1  ΗΖ ... So the triangle ZΡH is equilateral. Thus also the triangles ΗΣΘ, 
2  ΘΤΙ, ΙΥΚ, ΚΦΖ. But equal is also ΡΣ to the line ΗΘ, and as equal we have shown 
3  the lines ΡΗ and ΣΘ. So each of the triangles lying between (scil. the circles) is equilateral, 
4  because we had drawn the perpendiculars ΖΜ, ΗΝ, ΘΞ, ΙΟ, ΚΠ equal and parallel to the line XΛ. 
5  That is why XP is also (scil. equal to one) hexagonal side. But also ΨΧ was (scil. equal to a) decagonal side, so that ΨΡ and ΨΣ 
6  are pentagonal sides. Equal was <also> ΦΡ to side ΡΣ, therefore ΨΦΡ and ΨΡΣ are equilateral 
7  triangles. Therefore each of the triangles whose common apex is Ψ is also 
8  equilateral. Furthermore, let the distance 
9  ΛΩ be drawn as an extension of the distance ΛΨ, equal to the side of the decagon. 
11  Let the radius ΛΖ equal to the hexagon side be drawn from the centre of the circle Λ, but (scil. equal) to the decagon side was 
12  ΛΩ , so ΖΩ is a pentagon side. Therefore equilateral is 
13  the triangle ΩΖΗ. For the same reason, the spatial figure ΖΗΘΙΚΩ is formed by 
14  five equilateral triangles having their apex in Ω. 
15  Two decagonal sides and one hexagonal side together 
16  give the diameter of the given sphere cf. Euclid El. XIII 16 corollary, 
17  and the diameter of the given sphere squared is five times 
18  the radius of the given circle squared, 
19  drawn around the pentagon on which the icosahedron was built. (after Müller/Mau 1960)
