If a straight line meets the sides

** BC, CA, and AB **of a triangle

**ABC** in the points

** D, E, and F respectively **then

** (AE/EC)* (CD/DB)* (BF/FA) = 1.**
Prove the above theorem.

BTW the converse is also true.

I just noticed this problem from a few weeks ago, and I was initially puzzled. It is not possible for a straight line to cross all sides of a triangle at three distinct points. A straight line can only cross 0 or 2 sides (unless a straight line goes through A or B or C, which I am ruling out, because that makes D or E or F the same as A or B or C and the formula fails due to division by zero).

The only way this problem makes sense is if we are talking about extending the sides. In which case, 1 (or 3) of D, E, and F are outside triangle A, B, C.

Unfortunately, I do not have a proof, but at least I have a way to make the problem make sense.