DIVISION OF A CIRCLE, 229 



Scholium, 



This problem is conflrucled at page 342 of the Appendix to Division of an 

 Simpfon's Algebra, 2d Edition; and at page 140 of his Seleft j^ t ^ V^,* 

 Exercifes, 1ft Edition; but the conftrudions given by that having their 

 able geometrician do not mew the various limits of the quef- ^"'^es^a 

 tion with that degree of perfpicuity which appears in the pre- given ratio, 

 lent method. 



Lemma. 



Let A B C D be a fquare, (See Fig. 4.) from any two ad- 

 jacent fides of which, C B, CD, take the fegments T C, 

 C S, then will the rectangle of the remaining fegments BT 

 X S D = BCTl* + TCxCS-BCxCT-BCxCS. 



Demonjlration, 



Draw S G, T H, parallel to B C, C D, and let them inter,, 

 fea in F ;— 



Then the Wangle FTCS=TCxCS, 



and the reftangle FHAG = BTx S D, 



Butf H A G -f G B C S -f H F S D = the f qua re -ABCD; 



(Euc. l. II.) 



Add FTC S to both, 



ThenFHAG + GBCS-f TCDH = ABCD + FTCS; 



But. C D is equal to B C, 

 Therefore F H A G =ABCD + FTCS-BC x CT- 



BC x CS; 

 That is, B T x S D = Bey +TC x CS-BCxCT- 



B C x C S. Q. E. D. 



Proposition V. Problem. 



To divide A F B, a given arch of a circle, (See Fig. 5.) 

 into two parts, A F, and F B ; fo that the reftangle of their 

 cofines may be equal to a given fquare, k x k ? 



Conjiruclion, 



Join A B, and from the center, O, draw O Z perpendi- 

 cular to A B ; in ZO take Z V equal to the given line, k, 

 and join B V ; draw the diameter, H h, para llel to A B, and 

 divide it in I fo as to make HI x lh = B V) 2 ; from I draw 

 I Q perpendicular to A B, and when produced let it meet the 

 given arch in F -, then will A F, F B, be the required arches. 



PemonJlration 9 



