Let $ABC$ be an acute triangle with $| AB | > | AC |$. Let $D$ be the foot of the altitude from $A$ to $BC$, let $K$ be the intersection of $AD$ with the internal bisector of angle $B$, Let $M$ be the foot of the perpendicular from $B$ to $CK$ (it could be in the extension of segment $CK$) and$ N$ the intersection of $BM$ and $AK$ (it could be in the extension of the segments). Let $T$ be the intersection of$ AC$ with the line that passes through $N$ and parallel to $DM$. Prove that $BM$ is the internal bisector of the angle $\angle TBC$