Solution. Since $m(\angle D F H)=90^{\circ} \Rightarrow D H$ is the diameter of the circumcircle of triangle $D F H$. The circumcircle of triangle $D F H$ intersects the segment $(A D)$ at point $N$, therefore, $m(\angle D N H)=90^{\circ}=m(\angle A N H)$. From here we obtain that line $N H$ is perpendicular to line $A D$. Let $K$ be the foot of the altitude from vertex $C$, and $E$ be the foot of the altitude from vertex $B$. It follows that quadrilateral $A E H K$ is inscribed in the circle with diameter $A H$. From here we obtain that $m(\angle E H K)=180^{\circ}-m(\angle A)=m(\angle B H C)$. Since $m(\angle A N H)=90^{\circ}$ and $A B>A C$, it follows that point $N$ is the point where the circumcircle of triangle $A B C$ intersects the circumcircle of quadrilateral

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$A E H K$. Let $P$ be the point where line $N H$ intersects the circumcircle of triangle $A B C$. Then we obtain that $m(\angle B P C)=180^{\circ}-m(\angle A)=m(\angle B H C)$. Since $m(\angle A N P)=90^{\circ}$, it follows that segment $A P$ is the diameter of the circumcircle of triangle $A B C$. From here we obtain that $m(\angle A B P)=90^{\circ}$, hence $B P \| C H$. Therefore, quadrilateral $B P C H$ is a parallelogram, in which segment $P H$ is a diagonal. Thus, line $N H$ passes through the midpoint of side $(B C)$, q.e.d.