428 A Demonstration of 
18 therefore TH? =iH .bH. Whieh is the 
eighteenth Proposition. 
_ Or describing the semi-circle that passes 
through F, T, », cutting FE int, we have 
Fo: FT:: FT: oe and FG : Fg : : Ft ;, 
Fo; «. FT? = - Ft. — ¥ Pro- 
position 18. 
Join d t cutting TT’ in z, aT souidbinig 
the circle in T being first drawn, then Fd 
= dT the angle dFT = FTT’, the trian- 
gles FdT, TFT’ equiangular, and Fd: 
FT: : FT : TT’; therefore Fd . TT’ =FG 
-Fn— Fe. Feand Fd: Ft :: Fg: TT 
:: ag: gt, zg. TT’ = Fg. gt = (because 
T,t, T’, Fare ma circle) Tg.gT’= Tg. 
Prov. XXIII. If AB be the diameter of a circle and 
CD perpendicular thereto meeting it in C, and from the 
points A and B be inflected AE, BE to any point E in 
the circumference, meeting CD in F and G; I say the 
rectangle GCF is equal to the rectangle ACB. 
Prop. XXIV. In AB the diameter of a circle let two 
points C and D be taken such that AC; CB: : AD: DB, 
and the point D be within the circle, and DE be perpendi- 
cular to AB, meeting the circumference in E and F, and 
Jet through C any line be drawn meeting the same inG 
and H, and from the points G and H let GN, HN be in- 
flected to any point in the same N, andlet them meet DE 
in M and L; [I say the rectangle LDM is equal to the 
square of DE. : 
Prop. XXV. Let AB he the diameter of a circle and 
