to be perpendicularly erected to the superficies
of BCD. Seing that the portion of the playne remayning in the sphere, is called their common section: the sayd section shall be a circle, as be∣fore is proued.
And the common section of the sayd playne, and the greatest circle BCD, (which is BE by supposition) shall be the diameter of the same cir∣cle, as we will proue. For, let that circle be BL∣EM. Let the center of the sphere A, be the point H: which H, is also the cēter of the circle BCD, because BCD is the greatest circle in A contey∣ned. From H, the center of the sphere A, let a line perpendicularly be let fall to the circle BL∣EM. Let that line be HO: and it is euident that HO shall fall vpon the common section BE, by the 38. of the eleuenth. And it deuideth BE, into two equall parts, by the second part of the third proposition of the third booke: by which poynt O, all other lines drawne in the circle BLEM, are, at the same pointe O, deuided into two e∣quall parts. As if from the poynt M, by the point O, a right line be drawne one the other side com¦ming to the circumference, at the poynt N: it is manifest that NOM is deuided into two equall partes at the poynt O: by reason, if from the cen∣ter H, to the poyntes N and M, right lines be drawne, HN and HM, the squares of HM, and HN are equall: for that all the semidiameters of the sphere are equal: and therefore their squares are equall one to the other: and the square of the perpendicular HO, is common: wherefore the square of the third line MO is equall to the square of the third line NO: and therefore the line MO to the line NO. So therefore is NM equally deuided at the poynt O. And so may be proued of all other right lines, drawne in the circle BLEM, passing by the poynt O, to the circumference one both sides. Wherefore O is the center of the circle BLEM: and therefore BE passing by the poynt O is the diameter of the circle BLEM. Which circle (I say) is equal to FKG: for by construction BE is e∣quall to FG: and BE is proued the diameter of BLEM, and FG is by supposition the diameter of the circle FKG: wherefore BLEM is equall to FKG the circle geuen: and BLEM is in A the sphere geuē. Wherfore we haue in a sphere geuen coapted a circle equall to a circle geuen: which was to be done.
A Corollary.
Besides our principall purpose, in this Probleme, euidently demonstrated, this is also made mani∣fest: that if the greatest circle in a Sphere, be cut by an other circle, erected vpon him at right angles, [ 1] that the other circle is cut by the center, and that their common section is the diameter of that other [ 2] circle: and therefore that other circle deuided is into two equall partes. [ 3]