e 29. 6.

f 19. 5.

appears from Prop. 24. dat. by making as KM to KL, so GD to HB.

To the given straight line BH apply e 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 is to CD, so is HB to GD, each of these ratios being the same with the ratio of LK to KM; therefore f AH is to CG, as (AB to CD, that is, as) LK to KM; and as CG is to EF, so is KM 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 8 the rectangle LK, KO to the rect

angle KN, KO: And by the construction, the rectangle BA, h 14. 5. AH is equal to LK, KO: Therefore 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.

g 1. 6.


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

if a given straight line be taken from BC, the ratio of the rea 24. dat. mainder to BD is also given; therefore a the excess of the first

A B above a given straight line has a given ratio to the third
BD: Let AE be that given straight line, therefore the remain-

der EB has a given ratio to BD: Let BD be placed at right b 44. dat. angles to EB, and join DE; then the triangle EBD is b given

in species; wherefore the angle BED is given. Let AE which

is given in magnitude, be given also in position, as also the c 32. dat. point E, and the straight line ED will be given in position :

Join AD, and because the sum of the squares of AB, BD, that d 47. 1. is d, the square of AD is given, therefore the straight line AD e 34. dat. is given in magnitude; and it is also given 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 givenf: And the straight line DB, which is at right angles to f 28. dat. AB, is given 8 in position, and AB is given in position, there- g 33. dat. fore f the point B is given : And the points A, D are given, wherefore the straight lines AB, BD are given : And the ra- h 29. dat. tio of AB to BC is given, and therefore i BC is given.

i 2. dat.

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 whicb is to be taken from BC, and let the ratio which the re

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H K mainder is required to have to BD be the given ratio of HG to GL, 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 may 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 may 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, k 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 equald to the square of AD or AN, d 47. 1. 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.


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