434 A Demonstration of 
DHG, GH is parallel to AB. Take FY 
= Fl, then F’ . FK = FT? = FD . FH 
— Fq . FN, two lines FN’, KN’ being in- 
‘flected to any point N’ in the circle cut- 
ting it ing and T’, draw Iq cutting the 
circle again in T, then the points  K q 
N’ are in a circle because Fl’, FK = Fq. 
. FN’, therefore the angle Fl’q = FN’K 
= q TT’. consequently 'T'T’ is parallel to 
FK. Hence if ET touch a circle in T, 
and any line l’F be drawn, and therein be 
taken two points I’ and K on the same side 
of F such that the rectangle /FK may be 
equal to the square of FT, and from F 
any line be drawn to meet the circle in q 
and N’, and lq, KN’ be drawn meeting 
the circle again in T and T’, TT’ being 
31 drawn will be parallel to ’'F. Which 1s 
Proposition the 31st. 
circle in E and F, and BE, BF be drawn meeting the circle 
againin Gand H; then the points A, G, H, are in a right 
line. 
Prop. XXXV. From the extremes of AB let two lines 
AC, BD be drawn to touch a circle in Cand D, and in AB 
let a point E be taken onthe same side of A with B such 
that the rectangle BAE may be equal to the square of AC, 
and also in AB another point F on the same side of B with 
‘E suchthat the rectangle EBF may be equal to the square 
of BD, and through A any line be drawn meeting the 
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