APOLLONIUS, HERON, PHILON OF BYZANTIUM 263 
and move it about D until the edge intersects AB, AC pro 
duced in points (F, G) which are equidistant from E. 
Or (Pinion) place a ruler so that it passes through D and 
turn it round D until it cuts AB, AG produced and the circle 
about ABDG in points F, G, H such that the intercepts FI), 
HG are equal. 
Clearly all three constructions give the same points F, G. 
For in Pinion’s construction, since FD — HG, the perpendicular 
from E on DH, which bisects DH, must also bisect FG, so 
that EF = EG. 
We have first to prove that AF. FB = AG. GC. 
(a) With Apollonius’s and Heron’s constructions we have, if 
K be the middle point of AB, 
A F. FB + BK 2 = FK 2 . 
Add KE 2 to both sides ; 
therefore AF.FB + BE 2 = EF 2 . 
Similarly A G. GC + CE 2 = EG 2 . 
But BE = CE, and EF = EG ; 
therefore AF. FB = AG. GC. 
(b) With Pinion’s construction, since GH = FD, 
HF.FD = DG.GH. 
But, since the circle BDHG passes through A, 
HF. FD = AF. FB, and DG. GH = AG. GC; 
therefore AF .FB = AG . GC. 
Therefore FA : AG = G G : FB. 
But, by similar triangles, 
FA : AG — DC : CG, and also = FB : BD ; 
therefore DC : CG = CG : FB = FB : BD, 
or AB : CG = CG : FB = FB : AG. 
The connexion between this solution and that of Menaech- 
mus can be seen thus. We saw that, if a:x = x:y = y :b, 
x 2 — ay, y 2 = hx, xy = ah, 
which equations represent, in Cartesian coordinates, two 
parabolas and a hyperbola. Menaechmus in effect solved the