Convex hexagon with AD=BC+EF, BE=AF+CD, CF=DE+AB
Source: Kazakhstan international contest 2006, Problem 6
January 22, 2006
inequalitiesgeometryparallelogramEulergeometric transformationreflectiontriangle inequality
Problem Statement
Let be a convex hexagon such that AD \equal{} BC \plus{} EF, BE \equal{} AF \plus{} CD, CF \equal{} DE \plus{} AB. Prove that:
\frac {AB}{DE} \equal{} \frac {CD}{AF} \equal{} \frac {EF}{BC}.