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Prof. Twining on rallelogram 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 z, (m and 
z 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). But, since FBG=EBC, 2BH=D7;; also AB, BC in- 
clude the angle m-+14A’, and DB, BE include m—1A’. We have, 
therefore, the diagonal pertaining to the sides with the included 
angle m-+1A’ equal to that with the included angle mA‘, aug- 
mented in the ratio of 1 to z, 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’4, 
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. Bat 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- 
edangle z’. Take nz, nz’, such entire multiples of z and 2/ 
that one multiple shall exceed, while the other shall not equal 
two right angles; which, on the supposition that z and z’ have 
