NICOMEDES 
261 
Then from the point F draw FHK cutting GH and EC 
produced in H and K in such a way that the intercept 
UK = CF = AD. 
(This is done by means of a conchoid constructed with F as 
pole, CH as £ ruler and ‘ distance ’ equal to AD or GF. This 
M 
conchoid meets EG produced in a point K. We then join FK 
and, by the property of the conchoid, HK — the ‘ distance ’,) 
Join KL, and produce it to meet BA produced in M. 
Then shall GK, MA be the required mean proportionals. 
For, since BG is bisected at E and produced to K, 
BK.KG + GE 2 = EK\ 
Add EF 2 to each; 
therefore BK. KC + CF 2 = KF 2 . (1) 
Now, by parallels, MA : AB = ML: LK 
= BG: OK. ' 
But AB = 2 AD, and BG = \ GG; 
therefore ( MA : AD = GG: GK 
= FH: HK, 
and, componendo, MD: DA — FK: HK. 
But, by construction, DA = HK; 
therefore MD = FK, and MD' 1 = FK 2 .