Lawson's Geometrical Theorems. 419 
points D and n, or Fn: FH: : FD: FG. 1 
Which is the first Proposition. 
On the centre n, describe the semi-circle 
DEH, join FE, cutting the circle again 
in h, erect Gg perp. to FH, cutting FE 
in g, and the circle in T ; then since 
FG : FD: : FH: Fn, therefore by divi- 
sion, FG: FD: :GH: Dn = Hh, and 
FG :GH:: DG: Gn, or DG.GH = 
FG . Gn, but DG .GH = TG’, therefore 
FG . Gn = TG’; consequently the points 
F, T, n are in a semi-circle, and FT a tan- 
gent tothe circle DEH. Which is the 15 
Jifteenth Proposition. 
Produce EG, hG till they cut the circle 
drawn parallel to BC, and any line drawn through D to 
meet A B, AC, AE in F, G, H; thenIsay GD: DF:: 
GH: HF, 
Prop. IV. If in AB the diameter of a circle two points 
C and D be taken such that AC: CB: : AD: DB, and 
through the point D any line be drawn to meet the circle 
in E and F, and CE, CF be joined; then I say EC : CF 
:: ED: DF. 
Pror. V. If the base BC of a triangle be bisected in D, 
and through the vertex A a parallel thereto be drawn, and 
from D a perpendicular to BC be drawn to meet the pa- 
rallel in E, and through D any line be drawn to meet AB, 
AC in F and G, and EF, EG be joined; then I say EF : 
EG: : FD: DG. 
Psor. VI. If in the line AB be taken two points C and 
3G2 
