Question 6
(POL) Let $ABCDEF$ be a convex hexagon such that $\angle B+\angle D+\angle F=360^\circ$ and
$$\frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{EF}{FA}=1.$$
Prove that
$$\frac{BC}{CA}\cdot\frac{AE}{EF}\cdot\frac{FD}{DB}=1.$$
$$\frac{AB}{BC}\cdot\frac{CD}{DE}\cdot\frac{EF}{FA}=1.$$
Prove that
$$\frac{BC}{CA}\cdot\frac{AE}{EF}\cdot\frac{FD}{DB}=1.$$