THE COLLECTION. BOOK VIII 
439 
Then he proves that, if we join AB, AB is equal to the length 
of the side of the hexagon required. 
Produce BG to D so that BI) = BA, and join DA. ABD 
is then equilateral. 
Since EB is a tangent to the segment, AE.EG — EB 2 or 
AE\EB = EB :EC, and the triangles EAB, EBG are similar. 
Therefore BA 2 : BG 2 = AE' 1 : EB* = AE:EC =9:4; 
and BG = %BA = %BD, so that BG = 2CD. 
But CF= 2GA ; therefore AG:GF= DC: GB, and AD, BF 
are parallel. 
Therefore BF:AD = BG:GD = 2 :1, so that 
BF= 2AD=2AB. 
Also Z FBG = ZBDA = 60°, so that ¿ABF = 120°, and 
the triangle ABF is therefore equal and similar to the required 
triangle NLO. 
Construction of toothed wheels and indented screws. 
The rest of the Book is devoted to the construction (1) of 
toothed wheels with a given number of teeth equal to those of 
a given wheel, (2) of a cylindrical helix, the cochlias, indented 
so as to work on a toothed wheel. The text is evidently 
defective, and at the end an interpolator has inserted extracts 
about the mechanical powers from Heron’s Mechanics.