The proof goes as follows:
1.) Begin with a triangle ABC as in Figure 8, with the line through
D, E, and F where D, E, and F are points on the sides of DABC.
Now we must prove that BD/CD*CE/AE*AF/BF=1.
2.) Construct the segment parallel to DF through C and let G be
the point of intersection of the constructed segment and the side
AB.
3.) So we have triangle ACG similar to triangle AEF and triangle
BDF similar to triangle BCG.
So CE/AE=GF/AF and BD/CD=BF/GF.
4.) So CE/AE*BD/CD*AF/BF=GF/FA*BF/GF*AF/FB=1