HIPPOCRATES’S QUADRATURE OF LUNES 191 
which are one third of the circles respectively, are from this 
point of view quite harmless. 
Only the transition to the view of similar segments as 
segments ‘ containing equal angles ’ remains to be explained. 
And here we are in the dark, because we do not know how, for 
instance, Hippocrates would have drawn a segment in one 
given circle which should be ‘ the same part ’ of that circle 
that a given segment of another given circle is of that circle. 
(If e.g. he had used the proportionality of the parts into which 
the bases of the two similar segments divide the diameters 
of the circles which bisect them perpendicularly, he could, 
by means of the sectors to which the segments belong, have 
proved that the segments, like the sectors, are in the ratio 
of the circles, just as Rudio supposes him to have done; and 
the equality of the angles in the segments would have followed 
as in Rudio’s proof.) 
As it is, I cannot feel certain that the sentence Sib kou ktX. 
‘ this is the reason why similar segments contain equal angles ’ 
is not an addition by Simplicius. Although Hippocrates was 
fully aware of the fact, he need not have stated it in this 
place, and Simplicius may have inserted the sentence in order 
to bring Hippocrates’s view of similar segments into relation 
with Euclid’s definition. The sentence which follows about 
‘angles of’ semicircles and ‘angles of’ segments, greater or 
less than semicircles, is out of place, to say the least, and can 
hardly come from Eudemus. 
W e resume Eudemus’s account. 
‘ After proving this, he proceeded to show in what way it 
was possible to square a lune the outer circumference of which 
is that of a semicircle. This he effected by circumscribing 
a semicircle about an isosceles right-angled triangle and 
(circumscribing) about the base [= describing on the base] 
a segment of a circle similar to those cut off by the sides.’ 
[This is the problem of Eucl. III. 33, 
and involves the knowledge that similar 
segments contain equal angles.] 
‘Then, since the segment about the 
base is equal to the sum of those about 
the sides, it follows that, when the part 
of the triangle above the segment about the base is added 
to both alike, the lune will be equal to the triangle.