APPENDIX 
559 
But (arc A/S'P) = OT, by hypothesis; 
therefore it was necessary to prove, alternando, that 
(3) F'R: (arc RP) < RO : OT, or PO : OT, 
i.e. < PM: MO, where OM is perpendicular to SP. 
Similarly, in order to satisfy (2), it was necessary to 
prove that 
(4) FQ: (arc PQ) > PM: MO. 
Now, as a matter of fact, (3) is a fortiori satisfied if 
F'R: {chord RP) < PM: MO ; 
but in the case of (4) we cannot substitute the chord PQ for 
the arc PQ, and we have to substitute PG', where G' is the 
point in which the tangent at P to 
the circle meets OQ produced; for 
of course PG' > (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 
cutting AB in F, and if PB produced cuts OT, the parallel to 
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: PB diminishes continually, 
while it is always less than OB : 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' always lies on the side 
of TP towards 0.