orl 03.11.2005 22:49 In the convex pentagon $ ABCDE,$ the sides $ BC, CD, DE$ are equal. Moreover each diagonal of the pentagon is parallel to a side ($ AC$ is parallel to $ DE$, $ BD$ is parallel to $ AE$ etc.). Prove that $ ABCDE$ is a regular pentagon.
Tiks 06.11.2005 13:27 Let $DE$ meet $AB$ at $M$ and $DC$ meet $AB$ at$N$. $EC$ parallel to $AB$=>$EC$ parallel to$MN$,but $ED=DC$=>$DM=DN$=> $EM=CN$. $AE$paralel to $BD$ =>$MA/AB=ME/ED$and $BC$paralel $AD$=>$NB/AB=NC/CD$,but $ME/ED= NC/CD$=>$MA/AB=NB/AB$=>$MA=NB$ $MA=NB$,$EM=CN$and$<EMA=<BNC$=>$AE=BC$ Respectiveli $BC=AB$=>$BA=ED$=>$MA=ME$=>$<A=<E$ Respectiveli $<A=<B=<C=<D=<E$ and $AB=BC=CD=DE=EA$so $ABCDE$ is regular.