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 4 of 4: Join the inequalities

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In plain words

The straight segment is no longer than the route through $G$ and $H$ .

CF=C'F'\le C'G+GH+HF'\le AG+GB+GH+DH+HE

Detailed analysis

The triangle inequality gives $CF=C'F'\le C'G+GH+HF'$ . Substituting the two Ptolemy bounds yields $CF\le AG+GB+GH+DH+HE$ , which is the required result.