1.) Extend the lines BE and CF beyond the triangle until they meet GH, the line through A parallel to BC (see Figure 2). There are now several pairs of similar triangles: AHF and BCF, AEG and BCE, AGK and BDK, CDK and AHK (where K is the intersection point of AD, BE, and CF). From these triangles in that order we have the following ratios:

a. AF/FB=AH/BC

b. CE/EA=BC/AG

c. AG/BD=AK/DK

d. AH/DC=AK/DK

2.) From c and d we conclude that AG/BD=AH/DC, and after doing a little algebra, we have:

e.) BD/DC=AG/AH.

Now multiplying a, b, and e we have

AF/FB*BC/DC*CE/EA=AH/BC*BC/AG*AG/AH=(AH*BC*AG)/(BC*AG*AH)=1

So, if the lines AD, BE, and CF intersect at the point K, then
AF/FB*BD/DC*CE/EA=1 holds. So the fact that AD, BE, and CF intersect
at one point is sufficient for this condition to hold. We must
now prove that it is not only sufficient but is also necessary.
In other words, we must now prove that if AF/FB*BC/DC*CE/EA =
1 holds then AD, BE, and CF are concurrent.

3.) So assume that K is the point of intersection of BE and CF
and draw the line AK until its intersection with BC at a point
D'. Then, from part 1 of the theorem, we haveAF/FB*BD'/D'C*CE/EA=1.
However, it is given that AF/FB*BD/DC*CE/EA=1. So combining the
two we have (BD'+D'C)/D'C=(BD+DC)/DC. This gives us BD'/D'C +1=BD/DC
+ 1 or BD'/D'C=BD/DC. Finally we have BC/D'C=BC/DC which implies
that D'C=DC. So D' and D are the same point and the proof is complete.

Comment from Jerome Shipman (e-mail 04 Jun 06):

I am sure that your students have no difficulties with your proof, but I had a bit of a problem with the necessary part. If the hypothesis AF/FB*BD/DC/CE/EA=1 is called (2), you say in effect on Page 2 "Combining (1) and (2) we have ...." "Combining" does not have a very precise meaning. Setting the left hand side of (1) equal to the left hand side of (2), since both right hand sides are 1, gives BD'/D'C=BD/DC, which is what you want. Finally you say "Finally we have BC/D'C=DC/DC" which implies that D'C=DC. So D' and D are the same point ..." which is no doubt obvious to your students. But I had first to assume that D' falls to the left of D, in which case BD'/D'C < BD/DC, a contradiction, or that D' falls to the right of D, in which case BD'/D'C > BD/DC, also a contradiction, so it must be the case that D' falls on D, your assertion that D' and D are the same point.