In triangle $ ABC$, one has marked the incenter, the foot of altitude from vertex $ C$ and the center of the excircle tangent to side $ AB$. After this, the triangle was erased. Restore it.
1. **Identify the given points and their significance:**
   - \( I \): Incenter of \(\triangle ABC\)
   - \( I_c \): C-excenter of \(\triangle ABC\)
   - \( H \): Foot of the altitude from vertex \( C \) to side \( AB \)

2. **Define the foot of the C-angle bisector:**
   - Let \( V \) be the foot of the C-angle bisector.

3. **Use the harmonic division property:**
   - Since \((C, V, I, I_c) = -1\), the circle with diameter \(\overline{CV}\) passes through \( H \) and is orthogonal to the circle with diameter \(\overline{II_c}\).

4. **Construct the circle \(\mathcal{Q}\):**
   - The circle \(\mathcal{Q}\) passes through \( C, H, V \) and belongs to the orthogonal pencil defined by the circle \((K)\) with diameter \(\overline{II_c}\).
   - The center of \(\mathcal{Q}\) lies on \(\overline{II_c}\).

5. **Determine the power of point \( K \) with respect to \(\mathcal{Q}\):**
   - The power of \( K \) to \(\mathcal{Q}\) equals the square of the radius \( KI \).
   - Draw line \( KH \) and construct point \( P \) on the ray \( KH \) such that \( KP \cdot KH = KI^2 \).

6. **Find the center of \(\mathcal{Q}\):**
   - The intersection of the perpendicular bisector of \(\overline{PH}\) with the line \(\overline{II_c}\) gives the center of \(\mathcal{Q}\).

7. **Draw the circle \(\mathcal{Q}\):**
   - Once the circle \(\mathcal{Q}\) is drawn, we can easily find points \( C \) and \( V \).

8. **Determine the vertices \( A \) and \( B \):**
   - The line \( VH \) intersects the circle \((K)\) at the desired vertices \( A \) and \( B \).

The final answer is \( \boxed{ \triangle ABC } \).