2*$ RESOLVTIONES VROBLEMATVM 



m I centrum ducendi arcus circularis HB, atqne in Cf 

 centrum erit defcribcndi arcus alterius circularis AH; 

 Confhtt ita de conftructionis huius ratione Geometrica 

 petfefta; deinde arcus A H perfe&e coincidet cum pila 

 A P, quoniam et ad A P , et ad A H , perpendicularis 

 eft radius A C , qnod fimiii modo de arcu HB proba^ 

 tur. Du&us eft igitnr , quod Blon-Iellus fieri pofife negauit , 

 fuper piias quascunque arcus circularis duplex ? GeometriT 

 ce et omni poplite carens. 



§. XXXIX. Solutio problematis Algebraica haes 

 efie poteft. Sint DCn^, D I =y , et , demiffa per- 

 pendiculari BK, DAz:^, BK~£, DKzz^BDrz^ 

 atque erit , ob triangula fimilia BKD, IGD, BD(?): 

 DIQ)=:BK(^): IG(^- "); etpariterBD(^):DI(j/) 

 = DK(*):DG(¥); vnde GC^x- ? , et CI~ 



= V (/-+■ *' ~ ~ 22 ') ' ob b% ~*- c * ~ **'• Exinde debet 



effe CA=jp + «= C H — C I -f- I B z= V (/+ Jrt 



2 -~^ ) 4- * —J ; aut vero , ponendo m^ze—a , debet 



effe V (y -*-.** — ^—^ ) rr # ~fc-X*- ;/* ; vnde prodit 



\m e — emx 



y ~ em-e + c.x ' CX qU ° pntet ' infinitas "folutiones a& 



mittere hoc. problema , vti fuperius (§. XXIIL) ; qua- 



rum vnicam dabo , affumendo nempe , vti modo. ante 



- m e 

 feei ,x — D C r=r m \ hinc fit ergo y~ ' — ; eft autem 



c - 



ger ; conftru&ionem traditam, ob triangula fimilia. DKB 



et 



