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Answer to Question #7704 in Geometry for Simone
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
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Comments
We used lemma for polygons
How did you get this answer? Is there a theorem or something I need to prove this.
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