In a convex hexagon $ABCDEF$, triangles $ACE$ and $BDF$ have the same circumradius $R$. If triangle $ACE$ has inradius $r$, prove that
\[ \text{Area}(ABCDEF)\le\frac{R}{r}\cdot\text{Area}(ACE).\]
if the triangle have all the angles $ \leq 90 $
let be $ O $ the circumcenter of $ BFD $.
then $ O $ will be in the interior of $ BDF $ and then $ 2S[AOBC] \leq BO*AC $ and adding these inequalities we'll obtain the inequality.
if the triangle has $ B>90 $ then taking $ O_{1} $ the circumcenter and $ O $ such that $ FO=DO=R $ then making the same inequalities like in the previous case for $ O $ we'll obtain the inequality also using that $ BO_{1}\geq BO $.