COURSE OF PROJECTIVE GEOMETRY

§ 26:

*O96*

Apply Pascal to hexagon

ABC

BAD .

*O97*

Let *P '* be the point from which the triangles are perpective. We have to prove *P '*, *X*, *Y* are collinear and *P '*, *X*, *Z* are collinear.

Apply Pacal to
*A*_{2}*C*_{1}*B*_{2}
*B*_{1}*P*_{ }*A*_{1}

and to
*B*_{1}*C*_{2}*A*_{2}
*P*_{ }*A*_{1}*C*_{1}

*O98*

If *A*, *B*, *C*, *D*, *E* are the given points, and *X* the second intersection point of *J* and an arbitrary line *l* through *A*,
then we construct *X* by applying Pascal to *A**B**C**D**E**X*.

So let*P* := *AE***.***BD*, and *Q* := *AX***.***CD* = *l***.***CD*, and *t* := *PQ*. Then *R* := *BX***.***EC* =
*t***.***EC*. So *X* := *l***.***BR*.

*O99*

For *X* on *J*, let *x* be the tangent to *J* in *X*, so the polar line of *X*.

For *X*, *Y* in {*A*, *B*, *C*} is *x***.***y '* the pole of *XY '* , so *x***.***y '* - *x '***.***y* the polar line of
*XY '***.***YX '*.

So the point *P* that lies on the three lines *x***.***y '* - *x '***.***y*, is the pole of the line *p* on which the three points *XY '***.***YX '*
are lying.

*O100*

Let *A*, *B*, *C*, *D* be the four given points, and *a* the given tangent. Construct the tangent in *B* by applying Pascal to
*A**B**C*
*D**A**B*.

So: let *P* := *AA***.***BD* = *a***.***BD*, *Q* := *AB***.***CD*, *t* := *PQ*.

Then the tangent to *J* in *B* goes through *t***.***AC*.

*O101*

Ad *O88*: apply Pascal to
*A**B**C*
*D**A**E*.

Ad *O90*: Construct the Brianchon point *P* of hexagon
*a**b**c*
*c**a**b*.

The point of contact we have to construct, *c***.***c* = *C*, is collinear with *P* and *a***.***b*.

*Note:* The order of the points in the hexagon is important:

has a Pascal line which goes through *AE.BD*, *AF.CD*, *BF.CE*.

If you didn't know this yet, you have to study the answers again.