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