[ 926 ] 
In the plain triangle ABC, if on the centers A 
and C be defcribed circles through B, cutting AC in 
Eh I' 5 G, and D, the perpendicular B H being let fall 
on A C, E H x H F is equal toGHxHD. There- 
fore H F : GH ; : HD : HE, and by compofition 
GF : GFI : : ED : HE, and;: ED — GF : EG; 
alfo HF : GF : ; HD : ED, and : : DF : E D- GF. ’ 
Moreover ED is = AB -f AC + CB, EG = 
ED - 2CB, FD--ED — 2 A B, ED — GF 
= EG -f FD = 2ED — 2CB — 2 AB = 2 AC. 
Hence GF is to GH as 2 AC to EG, whence 
GF x EG is = 2 AC xGH; and D II is to E D 
as D F to 2 A C, that is, 2 A C x D H will be — 
E D x D F. 
Thus 2 AC : DG 2 AC x DG : DG’} 
: : 2 AC x GH (E G x GF ) : D GH (GB«») and 
A C x C B : r ad.q ; : 4. E D — C B x i E FT— AC 
: lin. ~ A C h|b 
Likewife 2 A C : D G ( ; : 2 A C x DG : D GO 
:: 2 ACxDH(EDxDF):GDH(DB^)j therefori 
AC x CB : rad.'i ; ; FE D x FE D - A B : cof. F A C B| q . 
in th e laft place F E D x ED — A B 
: t E D C B XJ-ED — AC ; ; rad.* 1 : tang. F A C B] q . 
J * This 
