Base on the power from $H$ to the circle $(\omega)$, $HD \cdot HE = HB \cdot HC = AH^2$. Moreover, $HD + HE = DE = AH\sqrt{5}$. Thus, the lengths $HD, HE$ will be the solutions of the quadratic equation
$$x^2 - AH\sqrt{5} \cdot x + AH^2 = 0.$$
Since $D$ is inside triangle $ABC$, $HD < AH$, entails $HE > AH$ and will have $HE > HD$. From there, solving the above equation, we get
$$HD = \frac{-1 + \sqrt{5}}{2}AH \text{ and } HE = \frac{1 + \sqrt{5}}{2}AH.$$
Hence,
$$AD = AH - HD = \frac{3 - \sqrt{5}}{2}AH \text{ and } AE = AH + HE = \frac{3 + \sqrt{5}}{2}AH.$$
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It follows that
$$AD \cdot AE = \left(\frac{3 - \sqrt{5}}{2}\right) \left(\frac{3 + \sqrt{5}}{2}\right) AH^2 = AH^2.$$
Furthermore, according to the power from $A$ to the circle $(\omega)$, $AD \cdot AE = AM \cdot AB = AN \cdot AC$. Therefore $AH^2 = AM \cdot AB = AN \cdot AC$, which implies that $HM \perp AB$, $HN \perp AC$ so the quadrilateral $AMHN$ is a rectangle. Therefore, the circle circumscribing triangle $HMN$ is also the circle with diameter $AH$ so it will be tangent to $BC$. $\square$