HIPPOCRATES’S QUADRATURE OF LUNES 195 
‘ The fact that this lime (is one which) has its outer circum 
ference less than a semicircle he proves by means of the fact 
that the angle [EKG] in the outer segment is obtuse. 
‘And the fact that the angle EKG is obtuse he proves as 
follows.’ 
[This proof is supposed to have been given by Eudemus in 
Hippocrates’s own words, but unfortunately the text is con 
fused. The argument seems to have been substantially as 
follows. 
By hypothesis, 
Also 
EF 2 =*EK 2 . 
BK 2 > 2BF 2 (this is assumed; we shall 
consider the ground later) ; 
EK 2 > 2 KF 2 . 
EF 2 = EK 2 + \ EK 2 
or 
Therefore 
so that the angle EKF is obtuse, and the segment is less than 
a semicircle. 
How did Hippocrates prove that BK 2 > 2 BF 2 1 The manu 
scripts have the phrase ‘ because the angle at Fis greater’ (where 
presumably we should supply opOrjs, ‘than a right angle’). 
But, if Hippocrates proved this, he must evidently have proved 
it by means of his hypothesis EF 2 = f EK 2 , and this hypo 
thesis leads more directly to the consequence that BK 2 > 2 KF 2 
than to the fact that the angle at F is greater than a right 
angle. 
We may supply the proof thus. 
By hypothesis, EF 2 = f KB 2 . 
Also, since A, E, F, G are concyclic, 
EB. BF = AB.BG 
= KB 2 , 
EF. FB + BF 2 = KB 2 
= § EF 2 . 
or 
It follows from the last relations that EF > FB, and that 
KB 2 > 2BF 2 . 
The most remarkable feature in the above proof is the 
assumption of the solution of the problem ‘ to place a straight 
o 2