THE COLLECTION. BOOK VII 
415 
The problem is reduced to a problem contained in Apollo 
nius’s Determinate Section thus. 
Suppose the problem solved by the semicircle DEF, BE 
being equal to AD. Join E to the centre G of the semicircle 
DEF. Produce DA to H, making HA equal to AD. Let K 
be the middle point of DC. 
Since the triangles ABC, GEG are similar, 
AG 2 : GG 2 = BE 2 :EC 2 
= AD 2 : EG 2 , by hypothesis, 
= AD 2 : GG 1 — DG 2 (since DG = GE) 
= AG 2 —AD 2 : DG 2 
= HG.DG-.DG 2 
= EG: DG. 
Therefore 
HG:DG = AD 2 :GC 2 -DG 2 
= AD 2 : 2 DG . GK. 
Take a straight line L such that AD 2 = L . 2DC; 
therefore HG : DG = L: GK, 
or HG . GK = L . DG. 
Therefore, given the two straight lines HD, DK (or the 
three points H, D, K on a straight line), we have to find 
a point G between D and K such that 
HG. GK = L. DG, 
which is the second epitagma of the third Problem in the 
Determinate Section of Apollonius, and therefore may be 
taken as solved. (The problem is the equivalent of the