A history of Greek mathematics - Wilbourhall.org

APPENDIX 559
But (arc ASP) = OT,by hypothesis ;
therefore it was necessary to prove, alter nando, that
(3) F'R :
(arc RP) < RO :
i.e.
OT, or PO : OT,
< PM:MO, where OM is perpendicular to SP.
Similarly, in order to satisfy (2), it was necessary to
prove that
(4)
FQ:(sxgPQ) > PM:MO.
Now, as a matter of fact,
F'R :
but in the case of (4)
(3) is a fortiori satisfied if
(chord RP) < PM:MO
;
we cannot substitute the chord PQ for
the arc PQ, and we have to substitute PG\ where G' is the
Fig. 1.
point in
which the tangent at P to
the circle meets OQ produced ; for
of course PG f > (arc PQ), so that (4)
is a fortiori satisfied if
FQ:PG'>PM:MO.
It is remarkable that Archimedes
uses for his proof of the'two cases Prop.
8 and Prop. 7 respectively, and makes
no use of Props 6 and 9, whereas
the above argument points precisely to the use of the figures
of the two latter propositions only.
For in the figure of Prop. 6 (Fig. 1), if OFP is any radius
PB produced cuts OT, the parallel to
cutting AB in F, and if
AB through 0, in H, it is obvious, by parallels, that
PF : (chord PB) = OP :
PH.
Also PH becomes greater the farther P moves from B
towards A, so that the ratio PF :
while it is always less than OB :
PB diminishes continually,
BT (where BT is the tangent
at B and meets OH in T), i.e. always less than BM : MO.
Hence the relation (3)
is always satisfied for any point R' of
the spiral on the backward ' ' side of P.
But (3) is equivalent to (1), from which it follows that F'R
is always less than RR', so that R f always lies on the side
of TP towards 0.