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 1 of 4: Identify the equilateral triangles

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In plain words

Equal consecutive sides and a 60-degree angle create equilateral triangles.

BCD\text{ and }AEF\text{ are equilateral}

Detailed analysis

From $AB=BC=CD$ and $\angle BCD=60^\circ$ , triangle $BCD$ is equilateral. Similarly, $AEF$ is equilateral from $DE=EF=FA$ and $\angle EFA=60^\circ$ . Consequently $BA=BD$ and $EA=ED$ , so reflection in $BE$ swaps $A$ with $D$ .