THEOREM IN GEOMETRY OF POSITION. 659 
4. Or thus: when a set of points are joined so that two, and only two, 
joining lines meet at each point, these lines must obviously form one or more 
closed polygons. Hence, in the case before us, by limiting the selection to two 
out of the three lines drawn to each point, we can always, in many different ways, 
forma polygon or polygons. If the number of sides in each of these is even, 
the main proposition is at once proved ; for the alternate sides of the polygons 
belong to two of the three groups—the unused lines forming the third group. 
Such solutions must evidently be possible in all cases, with the exception of 
that already excluded. ‘This knowledge, however, does not at once help us to 
a practical solution of the problem in any particular case. We must, therefore, 
look at the result more generally. 
If the selection we have made gives more than one polygon, two or other 
even number of them may have an odd number of sides each. Suppose there 
are but two. If these be connected by one line only, we have the excepted 
case above. If they be connected by three, or a larger odd number of lines, 
we may always proceed as is indicated in figs. 6. 6a shows the two odd-sided 
polygons. 60 and 6c show how, neglecting the points C and C’, we form even- 
sided polygons passing through them and including AB and A/D’ respectively. 
Finally, 6d shows the result when the two latter figures are joined. Thus the 
proposition is proved by actually effecting the decomposition into polygons of 
an even number of sides. Hence it is true for any even number of points (the 
excepted case excluded) if it is true for smaller even numbers of points. But 
it is obviously true for two, for four, and for six, points. 
5. Another mode of reaching the same conclusion, is to pass from a case of 
2n points to one of 2n+2 by drawing a new line terminating in any two sides 
of one of the even-sided polygons of the former case (§ 2). That polygon 
remains even-sided, but its sides must be relettered; and then we have one 
or more solutions of the new case. 
In fact, by temporarily suppressing, two by two, points and their joining line 
(always taking care that the figure left shall not belong to the excepted case) 
we can reduce any case, however complex, to the four poimts for which the 
proposition is always true. [Or we may suppress one line, and divide the 
figure into two odd-sided polygons passing respectively through its ends. On 
restoring the line, these two polygons give a solution. | 
6. Practically, in every case, the simplest mode of solution is to begin at 
any point, and go through all (through some, perhaps, more than once) till we 
return to the starting-point. Then treat, as not gone over, all the lines which 
have been gone over an even number of times. This process is very easily 
learned by trial, the only special rule to be attended to being that we must 
never isolate a point. Should two odd-sided polygons be thus obtained, we 
may either begin afresh :—or go over a second time, attending to the above 
