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 $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$.