194 THE ELEMENTS DOWN TO PLATO’S TIME 
e Let EG he drawn parallel to AB, and let {straight lines) 
he drawn joining K to E and F. 
‘ Let the straight line [KF] joined to F and produced meet 
EG in G, and again let {straight lines) he drawn joining 
B to F, G. 
‘ It is then manifest that BF produced will pass through 
[“fall on”] E [for by hypothesis EF verges towards BJ, and 
BG will he equal to EK.’ 
[Simplicius proves this at length. The proof is easy. The 
triangles FKC, FBG are equal in all respects [Enel. I. 4]. 
Therefore, EG being parallel to KB, the triangles EOF, GDF 
are equal in all respects [Enel. I. 15, 29, 26]. Hence the 
trapezium is isosceles, and BG = EK. 
‘This being so, I say that the trapezium EKBG can be 
comprehended in a circle.’ 
[Let the segment EKBG circumscribe it.] 
‘ Next let a segment of a circle be circumscribed about the 
triangle EFG also; 
then manifestly each of the segments [on] EF, FG will be 
similar to each of the segments [on] EK, KB, BG.’ 
[This is because all the segments contain equal angles, 
namely an angle equal to the supplement of EGK.] 
‘This being so, the lune so formed, of which EKBG is the 
outer circumference, will be equal to the rectilineal figure made 
up of the three triangles BEG, BFK, EKF. 
‘ For the segments cut off from the rectilineal figure, on the 
inner side of the lune, by the straight lines EF, FG. are 
(together) equal to the segments outside the rectilineal figure 
cut off by the straight lines EK, KB, BG, since each of the 
inner segments is \\ times each of the outer, because, by 
hypothesis, EF 2 {= FG 2 ) = %EK 2 
[i.e. 2 EF 2 =ZEK 2 , 
= EK 2 + KB' 2 + BG 2 ]. 
‘ If then 
(lune) = (the three segmts.) + {(rect. fig.) — (the two segmts.)}, 
the trapezium including the two segments but not the three, 
while the (sum of the) two segments is equal to the (sum 
of the) three, it follows that 
(lune) = (rectilineal figure').