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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 a 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 • 14 Dat. isb given in species; wherefore the angle BED is given :

Let AE which is given in magnitude, be given also in po

sition, 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 d 47. 1. squares of AB, BD, that is, the square of AD is given,

therefore the straight line AD is given in magnitude; and 34 Dat. it 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, 128 Dat. given in position, cut one another, is givenf: And the • 39 Dat. straight 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 : And the points A, D, are given, whereb 29 Dat fore h the straight lines AB, BD are given: And the ratio of i q Dat. AB to BC is given, and thereforei 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

L
D

A

E BN M. C F G H K line which is to be taken from BC, and let the ratio which the 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

1

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 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; whereforek, AB is to BC, as AE to HK, that is, as FG to GH, * 22. 5. 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 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.

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