Denote by $\omega$ the circumcircle of $\triangle AHF$.
The key idea in the problem is to introduce a new point $X$ which we define as the second intersection of $DN$ and $\omega$. We now note that the $\angle JAD = \angle CAD = 90^\circ \pm \frac{\alpha}{2}$ where $\alpha = \angle CAB$. As $AD$ is an external bisector of $\angle CAB$.
The $\pm$ signs depend on the picture and student shouldn't be deduced any points for not noticing this.
Hence we have either $\angle JAD = \angle BAD$ or $\angle JAD + \angle IAD = 180^\circ$ so in both cases $DI = DJ$.
Now as $N$ is midpoint of $IJ$ this means that $DN$ is bisector of $IJ$ and hence passes through the centre of the. This shows that $DX$ is a diameter of $\omega$ and $EH \parallel IJ$.
We also notice that $\angle EAD = 90^\circ$ as angle between bisectors and $\angle XAD = 90^\circ$ as $DX$ is a diameter. Hence $X, A, E$ are collinear.
Now this gives us $\angle DHE = \angle XHE = 90^\circ$ and $\angle XFE = \angle DFE = 90^\circ$ as $DX$ is a diameter of $\omega$ and finally again $\angle EAD = 90^\circ$. All this gives us that quadrilaterals $XFEH$ and $ADEH$ are cyclic.
Final step is to use some angle chasing to get $\angle AHE = \angle ADH = \angle AXF = \angle EXF = \angle EHF$ where first, second and fourth equalities are due to cyclicity of $ADEH$, $ADXF$ and $XFEH$ respectively. Also $\angle DFH = \angle EFH = \angle EXH = \angle AFD = \angle AFE$ where the second and fourth equalities are due to cyclicity of $XFEH$ and $ADXF$. This shows $E$ is the incenter of $\triangle AFH$ as desired.