Solution:

In order to prove that $SM$ and $TN$ are parallel, it suffices to prove that both of them are perpendicular to $ST$. Due to symmetry, we will provide a detailed proof of $SM \perp ST$, whereas the proof of $TN \perp ST$ is analogous. In this solution we will use the following notation: $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$.

We first observe that, due to the tangency condition, we have
$$\angle SHB = \angle HCB = 90^\circ - \beta$$
Combining the above with
$$\angle SBH = \angle ABH = 90^\circ - \alpha$$
we get
$$\angle BSH = 180^\circ - (90^\circ - \beta) - (90^\circ - \alpha) = \alpha + \beta = 180^\circ - \gamma$$
from which it follows that $\angle AST = \gamma$.

Since $AH = HD$, $H$ is the midpoint of $AD$. If $K$ denotes the midpoint of $AB$, we have that $KH$ and $BC$ are parallel. Since $M$ is the midpoint of $BH$, the lines $KM$ and $AD$ are parallel, from which it follows that $KM$ is perpendicular to $BC$. As $KH$ and $BC$ are parallel, we have that $KM$ is perpendicular to $KH$ so $\angle MKH = 90^\circ$. Using the parallel lines $KH$ and $BC$ we also have
$$\angle KHM = \angle KHB = \angle HBC$$
Now,
$$\angle HMK = 90^\circ - \angle KHM = 90^\circ - \angle HBC = 90^\circ - (90^\circ - \gamma) = \gamma = \angle AST = \angle KSH$$
so the quadrilateral $MSKH$ is cyclic, which implies that $\angle MSH = \angle MKH = 90^\circ$. In other words, the lines $SM$ and $ST$ are perpendicular, which completes our proof.

Solution:

We will refer to the same figure as in the first solution. Since $CH$ is tangent to the circumcircle of triangle $BHC$, we have
$$\angle SHB = \angle HCB = 90^\circ - \angle ABC = \angle HAB$$
From the above it follows that triangles $AHB$ and $HSB$ are similar. If $K$ denotes the midpoint of $AB$, then triangles $AHK$ and $HSM$ are also similar. Now, observe that $H$ and $K$ are respectively the midpoints of $AD$ and $AB$, which implies that $HK \parallel DB$, so
$$\angle AHK = \angle ADB = 90^\circ$$
Now, from the last observation and the similarity of triangles $AHK$ and $HSM$, it follows that
$$\angle HSM = \angle AHK = 90^\circ$$
Due to symmetry, analogously as above, we can prove that $\angle HTN = 90^\circ$, implying that both $SM$ and $TN$ are perpendicular to $TS$, hence they are parallel.