CONCERNING TERRESTRIAL MAGNETISM. 
79 
Lemma II. Theorem. — If lines be inflected to a point without the circle TNU from 
T and II, the ratio TN' : N'U will be less than the ratio T N : N U ; but if to a point 
within the circle , the ratio will be greater. (Plate X. fig. 1 .) 
First. Let N' be without the circle. This divides itself into two cases. 
1. Where N' is on the same side of M R (drawn from M at right angles to T U) as 
the points C and D are. 
Join TN', N'U, and let N'U cut the circle in N. Join also TN and draw N'S 
parallel to TU, meeting TN in S, and produce TN to Q. 
Then, by parallels, the angles QTU, TN'S are equal. And since N'TU is less 
than a right angle, (for it is also an angle of the triangle T M R, of which T M R is a 
right angle, the point N' being by hypothesis on the other side of R M from T,) the 
angle Q T U is greater than a right angle ; and hence TN'S is greater than a right 
angle, and consequently N' S T is less than a right angle, and, a fortiori , less than 
T N' S. The line T S is therefore greater than T N'. 
But by similar triangles TN : NU :: TS : N'U. Hence since the line TN' is less 
than TS, the ratio TN' : N'U is less than the ratio TS : N'U, and hence less than 
the ratio T N : N U. 
2. Let N' and C D be on opposite sides of M R. Join N' U cutting M R in R' and 
join TU'. Then the angles R'TU, R'UT are equal. But N'TU is greater than 
R' T U, and hence greater than N' U T. The side N' U of the triangle N' T U is there- 
fore also greater than N' T. Hence the ratio N' T : N' U is a ratio of less inequality, 
whilst the ratio TN : NU is by hypothesis a ratio of greater inequality. The ratio 
T N' : N' U is therefore, in this case also, less than the ratio T N : N U. 
Secondly. Let the point N' lie within the circle. Produce U N' to meet the circle 
at N and join N T. Draw N' S parallel to T U, to meet T N at S. 
Then it may be proved as before that TN' is greater than TS. And by similar 
triangles TN : NU : : TS : NU'. But since T N' is greater than T S, the ratio 
T S : N' U, that is the ratio T N : N U, is less than the ratio TN':N U. 
Scholium. — The antecedent of the lines in the expression of the ratio are considered 
to be drawn from the more distant point T : but these conditions will be reversed 
when the order of the terms is reversed. 
Lemma III. Problem. — From two given points T and U (Plate X. figg. 2. and 3.) to 
infect lines to meet in a right line given by position , so that their ratio shall be the least 
or greatest possible. 
This problem is divisible into two cases according as the intersection S of the given 
line H S with the line T U drawn through the given points T and U, is on the same 
side of M with the antecedent or with the consequent of the lines which are in the 
required ratio. As, however, both cases are constructed by the same operation, it 
will be more convenient to give the analysis of them in juxtaposition by means of 
parallel vertical columns. Also for convenience of comparison, I shall employ the 
same letters in both cases, merely accentuating one set for the sake of distinction. 
