The Laws of Statical Equilibrium. 389 
line DB perpendicular to PV, will cause it 
so to revolve. 
Art. 15. Through PI draw P paedicl to 
DB, meeting AB produced in I and make 
IK parallel to CB, meeting PT produced in 
R; also from D, draw DM, DN, perpendi- 
- cular to AB, CB; then the triangles DMB, 
PSI are similar, as are also DNB, PRI; 
thereforeas BD: IP:: DM: PS; andas BD: 
IP:: DN: PR; henceasDM:PS::DN: PR; 
and multiplying by F and G we have, as 
FxDM:FxPS::GxDN:GxPR; but Fx 
DM=Gx DN, by Art. 12 ; consequently Fx. 
PS=GxPR; therefore Fx PS—GxPT= 
Gx PR--Gx PT=Gx TR; but by trigono- 
metry TR=IBxsine RIB or ABC; and 
PV=IBxsine PIB or of DBA; hence as 
PV:RT ::sineZ ABD: sine CBA; but as sine 
ABD :sine CBA::G:K, by Art. 11; con- 
sequently as PV: RT::G@:K; and PVxXK 
=RTxG=PSxF—PT*xG:; that is, the mo- 
mentum of K=the difference of the mo- 
menta of F and G. Had P been situated 
out of the angle ABC, the momentum of 
K=the sum of the momenta of F and G.— 
If aB, cB, &c. be the direction of forces 
F’, G’, dB the direction of their equivalent 
K’, and /B the direction of L the equivalent 
of K and K’, draw Ps, Pt, Pv, Pw, perpen- 
