*81. Tangents to a conic.* If now this figure be projected to a point outside the plane of the circle, and any section of the resulting cone be made by a plane, we can easily see that the system of rays tangent to any conic section is a pencil of rays of the second order. The converse is also true, as we shall see later, and a pencil of rays of the second order is also a set of lines tangent to a conic section. *82.* The point-rows _u_ and _u’_ are, themselves, lines of the system, for to the common point of the two point-rows, considered as a point of _u_, must correspond some point of _u’_, and the line joining these two corresponding points is clearly _u’_ itself. Similarly for the line _u_. *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: