198 On the ORIGIN and 
fied ; the inveftigation depending on a Jemma fimilar to that 
which is prefixed to the preceding propofition. 
LEMMA MIL Vic. 12. 
Ir 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, wz. 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. 
Tue 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. 
Bur 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 
imilar to the triangle GDH, and is given in {pecies. 
Draw GM perpendicular to CF, and GN making the angle MGN equal to the angle 
FGK or GDH, and let GM be to GN in the given ratio of FG toGK, or of GD to DE. 
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 aright angle. But fince the ratio of MG to GN is given, and alfo of MG to GC, the 
triangle CGM being given in {pecies, 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 {pecies. 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 
