European Mathematical Cup

Denote by $\omega$ the circumcircle of $△AHF$. 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 $△AFH$ as desired.