Question 11
G5 (NZL), IMO5. Let $ABCDEF$ be a convex hexagon with $AB=BC=CD$, $DE=EF=FA$, and $\angle BCD=\angle EFA=\pi/3$ (that is, $60^\circ$). Let $G$ and $H$ be two points interior to the hexagon such that angles $AGB$ and $DHE$ are both $2\pi/3$ (that is, $120^\circ$). Prove that $AG+GB+GH+DH+HE\ge CF$.