Two Parallel Lines Give Rise to a Third Parallel Naturally Consider the  shown. Here $A, B, C$ and $A', B', C'$ are points in the plane such that $AB$ is parallel to $A'B'$ and $AC$ is parallel to $A'C'$. Further, the points $B, C, B', C'$ are collinear. Let $AC$ intersect $A'B$ at $P$ and let $A'B'$ intersect $AC'$ at $Q$.

Then $PQ$ is parallel to $BC$.

This problem is certainly amenable to attack via coordinates. But I feel that is an elegant geometric argument hiding in plain sight. In fact, I think somehow projective geometry can be used here, though I am a toddler at projective geometry.
 A: Your intuition about projective geometry is right. This is an instance of Pappus' theorem, with one of the nine lines of the configuration being the line at infinity. Keeping one line at infinity, the combinatorics from your configuration can also be drawn like this:

Note that $B'$ and $C$ are rather irrelevant: they are the intersection of two lines, but since any two lines intersect (the parallel ones at infinity), there is nothing significant about these points. So the statement goes like this: When $AB\parallel QA'$ and $A'C'\parallel PA$ then $BC'\parallel PQ$.
The corresponding projective situation would look like this:

I started out with a projective version of your input drawing, i.e. with parallels replaced by intersections on a common line. After eliminating the superfluous points, I counted nine points and nine lines. That alone was enough to identify the underlying theorem: there is no other theorem in projective incidence geometry with this few points. Desargues takes ten points and ten lines.
