1. Identify the key points and properties:
- Let $I$ be the incenter of $\triangle ABC$.
- The perpendicular bisector of $BC$ intersects the angle bisector of $\angle BAC$ at $S$.
- The angle bisector of $\angle CBA$ intersects the angle bisector of $\angle BAC$ at $T$.

2. **Establish the cyclic nature of points $C, I, S, T$:**
- We need to show that $C, I, S, T$ lie on a circle. This can be done by proving that the power of point $I$ with respect to the circle passing through $C, S, T$ is equal to the power of point $I$ with respect to the circle passing through $C, S, T$.

3. Use the Sine Rule and Angle Bisector Theorem:
- Let $IC$ intersect $TS$ at point $R$. Assume $IR = 1$.
- Since $I$ is the incenter, $BR = RC$.
- Using the Sine Rule in the relevant triangles, we get:
$$TR = \frac{\sin\left(\frac{B+C}{2}\right)}{\cos\left(\frac{B}{2}\right)}$$
$$RS = \frac{\sin\left(\frac{A+C}{2}\right)}{\sin\left(\frac{B-C}{2}\right)}$$
$$BR = \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{B-C}{2}\right)}$$

4. Verify the cyclic condition:
- We need to check if the equation $IR \cdot RC = TR \cdot RS$ holds:
$$IR \cdot RC = TR \cdot RS$$
Substituting the values, we get:
$$\frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{B-C}{2}\right)} = \frac{\sin\left(\frac{B+C}{2}\right)}{\cos\left(\frac{B}{2}\right)} \cdot \frac{\sin\left(\frac{A+C}{2}\right)}{\sin\left(\frac{B-C}{2}\right)}$$
Simplifying, we get:
$$\cos\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{B+C}{2}\right) \cdot \sin\left(\frac{A+C}{2}\right)}{\cos\left(\frac{B}{2}\right)}$$
$$\cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) = \sin\left(\frac{B+C}{2}\right) \cdot \sin\left(\frac{A+C}{2}\right)$$
Using the identity $\sin(90^\circ - x) = \cos(x)$, we get:
$$\cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) \cdot \cos\left(90^\circ - \frac{B}{2}\right)$$
This is obviously true.

5. Conclusion:
- Since the equation $IR \cdot RC = TR \cdot RS$ holds, the points $C, I, S, T$ must be cyclic.

$\blacksquare$