Extend $CA$ to $E$ so that $AE=AD.$ As a result, $\triangle EAD$ is isosceles with $\angle AED=\angle A/4.$ However, $\triangle ECB$ is an isosceles and right triangle and thus $\angle CEB=\angle CB!=45^\circ.$ Also, $\triangle CDE$ and $\triangle CDB$ are congruent (because $CB=CE,$ and $\angle ECD=\angle BCD=45^\circ$) and thus $ED=DB.$ Therefore, $\triangle EDB$ is isosceles and thus $\angle DEB=45-\angle A/4=\angle DBE=45^\circ-\angle B/2.$ As a result, $\angle A=2\angle B$ and because $\angle A+\angle B=90^\circ,$ we finally get $\angle A=60^\circ$ and $\angle B=30^\circ.$ This means that $AC/AB=1/2.$ With $AN$ being a bisector, we get $CN/AC=NB/AB,$ or $NB=(AB/AC)\cdot CN=4.$

Take $E$ the reflection of $A$ in $CD$, i.e. $E\in BC$; by symmetry, $CE=AC, DE=AD, \angle CED=\angle CAD=\frac{\widehat{BAC}}{2}$ and, from the given relation, we get $BE=DE\iff \angle EBD=\frac{\widehat{BAC}}{4}$, consequently $\widehat{BAC}=2\widehat{ABC}$, next as above. Best regards, sunken rock