840] 
ON MASCHERONI S GEOMETRY OF THE COMPASS. 
317 
and find y such that the distances Cy, dy are = Dd, DC respectively; 7 is in a line 
with cC, and we have CydD a parallelogram. Find cS a fourth proportional to 
cy, cd, cC, and with centres c, C respectively and radii each = cS describe circles cutting 
in the point 8; this will be the required intersection of the two lines. In fact, the 
required point S will be the intersection of the two lines CD, cd: supposing these 
lines each of them drawn, and also the lines cCy and dy, we have DC parallel to 
dy, that is, the triangles cdy, cSC are similar to each other or cy : cd :: cC : cS: viz. 
the distance cS having been found by this proportion, and the point 8 found as the 
intersection of the two circles, centres c and C respectively, the point S so determined 
is the required point of intersection of the given lines AB and CD. 
If a circle be given without its centre being known, then taking any three points 
A, B, C on the circle, and a pair of counter-points D, E of the line AB, and also 
a pair of counter-points F, G of the line AC, we have obviously the centre of the 
given circle as the intersection of the lines DE and FG; and the centre can thus be 
found with the compass only. 
It is proper to remark that the problems considered in the present paper are those 
connecting the theory with ordinary geometry, not the problems which are most readity 
and elegantly solved with the compass only: a large collection of these are contained in 
the work, and in particular the twelfth book contains some interesting approximate solu 
tions of the problems of the quadrature of the circle, the duplication of the cube, and 
other problems not solvable by ordinary geometry. 
Cambridge, March 19, 1885.