Prof. Twining on the Parallelogram of Forces. 325 



included angles ABE, DBC, be, each, a multiple by m of the an- 

 gle EBC, which I call A', and which I will suppose to be such as 

 to have the diagonal of EBC, when completed, equal to *, (m and 

 x having the same numerical values as in the former paragraph.) 

 Join FG, DE and AC, which are evidently parallel ; and inter- 

 sect these parallels by BH, which bisects FBG. Since EF and 

 DG equal BA and BC, and intersect at the same angle, it is plain 

 that the part of BH intercepted between the parallels FG and 

 DE equals the part between AC and the point B. But this last 

 is half the diagonal of AB, BC, if completed; and the part of BH 

 between DE and B is half the diagonal of DB, BE, if completed. 

 Doubling, we deduce, therefore, 2BH = diag. (AB, BC) + diag. 

 (DB, BE). Bu t, sinc e FBG = EBC, 2BH= D^; al so AB, BC in- 

 clude the angle m + lA', and DB, BE include m— 1A'. We have, 

 therefore, the diagonal pertaining to the sides with the included 

 angle m + 1 A' equal to that with the included angle mA', aug- 

 mented in the ratio of 1 to %, diminished by that with the inclu- 

 ded angle m — 1A'. 



The law of formation is the same, therefore, both for resultants 

 and diagonals ; so that, if both the forces and the sides of the 

 parallelograms are represented by unity, and the former, acting at 

 the angle A, have a resultant represented in intensity by the di- 

 agonal pertaining to the latter, when their included angle is A', 

 then, also, would the resultant of the same forces, acting at the 

 angle mA, be represented by the diagonal pertaining to the sides 

 with the included angle mA'. And the converse is evidently 

 true. But whether A=A' remains to be shown. 



For this purpose, I again resume the figure first used. Let the 

 forces BA, BC — each equal to unity — acting at the given angle 

 ABC, have a resultant represented in value by the diagonal of "a 

 parallelogram, whose two adjacent sides DB, BE — each equal to 

 1 — include the unknown angle DBE, less or greater than ABC. 

 Take of ABC an undetermined exact part or measure z ; also of 

 DBE a like proportional part z'. Then, by the converse of the 

 proof already given, it is evident that the two given forces, act- 

 ing at the angle z, would have a resultant represented by the 

 diagonal pertaining to the given sides DB, BE, having the inclu- 

 ded angle z' . Take nz, nz', such entire multiples of z and z' 

 that one multiple shall exceed, while the other shall not equal 

 two right angles ; which, on the supposition that z and z 1 have 



