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FDA, ABE must be equal, and therefore the quadrilateral 
BEFD can be inscribed in a circle, and therefore the angles 
FEA, DBA are equal, and the corresponding angles in the 
frame 7'QS, SPV are equal, and therefore PT’ is equal to QT. 
If, therefore, O is in one diagonal of the rhombus, 7’ must 
be in the other diagonal. 
The diagram of stress is completed by drawing HC parallel 
to BD, and DC parallel to BH, and joining FC. 
This diagram therefore consists of a parallelogram BDCE, a 
diagonal ED, a point Fin the circle passing through FBD, and 
four lines drawn from F to the angles of the parallelogram. 
If we now begin with the diagram of stress, and proceed to 
construct a frame reciprocal to it, the form of the frame will 
be different according to the cyclical direction in which the 
sides of the rhombus PRQS are lettered. Ifin the one case we 
have the points O and T both outside the rhombus as in fig. 1, 
in the other O and T will both be within the rhombus as in 
fiz. 3. The stresses in the corresponding pieces of fig. 1 and 
fig. 3 are all equal if they are equal in any pair of them. 
If in the frames represented in fig. 1 and fig. 3, we con- 
sider that the pieces OS and 7'P cross one another at V without 
intersecting, we have six points O, P, Q, R, S, T joined by nine 
lines. Now in general if p points in a plane are joined by 2p—3 
lines the figure is simply stiff, that is to say the form of the 
figure is determined by the lengths of the lines, and there are 
no necessary relations between the lengths of the lines. 
But in Peaucellier’s linkage the length of any line, as OT, is 
determined when those of the other eight are given. For if a 
is the length of a side of the rhombus, b the length of either arm 
OR or OS,c the length of either arm TP or TQ, then if OT =d, 
d’=0'+c—a’. 
Hence if any one of the nine pieces of the linkage be re- 
moved, the motion of the remaining eight will be the same as 
