In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, and $BN\perp AN$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then find $MN$.

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ \dfrac{5}{2}\qquad\textbf{(C)}\ \dfrac{5}{2}-\sin\theta\qquad\textbf{(D)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right)$
Extend $BN$ to meet $AC$ at $Q$. Then $\triangle BNM \sim \triangle BQC$, 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 $\triangle ANB$ and $\triangle ANQ$ share side $AN$, $\triangle ANB \cong \triangle ANQ$. Thus $AQ=14$, and so $MN=\frac{19-AQ}{2}=\frac{5}{2}$, hence our answer is $\fbox{B}$.