f. 28. Dat. position cut one another is given f. and the straight line DB which g. 33. Dat. is at right angles to AB is given & 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 1. 2. Dat. ratio of AB to BC is given, and therefore i 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 to BD be the given ratio of HG to GL, and place GL at

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right angles to FH, and join LF, LH. Next, as HG is to GF, so make IIK 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. from the center 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 aequali, as EB to BM, so is (FG to GH, 12. 5. and so is) AE to HK or MC; wherefore k AB is to BC, as AE to

HK, that is, as FG to GH, that is, in the given ratio. and from the ítraight 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 d to the square of 2:47. 1. AD or AN which is the given space. Q. E. D.

I believe it would be in vain to try to deduce the preceeding Construction from an Algebraical Solution of the Problem.

F I N I S.

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