ratio of HG to LG, and place GL at right angles to FH, and join LE, LH: Next, as HG is to GF, so 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, BD is required to be equal; and make the angle NED equal to the angle GFL; and from the centre A, at the distance AN, describe a circle, and let its circumference meet ED in D, and draw DB perpendicular to AN and DM making the angle BDM equal to the angle GLH. Lastly, produce BM to C, so that MC be equal to KH; then is AB the first, BC the second, and BD the third of the straight lines that were to be found. For the triangles EBD, FGL, as also DBM, LGH being equiangular, as EB, to BD, so is FG to GL: and as DB to BM, so is LG to GH; therefore, ex æquali, as EB to BM, so is (FG to GH, and so is) AE to HK or MC; wherefore, AB is to BC, as AE to HK, that is, as FG to GH, that is, in the given ratio and from the straight line BC taking MC, which is equal to the given straight line HK, the remainder BM has to BD the given ratio of HG to GL: and the sum of the squares of AB, BD is equal to the square of AD or 4 47. 1. AN, which is the given space. Q. E. D. I believe it would be in vain to try to deduce the preceding construction from an algebraical solution of the problem. END OF THE NOTES TO THE DATA. |