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, a1, a2, a3, a4, which meet b in four harmonic points. These points project to [pg 51] 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 a1, a2, a3, a4 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.
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: