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;*