[642 
fc is clear from the 
at once infer the 
one variable, 
infinitesimal arc 00', 
OM drawn from 0 
And then, writing 
a cos (X + M) = Xa, 
e the equation is 
n between a, b, c, 
respectively ; this 
-A 2 ) 
-A 2 
= 0 : 
V)+f\ 
THE SIDES OF A QUADRANGLE. 
35 
But in virtue of u = 0, we have 
((7a 2 + ±B) 2 = (7 (Ca 4 + Ba 2 + A) + \ (B 2 — AC), 
that is, ^ = *J(B- — 4<AC); and here B 2 — 4AC is a quartic function of f 2 , which is 
easily seen to reduce itself to the form 
r-- (g+¿) 2 / 2 -(9- /0 2 / 2 - (b + c) 2 / 2 -(b- cf. 
The coefficients of bdb, cdc, &c., are given as expressions of the like form; substituting 
their values, the differential relation is 
V{/ 2 -(9 + h) 2 f 2 ~(g~ A) 2 / 2 — (b + c) 2 / 2 -(b- c) 2 } ada + &c. = 0, 
which is, in fact, the foregoing result. 
It is right to 
notice that there are in 
all 16 lineal 
factors, 
f + g + h, 
b + c+f, 
c + a+g, 
a + b + h 
sa yd, / , 
9 
A 
~f+9 + 
— b + c +/, 
- c + a + g, 
— a + b + h 
d' , 
9' > 
A' 
f-g + h, 
b-c+f, 
c — a + g, 
a —b + h 
d", /", 
g" > 
h" 
f+g-h, 
b + c-f 
c + a- g, 
a+ b — h 
d!", /"', 
g'"> 
h m 
and this being so, the coefficients of ada, bdb, cdc, fdf, gdg, hdh, are 
Y(dd'd"d"' • //'/"/")> - W • hh'h"h"' ), 
V(«T'. gg'g"g'" ), - V(AA'A"/r . fff'f" \ 
*J(dd'd"d r " . hh'h"h'” ), - fff'f" ■ 99'9"9"'), 
respectively. 
We may imagine the quadrilateral ZOXY composed of the four rods ZO, OX. 
XY, YZ (lengths c, a, h, f as before) jointed together at the angles, and kept in 
equilibrium by forces B, G acting along the diagonals 0 Y (= b), ZX (= a) respectively 
We have c, a, h, f given constants, and the relation 0 (a, b, c, f g, h) = 0, which 
connects the six quantities is the relation between the two variable diagonals (g, b) ; 
by what precedes, the differential relation fig . dg + fib . db = 0 is equivalent to 
ClBbdb — CAgdg = 0. By virtual velocities we have as the condition of equilibrium 
Bdb + Gdg = 0 ; hence, eliminating db, dg we have 
B __ G_ 
ilBb ~ CAg ’ 
or, say 
b ' g~ AXYO . AZYO ‘ HXYZ.MXO ’ 
viz. the forces, divided by the diagonals along which they act, are proportional to the 
reciprocals of the products of the two pairs of triangles which stand on these diagonals 
respectively. The negative sign shows, what is obvious, that the forces must be, one 
of them a pull, the other a push. 
5—2