HIPPOCRATES’S QUADRATURE OF LUNES 193 
[CD] which is subtended 1 by the said (greatest) side [BD] 
together with the diagonal [BG] ’ [i.e. BD 2 < BG 2 + CD 2 ]. 
‘Therefore the angle standing on the greater side of the 
trapezium [Z BCD] is acute. 
‘ Therefore the segment in which the said angle is is greater 
than a semicircle. And this (segment) is the outer circum 
ference of the lune.’ 
[Simplicius observes that Eudemus has omitted the actual 
squaring of the lune, presumably as being obvious. We have 
only to supply the following. 
Since BD 2 = 3 BA 2 , 
(segment on BD) = 3 (segment on BA) 
= (sum of segments on BA, AC, CD). 
Add to each side the area between BA, AG, CD, and the 
circumference of the segment on BD, and we have 
(trapezium ABDG) = (lune bounded by the two circumferences).] 
L 
‘A case too where the outer circumference is less than 
a semicircle was solved by Hippocrates, 2 who gave the follow 
ing preliminary construction. 
‘ Let there he a circle with diameter AB, and let its centre 
he K. 
e Let CD bisect BK at right angles; and let the straight 
line EF he so placed between CD and the circumference that it 
verges towards B [i.e. will, if produced, pass through B], while 
its length is also such that the square on it is 1-| times the square 
on (one of) the radii. 
1 Observe the curious use of vnordveiv, stretch under, sub&nd. The 
third side of a triangle is said to be ‘ subtended ’ by the other two 
together. 
2 Litei'ally ‘ If (the outer circumference) were less than a semicircle, 
Hippocrates solved (KareirKevaa-ev, constructed) this (case). 1 
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