the line GC be treble to the line HC, by the 9. of the sixt. Now I say, that that which is contained vnder the lines AF and BH, is equall to the pentagon inscribed in the circle ABC. Draw a right line from the point B to the point D. Now forasmuch as the line AD is double to the line DF, there∣fore the line AF is sesquialter to the line AD.
Againe, forasmuch as the line GC is treble to the line CH, therefore the line GH is double to the line CH. Wherefore the line GC is sesquialter to the line HG. Wherefore as the line FA is to the line AD, so is the line GC to the line GH. Wherefore (by the 16. of the sixt) that which is contained vnder the lines AF & HG, is equall to that which is contained vnder the lines DA and GC. But the line GC is equall to the line BG (by the 3. of the third). Wherfore that which is contained vnder the lines AD and BG, is e∣quall to that which is contained vnder the lines AF and GH. But that which is contained vn∣der the lines AD and BG, is equall to two such triangles as the triangle ABD is (by the 41. of the first). Wherefore that which is contained vnder the lines AF and GH, is equall to two such triangles as the triangle ABD is. Wherefore that which is contained vnder the lines AF and GH
••iue times, is equall to ten triangles. But ten triangles are two pentagons. Wherefore that which is contained vnder the lines AF and GH fiue times, is equall to two pentagons. And forasmuch as the line GH is double to the line HC, therefore that which is contained vnder the lines AF and GH, is double to that which is contained vnder the lines AF and HC (by the 1. of the sixt). Wherefore that which is contained vnder the lines AF and CH twise, is equall to that which is contained vnder the lines AF and GH once. Take eche of those parallelogrammes fiue times. Wherefore that which is contained vnder the lines AF and HC ten times, is equall to that which is contained vnder the lines AF & GH fiue times, that is, to two pentagons. Wherefore that which is contained vnder the lines AF and HC fiue times, is equall to one pentagon. But that which is contained vnder the lines AF and HC fiue times, is equall (by the 1. of the sixt) to that which is contained vnder the lines AF and HB, for the line HB is quintuple to the line HC (as it is easie to see by the con∣struction) and they are both vnder one & the selfe same altitude, namely, vnder AF. Wher∣fore that which is contained vnder the lines AF and BH, is equall to one pentagon.
This being proued, now let there be drawne a Circle comprehending both the Pentagon of a Dodecahedron, and the triangle of an Icosahedron, being both described in one and the selfe same Sphere.
LEt the circle be ABC. And in it describe as before, two sides of an equilater pentagon, namely BA and AC•• and draw a right line from the point B to the point C: and take the centre of the circle and let the same be E. And from the point A to the point E draw a right line AE: and extend the line AE to the point F. And let it cut the line BC in the point K. And let the line AE be do∣ble to the line EG, & let the line CK be treble to the line CH, by the .9. of the sixth. And frō the point G raise vp (by the .11. of the first) vnto the line AF a perpendicular line GM: and extend the line GM directly to the point D. Wherfore the line MD is the side of an equi∣liter triāgle, by the corollary of the .1••. of the thirtenth: draw these right lines AD and AM. Wherfore ADM is an equilater triangle. And for as much as that which is contained vn∣der the lines AG and BH is equal to the pentagon (by the former assump••) and that which