($\Rightarrow$) Let $S$ be the intersection of diagonals $AD, BE$ and $CF$. Since $ABCDEF$ is a convex hexagon inscribed in a circle we have $\triangle ABS\sim\triangle EDS,\ \triangle BCS\sim\triangle FES,\ \triangle CDS\sim\triangle AES$, hence $$\frac{AB}{BC}\cdot \frac{CD}{DE}\cdot \frac{EF}{FA} =\frac{AB}{DE}\cdot \frac{EF}{BC}\cdot \frac{CD}{FA}= \frac{SB}{SD} \cdot\frac{SE}{SC}\cdot \frac{SD}{SF} =\frac{SB\cdot SE}{SC\cdot SF}=1.$$($\Leftarrow$) Let $S$ be the intersection of diagonals $CF, BE$. Denote $G\neq A$ as the second intersection of line $AS$ with circumscirbed circle. Because $ABCDEF$ is a convex hexagon point $G$ lies on arc $CE$ that doesn't contain point $A$. As previously we get $$\frac{AB}{BC}\cdot \frac{CG}{GE}\cdot \frac{EF}{FA} =1.$$Hence $$\frac{CG}{GE}=\frac{CD}{DE}.$$We will prove that function $$f(X)=\frac{CX}{XE}$$is strictly increasing when we move point $X$ on arc $CE$ that doesn't contain point $A$ from point $C$ to point $E$. After denoting by $O$ the circumcenter it's equivalent with proving $$f(X)=\frac{\sin\frac{\angle COX}{2}}{\sin\frac{\angle XOE}{2}}.$$In other words we need to prove the property for $$f(x)=\frac{\sin x}{\sin(a-x)}$$where $x\in(0;a)\wedge 0<a<\pi$. There are two ways: $$f'(x)=\frac{\sin(a)}{\sin^2(a-x)}>0$$or $$\forall_{x,y\in(0;a)}f(x)>f(y)\iff \sin(x-y)\cdot\sin(a)>0\iff x-y>0.$$Hence $G=D$ QED.