Let say that IC intersects TS at point R. Let assume IR = 1. Obviosly BR = RC. If we write angles and applying sine theorem, we get this 3 equaitons: $$TR = \frac{sin(\frac{B+C}{2})}{cos(\frac{B}{2})}$$$$RS = \frac{sin(\frac{A+C}{2})}{sin(\frac{B-C}{2})}$$$$BR = \frac{cos(\frac{A}{2})}{sin(\frac{B-C}{2})}$$Let check if this equation true: $$IR.RC =_? TR.RS$$$$\frac{cos(\frac{A}{2})}{sin(\frac{B-C}{2})} =_? \frac{sin(\frac{B+C}{2})}{cos(\frac{B}{2})}\frac{sin(\frac{A+C}{2})}{sin(\frac{B-C}{2})}$$$$cos(\frac{A}{2}) =_? \frac{sin(\frac{B+C}{2})sin(\frac{A+C}{2})}{cos(\frac{B}{2})}$$$$cos(\frac{A}{2})cos(\frac{B}{2}) =_? sin(\frac{B+C}{2})sin(\frac{A+C}{2})$$$$cos(\frac{A}{2})cos(\frac{B}{2}) =_? sin(90-\frac{A}{2})sin(90-\frac{B}{2})$$Which is obviously true. Therefore, the equation, $IR.RC = TR.RS$, is true. So, these 4 point must be cyclic.

We have $\angle CIS = \angle CAI + \angle ACI = \frac{\angle A}{2} + \frac{\angle C}{2} = \frac{180 - \angle B}{2} = 90 - \frac{\angle B}{2}= 90 - \angle TBC = 90 - \angle TCB = \angle CTS$ Thus, $CTIS$ is cyclic.