316 
ON MASCHERONI S GEOMETRY OF THE COMPASS. 
[840 
and by construction, (EG) 2 = (OF) 2 ; that is, 
(EG) 2 = (CE) 2 + rad. 2 , = (CE) 2 + (GG) 2 ; 
viz. CEG is a right-angled triangle; that is, CG is at right angles to BCE, or the 
line CGF bisects the arc AB in G. 
For the second problem, (2), to find the intersection of two given lines, we require 
the solution of the problem to find a fourth proportional to three given distances, and 
this immediately depends on the following problem. 
Problem. In two concentric circles to place chords subtending equal angles. If AB 
(fig. 2) be a chord of the one circle, then with centres A and B respectively, and 
Fig. 2. 
an arbitrary radius AA' = BB', describing circles to cut the larger circle in the points 
A' and B' (A' is either of the two intersections and B' is the intersection lying in 
regard to CB as A' lies in regard to CA) then clearly Z ACB = Z A'CB'. And hence 
also we have the following problem. 
Problem. To find a fourth proportional to three given distances. We have only 
to take as the given distances the two radii CA, CA' and the chord AB; and then 
from the similar triangles ACB, A'CB', we have CA : CA':: AB : A'B', viz. A'B' is a 
fourth proportional to CA, CA', AB. 
We have now the solution of 
(2) To construct the intersection of two given lines AB and CD (fig. 3). Find 
Fig. 3. 
B- 
e the counter-point of C, and d the counter-point of D, in regard to the line AB :