ERATOSTHENES 
259 
Suppose the frame bounded by the parallels AX, EY. The 
initial position of the triangles is that shown in the first figure, 
where the triangles are A MF, MNG, NQH. 
In the second figure the straight lines AE, DH which are 
parallel to one another are those between which two mean 
proportionals have to be found. 
In the second figure the triangles (except AMF, which 
remains fixed) are moved parallel to their original positions 
towards AMF so that they overlap (as AMF, M'NG, N'QH), 
XQH taking the position N'QH in which QH passes through D, 
and MNG a position M'NG such that the points B, G where 
MF, M'G and NG, N'H respectively intersect are in a straight 
line with A, D. 
Let AD, EH meet in K. 
Then EK:KF= A K: KB 
= FK: KG, 
and EK: KF = AE: BF, while FK :KG = BF: CG; 
therefore AE: BF = BF: CG. 
Similarly BF: CG — CG: DH, 
so that AE, BF, CG, DH are in continued proportion, and 
BF, CG are the required mean proportionals. 
This is substantially the short proof given in Eratosthenes's 
s 2