Problem

Source: Regional Olympiad - Federation of Bosnia and Herzegovina 2015

Tags: geometry, parallelogram



In parallelogram $ABCD$ holds $AB=BD$. Let $K$ be a point on $AB$, different from $A$, such that $KD=AD$. Let $M$ be a point symmetric to $C$ with respect to $K$, and $N$ be a point symmetric to point $B$ with respect to $A$. Prove that $DM=DN$