Question #7704

The lengths of the sides of a plane hexagon ABCDEF (not necessarily convex) satisfy that 2 AB= BC, 2 CD = DE, and 2EF =FA. Prove that AF/CF + CB/EB + ED/AD >= 2

Expert's answer

2 AB = BC, 2 CD = DE, и 2EF= FA

2EF/CF+2AB/EB+2CD/AD>=2

DE>DA>DC

FA>FC>FE

BC>BE>BA

AF / CF + CB / EB + ED / AD> = 2

2EF/CF+2AB/EB+2CD/AD>=2

DE>DA>DC

FA>FC>FE

BC>BE>BA

AF / CF + CB / EB + ED / AD> = 2

## Comments

Assignment Expert27.03.12, 18:22We used lemma for polygons

Simone23.03.12, 21:13How did you get this answer? Is there a theorem or something I need to prove this.

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