In the diagram, $ABCDEF$ is a hexagon with $AB=BC=CD$ and $DE=EF=FA$. Angles $BCD$ and $EFA$ both equal $60°$. $G$ and $H$ are two points taken from inside the hexagon such that angles $AGB$ and $DHE$ both equal $120°$. Prove that $AG+GB+GH+DH+HE\ge CF.$
I tried to construct equilateral triangles by using the angle bisector of the two $120°$ angles, but still could not link anything to the conclusion. I also know that the conclusion is equivalent to $AB+GH+DE\ge CF$ by Triangle Inequality. Please advise.
$|AB| = |BD|,\ |DE|=|AE|$ Hence we have $$ \Delta ABG \sim \Delta DBG',\ \Delta DHE \sim \Delta AH'E$$ s.t. $|GH|=|G'H'|$
And we have $G'',\ H''$ s.t. $$ \Delta BDG' \sim \Delta CDG'',\ \Delta AEH' \sim\Delta AFH''$$
Hence clearly $$|CG''| + |G''G'|+|G'H'| + |H'H''| + |H'' F|\geq |CF|$$