BALKAN-SL 2009 G2

If $ABCDEF$ is a convex cyclic hexagon, then its diagonals $AD$ , $BE$ , $CF$ are concurrent if and only if $\frac{AB}{BC}\cdot \frac{CD}{DE}\cdot \frac{EF}{FA}=1$ .

Alternative version. Let $ABCDEF$ be a hexagon inscribed in a circle. Then, the lines $AD$ , $BE$ , $CF$ are concurrent if and only if $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$ .