drawe these lines IN, IT, TN, ET. And
forasmuche as in the perpendiculars IT & IN are the centres of the circles which containe the equilater triangles IBG, & IGD, by the corollarye of the first of the thirde. Let those centres be the points S and O. And draw the line SO. Deuide the line TB the half of BG the side of the Icosahedron by an extreme and meane proportion in the poynt R, by the 30. of the sixth, and let the lesse segment therof be RB. And forasmuch as the line SO cou¦pleth the centres of the triangles IBG, & IGD, it is by the 5. of the fiuetenth, the side of the Dodecahedrō inscribed in the Icosahedron, whose side is the line BG. From the side BG take away
••R the lesse segment of the halfe side. And from the residue GR take away the third part GV (by the 9. of the sixth.) Then I say that the residue RV is equal to SO the side of the Dodecahedron inscribed. For forasmuch as the perpendicular EN is in the poynt C deuided by an extreme and meane proportion, by the corollary of the first of the fourtenth, and the greater segment therof is the line EC, and vnto the line EC the line CM is equal, by the corollary of the 12. of the thirtenth: wherefore the line EC is to the line CN, as the line CM is to the same line CN, by the 7. of the fiueth. But as the line EC is to the line CN, so is the whole line
••N, to the greater segment E∣C, by the 3. diffinition of the sixth. Wherefore (by the 11. of the fiueth), as the whole line EN is to the greater segment EC, so is the line CM to the line CN. Wherefore the line CM, is deuided by an extreme and meane proportion in the poynt N, namely, is deuided like vnto the line EN, by the 2. of the fourtenth. Wherfore the line EM excedeth the line EN by the lesse segment of his halfe, namely, by MN. And forasmuche as EGD is the triangle of an equilater and equiangle pentagon ABGDF, and ETN is likewise the triangle of the like pentagon inscribed in the pentagon ABGDF: Therefore by the 20. of the sixth, the triangle ETN is like to the triangle EGD
•• Wherefore as the line EG is to the line EN, so by the 4. of the sixth, is the line GD to the line NT. Wherefore the line GD (or BG which is equal vnto it) excedeth the line NT by the lesse segment of the halfe of BG. For the line EG did in like sort excede the line EN. But that lesse segment is the line BR. Wherefore the residue RG is equal to the line TN. And forasmuch as IBG is an equilater triangle: the perpendicular ST shalbe the halfe of the line SI which is drawne from the centre, by the corollary of the 12. of the thirtenth: wher∣fore the line IT excedeth the line IS by his third part. And forasmuche as the line SO which coupleth the sections, is a parallel to the line TN, by the 2. of the sixth. For the equal perpendiculars IT, and IN are cut like in the poynts S & O: therfore the triangles ITN & I
••O, are like by the corollary of the second of the sixth. Wherfore as the line IT is to the line IS, so by the 4. of the sixth is the line TN to the line SO. But the line IT excedeth the line IS by a third part: wherfore the line TN, excedeth the line SO by a third part: but the line TN is proued equal to the line RG. Wherfore the line RG exce∣deth the line SO by a third part of himself, which is GV. Wherfore the residue RV, is equal to the line SO, which is the side of the dodecahedron inscribed in the Icosahedron, whose side is the line BG. If therfore halfe of the side of an Icosahedrō, be deuided by an extreme & meane proportion: and if the lesse segment therof be taken away from the whole side, and againe from the residue be takē away the third part: that which remaineth shall be equal to the side of the dodecahedron inscribed in the same Icosahedron.