il6 ESSAYS AND OBSERVATIONS 



very excentric ; granting the qua- 

 drature of the circle, it is required 

 to draw a Hne, DE, meeting the 

 circle in E, fo that the femicircle 

 may be to the fedlor BDE in a 

 given ratio, fuppofe that of^ to (j. 



Suppose the problem folved ; and let the 

 femicircle be to the fedor BCF as ^ is to q-, 

 join DF, and draw CG parallel to DF meet- 

 ing FG a tangent to the circle at F in G ; 

 join DG, and let CG, DG meet the circle 

 in H, K : join DH, FH; draw CM parallel 

 to DH meeting HM a tangent to the circle 

 at H in M J draw DL perpendicular to CH 

 meeting CH in L ; join LE, and let FN 

 perpendicular to CH meet CH in N. 



Because the femicircle is to the fedlor 

 BDE as/> is to q, that'is, as the femicircle to 

 the fedor BCF, the fedor BDE will be equal 

 to the fe^or BCF : and becaufe the fedor 

 BCH is common to both, the fpace HCDE 

 will be equal to the fedor HCF ; but be- 

 caufe CH, DF are parallel, the triangle 

 PCH is equal to the triangle FCH 3 there- 

 fore 



