386 The Lans of Statical Equilibrium. 
is parallel to AP and CR, therefore as PB 
is to BR, so is AM to MC; hence AM= 
MC. (Eucl. VI. and 2.) complete the pa- 
rallelogram ADCB, and BM produced will 
evidently pass through the angle D; therefore 
the diameter BD, gives the direction of the 
equivalent K. Again make AT perpendi- . 
cular to BD, and as BTAP is a parallelo- 
gram BT = AP, but AB represents F’, there- 
fore AP or BT represents that part of F 
which acts in the direction DB by Art. 7 and 
8; for the same reason CR represents the 
part of G which acts in DB ; but the triangle 
CBR and DAT are similar and equal, con- 
sequently BT + TD or BD expresses K in 
magnitude, which it also represents in di- 
rection. 
Art. 10. The angle ADB = angle CBD; 
therefore their sines are equal; and we have 
AB x sine ABD= BC x sine CBD ; hence 
as AB: CB:: sine CBD: sine ABD; but 
as AB: CB:: F: G; hence as F: G :: sine 
CBD: sine ABD; consequently if F=G, 
DB bisects the angle ABC; which is as- 
sumed as an axiom by the French analysts, 
Art. 11, Sine of angle BAD or. of 
BCD = sine of ABC; therefore BD = 
BA X sine ABC CB * sine ABC, 
~ sineDBC sine ABD__* Bence 
