Solution:
Let $AB$ be a diameter of $k$. Since $\text{D H A = D C H =}$, $\text{C H B = C D H =}$, $\text{D M H = 2}$, $\text{C M H = 2}$ and $\text{M D C = 90 - -}$, the Sine theorem for $\triangle D M O$ gives
$$MO = \frac{r \cos (\alpha + \beta)}{\cos (\alpha - \beta)}$$
where $r$ is the radius of $k_{1}$ and $O = MH \cap CD$. On the other hand, if $MH \cap k = P$,

then
$$\text{MAP = 1 2 MP = 1 2 ( DM + CP ) = DOM = 90 - +}$$
Applying the Sine theorem for $\triangle APM$ we get that $2r = MP = 2R \cos (\alpha - \beta)$, where $R$ is the radius of $k$. The Sine theorem for $\triangle DMC$ implies that
$$r = MC = 2R \cos (\alpha + \beta)$$
and hence $2 \cos (\alpha + \beta) = \cos (\alpha - \beta)$. Now (1) shows that $O$ is the midpoint of $MH$.

Conversely, it is easy to see that if $O$ is the midpoint of $MH$, then $H$ is the midpoint of $MP$, i.e. $AB$ is a diameter of $k$.

Alternative solution.
Let $AB$ be a diameter of $k$ and $MH$ meets $k$ for (the) second time at the point $P$. It is clear that $MH = HP = r$, where $r$ is the radius of $k_{1}$. Consider the inversion with respect to the circle with center $M$ and radius $r$. Then $P$ is the image of the midpoint $T$ of $MH$. The image of $k$ is the line $DC$. Hence the image of $P$ lies on $DC$, i.e. $T \in DC$.

Considering the same inversion also implies easily the converse statement.