£30 DIVISION OF A CIRCLE. 



Demonfiration. 

 DivMon*fan Let FI meet the circle again in G, draw the diameters 



arch of a circle A j^ B j^ and the fines F S, F T ; 

 having their ' Then the cofine O S = O A — A S, 



fines, orcofincs, an d the cofine O T = O B -" B T = O A - B T ; 



gfvenratio! Hence SO X OT = AO^-f- AS xBT-AOxAS- 

 AOx BT; 



But A S x B T = Fq)*, by Prop. II. 



Therefore, : 



SO x OT=Xo) 2 4-Tq\»~ AOxAS-AOxBT. 

 Again, KS = 2AO~AS, 

 and DT = 2AO-BT; 

 Hence, K S x DT = 4 AOY +ASxBT-2AOx 

 AS-2AOxBT; 



But K S x D T = q"g)% by Cor. 3. Prop. II. 



Theref ore, 



4TA15V+TQV- 2AOxAS-2AOxBT=QG>;, 



, FQl 2 Q"G> 



Hence, AO x A S -f AOxBT = 2XoV + --~ - -^— t 



ButS0xOT = AO 2 +FQ*-AOxAS-AOxBT 



FQ = FI-IQ = BP-OZ, 



andGQ = GI-HQ:=BV-f-OZ; 



FQ a GO* s s 



Hence t-±- + H^L = BVY + 02) s ; 



2 2 } 



• Confequently, S O x O T ~!Tv]* + O^) 2 - A3] * ; 



But a51* -~ol£ 8 = b oy -"o2|* ="bB* 5 



Therefore S O x OT:=fBV\ 2 - "BZ] * zzVl!\ *, 

 (Euc. 47. I.) r= k x k, by conftruction. Q. E. D. 



Limitation. — If Z V be greater than Z O, B V will be 

 greater than BO; i. e, F Q will be greater than H h, which 

 is impoffible, Euc. 15. III. therefore Z V, ork, cannot ex- 

 ceed Z O. 



Proposition VI. Problem. 



To divide A F B, a given arch of a circle, (See Fig. 6.) 

 into two parts, fo that the fum of their verfed fines may be 



equal to a given right line, u f 



ConftruBion. 



