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
1
Expert's answer
2012-03-22T11:43:55-0400
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
Numbers and figures are an essential part of our world, necessary for almost everything we do every day. As important…
APPROVED BY CLIENTS
Finding a professional expert in "partial differential equations" in the advanced level is difficult.
You can find this expert in "Assignmentexpert.com" with confidence.
Exceptional experts! I appreciate your help. God bless you!
Comments
We used lemma for polygons
How did you get this answer? Is there a theorem or something I need to prove this.
Leave a comment