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 DABG. 
Draw the tangent OF to the quadrant at G. With G as 
centre and radius GA draw the circle AEF meeting GF in F. 
Then the sector GA F is equal to the sector ADC (since 
GA 2 = 2 AD 2 , while ¿ACF= \ L 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 GAF). 
Let KL be a (small) fraction, say 1 /nth, of the circum 
ference of the circle KLM, and let HPL be the quadrant of the 
great circle through H, L meeting the spiral in F. Then, by 
the property of the spiral, 
(arc HP): (arc HL) = (arc KL): (circumf. of KLM) 
= 1: n.' 
Let the small circle NPQ passing through P be described 
about the pole H. 
Next let FE be the same fraction, 1 /nth, of the arc FA 
that KL is of the circumference of the circle KLM, and join EG 
meeting the arc ABC in B. With C as centre and GB as 
radius describe the arc BG meeting GF in G. 
Then the arc GB is the same fraction, 1 / nth, of the arc 
CBA that the arc FE is of FA (for it is easily seen that 
LFGE — BBC, while LFGA = \LCPA). Therefore, since 
(arc CBA) = (arc HPL), (arc GB) = (arc HP), and chord GB 
— chord HP.