Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively. Let's investigate this situation paying attention to the length of the sengments we just constructed.
After some playing around, we can see that it appears (AF)(BD)(CE) = (BF)(CD)AE). Let's give a formal proof! We begin by dropping altitudes from the point P to each side, and also an altitude from one vertex of ABC to the opposite side. We can now continue without loss of generality.
One thing to consider is the case where the point P we choose actually lies outside of the triangle. What might this look like? Lets take a look at the sketch below which allows for this case.
We can see that it appears that our original conjecture would still hold. And in fact, the proof is somewhat similar to prove in this case.
[GSP SKETCH]