<Tri-2023-2034>, hint

Consider the homothety centered at C that takes the incenter to the C-excenter. Note that the tangent at D to the incircle is parallel to AB, hence the homothety takes D to S (since the tangent at S to the excircle is also parallel to AB). It is well known that AT = SB in this case (they both equal s - a, where a is the semi-perimeter).