232 DIVISION OF A CIRCLE." 



Conjlruction, 

 Ptvifion of an Draw the radii A O, OB, and the tangents AG, BE, in 



fntotwoparts^ Wh ' ch take A S ' B T ' each e( l uaI t0 the half ot " W J dfaW 



having their ' S N, T N, parallel to AO, OB; and through their inter- 

 fmes, or cofmes, feflion, N, draw N F, parallel to ST, to meet the arch in 

 given ratio.' ?» then A F, F B, are the parts required. 



Demonftration. 



Draw F K, F M, parallel to N S, N T, and let them meet 

 S T in K, M, and A G, B E, iii P. R ; then it is eafily 

 proved that the triangles K F M, S N T, are equal and 

 fimilar, and that K M zz S T ; confequently SKrTM. 



But the angles K P S, M R T, are right, being equal to 

 the angles O A G, O B E, by conftru&ion ; and the angles 

 K S P, M T R, are equal ; therefore the triangles PSK, 

 R T M, are equiangular, they are therelore equal, (Euc. 4. 

 VI.) becaufeSKzzTM; confequently S P zz R T ; there- 

 fore A P -f B R zz A S + B T zz w. 



But the fum of the fines of A F, FBzzAP + BR; this 

 fum is therefore equal to w. 



Limitation. — Join A B, which will be parallel to S F, alfo 

 let the radius O C bifed the angle A O B, when properly 

 produced, or not, it will pafs through the point N. Now if 

 N be in O C produced, N F, being parallel to ST, or A B, 

 will not meet the circle ; on the other hand, if N lie beA 

 tween O and A B, F will be in the oppofite fegment of the 

 circle, confequently the construction is impoffible, unlefs N 

 fall between C and the line A B, or in the verfed fine of 

 half the given arch : Thefe things being premifed, it will be 

 eafily perceived that the fine of the arch A F B is the lefs 

 limit of the problem, and twice the fine of A C its greater 

 Y-nit. 



Concerning 



