Carolus Renaldinus, from whose Treatise de Resol. & Com∣pos. Math. Lib. 2. we have the present Problem, proceeds to solve it in another way, changing it plainly into another Pro∣blem: viz. he observes, 1. That the angles FAD (see n. 1. of our 61. Fig.) and FGD, since both are right ones on the same common base FD, are in circle. Hence he infers, 2. (by vertue of the Coroll. of the 26. Prop. 3. Eucl.) that the □ □ DCG and ACF are equal, and consequently CD, CA, FC and CG are four continued proportionals. Then he observes, 3. That GD is the excess of the first of these proportionals above the fourth CG, and AF is the excess of the second AC above the third CF; and so, since 4. the rectangle of AF and GD is = to the square of the mean proportional FG (for AF, FG, GD, are supposed to be continual proportionals) and this □ FG is the excess, by which the square of the third CF ex∣ceeds the square of the fourth CG; now the present Problem will be 5. reduc'd to this other: Having two right lines (CD and CG) given to find two such mean proportionals (AC and FC) that the ▭ of the excess of the first above the fourth (viz. of FA into GD) shall be equal to the excess,
Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.
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- Title
- Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.
- Author
- Sturm, Johann Christophorus, 1635-1703.
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- London :: Printed for Robert Knaplock and Dan. Midwinter and Tho. Leigh,
- 1700.
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- Mathematics -- Early works to 1800.
- Geometry -- Early works to 1800.
- Algebra -- Early works to 1800.
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http://name.umdl.umich.edu/A61912.0001.001
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"Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S." In the digital collection Early English Books Online 2. https://name.umdl.umich.edu/A61912.0001.001. University of Michigan Library Digital Collections. Accessed June 20, 2025.
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Page 92
by which the square of the third (FC) exceeds the square of the fourth (CG) viz. by the square FG.
Wherefore instead of the former he solves this latter Pro∣blem, putting for CG, b, for GD, c, so that the first of the given lines CD shall be = b+c, and the other GD=b; calling the first mean proportional AC, x; and thence deno∣minating the latter 〈 math 〉〈 math 〉 (viz. multiplying the fourth by the first, and dividing the product by the second) and more∣over he finds the excess of the first (b+c) above the fourth (b) to be c, and the excess of the second (x) above the third 〈 math 〉〈 math 〉 to be 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉; so that the □ □ of these two excesses is 〈 math 〉〈 math 〉, and because the □ of the third FC is = 〈 math 〉〈 math 〉, having substra∣cted bb=□GC, there is given the □FG = 〈 math 〉〈 math 〉 = ▭ of the excesses we just now found. So that now you'l have the Equation 〈 math 〉〈 math 〉 &c.
We also endeavour'd to find another Solution, by finding an Equation from the line BD (n. 1. Fig. 61.) as which might be twice obtain'd by means of the two right-angled Triangles FAD and FGD, since it is the hypothenusa of both. But here, besides the former denominations of our Solution we must first give a denomination to the line AD, by making as CF to FG so CD to AD 〈 math 〉〈 math 〉, &c.
But whosoever shall prosecute this Solution of ours, or that of Renaldinus to the end, will find much more labour and dif∣ficulty in either, than in the first we have given.