430 A Demonstration of 
mE — ,GE= GEE’ = GHE= thE, 
therefore 2n bisects the angle hnE, and be- 
cause nh = nH, therefore ht = tH is per- 
pendicular to na, therefore the same circle 
passes through h, and ha = Ea is perpendi- 
cularto hn, and Ge .e*=hg .gH =Teg. 
 gTY, add Gg* and Gg . GAa=Tg. gT’ 
21 + Gg? = TG*. Which is Proposition 
21st. both cases. 
Dracram III. 
Through h’ draw EK cutting the per- 
pendicular through F in K, and produce 
FE’ till it cuts it in 1, then the triangles 
ing the circumference in Dand N, in CD let be taken 
two points E and F on the same side of C with D such 
that the rectangle ECF may be equal to the square of CD, 
and from the points E and F let EG, FG be inflected to 
any point Gin the circumference, meeting the same in H 
and K, and let HK when drawn meet the diameter AB. in 
L; then I say that AL: LB: : AC : CB. 
Prove. XXVIII. In AB the diameter of a circle pro- 
duced let be taken the point C, and CD be perpendicular 
to AB, and therein be taken two points E and F on differ- - 
ent sides of C such that the rectangle ECF, may be equal 
to the rectangle ACB, and from the points E and F let 
EG, FG be inflected to any point G in the circle, meeting 
the samie in H and K, and Jet HK when drawn meet 
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