[436] 



AC, BL. Here, FL being to FB as 2 AFL to 

 aAFBorEFC, and 2AFLlefsthan AL*-~AFf, 

 by converfion, the ratio of LB to FB will be greater 

 than the ratio of ELC to EFC; therefore, as be- 

 fore, the re&angie under the fine of BAG and the 

 tangent of ABG is greater than that under the fine 

 of BAK and the tangent of ABK. 



Corollary i. 



BF is equal to the tangent of the circle from the 

 point B; therefore, BF is the tangent, and AB the 

 iecant, to the radius AC, of the angle, whofe cofine 

 is to the radius as AC to AB. Therefore, AF is- 

 the tangent, to the fame radius, of half the comple- 

 ment of that angle j and A F is alfo the cofine of 

 the angle BAG to this radius* 



C o r o l. 2. 



The fine of the angle compofed of the comple- 

 ment of AGB, and twice the complement of ABG,. 

 is equal to three times the fine of the complement 

 of AGB. Let fall the perpendicular AH (Fig. 2.), 

 cutting the circle in I ; continue GF to K, and draw 

 AK^ Then BF' = EBC=:GBL. Therefore, 

 GB : BF :: BF : BL, and the triangles GBF, 

 F B L are fimilar. Confequently F L is perpendi- 

 cular to G B, and parallel to AH; whence G H 

 being equal to HL, GM is equal to MF, and 

 M K equal to three times G M. 



Now, the arc IK = 2TC-kGI; and the angle 

 IAK~ 2IAC4-GAI; alfo G Mis to MR as 



the 



