Let line \( AT \) intersect \( BC \) at \( X \). Note that the given angle condition implies that \( \angle DTA = \angle AXB \). The key to solving the problem is to construct \( T' \), the reflection of \( T \) across point \( S \). This construction is motivated by considering the farthest point \( A \) can reach along line \( TX \).
Since \( AD < DT' \), \( A \) must lie on the interior of segment \( ST \). Additionally, because \( \angle DT'X = \angle T'XB \), we conclude that \( DT' \parallel CB \). Furthermore, since \( DT' = CB \), quadrilateral \( DT'BC \) must be a parallelogram.
To obtain the desired result, we apply the triangle inequality in \( \triangle ABT' \):
\[
AB + AT' \geq BT'
\]
which simplifies to:
\[
AB + ST - AS \geq CD.
\]
Attachments: