A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

ON SPIRALS 71
Let OF meet the spiral in Q f .
Then we have, alternando, since
FQ:QO = PQ:OU
PO = QO,
< (arc PQ) : (arc ASP), by hypothesis and a fortiori.
Componendo, FO :Q0 < (arc ASQ) :
< OQ':OP.
(arc ASP)
But QO = OP ; therefore FO < OQ' ;
which is impossible.
Therefore OT is not greater than the arc ASP.
*
72 ARCHIMEDES
Let OF'G meet the spiral in R'.
Then, since
F'R:RO= GP:OV
PO = RO, we have, alter nando,
> (arc PR) : (arc ASP), a fortiori,
whence F'O : RO < (arc ASR) :
< OR: OP,
(arc ASP)
so that F'O < OR'; which is impossible.
Therefore OT is not less than the arc ASP. And it was
proved not greater than the same arc. Therefore
0T= (arc ASP).
As particular cases (separately proved by Archimedes), if
P be the extremity of the first turn and c 2
the circumference
of the first circle, the subtangent = c x
;
if P be the extremity
of the second turn and c 2
the circumference of the 'second
circle', the subtangent = 2c 2 ; and generally, if c n be the
circumference of the nth circle (the circle with the radius
vector to the extremity of the nth turn as radius),
II.
Next suppose, if possible, that OT < arc ASP.
Measure OF along OT such that OV is greater than OT but
less than the arc ASP.
Then the ratio PO : OV is less than the ratio PO : OT, i.e.
than the ratio of \PS to the perpendicular from on PS;
therefore it is possible (Prop. 8) to draw a straight line OF'RG
meeting PS, the circle PSA, and the tangent to the circle at P
in F\ R, G respectively, and such that
F'R:GP±=PO:OV.
the subtangent
to the tangent at the extremity of the nth turn = nc n
.
If P is a point on the nth turn, not the extremity, and the
circle with
as centre and OP as radius cuts the initial line
in K, while p is the circumference of the circle, the subtangent
to the tangent at P = (w,— l)p + arc iTP (measured
'
forward ')}
The remainder of the book (Props. 21-8) is devoted to
finding the areas of portions of the spiral and its several
turns cut off by the initial line or any two radii vectores.
We will illustrate by the general case (Prop. 26). Take
OB, OC, two bounding radii vectores, including an arc BG
of the spiral. With centre and radius OC describe a circle.
Divide the angle BOG into any number of equal parts by
radii of this circle. The spiral meets these radii in points
P, Q ... F, Z such that the radii vectores OB, OP, OQ ... OZ, OC
1
On the whole course of Archimedes's proof of the property of the
subtangent, see note in the Appendix.