450 _ A Demonstration of 
— FAR = AFE, and the triangles CFB, 
AEF similar, consequently AF’: AE= 
50 CB: : CB: BF. Which is Proposi- 
tion 50. 
If CG be drawn perpendicular to EB 
and EL parallel to CG cutting AC in L; 
then because the angle FEB = CAB, AC 
is parallel to EF, consequently LE = co 
57 — EO = BO. Which is Proposition 
the 57th. 
Dracram XII. 
Erect DC perpendicular to the centre 
D of the semi-circle ACB; join AC, CB, 
which produce till it meets a parallel FE 
to CD in F, that cuts the semi-circle in 
H and AC in G; then because AE = GE, 
and FE = EB, and AE. EB = HE’ 
60 = FE. GE, therefore GE : HE: : HE: 
FE. Which is Proposition the 60th. 
——_— 
Prop. LVIII. If BD bisect the vertical angle B of a 
triangle ABC and meet the base in D, and if with either 
of the other angular points A or C as center and the adja- 
cent segment of the base as radius a circle be described to 
cut BD again in E; then I say that BE isto BD as that seg- 
ment used as a radius is to the other. 
Prop. LIX. If BD bisect the vertical angle B of the 
