SAMC

Denote $B{B}^{\prime }$ the diameter corresponding to point $B$ and consider $\alpha$ the angle $\widehat{AB{B}^{\prime }}$. Writing the power of $T$ with respect to the circle, we get $$\begin{array}{c}TE\cdot TD=TB\cdot T{B}^{\prime }=TB\cdot (2R-TB)=2R\cdot \frac{BS}{\cos \alpha }-T{B}^2\\ =2R\cdot \frac{BS}{\cos \alpha }-BC\cdot BS\end{array}$$ where $R$ is the radius of the circle.

We have $AS\cdot BC=TE\cdot TD$ if and only if $$AS\cdot BC=2R\cdot \frac{BS}{\cos \alpha }-BC\cdot BS,$$ that is equivalent to $$2R\cdot \frac{BS}{\cos \alpha }=AS\cdot BC+BC\cdot BS=BC\cdot (AS+BS)=BC\cdot AB.$$ It follows that the desired relation is equivalent to $$BC\cdot \frac{AB}{2R}=\frac{BS}{\cos \alpha },$$ that is $BC\cdot \cos \alpha =\frac{BS}{\cos \alpha }$, hence $\frac{BS}{BC}={\cos }^2\alpha$.

On the other hand, it is clear that $$\cos \alpha =\frac{BS}{BT}\text{ and }\cos \alpha =\frac{BT}{BC}.$$ Multiplying these relations we get ${\cos }^2\alpha =BS/BC$ and we are done.