PROJECTIVE GEOMETRY COURSE

§ 26: *The theorem of Pascal*

*Proposition:* Let *A*, *B*, *C*, *A '*, *B '*, *C '* be points that lie on a non-degenerated conic *J*.

Then *P* := *AB '***.** *A 'B*, *Q* := *BC '***.** *B 'C* and *R* := *AC '***.** *A 'C* are lying on one and the same line (the *pascal line*
of hexagon *ABCA 'B 'C '*).

*Proof:* *J* is the conic generated by the projectivity φ from *A* onto *C* with φ(*AA '* ) = *CA '*, φ(*AB '* ) = *CB '*,
φ(*AB* ) = *CB*. Since *C '* lies on this conic, φ(*AC '* ) = *CC '*.

Let *D* := *A ' B* **.** *AC '*, *E* := *A ' C* **.** *BC '*, *l* := *A ' B*, *m* := *BC '*. Then we have:
*l*(*A '*, *P*, *D*, *B*) ∧^{=} *A*(*AA '*, *AP*, *AD*, *AB*) = *A*(*AA '*, *AB '*, *AC '*, *AB*)
_{∧}^{_} *C*(*CA '*, *CB '*, *CC '*, *CB*) = *C*(*CE*, *CQ*, *CC '*, *CB*) ∧^{=}
*m*(*E*, *Q*, *C '*, *B*), so *l*(*A '*, *P*, *D*, *B*) _{∧}^{_} *m*(*E*, *Q*, *C '*, *B*).

Apparently, *B* is fixed point of the projectivity ψ from *l* onto *m* with ψ(*A '* ) = *E*, ψ(*P* ) = *Q*, ψ(*D* ) = *C '*.

According to *O40*, ψ is a perspectivity with center *A ' E* **.** *DC '* = *R*. Since ψ(*P*) = *Q*, *P*, we see *Q* and *R* lie on one and the
same line.

*Note:* The proposition also holds in the case that one, two or three pairs of points coincide. The line connecting two coinciding points is the tangent in that point.

*Note:* Pascal proved his theorem by first proving it for a circle, and then projecting the whole configuration on an other plane.

*Note:* The dual theorem is called theorem of *Brianchon*:

*Proposition:* Let *a*, *b*, *c*, *a '*, *b '*, *c '* be tangents to a non-degenerated conic *J*.

Then the connecting lines *p* := *ab '***.** *a 'b*, *q* := *bc '***.** *b 'c* and *r* := *ac '***.** *a 'c* go through one and the same point
(the *Brianchon point* of hexagon *abca 'b 'c '*; the intersection point of two coinciding tangents is the contact point on that line).

*Problems:*

*O95* Prove the theorem of Brianchon by dualising the proof of the theorem of Pascal.

*O96* Let *A*, *B*, *C* and *D* be four distinct points of a non-degenerated conic *J*. Prove the intersection point of the tangents in *A* and *B*
is collinear with the two diagonal points of quadrangle *ABCD* that don't lie on *AB*, without using the theory of poles and polar lines.

*O97* The vertices of two point perspective triangles *A*_{1}*B*_{1}*C*_{1} and
*A*_{2}*B*_{2}*C*_{2} are lying on a conic *J*.
*P* is a point of *J* that doesn't coincide with one of these six points. *X* := *P**A*_{2}**.** *B*_{1}*C*_{1},
*Y* := *P**B*_{2}**.** *A*_{1}*C*_{1}, *Z* := *P**C*_{2}**.** *A*_{1}*B*_{1}.

Prove that *X*, *Y*, *Z* and the center of perspectivity of the given triangles are collinear.

*O98* Construct a sixth point of the conic through five freely situated points, using the theorem of Pascal. Likewise a sixth line of the conic of lines determined by five
freely situated lines.

*O99* Prove that the Pascal line of a hexagon whose vertices are lying on a conic *J* is the polar line of the Brianchon point of the corresponding hexagon of the tangents in these six points.

*O100* Of a conic *J* four points and the tangent in one of these points are given. Construct the tangents in the other points. Also solve the dual problem.

*O101* Solve the problems of §24 using the theorems of Pascal and Brianchon.