8.  Use the last exercise to solve the problem:  Given five points, Q, R, S, C, C’, on a conic, to draw the tangent at any one of them.

9.  State and prove the dual of problem 7 and use it to prove the dual of problem 8.

10.  If a transversal cut two tangents to a conic in B and B’, their chord of contact in A, and the conic itself in P and P’, then the point A is a double point of the involution determined by BB’ and PP’.

11.  State and prove the dual of problem 10.

12.  If a variable conic pass through two given points, P and P’, and if it be tangent to two given lines, the chord of contact of these two tangents will always pass through a fixed point on PP’.

13.  Use the last theorem to solve the problem:  Given four points, P, P’, Q, S, on a conic, and the tangent at one of them, Q, to draw the tangent at any one of the other points, S.

14.  Apply the theorem of problem 9 to the case of a hyperbola where the two tangents are the asymptotes.  Show in this way that if a hyperbola and its asymptotes be cut by a transversal, the segments intercepted by the curve and by the asymptotes respectively have the same middle point.

15.  In a triangle circumscribed about a conic, any side is divided harmonically by its point of contact and the point where it meets the chord joining the points of contact of the other two sides.

CHAPTER IX — METRICAL PROPERTIES OF INVOLUTIONS

[Figure 39]

FIG. 39

141.  Introduction of infinite point; center of involution. We connect the projective theory of involution with the metrical, as usual, by the introduction of the elements at infinity.  In an involution of points on a line the point which corresponds to the infinitely distant point is called the center of the involution.  Since corresponding points in the involution have been shown to be harmonic conjugates with respect to the double points, the center is midway between the double points when they exist.  To construct the center (Fig. 39) we draw as usual through A and A’ any two rays and cut them by a line parallel to AA’ in the points K and M.  Join these points to B and B’, thus determining on AK and AN the points L and N. LN meets AA’ in the center O of the involution.

142.  Fundamental metrical theorem. From the figure we see that the triangles OLB’ and PLM are similar, P being the intersection of KM and LN.  Also the triangles KPN and BON are similar.  We thus have

OB :  PK = ON :  PN

and

OB’ :  PM = OL :  PL;