DIVISION Or A CIRCLE* . 22? 



Therefore, as AF : FB: : PF : FQ. Divifion of an 



Alfo, the triangles QFA, RFB, are equiangular, for the ™££*™£ 



fame reafons. having their 



Therefore, as A F : F B : : F Q : F R. fincs > ° r cofines, 



Confequently, as P F : F Q : : F Q : F R, (Euc. 11. V.) g [ ve v n ™* w * 

 And P.F x F R = F Q>, (Euc. 14. VI.) 



Q. E. 1° D. 

 Again, by the fame triangles, as F A : F B : : A P : B Q, 

 and asFA:FB::AQ:BR; 

 hence, as A P : B Q : : A Q : B R, 

 Whence A P x BR = B Q x A Q; 

 Q. E. 2° D. 

 Corol. I. Produce the perpendicular FQ till it meets the 

 circumference again in G, and PA x RB = FQ x QG: 

 For PA x RB = AQ x QB by the propofition; but 

 AQ x QB = FQ x QG, (Euc. 35. III.) 



Corol. 2, If the lines P F, R F, meet the circle again in M 

 and N, then will PMxRN =*Q"gV : 



For~AP} a = F P x P M, and~BR) 2 = F R x R N, (by 

 Euc. 36. III.) 

 Therefore, as~AP) 2 :FP*PM::FRxRN :""br\ 4 : 



But~XP\ 2 :FQxQG::FQ x QG C$VK 

 by Corol. 1 . 



And P F : F Q : : F Q : F R, by the propofition. 

 Therefore, 



P F x P M : F Q x P M : : F Q x R N : F R x R N. 

 Hence, 



FQxQG:FQxPM::FQxRN:FQxQG. 

 Confequently, PM x R& = QG>. 

 CoroL 3. Draw the diameters A K, B D, and make F S, 

 FT perpendicular to A K, B D ; then A K x B T (the red- 

 angle of the verfed fines) = Fq) 2 ; S F x FT (the red- 

 angle of the fines) = A Q x QB; andSKxTD (the 

 re&angle of the fupplementary verfed lines) = QG) a . Thefe 

 things follow from Props. I. and II. 



Proposition III. Problem. 



« To divide a given arch of a circle (A B) into two parts (A F, 

 f B), fo that the redangle of their verfed fines (A S, B T} 

 may be equal to a given magnitude, or fquare, (m x mj. 



Q 2 Conjlruaion. 



