190 Proceedings of Royal Society of Edinburgh. [sess. 
sGA = (m + l)sDA = (m + l)sDF ; but sGA is also equal tosGB = 
msCB = msCF, .*. (m + l)sDF = msCF, and (m + l)DF = mCF (5). 
Hence (m + 1)DF - mDF = mCF - mDF = m(CF - DF) = mCD • and 
(m + 1)DF - mDF=DF DF = mCD. That FB = mAB may be 
sliown in the same way as that GB = mCB. 
(2) FG is bisected in K. Take KB as the direction of motion, 
then sGK = sGD = ms AD -- msAD = msCE = msGD — sFD = sFK, 
. \ since sGK = sFK we have GK = FK. 
(3) AC is parallel to FG. Because GD = mDA . \ GA = mDA + 
DA = (m+l)DA. In the same way FC = (m + 1)DC, and taking 
AC as the direction of motion, sGA = (m+ l)sDA = (ra + l)sDC = 
sFC, since sGA = sFC, AC is parallel to FG. 
Let AB, CD, EF, GK be four straight lines, and AB-CD = EF-GK. 
If CD and GK are generating lines, and equally inclined to their 
respective directions of motion, we plainly have ABsCD = EFsGK ; 
and conversely, if ABsCD = EFsGK then AB-CD = EF-GK. It is 
easy to see that this truth is general whatever number of factors 
we may have, and that if A-B-C = D«E-F then A-BsC = D-EsF = 
A-CsB = D-FsE = C-BsA = F-EsD. 
The converse is also evidently true, that if A-BsC = D-EsF, and 
the generating lines C, F be equally inclined to their respective 
directions of motion, then A-B-C = D-E-F. 
For the sake of brevity it will be convenient to use some symbol 
to indicate the direction of motion. In the expression sAB(CD) 
the symbol is (CD), which indicates that the direction of motion of 
the generating line AB is CD. 
To employ the properties of generating lines in geometrical de- 
monstration, it is necessary to observe those angles which are equal, 
and to remember that if the angle made by the generating line with 
the direction of motion be changed into its supplement the space 
generated is unaltered. 
* CA and CB (fig. 4) are two tangents at the extremities of the chord 
AB of the circle ABF, and CF a secant cutting the convex circumfer- 
ence in D, the chord AB in E, and meeting the concave circumfer- 
ence in F ; to show that CF is divided harmonically in D and E. 
Join FA, FB, BD. It is easy to see that the angle DBE = DFA in 
the same segment, CBD = BFE in the alternate segment, and FAC 
the supplement of EBF. 
