+++++++++++++++++++++++++++
Two non-parallel planes meet along a line. Two parallel planes now meet
along their common ideal line. A plane and a line (that does not
lie in the plane) must meet, either where the line pierces the plane.
Or, if the line is parallel to the plane, they meet "at infinity", i.e.
the ideal point of the line lies on the ideal line on the plane.
Two lines, however, need not meet at all, either at any real point
or at any ideal point. Such lines are called 'skew'.
Can you figure out what it means for a real and an ideal line
to be skew?
pass:[

]
Kepler's peculiar geometry had already been discovered,
practically speaking, by the Rennaissance artists in
their study of perspective. The 'horizon',
particularly familiar for residents of Illinois, is an ideal line. In
perspective, the horizon is drawn as a line in the painter's canvas.
Later, these ideas were adopted by mathematicians as
'projective geometry'. It differs from the traditional Euclidean
geometry in many ways. Most notably, there is greater logical
symmetry between points and lines.
In projective geometry, any two points are joined by a line ( as they
do in Euclid's geometry)
and any two lines (in a plane) have a point in common (non-Euclidean).
In projective 3-space, any two points are joined by a line (Euclidean),
and any two planes meet along line (non-Euclidean).
To exercise this expanded way of thinking we re-examine Desargues'
theorem, but allowing for certain points to be ideal.
Desargues Theorem
-----------------
The figure we usually draw to illustrate Desargues Theorem looks
like the following. But you should experiment with figures where
the red and green triangles are in more general positions along the
three lines.
[frame="none",cols="^",valign="middle",grid="all"]
|===================
|pass:[
+++++++++++++++++++++++++++
Consider four points pass:[$ W = (AC)(BD), X=(AB)(DE), Y=(BC)(EF),
Z = (CD)(FA) $]. The theorem says that if pass:[$ W, X, Y $] are
ideal, then pass:[$ Z $] must also be ideal. In particular,
pass:[$ (XYZ) = \infty $].
pass:[

]
.Exercise.
==================
Use KSEG or ruler and compass to investigate the case that
pass:[$ W $] is a real point. Is it still true? If both pass:[$ X, Y $]
be real too, is it still true that pass:[$ (XYZ) $].
==================
Pascal's Theorem in KSEG
------------------------
Having extended the concept of 'collinearity' to include ideal and
real points in the plane, you should revisit Pascal's generalization of
Pappus' theorem we mentioned earlier. Pascal was so amazed by the
result he discovered, he gave it a fancy name:
[frame="none",cols="^",valign="middle",grid="all"]
|===================
|pass:[