A history of Greek mathematics - Wilbourhall.org

)
THE COLLECTION. BOOK IV 383
a certain ratio shown in the second figure where ABC is
a quadrant of a circle equal to a great circle in the sphere,
namely the ratio of the segment ABC to the sector DABC.
Draw the tangent CF to the quadrant at C. With C as
centre and radius CA draw the circle AEF meeting CF in F.
Then the sector CAF is equal to the sector ADC (since
CA 2 = 2 AD 2 ,
while Z ACF = \ Z ADC).
It is
required, therefore, to prove that, if S be the area cut
off by the spiral as above described,
S: (surface of hemisphere) = (segmt. ABC) :
(sector CAF).
Let KL be a (small) fraction, say I /nth, of the circumference
of the circle KLM, and let HPL be the quadrant of the
great circle through H, L meeting the spiral in P. Then, by
the property of the spiral,
(arc HP) :
(arc HL) = (arc KL) :
= l:n.
(circumf . of KLM
Let the small circle NPQ passing through P be described
about the pole H.
Next let FE be the same fraction, \/nth, of the arc FA
that KL is of the circumference of the circle KLM, and join EC
meeting the arc ABC in B. With C as centre and CB as
radius describe the arc BG meeting CF in G.
Then the arc CB is the same fraction, 1/^th, of the arc
CBA that the arc FE is of FA (for it is easily seen that
IFCE = \LBDCy while Z FCA = \LCDA). Therefore, since
(arc CBA) = (arc HPL), (arc CB) = (arc HP), and chord CB
= chord HP.