Solution 1. Place $A$ at $(0, 1)$, $B$ at $(b, 0)$ and $C$ at $(c, 0)$ with $b \neq c$. Then the coordinates of $D$ are $(c/2, 1/2)$. The length of $BD$ is thus $\sqrt{(b-c/2)^2 + 1/4}$. This has a minimum value of $1/2$ when $b=c/2$ and $b \neq 0$. There is no upper bound.

Solution 2. Let the position of the point $A$ and the line $BC$ be fixed, but consider the points $B$ and $C$ to be variable. As the distance of $A$ from the line $BC$ is equal to $1$, the midpoint $D$ of $AC$ lies on the line $L$ which is parallel to $BC$ and has distance $1/2$ from $A$ and from $BC$. Therefore, the minimal distance between $B$ and $D$ is $1/2$ and for any given $C$ this is achieved when $BD$ is perpendicular to $BC$, provided that $AC$ is not perpendicular to $BC$. There is no upper bound for $|BD|$ because $B$ can lie as far away as we like.