LDMJ is a circle centered at O. Point K, on DJ, bisects chord LM. DSJ is another circle drawn using DJ as diameter. If $\alpha = 90^0$, then KS = KL.
This can be proved by applying “power of a point” couple of times.
Question:- Is the converse true? (i. e. If KS = KL, will $\alpha$ be $90^0$?)
The circle center $K$ radius $KL$ intersects the circle with diameter $DJ$ at two points $S_1$ and $S_2$, which are symmetry about $DJ$. By your argument (when $\alpha=90^\circ$), one of these $S_1$ and $S_2$ is $S$. It follows that $K$ is the middle point of $S_1S_2$ and we are done.