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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, BL 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, GH, that is) of the straight line GL to FE, wbich ratio of DB to BA 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 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 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 composition, 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 rectangle HG, GO is to HG, GL, as the square of BC 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 consequently found by taking a mean proportional betwixt HG and GO, as is said in the construction: and because it was shown that GO is to GL, as BC to BD, and that now the three first are found, the fourth BD is found by 12. 6. It was likewise shown that LG is to FG, or GK, as DB to BA, and the three first are now found, and thereby the fourth BA. Make the angle ABC equal to EFG, and complete the parallelogram of which the sides are AB, GC, and the construction is finished; the rest of the composition contains the demonstration.

As the propositions from the 13th to the 28th may be thought by beginners to be less useful than the rest, because they cannot so readily see how they are to be made use of in the solution of problems; on this account the two following problems are added, to show that they are equally useful with the other propositions, and from which it may be easily judged that many other problems depend upon these propositions.

PROBLEM I. 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 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 EF is given: therefore (24. dat.) the excess of the first AB above a given straight line has a given ratio to the third EF; let BH be that given straight line; therefore AH, the excess of AB above it, has a given ratio to EF; A H B and consequently (1. 6.) the rectangle BA, AH,

has a given ratio to the rectangle AB, EF, which C G D last rectangle is given by the hypothesis ; there. fore (2. dat.) the rectangle BA, AH is given, and E F BH the excess of its sides is given; wherefore the sides AB, AH are given (85. dat.): and because K NM L O the ratios of AB to CD, and of AH to EF are - - - given, CD and EF are (2. dat.) given.

'The Composition. Let the given ratio of KL to KM be that which AB is required to have to CD; and let DG 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 EF; also let the given rectangle NK, KO be that to which the rectangle AB, EF is required to be equal : find the given stright line BH which is to be taken from AB, which is done, as plainly appears from prop. 24. dat. by making as KM to KL, so GD to HB. To the given straight line BH apply (29. 6.) a rectangle equal to LK, KO exceeding by a square, and let BA, AH be its sides: then is AB the first of the straight lines required to be found, and by making as LK to KM, so AB to DC, DC will be the second: and lastly, make as KM to KN, so 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 (19. 5.) AH is to CG, as (AB to CD, that is, as) LK to KM; and as CG to EF, so is KL to KN: wherefore, ex æquali, as AH to EF, so is LK to KN: and as the rectangle BA, AH to the rectangle BA, EF, so is (1. 6.) the rectangle LK, KO to the rectangle KN, KO: and by the construction, the rectangle BA, AH is equal to LK, KO: there

fore (14. 5.) the rectangle AB, EF is equal to the given rectangle NK, KO: and AB has to CD the given 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.

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 on 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 remainder to BD is given; therefore (24. dat.) 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 is (44. dat.) given in species; wherefore the angle BED is given : let AE, which is given in magnitude, be given also in position, as also the point E, and the straight line ED will be given (32. dat.) in position : join AD, and because the sum of the squares of AB, BD, that is (47. 1.), the square of AD is given, therefore the straight line AD is given in magnitude; and it is also given (34. dat.) in position, because from the given 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 position cut one another, is given (28. dat.): and the straight line DB which is at right angles to AB is given (33. dat.) in position, and AB is given in position, therefore (28. dat.) the point B is given: and the points A, D are given, wherefore (29. dat.) the straight lines AB, BD are given ; and the ratio of AB to BC is given, and therefore (2. dat.) 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 remainder is required to have

A E B N M C F G H K to BD, be the given ratio of HG to LG, and place GL at right angles to FH, and join LF, 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 HK; 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 (12. 5.) 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 (47. 1.) to the square of AD or 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.

FINS.

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PL A NE AND SPHERICAL

TRIGONOMETRY.

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