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

Figure 2

 

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/EC 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'CA*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.

 

Return to Essay