THE AN A LEMMA OF PTOLEMY 
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vertical; and draw FG perpendicular to BA, and ET to OZ. 
Join EG, and we have FG = SR, GH = FS = ET. 
We now represent SF in a separate figure (for clearness 5 
sake, as Ptolemy uses only one figure), where B'Z'A' corre 
sponds to BZA, P' to P and O'M' to OM. Set off the arc 
P'S' equal to CS (= 90° — i), and draw S'F' perpendicular 
to O'M'. Then S'M'= SM, and S'F'= SF; it is as if in the 
original figure we had turned the quadrant MSC round MO 
till it coincided with the meridian circle. 
In the two figures draw IFK, I'F'K' parallel to BA, B'A', 
and LFG, L'F'G' parallel to OZ, O'Z'. 
Then (1) arc ZI = arc ZS — arc (90° — $F), because if we 
turn the quadrant ZSV about ZO till it coincides with the 
meridian, S falls on I, and V on B. It follows that the 
required arc >S'F = arc B'l' in the second figure. 
(2) To find the arc VC, set off G'X (in the second figure) 
along G'F' equal to FS or F'S', and draw O'X through to 
meet the circle in X'. Then arc Z'X' — arc VC; for it is as if 
we had turned the quadrant BVC about BO till it coincided 
with the meridian, when (since G'X = FS = GH) H would 
coincide with X and V with X'. Therefore BV is also equal 
to B'X'. 
(3) To find QG or ZQ, set off along T'F' in the second figure 
T'Y equal to F'S', and draw O'Y through to Y' on the circle. 
Then arc B'Y' — arc QG; for it is as if we turned the prime 
vertical ZQG about ZO till it coincided with the meridian, 
when (since T'Y = S'F' = TE) E would fall on Y, the radius 
OEQ on O'YY' and Q on ,F'. 
(4) Lastly, arc BS = arc BL = arc B'L', because S, L are 
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