As usual, we denote the angles $\angle BAC$, $\angle ABC$ and $\angle BCA$ by $\alpha$, $\beta$ and $\gamma$, respectively. The center of the circumcircle is denoted by $O$, see Figure 1.

Figure 1: Problem 2
By assumption, we have $\beta < \gamma$. Let $E$ be the point on the circumcircle of $ABC$ diametrically opposite to $A$.
By the inscribed angle theorem, we have $\angle AOB = 2\gamma$. By definition, $S$ is the intersection of the angular bisector of $\angle CAB$ and the circumcircle. We note that
$$\angle EAS = \angle BAS - \angle BAO = \frac{\alpha}{2} - \frac{1}{2}(180^\circ - \angle AOB) = \frac{\alpha}{2} - \frac{1}{2}(180^\circ - 2\gamma) = \frac{\alpha}{2} + \gamma - 90^\circ$$
Since we also have
$$\angle SAT = \angle SAC - \angle DAC = \frac{\alpha}{2} - (90^\circ - \angle ACD) = \frac{\alpha}{2} + \gamma - 90^\circ$$

it therefore follows that $\angle EAS = \angle TAS$ holds. Since we also have $AE = AT = 2R$, triangles $ASE$ and $AST$ are congruent, and therefore $\angle AST = \angle ASE$ follows. Since $AE$ is a diameter of the circumcircle, we have $\angle ASE = 90^\circ$, and the claim is proven.