the increments will be Vlada — Cgdg = 0 (as to the negative sign it is clear from the 
figure that a, g will increase or diminish together): and we thence at once infer the 
general relation. 
We have consequently to prove that, considering a and b as alone variable, 
Aada + Bbdb = 0 ; 
or, what is the same thing, 
-bdb = XOZ : YOZ. 
The points XYZ remain fixed; but 0 moves through the infinitesimal arc 00', 
centre Z, which may be considered as situate in the right line OM drawn from 0 
at right angles to ZO, and meeting XY produced in the point M. And then, writing 
for a moment z OXY—X, zOYX=Y, zOMY=M, we find at once 
that is, 
da = 00' cos (X + M), 
- db = 00'cos (Y-M)- 
da cos (X + M) 
ada _ a cos (X + M) 
db cos (F — M) ’ 
bdb b cos ( F — M) ’ 
But drawing Xa., Y/3 each of them at right angles to ZO, we have a cos (X + M) = Xa, 
b cos (F — M) = F/3, and evidently XOZ : YOZ = Xa : Y/3; whence the equation is 
ada XOZ ...... . * , 
~~ bdb ~ YOZ ’ W “ 1C ^ 1S required relation. 
For the analytical proof, it is to be observed that the relation between a, b, c, 
f, g, h is a quadric relation in the quantities a 2 , b 2 , c 2 , f 2 , g 2 , h 2 respectively; this 
may be written 
1 
by + by + c 2 /d + cdi 2 - (b 2 + c 2 ) g 2 h 2 - (g 2 + h 2 ) b 2 c 2 
~ i b ~ ~ ° 2 ) iff 2 ~ ¿ 2 ) 
+ (b 2 -h 2 ) (c 2 — g 2 ) 
— b 2 — c 2 - <f - h 2 
+ 1 
+ 1 
say for a moment this is A + Bo? + Oa 4 = 0, where 
A = b 2 g 4 + ¥g 2 + c 2 h* + c*h 2 + f 2 (b 2 — h 2 ) (c 2 — g 2 ) 
— (b 2 + c 2 ) g 2 h? — (g 2 + li 2 ) b 2 c 2 
B = -(b 2 - c 2 ) (g 2 - h?) +/ 2 (- b 2 -c 2 -g 2 -li 2 ) + f\ 
/ 2 ; 
then we have as usual 
But 
that 
easily 
The 
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which 
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by wl 
VIBbdl 
Bdb + 
or, say 
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