smc

Extend $BN$ to meet $AC$ at $Q$. Then $△BNM\sim △BQC$, so $BN=NQ$ and $QC=19-AQ=2MN$. Since $\angle ANB=90^{\circ }=\angle ANQ$, $\angle BAN=\angle NAQ$ (since $AN$ is an angle bisector) and $△ANB$ and $△ANQ$ share side $AN$, $△ANB\cong △ANQ$. Thus $AQ=14$, and so $MN=\frac{19-AQ}{2}=\frac{5}{2}$, hence our answer is $\boxed{\frac{5}{2}}$.