442 A Demonstration of 
AC respectively, and then another cir- 
ele through G, F, C cutting AG in b, 
and jom LF, GC. Then the zAFL = 
AGC, ALF = ACG = the supplement 
of ABG, therefore ABG = DLF, and 
ADF — BAG by construction ; conse- 
quently the triangles DLF, ABG are 
similar, and AG: BA:: DF ~ AE: DL 
and AG: AC: : AF; AL; but DL + AL 
— AD, therefore BA. AE + CA. AF 
~ AG. (DL+ AL) = AG .AD=GD. 
36 AD + AD? = BD. DC + AD*. Which 
is Proposition 36. 
Dracram VII. 
The construction being as in the Propo- 
sition, through B draw BF, join FC, 
make EH = EF, and join AH; then by 
liypothesis AC: FC: : FC: BC, there- 
drawn to meet the circumference in D, anda perpendicalar 
DE be demitted; then I say that the square of the line 
AC is equal to the square of the line CD together with the 
rectangle under the sum of the distances of C from A and 
C from Band the line AE, when C is taken in the diameter 
AC produced; but equal to the square of CD together 
with the rectangle under the difference of the distances of 
C from A and'C from Band the same line AE, when vain is 
taken in the diameter itself. 
