83.  Determination of the pencil. We now show that it is possible to assign arbitrarily three lines, _a__, __b__, and __c__, of __ the system (besides the lines __u__ and __u’__); but if these three lines are chosen, the system is completely determined._

This statement is equivalent to the following: 

Given three pairs of corresponding points in two projective point-rows, it is possible to find a point in one which corresponds to any point of the other.

We proceed, then, to the solution of the fundamental

PROBLEM. Given three pairs of points, _AA’__, __BB’__, and __CC’__, of two projective point-rows __u__ and __u’__, to find the point __D’__ of __u’__ which corresponds to any given point __D__ of __u__._

[Figure 20]

FIG. 20

On the line a, joining A and A’, take two points, S and S’, as centers of pencils perspective to u and u’ respectively (Fig. 20).  The figure will be much simplified if we take S on BB’ and S’ on CC’. SA and S’A’ are corresponding rays of S and S’, and the two pencils are therefore in perspective position.  It is not difficult to see that the axis of perspectivity m is the line joining B’ and C.  Given any point D on u, to find the corresponding point D’ on u’ we proceed as follows:  Join D to S and note where the joining line meets m.  Join this point to S’.  This last line meets u’ in the desired point D’.

We have now in this figure six lines of the system, a, b, c, d, u, and u’.  Fix now the position of u, u’, b, c, and d, and take four lines of the system, a_1_, a_2_, a_3_, a_4_, which meet b in four harmonic points.  These points project to D, giving four harmonic points on m.  These again project to D’, giving four harmonic points on c.  It is thus clear that the rays a_1_, a_2_, a_3_, a_4_ cut out two projective point-rows on any two lines of the system.  Thus u and u’ are not special rays, and any two rays of the system will serve as the point-rows to generate the system of lines.

84.  Brianchon’s theorem. From the figure also appears a fundamental theorem due to Brianchon: 

If _1__, __2__, __3__, __4__, __5__, __6__ are any six rays of a pencil of the second order, then the lines __l = (12, 45)__, __m = (23, 56)__, __n = (34, 61)__ all pass through a point._

[Figure 21]

FIG. 21

85. To make the notation fit the figure (Fig. 21), make a=1, b = 2, u’ = 3, d = 4, u = 5, c = 6; or, interchanging two of the lines, a = 1, c = 2, u = 3, d = 4, u’ = 5, b = 6.  Thus, by different namings of the lines, it appears that not more than 60 different Brianchon points are possible.  If we call 12 and 45 opposite vertices of a circumscribed hexagon, then Brianchon’s theorem may be stated as follows: