198 On the ORIGIN and 



lied ; the inveftigation depending on a lemma fimilar to that 

 which is prefixed to the preceding propofition. 



LEMMA II. Fig. 12. 



If two triangles ABC, DEF, fimilar to a given triangle, be 

 placed with their angles on three ftraight lines given in po- 

 fition, fo that the equal angles in both the triangles may 

 be upon the fame ftraight lines, the ratios of the fegments 

 of thefe ftraight lines, intercepted between the two tri- 

 angles, that is, of AD, BE and CF, are given *. ' 



PROP. 



* Demonstration. Complete the parallelogram under AC and AD, -viz. AG 



and on DG defcribe the triangle DGH, fimilar and equal to the triangle ABC. Join 

 FG, BH and HE. Through G alfo, draw GK, equal and parallel to HE, and join CK; 

 CK will be equal and parallel to BE, and the triangle CGK equal to the triangle BHE. 

 The angle GCK is therefore given, being equal to the given angle HBE ; and the angle 

 GCF being given, the angle FCK is alfo given. 



The triangles DHE, DGF are fimilar; for the angles FDE, GDH being equal, the 

 angles FDG, EDH are likewife equal ; and alfo, by fuppofition, FD being to DE as GD 

 to DH, FD is to DG as DE to DH. The angle FGD is therefore equal to the angle 

 EHD, and FG is alfo to EH, or to KG, as FD to DE, or as GD to DH. 



But if GL be drawn parallel to HD, the angle KGL will be equal to the angle EHD 

 that is, to the angle FGD, and therefore the angle KGF to the angle LGD or GDH • 

 and it has been fhewn, that FG is to GK as GD to DH; therefore the triangle FGK is 

 fimilar to the triangle GDH, and is given in fpecies. 



Draw GM perpendicular to CF, and GN making the angle MGN equal to the an^le 

 FGK or GDH, and let GM be to GN in the given ratio of FG to GK, or of GD to DH. 

 Join CN and NK. Then, becaufe MG:GN::FG:GK, MG:FG::GN:GK ; and the an- 

 gle MGF being equal NGK, the triangles MGF, NGK are fimilar, and therefore GNK 

 is a right angle. But fince the ratio of MG to GN is given, and alfo of MG to GC the 

 triangle CGM being given in fpecies, the ratio of GC to GN is given, and CGN being 

 alfo a given angle, becaufe each of the angles CGM, MGN is' given, the triangle CGN 

 is given in fpecies, and confequently the ratio of CG to CN is given. The angle NCK 

 is therefore given ; and the angle CNK is likewife given, each of the angles CNG, GNK 

 being given, therefore the triangle CNK is alfo given in fpecies. The ratio of CN to CK 

 is therefore given, and fince the ratio of CN to CG is alfo given, the ratio of CG to CK 



