Problem 5

Let

ABCDEF

be a convex hexagon with

AB=BC=CD

and

DE=EF=FA

, such that

\angle BCD=\angle EFA=60^\circ

. Suppose that

G

and

H

are points in the interior of the hexagon such that

\angle AGB=\angle DHE=120^\circ

. Prove that

AG+GB+GH+DH+HE\ge CF

Step 2 of 4: Reflect the endpoints

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In plain words

The reflection converts the original equilateral triangles into ones based at $A$ and $D$ .

C'=\operatorname{ref}_{BE}(C),\quad F'=\operatorname{ref}_{BE}(F)

Detailed analysis

Reflect $C$ and $F$ in $BE$ , calling the images $C'$ and $F'$ . Since $A,D$ are exchanged, $C'AB$ and $F'DE$ are equilateral, and reflection preserves distance: $C'F'=CF$ .