Consider a hexagon inscribed in a circle of radius $R$, and satisfying: $A B=$ $B C=C D=a$, and $D E=E F=F A=b$. Calculate $R$ in terms of $a$ and $b$.
It follows from these equalities that the arc $C E$ containing $D$ is equal to one third of the entire circle, so the angle $\widehat{C D E}=120$. Therefore, $C E=R \sqrt{3}$ and according to Al Kashi, $C E^{2}=C D^{2}+$ $D E^{2}+C D . D E$, which ultimately gives: $R^{2}=\frac{a^{2}+b^{2}+a b}{3}$.