A history of Greek mathematics - Wilbourhall.org

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THE COLLECTION. BOOK VIII 439
Then he proves that, if we join A B, A B is equal to the length
of the side of the hexagon required.
Produce BC to D so that BD = BA, and join DA. ABD
is then equilateral.
Since EB is a tangent to the segment, AE.EC — EB 2 or
AE: EB = EB : EC, and the triangles EAB, EBC are similar.
Therefore BA 2 : BC 2 = AE 2 : EB* = AE'.EC = 9 : 4
and BC = %BA = §52), so that £6' = 2 CD.
But Ci^= 2C.4 ; therefore AC:CF= DC:CB, and 47), BF
are parallel.
Therefore Itf 7 : AD
= BC.CD = 2 :
1, so that
BF=2AD = 2AB.
Also £FBC= A BDA = 60°, so that ZARF= 120°, and
the triangle J.I?i^is therefore equal and similar to the required
triangle NLO.
Construction of toothed ivheels 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.