Let $ ABC$ be an acute triangle, $ CC_1$ its bisector, $ O$ its circumcenter. The perpendicular from $ C$ to $ AB$ meets line $ OC_1$ in a point lying on the circumcircle of $ AOB$. Determine angle $ C$.
1. **Identify the given elements and their relationships:**
   - Let \( ABC \) be an acute triangle.
   - \( CC_1 \) is the angle bisector of \( \angle ACB \).
   - \( O \) is the circumcenter of \( \triangle ABC \).
   - The perpendicular from \( C \) to \( AB \) meets line \( OC_1 \) at point \( X \), which lies on the circumcircle of \( \triangle AOB \).

2. **Define the intersection points:**
   - Let \( \{X\} = OC_1 \cap h_c \), where \( h_c \) is the altitude from \( C \) to \( AB \).
   - Let \( K \) be the intersection point of the line \( CC_1 \) with the circumcircle of \( \triangle ABC \).

3. **Consider the inversion:**
   - Perform an inversion \( I \) with pole \( O \) and power \( CO^2 \).
   - Under this inversion, \( I(C_1) = X \), implying that \( COKX \) is a cyclic trapezoid.

4. **Analyze the cyclic trapezoid:**
   - Since \( COKX \) is a cyclic trapezoid, \( AB \) is the perpendicular bisector of \( CX \).

5. **Use the cyclic nature of \( AOBX \):**
   - Since \( AOBX \) is cyclic, we have:
     \[
     180^\circ = \angle AOB + \angle AXB
     \]
   - Given that \( \angle AOB = 2\angle C \) (since \( O \) is the circumcenter and \( \angle AOB \) is twice the angle at \( C \) in the triangle), and \( \angle AXB = \angle C \) (since \( X \) lies on the circumcircle of \( \triangle AOB \)), we can write:
     \[
     180^\circ = 2\angle C + \angle C
     \]
     \[
     180^\circ = 3\angle C
     \]

6. **Solve for \( \angle C \):**
   - Dividing both sides by 3, we get:
     \[
     \angle C = \frac{180^\circ}{3} = 60^\circ
     \]

The final answer is \( \boxed{60^\circ} \).