In a parallelogram $ABCD$ the trisectors of angles $A$ and $B$ are drawn. Let $O$ be the common points of the trisectors nearest to $AB$. Let $AO$ meet the second trisector of angle $B$ at point $A_1$, and let $BO$ meet the second trisector of angle $A$ at point $B_1$. Let $M$ be the midpoint of $A_1B_1$. Line $MO$ meets $AB$ at point $N$ Prove that triangle $A_1B_1N$ is equilateral.