200 On the ORIGIN and 
drilateral ABLC; join BH and LN, and it is evident, that the 
three lines GG, BH and LN are all equal, and parallel to AD, and 
are all given in pofition. Join alfo AL, DN, DM, MN and FG. 
BrEcAUsE the two quadrilaterals DEMF, DHNG are fimilar, 
the angle FDM is equal to the angle GDN, and therefore the 
angle GDF to the angle NDM. For the fame reafon alfo, 
GD: DF:: ND: DM, and therefore the triangles GDF, NDM are 
fimilar, and the angle FGD equal to the angle MND, and 
FG: MN:: GD: DN, fo that FG has a given ratio to MN. 
But becaufe the triangles ABC, DEF are fimilar, CG has a 
given ratio to CF, (Lem. 2.) fo that the angle GCF being given, 
the triangle CGF is given in fpecies, and FG has to GC a given 
ratio; now, FG waé alfo fhewn to have to MN a given ratio; 
therefore MN has a given ratio to CG, that is, to LN. 
Acatn, fince the triangle CGF is given in fpecies, the angle 
CGF is given, and CGD being alfo a given angle, the angle 
FGD is given, and therefore MND, which is equal to it. But 
the angle LND is given, therefore the angle LNM is given; 
and it was fhewn, that MN has a given ratio to NU, therefore 
the angle MLN is given ; now, the point L, and the line LN, 
are given in pofition; therefore LM is alfo given in pofition, 
which was to be found. 
Tue conftrudtion for finding LM is obvious. Take A and 
D, two given points in one of the lines given in pofition, and 
place the two triangles ABC, DEF fimilar to the given triangle 
abc, fo that two of their equal angles may be at A and D, and 
the other equal angles on the lines BE and CF, (Lem. 2. Cor.). 
On BC and EF, defcribe the triangles BLC, FEM, fimilar to the 
triangle cbl; if LM be drawn, it will be the line required. 
From the analyfis it alfo follows, that the quadrilaterals de- 
feribed with their angles on the four ftraight lines given in po- 
fition, as fuppofed in the Porifm, will intercept between them 
fegments of thefe lines, having given ratios to one another. 
. 37 
