420 A Demonstration of 
again in h’, EH’, and TG to T’; join FT’, 
which must be equal to FT, and join EE, 
cutting FH in f; then because the angle 
hEh’ = hE‘h’, and EhE’ = El’£,, therefore 
WwhE’ = EE‘, hE’ = bE, bh’ ’ = hE; also 
because hGE = hGH’, ... Gh = Gh’, GE 
= GE’; but GT — GT’, therefore KE’, as 
also hh’, is parallel to T'T’; and because 
FhE is a right line, .-. Fh’E’is aright line 
and the triangles TGH, T’GEH’ equal and 
similar, and therefore gEG, g’H’G are so; 
consequently the angle EGg = H’Gg’ = 
hGg, and gE : hg: : EG: Gh= Gh’:: 
Ef : hq: : EE’: hh’:: FE: FH. Hence 
if in DH the diameter of a circle, two 
Dsuch that AC: CB: : AD: DB, and AE, BF be drawn 
perpendicular to AB, and through the point C be drawn 
any line to meet AE, BF in G and H, and DG, DH be 
joined; then I say that DG: DH: : GC: CH. 
Prop. VII. Hin the diameter of a circle AB be taken 
any point C, and CDE be drawn meeting the circle in D 
and E, and DF be perpendicular to AB meeting it in F, and 
the circle again in G, and EG be joined meeting AB in H; 
I say that AC: CB: : AH: HB. 
Also, as the converse, that if in the diameter AB two 
points be taken as C and H such that AC: CB :: AH: HB, 
and from the points C and H two lines CE, HE be inflected 
to any point of the circumference E meeting the same 
again in D and G; when DG is drawn, it will be perpen- 
dicular to AB. 
