153 



in V. Then by the lemma it eafily appears that K, M, C are in one 

 right line, and alfo that m is a right angle. 



Becaufe FCK and fCk are each right angles, therefore ¥f = K^. 

 Alfo if Fw be drawn parallel to the tangent at E, the triangle F/w 

 is ultimately fimilar to the triangle CLK ; for CL is parallel to the 

 tangent at E, becaufe FCK is a right angle, and therefore CE and 

 CL are conjugate to each other ; and alfo CK is parallel to a tang. pig. ^j, 

 at F. 



Therefore ultimo Ff-Kk : :Fzu ::CK: CL=CM by Iemma= 



Alfo ultimo' Mv : Kk I : CM : CK 



confequently ultimo M-y^Fw. But ultimo 

 Mv*=TM^tm+a.rc Tt and Fw = Ee, therefore 



ultimo Tt — Ee = tm-MT or ultimo the increment of BT — increment 

 of AE = increment of tang. MX. Therefore as thefe magnitudes begin 

 together the arc BT— arc AE = T M. (^ E D. 



Cor. If CO be taken a mean proportional between AC and CB, 

 and be produced to meet the circle in F, then drawing the ordinate 

 FG, the point of interfeftion E will divide the elliptic quadrant, fo 

 that BE— AE = AC— BC. 



Demonjiration. 



By this conflruction CE = the femiconjugate to CO (cor. to lemma) 

 and therefore CE=CT, and therefore BE=BT. Whence TM=tang, 

 at E= (by cor. lemma) AC— BC. Confequently by the theorem, arc 

 BE— arc AE=AC— BC. q^ E. D. 



Vol. IX. ( U ) 



• Let the tang. TM, tm interfea in n. Then M-b=MT— t.T=MT+T«— »«. F'g- S- 

 But ultimo 'vn=:mn=tm—tii therefore ultimo Mv=TM—m+Tn+nt=\iltimo TM— 

 /m+arc Tt. 



