1889 - 90 .] Mr David Maver on a Geometrical Method. 189 
4. If AB = CD and sAB = sCD, or if p AB = ^CD and ^sAB = 
gsCD then AB and CD are equally inclined to their respective 
directions of motion. 
5. If sAB = sCD and AB and CD are equally inclined to their 
respective directions of motion then AB = CD, or generally if 
psAB = gsCD then £>AB = gCD. 
It would appear, therefore, that of the three conditions, equality 
of straight lines, equality of the spaces generated by these lines, and 
equality of their inclinations to their respective directions of motion, 
if any two are given the third is also given. 
There are other properties of generating lines which might be 
noticed, but to do so, and show their application to the solution of 
geometrical theorems, would extend this paper to too great a length. 
We shall therefore proceed to apply what has already been gone 
over. 
BA and BC (fig. 1) are two tangents to the circle EAC whose 
centre is D. If the diameter CDE be drawn and EA and DB joined, 
EA shall be parallel to DB. It is easily seen that the angle ABD 
= CBD. Let DB be the direction of motion then, sED = sCD = 
sCB = sAB, .*. since sED = sAB, EA is parallel to DB (2). 
Let the sides AB, BC (fig. 2) of the triangle ABC be bisected by 
the straight lines CE, AF intersecting in G, then shall CG = 2EG 
AG = 2FG; and if BG be drawn and produced to D, AD = CD. 
Let AF be the direction of motion, then sCG = sCF(l) = sBF = sBA = 
2sEA = 2sEG, . *. since sCG = 2sEG, CG = 2EG (5). In the same 
way AG = 2FG. Again, if DB be the direction of motion, we have 
sAD = sAG = 2sFG = 2sFB = sCB = sCD, . *. since sAD = sCD, AD = 
CD. 
ABCD (fig. 3) is a quadrilateral, and the diagonal BD bisects the 
diagonal AC in E. If the opposite sides be produced to meet at 
F and G we have the following results: — (1) If GD = mAD then 
shall GB = mCB, FD = mCD, and FB = mAB. (2) If BD be pro- 
duced it will bisect FG in K. (3) AC is parallel to FG. 
(1) TakeDB as the direction of motion, then because GD = mAD 
.*. sGD = ms AD = msAE = msCE = msCB ; but sGD = sGB, .*. sGB = 
msCB, and GB = mCB (5). 
To show that DF = mCD. Because GD = mDA .-. GA = mDA + 
DA = (m + l)DA. Taking FB as the direction of motion, we have 
