A Probleme. 9.
A Sphere being geuen, to geue an other Sphere, to whose Sphericall superficies, the superficies Sphe∣ricall of the Sphere geuen shall haue any proportion, betwene two right lines geuen.
Suppose A, to be a sphere geuen, and the proportion geuen, to be that, which is betwene the right lines X and Y. I say that a sphere is to be g••uen to whose sphericall superficies, the superficies sphericall of A, shall haue that proportion which X hath to Y. Let the greatest circle, conteyned in A the sphere be the circle BCD. And by the probleme of my additions, vpon the second proposition of this booke, as X is to Y, so let the circle BCD be to an other circle found, let that other circle be EFG: and his dia∣meter EG. I say that
the sphericall superfici∣es of the sphere A, hath to the sphericall super∣ficies of the sphere, whose greatest circle is EFG, (or his equall) that proportion, which X hath to Y. For (by con¦struction) BCD is to E∣FG, as X is to Y: and by the theoreme next be∣fore
•• as BCD is to
••F∣G, so is the spherical su¦perficies of A (whose greatest circle is BCD, by supposition) to the sphericall superficies, of the sphere, whose grea∣test circle is EFG: wherefore, by the 11. of the fifth as X is to Y: So is the sphericall su∣perficies of A, to the sphericall superficies of the sphere, whose greatest circle is EFG: wherefore, the sphere whose diame∣ter is EG, (the diameter also of EFG) is the sphere, to whose sphericall superficies, the sphericall su∣perficies of the sphere A, hath that proportion which X hath to Y. A sphere being geuen therefore, we haue geuen an other sphere, to whose sphericall superficies, the superficies sphericall of the sphere geu
•• hath any proportion geuen, betwene two right lines: which ought to be done.