Lawson's Geometrical Theorems. 443 
fore the 7 F BC = AFC, CFB = FAC 
= HFA, but the -AHF — BFH — FBC 
= AFC, therefore HAF = FCA, and the 
triangles HFA, FCA similar, conse- 
quently HF = 2EF: AF:; AF : AC, 37 
Which is the thirty-seventh Proposition. 
Moreover, the two triangles AFC, AFB 
have the angle A common, and the angle 
AFC = FBC = A + AFB, and taking 
FI = BF, the angle AIF=FBC = AFC, 
therefore AC: FC: : AF: FI = FB:: 
(drawing GK parallel to FB) AG : GK. 40 
Which is Proposition Av. 
As to the two intermediate Propositions 
viz. the 38th and 39th; since the double 
area of any regular polygon is = the con- 
tinual product of the radius of the in- 
scribed circle, the number of sides and. 
the length of one side; and if it be di- 
vided into triangles equinumerous with 
Prop. XLVIJI. If ‘from one angle A ofa rectangle 
ABCD a line be drawn to cut the two opposite sides BC, 
DC, the former in F, and the latter produce in E; then I 
say that the rectangle EAF is equal to the sum of the rect 
angles EDC, CBF. ; 
Prop. XLVI. If a rectangle be inscribed in a right- 
angled triangle, so that one of its angles coincide with the 
angle of the triangle; then I say that the rectangle under 
the segments of the hypothenuse is equal to the sum of 
3K 2 
