IMO 1995 - Problem 5

Problem: Let $ABCDEF$ be a convex hexagon with $AB=BC=CD,DE=EF=FA$ and $\angle{BCD}=\angle{EFA}=\frac{\pi}{3}$. Let $G$ and $H$ be two points interior to the hexagon such that angles $AGB$ and $DHE$ are both $\frac{2\pi}{3}$. Prove that: $AG+GB+GH+DH+HE\ge CF$.
Solution: Note that $BCD$ and $EFA$ are equilateral triangles. Hence $BE$ is an axis of symmetry of $ABDE$. Reflect $BCD$ and $EFA$ about $BE$ to $BC'A$ and $EF'D$ respectively. Since $\angle{BGA}=180^0-\angle{AC'B}$, $G$ lies on the circumcircle of $ABC'$. By Ptolemy's Theorem, $AG+GB=C'G$. Similarly, $DH+HE=HF'$. It follows that:
$CF=C'F'\le C'G+GH+HF'=AG+GB+GH+DH+HE,$ with equality if and only if $G$ and $H$ both lie on $C'F'$
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf