432 A Demonstration of 
kN = Tk. kT’ = TG* ~Gk?; therefore 
24 TG? = Gk> + Lk . Gk = LG . Gk. 
Which is the twenty-fourth Proposition. 
Since (by the first) Fn: FH:: FD: 
FG::Fn:Ho::FD: DG:: Fn—FD: 
25 Hn —DG or Fn: Hn: : Hn: Gn. Which 
is the twenty-fifth Proposition. . 
Produce ET, ET’ till they cut the per- 
pendicular in A and fF; then the 2 TFA = 
FFT’ and the angle ATF made with the 
tangent is equal to the angle T'T’E in the 
segment — AFE, therefore the triangles 
a TE, FFT’ are similar, therefore FT: 
FY’ = FT :: FT: F4, and ¥72 . FA 
line AE be drawn and therein be taken two points Eand F 
on the same side of A such that the rectangle EAF 
may be equal to the square of AB, and from A any line be 
drawn to meet the circle in C and D, and EC, FD be 
drawn meeting the circle again in G and H; GH being 
drawn will be parallel to AE, 
Prov. XXXII, Through any point A within a circle 
jJeta line be drawn meeting it in Band E, and therein two 
points F and G be taken such on different sides of A that 
the rectangle FAG may be equal to the rectangle BAE, 
and through A any line be drawn meeting the circle in 
C and D, and FC, GD being drawn to meet the circle 
again in H and K; then HK being drawn will be parallel 
to AB. : 
Prop, XXXIII. Let AB ‘be a line without a circle, 
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